# Sharif Haason — full writing corpus This is the complete public writing archive. Articles are included as rendered HTML fragments so their MathML remains intact. The canonical page for each article and its structured JSON representation are included in the metadata block. ## A New Chapter: Graduation and a New Role at Monark - URL: https://sharifhsn.dev/blog/linkedin-2026-04-01-a-new-chapter-graduation-and-a-new-role-at-monark/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2026-04-01-a-new-chapter-graduation-and-a-new-role-at-monark.json - Description: I’m thrilled to make two major announcements: I have completed my Master’s in Financial Engineering from Stevens Institute of Technology, and I am starting a new role at Monark… - Date: 2026-04-01 - Exact published timestamp: 2026-04-01T14:42:17.714Z - Topics: Career, Personal - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/feed/update/urn:li:activity:7445118347752402944/ - Content source format: typst

I’m thrilled to make two major announcements: I have completed my Master’s in Financial Engineering from Stevens Institute of Technology, and I am starting a new role at Monark Markets as a Product Engineer!

I’d like to specially thank Dr. Ionut Florescu, Dragos Bozdog, Zheng Xing, and Zhenyu Cui for their mentorship at Stevens, as well as my coworkers at the Hanlon Financial Systems Center at Stevens Institute of Technology, my professors, my classmates in my program, and everyone else at Stevens who has helped me on my journey. Additionally, I never could have done this without the support of my friends and family, who I cherish always.

I am excited to take on new challenges at Monark Markets and I believe their product is a complete hit in the making. If anyone is interested in investing in private markets like SpaceX or Anthropic through retail brokerages, please look into what we’re doing.

## The December Fed Meeting: What to Watch - URL: https://sharifhsn.dev/blog/linkedin-2025-12-10-the-december-fed-meeting-what-to-watch/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2025-12-10-the-december-fed-meeting-what-to-watch.json - Description: The December Fed meeting is today! Despite the worrying lack of data, the market is confident that rates will be cut by 25 bps, with the FedWatch tool giving a 87.6% chance for a… - Date: 2025-12-10 - Exact published timestamp: 2025-12-10T11:00:16.071Z - Topics: Central Banking, Monetary Policy - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/feed/update/urn:li:activity:7404475031784730625/ - Content source format: typst

The December Fed meeting is today! Despite the worrying lack of data, the market is confident that rates will be cut by 25 bps, with the FedWatch tool giving a 87.6% chance for a cut.

With PCE inflation at expectations and no surprises from JOLTS, all indicators are pointing to a cut. But there’s more that meets than eye to this situation.

This is Jerome Powell’s last meeting as Fed Chair, with his successor already decided, though not announced. Wall Street is speculating that Kevin Hassett, current director of the National Economic Council, will replace Powell. He is qualified, which is more than can be said for some of Trump’s other appointments, but he is also a MAGA loyalist, which may present issues for the Fed’s neutrality going forward.

While a cut is expected, close attention should be paid towards Powell’s rhetoric during his remarks and responses to Q&A after the meeting, as they will signal what the rest of the FOMC is thinking about the economy. This could be the last meeting before those carefully chosen words are replaced by partisan rhetoric.

I’ll close this post out with a commendation of Jerome Powell. He fearlessly commanded the Fed through the COVID crisis, the subsequent inflation, and immense pressure from the White House, never wavering in his resolve. The Fed is worse off without him at the helm.

What do you think? Will the Fed cut rates?

## Why 50-Year Mortgages Fall Short - URL: https://sharifhsn.dev/blog/linkedin-2025-11-15-why-50-year-mortgages-fall-short/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2025-11-15-why-50-year-mortgages-fall-short.json - Description: Federal Housing Finance Agency director Bill Pulte recently floated a big potential change to the mortgage market: 50 year mortgages. Unfortunately, these aren’t realistic.… - Date: 2025-11-15 - Exact published timestamp: 2025-11-15T14:30:01.051Z - Topics: Housing, Public Policy - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/feed/update/urn:li:activity:7395468120376627200/ - Content source format: typst

Federal Housing Finance Agency director Bill Pulte recently floated a big potential change to the mortgage market: 50 year mortgages. Unfortunately, these aren’t realistic.

Typical mortgages last 30 years. Extending them to 50 years would have the benefit of lower monthly payments for borrowers. However, the idea has several issues.

The basic idea is a non-starter from the mortgage holder point of view. They need to hedge away their interest rate risk, or “duration”. If rates rise, then the value of the loan goes down. Therefore, the mortgage holder will short Treasury bonds of the same maturity to make themselves duration neutral. If interest rates move up, then both the loan and the treasury bond will lose value equally if balanced appropriately. The mortgage holder will only profit off of the spread between the Treasury yield and their mortgage rate, and be protected from moves in the interest rate.*

But a 50 year Treasury bond doesn’t exist, and it’s doubtful that one would ever be issued. There is enough demand for lower terms that a 50 year Treasury bond would carry an illiquidity premium, on top of the extreme inflation premium. Unless the national deficit balloons massively, there won’t be a funding need for it.

Even if this bond did exist, a 50 year mortgage would be horrible for borrowers, causing them to pay significantly more interest over the period of the mortgage. See the image below for a comparison of a $400k principal mortgage. The 50 year rate is guaranteed to be higher than the 30 year rate to compensate for the increased duration, credit, and inflation risk. Paying an extra $500k in interest is extreme, and dangerous when most people don’t really understand interest to begin with.

Pulte has pulled back from his enthusiasm on the 50 year mortgage after the market responded negatively. However, given the explicit cosign of President Trump, we might continue to hear about this… or they’ll move onto a new product, like portable mortgages. But that’s a subject for another day.

Read more at Bloomberg: https://lnkd.in/edU-T2Zx Trump’s 50-Year Mortgage Loses Steam as Industry Questions Costs

*Treasury bonds do not make a perfect hedge, because mortgages can be refinanced, introducing negative convexity (short gamma) into the asset. Portfolio managers will long gamma (e.g. a payer swaption) to neutralize this cost. The same liquidity problems exist.

## Jefferies and First Brands: What the Trade Debt Means - URL: https://sharifhsn.dev/blog/linkedin-2025-10-09-jefferies-and-first-brands-what-the-trade-debt-means/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2025-10-09-jefferies-and-first-brands-what-the-trade-debt-means.json - Description: “Jefferies Fund Has $715 Million in First Brands Group, LLC’ Trade Debt”… but what does that exactly mean? This Bloomberg headline is causing market turmoil about Jefferies, but… - Date: 2025-10-09 - Exact published timestamp: 2025-10-09T14:28:33.464Z - Topics: Credit, Financial Markets - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/feed/update/urn:li:activity:7382059401983197184/ - Content source format: typst

“Jefferies Fund Has $715 Million in First Brands Group, LLC’ Trade Debt”… but what does that exactly mean?

This Bloomberg headline is causing market turmoil about Jefferies, but let’s cut through the noise and disambiguate what kind of exposure Jefferies actually has here. [1]

First, a quick summary of the story so far. First Brands is an auto parts conglomerate which expanded aggressively by taking on debt through trade financing, which allows the company to be immediately compensated for future receivables by a fund (in this case, Point Bonita Capital) which takes on that risk.

When Jefferies pitched a $6 billion refinancing deal to First Brands in July, it triggered scrutiny into First Brands’ books, causing its slow implosion into bankruptcy. (The full story of First Brands’ bankruptcy is still unfolding and is a very interesting read [2])

Jefferies was quick to open their books and clear up their exposure to First Brands. In the process, they showed the kind of decisions that led to this bankruptcy. In one instance, Point Bonita accepted “side letter” deals with First Brands which allowed them to raise funds at a higher interest rate beyond what the loans allowed, increasing financing costs for Point Bonita. [3] This event shows the risk of trade financing, especially in the dark world of private credit.

I’ve created a diagram showing how Jefferies is exposed. In short, they have a $113MM equity stake in Point Bonita Capital, which is the fund exposed to the eye-popping $715MM in receivables. But according to analysts at Morgan Stanley, estimated losses to Jefferies top out at about $45MM, which is definitely manageable.

Some investors, like BlackRock, have pulled out of Point Bonita as a result, causing Jefferies stock to tumble by ~8% yesterday [4].

But I think this fear is unwarranted. Jefferies is in the market for high risk, high reward borrowers, and blowups like this are a cost of doing business. Headline hysteria might cause their stock price to fluctuate, but Jefferies’ meteoric rise from boutique to multinational bank is no accident. When you play with fire, you’re bound to get singed now and then… but I say “let them cook!”

What do you think? How big of a deal is the First Brands’ bankruptcy for Jefferies?

[1] https://lnkd.in/eD89yvGd Jefferies Fund Has $715 Million in First Brands’ Trade Debt

[2] https://lnkd.in/e-MkP49W First Brands Creditor Seeks Investigation of “Vanished” Cash

[3] https://lnkd.in/e-MWFVhs First Brands’ Go-To Bank Jefferies Opens Books as Fallout Builds

[4] https://lnkd.in/eQfcRXFi BlackRock Seeks Cash From Jefferies Fund Exposed to First Brands

## Why Rate Cuts Won't Fix a Labor Supply Shock - URL: https://sharifhsn.dev/blog/linkedin-2025-09-09-why-rate-cuts-won-t-fix-a-labor-supply-shock/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2025-09-09-why-rate-cuts-won-t-fix-a-labor-supply-shock.json - Description: Interest rate cuts won’t relieve a labor supply shock: with lessons from energy and housing crises. After today’s stunning job revision of −911,000 jobs for March 2025, the stock… - Date: 2025-09-09 - Exact published timestamp: 2025-09-09T19:37:25.419Z - Topics: Monetary Policy, Labor Markets - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/feed/update/urn:li:activity:7371265494668156929/ - Content source format: typst

Interest rate cuts won’t relieve a labor supply shock: with lessons from energy and housing crises.

After today’s stunning job revision of −911,000 jobs for March 2025, the stock market rallied. The S&P index is hitting all time highs, as it has been over the past two weeks. That might seem like a contradiction, but it’s because the BLS data is used to guide the decisions of the Federal Open Market Committee in raising or lowering the interest rate. A bad jobs report will push the FOMC into lowering rates.

Lower interest rates make it cheaper to borrow money across all sectors of the economy. This makes companies more likely to expand and therefore hire more workers. That’s the thinking on how rate cuts counteract a bad labor market. But there’s a problem with how this applies to the current macro environment.

Think like a company. Just because it’s cheaper to borrow money doesn’t it mean make sense for to take on new ventures. If there’s nobody to hire, then it doesn’t matter how much money you have. Monetary policy is fundamentally a demand-side policy.

Retiring boomers, decreased immigration, and general uncertainty over federal policy have made companies cautious in taking risks dependent on labor.

The same situation happened in the 1970s during the energy crisis. When geopolitical instability caused supply shocks in oil, the US economy entered “stagflation”, where both inflation and unemployment were increasing. The only solution was the “Volcker shock”, when Fed Reserve head Paul Volcker sharply rose interest rates from 11.2% to 20% in order to crush the runaway inflation with a recession.

And last year, despite rate cuts of a total 1%, interest rates on mortgages remained high, increasing from a high 6% to 7.1%. Typically, mortgage rates decrease when the Fed Funds rate does, as lenders try to attract new business with lower rates while existing homeowners refinance at lower rates. But the supply constraints caused by slowdowns in building homes meant that even if it was cheap to buy a home, people wouldn’t be able to find one.

With the increased media scrutiny on the Federal Reserve, along with the president’s statements about them, their decisions matter more than ever. On September 17, we will see whether they cave to these expectations.

https://lnkd.in/gjBtmekN

## Oil Prices and the Energy Provisions in the One Big Beautiful Bill - URL: https://sharifhsn.dev/blog/linkedin-2025-07-03-energy-provisions-one-big-beautiful-bill/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2025-07-03-energy-provisions-one-big-beautiful-bill.json - Description: Great video about oil prices from experts in commodities. I think Wasif Latif’s point about the price being right for oil is really significant for the implications of the energy… - Date: 2025-07-03 - Exact published timestamp: 2025-07-03T18:00:26.040Z - Topics: Energy Policy, Fiscal Policy - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/feed/update/urn:li:activity:7346598711562579969/ - Content source format: typst

Great video about oil prices from experts in commodities.

I think Wasif Latif’s point about the price being right for oil is really significant for the implications of the energy provisions of the One Big Beautiful Bill set to pass in the House soon.

Subsidies make sense for a growing innovative industry like solar, not so much for oil where geopolitical tail risk surrounding a massive supply is the biggest concern.

Oil companies are smart. They’re not going to just jump up and bite at oil subsidies and expand infrastructure when they know that the supply and demand factors don’t favor them.

## Neural Networks Aren't Impressive on Their Own - URL: https://sharifhsn.dev/blog/linkedin-2025-06-07-neural-networks-aren-t-impressive-on-their-own/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2025-06-07-neural-networks-aren-t-impressive-on-their-own.json - Description: Neural networks AREN’T impressive… in the context of data analysis, by themselves. With the explosion in discourse about AI, traditional neural networks have also gained cultural… - Date: 2025-06-07 - Exact published timestamp: 2025-06-07T19:26:59.305Z - Topics: AI, Machine Learning - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/feed/update/urn:li:activity:7337198409189150720/ - Content source format: typst

Neural networks AREN’T impressive…

in the context of data analysis, by themselves.

With the explosion in discourse about AI, traditional neural networks have also gained cultural prestige. LLMs do use neural networks as their underlying architecture, as do other forms of machine learning.

At first blush, neural networks can accomplish some pretty cool results. For any input and output, neural networks can create a mapping that works effectively.

The obvious problem is that without mitigation, neural networks will massively overfit the training data, and therefore be useless when applied to new data.

But even when neural networks successfully fit test data, they are limited in what they can do.

Sure, if your only goal is to predict outputs from an input, neural networks can do that for you.

But neural networks (and nonparametric models in general) don’t give a lot of general insights into relationships in your data. If you tried to approximate the relationship with a polynomial, you would get an ugly, long expression where the numbers don’t mean anything.

Compare to linear regression. You get a simple number of parameters on the same order of your features, and each number directly expresses the relationship of the feature with the label.

Obviously, linear regression isn’t feasible for expressing nonlinear relationships, which are common in data. And there’s no parametric model that fits all types of data.

That’s kind of the point, though. By finding a model that fits your data, you’re inherently saying something about the structure of the relationships in your data, which is often useful information.

Neural networks are the lowest common denominator. You can put something in, and get something out, but for the real questions about your data, there’s only a black box. They can be useful as a starting point. But they can’t be the end of your data analysis.

What do you think? Are neural networks useful in data analysis?

## The Case for Structured Errors in Rust - URL: https://sharifhsn.dev/blog/linkedin-2025-05-28-the-case-for-structured-errors-in-rust/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2025-05-28-the-case-for-structured-errors-in-rust.json - Description: I found this article on structured errors insightful. Although it is focused on Rust, the explanations of the direct benefits of encoding information about the failure surface of… - Date: 2025-05-28 - Exact published timestamp: 2025-05-28T18:20:44.156Z - Topics: Rust, Software Engineering - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/feed/update/urn:li:activity:7333557857549783040/ - Content source format: typst

I found this article on structured errors insightful. Although it is focused on Rust, the explanations of the direct benefits of encoding information about the failure surface of code are succinct and persuasive.

“Custom error types allow you to see all potential failure modes of a function at a glance, without (recursively) inspecting its implementation or maintaining fragile hand-written docs that can’t be fully trusted anyway. In a code review, you can easily notice when some error variant doesn’t make sense and should be handled locally, or comes from an action that shouldn’t be performed at all. Interfaces become more descriptive.”

https://lnkd.in/esm4w9WF

## Rust's `subsecond` Crate Brings Hot Reloading - URL: https://sharifhsn.dev/blog/linkedin-2025-05-28-rust-s-subsecond-crate-brings-hot-reloading/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2025-05-28-rust-s-subsecond-crate-brings-hot-reloading.json - Description: Rust’s subsecond crate is a game changer for application and game development in Rust! Its innovative hotpatching method enables hot reloading for select libraries. By far the… - Date: 2025-05-28 - Exact published timestamp: 2025-05-28T18:00:05.331Z - Topics: Rust, Developer Tools - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/feed/update/urn:li:activity:7333552661541117952/ - Content source format: typst

Rust’s subsecond crate is a game changer for application and game development in Rust! Its innovative hotpatching method enables hot reloading for select libraries.

By far the biggest drawback of Rust is its compilation time. Although it has improved over the years, it is far slower than any other language. Even an extra wait of a few seconds can break developer flow.

This is, unfortunately, by design. C/C++ are able to compile quickly because they are generally split up into source code and header files which describe the functions individually, so the compiler can compile them in parallel if they don’t depend on each other.

Rust, however, treats the entire package as the compilation unit. Library authors have to split up their code, often awkwardly, in order to achieve faster compilation times.

Dioxus Labs have finally engineered the solution for their GUI library Dioxus. Their innovative subsecond crate compiles only the part of the code that has changed and patches the change into the existing binary, significantly reducing compilation time for small changes. This is common when designing UIs, when shifting elements and changing small details.

Most significantly, subsecond is available on ALL major platforms, which has never been achieved by ANY hot patching system for a compiled language.

The developers behind the game engine Bevy were able to integrate subsecond into Bevy and hot patch game development with 500 ms compilation times, basically as fast as you can press Ctrl + S!

Currently, subsecond is too bespokely designed to be used widely throughout the ecosystem. However, the Dioxus team is actively working to improve subsecond, and hopefully it will become an option for developers working on large, complex libraries.

subsecond hot reloading will be included in the upcoming release of Dioxus 0.7, and likely Bevy 0.17. The Dioxus team will be presenting technical details about subsecond at Seattle RustConf from September 2-5.

Dioxus release post: https://lnkd.in/eJfHXWJH

Example of Bevy running subsecond: https://lnkd.in/e457vQX9

## What a U.S. Credit Rating Downgrade Means - URL: https://sharifhsn.dev/blog/linkedin-2025-05-17-what-a-u-s-credit-rating-downgrade-means/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2025-05-17-what-a-u-s-credit-rating-downgrade-means.json - Description: Moody’s has downgraded the US’s credit rating from the best (Aaa) to second-best (Aa1), sending the markets into a buzz. But what does it mean to downgrade a credit rating? All… - Date: 2025-05-17 - Exact published timestamp: 2025-05-17T15:52:46.770Z - Topics: Fiscal Policy, Credit Ratings - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/feed/update/urn:li:activity:7329534356572508160/ - Content source format: typst

Moody’s has downgraded the US’s credit rating from the best (Aaa) to second-best (Aa1), sending the markets into a buzz. But what does it mean to downgrade a credit rating?

All bonds have a property called credit risk. This is the probability that the seller of the bond will default on their debt and not pay it back in full or on time. This goes for all kinds of debt, whether it’s a corporate bond, a mortgage, or credit card debt.

The quantification of credit risk is important for pricing financial instruments. The reason that the interest rate on a credit card is much higher than a mortgage is because lenders know that people are much more likely to default on credit card debt than mortgages, and want to be compensated more for taking on that risk. (A mortgage also has a lower interest rate because it’s secured debt, but I’ll ignore that for the purposes of this explanation)

For corporate and sovereign (fancy word for national) bonds, this risk is expressed through credit ratings. Ratings agencies like Moody’s, S&P, and Fitch look through financial statements, the macroeconomic outlook, and historical data to determine the credit risk of a debt issuer. They will then issue ratings based on their assessment, which can then be directly translated into an equivalent hazard rate (probability of default over a given period) for the purpose of pricing credit derivatives.

This downgrade reflects Moody’s belief that, as a result of mounting US debt, they have increased in credit risk, a belief shared by S&P and Fitch, who have both already downgraded the US in a similar way.

A Saturday surprise like this will no doubt impact markets soon. Previous downgrades have led to yield spikes and market rallies. The 10-year yield is already going up, and we’ll see how the markets will look on Monday.

What do you think? Does the US deserve this downgrade?

From Bloomberg: https://lnkd.in/e9yQ9u2E

## Rust at 10: Why I've Followed the Language - URL: https://sharifhsn.dev/blog/linkedin-2025-05-16-rust-at-10-why-i-ve-followed-the-language/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2025-05-16-rust-at-10-why-i-ve-followed-the-language.json - Description: Rust turns 10! In honor of this momentous anniversary, I’d like to reflect on what Rust means to me, and why I have followed it for so long. I first heard about Rust in 2021,… - Date: 2025-05-16 - Exact published timestamp: 2025-05-16T17:38:11.094Z - Topics: Rust, Programming Languages - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/feed/update/urn:li:activity:7329198494844411905/ - Content source format: typst

Rust turns 10! In honor of this momentous anniversary, I’d like to reflect on what Rust means to me, and why I have followed it for so long.

I first heard about Rust in 2021, when I was in my third year of my computer science degree. At this point, I had taken and was taking multiple classes in C, like Computer Architecture, Systems Programming, and Operating Systems. In each of these classes, I had been immensely frustrated with the memory errors I would get and their inscrutability. To this day I have a PTSD reaction to the phrase “Shadow bytes around the buggy address”.

Rust seemed like the perfect solution. Memory safety with all the same low-level interaction? Sign me up!

But it became much more than that to me. I noticed that algebraic data types, which I had learned in OCaml in my PrinProg class, were present in Rust, and I was surprised by how useful they were in modeling state, having only used them in a toy context. From generics to the module system, everything about Rust seemed to have learned from the lessons of previous programming languages and improved on them.

Rust was also constantly evolving, which made it exciting to follow. Every six weeks there was a new release, which would invariably have some new feature that people had been requesting. Core libraries like async, error handling, and time objects were shifting in prominence, constantly adding new features. Large projects like the game engine bevy and the web server axum would wow the community on every new version.

At the same time, Rust was being slowly integrated into the business world. The first big news was a blog post from Discord describing how they used Rust to greatly optimize a service they had previously relied on Go for. Microsoft, Google, and Amazon all began to trickle out announcements of how they used Rust to optimize heavily used core functionality.

The biggest deal by far was the announcement of Rust for Linux, where the Linux kernel would, for the first time, allow a language other than C (or Assembly) into the source code. Although it has pretty much only been used for drivers so far, the symbolism of being included in Linux cannot be overstated.

Ten years later, the third-party ecosystem has stabilized significantly, and more and more companies are considering Rust as the developer base grows. Only time will tell where Rust goes from here, but I can safely say that we’ll only hear more about it.

More reflections on Rust from well-known community members:

★ Graydon Hoare, creator of Rust: https://lnkd.in/eagHb3Xc ★ Steve Klabnik, a founding member: https://lnkd.in/esJy_PZV ★ Niko Matsakis, compiler theorizer: https://lnkd.in/eKQytR2Z

## Copulas and Portfolio Risk - URL: https://sharifhsn.dev/blog/computational-methods-week-14/ - Structured data: https://sharifhsn.dev/api/posts/computational-methods-week-14.json - Description: Ending early today because the Pacers are playing tonight Generally speaking, the Basel 2 , Basel 3, and current Basel 3.5, all these rules and regulations require that banks… - Date: 2025-04-29 - Exact published timestamp: 2025-04-29 - Topics: Computational Methods, Copulas, Value at Risk, Expected Shortfall - Categories: Computational Methods - Source: Computational Methods in Quantitative Finance - Source URL: None - Content source format: typst

Ending early today because the Pacers are playing tonight

Copulas

Why do we need this?

First let’s talk about:

Generally speaking, the Basel 2, Basel 3, and current Basel 3.5, all these rules and regulations require that banks assess two things. Any unit that owns funds that don’t belong to them (banks, basically), they need to assess risk. What exactly does that mean?

The class 535, 635, and 636 are all about this. 636 will be offered this fall.

At the core, regulators don’t understand any of it. What they want to know is, if the economy goes to crap, how much will you lose? So how do you come up with this number?

There’s two basic numbers: VaR, or Value at Risk, and ES, Expected Shortfall

One of my former students is head of globalr isk at Barclays, and every day he has to come up with these numbers. Basically he aggregates the positions of every desk at Barclays. Derivatives, asset management, whatever. Each has a risk associated with it. VaR is supposed to be a quantile. Imagine you have a distribution for the value of your assets tomorrow. It’s actually in 10 days that Basel demands.Obviously, I know the value of assets today, by marking to market. I project in ten days what are the possible things that could happen to all my divisions. My derivatives desk has a distribution, another desk has another one, so I have a whole bunch of random variables. Now I need to combine them. I have tons of distributions, how do I put them together?

For example, 10% of my bank is invested in derivatives. So I should take the risk from that and multiply by 10%, right? Wrong. Because things don’t move independently. We could do such aggregation only then. If the derivatives desk goes down because Trump said something dumb, th equities desk will also go down, and so will bonds. Things are related.

That’s the purpose of copulas. These are very much used in risk management.

VaR can only be calculated for a 1D r.v. You need to have a typical marginal distribution. Let’s say you get this distribution, of values with probabilities. Let’s say I look at the value where the probability is 0.01. If I look at the value I have now, and subtract the value there, that resulting value is the loss. The amount of money I will lose if this thing happens, which is worst case scenario, with a 1% chance of happening. This particular one is $𝑉 𝑎 𝑅_{99}$ . You can obviously calculate different percentiles.

Expected Shortfall

If you are in this region, you are going to lose money. If my estimation is correct, then this thing is going to happen. If I look at the next 100 days, and do this distribution every single day, then I expect to see one of these events every 100 days.

But if it does happen, exactly how much money will I use? I don’t know which point I’m in, just the region. So to determine that value, you calculate ES.

ES is nothing more than

𝔼 [Loss | 𝑋 < VaR]

Basically it’s just the expected value of the random variable when it’s less than VaR.

This is important, there’s an entire area of risk management dedicated to this. The trader cannot take risks because the risk manager won’t let them, because of this stuff. You can’t take crazy risks.

Marginal vs Joint Distribution

So what is the problem? This is calculated on a daily thing. A natural thing to do is to look back 100 days, have a historical look at each desk. That gives me distributions that are reasonable for each of those r.v.s. If I have a 1D r.v., with 100 observations, I can make a pretty accurate histogram.

For a 2D rv, I suddenly need a lot more observations. And for more, it’s impossible.

So although you can estimate the distribution of the marginals very easily, you can’t do that for the joint. You would need mountains of data, which is not possible. And the problem is if you go too far in the past, you end up with irrelevant data that is nonstationary.

The question is, can we do that?

You can do this in one case, and only one case. And that’s if $𝑋_{1}, \dots, 𝑋_{𝑑}$ are independent. Then the joint distribution is the product of the marginals. This holds for both PDF and CDF.

However, irl, nothing is independent. In the 1940s, there’s an old theorem by Sklar:

Let $𝑋_{1}, \dots, 𝑋_{𝑑}$ be random variables with CDFs $𝐹_{1}, \dots, 𝐹_{𝑑}$ . (These are one dimensional CDFs)

Let

𝐹 (𝑥_{1}, \dots 𝑥_{𝑑}) = ℙ (𝑋 \leq 𝑥_{1}, \dots, 𝑥_{𝑑} \leq 𝑥_{𝑑})

There exists a function $𝐶$ which follows

ℝ^{𝑑} \to [0, 1]

That this function

𝐶 (𝐹_{1} (𝑥_{1}), 𝐹_{2} (𝑥_{2}), \dots 𝐹_{𝑑} (𝑥_{𝑑}) = 𝐹 (𝑥_{1}, \dots 𝑥_{𝑑})

In the case of independent r.v.s, then it’s just the product. But maybe we have to do something further.

Then if you have the joint pdf $𝑓_{1} (𝑥_{1}, \dots, 𝑥_{𝑑})$ , then you also have a similar

𝑓 (𝑥_{1}, \dots, 𝑥_{𝑑}) = 𝑐 (𝐹_{1} (𝑥_{1}), \dots 𝐹_{𝑑} (𝑥_{𝑑})) 𝑓_{1} (𝑥_{1}) \dots 𝑓_{𝑑} (𝑥_{𝑑})

And this is just how derivatives work, this is the chain rule, where

𝑐 = \frac{𝑑}{𝑑 𝑥_{1}} \frac{𝑑}{𝑑 𝑥_{2}} \dots \frac{𝑑}{𝑑 𝑥_{𝑑}} 𝐶

There’s a second property that $𝐶$ is unique on $[0, 1]^{𝑑}$ . Basically, the function $𝑐$ connects the marginals, which are probabilities $[0, 1]$ . So you limit yourself to the space of distributions, which are the

ℝ \to [0, 1]

, then we are unique.

This is pretty powerful.

Issues with Copulas

There is a non obvious issue: time stationarity.

This is not even mentioned in most texts. When you estimate this, you need to estimate the marginals. You need a mountain of data to have these two things. You calculate the copula which connects them. Let’s say you have 2001-2006 and estimate the copula during that period. In order to make forecasts about the relationship in the future, you’re assuming that the relationship stays the same. This is the reason for the 2008 financial crisis. The CDOs are estimated assuming the functions doesn’t change, which is pretty stupid. Things don’t seem correlated right now, but they were, even though the historical data didn’t show it.

The second big issue: what EXACTLY is this copula?

We know this function exists, but we don’t know how to get to it. It’s like God.

We know God exists, but if we go by Christian religion it’s this god, if it’s Islam it’s that god, etc.

Copula Families

I’ll say that a copula function, sorta, looks like this. The only quantity I know is that it’s defined from $[0, 1]$ to $[0, 1]$ . So if I have a function that does this, I should be okay.

If you take 1D CDF, then what does it do? This is a function

𝐹 : ℝ \to [0, 1]

. It takes a random variable, and gives it a probability.

What about

𝐹^{- 1}

, the quantile function? This is defined

𝐹^{- 1} : [0, 1] \to ℝ

All copulas are based on this. They use the cdf and the inverse. The most popular one, we have already seen. The simplest one is called

Gaussian Copula

The whole thing is that the function needs to associate

ℝ^{𝑑} \to [0, 1]

. We have the multivariate normal function

Φ_{Σ}

CDF with mean 0 and covariance matrix $Σ$ . The cov matrix is positive, semidefinite, and symmetric.

Φ_{Σ} (𝑢_{1} \dots 𝑢_{𝑑}) = \int_{- \infty}^{𝑥_{1}} \int_{- \infty}^{𝑥_{𝑑}} \frac{1}{\sqrt{(2 𝜋)^{𝑑} det Σ}} 𝑒^{- \frac{1}{2} 𝑢^{𝑇} Σ^{- 1} 𝑢} 𝑑 𝑢_{1}, \dots 𝑑 𝑢_{𝑑}

$𝑢$ is a line vector.

This has to be between 0 and 1.

But this is defined for R.

So I’m going to write the gaussian copulas as

𝐶^{Gaussian} (𝑥_{1} \dots 𝑥_{𝑑}) = Φ_{Σ} (Φ^{- 1} (𝑥_{1}), \dots, Φ^{- 1} (𝑥_{𝑑}))

This is really complicated. The $𝐹$ s are the $𝑥$ s, you plug in the marginals, and that gives you the copula function. The important concepts are the multivariate Gaussian with given $𝜎$ , the $Φ^{- 1}$ inverse CDF, and you plug in the CDFs of the marginals. You look at your data, obtain the marginals, and obtain the joint. And fit the function to the joint. The only thing to estimate here is $Σ$ .

Student t Copula

There is another thing called the Student t copula. Remember the Cholesky decomposition? It’s basically used for this. You use the data, you use the numbers you get, and calculate the joint distribution. The correlated numbers you plug into the copulas, and you get the thing correlated. Fancy way of doing the homework.

It’s an extension that uses the fact that

𝑡 = \frac{𝑁 𝑜 𝑟 𝑚 𝑎 𝑙}{Χ^{2}}

You obtain $𝜒^{2}$ by squaring the normal.

This is useful.

Gaussian copula is nice and easy to work with.

Archimedean Copulas

This is an entire class. We have Gaussian, and student t which inherits from Gaussian processes. (By the way, if you want to know more about Gaussian processes, take Florescu’s class).

The whole point is that you’re going with CDFs, and inverse CDFs. Remember a copula goes from $[0, 1]^{𝑑}$ to $[0, 1]$ . It is defined like this

𝐶 (𝑢_{1}, \dots 𝑢_{𝑑}) = Ψ^{- 1} (Ψ (𝑢_{1}) + \dots + \dots Ψ (𝑢_{𝑑}))

where

Ψ : [0, 1] \to [0, \infty)

and

Ψ (1) = 0

This $Ψ$ function is kinda like inverse CDF.

Importantly, $Ψ^{- 1}$ is monotonic. And it follows that

$Ψ^{- 1} (0) = 1$

$Ψ$ goes from $[0, 1]$ to $\infty$ , and the inverse goes the other way.

And there’s a property that $Ψ^{- 1}$ starts at 1, and keeps going down over to $\infty$ .

Product Copula

You take the

Ψ (𝑢) = - ln 𝑢

- ln 𝑢 = 𝑦

𝑢 = 𝑒^{- 𝑦}

So that’s the inverse

Ψ^{- 1} (𝑦) = 𝑒^{- 𝑦}

Remember this starts at 1 at $𝑥 = 0$ , then decreases towards $\infty$ .

The archimedean copula function is the sum of $Ψ$ . So what is the copula in this case?

𝐶 (𝑢_{1}, \dots 𝑢_{𝑑}) = Ψ^{- 1} (Ψ (𝑢_{1}) + \dots + Ψ (𝑢_{𝑑}))

= 𝑒^{- (- ln 𝑢_{1} - ln 𝑢_{2} - \dots - ln 𝑢_{𝑑})} = 𝑒^{ln (𝑢_{1} \dots 𝑢_{𝑑})}

which ends up being

= 𝑢_{1} 𝑢_{2} \dots 𝑢_{𝑑}

You take each cdf and multiply them, which is numbers between 0 and 1.

It’s kind of useless because it doesn’t have a parameter, so you can’t fit it.

Clayton

These are the most used ones

Ψ (𝑢) = \frac{1}{𝜃} (𝑢^{- 𝜃} - 1)

Basically you construct these, by picking something. Someone came up with a function such that the inverse looks like this, then they pick $𝜃$ such that they have a parameter to fit something, where $Ψ (1) = 0$ .

They actually are simple, and then they just fit the properties. SOmoene tried it, now everyone uses it.

The inverse is

Ψ^{- 1} = (𝜃 𝑦 + 1)^{- \frac{1}{𝜃}}

𝐶^{Clayton} (𝑢_{1} \dots 𝑢_{𝑑}) = (𝜃 (Ψ (𝑢_{1}) + \dots + Ψ (𝑢_{𝑑})) + 1)^{\frac{1}{𝜃}}

Then just plug in the $Ψ$ term.

= (𝜃 (\sum_{𝑖 = 1}^{𝑑} \frac{1}{𝜃} (𝑢_{𝑖}^{- 𝜃} - 1)) + 1)^{\frac{1}{𝜃}}

= (\sum_{𝑖 = 1}^{𝑑} 𝑢_{𝑖}^{- 𝜃} - 𝑑 + 1)^{\frac{1}{𝜃}}

This formula IS NOT ONLINE, only for bivariate copula. All the formulas online are bivariate

I have four more pages of crap for Frank and Gumbel. But…

Frank

This is a generator

Ψ (𝑢) = - log (\frac{𝑒^{- 𝜃 𝑢} - 1}{𝑒^{- 𝜃} - 1})

Gumbel

Ψ (𝑢) = (- log 𝑢)^{𝜃}

And if you understand the principles, you can come upw ith your own. you can notice that you have logs everywhere, because it’s very nice because when you exponentiate it disappears. But it could be something like cosine! The only requirement is that the inverse function is convex.

The $𝜃$ is the parameter fit. The way you fit $𝜃$ , the relationship is

𝐶 (𝐹_{1} (𝑥_{1}), \dots 𝐹_{𝑑} (𝑥_{𝑑})) = 𝐹 (𝑥_{1} \dots 𝑥_{𝑑})

You’re just inventing something to fit the copula, something close.

It’s a high barrier of entry because the formulas seem complicated.

Why Do We Care

We need to estimate the wealth and know the risks, and come up with a number.

But how do we actually use this stuff?

I used to do a lecture on generating random variables. Now I don’t. If you’re interested in more detail, read the Handbook of Probability. Chapter 6 proves this. If you take a random vairalbe and plug it into a cdf, it generates a uniform random variable.

This tells you how to generate any rv you want.

All you have to do is the inverse cdf method.

𝑈 \sim Uniform (0, 1)

and you can get the inverse cdf

𝐹^{- 1} (𝑈)

has the same distribution as X.

If you can calculate the CDF, and the inverse CDF, you can generate random uniform, plug it into it, and then you get a random.

The problem is the normal CDF, because the normal CDF is not invertible. But a lot of other rvs work.

The problem is that it only works for 1D rvs. That’s why this copula is pretty useful.

Recall that the joint distribution cdf is equivalent to the copula of the marginal cdfs. The methodology will work for the marginal cdfs, so you can apply the copulas to get to the joint cdf.

Algorithm

First you estimate C.

1: First you generate many uniforms iid $𝑢_{1}, \dots, 𝑢_{𝑑}$

2: For each such point, you calculate the $𝐶 (𝑢_{1}, \dots, 𝑢_{𝑑})$ for all generated $𝑢_{𝑖}$ s.

3: Calculate marginals densities

4: I now have all these probabilities that I generate $𝑥_{1}, \dots, 𝑥_{𝑑}$ .

My $𝑥_{𝑖}$ CDF can be used to generate numbers, I have $𝑑$ distributions, I have a number between 0 and 1, and extract the corresponding $𝑥_{𝑖}$ , which gives me a generated number. That will calculate not only VaR but also ES with respect to these distributions.

There could be an easier version of this, which is what Marina will do. This is just how I would do it.

## Counterparty Risk-Weighted Assets - URL: https://sharifhsn.dev/blog/counterparty-risk-weighted-assets/ - Structured data: https://sharifhsn.dev/api/posts/counterparty-risk-weighted-assets.json - Description: The syllabus records “Risk Weighted Assets: Counterparty RWAs. Calculation and modeling,” taught by Juan. The personal course export contains the scheduled topic but no worked… - Date: 2025-04-28 - Exact published timestamp: 2025-04-28 - Topics: FX, Counterparty Risk, Risk-Weighted Assets - Categories: FX - Source: FE-635 \| Risk Engineering - Source URL: None - Content source format: typst

Week 12 — 28 April 2025

The syllabus records “Risk Weighted Assets: Counterparty RWAs. Calculation and modeling,” taught by Juan.

Source boundary

The personal course export contains the scheduled topic but no worked RWA calculation or lecture transcript. The previous capital-model explanation was an LLM summary, not a recovered note, so it has been removed rather than presented as class material.

## Multi-Name Credit Risk - URL: https://sharifhsn.dev/blog/advanced-derivatives-week-13/ - Structured data: https://sharifhsn.dev/api/posts/advanced-derivatives-week-13.json - Description: This is going to be a basic introduction, this is a complex topic to cover comprehensively. The Gaussian latent-variable framework and conditional-independence construction are… - Date: 2025-04-24 - Exact published timestamp: 2025-04-24 - Topics: Credit, Copulas, Dependence Modelling, Multi-Name Credit Risk - Categories: Credit - Source: Advanced Derivatives - Source URL: None - Content source format: typst

Copulas

This is going to be a basic introduction, this is a complex topic to cover comprehensively.

Limitations of Multi-Name Latent Variables

The Gaussian latent-variable framework and conditional-independence construction are developed in O’Kane’s Chapter 13 (pp. 241–258). Its one-factor version is useful for simulating default times and calibrating thresholds to single-name hazard rates, but one common factor constrains how the names can move together.

The Multi-Name latent variable model is single factor. There is one variable which is seen by the other credits. They each have their own idiosyncratic component. In terms of properties, simulations, that’s it. We looked at conditional hazard rate and calculation of the portfolio of loss. Last assignment is related to this particular topic.

We can model the correlation, we can simulate, we can calibrate, the time-dependent threshold, based on the hazard rate and survival probabilities.

How can we extend this model?

There’s limitations on the one-factor correlation structure.

In one-factor, we have a number of correlation parameters $𝑁_{𝑐}$ . Recall that the correlation between two credits $𝑖$ and $𝑗$ is $𝛽_{𝑖 𝑗}$ . We have the constraint that the absolute value of each correlation parameter is less than or equal to $1$ .

The entire correlation matrix has $\frac{𝑁_{𝑐} (𝑁_{𝑐} - 1)}{2}$ correlation parameters, which is a lot.

The question we can ask: Does one factor structure prevent the modeling of groups, or sectors? The model may be insufficient.

We can look at some examples. Consider a simple portfolio of $4$ credits grouped in two sectors. How can we represent such groups? We could take $𝛽_{1} = 𝛽_{2} = 𝛽_{𝑎}$ . These two stocks will have the same exposure to the market variable $𝑧$ , and vice versa for the other sector $𝛽_{3} = 𝛽_{4} = 𝛽_{𝑏}$ .

In a single factor model, the correlation is the product of two correlation parameters for two credits.

𝐶 = [\begin{matrix} 1 & 𝛽_{𝑎}^{2} & 𝛽_{𝑎} 𝛽_{𝑏} & 𝛽_{𝑎} 𝛽_{𝑏} 𝛽_{𝑎}^{2} & 1 & 𝛽_{𝑎} 𝛽_{𝑏} & 𝛽_{𝑎} 𝛽_{𝑏} 𝛽_{𝑎} 𝛽_{𝑏} & 𝛽_{𝑎} 𝛽_{𝑏} 1 & 𝛽_{𝑏}^{2} 𝛽_{𝑎} 𝛽_{𝑏} & 𝛽_{𝑎} 𝛽_{𝑏} & 𝛽_{𝑏}^{2} & 1 \end{matrix}]

If you split this into four square matrices, then you’ll see the sub-correlation matrix for each group. The correlation between the two groups.

Let’s assume that beta_a > beta_b > 0. One is more exposed to the market than the other. This also implies that

𝑐 = 𝛽_{𝑎}^{2} \geq 𝛽_{𝑎} 𝛽_{𝑏} \geq 𝛽_{𝑏}^{2}

For example, take $𝛽_{𝑎} = 0.8$ and $𝛽_{𝑏} = 0.5$ . The within-sector correlations are $𝛽_{𝑎}^{2} = 64 %$ in sector A and $𝛽_{𝑏}^{2} = 25 %$ in sector B. The cross-sector correlation is $𝛽_{𝑎} 𝛽_{𝑏} = 40 %$ . Sector B is therefore more correlated with sector A than it is with itself. That is the one-factor limitation: every correlation must be the product of two loadings on the same market factor.

Sector B is not fully captured.

So we can try to extend the model, make it more complex, and accommodate this particular scenario.

In this case, we can consider a two-factor version of the model.

For some credit i, we have

𝐴_{𝑖} = 𝛽_{1 𝑖} 𝑍_{1} + 𝛽_{2 𝑖} 𝑍_{2} + \sqrt{1 - 𝛽_{1 𝑖}^{2} - 𝛽_{2 𝑖}^{2}} 𝜖_{𝑖}

𝐴_{𝑗} = 𝛽_{1 𝑗} 𝑍_{1} + 𝛽_{2 𝑗} 𝑍_{2} + \sqrt{1 - 𝛽_{1 𝑗}^{2} - 𝛽_{2 𝑗}^{2}} 𝜖_{𝑗}

Similar to the previous formulation, but with one more factor.

The linear combination of standard normal is standard normal, so that doesn’t change.

The latent variable A_j corresponds still to the time-dependent threshold for each credit.

We can calculate the correlation between these assets in a similar kind of matrix. For each element:

𝑐_{𝑖 𝑗} = 𝛽_{1 𝑖} 𝛽_{1 𝑗} + 𝛽_{2 𝑖} 𝛽_{2 𝑗}

We can revisit the matrix in the following case.

In sector A, the credits have a first factor weight and second factor weight beta_1a and beta_2a.

Sector B is similar with beta_1b and beta_2b

This is more generic, you would say this group has correlation to these factors.

We’ll assume the first credits are from sector A.

That would be the correlation between credit a in sector a and b in sector b

𝑐 = [\begin{matrix} 1 & 𝛽_{1 𝑎}^{2} + 𝛽_{2 𝑎}^{2} & 𝛽_{1 𝑎} 𝛽_{1 𝑏} + 𝛽_{2 𝑎} 𝛽_{2 𝑏} & 𝛽_{1 𝑎} 𝛽_{1 𝑏} + 𝛽_{2 𝑎} 𝛽_{2 𝑏} 𝛽_{1 𝑎}^{2} + 𝛽_{2 𝑎}^{2} & 1 & 𝛽_{1 𝑎} 𝛽_{1 𝑏} + 𝛽_{2 𝑎} 𝛽_{2 𝑏} & 𝛽_{1 𝑎} 𝛽_{1 𝑏} + 𝛽_{2 𝑎} 𝛽_{2 𝑏} \end{matrix}]

(Bottom half is same as top half, mirrored)

For example, give sector A loadings $(𝛽_{1 𝑎}, 𝛽_{2 𝑎}) = (0.8, 0)$ and sector B loadings $(𝛽_{1 𝑏}, 𝛽_{2 𝑏}) = (0, 0.5)$ . The within-sector correlations are then 64% for A and 25% for B, while the cross-sector correlation is the dot product $0.8 \cdot 0 + 0 \cdot 0.5 = 0$ . Separate factors can represent strong dependence within each group and weaker dependence across groups. More generally, the correlation between two credits is the dot product of their factor-loading vectors.

In general, M-factors are needed to model M-sector portfolio. Three sectors? Three factors.

The simulation is very simple, but instead of generating just a Z_1, we generate Z_1 and Z_2. It becomes a little more complex, with the number of parameters, compared to a single factor. We go from n(n-1)/2, to a lot more. If you want to implement these, you need to calibrate the exposure to the parameters. So the complexity becomes prohibitive, and that’s a limitation.

It has some applicability, but in general, the more typical way is through copula functions.

Copulas (Finally)

You can model default time dependencies, correlations for default times. We also will have measures of dependency. We will look at main copulas for credit modeling, and Monte Carlo pricing of correlated products with copulas.

We’ll look at some properties, and dependency limits, and some relations between copulas and the model we just discussed. And other ways of modeling dependence than correlation. If we have time, we’ll have some code to visualize copulas.

Next week, I have a section for estimation of parameters of copulas, more complex. But for the final exam, copulas will not be included.

The latent variable model can also be expressed as a Gaussian copula, so they are related.

Definition: An N-dimensional copula function C is a multivariate cumulative distribution function with N uniform marginals with probabilities u_1, u_2, … u_N.

This is just the multivariate cdf. One of the characteristics is that it has uniform marginals. So we can say that thIS IS THE JOINT PROBABILITY

ℙ (\hat{𝑈_{1}} \leq 𝑢_{1}, \hat{𝑈_{2}} \leq 𝑢_{2} \dots \hat{𝑈_{𝑁}} \leq 𝑢_{𝑁}) = 𝐶 (𝑢_{1}, 𝑢_{2} \dots 𝑢_{𝑁})

This is the definition of such copulas where

\hat{𝑈_{𝑖}} \sim Uniform (0, 1)

Dependent Structure of Default Times of N Credits

Let $𝑞_{𝑖 (𝑡)} = 𝑃 (𝜏_{𝑖} \leq 𝑡)$ be credit $𝑖$ ’s cumulative default probability by time $𝑡$ . The default-time copula is

$𝐶 (𝑞_{1} (𝑡_{1}), 𝑞_{2} (𝑡_{2}), \dots, 𝑞_{𝑁 (𝑡_{𝑁})}) = 𝑃 (𝜏_{1} \leq 𝑡_{1}, 𝜏_{2} \leq 𝑡_{2}, \dots, 𝜏_{𝑁} \leq 𝑡_{𝑁})$ .

It expresses the joint probability of defaults by their marginal default probabilities. A copula is a multivariate distribution with uniform marginals; it separates the choice of each name’s marginal default curve from the choice of dependence across names. For continuous marginal distributions, Sklar’s theorem gives a unique copula representation:

$𝐻 (𝑥_{1}, \dots, 𝑥_{𝑁}) = 𝐶 (𝐹_{1} (𝑥_{1}), \dots, 𝐹_{𝑁 (𝑥_{𝑁})})$ .

Setting all but one input to one recovers that name’s marginal, for example $𝐶 (1, \dots, 𝑢_{𝑘}, \dots, 1) = 𝑢_{𝑘}$ . If any default probability is zero, the joint-default probability is zero. The same construction works in $𝑁$ dimensions; the bivariate case is a convenient way to display it.

For the Gaussian latent-variable model, the bivariate default probability is

$𝑃 (𝜏_{𝑖} \leq 𝑡_{𝑖}, 𝜏_{𝑗} \leq 𝑡_{𝑗}) = Φ_{2} (Φ^{- 1} (𝑞_{𝑖 (𝑡_{𝑖})}), Φ^{- 1} (𝑞_{𝑗 (𝑡_{𝑗})}); 𝜌_{𝑖, 𝑗})$ .

Equivalently, its Gaussian copula is $𝐶_{𝜌}^{𝐺 𝐶} (𝑢, 𝑣) = Φ_{2} (Φ^{- 1} (𝑢), Φ^{- 1} (𝑣); 𝜌)$ . This is the expression that was missing from the original note at this point (O’Kane, Chapter 14, pp. 261–263).

Dependency Limits

Latent variable model has dependence limits. We kind of look to see when we have positive and negative dependence of names in the portfolio. This is not equal to the correlations, but we can think a little bit in that way.

When we have an expression for a copula, (the joint probability distribution), it’s good to look at these properties to see the maximum and minimum dependence, and then how to measure them, based on differences dependency metrics.

Independence

The simplest is independence.

The expression for such an independence copula is

𝐶 (𝑢_{1}, 𝑢_{2}, \dots 𝑢_{𝑁}) = 𝑢_{1} 𝑢_{2} \dots 𝑢_{𝑁} = \prod_{𝑖 = 1}^{𝑁} 𝑢_{𝑖}

If they’re independent, the joint probability distribution is the product of the marginals.

Perfect Positive Dependence

Perfect Negative Dependence

We can then state the Fréchet-Hoeffding Theorem:

For any copula $𝐶 : {[0, 1]}^{𝑁} \to [0, 1]$

and any realization $(𝑢_{1}, \dots 𝑢_{𝑁}) \in {[0, 1]}^{𝑁}$ , the following bounds hold:

𝑤 (𝑢_{1}, 𝑢_{2} \dots 𝑢_{𝑁}) \leq 𝐶 (𝑢_{1}, 𝑢_{2} \dots 𝑢_{𝑁}) \leq 𝑀 (𝑢_{1}, \dots 𝑢_{𝑁})

Basically this theorem says if you have a copula with a particular realization, this copula is bounded above and below by two functions M and w.

w the lower Fréchet-Hoeffding bound is

𝑤 (𝑢_{1}, \dots 𝑢_{𝑁}) = (1 - 𝑁 + \sum_{𝑖 = 1}^{𝑁} 𝑢_{𝑖})_{+}

and upper bound is

𝑀 (𝑢_{1}, \dots 𝑢_{𝑁}) = min (𝑢_{1}, \dots 𝑢_{𝑁})

In the bivariate case where we only have two variables u and v, we have

max (𝑢 + 𝑣 - 1, 0) \leq 𝐶 (𝑢, 𝑣) \leq min (𝑢, 𝑣)

Upper bound is perfect positive, lower bound is perfect negative.

This result holds for any copula.

Ideally we would like to estimate this copula with some data. We need to make a choice for this expression, and estimate those parameters.

But the question is, this is a representation of the original joint probability distribution, but is this a unique representation? Can we have different copulas for this?

We have one more theorem, Sklar’s Theorem.

We know by definition, the copula is a multivariate distribution function.

Any such function can be written as a copula, and its representation is unique, the marginal distributions are continuous. If you have discrete distributions, then uniqueness is not guaranteed.

Let’s consider more generally, any N-dimensional distribution function H with marginal distributions F_1, F_2, F_N for random variables x_1, x_2, … x_N. H is the joint probability distribution.

Because this is multi-dimensional, this will correspond to

𝐻 (𝑥_{1}, 𝑥_{2}, \dots 𝑥_{𝑁}) = ℙ (𝑋_{1} \leq 𝑥_{1}, 𝑋_{2} \leq 𝑥_{2} \dots, 𝑋_{𝑁} \leq 𝑥_{𝑁})

We can write this as the probability that the marginal function F_1

= ℙ (𝐹_{1} (𝑋_{1}) \leq 𝐹_{1} (𝑥_{1}), 𝐹_{2} (𝑋_{2}) \leq 𝐹_{2} (𝑥_{2}), \dots 𝐹_{𝑁} (𝑋_{𝑁}) \leq 𝐹_{𝑁} (𝑥_{𝑁}))

We basically want to transform this to get a copula function, to get here from a generic N-dimensional function.

Let $\hat{𝑈_{𝑖}} = 𝐹_{𝑖} (𝑋_{𝑖})$ and the realization $𝑢_{𝑖} = 𝐹_{𝑖} (𝑥_{𝑖})$ The marginal will take values between 0 and 1, it’s a cdf.

The original function

𝐻 (𝑥_{1}, 𝑥_{2} \dots 𝑥_{𝑁}) = ℙ (\hat{𝑈_{1}} \leq 𝑢_{1}, \dots \hat{𝑈_{𝑁}} \leq 𝑢_{𝑁})

𝐻 (𝑥_{1}, 𝑥_{2} \dots 𝑥_{𝑁}) = 𝐶 (𝐹_{1} (𝑥_{1}), 𝐹_{2} (𝑥_{2}) \dots 𝐹_{𝑁} (𝑥_{𝑛}))

We know that C is a unique copula if these Fs are continuous.

And in reverse,

𝐶 (𝑢_{1}, 𝑢_{2} \dots 𝑢_{𝑁}) = 𝐻 (𝐹_{1}^{- 1} (𝑢_{1}), 𝐹_{2}^{- 1} (𝑢_{2}), \dots 𝐹_{𝑁}^{- 1} (𝑢_{𝑁}))

The relationship between these two is that the copula is going to separate the choice of the marginals from the choice of the dependency structures. It decouples these things. We can take any function and its copula representation, and calibrate it to the names of the portfolio independent of the dependency structure based on the way I want to assign dependency to this portfolio.

We might want to add some dynamics to this dependencies, the difference between credit default so we can calculate the conditional survival/default probability. We could impose this by choosing a different dependency structure.

Alternative to Dependence Structure of Default Times

The survival copula uses survival probabilities $𝑠_{𝑖 (𝑡)} = 1 - 𝑞_{𝑖 (𝑡)}$ :

$\hat{𝐶} (𝑠_{1} (𝑡_{1}), \dots, 𝑠_{𝑁 (𝑡_{𝑁})}) = 𝑃 (𝜏_{1} > 𝑡_{1}, \dots, 𝜏_{𝑁} > 𝑡_{𝑁})$ .

In the bivariate case, the default and survival copulas are related by

$\hat{𝐶} (𝑢, 𝑣) = 𝑢 + 𝑣 - 1 + 𝐶 (1 - 𝑢, 1 - 𝑣)$ .

For the Gaussian latent-variable model, default thresholds are jointly normal. The probability that both thresholds have been crossed is

$𝑃 (𝜏_{𝑖} \leq 𝑡_{𝑖}, 𝜏_{𝑗} \leq 𝑡_{𝑗}) = Φ_{2} (Φ^{- 1} (𝑞_{𝑖 (𝑡_{𝑖})}), Φ^{- 1} (𝑞_{𝑗 (𝑡_{𝑗})}); 𝜌_{𝑖, 𝑗})$ .

Thus the Gaussian copula is

$𝐶_{𝜌}^{𝐺 𝐶} (𝑢, 𝑣) = Φ_{2} (Φ^{- 1} (𝑢), Φ^{- 1} (𝑣); 𝜌)$ .

The standard bivariate normal is unchanged when both variables are multiplied by $- 1$ . As a result, the Gaussian default and survival copulas have the same form. For a non-degenerate Gaussian copula, extreme joint events become less likely relative to their marginal tails as the thresholds move farther out: its upper- and lower-tail dependence coefficients are zero (O’Kane, Chapter 14, pp. 261–263, 269–270).

Measuring Dependence

An important property of this is dependence, so the question is how do you measure?

There are many ways.

The most popular way is Pearson Linear Correlation.

𝜌_{𝑃} = \frac{𝔼 [𝑋 𝑌] - 𝔼 [𝑋] 𝔼 [𝑌]}{\sqrt{𝔼 [𝑋^{2}] - 𝔼 {[𝑋]}^{2}} \sqrt{𝔼 [𝑌^{2}] - 𝔼 {[𝑌]}^{2}}}

Advantages: easier to understand. Everyone understands correlation. It is also invariant under linear transformation. Say we change the variables by scaling and shifting (multiplying and adding), it doesn’t affect the correlation.

For jointly Gaussian variables, zero correlation does imply independence.

However, there is a big disadvantage: linear correlation being 0 DOES NOT imply independence in general

Sometimes the correlation is abused. It is a measure of linear dependence between variables. But just because there’s no linear dependence, doesn’t mean there are other kinds of dependence.

Here is a five-point counterexample. Let $𝑋$ take the equally likely values $- 2, - 1, 0, 1, 2$ , and set $𝑌 = 𝑋^{2}$ . Then $𝐸 [𝑋] = 0$ , $𝐸 [𝑌] = 2$ , and $𝐸 [𝑋 𝑌] = 𝐸 [𝑋^{3}] = 0$ , so

$Cov (𝑋, 𝑌) = 𝐸 [𝑋 𝑌] - 𝐸 [𝑋] 𝐸 [𝑌] = 0$ .

Yet $𝑌$ is completely determined by $𝑋$ . The variables are dependent, but their linear correlation is zero. This illustrates why Pearson correlation alone does not describe general dependence (O’Kane, Chapter 14, §14.3, pp. 264–265).

Rank Correlation

Given $𝑛$ observations $(𝑋_{𝑖}, 𝑌_{𝑖})$ , let $𝑅_{𝑖}$ and $𝑆_{𝑖}$ be the ranks of $𝑋_{𝑖}$ and $𝑌_{𝑖}$ . A pair of observations is concordant when their order agrees:

$(𝑋_{1} - 𝑋_{2}) (𝑌_{1} - 𝑌_{2}) > 0$ .

It is discordant when the product is negative. Rank measures compare these orderings, so a monotone transformation of either variable does not change them.

For example, if five observations have identical ranks in both variables, then all $(\binom{5}{2}) = 10$ pairs are concordant and none are discordant. Kendall’s tau and Spearman’s rho are both $1$ . This spells out the perfectly ordered example in the notes (O’Kane, Chapter 14, §14.4, pp. 265–268).

Kendall’s Tau

𝜏 = \frac{𝑐 - 𝑑}{𝑐 + 𝑑}

where c is the number of concordant pairs, and d is the number of discordant pairs.

Then the total number of pairs is c + d, which n choose 2 = n(n-1)/2.

In the previous case, all of these are concordant, so this is 1/1.

And if it’s discordant, that’s −1.

The sample estimator of Kendall’s τ is

𝜏 = \frac{2 \sum_{𝑖 = 1}^{𝑛 - 1} \sum_{𝑗 = 𝑖 + 1}^{𝑛} sign ((𝑥_{𝑖} - 𝑥_{𝑗}) (𝑦_{𝑖} - 𝑦_{𝑗}))}{𝑛 (𝑛 - 1)}

So basically you’re enumerating all these pairs, and then dividing it by the known bottom, and flipping the 2.

For continuous variables, independence gives $𝜏 = 0$ , perfect positive dependence gives $𝜏 = 1$ , and perfect negative dependence gives $𝜏 = - 1$ .

For a continuous x, y, Kendall’s tau should be

𝜏_{𝑥 𝑦} = 4 \int_{0}^{1} \int_{0}^{1} 𝐶 (𝑢, 𝑣) 𝑑 𝐶 (𝑢, 𝑣) - 1

Spearman’s Rho

Spearman’s rho is Pearson correlation applied to the ranks. With no ties, it can be calculated directly from the rank differences:

𝜌_{𝑆} = \frac{\sum_{𝑖 = 1}^{𝑛} (𝑅_{𝑖} - \bar{𝑅}) (𝑆_{𝑖} - \bar{𝑆})}{\sqrt{\sum_{𝑖 = 1}^{𝑛} (𝑅_{𝑖} - \bar{𝑅})^{2}} \sqrt{\sum_{𝑖 = 1}^{𝑛} (𝑆_{𝑖} - \bar{𝑆})^{2}}} = \frac{12 \sum_{𝑖 = 1}^{𝑛} (𝑅_{𝑖} - \bar{𝑅}) (𝑆_{𝑖} - \bar{𝑆})}{𝑛 (𝑛^{2} - 1)}

Furthermore, for a continuous x and y.

𝜌_{𝑆} = 12 \int_{0}^{1} \int_{0}^{1} 𝐶 (𝑢, 𝑣) 𝑑 𝑢 𝑑 𝑣 - 3

This is important because for each copula it’s going to be dependent, as a function of those parameters, it’s going to have different values in the Pearson, Kendall or Spearman value.

There are some advantages to rank correlation, which is that it’s able to capture nonlinear dependence, but it has some disadvantage as well.

Tail Dependence

Rank correlation records whether observations tend to move in the same or opposite order, but it does not measure the magnitude of extreme co-movements. Tail dependence asks whether one variable remains extreme when the other is already in an extreme tail.

For variables with marginal CDFs $𝐹_{𝑋}$ and $𝐹_{𝑌}$ , the upper-tail dependence coefficient is

$𝜆_{𝑈} = {lim}_{𝑢 \to 1} 𝑃 (𝑌 > 𝐹_{𝑌}^{- 1} (𝑢) | 𝑋 > 𝐹_{𝑋}^{- 1} (𝑢))$ .

The lower-tail coefficient is

$𝜆_{𝐿} = {lim}_{𝑢 \to 0} 𝑃 (𝑌 \leq 𝐹_{𝑌}^{- 1} (𝑢) | 𝑋 \leq 𝐹_{𝑋}^{- 1} (𝑢))$ .

Here $𝑢$ is a quantile level. In credit-risk terms, joint defaults correspond to the lower tail of latent asset values, while large portfolio losses correspond to the upper tail of losses. A positive coefficient means that an extreme observation in one variable has a non-vanishing limiting probability of being accompanied by an extreme observation in the other. The tail plot in the notes is illustrating this limiting behavior.

The Gaussian copula has zero upper- and lower-tail dependence for correlations strictly between $- 1$ and $1$ . It can match ordinary dependence while understating the chance of joint extremes; a copula with tail dependence is needed when that behavior matters (O’Kane, Chapter 14, §§14.5–14.6, pp. 269–270).

## Neural Networks and Credit Ratings - URL: https://sharifhsn.dev/blog/computational-methods-week-13/ - Structured data: https://sharifhsn.dev/api/posts/computational-methods-week-13.json - Description: This is an idea that was created in the 50s, exploded in the 90s, where it was called neural networks. A lot of development on the brain happened, so ANN was a term to… - Date: 2025-04-22 - Exact published timestamp: 2025-04-22 - Topics: Computational Methods, Neural Networks, Backpropagation, Credit Ratings, Volatility Estimation - Categories: Computational Methods - Source: Computational Methods in Quantitative Finance - Source URL: None - Content source format: typst

Artificial Neural Networks (ANN)

This is an idea that was created in the 50s, exploded in the 90s, where it was called neural networks. A lot of development on the brain happened, so ANN was a term to distinguish from the completely different human neural network. There were a lot of movies about AI, grants given, and they all failed. In 2000, every grant would be rejected because nobody believed it was intelligent.

The problem is this stuff is very prone to overfitting. It fits the data you have really well, but trying to predict something else will fail miserably. It cannot move outside that realm. In 2020, they used this to learn the vocabulary. It’s based on the same technology, they put several neural networks together and learned the English language. Turn sout that it works because you can overfit that no problem, because everything you can ask fits in the sample. That’s the principle of generative language. There are some morons like Sam Altman and Elon who don’t understand the technology and think it’s really intelligent.

You can only do well in your overfitted universe. You can’t do well outside of that. GPT-5 was supposed to do judgement and reasoning, but it can’t do this on the limits of technology, a retrieval augmented mechanism.

Today, we’re not talking about any of that. This course is about the bottom of it all, how exactly do you bild a neural network?

ANN is a very general framework, everything that has this kind of mechanics, with lots of designs. LSTM (long short term memory) is an example.

Multilayer Perceptron

The simplest possible ANN is this, which is inspired by brain function.

Today we will describe just a single layer, but I’m going to explain how it works completely.

At the core, you have

𝑥 \in ℝ^{𝑑}

. This is the number. You cannot do any of this with qualitative values. This is why LLMs are built on embeddings, which are numbers. And you also have

𝑦 \in ℝ^{𝑘}

You observe $𝑥$ and $𝑦$ , and you want to know the connection. If I input $𝑥$ , how do I get $𝑦$ ? The oldest problem: I have some function $𝑦 = 𝑓 (𝑥)$ , and I want to know what it is.

This is the one variable description. With multiple variables, you start talking about regressions, ANOVA, etc. These are all functions that relate x to y. But all of these things in typical regression are linear, which means f is a linear function.

What does linear mean exactly?

𝑓 (𝑥) = 𝑎 𝑥

In terms of matrices, if

𝑥 \in ℝ^{𝑑}

, then

𝑓 (𝑥) = 𝐴 𝑥

where A is a matrix

𝐴 \in 𝑀_{𝑘 \times 𝑑}

times the x vector

𝑑 \times 1

The whole problem here is to find the A that will give me the output y.

This is distinct from an affine function. This linear function is constrained so that $𝑥 = 0$ always gives me 0, $(0, 0)$ is valid.

In general, affine looks like

𝑓 (𝑥) = 𝐴 𝑥 + 𝑏

This b makes it affine. Instead of estimating A matrix and b vector, I estimate both. This is regression, nothing fancy here, I’m just making it look complicated, but it’s really nothing. There is a point in me doing this. These are the building blocks for the neural network.

The next question in the 50s was this. What is the idea? I am going to somehow measure the distance between my output $𝑦$ and the predicted thing

\hat{𝐴} 𝑥 + \hat{𝑏}

. This is kind of like a difference for $𝑦_{observed} - 𝐴 𝑥 + 𝑏$ , where $𝐴$ and $𝑏$ are observed. I somehow want to minimize this distance. What does that mean? The simplest thing is Euclidean distance. These are all points in

ℝ^{𝑘}

, so you can take the sum of squares. You can also take absolute values, but it doesn’t work that well because it’s not differentiable. The minimization happens when you take the derivative equal to 0. Once I have the minimization, I can obtain the value $\hat{𝐴}$ and $\hat{𝑏}$ . In statistics we take least squares regression, that’s what this is.

Remember that $𝐴 𝑥 + 𝑏$ is $𝑓 (𝑥)$ . I am literally taking the difference between $𝑦$ and $𝑓$ and finding out $𝑓$ . But what if the function $𝑓$ is nonlinear? Then I’m screwed. The least squares is very simple to derive because the other term will disappear when you take the derivative of a particular term.

The advantage in the 80s was to come up with this function.

The Idea

Here are the things we have to work with:

𝑥_{1}^{1} \dots 𝑥_{𝑑}^{1}, 𝑦_{1}^{1} \dots 𝑦_{𝑘}^{1}

Typical the models we use are output one dimensional, but nothing prevents you from making it k-dimensional

And then all the way up to

𝑥_{1}^{𝑛} \dots 𝑥_{𝑑}^{𝑛}, 𝑦_{1}^{𝑛} \dots 𝑦_{𝑘}^{𝑛}

I have n observations. I have 40 people in the class and I start measuring their x values, like height. Then I monitor something internal to predict, their cholesterol level.

There was a commercial which was the IBM intelligent machines. In Manhattan, for this coffee place, they discovered as they train people to order more puffs, then the IBM machine looked into the data, and they sold more puffs. I hated that commercial because it implied that the machine somehow went int o the data nad saw that connection. You are literally looking at all the k products. First of all, it has to be logical. Machines cannot discover connections. You the researcher will hypothesize the connection, and the machine will check it for you.

The question is, how do we relate x to y? This is howt he neural network works.

(Full diagram in notes)

Let me take the components

$𝑥_{1}, \dots, 𝑥_{𝑑}$

This is for a generic input. We talk about random variables in financial engineering. These are observations. We are making a relationship between the random variables in the input and the output

$𝑦_{1}, \dots, 𝑦_{𝑘}$ .

How do you make this connection? We are going to form this hidden layer. In the simplest case, just one layer. We will call it

$ℎ_{1}, \dots, ℎ_{𝑒}$ .

I’m going to make a nonlinear activation function. I’ll combine all of the $𝑥$ s into a nonlinear relationship with the $ℎ$ s. Then I’ll combine these $ℎ$ s in a nonlinear relationship with the $𝑦$ s. How do you do this? First you take for every $ℎ_{𝑖}$ , you consider the function $𝑔_{1}$ . The difference between linear and affine, we will construct an affine relationship. I will add an $𝑥_{0} = 1$ here. This is a free term, this constant. I will add some weights which connect these $𝑥$ s to the hidden layers.

$𝑤_{01}^{1}, 𝑤_{11}^{1}, \dots, 𝑤_{𝑑 1}^{1}$ .

This $ℎ_{𝑖}$ will be a function of

ℎ_{𝑖} = 𝑔_{1} (\sum_{𝑗 = 0}^{𝑑} 𝑤_{𝑗 𝑖}^{1} 𝑥_{𝑗})

This relationship is really not important. I am literally taking a linear relationship, this function $𝑔_{1}$ , and I’m applying the affine to it

= 𝑔_{1} (𝑤_{0 𝑗} + 𝑤_{𝑖 𝑗} 𝑥_{1} + \dots + 𝑤_{𝑑 𝑗} 𝑥_{𝑑}

I’m taking all of the inputs times this weight. And that’s just for one node.

I need to do it for all layers because I will combine them when I reach $𝑦$ .

The $𝑦$ will be another function $𝑔_{2}$ .

We will add another $ℎ_{0} = 1$ at the top, a constant. So that this function will be

𝑦_{𝑘} = 𝑔_{2} (𝑤_{0 𝑘}^{2} + 𝑤_{1 𝑘}^{2} ℎ_{1} + \dots + 𝑤_{𝑒 𝑘}^{2} ℎ_{𝑒}

Literally, you are taking a linear combination of the weights, times the nodes, and you put it into the activation function, and put it into the nex tithing. At the end, you have these output y from a mountain of inputs.

So what do we do next?

Activation Functions

We mentioned $𝑔_{1}$ and $𝑔_{2}$ . What are those? These are called activation functions. If you’ve taken statistics, or FA590, at the end of the class, you do something called logistic regression. You are associating real numbers with probability. Because you need to map them continuous to continuous, since you can’t really map from continuous to discrete. Instead of mapping $𝑥$ into the outcome, I will map $𝑥$ into something which maps into $[0, 1]$ , this reduced region. The typical activation functions are the following. Technically you can use anything, but you will usually depend on one of seven fundamental functions, something every math student learns.

Exponential, tirgonometric, polynomial, etc.

(Graphs in notes)

Examples include:

hyperbolic tangent:
$𝑔 (𝑥) = tanh (𝑥) = \frac{𝑒^{𝑥} - 𝑒^{- 𝑥}}{𝑒^{𝑥} + 𝑒^{- 𝑥}}$
This maps into $[- 1, 1]$ If x is multidimensional, then replace this with something multidimensional You can’t use cosine because it will explode at some values. This is what these things are trying to do. If you have x in the left region, it maps into a negative value, and the right is positive. So you can literally by playing with the weights, you can make your resulting point, which is a linear combination of inputs times weights, you can make it be either left or right. You can guide your output to y to be more positive or more negative.
logistic:
$𝑔 (𝑥) = \frac{1}{1 + 𝑒^{- 𝑥}}$
It’s kinda similar to tanh, but the difference is that it goes into $[0, 1]$ . The reason it’s called logistic and it’s used in logistic regression, because you’re mapping real numbers into probabilities, and this is the correct domain for probabilities.
Rectified Linear Unit ReLU:
$𝑔 (𝑥) = max (𝑥, 0) = 𝑥_{+}$
This is used in machine learning quite extensively. My mom decided to call me Ionut, so I’m only known by that one name. But the computer science decided to call this The ReLU gate is used to delete negative numbers. It’s used a lot in computer engineering.
Softplus:
$𝑔 (𝑥) = log (1 + 𝑒^{𝑥})$
ReLU is called xplus. But it has a problem at 0, where it’s not derivable. So this makes it smooth. It’s nonzero everywhere and derivable, but it has the same general form.

This is the activated function, you can use it in both places, construct it however you like. It’s really important that you don’t do it with trial and error. People fire up PyTorch and use defaults, and think it’s good. Obviously each of these activation functions have their own meanings. You need to know when to use one or the other.

Typically, the probability stuff uses logistic for the y. Typically, you want to predict some kind of number. Sometimes SoftReLU is used, sometimes regular ReLU, I don’t know which one is better. Depends on the output’s relationship with the input.

The Weights

How many weights, and how do we estimate them?

How many weights is pretty simple. I have $𝑑$ input, and $𝑘$ output. In that case, it depends on this internal layer. We’ll say the hidden layer has $𝑒$ nodes. Then I have $𝑑 + 1$ inputs (adding the affine part). Each of those $𝑑 + 1$ , I weight each of them to connect to each of the $𝑒$ nodes. That’s $(𝑑 + 1) 𝑒$ . That’s just for $𝑤_{1}$ . Now I have to do the same thing to connect with $𝑘$ output $𝑦$ s. $𝑒 + 1$ affine term, and $𝑘$ of it. $𝑤_{2}$ has $(𝑒 + 1) 𝑘$ terms. So it’s total

$(𝑒 + 1) 𝑘$

We’ll say x is 20 dimensional, we have 20 parameters, and y is one dimensional, with one output.

What is the typical number of nodes in a hidden layer? 64

‘

That would make this calculation 1409 weights. That’s an immense number of weights. In regression, if we have 20 inputs, then we have 22 parameters. If this is a class, and I add class characteristics, and I have years, and I have only 1200 cases, and I fit a neural network, what’s going to happen? I will get a perfect fit. Not only that, it will not be unique, because there are too many weights compared to observations. What about 10k cases? It will still be overfit. You will need millions of observations. That’s why when you’re doing discriminants between images, you need thousands. There are advanced methods to deal with this.

If you want to create a model of a person (this is an old thing, the computer vision problem). When I ame to Stevens, I did CV research for a while, in it searly days. When you plugged in 20 algorithms, 19 would not even work, just give wrong answers. They were based on this overfitting data that you have. Today is 20 years later, very mature CV, things that appear and disappear, all these techniques. You can get the shape from shading problem by lighting it differently, and get a model of a person. A friend of mine has this problem 2D to 3D. All of this is set in this particular domain.

Backpropagation

This is complicated, I’m not going to explain in detail. But I will explain the fundamentals.

We have these weights that we’re trying to estimate. I know if I put my weights with my function, I get a candidate.

So we get something called a Loss Function. We already have one here, which is the least squares.

There are two cases, if y is discrete or continuous. For the height of a student it’s continuous. If you output a gender, it’s discrete.

Squared error loss is good if it’s continuous:

\sum_{𝑖 = 1}^{𝑛} (𝑦_{𝑖} - \hat{𝑦_{𝑖}})^{2}

${\hat{𝑦}}_{𝑖}$ is the output of the neural network.

\sum_{1}^{|} 𝑦_{𝑖} - 𝑔_{2} | \sum 𝑤_{𝑖}^{1} 𝑔_{1} (\sum 𝑤^{2} 𝑥))^{2}

This function contains all 1000 observations, and minimizes it.

The backpropagation does not minimize all the 1000. It minimizes $𝑤_{2}$ first, then it minimizes $𝑤_{1}$ with respect to that, propagating the weights backwards, and then doing it again.

This is the simple part. The complicated part…

I learned in my other class that I should use mean squared error. Who gives a shit. Some of you may say that some are more important than others. You can definitely do that. You might say it’s very important to fit particular ys, and make it a weighted sum. But most of the time, it basically looks like this, with some kind of Euclidean distance.

What do you do in the discrete case?

They use something called Cross-Entropy which is bullshit. The loss function for the discrete variables is called this, and it’s hard to explain.

So first I will explain something simpler.

If we have discrete, we are going to output either male or female. We’re going to create two nodes. We will have the probability of one and one minus the probability of the other. If you have multiple outputs, you use a hot-cold encoder, which maps this into a multi-dimensional space. The upshot is that you have to compare the observed discrete distribution and what you output, which is going to be probability. The function mapping is of numbers. In any of these ML techniques, you will take those numbers and calculate the probability distribution out of them.

Then how do you compare this probability distribution from the model, with the true probability distribution. Let’s say I have three outputs, green yellow red. If my object that I have is a combination of them, I will write down the combination. That’s the distribution I’m looking for. Most of the time, the categorial thing I’m outputting is one of them. 30s, 20s, teens. When I’m looking at one picture, the actual observed will be 0 0 1 0 0. The probability distribution is 1 that belongs to this category and 0 to not others. So we nee dto calculate the difference between the output and this distribution.

We will use cross entropy, but it’s hard to understand so we’re not using this.

“If p and q are two probability distributions, and they have the same support. The rv is characterized by probability and outcome. The support is the domain of outcomes.”

For continuous, this is not useful, because we have MSE. But it can be useful for discrete.

𝐻 (𝑝 𝑞) = - 𝔼^{𝑝} (log 𝑞) = - \sum_{𝑖 = 1}^{𝑘} (log 𝑞_{𝑖}) 𝑝_{𝑖}

What does that mean? It’s outcome times probability. It’s kind of interesting. They have to be the same outcome to calculate this. So how does this work in terms of ML techniques. I have three outcomes: red, green, blue. One of the distributions is going to be 1 0 0, the other one from the neural network will be a b c, which corresponds to red green blue. So we have to pair these.

That’s the cross-entropy.

Instead we will look at Kullback-Leibler Divergence. This is defined as

𝔼^{𝑃} (log \frac{𝑃}{𝑄}) = \sum_{𝑖 = 1}^{𝑘} log \frac{𝑝_{𝑖}}{𝑞_{𝑖}} 𝑝_{𝑖} = \sum 𝑝_{𝑖} log 𝑝_{𝑖} - \sum 𝑝_{𝑖} log 𝑞_{𝑖}

IT’s pretty much the same thing as entropy in practice. The first term is only in p, which is typically a constant, it’s what I know.

This is simple to understand. If $𝑝_{𝑖}$ is close to $𝑞_{𝑖}$ , this is close to 1. The logarithm of 1 is 0. So it’s basically it’s a bunch of small parts of 0. This looks like a distance between $𝑝$ and $𝑞$ . But it’s a divergence, not a distance.

Distance has three properties. distance should be commutative, reflexive, and additive (triangle inequality). This only follows the first two. If something is equal to 0, then the two are the same. But, if something is tiny different from 0, and in other experiment you have the same tiny number, you cannot say that this cross function is the same. The distance itself is meaningless. Distance, I can measure 300 km here, 300 km in Romania, which is the same. Divergence, hell no. ou measure something here, the Trump tariff will be different than in Romania, because that’s not a distance.

This is neglected in all of CS literature. It’s a lot of wishy-washy here.

These numbers are special numbers, they’re between 0 and 1, and all sum up to 1. That’s why you can’t use MSE. If one is distant, the other two are close. In the most common situation, the output is one and a bunch of 0s.

It turns out that thing that maximizes it will give you 1 for each of the outcomes. That would be completely overfitting.

READ ABOUT LOGISTIC REGRESSION.

The way that it’s done is the same as it’s done with this, except you’re using logistic vs Kullback-Leibler or Cross Entropy.

Logistic Regression

$𝑞_{𝑖}$ is the output of the function you get from the node.

𝑞_{𝑖} = \frac{1}{𝑞 + 𝑒^{- 𝑤^{𝑖} 𝑥}}

If you’re going directly from $𝑥$ to $𝑦$ without a hidden layer in between. There are weights that correspond exactly to this particular outcome.

Typically you take $𝑞_{𝑖}$ as

𝑞_{𝑖} = 𝑔_{1} (𝑞_{1}, \dots)

For the output, you want to make sure bigger is bigger and lower is lower.

A case study in corporate credit rating

Basically the idea is, we have a credit rating. A company raises money by issuing stock, or debt. A debt is a loan, and repayment of the loan depends on credit rating. If the rating is high, you can get a low repayment rate. Who does the rating? S&P, Fitch, and Moody’s. But there are many more smaller ones, DB Morningstar, that hire a lot of our students. In India, these companies, do not operate there. You have consultant companies which work with S&P to help with their models, and move them to India to use them there. Well, at least they do this in Europe. The bank will hire a consultant that implements a Moody’s model and gives you the answer, it’s very bad.

In practice, you have all the ratings.

We looked at 16 years, some companies started more recently, some later. 62 companies, with 297 financial features, 297 xs. One random variable is the output, but it’s all 24, because you want to create an output for each of the likely outcomes. This is basically talking about the various techniques we used.

How do we evaluate algorithms? How good is it? Once you train you get the weights. So you have to do cross validation to test it.

Typically we have two data sets, training and testing. You typically have three regions in finance. In CS, you split into training and testing. Testing you just use for output, that’s it. Testing you split into this K-fold cross validation techniques. You’re always dealing with hyperparameters, should I have one hidden layer or multiple, 64 nodes, or 128, or 50, how many? The way to determine this is to split the data into 10 pieces, use 9 pieces to construct your weights, and then use that to predict the 10th part, and then see how good it is. Then you pick your hyperparameters which you can use. Estimation of actual parameters is done with the other side of the data. Now you have your weights, you can plug in your features, and get a credit rating.

You use these metrics, which have nothing to do with cross entropy baloney which is used to calculate weights. Once you have the weights, you’re concerned with accuracy. I have my company which is rated A, which has an output number. Is it the same as the observed number? You can do accuracy for the entire dataset. I have 100 points, and 50 of them I guessed the credit rating exactly, the other 50 I did wrong. So my accuracy is 50%.

We also have something called recall. Of those 100 companies, 20 were AA. Of those, the model only predicted 15 of them. 5 of them were predicted as being something else. That’s recall. Recall can only be done per category, and accuracy is done overall.

Precision is the same as accuracy, but only for those categories. You have 100 observations, accuracy for 100 points. My algorithm predicted 25 AA. There were 20 AA. Of those 20, 15 were actually AA. My precision is 15/25, my recall is 15/20.

Math is in corporate credit rating slides for Precision and Recall.

You have two numbers, how many of them were guessed correctly, and how many of the guesses were correct. Which one is better to describe this? The F1 score is the harmonic mean of precision and recall. There are 3 means, arithmetic, geometric, and harmonic. Arithmetic is largest, and harmonic is smallest, always. This is as conservative as possible, so it picks harmonic, the smallest possible average of the two.

This is for only one category.

You can do macro-averaging, regular averaging, or micro-averaging, which is the same as accuracy.

There’s not that much complication here. You just need to understand the fundamentals.

The skill for interviews: you need to explain the model. GIVE EXAMPLES.

Shanshan’s Volatility Estimation

Goals: show that ANN can model complex nonlinear relationship, like a function with no closed-form solution

And how to do this with PyTorch

Training and testing process, will be completely

We set up with 5 input features, one output .

Hyperparameters we set up ourselves. There are two hidden layers with 64 nodes in each layer, ReLU is used.

Hidden layer can choose different activation functions. ReLU is common. For output layer, I used ReLU because volatility is always positive, and can be $> 1$ .

The $𝑊_{1}$ and $𝑏_{1}$ parameters are the initialized parameters, which will be updated during training.

Defining PyTorch neural network is done by extending nn.Module in a class.

You define a function Sequential, with a series of nn.Linear and the activation (nn.ReLu). The number of input and output features for each layer is given as parameters to Linear each layer.

Then you run the sequential for a given input.

Dataset prep:

you do this beforehand to understand the data.

training data and test data are split to evaluate the model

convert the dataset into tensor, wrap in Dataset class, allows you to use DataLoader.

You can do batches for efficiency, and shuffle training data for automatic cross validation.

Use Adam optimizer to automatically adjust learning rates, faster than gradient descent.

Compute training MSE loss to monitor progress

Evaluation MSE loss for overfitting

Testing MSE loss for results

How to prevent overfitting? When it learns the noise and random fluctuations instead of general pattern.

Use an evaluation set during training. If training loss and evaluation loss diverge too much, that’s a problem.

Ways to deal:

Early stopping. Orange line stays, but blue line is decreasing. Then stop the epochs as the point when the blue line passes the orange line.
Regularization. Penalize large weights. If the model is very complex, it has the space to learn the noise and random fluctuations rather than general pattern. Adam has (weight_decay=1e-5)

## XVA and Counterparty Risk - URL: https://sharifhsn.dev/blog/xva-and-counterparty-risk/ - Structured data: https://sharifhsn.dev/api/posts/xva-and-counterparty-risk.json - Description: The syllabus records “Introduction to X-VAs: CVA/FVA,” taught by Yee Han. The personal course export has the topic and date, but no worked XVA lecture notes. The former CVA/FVA… - Date: 2025-04-21 - Exact published timestamp: 2025-04-21 - Topics: FX, Counterparty Risk, XVA, CVA, FVA - Categories: FX - Source: FE-635 \| Risk Engineering - Source URL: None - Content source format: typst

Week 11 — 21 April 2025

The syllabus records “Introduction to X-VAs: CVA/FVA,” taught by Yee Han.

Source boundary

The personal course export has the topic and date, but no worked XVA lecture notes. The former CVA/FVA explanation was generated from general knowledge rather than recovered class notes, so this entry now records only what the source supports.

## Why Federal Reserve Independence Matters - URL: https://sharifhsn.dev/blog/linkedin-2025-04-18-why-federal-reserve-independence-matters/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2025-04-18-why-federal-reserve-independence-matters.json - Description: Trump is threatening the independence of the Federal Reserve. But why does this even matter? And what does the Fed even do? The role of central banks in the economy is a… - Date: 2025-04-18 - Exact published timestamp: 2025-04-18T16:23:13.160Z - Topics: Central Banking, Monetary Policy - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/feed/update/urn:li:activity:7319032768905064449/ - Content source format: typst

Trump is threatening the independence of the Federal Reserve. But why does this even matter? And what does the Fed even do?

The role of central banks in the economy is a privileged one. They have the ability to manipulate the supply of money, and more importantly, they set the overnight interest rate that banks will lend money to each other at, which cascades into the interest rate for borrowing money across the whole economy.

The Federal Reserve has a dual mandate, which Chairman Powell stresses in every meeting: to meet target inflation and unemployment.

Generally, lowering interest rates lowers unemployment. When businesses can borrow money more cheaply, they’re more willing to expand and hire. However, this also increases inflation. Demand for labor comes with a demand for everything else, and if supply can’t meet it, prices will rise.

This was seen most dramatically when the Fed dropped interest rates to zero after the COVID pandemic. It prevented mass layoffs, but the increased demand from consumers combined with a lack of supply from pandemic-era supply chains led to record inflation.

The Fed’s decisions reverberate through the market, and a lot of money rides on predicting this. In fact, you can see exactly what the market thinks the future interest rates will be by the Fed Funds Futures curve.

CME’s FedWatch tool provides a nice visualization of this in probabilistic terms. As you can see in this graph, the market currently thinks that the Fed will continue to pause interest rate changes at the next meeting in three weeks.

This is because the Fed is conservative. The organization is composed of experts who deeply understand the impacts of their decision on the economy, and will not threaten it for short-term economic reasons, especially in times of instability.

The same cannot be said for politicians. They are motivated to keep their constituents happy in time for the next election. This biases them in favor of lowering interest rates, which will heat up the economy in the short term, even if it might cause inflation in the long term.

If the Fed has to bow to political pressure, the probabilities you see would shift dramatically to the left, since Trump has been loudly proclaiming that he wants to see interest rates cut.

Such a move would be completely unprecedented and would undermine trust in the dollar as a whole. The falling stock market, rising treasury yields, and falling dollar all imply that institutions are already seeking to exit US markets, and this would only exacerbate this process.

Do you think Trump will fire Powell? Tell me in the comments.

Read more at Bloomberg: https://lnkd.in/gBQviQzy

## Dependence and Copulas - URL: https://sharifhsn.dev/blog/advanced-derivatives-week-12/ - Structured data: https://sharifhsn.dev/api/posts/advanced-derivatives-week-12.json - Description: Lecture notes on minimum, independent, and maximum dependence in a one-factor credit model. - Date: 2025-04-17 - Exact published timestamp: 2025-04-17 - Topics: Credit, Dependence, Copulas - Categories: Credit - Source: Advanced Derivatives - Source URL: None - Content source format: typst

Dependence cases

The one-factor model associates each credit with a latent asset value

$𝐴_{𝑖} = 𝛽_{𝑖} 𝑍 + \sqrt{1 - 𝛽_{𝑖}^{2}} 𝜀_{𝑖}$ ,

where $𝑍$ is the common market factor and each $𝜀_{𝑖}$ is an independent idiosyncratic factor. The coefficients $𝛽_{𝑖}$ determine how much each credit is exposed to the common factor; the asset correlation of two credits is $𝜌_{𝑖, 𝑗} = 𝛽_{𝑖} 𝛽_{𝑗}$ . The Lecture 9 notes introduce this model on PDF page 1.

Let $𝑝_{𝑖 (𝑇)} = 𝑃 (𝜏_{𝑖} \leq 𝑇)$ be credit $𝑖$ ’s default probability by time $𝑇$ . Changing the signs and sizes of the market-factor loadings gives three limiting dependence cases.

Minimum dependence

The smallest joint-default probability comes from opposite exposures to the same market factor. In the limit, set $𝛽_{𝑖} = 1$ and $𝛽_{𝑗} = - 1$ . Both idiosyncratic components disappear and the asset correlation is $𝜌_{𝑖, 𝑗} = - 1$ . The joint-default probability is

$𝑃 (𝜏_{𝑖} \leq 𝑇, 𝜏_{𝑗} \leq 𝑇) = max (𝑝_{𝑖 (𝑇)} + 𝑝_{𝑗 (𝑇)} - 1, 0)$ .

It is zero whenever the two default probabilities sum to at most one. If their sum exceeds one, the joint probability is the excess over one.

Independence

Set $𝛽_{𝑖} = 𝛽_{𝑗} = 0$ . Each asset value is then only its own idiosyncratic factor, the correlation is zero, and the default events are independent:

$𝑃 (𝜏_{𝑖} \leq 𝑇, 𝜏_{𝑗} \leq 𝑇) = 𝑝_{𝑖 (𝑇)} 𝑝_{𝑗 (𝑇)}$ .

Maximum dependence

Set $𝛽_{𝑖} = 𝛽_{𝑗} = 1$ . Both credits have the same asset value $𝑍$ and their asset correlation is $1$ . The joint-default probability is the smaller of their marginal default probabilities:

$𝑃 (𝜏_{𝑖} \leq 𝑇, 𝜏_{𝑗} \leq 𝑇) = min (𝑝_{𝑖 (𝑇)}, 𝑝_{𝑗 (𝑇)})$ .

Lecture 9, slide 5 (PDF p. 5), captions its graph “The joint default probability as a function of the asset correlation (4 cases).”

Conditional hazard and portfolio loss

The April 17 handwritten lecture notes then condition on a realized common factor $𝑍 = 𝑧$ . With a flat deterministic conditional hazard rate, they write conditional survival as

$𝑆_{𝑖 (𝑇 | 𝑧)} = 𝑒^{- 𝜆_{𝑖 (𝑇 | 𝑧)} 𝑇} = Φ (\frac{𝛽_{𝑖} 𝑧 - 𝐶_{𝑖 (𝑇)}}{\sqrt{1 - 𝛽_{𝑖}^{2}}})$ ,

so that

$𝜆_{𝑖 (𝑇 | 𝑧)} = - \frac{1}{𝑇} ln Φ (\frac{𝛽_{𝑖} 𝑧 - 𝐶_{𝑖 (𝑇)}}{\sqrt{1 - 𝛽_{𝑖}^{2}}})$ .

Here $𝐶_{𝑖 (𝑇)}$ is the default threshold calibrated from the single-name survival curve. Conditioning on $𝑧$ makes the credits independent: compute the portfolio-loss distribution at that factor value, then integrate over the common factor to obtain the unconditional distribution. The notes state this procedure on PDF page 4.

Lecture 9, slide 6 (PDF p. 6), captions the conditional hazard-rate plot: “Plot of conditional hazard rate distribution for $𝛽 = 0, 0.2, 0.4$ ,” assuming an unconditional hazard rate of 2% and a one-year horizon. Slide 3 (PDF p. 3) shows the time dependence of survival probabilities and default thresholds for issuer curves with flat deterministic hazard rates.

## Euler–Milstein and Monte Carlo Extensions - URL: https://sharifhsn.dev/blog/computational-methods-week-12/ - Structured data: https://sharifhsn.dev/api/posts/computational-methods-week-12.json - Description: We will expand on the variance reduction techniques, we will repeat a couple things from last week. But before we do that, we should talk about Euler Milstein or Euler Maruyama.… - Date: 2025-04-15 - Exact published timestamp: 2025-04-15 - Topics: Computational Methods, Euler–Milstein, Monte Carlo, Control Variates, Correlated Processes - Categories: Computational Methods - Source: Computational Methods in Quantitative Finance - Source URL: None - Content source format: typst

Euler Milstein

We will expand on the variance reduction techniques, we will repeat a couple things from last week.

But before we do that, we should talk about Euler Milstein or Euler Maruyama.

𝑃 𝑒 𝑜 𝑝 𝑙 𝑒 𝑘 𝑒 𝑒 𝑝 𝑡 𝑎 𝑙 𝑘 𝑖 𝑛 𝑔 . 𝐼 𝑡 ℎ 𝑢 𝑟 𝑡 𝑠 𝐹 𝑙 𝑜 𝑟 𝑒 𝑠 𝑐 𝑢 𝑏 𝑒 𝑐 𝑎 𝑢 𝑠 𝑒 ℎ 𝑒 ℎ 𝑎 𝑠 𝑎 𝑑 𝑖 𝑠 𝑒 𝑎 𝑠 𝑒 𝑐 𝑎 𝑙 𝑙 𝑒 𝑑 𝐴 𝐷 𝐻 𝐷 .

This is a better approximation. I’m going to show you this because it involves applying Itô’s formula a bunch.

Here is the stochastic process we can approximate. Euler works for any stochastic process, but Milstein has to be homogeneous, like so:

𝑑 𝑋_{𝑡} = 𝛼 (𝑋_{𝑡}) 𝑑 𝑡 + 𝛽 (𝑋_{𝑡}) 𝑑 𝑊_{𝑡}

These coefficients are not functions of time, so they are homogeneous.

The idea is to use Itô for both $𝛼$ and $𝛽$ . They are functions of stochastic process $𝑋_{𝑡}$ . Therefore I can use Itô for both.

I get

𝑑 𝛼 (𝑋_{𝑡}) = 𝛼^{'} (𝑋_{𝑡}) 𝑑 𝑋_{𝑡} + \frac{1}{2} 𝛼^{″} (𝑋_{𝑡}) (𝑑 𝑋_{𝑡})^{2}

This is Itô, so we are lacking the dt, that term would screw up our calculations.

Now we substitute $𝑋_{𝑡}$ here, so we get

The dX^2 gets eliminated

= 𝛼^{'} (𝑋_{𝑡}) 𝛼 (𝑋_{𝑡}) 𝑑 𝑡 + 𝛼^{'} (𝑋_{𝑡}) 𝛽 (𝑋_{𝑡}) 𝑑 𝑊_{𝑡} + \frac{1}{2} 𝛼^{″} (𝑋_{𝑡}) 𝛽^{2} (𝑋_{𝑡}) 𝑑 𝑡

If I just group these terms and drop $𝑋_{𝑡}$ all over the place to make it easier to write:

= (𝛼^{'} 𝛼 + \frac{1}{2} 𝛼^{″} 𝛽^{2}) 𝑑 𝑡 + 𝛼^{'} 𝛽 𝑑 𝑊_{𝑡}

Now, if we do the same calculation/derivation for $𝛽$ , then we basically get something very similar. We’re not going to go through the entire calculation. You can derive this yourself,

𝑑 𝛽 (𝑋_{𝑡}) = (\dots) 𝑑 𝑡 + \dots + 𝑑 𝑊_{𝑡}

Writing in integral from,

Now we substitute $𝛼$ and $𝛽$ from the same formula. So there are two integrals:

X\_{t+\\Delta t} \- X\_t \= \\int\_t^{t+\\Delta t}

When we substitute, we will get four terms, two terms for $𝛼$ and $𝛽$ each.

We will get terms such as

𝑑 𝑠 𝑑 𝑢 \sim 𝑂 (Δ 𝑡^{2})

We will change the letters to make sure they’re right, based on what our dummy variables are

𝑑 𝑠 𝑑 𝑊_{𝑢} \approx 𝑑 𝑢 𝑑 𝑊_{𝑠} \sim 𝑂 (Δ 𝑡^{\frac{3}{2}})

Because $𝑑 𝑠$ is order $Δ 𝑡$ , and $𝑑 𝑊_{𝑡}$ is order $\sqrt{Δ 𝑡}$

Then there is

𝑑 𝑊_{𝑢} 𝑑 𝑊_{𝑠} \sim 𝑂 (Δ 𝑡)

For the reason just discussed.

Once I substitute everything, I will neglect the first two orders, and will be left with terms with just du and ds

𝑑 𝑢

𝑑 𝑠

The existing equation then simplifies a lot

𝑋_{𝑡 + Δ 𝑡} = 𝑋_{𝑡} + 𝛼 (𝑋_{𝑡}) Δ 𝑡 + 𝛽 (𝑋_{𝑡}) Δ 𝑊_{𝑡} + \int_{𝑡}^{𝑡 + Δ 𝑡} \int_{𝑡}^{𝑢} 𝛽_{𝑠}^{'} 𝛽_{𝑠} 𝑑 𝑊_{𝑠} 𝑑 𝑊_{𝑢}

And there’s another integral term you can see here.

You need to take the increments of Brownian motion and such, but then you can show that the integral term is equal to the

= 𝛽_{𝑡}^{'} 𝛽_{𝑡} \frac{1}{2} (Δ 𝑊_{𝑡} - Δ 𝑡)

This gives the Euler Milstein scheme.

The $𝑋_{𝑡 + Δ 𝑡}$ is what you’re approximating. The first part is the regular Euler, and then the integrals are the Milstein part.

If the model

𝑑 𝑋_{𝑡} = 𝛼 (𝑋_{𝑡}) 𝑑 𝑡 + 𝛽 (𝑋_{𝑡}) 𝑑 𝑊_{𝑡}

, then the Euler Milstein scheme is

𝑋_{𝑡 + Δ 𝑡} = 𝑋_{𝑡} + 𝛼 (𝑋_{𝑡}) Δ 𝑡 + 𝛽 (𝑋_{𝑡}) Δ 𝑊_{𝑡} + \frac{1}{2} 𝛽^{'} (𝑋_{𝑡}) 𝛽 (𝑋_{𝑡}) (Δ 𝑊_{𝑡}^{2} - Δ 𝑡)

And by $𝑊$ you introduce a normal variable multiplied by

𝑍 \sim 𝑁 (0, 1)

. The $Δ 𝑊$ is created by

𝑍 \sqrt{Δ 𝑡}

. The other one is

(𝑍^{2} - 1) Δ 𝑡

when the thing factors. For the same increment, you put it in two places, not just one places.

However, there is a possible issue, which might be the derivative. If the $𝛽$ function, the volatility part, if it’s complicated, how do you calculate the derivative? That might be hard. There’s another way to deal with this, a scheme called Runge-Kutta, a generalization where you calculate the Euler part, and plug it into the $𝛽$ , then you calculate the finite difference as an approximation of the derivative.

One more thing to mention. Which I shouldn’t, because it’s from the homework. Let’s have an example.

We have a process

𝑑 𝑌_{𝑡} = 𝜅 (\bar{𝑦} - 𝑌_{𝑡}) 𝑑 𝑡 + 𝜎 \sqrt{𝑦_{𝑡}} 𝑑 𝑊_{𝑡}

This is the CIR process.

In this process, I have my

𝛼 (𝑥) = 𝜅 (\bar{𝑌} - 𝑥)

Then you have $𝛽$

𝛽 (𝑥) = 𝜎 \sqrt{𝑥}

If we now substitute in this formula, we have to calculate the derivative.

𝛽^{'} (𝑥) = \frac{𝜎}{2 \sqrt{𝑥}}

Now if we do Euler Milstein:

𝑌_{𝑡 + Δ 𝑡} = 𝑌_{𝑡} + 𝜅 (\bar{𝑌} - 𝑌_{𝑡}) Δ 𝑡 + 𝜎 \sqrt{𝑌_{𝑡}} Δ 𝑊_{𝑡} + milstein correction: 𝜎 \sqrt{𝑌_{𝑡}} \frac{𝜎}{2 \sqrt{𝑌_{𝑡}}} (Δ 𝑊_{𝑡}^{2} - Δ 𝑡)

Here, the two $\sqrt{𝑌_{𝑡}}$ terms cancel, so it becomes $𝜎^{2} / 2$ .

The other thing to show is that if you look for example, this process.

𝑑 𝑋_{=} 𝑒^{𝑋_{𝑡}} cos 𝑋_{𝑡} 𝑑 𝑡 + 0.7 𝑑 𝑊_{𝑡}

Here, you have this constant

𝛽 (𝑋_{𝑡}) = 0.7

𝛽^{'} (𝑋_{𝑡}) = 0

Therefore there is no Euler-Milstein correction, because it relies on multiplication by the derivative. Like for example in GBM.

Variance Reduction Redux

More of the idea to use on HW 4.

One additional note will be given on antithetic variate.

We discussed the CLT and the basis of the whole thing, and the variance being smaller gives you better estimates.

However, this is more complicated than that.

The better method is not always the one with smaller variability of the sample paths.

We also need to consider the time to generate paths. If you have a Monte Carlo technique with less variability, but it takes one minute to generate a path, then it could be that another method with much higher variability that can generate 1 per second, is better, just by raw brute force. The paper mentioned is really good, the basis of the Monte Carlo book, which expands on the paper. Boyle is a Canadian professor from University of Waterloo, in 1997 they met, he did a summer school. He was drunk all the time in the morning lectures. Glasserman is a friend of the show as well. Third guy Brodie sucks.

If method 1 has variance $𝜎_{1}^{2}$ , and $𝑏_{1}$ , they have a term called “work”, which could be time, but could be other stuff, $𝑏_{1}$ is the work to generate one replication of the final parameter, the one you’re trying to estimate, then we need to look at

𝜎_{1}^{2} 𝑏_{1} < 𝜎_{2}^{2} 𝑏_{2}

He has a better expression, but I’m showing the simple thing. Why multiplication? There’s a reason

If you look at this perspective, and if you rewrite this as

\frac{𝜎_{1}^{2}}{𝜎_{2}^{2}} < \frac{𝑏_{2}}{𝑏_{1}}

That’s when you would prefer one method over another.

Antithetic note:

Reminder: The way it works is you create one normal for one path, and use the negative of that normal for another path.

For antithetic variates, one replicate is

\frac{𝑐_{𝑖} + 𝑐_{𝑖}^{𝑎}}{2} = \bar{𝐶_{𝑖}}

The actual estimate is the average of the 2. This is not necessarily important for the final pricing, because if you take the general average, it doesn’t matter, but it’s important for the variance estimate, for confidence intervals.

This is because these paths are not independent, they’re very related. So you would use something like

Variance = \frac{1}{𝑛 - 1} \sum_{𝑖 - 1}^{𝑛} (\bar{𝐶_{𝑖}} - \bar{𝐶})^{2}

This is important because each sequence of n is independent.

You might think I should iterate over 2n elements instead of n. and do both the regular and antithetic.

But this one will show less than it really is.

Control Variates for the Asian Option

Glasserman’s explanation is better than mine, I will follow him.

The one with $Δ$ -hedging and Asian option. There’s nothing wrong with $Δ$ -hedging or Asian option, it’s just that it needs to be explained where it’s coming from.

The control variate for Monte Carlo, what it does, it should be called “use what you know”. You know, for example, that the option that $Δ$ -hedges, you’re using the concept that if you do this very fast, the two values should be the same.

Similarly, for the Asian option, we’re using what we know.

𝑃_{𝐴}

is the price of an Asian option based on Arithmetic average, which is what is encountered in practice.

𝑃_{𝐺}

is geometric.

This $𝑃_{𝐺}$ has a formula, and for any variation on Asian options it’s very easy to get a geometric formula.

We’ll say we have a formula, if you give me characteristics and $𝜇$ and $𝜎$ , you get an exact number.

With this, let

\hat{𝑃_{𝐴}}

be the value calculated using a single path.

It doesn’t matter what you use, just one single path, where you calculate the value of the Asian option based on this average of all these paths.

And we’ll do the same for geometric

\hat{𝑃_{𝐺}}

We don’t need this for geometric because we have the formula.

We are using the same path for both of these.

I do know that

𝔼 [\hat{𝑃_{𝐴}}] = 𝑃_{𝐴}

On expectation, I get the true vlau eof my option.

I also know

𝔼 [\hat{𝑃_{𝐺}}] = 𝑃_{𝐺}

If we subtract, we get

𝑃_{𝐴} - 𝑃_{𝐺} = 𝔼 [\hat{𝑃_{𝐴}} - \hat{𝑃_{𝐺}}]

This gives you a very natural kind of estimate.

𝑃_{𝐴} = 𝑃_{𝐺} + 𝔼 [\hat{𝑃_{𝐴}} - \hat{𝑃_{𝐺}}]

We can create a Monte Carlo path using this control variate:

{\hat{𝑃_{𝐴}}}^{𝑐 𝑣} = \hat{𝑃_{𝐴}} - \hat{𝑃_{𝐺}} + 𝑃_{𝐺}

If I create a new path and average this, then it should give me on average this difference

You can write it like this

= \hat{𝑃_{𝐴}} + (𝑃_{𝐺} - \hat{𝑃_{𝐺}})

That parentheses statement is the control variate. I have my original path with Euler Milstein, then I control it.

For every path, I obtain the difference between the true value and the value of geometric from that particular path. But is this better than just using

\hat{𝑃_{𝐴}}

If we want, we can calculate the variance. So the variance should be less. And remember that

𝑃_{𝐺}

is just a number, a constant with no variance.

𝕍 [{\hat{𝑃_{𝐴}}}^{𝑐 𝑣}] = 𝕍 [\hat{𝑃_{𝐴}}] + 𝕍 [\hat{𝑃_{𝐺}}] - 2 Cov (\hat{𝑃_{𝐴}}, \hat{𝑃_{𝐺}})

Everything after the first term should be negative to give me a better variance. The covariance should be greater than the variance. It’s only worth it if the covariance is large.

This brings the next idea. This is the original term plus this term. I can control the size of the difference, which puts a $𝛽$ on the coefficient. That $𝛽$ allows me to make the thing smaller. This is all specific to the Asian option, where there is this arithmetic and geometric thing. But there is no assumption about the stochastic model.

$𝛽$ means I’m going to parameterize this:

{\hat{𝑃_{𝐴}}}^{𝛽} = \hat{𝑃_{𝐴}} + 𝛽 (𝑃_{𝐺} - \hat{𝑃_{𝐺}})

I can play around with $𝛽$ in such a way that the resulting variance is the smallest. What is that? We can do the same exact calculation, and minimize the result with respect to $𝛽$ .

𝕍 [{\hat{𝑃_{𝐴}}}^{𝛽}] = 𝕍 [\hat{𝑃_{𝐴}}] + 𝛽^{2} 𝕍 [\hat{𝑃_{𝐺}}] - 2 𝛽 Cov (\hat{𝑃_{𝐴}}, \hat{𝑃_{𝐺}})

This is a quadratic expression. It’s a parabola, so the smallest value is in the vertex. The position of the vertex is obtain for

𝛽_{min} = - \frac{𝑏}{2 𝑎}

Where the varG is a, and the cov stuff is b.

= - \frac{- 2 Cov (\hat{𝑃_{𝐴}}, \hat{𝑃_{𝐺}})}{2 𝕍 [\hat{𝑃_{𝐺}}]}

This cancels to:

= \frac{Cov (\hat{𝑃_{𝐴}}, \hat{𝑃_{𝐺}})}{𝕍 [\hat{𝑃_{𝐺}}]}

If you have two variables, the regression is the covariance divided by the variance, so this is the formula for market $𝛽$ /regression.

So if we regress

\hat{𝑃_{𝐴}} = 𝛼 + 𝛽 \hat{𝑃_{𝐺}} + 𝜖

, the $𝛽$ is the $𝛽$ .

Last week, we learned this was true, but now we know how to get it.

This control variate thing is basically

{\hat{𝑃_{𝐴}}}^{𝛽} = \hat{𝑃_{𝐴}} + \hat{𝛽} (𝑃_{𝐺} - \hat{𝑃_{𝐺}})

Now we have another problem.

This is kinda screwed up. Because you’re using the same paths to estimate $𝛽$ , and the same path to estimate the value of the option. That introduces a bias, and this is calculated in the Glasserman paper.

Typically you have $𝑛$ paths, and you set $𝑛_{1}$ paths out to do regression. Then you use $𝑛 - 𝑛_{1}$ paths for calculation.

The advantage here is to do a proper regression, you don’t need a lot of observations, 100 would be plenty. But for Monte Carlo, you need hundreds of thousands.

Hopefully this is more clear, and what I hope you get is that he used the particular relationship that exists in the Asian option, to reason through the whole thing.

I’m going to make another expansion to this. We can introduce more control!!

It’s not really necessary because this existing control variate already gives a good estimate, but this shows you can introduce as many control as you like!

Under risk-neutral equivalent martingale measure, we have

𝑆_{0} = 𝔼^{𝑄} [𝑆_{𝑇} 𝑒^{- 𝑟 𝑇}]

If you take the stock price as a martingale, and discount it back, you should get $𝑆_{0}$ . Nothing new.

This brings up another way to control.

We use the following, with $𝛽_{1}$ being our original control variate.

{\hat{𝑃_{𝐴}}}^{𝑐 𝑣} = \hat{𝑃_{𝐴}} + 𝛽_{1} (𝑃_{𝐺} - \hat{𝑃_{𝐺}}) + 𝛽_{2} (𝑆_{0} - \hat{𝑆_{𝑇}} 𝑒^{- 𝑟 𝑇})

Now you have the path, you know what $𝑆_{𝑇}$ is, and $𝑆_{0}$ is a constant.

You have the same kind of regression of

\hat{𝑃_{𝐴}} = 𝛼 + 𝛽_{1} \hat{𝑃_{𝐺}} + 𝛽_{2} \hat{𝑆_{𝑇}} + 𝜖

The constants don’t matter because you’re doing a regression, it just changes the y-intercept, doesn’t impact the $𝛽$ .

Technically, it’s $𝜎 \sqrt{𝑡}$ , which is the confidence interval size, the diffusion size. It’s an estimate, work could refer to other things.

Moment Matching Method

This is simple to use, so I’ll mention it, even though it’s useless.

If we have

𝑍_{1}, \dots 𝑍_{𝑛} \sim 𝑁 (0, 1)

They should have theoretical mean 0, but the sample mean is not 0.

The idea is to modify the sample to match the theoretical moments.

The reason I actually have never taught this method is because it’s kinda stupid, as a statistician.

Also, from the raw power of this method, it doesn’t do better than straight Monte Carlo. It does better when you pair it with control variate, where the power comes from the control variate.

In the example here, the random variables don’t have mean 0, so instead use

𝑍_{1} - \bar{𝑍}, 𝑍_{2} - \bar{𝑍}, \dots

These now all have mean 0, but now they’re correlated. So it creates problems when estimating stdev, it becomes bad, very tricky for estimating confidence intervals.

In the example, let’s say I’m going to price a European option based on GBM. When you do GBM, you don’t have to do all the intermediate steps, you can do all in one step. Because the terminal value

\tilde{𝑆_{𝑇}} (𝑖) = 𝑆_{0} 𝑒^{(𝑟 - \frac{𝜎^{2}}{2}) 𝑇 + 𝜎 \sqrt{𝑇} {\tilde{𝑍}}_{𝑖}}

After you modify the normal variable with the thing I said.

This is fine for European options, but it won’t work for path-dependent options.

Confidence intervals are hard to obtain.

This is the first order moment matching. You can also do second order moment matching. Say we want to create

𝑁 (𝜇_{𝑍}, 𝜎_{𝑍}^{2})

I’m trying to create numbers that are normal with this particular target. The usual thing to do here is

𝑍_{𝑖} \sim 𝑁 (0, 1) \to 𝜎_{𝑍} 𝑍_{𝑖} + 𝜇_{𝑍}

Multiply to create the desired distribution.

But the numbers in the sample will have their own sample mean and stdev. So you have to modify this as

\tilde{𝑍_{𝑖}} = \frac{𝜎_{𝑍}}{𝑆_{𝑍}} (𝑍_{𝑖} - \bar{𝑍}) + 𝜇_{𝑍}

Each number is modified by these two numbers $𝑆_{𝑍}$ and $\bar{𝑍}$ , where $𝑆_{𝑍}$ is the sample stdev.

𝑆_{𝑍} = \sqrt{\frac{1}{𝑛 - 1} \sum (𝑍_{𝑖} - \bar{𝑍})^{2}}

For the random variables in the sample, they will have the desired distribution.

Like I said, this is a method from the 90s. I never liked it because it’s slower. All these modifications…

And in order to estimate the sample $𝜇$ and stdev, I have to do all of the simulations first, then calculate the samples, then plug them back, so it’s SLOWER than doing it all at once.

And generally, from my experience, improvement is marginal, it’s not particularly useful.

Is the sample calculated in one path or all paths? It depends. In this example, you need all the paths. But you can do it for one path if you like.

Computationally, when you have numbers that you’re observing, if I have to estimate an average or variance of a certain number of them, theoretically, I can do this.

The idea of the GPU is that you have a lot of matrices and you do these calculations really fast. There is a memory of GPU and a memory of CPU. You want to keep everything in GPU memory as much as possible. The problem is that this method aggregates the paths, and does it again, which is impossible with a GPU.

Additional Stuff

Stratified sampling is not the best explained here. It is a statistics technique, check any statistics book they will explain better there.

Importance sampling, same deal. I have a chapter in my book about importance sampling which is way better.

Conditional MC: very well explained in the paper

Low discrepancy sequences. (Quasi-Monte Carlo techniques). It’s not explained well in the paper, but you can look it up and read better papers.

Briefly:

when you start generating one dimensional random variables, we’ll say for a uniform distribution, we can use testers to see if the numbers are actually uniform.

Then you can generate pairs, two at a time $(𝑋_{1}, 𝑋_{2})$ and plot them. You should definitely do this experiment with a random number generator. It would not look uniform at all. Human mind when you say uniform, thinks that it’s perfectly spread out. The point of the quasi generator is to spread out on purpose, to get a perfect uniform distribution. It’s okay as an exhaustive search method. But if you need more points, you have to quadruple them to have the same spread everywhere. The more detail you want, the finer quasi becomes, and it gets a lot slower.

But there are circumstances in which this is useful, and people in engineering that don’t understand randomness like this thing.

Chapter 5 is about estimating American options using Monte Carlo simulations.

subjective probability

George Calhoun sent an article to me in Nature

Trump’s election is a subjectiv eprobabiolity, it matters what people’s perceptions are.

A stock, TSLA. It’s been going down, so you’r wondering if it should keep going down or should it stabilize? And we don’t know because the stock is not the value of the company, it reflects the perception of people about the company.

It’s the same as poker. Game theory is so close to probability.

When people lose concentration, they react poorly.

Generating Correlated Processes

If you have multiple stock processes that you want to generate that are correlated, how do you do that?

Each asset looks like

𝑑 𝑋_{𝑡}^{𝑖} = 𝑓 (𝑋_{𝑡}^{𝑖}) 𝑑 𝑡 + 𝑔 (𝑋_{𝑡}^{𝑖}) 𝑑 𝑊_{𝑡}^{𝑖}

Each is a separate stochastic process, but they don’t move their own way at random.

This Brownian motion is correlated.

If we take the notation

𝑋_{𝑡}

the vector of components, and same for f(x). This is the simpler case. You can have a more complicated thing. Each function could depend on all the other ones, it doesn’t have to be driven by just one variable.

g(x) is a matrix dxd size, where dW is dx1, the vector of Brownian motion.

When you do the Monte Carlo simulation, the Brownian motion is what you’re interested in simulating.

We will assume that these are correlated. With this notation, we have

𝑑 𝑋_{𝑡} = 𝑓 (𝑋_{𝑡}) 𝑑 𝑡 + 𝑔 (𝑋_{𝑡}) 𝑑 𝑊_{𝑡}

It’s more complicated because it’s a collection of multiple integrals on dW, but we will write like this for conciseness.

We have the covariance matrix

Cov (Δ 𝑊_{𝑡}) = Σ Δ 𝑡

This is Brownian motion. So what exactly is the covariance matrix? The components are the covariances of each of the components.

[\begin{matrix} 1 & 𝜌_{12} & 𝜌_{13} & 𝜌_{1 𝑑} \end{matrix}]

Symmetric and positive definite matrix!

Positive is vector times matrix times vector transpose, which is a 1x1 number. Positive definite means that for a vector u, this is always greater than 0. This is very important because you can take a vector multiply with vector which will give us the variance, which will be always positive.

For example, take Heston

𝑑 𝑆_{𝑡} = 𝑟 𝑆_{𝑡} 𝑑 𝑡 + \sqrt{𝑌_{𝑡}} 𝑆_{𝑡} 𝑑 𝑊_{𝑡}^{1}

𝑑 𝑌_{𝑡} = 𝛼 (\bar{𝑌} - 𝑌_{𝑡}) 𝑑 𝑡 + 𝜎 \sqrt{𝑌_{𝑡}} 𝑑 𝑊_{𝑡}^{2}

If I want to simulate this, which is part of homework, how would you do this?

Generate two random numbers which are correlated with $𝜌$ .

For two, it’s very simple.

We need

𝑋_{1}, 𝑋_{2} \sim 𝑁 (0, 1)

, such that

Corr (𝑋_{1}, 𝑋_{2}) = 𝜌

Then we can multiply by $\sqrt{Δ 𝑡}$ and everything will work.

We will start with uncorrelated $𝑍_{1}, 𝑍_{2} \sim 𝑁 (0, 1)$ . I’ll take $𝑋_{1} = 𝑍_{1}$ .

Then, I’ll take $𝑋_{2} = 𝜌 𝑍_{1}$ . We can do this because the variance of $𝑍_{1}$ is 1, and covariance of $𝑍_{1}$ and $𝑍_{2} = 0$ ;

The combination has to have variance of 1, and if we’re combining linear combinations, then it’s normal.

So total value is

𝑋_{2} = 𝜌 𝑍_{1} + \sqrt{1 - 𝜌^{2}} 𝑍_{2}

If you understand the principle, what do I do when I want to do three? $𝑋_{1}, 𝑋_{2}, 𝑋_{3}$

Take the same idea

𝑋_{1} = 𝑍_{1}

𝑋_{2} = 𝜌_{12} 𝑍_{1} + \sqrt{1 - 𝜌_{12}^{2}} 𝑍_{2} ? ? ? ?

𝑋_{3} = 𝜌_{13} 𝑍_{1} ? ? ? ?

It becomes tricky. So is there a method to do this?

There is! You should know where this is coming from

We have a vector $𝑋$ which is a lot of $𝑋$ s. We are interested in covariance, which is distinct from correlation. Our covariance matrix will be one diagonal, where we multiply by $\sqrt{Δ 𝑡}$ . The values on the diagonal of the matrix are the variance, and the rest are covariance.

So how do I generate a vector with this covariance structure? Here is the idea.

If $𝑋$ is a random vector, (more details you could talk about in 540) with mean $𝜇$ , componentwise for each element in vector, then

Cov (𝑋) = 𝔼 [(𝑋 - 𝜇) (𝑋 - 𝜇)^{𝑇}]

Take $𝑌 = 𝐴 𝑋$ . $𝐴$ is a matrix, but it can be ANY DIMENSION $𝑛 \times 𝑑$ . We can transform 4 components into 15 components with different linear combinations.

The basis of the whole method is this:

Cov (𝑌) = 𝔼 [(𝑌 - 𝜇_{𝑌}) 𝑌 - 𝜇_{𝑦})^{𝑇}]

And we know that the expectation of Y from linear combination si

𝔼 [𝑌] = 𝐴 𝔼 [𝑋]

Because of this, we get

= 𝔼 [(𝐴 𝑋 - 𝐴 𝜇_{𝑋}) (𝐴 𝑋 - 𝐴 𝜇_{𝑋})^{𝑇}]

And you can factor this. Remember that the order is very important.

= 𝔼 [𝐴 (𝑋 - 𝜇_{𝑋}) (𝐴 𝑋 - 𝐴 𝜇_{𝑋})^{𝑇}]

And we can also do this in the transpose

= 𝔼 [𝐴 (𝑋 - 𝜇_{𝑋}) {(𝑋 - 𝜇_{𝑋})}^{𝑇} 𝐴^{𝑇}]

And the inner product is the covariance matrix! So it ends up being this

= 𝐴 Σ 𝐴^{𝑇}

CHOLESKY DECOMPOSITION

Take the $𝑍$ vector of iid normals, which are $𝑁 (0, 𝐼_{𝑑})$ . On the diagonal, you have 1 correlation.

Find a matrix A such that

𝐴 𝐴^{𝑇} = Σ

, the desired covariance.

The reason why you do this is because you have A, and you multiply it, and you get the thing you need.

Cholesky uses eigenvalues and eigenvectors, but it does exactly this.

Covariance vs Correlation

How do you get from the covariance matrix to correlation?

𝜌_{𝑖 𝑗} = \frac{𝜎_{𝑖 𝑗}}{𝜎_{𝑖} 𝜎_{𝑗}}

How do you go from this method to other methods and vice versa?

In terms of matrix operations, it’s not complicated.

If D is the diagonal matrix of the variances (just the diagonal), then the correlation matrix is this. Inverse is regular inverse because it’s just a diagonal.

= {\sqrt{𝐷}}^{- 1} \times Σ ({\sqrt{𝐷}}^{- 1})^{𝑇}

And the transpose is actually the same thing.

And you do non-inverse to get from correlation to covariance.

Not commutative, so be careful.

Markov Chain Monte Carlo

Left unsaid

## Vietnam and the Tariff Shock - URL: https://sharifhsn.dev/blog/linkedin-2025-04-14-vietnam-and-the-tariff-shock/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2025-04-14-vietnam-and-the-tariff-shock.json - Description: The biggest loser of the tariffs, the tragedy of the “miracle on the Mekong”: Vietnam 🇻🇳 When President Trump revealed the list of tariffs on “Liberation Day”, one country stood… - Date: 2025-04-14 - Exact published timestamp: 2025-04-14T15:40:44.045Z - Topics: Tariffs & Trade, Trade Policy - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/feed/update/urn:li:activity:7317572525679398915/ - Content source format: typst

The biggest loser of the tariffs, the tragedy of the “miracle on the Mekong”: Vietnam 🇻🇳

When President Trump revealed the list of tariffs on “Liberation Day”, one country stood out: Vietnam. They were hit by 46% tariffs, the second highest out of all of them. But to really understand the scale of this mistake here, you have to walk back in Vietnamese history.

Vietnam was destroyed after its war with the United States. It was run as a socialist nation for a decade, with strict agricultural collectivization and no free market trade. This led, predictably, to mass famine and slow growth. It seemed that there was no hope left for this country. The Soviet Union was their only lifeline, and the economic support from them dried up within only a few years.

Then, in 1986, as the Soviet Union was on the verge of collapse, the Vietnamese government passed the Đổi Mới reforms. They made sweeping changes to the organization of the economy, allowing private enterprise. Vietnam opened itself up to the world, encouraging foreign direct investment and trade with the West.

The success of these reforms led Vietnam to grow its GDP at a rate of 4-6% in the 1990s. Vietnam’s socialist policies led them to redistribute these gains across the nation, leading to massive reductions in poverty. Today, Vietnam is akin to a middle-income nation like Mexico or Turkey, a far cry from its war-ravaged, famine-stricken past.

Despite the trauma of war, Vietnam has significantly warmed its relationships with the U.S. in the past decades. Indeed, they are far friendlier to us than they are to China. The U.S. has taken advantage of this and shifted much of our manufacturing from China to Vietnam, reducing our dependence on China.

This would seem like a win-win scenario. Not according to Trump. He saw our close trade relationship with Vietnam as a threat to American manufacturing, and slapped them with massive tariffs as a result. All of those years of careful diplomacy and relationship strengthening between the two countries are circling the drain.

Although the tariffs have been paused, nobody is sure what will happen next. Vietnam is in talks with China and have already agreed to some cooperation agreements. [1] This can only be bad news for the U.S.

I visited Vietnam last year. It’s a beautiful nation, with delicious food, warm people, and a fascinating culture. On the flight there and leaving, they played a beautiful song called “Hello, Vietnam” on the speakers [2]. It’s sung by a child of Vietnamese expats, who wishes to see the country where she originated. It’s an anthem for Vietnamese immigrants, and paints a beautiful picture of Vietnam from a Western perspective.

Unfortunately, it seems like the U.S. has decided to say “Goodbye, Vietnam”.

[1] https://lnkd.in/eeX6mP2B

[2] https://lnkd.in/e653TUpg

## The Big Short and the Boring Cause of the 2008 Crisis - URL: https://sharifhsn.dev/blog/linkedin-2025-04-12-the-big-short-and-the-boring-cause-of-the-2008-crisis/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2025-04-12-the-big-short-and-the-boring-cause-of-the-2008-crisis.json - Description: I rewatched The Big Short last night, and realized that they sensationalized a pretty boring reason that the 2008 crisis happened. When the housing bubble burst and the Great… - Date: 2025-04-12 - Exact published timestamp: 2025-04-12T18:25:14.194Z - Topics: Financial Markets, Housing - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/feed/update/urn:li:activity:7316889148353630210/ - Content source format: typst

I rewatched The Big Short last night, and realized that they sensationalized a pretty boring reason that the 2008 crisis happened.

When the housing bubble burst and the Great Recession occurred in 2008, everyone wanted to find the individuals to blame. The easiest targets were the Wall Street bankers who made the bad bet that people wouldn’t default on their mortgages, even as riskier and riskier subprime loans were issued.

The Big Short is the most popular movie depicting these events. It focused on the maverick investors who realized that there was a bubble and bet against the housing markets. The villains of this movie are the bankers who created and sold the “CDO” or “collateralized debt obligation”, which is what caused the housing crisis to become an economic disaster.

The movie explains the CDO like this: imagine a chef buys some fresh fish which doesn’t sell. He can throw the unsold fish into a stew, which becomes a whole new thing and is now palatable.

In this case, the unsold fish are poorly rated BBB bonds, the stew is the CDO (a portfolio of bonds), and the new palatability is a AAA rating.

But why would ratings agencies rate a bunch of terrible bonds packaged together as AAA?

The movie frames this as a corrupt racket. The main characters ask their colleague at S&P why these securities are being rated AAA, and she admits that if they don’t rate them AAA, then the banks will go down the block to Moody’s and get a better rating there. They have essentially become a ratings shop.

The reality of the situation was that the ratings agencies were using a flawed model, the Gaussian copula. In essence, the model assumed that bond default probabilities were uncorrelated.

Any good portfolio manager knows that the key to risk management is diversification. By packaging together uncorrelated assets, you can reduce the risk of all of them going down at once. So in theory, if you have a lot of uncorrelated BBB bonds, the likelihood of many of them defaulting is fairly low.

You can see that in this graph. The probability of portfolio loss exceeding 10% is vanishingly small for a portfolio of uncorrelated bonds. As correlation increases, such a risk becomes more and more likely.

In this case, the likelihood of bond default wasn’t uncorrelated. Unscrupulous mortgage brokers were issuing subprime loans en masse, making it likely that all of these people would default on their mortgages.

In the wake of the crisis, on top of stricter banking regulations by Dodd-Frank, ratings agencies have adopted more complex models that incorporate bond default correlation, meaning that they’re unlikely to make the same mistakes.

The narrative of a ratings shop might play well to a lay audience, but it doesn’t fit with reality.

Did you know about this flaw, or did you assume that The Big Short had it right? Let me know in the comments.

## Romania's Export-Driven Debt Strategy as a Warning - URL: https://sharifhsn.dev/blog/linkedin-2025-04-11-romania-s-export-driven-debt-strategy-as-a-warning/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2025-04-11-romania-s-export-driven-debt-strategy-as-a-warning.json - Description: Trump’s policy of reducing the national debt through an export-driven economy has parallels to 1980s Romania—and it doesn’t end well. President Nicolae Ceaușescu was elected to… - Date: 2025-04-11 - Exact published timestamp: 2025-04-11T15:38:23.906Z - Topics: Sovereign Debt, Fiscal Policy - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/feed/update/urn:li:activity:7316484774297038848/ - Content source format: typst

Trump’s policy of reducing the national debt through an export-driven economy has parallels to 1980s Romania—and it doesn’t end well.

President Nicolae Ceaușescu was elected to lead the Socialist Republic of Romania in 1965, and became a totalitarian dictator in 1967. Among other seemingly anti-Soviet policies, his decision not to invade Czechoslovakia in 1968 made him popular in the West. Because of this, they generously lent money to Romania for the purpose of industrialization in the 1970s.

However, due to internal corruption and the energy crisis of the late 1970s, Romania was unable to pay its debts. Like many other developing countries, it requested a line of credit from the IMF to pay off its debts.

Unusually, however, Romania refused to negotiate with its creditors to restructure their debt and have them take a haircut, as those other developing countries did. They resolved to pay off their national debt on schedule and as soon as possible.

For this purpose, Romania cut imports significantly and expanded exports. This austerity policy led to significant reductions in standards of living, with food and energy shortages throughout the 1980s.

The export-driven economy further exacerbated this. Take the example of a shoe factory, since Romania is known for producing the best shoes in the world. All the most high quality leather would be reserved for shoes that were designated for exports. Shoes that were sold in Romania itself only used the leather that wasn’t fit for export. The economic crisis led to further corruption, so people would steal the best domestic leather off the lines, leading to even worse quality for Romanian shoes on the market.

In the end, Romania eliminated its national debt in 1989. But that same year, the political unrest caused by the impending fall of the Soviet Union led to revolution in Romania. The Romanian Revolution was the only one in the Eastern Bloc to turn violent, ending in the conviction and execution of Ceaușescu and his wife.

The conclusion seems to be that the sharp drop in standard of living due to austerity was what led to the more severe consequences for the leader who put them into place.

What do you think? Is Romania’s example instructive for the US?

(Special thanks to Dr. Ionut Florescu for the detail on austerity in 1980s Romania)

## China's Tariff Strategy and the One-Punch Proverb - URL: https://sharifhsn.dev/blog/linkedin-2025-04-10-china-s-tariff-strategy-and-the-one-punch-proverb/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2025-04-10-china-s-tariff-strategy-and-the-one-punch-proverb.json - Description: An old Chinese proverb contains the key to understanding their tariff strategy: 打得一拳开免得百拳来 “throw out one punch now to avoid a hundred punches in the future” What does it mean?… - Date: 2025-04-10 - Exact published timestamp: 2025-04-10T13:30:12.308Z - Topics: Tariffs & Trade, Trade Policy - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/feed/update/urn:li:activity:7316090125531111425/ - Content source format: typst

An old Chinese proverb contains the key to understanding their tariff strategy: 打得一拳开免得百拳来

“throw out one punch now to avoid a hundred punches in the future”

What does it mean? By striking hard all at once, you can avoid the pain and tedium of striking many times later.

When President Trump placed a pause on the Liberation Day tariffs, the entire world breathed a sigh of relief. He promised that the tariffs would go into effect only on countries who had retaliated, which at this point was only China. China responded by escalating its tariffs to 84%.

China is the only country refusing to back down from Trump’s threats. This will devastate both economies. Last year, China imported $143.5 billion of goods per year from the U.S., and the U.S. imported $438.9 billion of goods from China. So why is China doing this?

Trump’s tariff policy has been, to put it mildly, volatile. Nobody is really sure what he will do next, and whether he will stick to his promises. By forcing the issue and escalating now, China is forcing Trump to deal with the economic reality of the tariffs.

And unlike China, the U.S. is a democracy, where the leaders must listen to the citizens, who are loudly responding to the tariffs. Although the stock market has rallied in response to the 90-day pause, if the proposed 125% tariffs actually go into effect on China, we can expect to see further dips in the market as businesses are crushed by expenses.

If China had backed down, we might have seen months and months of additional uncertainty around tariffs, with Trump pulling them in and out based on his whims. But with the pain that is sure to come now, Trump will have to face facts about the consequences of his policies, and hopefully reconsider his beliefs.

What do you think? Is China’s policy the right way to go?

(Special thanks to Ke Ren for teaching me this proverb, which has gone viral on Chinese social media)

## Structured Credit and CDOs - URL: https://sharifhsn.dev/blog/advanced-derivatives-week-11/ - Structured data: https://sharifhsn.dev/api/posts/advanced-derivatives-week-11.json - Description: Last day of class is Thursday, May 8. The final exam is going to be posted then, take-home exam. It’ll be posted like an assignment. Will be posted in evening around 6:30, you’ll… - Date: 2025-04-10 - Exact published timestamp: 2025-04-10 - Topics: Credit, CDOs, Structured Credit, Tranches, Synthetic CDOs - Categories: Credit - Source: Advanced Derivatives - Source URL: None - Content source format: typst

Final

Last day of class is Thursday, May 8.

The final exam is going to be posted then, take-home exam. It’ll be posted like an assignment.

Will be posted in evening around 6:30, you’ll have Friday and Saturday night to do this.

Questions will be theoretical.

Multi-names

Last lectures we discussed the mechanics of single-name derivatives like the credit default swaps, and the calculation of the implied hazard rates. This is what is calculated in the assignments, along with default/survival probabilities.

Now we will discuss the multi-name credit derivative, mainly the CDO: Collateralized Debt Obligation.

These are securities whose payments are linked to the incidence of default of an underlying portfolio of credit risky assets. It’s based on not just one, but a portfolio of risky assets.

The most interesting component of this is modeling the dependencies in the portfolio, and then calculating the fractional loss in the portfolio, and essentially pricing. So you have a basket of such credit-linked assets, then the question is, if you want to sell a portion of this portfolio, in shares, what should be that price? Basically you need to calculate the cash flows, as well as expected loss. Same idea as fair price of two legs as in CDS.

Some of these instruments are more complex, but in general, we can consider asset-backed securities. They are formed from loans, bonds, mortgages, and so forth. Usually, the income from the assets is tranched. A MBS has claims on money generated from these pools, cash flows that these mortgages generate from regular payments. If you aggregate these together, you get cash flows. These will be distributed to the owners of the pool.

MBS that are created by adding many such mortgages and selling shares on the revolving pool. For example, such security is secured by the underlying mortgages.

This kind of instrument is obviously more convenient to put in a portfolio than a single-name. If you want a single bond, you are dependent on the probability of default on the counterparty. If you have two bonds, then you have two probabilities of default. You might have probability $𝑝$ of default, and the probability of both defaulting is $𝑝_{1} 𝑝_{2}$ (absent some dependencies between bond defaults). It’s an expectation that helps in constructing such pools.

We can calculate under different assumptions. Let’s say we assume that you have no correlations for them, that would be a simple formula. You would want to get a distribution, the probability that you have 1, 2, n names default. If you have the probability of default, if you assume that’s a homogeneous portfolio where all bonds have same probability of default. Then you would enumerate the number of ways that you can have 3 bonds default in the portfolio, which forms the distribution.

Some assumptions, it helps in some way to have diversification. But many times, these mortgage bonds wouldn’t sell because their credit rating was too bad, like DDD. You can have CMOs, which consist of multiple pools of securities, which are tranches (or slices). The idea is that even if it’s a homogeneous universe of very low-rated loans, like BBB, if you have the securitization of this and construct the CMO and CDO in such a way that they would have AAA rating. We will look at such a mechanism.

Waterfall

Lecture 8, slide 2 (PDF p. 2), says that securities are created from “a portfolio of loans, bonds, credit card receivables, mortgages, auto loans, aircraft leases, music royalties, etc.” Usually, “the income from the assets is tranched.” It defines a waterfall this way: “income is first used to pay the promised return to the senior tranche, then to the next most senior tranche, and so on.” Slide 4, titled “The Waterfall,” draws the asset cash flows passing through senior, mezzanine, and equity tranches.

Slide 3 (PDF p. 3) shows assets with $100 million principal transferred to an SPV, which issues these claims:

Tranche	Principal	Return
Senior	$80 million	LIBOR + 60 bp
Mezzanine	$15 million	LIBOR + 250 bp
Equity	$5 million	LIBOR + 2,000 bp

Basically, the waterfall, which is also called the structure subordination, this describes the scheduled coupon and principal payments from different securities. It also describes the losses. The arrangement of this pool will generate cash flows. This cash flow is used first to pay the senior tranche. Then after the senior tranche is completely paid, the cash flows go to mezzanine, then equity. So as long as you have no defaults, this system works. But when you have a loss, the loss is first taken by the equity tranche, then the mezzanine, and then the senior tranche. This structure of subordination or waterfall creates advantages and disadvantages for the equity tranche. This tranche is far riskier than the other tranches. The idea behind the whole mechanism is to have many market participants. The senior tranche has AAA rating. The rating agencies will rate these particular tranches. The magic of it was to take something rated BBB and turn it into something rated AAA. You could take the mezzanine tranche from multiple CDOs and get a new CDO from that. Things got kind of complicated.

2008

Along with the increase in prices of homes, the MBS market reached $4 trillion. Wall Street issued something like $700 billion in CDOs. This is a kind of interesting instrument. There were many problems. The mortgages given in the first place had various loose criteria. People would get houses without giving a down payment. The rating agencies also had a problem. The modeling of the dependencies of the names made the assumption that these were independent. It’s easy to calculate, but typically these names are not independent. You can argue that certain regions have high correlations, natural causes and such, so they are correlated more locally than all over the United States.

During the crisis, these defaults were kind of correlated, so more defaults generated more defaults.

They tended to use Gaussian models, which do not have tail dependence, which was the main issue. This kind of behavior wasn’t modeled, this possibility of loss.

Synthetic CDOs

Lecture 8, slide 5 (PDF p. 5), states: “A cash CDO is an ABS where the underlying assets are debt obligations”; “a synthetic CDO involves forming a similar structure with short CDS contracts.” It continues: “In a synthetic CDO most junior tranche bears losses first. After it has been wiped out, the second most junior tranche bears losses, and so on.”

The originator of a synthetic CDO can select a portfolio of companies that have a certain maturity structure, and look at the CDS for the companies. The notional is the total notional of the CDS contracts.

Slide 6 (PDF p. 6) gives the tranche boundaries and spreads:

Tranche	Loss allocation	Spread
Equity	Losses up to 5% of total CDS notional	1,000 bp
Mezzanine	Losses between 5% and 20%	200 bp
Senior	Losses over 20%	10 bp

Slide 7 (PDF p. 7) states: “The income is paid on the remaining tranche principal.” Its example says that when losses reach 8% of the principal underlying the CDSs, tranche 1 has been wiped out and tranche 2 earns its promised 200 bp spread on 80% of its principal. The remaining 80% follows because the 5%–20% mezzanine tranche has absorbed 3 percentage points of loss out of its 15-point width.

Single Tranche Trading

Lecture 8, slide 8 (PDF p. 8), says single-tranche trading “involves trading tranches of portfolios of CDSs without actually forming the portfolios”; “cash flows are calculated in the same way as they would be if the portfolios had been formed.” The protection buyer pays the tranche spread, and the protection seller pays losses falling inside the tranche’s interval.

Let $𝐴$ be the attachment point, $𝐷$ the detachment point, and $𝐿 (𝑇)$ the cumulative fractional loss on the reference portfolio by time $𝑇$ . The tranche width is $𝐷 - 𝐴$ . Its fractional loss is

$ℓ_{𝐴, 𝐷} (𝑇) = min (1, max (0, \frac{𝐿 (𝑇) - 𝐴}{𝐷 - 𝐴}))$

Before portfolio losses reach $𝐴$ , tranche loss is zero. As losses move from $𝐴$ to $𝐷$ , the tranche loses principal linearly. Once portfolio losses reach $𝐷$ , the tranche is fully written down. For original tranche notional $𝑁_{𝐴, 𝐷}$ , remaining notional is $𝑁_{𝐴, 𝐷} (1 - ℓ_{𝐴, 𝐷} (𝑇))$ . The protection payment at a default is the increase in tranche loss times its original notional; the premium payment is the contractual spread applied to the surviving notional for that accrual period. If the portfolio loss has not changed, the protection leg has no payment. Each credit’s portfolio loss is its exposure times loss given default, so recovery reduces the loss below the full exposure.

Slide 9 (PDF p. 9) introduces the example: “The best way to make the mechanics of an STCDO clear is to consider a specific STCDO deal.” It uses 100 reference credits with $10 million face exposure each, for $1 billion total notional, and a $30 million tranche with a 250 bp spread, 3% attachment point, 4% width, and five-year maturity. The slide reproduces Table 12.3, “Coupon and loss payments on the synthetic CDO example discussed in the text”:

Payment date	Number of defaults	Portfolio loss (%)	Tranche loss (%)	Notional ($m)	Coupon payment	Protection leg ($m)	Net flow
31 Mar 2007		0.00	0.0	30.00
30 Jun 2007	1	0.70	0.0	30.00	191 667		191 667
30 Sep 2007		0.70	0.0	30.00	191 667		191 667
31 Dec 2007	2	1.40	0.0	30.00	189 583		189 583
31 Mar 2008		1.40	0.0	30.00	189 583		189 583
30 Jun 2008	3	2.10	0.0	30.00	191 667		191 667
30 Sep 2008		2.10	0.0	30.00	195 833		195 833
31 Dec 2008	4	2.80	0.0	30.00	189 583		189 583
31 Mar 2009		2.80	0.0	30.00	183 333		183 333
30 Jun 2009	5	3.50	12.5	26.25	195 833	−3.75	−3 554 167
30 Sep 2009		3.50	12.5	26.25	165 885		165 885
31 Dec 2009	6	4.20	30.0	21.00	165 885	−5.25	−5 084 115
31 Mar 2010		4.20	30.0	21.00	132 708		132 708
30 Jun 2010	7	4.90	47.5	15.75	132 708	−5.25	−5 117 292
30 Sep 2010		4.90	47.5	15.75	99 531		99 531
31 Dec 2010	8	5.60	65.0	10.50	99 531	−5.25	−5 150 469
31 Mar 2011		5.60	65.0	10.50	66 354		66 354
30 Jun 2011	9	6.30	82.5	5.25	66 354	−5.25	−5 183 646
30 Sep 2011		6.30	82.5	5.25	33 542		33 542
31 Dec 2011	10	7.00	100.0			−5.25	−5 250 000
31 Mar 2012		7.00	100.0

Senior Tranche

This works very well for the equity and mezzanine tranche. The senior tranche requires a small modification and some care.

For a senior tranche whose detachment point is 100%, complete loss of the reference portfolio must mean complete loss of the tranche. The loss assigned to the tranche is the portfolio loss above attachment, capped at the tranche width; to express it as a fraction of tranche principal, divide by that width. Thus, when $𝐿 (𝑇) = 1$ and $𝐷 = 1$ , the numerator and width both equal $1 - 𝐴$ , and $ℓ_{𝐴, 1} (𝑇) = 1$ . The 44% noted in the example is a loss measured on the portfolio scale; after normalization by the senior tranche’s 44% width, it is a 100% loss of that tranche.

Correlation

Pricing a tranche requires the distribution of portfolio loss at each horizon: the probabilities of one, two, or more defaults by one year, five years, or another maturity. With credit $𝑖$ weighted by portfolio share $𝑤_{𝑖}$ and recovery rate $𝑅_{𝑖}$ , cumulative fractional portfolio loss is

$𝐿 (𝑇) = \sum_{𝑖 = 1}^{𝑁_{𝑐}} 𝑤_{𝑖} (1 - 𝑅_{𝑖}) 1_{𝜏_{𝑖} \leq 𝑇}$ .

The portfolio contains a discrete number of names, and exposures, default probabilities, and recoveries can differ across them. Its expected loss is

$𝐸 [𝐿 (𝑇)] = \sum_{𝑖 = 1}^{𝑁_{𝑐}} 𝑤_{𝑖} (1 - 𝑅_{𝑖}) 𝑄_{𝑖 (𝑇)}$ ,

where $𝑄_{𝑖 (𝑇)} = 𝑃 (𝜏_{𝑖} \leq 𝑇)$ . Holding those marginal default probabilities and recoveries fixed, changing dependence changes the shape and variance of the loss distribution, not its mean.

Slide 10 (PDF p. 10) captions its graph “The portfolio loss distribution for three levels of correlation.” The vertical axis is probability; the horizontal axis is portfolio loss (%); and the curves are labeled zero, medium, and high correlation. With the same marginal default probabilities, correlation changes the loss distribution’s shape and tail, and therefore the chance that losses reach a given tranche.

Slide 11 (PDF p. 11) calls a factor-based Gaussian copula “a popular approach” for defining correlations between times to default. It says that often all pairwise correlations and all unconditional default distributions are assumed to be the same, and that the market implies a pairwise correlation from market quotes. Slide 12 (PDF p. 12), titled “Cumulative Default Probability Conditional on Factor,” gives the conditional default probability:

$𝑄 (𝑡 | 𝐹) = Φ (\frac{Φ^{- 1} (𝑄 (𝑡)) - \sqrt{𝜌} 𝐹}{\sqrt{1 - 𝜌}})$ .

Given $𝐹$ , the number of defaults among $𝑛$ names is binomial:

$𝑃 (𝐾 = 𝑘 | 𝐹) = (\binom{𝑛}{𝑘}) {𝑄 (𝑡 | 𝐹)}^{𝑘} {(1 - 𝑄 (𝑡 | 𝐹))}^{𝑛 - 𝑘}$ .

The slide writes the conditional default-count probability as $\frac{𝑛!}{(𝑛 - 𝑘)! 𝑘!} {𝑄 (𝑡 | 𝐹)}^{𝑘} {[1 - 𝑄 (𝑡 | 𝐹)]}^{𝑛 - 𝑘}$ and says: “This enables cash flows conditional on $𝐹$ to be calculated. By integrating over $𝐹$ the unconditional distributions are obtained.”

Valuation of Tranches of Synthetic CDOs and Basket CDSs

Copulas provide a way to model dependence across the names when valuing synthetic CDO tranches and basket CDSs.

For a homogeneous portfolio, let $𝑄 (𝑡)$ be the cumulative probability of default by time $𝑡$ . Conditional on the common Gaussian factor $𝐹$ , the per-name default probability is $𝑞 (𝑡 ∣ 𝐹) = Φ (\frac{Φ^{- 1} (𝑄 (𝑡)) - \sqrt{𝜌} 𝐹}{\sqrt{1 - 𝜌}})$ , where $Φ$ is the standard normal CDF. For $𝑛$ names, the number of defaults is binomial conditional on $𝐹$ : $𝑃 (𝐾 = 𝑘 ∣ 𝐹) = (\binom{𝑛}{𝑘}) 𝑞 (𝑡 ∣ 𝐹)^{𝑘} (1 - 𝑞 (𝑡 ∣ 𝐹))^{𝑛 - 𝑘}$ . This makes cash flows conditional on $𝐹$ tractable; integrating over $𝐹$ gives the unconditional loss distribution.

But, this would be for the names that are not correlated. We can impose some dependency between the names in the portfolio. We will look at factor models, starting with single factor, then looking at multiple factors, and generalizing with copula models. Next week we will discuss this in more depth.

Single Factor Models

Lecture 9, slide 3 (PDF p. 3), captions its figure: “Time dependence of the survival probabilities and default thresholds $𝐶 (𝑇)$ for issuer curves with flat deterministic hazard rates shown.” Slide 5 (PDF p. 5) is captioned “The joint default probability as a function of the asset correlation (4 cases).”

The Gaussian Latent Variable Model

This factor model assigns a standard-normal latent asset value to each credit:

$𝐴_{𝑖} = 𝛽_{𝑖} 𝑍 + \sqrt{1 - 𝛽_{𝑖}^{2}} 𝜀_{𝑖}$ ,

where the common market factor $𝑍$ and the name-specific idiosyncratic factors $𝜀_{𝑖}$ are independent standard normals. The linear combination keeps each $𝐴_{𝑖}$ standard normal. Default by time $𝑇$ occurs when $𝐴_{𝑖}$ falls below a time-dependent threshold $𝐶_{𝑖 (𝑇)}$ .

If $𝑄_{𝑖 (𝑇)}$ is the cumulative default probability and $𝑆_{𝑖 (𝑇)} = 1 - 𝑄_{𝑖 (𝑇)}$ is the survival probability, then

$𝑄_{𝑖 (𝑇)} = 𝑃 (𝐴_{𝑖} \leq 𝐶_{𝑖 (𝑇)}) = Φ (𝐶_{𝑖 (𝑇)})$ , and $𝐶_{𝑖 (𝑇)} = Φ^{- 1} (𝑄_{𝑖 (𝑇)}) = Φ^{- 1} (1 - 𝑆_{𝑖 (𝑇)})$ .

This links the latent threshold to market data. A CDS survival curve or an implied hazard rate supplies $𝑆_{𝑖 (𝑇)}$ ; the threshold is then calibrated at each horizon. When $𝑆_{𝑖 (𝑇)} = 0.5$ , the threshold is zero. When survival is near one, $𝑄_{𝑖 (𝑇)}$ is small and the threshold lies far in the lower normal tail.

The asset value $𝐴_{𝑖}$ itself is unobserved and has no dynamics in this construction, which is why it is called latent. A simulated value of $𝐴_{𝑖}$ is held fixed while the calibrated threshold moves over time. Default occurs at the first crossing,

Default time $𝜏_{𝑖}$ is the first time $𝑡$ for which $𝐴_{𝑖} \leq 𝐶_{𝑖 (𝑡)}$ .

Repeating this draw gives a distribution of default times, from which one can estimate the mean and confidence intervals. The four cases in slide 5 (PDF p. 5) show how joint-default probability changes with asset correlation. With name-specific hazard rates and notionals, the same simulation produces the distribution of the number of defaults and portfolio loss at each horizon.

Next Week

Will be remote, most likely.

## Why Crashing the Economy Won't Work - URL: https://sharifhsn.dev/blog/linkedin-2025-04-09-why-crashing-the-economy-won-t-work/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2025-04-09-why-crashing-the-economy-won-t-work.json - Description: The secret plan behind crashing the economy—and why it can’t work 😴 The anticipation of and uncertainty around Trump’s tariffs have caused the stock market to plunge rapidly… - Date: 2025-04-09 - Exact published timestamp: 2025-04-09T01:35:45.991Z - Topics: Macroeconomics, Monetary Policy - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/feed/update/urn:li:activity:7315547943300681728/ - Content source format: typst

The secret plan behind crashing the economy—and why it can’t work 😴

The anticipation of and uncertainty around Trump’s tariffs have caused the stock market to plunge rapidly since Wednesday. Most people would view this as a bad thing. But it’s possible that this is the explicit goal of the Trump administration.

The national debt is the source of much criticism from the administration. The purpose of DOGE is to cut the federal government of “waste, fraud, and abuse”, which combined with tariff revenue, is purported to reduce our national debt. So far, these claims are illusory.

The national debt itself is not a big deal, but the interest payments are. They have ballooned in recent years, and much of the treasury debt is due to be rolled over at higher interest rates than when the bonds were initially issued.

These interest rates are related to the yield of the treasury bonds. The yield is inversely related to the price, since the actual payoff of the bond is fixed. When bonds are in high demand, the price rises, and the yields fall. If the yields are low when the bonds are refinanced, we could see hundreds of billions in savings on interest payments.

What might cause bonds to be in high demand? As a safe harbor, people want to buy them when they are reducing risk. When the stock market tanks, bonds should see higher demand as investors flee from equities to safer shores.

So that’s the idea: crash the stock market, increase demand for bonds, reducing yields, then refinance our debt at the lower interest rates.

There are some significant problems with this theory, however.

For one, it’s not coming true. Although 10 year yields briefly dipped right after Liberation Day, they have risen back to previous levels.

China also plays a strong role here. They hold a giant amount of treasury bonds. If they wanted to, they could sell those bonds en masse, massively increasing the supply, driving down the price and spiking yields. And with the way the US has been threatening them, it’s not an impossible scenario.

There’s also the strong possibility that with political volatility in the US, bonds will no longer remain a safe investment. Equities and bonds in other regions like the EU are becoming more attractive, especially as countries like Germany begin to rev up their economies to stop relying on the US.

Bottom line? Tariffs crashing the economy might be the real plan to reduce the national debt, but they won’t work compared to real political decisions like taxation and spending cuts.

## Monte Carlo and Variance Reduction - URL: https://sharifhsn.dev/blog/computational-methods-week-11/ - Structured data: https://sharifhsn.dev/api/posts/computational-methods-week-11.json - Description: We will cover this stuff in two different lectures. Today is Monte Carlo 1. The two theorems are the basis of the entire Monte Carlo simulation. The Law of Large Numbers (LLN) - Date: 2025-04-08 - Exact published timestamp: 2025-04-08 - Topics: Computational Methods, Monte Carlo, Variance Reduction, Control Variates, Asian Options - Categories: Computational Methods - Source: Computational Methods in Quantitative Finance - Source URL: None - Content source format: typst

Monte Carlo Methods

We will cover this stuff in two different lectures. Today is Monte Carlo 1.

The two theorems are the basis of the entire Monte Carlo simulation.

The Law of Large Numbers (LLN)

This says, very simply, given iid random variables $𝑋_{1}, \dots, 𝑋_{𝑛}$ , mean $𝜇$ , finite/bounded variance. If the variable has an infinite variance, it cannot converge because it goes all over the place. So that’s the one condition.

Then

There is zero probability I won’t converge.

\bar{𝑋} = \frac{\sum_{𝑖 = 1}^{𝑛} 𝑋_{𝑖}}{𝑛}

𝜇 = 𝔼 [𝑋]

How does it converge, though? It’s one thing to converge slowly, and another to be fast.

This is governed by the Central Limit Theorem (CLT) that you also learn about in probability and statistics.

This says that if you have the same situation $𝑋_{1}, \dots, 𝑋_{𝑛}$ iid, mean $𝜇$ , variance $𝜎^{2} < \infty$ . Bounded means I have unknown variable $𝑋_{𝑖}$ , variance keeps going up, $𝜎 \cdot 𝑛$ . Technically its’ finite, but it keeps going up.

then

$\bar{𝑋} \approx 𝑁 (𝜇, \frac{𝜎^{2}}{𝑛})$

This itself is kind of nonsensical, so we would say that the distribution of this transformed random variable approaches a normal distribution.

\frac{\bar{𝑋} - 𝜇}{𝜎 / \sqrt{𝑛}} \sim 𝑁 (0, 1)

But the approximation helps us understand it better:

If you have fifty distributions, and you look at the distribution of those fifty, it will look close to some normal distribution. If you increase it even more, it will shrink the variance, and it will get closer to your object the estimate. And this gives you a numerical measure of how close you are, it allows you to express a confidence interval.

At 95% C.I. $\bar{𝑋} \pm 1.96 \frac{𝜎}{\sqrt{𝑛}}$

Now technically we don’t know σ. So in practice we use the sample stdev S, which is

𝑆 = \sqrt{\frac{1}{𝑛 - 1} \sum_{𝑖 = 1}^{𝑛} (𝑋_{𝑖} - \bar{𝑋})^{2}}

Then if $𝑛$ is small, we use $𝑡_{𝑛 - 1}$ instead of $𝑁 (0, 1)$

But we’re never doing less than 10k simulations so this is irrelevant to us.

$𝑛$ degrees of freedom with greater than 100 $𝑛$ , it goes Normal.

What does this have to do with anything?

This is the key idea.

We’re going to obtain somehow values for my stock in the future. Based on those values in the future, I’m going to estimate the value of my derivative. I estimate one path, and I pretend that’ smy path in the future. If I know that’s the path in the future, I can use the terminal value and current values of the path to calculate the derivative, if I know the values of the path. That’s my observation $𝑋_{1}$ .

Then I’ll do another path, which will be $𝑋_{2}$ . And i’ll obtain all those r.v.s, and I can estimate $𝜇$ , the expected value of the derivative. I’m using this formula to estimate an expectation.

It works every time for this purpose because of the LLN.

The question is, how do we calculate these things, with mean $𝜇$ ? The real problem needs you to forecast what $𝑋_{𝑇}$ is in the future.

Step 1, most important step:

Hypothesize a model of evolution from $𝑋_{0}$ to $𝑋_{𝑇}$

I know where I am now, I’m at $𝑋_{0}$ .

I want to know what are my assumptions about the world that will lead me from now until T. THis is more complicated, because you need to place all these assumptions. Let’s say I generate a stock price, which we’ve been doing, I want to know what the price of the fixed income of the treasury bond is, which is repaid a year from now. The Treasury is AAA, never default. You know what the price is. Can I calculate the rate of this instrument that is issued for this price right now, then gives me $1 payoff at maturity. But in the Trump era, but we don’t know if it’s safe anymore. So we need to figure out what the risks are. Then you need to play in things that aren’t part of your model? How do you implement this? You can do jumps. Between now and next year, Trump will do something to collapse the market, which will be permanent not temporary. Last week was temporary, nobody cares. But a permanent thing where the economy is severely damaged, that makes it so I might not get this money I’m getting. Then you need to introduce jumps. That requires you to do Monte Carlo with jumps, which is on the homework.

But this Step 1 is the most important part.

If you are financial engineer, you need to be able to understand randomness and how it relates to real life. You need to relate real life to this Monte Carlo simulation. The methodology by itself is very straightforward, but you have to understand how it relates to the real world.

Step 2

Once you have the model, you match the model to data. In the simplest example, I look at a model for my stock price, using GBM. GBM has two parameters, drift and variance. I look at historical data and estimate a long term drift and variance. I will use those two parameters to generate data for the future. Then you have to estimate your parameters. Then you also have to identify a way to generate randomness. This is a very vague statement, but what I mean is if I have a model that I hypothesize, the noise is gamma. I need to have a programming language that generates random variables as gamma random variables. First you need to justify why you need gamma and not normal, etc. Then you can generate randomness.

We have this high frequency trading simulator stuff, and we use these zero intelligence agents to interact with real teams that are doing the trading. We need to worry about, how do we initialize these agents? How much money do we give them? We did some research that if you start with homogeneous agents vs heterogeneous agents, heterogeneous is much closer to reality. So how do you create this heterogeneous? We use the Dirichlet distribution and sample from it to initialize the wealth of the initial agents. We identify the characteristics that the simulation needs to have, then identify distributions and rvs that fit that characteristic.

Once you’re done with steps 1 and 2,

Step 3

is really simple. Create values for $𝑋_{𝑇}$ (iid), and that’s it.

Example with Call Option

Using $𝑋_{𝑇}^{1}$ , you can calculate the value of $𝐶_{1}$ as

𝐶^{1} = (𝑋_{𝑇}^{1} - 𝐾)_{+} 𝑒^{- 𝑟 𝑇}

Then you repeat this for all your samples.

Then we average from LLN:

\bar{𝑋_{𝑇}} = \frac{\sum_{𝑖 = 1}^{𝑛} 𝑋_{𝑇}^{𝑖}}{𝑛} \to 𝔼 [𝑋_{𝑇}]

These are all matching the expectation of the stock price:

\bar{𝐶} = \frac{\sum_{𝑖 = 1}^{𝑛} 𝐶^{𝑖}}{𝑛} \to 𝔼 [(𝑋_{𝑇} - 𝐾)_{+} 𝑒^{- 𝑟 𝑇}]

Furthermore, the CLT tells us how close we are. I can construct the confidence interval

Students get really confused about this. There are two $𝑛$ s. There’s the confidence interval for the mean, which gets closer to 0. But it also appears in the second one, because that one is for the actual random variable. Not $\bar{𝐶}$ , which is an average, but the actual $𝐶$ random variable.

You have the random variable which is the value of the call. If you have one path, you get an estimate, one estimate, one call value. That estimate, that rv, is going to have a mean, that will be the true value of the call, and it will have a certain variance, which is probably huge. The way you estimate that, you take a sample, maybe 50, and then maybe estimate this quantity. It will get closer to the stdev.

I have my variability, which could be $200. My original option, if I do the correct path, I’m within $200 of the real value. The more I have the n in Sdc, it will go to that 200 value, not 0.

This is plain vanilla Monte Carlo. What do we do with this?

This works really well for European type options. But it’s tricky for American. There is a method called least squares Monte Carlo on pg 210-216. It does some kind of tree method, with the path, creates a distribution of the future based on where each path is. It’s not really Monte Carlo.

It also works for Barrier Options. I mention this because we discussed them. The path needs to decide if it goes above or stays within a threshold.

Asian options also work. The payoff is determined by taking an average over the lifetime of an option. If it expires in one month, and it’s calculated daily, you take daily stock values, and then the value of the stock is an average. Or you could have an average vs a terminal value, there’s different versions. But in general, one of the terms is the average value of the stock price.

Monte Carlo can be used for ANYTHING. Nothing is financial inherently. If I want to examine deaths from a disease because Trump took us out of WHO, you can make an estimate of the contagion. This $𝑋_{0}$ could be the number of infected people currently. Then you could have some kind of dynamic of how the disease spreads, and then $𝑋_{𝑇}$ is the number of people infected at time $𝑇$ . Then you can simulate multiple things, and then calculate all sorts of derivative prices, with some “payoff” of how much money I pay in hospitals, based on how many people die of disease. Monte Carlo is used everywhere, particularly in biology.

Monte Carlo for SDEs

Everything in our areas uses SDE. This is on MF pg 273.

I’ve been saying pages because it was written by two people independently. Monte carlo is written by Mariani, and Monte Carlo written by Florescu.

This is a pretty general dynamic, you can multivariate, multidimensional. In our homework, we have to work out for Bonus how to apply this Monte Carlo to a multidimensional process (Heston).

𝑑 𝑋_{𝑡} = 𝛼 (𝑡, 𝑋_{𝑡}) 𝑑 𝑡 + 𝛽 (𝑡, 𝑋_{𝑡}) 𝑑 𝑊_{𝑡}

This is odne in the book so you can look it up.

How do you develop a Monte Carlo skill from SDE?

You can assume that $𝑋_{0}$ is some fixed point.

The first thing is to remember that this is the notation, the SDE doesn’t actually exist. The actual SDE is

𝑋_{𝑡} - 𝑋_{0} = \int_{0}^{𝑡} 𝛼 (𝑠, 𝑋_{𝑠}) 𝑑 𝑠 + \int_{0}^{𝑡} 𝛽 (𝑠, 𝑋_{𝑠}) 𝑑 𝑊_{𝑠}

Euler Discretization

Euler is a very old mathematician. There was no SDE when he was living. His method worked to discretize regular integrals, so we are using the same exact method, which is Euler’s method.

Euler discretization.

We will take $[0, 𝑇]$ divided into $𝑀$ .

Typically you generate millions of paths. The number of intervals you create has nothing to do with number of paths. Good estimation requires number of paths. But for this thing it doesn’t matter. You use whatever frequency you want. Estimate a month call option. Now you have to decide, do I want to generate every day? The time interval will be 1/252 or 1/365, and generate 25 or 30 for a month, but it’s your choice, and it doesn’t make a big difference.

The actual diffusion of your rv will be the same. It will matter if for example you estimate a barrier option. It’s about passing a threshold. If you have a lot of frequency, your rv might cross the barrier and you don’t know it.

We have $𝑚$ intervals and $𝑚 + 1$ points, $𝑡_{0}, 𝑡_{1}, 𝑡_{𝑚} = 𝑇$ .

$𝑋_{0}$ will start at fixed $𝑥_{0}$

Then our process is:

𝑋_{𝑖} = 𝑋_{𝑖 - 1} + 𝛼 (𝑡_{𝑖 - 1}, 𝑋_{𝑡_{𝑖 - 1}}) Δ 𝑡_{𝑖} + 𝛽 (𝑡_{𝑖 - 1}, 𝑋_{𝑡_{𝑖 - 1}}) Δ 𝑊_{𝑖}

IT does not really need to be equally spaced, it works with everything. You can take day 1, 2, 14.

You just need to be careful about time between observations.

The functions $𝛼$ and $𝛽$ are known. You choose the times. The only unknown noise is $𝑊$ . We know that increments are $𝑁 (0, Δ 𝑡)$ , so you can calculate this as $𝑍 \cdot \sqrt{Δ 𝑡}$ where $𝑍$ is $𝑁 (0, 1)$ .

99% of the errors are coming from the student using the N(0,1) directly for W, and it’s too large.

Always use the left hand point that you have.

This is one path, you store the final value. But if you’re pricing the barrier option, you don’t just need to know this, you need to know path.

The problem he gave us:

Monte Carlo with Jumps

Jumps allow us to introduce events that are unpredictable. You need to know they’re going to happen, but I don’t know when. I know that Trump will introduce tariffs again, because he’s a moron. We know that this is going to happen, but we don’t know why. We need this compound Poisson process.

We discussed two ways. Exponential times we accumulate,

or in our case (much better), since we have 0 to $𝑇$ and we know the time interval, we create the Poisson random variables, generated, with $𝜆 \cdot 𝑡$ as the expected number of events. Once you generate that, the Poisson tells you how many times our president will influence the financial market. Then you create these $𝑛$ uniform variables from 0 to $𝑇$ , and sort them, and those are the actual generated times when he says the stupid things.

Then how are the stupid things going to influence the markets? Last week, he said tariffs on Wednesday, and markets went down. I am not watching the market, so tell me when to buy. NVDA is overpriced, TSLA will go down, I can’t make up my mind. This is being an investment banker or trader, not FE.

We have to

Step 1

Identify our model.

𝑑 𝑋_{𝑡} = 𝛼 (𝑡, 𝑋_{𝑡}) 𝑑 𝑡 + 𝛽 (𝑡, 𝑋_{𝑡}) 𝑑 𝑊_{𝑡} + 𝑑 𝑌_{𝑡}

This can be ANYTHING, our choice, could be Black-Scholes, or something else. You just have to fit it to real data.

Where $𝑌_{𝑡}$ is a compound Poisson process. The sum of these jumps, where the time of these jumps happens according to an actual Poisson process.

We know how to generate the $𝛼$ - $𝛽$ part, same methodology. The only question is, what do I do with the jumps?

I’m going to have to generate for each path, a set of jumps, and then add them to the price process.

Generate $Poisson (𝜆 𝑇) \to 𝑘$ (every time a different $𝑘$ )

Once I know $𝑘$ , I generate $𝑇_{1}, \dots, 𝑇_{𝑘}$ times of the jumps for the $𝑌_{𝑡}$ process.

And these are in the interval $[0, 𝑇]$ . If you forgot how to do this, you take each $𝑇_{𝑖} \sim Uniform (0, 𝑇)$ and sort them.

What about jump amounts? $𝑌_{1}, \dots, 𝑌_{𝑘}$ are the size of the jumps

It depends how you model your stochastic process. Let’s say I’m modeling Donald Trump. Maybe he says something about tariffs, he goes down. So there’s no point generating a rv that goes up, I know it goes down. So I will generate it on a distribution of negative numbers, uniform(-0.5, −1). Let’s say the second jump will be positive, so do something like that. You need to adapt it to your case.

But let’s just say for simplicity it’s iid. Homework says you should use the Normal distribution.

Identify the intervals $Δ 𝑡$ which contain $𝑇_{1}, \dots, 𝑇_{𝑘}$ It’s possible that all the jumps happen really close to each other, in the same interval. I will just add all the jumps in the interval together.

The jumps happen in the interval.

Whenever you have the jump, you add the jump. If not, you don’t care.

The path

In the example in the homework, it uses returns, so it’s going to be $log 𝑆_{𝑡}$ plus other stuff, so be careful. Write down the math before you program anything. You might not be sure whether to multiply or to add.

This is the basic idea behind Monte Carlo.

Next week, he’s going to give us a better Monte Carlo method

One of the other problems is a variance deduction technique.

Euler Milstein is a better approximation than this one, but it only works for homogeneous processes, which are not time-varying i.e. don’t contain $𝑥$ in $𝛼$ or $𝛽$ .

Variance Reduction

Imagine I’m pricing an option, and I look at the price I’m getting, and I get a 95% CI which is $10 wide, with n = 100 paths. If you look at the difference, it’s $2 * 1.96 (𝑏 𝑒 𝑐 𝑎 𝑢 𝑠 𝑒 𝑖 𝑡^{'} 𝑠 𝑚 𝑎 𝑟 𝑔 𝑖 𝑛 𝑜 𝑓 𝑒 𝑟 𝑟 𝑜 𝑟) * \frac{\hat{𝜎}}{100} = 10$

This is about $10.

And then my boss tells me the price of the option is $3, to give him a better estimate.

So how many paths do I need to do to get this to $3? Sigma hat is going to change, but that sigma hat converges to the true σ, so it’s supposed to be kind of close. If I divide by 10, I get

2 \cdot 1.96 \frac{\hat{𝜎}}{\sqrt{10000}} = 1

I get $1 if I increase my sample size to 10000 paths.

So then I go to my boss and he tells me the price is between $2 and $3. DId you forget that these options are sold in multiples of 100? It’s $200 to $300. This is way too wide a margin!!! I want to get within 10 cents. So I have to multiply by 100, and now I get to 1 million. Now that’s a lot of paths! And then if I want 0.01, which is within $1 of the actual contract, I need 100M paths. And it’s because of this square root.

So what do we do? Variance Reduction techniques. Paul Glasserman is the expert at this. If you want to read more about this. He has this classic book Monte Carlo Methods in Financial Engineering (2003). Also a paper Monte Carlo methods for security pricing, 50 pages, but it does what the book does in a condensed way. This was written in 1997, so it’s old, but a lot of techniques still used today. Broadie sucks, by the way. This paper: Paper

Having more n is a better approximation, but it is not the solution, because it gets harder to reduce the confidence interval.

So can we calculate σ hat smartly?

The quantile doesn’t change (1.96), n doesn’t change, so how do I get a better σ hat. That’s wy these are considered Variance reduction techniques.

Antithetic Variate

Th doesn’t exist in Latin languages. English is confusing, sometimes you pronounce it differently. It’s the same writing but it’s different! Always with the tongue in your mouth. Like the Sylvester from Looney Tunes, apotheosis of using the th sound.

How does this work? This antithetic variate is based on a very simple idea. If Z ~ N(0, 1), then -Z is also N(0, 1). That’s the entire idea.

So how does this work? The problem with this path generation, what takes you time? The time generating the random variables. The idea is to generate the path with m increments. Z1, Z_2, Z_m, N(0,1) rvs. You can create a path using these rvs, and also -Z_1, -Z_2, -Z_m. For the same amount of numbers, you calculate two paths, where the second one is created for free.

Unfortunately, it is not the same as having 2m independent paths. Then my stdev would decrease by √2, old interval, divided by 1.41. But these paths are correlated. So the estimate for σ hat is going to grow. This is a worse estimate.

The net result is still better, but it’s not √2 better.

𝐶 = \frac{\sum_{𝑖 = 1}^{𝑚} 𝐶_{𝑖} + \sum_{𝑖 = 1}^{𝑚} 𝐶_{𝑖}^{𝑎}}{2 𝑚}

Really simple to implement, easy to do, but not that impactful. You do it because it’s free.

Then we will talk about the versatile method, which is more complicated.

Sidebar: ChatGPT Investing

I don’t know how to invest. I asked ChatGPT. Tell me which stocks have the largest return drop since April 2 (Liberation Day). What I have noticed is ChatGPT is completely useless. I can’t even calculate. I said give me a table of returns showing largest drop. He found the three journal articles that talk about this, limit to what other people have published. Interesting: how do the results of ChatGPT correlate with the market.

Control Variates

The idea of this, this is more complicated. There are other simpler methods. Antithetic is the simplest.

Control variate takes advantage of certain relations between derivatives.

If you know certain things about derivatives, you can use that to your advantage.

This first example is on page 207 of MF, well-explained. Related to delta hedging.

How does this work?

At t = 0, we, the option seller, receive C_0, the premium. This is the option price. The idea of delta hedging is that I pay the payoff C_T to the option buyer.

(Only works for European options).

If we held Δ, the derivative of call price with respect to stock, units of stocks at time any time t. Then we can replicate the option payoff C_T.

If I hold that Δ units of stocks, at time T, I’m going to end up with exactly with the payoff. If the payoff is 0, I have 0 units of stocks, if I have one, I get one.

The problem of course is that you can’t own something that moves all the time. How do I own a derivative? I own 2 shares, then it goes to 1.5 shares. Everyone hates this because you have to buy stock when it goes up, and sell when it goes down.

The fact is that we don’t buy and hold. We have to adjust this delta hedge all the time. So how does this work in practice?

Say we have time t_0 = 0, t_i = iΔt, Δt = T/m, where m is number of intervals

I’m going to take these equally spaced, option expires in 30 days, take every day.

I find this easier to explain my way than the way other people do it.

⬇️check the notes for this table

Time | Receive | Pay to hold Δ units of S_t

If I do this instantaneously, this is supposed to be 0. The Δ is matched exactly. If you make m reasonably high, it should be close to 0. But I can’t really add them together. Because these are cash flows at certain times, and you have to multiply them by the discount factor. You could either discount them all to present day, or compound them to time T. Multiply by e-rΔt, e-r2Δt, … or ermΔt, er(m-1)Δt, …

If we are compounded, then we cancel out most of these, and then we get

If the time increment is really tiny, I get an error. Then we will rewrite this whole expression.

𝐶_{0} = 𝐶_{𝑇} 𝑒^{- 𝑟 𝑇} - \sum_{𝑖 = 1}^{𝑚 - 1} \frac{𝜕 𝐶_{𝑖}}{𝜕 𝑆} (𝑆_{(𝑖 + 1) Δ 𝑡} 𝑒^{- 𝑟 Δ 𝑡} - 𝑆_{𝑖}) 𝑒^{- 𝑖 Δ 𝑡} + 𝜂 𝑒^{- 𝑟 𝑇}

And we will say η is approximately 0, noise that is discounted.

This expression says that my value for my option, equals, using my paths, the intermediate steps. I know the final value, and I subtract it by all these steps. The only thing I don’t know is Δ, the derivative of the option. If I knew that, I would know how to generate my path. This control variate gives me a way better estimate. The only problem is I don’t know is Δ… except I do! For Black-Scholes, Δ = N(d_1) (for a call). This d1 you can calculate with current value of stock price, and you know what it is based on your path.

The idea behind this control variate method, which is conceptually so much more complicated, once you derive it is much easier. You generate paths using the Euler method, as explained before.

The larger the model variability, the worse prediction. The standard error, the sigma over error, if you have no variance reduction, antithetic is some reduction. The control variate has a huge reduction in standard error. And this is typically the difference between financial engineer and computer scientist. You need to know what you’re doing. Monte Carlo is really bad, UNLESS you know what to do. And this is a lot of derivation.

If you don’t have Black-Scholes, you need to find the Δ somehow, and that might take a long time. The variance might look good, but it will take longer.

Asian Options

This is the Asian option calculation, the example from the Monte Carlo paper.

What is an Asian option? It’s an option whose payoff depends on an average price throughout the lifetime of the option.

It’s usually easier to solv things in continuous time, and not discrete time.

The most general type of Asian option:

There are two types:

Arithmetic

The payoff looks like this

\frac{1}{𝑇} \int_{0}^{𝑇} 𝑆_{𝑡} 𝑑 𝑡 - 𝑆_{𝑇} - 𝐾)_{+}

This is the most general type you can imagine. It doesn’t exist in practice. This is the one mathematician slike to solve because it’s simpler. The integral takes the entire stok price, which you assume is continuous, calculate an average here. It’s basically the area. What you’re doing is, you’re taking the side, and dividing it by T. You’re actually finding out the height such that the area of the rectangle is equivalent to the area under the curve (integral). That’s the arithmetic average. And, it is including S_T and K. Typically it is only one of them. I wrote it in the most possible general way. You can get even more general.

Geometric

The payoff here is written like

(𝑒^{\frac{1}{𝑇} \int_{0}^{𝑇} ln 𝑆_{𝑡} 𝑑 𝑡} - 𝑆_{𝑇} - 𝐾)_{+}

This is called geometric, I’ll explain why.

It’ll reduce to a geometic average on discrete formula. But this looks so ugly! For this, we have an analytical formula. But for arithmetic, we don’t have a formula.

The reason we have a formula is because ln S_T is normal, which means that this is an mgf, and we have a formula that we can calculate with. In practice, none of this exists, it’s theoretical baloney. But in practice, everything that is solved is arithmetic. Nobody understands geometric average. What is that?

In RL

And how does it look like in practice?

Monitor the stock price at certain times, from S_t_1… S_t_m. You’re going to observe the price every end of the day. This is defined int eh contract. The contract says, look at the average, it’s calculated using beginning of the day prices, the high during the day, low, average, etc. That’s a specification on the S_t_i.

The arithmetic Asian:

(\frac{\sum_{𝑖 = 1}^{𝑚} 𝑆_{𝑡_{𝑖}}}{𝑚} - 𝐾)_{+}

This is a call

This is the same as the integral. You can take T/m as the integrand, you approximate LHS, and that’s what you get. T cancels, increment is T/m.

For geometric,

If you use the previous formulation, it is

𝑒^{\frac{1}{𝑚} \sum_{𝑖 = 1}^{𝑚} ln 𝑆_{𝑡_{𝑖}}} - 𝐾)_{+}

This is replacing the integral with an approximation, t cancels. THis is equal to, if you cancel out all of these sums of logs, you get

= (\sqrt{𝑆_{𝑡_{𝑖}} \dots 𝑆_{𝑡_{𝑚}}} - 𝐾)_{+}

Except not square root, it’s mth root.

This is a lot harder to price than using the initial formula.

And geometric has a formula!

If you assume that the underlying is GBM, same as Black-Scholes.

Arithmetic does not have ea formula. But in the market, there are zero contracts that are sold using the geometric average, because no one understands it. But everyone trades using the arithmetic Asian option. So how do you price the arithmetic, using the geometric?

You use the control variate thing.

First, we generate a path i, which is S_t_0, S_t_i, S_t_m

you calculate the arithmetic premium A_i from the same formula as before:

(\frac{\sum_{𝑖 = 1}^{𝑚} 𝑆_{𝑡_{𝑖}}}{𝑚} - 𝐾)_{+} 𝑒^{- 𝑟 𝑇}

You can also calculate G_i the geometric premium from before:

= (\sqrt{𝑆_{𝑡_{𝑖}} \dots 𝑆_{𝑡_{𝑚}}} - 𝐾)_{+}

I have three different ways to do this.

One is one that I came up with, which you can’t find anywhere, and I don’t know if it works.

Two is given by controlv ariates.

You have all these A_i and G_i. It would make sense to do some kind of regression. I assume that there is a certain distance between A_i and G_i. A which is the true price, follows

𝐴 = 𝛼 + 𝛽 𝐺

This is a theory, but it must be true because these are two numbers, in general.

What you do is a regression between A_i and G_i.

𝐴_{𝑖} = 𝛼 + 𝛽 𝐺_{𝑖} + 𝜖_{𝑖}

The idea is to minimize the noise, you’re fitting a regression so there is some noise.

We can come up with three different estimates.

Estimate 1:

Don’t use any control anything. Just use A bar, just the average, regular Monte Carlo.

Estimate 2:

This is what control variate is supposed to do. I have a formula G. I don’t need any estimation for that. In that formula, which is a function f(S_0, K, T, r, σ). I don’t need any path, so I know what that number is, without doing any approximation.

I basically take A bar, and adjust it by the following adjustment.

\bar{𝐴} - \hat{𝛽} (\bar{𝐺} - 𝐺_{formula})

You can derive this, it’s not very complicated. That’s the other estimate. EXCEPT THAT, I don’t understand what’s wrong with

Estimate 3

It would make sense that the relationship between A and G should be the same as the theoretical relationship between real A and real G. So why don’t we just take

\hat{𝛼} + \hat{𝛽} * 𝐺_{theory}

These intrinsically depend on all the paths, so they contain information about the paths.

Pretty sure I’m going to give you a problem for this, cook it up on the final.

## Misinformation in Volatile Markets - URL: https://sharifhsn.dev/blog/linkedin-2025-04-07-misinformation-in-volatile-markets/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2025-04-07-misinformation-in-volatile-markets.json - Description: Misinformation becomes especially dangerous in volatile markets… 😱 After President Trump’s tariff announcement on “Liberation Day” last Wednesday, markets have roiled. The… - Date: 2025-04-07 - Exact published timestamp: 2025-04-07T19:31:32.998Z - Topics: Financial Markets, Misinformation - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/feed/update/urn:li:activity:7315093897339101184/ - Content source format: typst

Misinformation becomes especially dangerous in volatile markets… 😱

After President Trump’s tariff announcement on “Liberation Day” last Wednesday, markets have roiled. The easiest way to see this is measuring VIX. This index averages the implied volatility of SPX options expiring in a month. It is known as the “fear gauge” because it goes up when markets are uncertain, and therefore fearful. VIX typically hovers around 20-30, but after Liberation Day, it has spiked to 40-50.

Market participants are waiting with bated breath for any news on the tariffs, whether they will be permanent or temporary, or be exempted from certain markets.

So when a tweet went viral from someone named “Walter Bloomberg” declaring that Trump was considering a 90-day pause on tariffs, the markets reacted immediately.

The S&P 500 jumped up 300 points in just ten minutes, representing over $2.5 trillion in value. But this optimism wouldn’t last.

When the market realized that the tweet had no factual basis (it was a wild misinterpretation of Hassett’s remarks on Fox News days earlier), the market tanked back to its earlier levels.

And after the White House confirmed that the tweet was fake news, the S&P continued as it had before, albeit with more volatility than normal.

It’s a sharp reminder of how even the smartest traders on Wall Street can get taken in by seemingly-plausible BS on social media.

If they can be fooled, so can you.

Read more at Bloomberg: https://lnkd.in/eBRP9jZg

## Documentation Is the Key to Vibe Coding - URL: https://sharifhsn.dev/blog/linkedin-2025-04-05-documentation-is-the-key-to-vibe-coding/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2025-04-05-documentation-is-the-key-to-vibe-coding.json - Description: Documentation is the key to vibe coding success! The use of LLM AI code assistants (or “copilots”) like Github Copilot and Claude has exploded recently. Many non-programmers have… - Date: 2025-04-05 - Exact published timestamp: 2025-04-05T19:46:48.390Z - Topics: AI, Software Engineering - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/feed/update/urn:li:activity:7314372961040289792/ - Content source format: typst

Documentation is the key to vibe coding success!

The use of LLM AI code assistants (or “copilots”) like Github Copilot and Claude has exploded recently. Many non-programmers have taken advantage of this to begin coding their own projects without any past software engineering experience, just “coding on vibes”.

However, this is not as easy as it might sound. There’s a lot of basic principles and best practices to software engineering that copilots don’t steer you away from. If you make a design mistake, your copilot might double down on it and spit out a huge amount of code to solve a simple problem in a wrong, complex way.

Ultimately, the only solution is to actually know how to code, how to use your language, and the library you’re using. But there are some things you can do in the meantime to improve your output.

The key to understanding here is that LLMs are input/output machines. The output is shaped heavily by the information and context you give it. If you just give the prompt “Make a full stack calculator app in React”, it’s going to have to make a lot of assumptions. Some, even most of those assumptions will be fine. But ultimately you won’t know the difference, which will lead to pain later on when some of them aren’t.

Therefore, for any vibe coding project you’re embarking, detail extensively in a document all of the project requirements. Any bit of information you have in your head related to the project, write down. If you feed that into your copilot, it will be able to reference that information when it’s relevant and tailor its output to what you give it.

This might seem daunting, especially if you’re vibe coding something you don’t fully understand yourself. But LLMs can help with that, too! Before you start actually coding, ask the LLM what information it thinks you can provide to help with its output. With just fifteen minutes of preparation and research, you can save yourself hours of debugging later.

If you have any other vibe coding suggestions, please share in the comments!

## Rust 1.86: `get_disjoint_mut` - URL: https://sharifhsn.dev/blog/linkedin-2025-04-03-rust-1-86-get-disjoint-mut/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2025-04-03-rust-1-86-get-disjoint-mut.json - Description: Rust 1.86.0 has been released, with a very nice ergonomic feature to get around the borrow checker: get_disjoint_mut Rust’s borrow checker is the heart of its promise of memory… - Date: 2025-04-03 - Exact published timestamp: 2025-04-03T18:41:40.946Z - Topics: Rust, Programming Languages - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/feed/update/urn:li:activity:7313631796301099010/ - Content source format: typst

Rust 1.86.0 has been released, with a very nice ergonomic feature to get around the borrow checker: get_disjoint_mut

Rust’s borrow checker is the heart of its promise of memory safety. On a basic level, it ensures the following - when you obtain an immutable reference to an object, no mutable reference can be made to that object while that reference exists - when you obtain a mutable reference to an object, that reference is the only reference that exists

This creates a problem with slices, however. Imagine that you have a buffer, and you want to mutate only the last element, and pass around the rest of the buffer through immutable reference. You can take a mutable reference to the buffer to mutate the last element through get_mut, but you can’t take an immutable reference to the rest of the buffer through get. Even though you know you’re only mutating the last element, the borrow checker doesn’t know that. This can require unergonomic reshuffling of memory access, or in the worst case, unnecessary cloning.

The new get_disjoint_mut solves this problem. By returning mutable references to only the slices that you intend to mutate, you can have finer control over mutable memory access.

Read more details in the release notes. https://lnkd.in/e5Je6Mb3

## Credit Default Swap Valuation - URL: https://sharifhsn.dev/blog/advanced-derivatives-week-10/ - Structured data: https://sharifhsn.dev/api/posts/advanced-derivatives-week-10.json - Description: We discussed the mechanics of CDS a bit, and then started on the valuation. So let’s discuss the valuation in more details. We have two legs. The premium leg is the buyer, who… - Date: 2025-04-03 - Exact published timestamp: 2025-04-03 - Topics: Credit, Credit Default Swaps, CDS Valuation, Premium Leg, Protection Leg - Categories: Credit - Source: Advanced Derivatives - Source URL: None - Content source format: typst

Credit Default Swaps

We discussed the mechanics of CDS a bit, and then started on the valuation. So let’s discuss the valuation in more details.

We have two legs. The premium leg is the buyer, who buys protection against default for the underlying company. They have to make scheduled payments throughout the life of the CDS. So that if there is a credit event, there is a payment on the premium which has accrued. If there is a default in this period, you still have to pay the accrual.

In case of default, the protection leg will purchase the underlying bond at face value. That’s the other leg.

Valuation of the Premium Leg (O’Kane 6.5)

So we want to determine the spread of the CDS. That represents the percentage of the principal that we have to pay to purchase the CDS.

Let’s review a little bit what we did last time.

We defined the present value of $1 to be paid at $𝑡_{𝑚}$ , which cancels with zero recovery on default before $𝑡_{𝑚}$

This is a risky ZCB as opposed to riskless.

\tilde{𝑃} (𝑡, 𝑡_{𝑛}) = 𝔼 [𝑒^{- \int_{𝑡}^{𝑡_{𝑛}} 𝑟 (𝑠) 𝑑 𝑠} \cdot 𝕀_{𝜏 > 𝑡_{𝑛}}]

Furthermore, if you assume independence between short rate process and the time of default τ, which is a reasonable assumption,

then the expectation can be separated as a product because the joint expectation can be written in such a way.

We can take the survival probability:

\tilde{𝑃} (𝑡, 𝑡_{𝑛}) = 𝑃 (𝑡, 𝑡_{𝑛}) 𝑄 (𝑡, 𝑡_{𝑛})

Then you can write the present value of premium leg as

𝑆_{0} \sum_{𝑛 = 1}^{𝑁} Δ (𝑡_{𝑛 - 1}, 𝑡_{𝑛}) 𝑄 (𝑡, 𝑡_{𝑛}) 𝑃 (𝑡, 𝑡_{𝑛})

where Q(t, T) is the survival probability at time t of the reference entity to T.

The Δ(t, T) is the day count fraction between T and t.

This is applied to some principal.

This is the expected payment.

Premium Accrual

Another component of the premium leg is the premium accrued.

The amount of premium accrued at default is contingent. The price today of $1 paid at default which occurs at $[𝑠, 𝑠 + 𝑑 𝑠]$ is given by

𝑃 (𝑡, 𝑠) [- 𝑑 𝑄 (𝑡, 𝑠)]

I want to calculate the accrued premium for an infinitesimal amount, then integrate, because we don’t actually know when the default is going to happen.

So what is such premium accrued?

This is the expected such contingent payment, so it is equivalent to the expected default in that particular interval. So we’re going to get an amount with some probability of default, then we multiply the probability of default in that small interval. So that survival probability is decreasing, an infinitesimal change in Q will correspond to the probability of default between t and s.

Then we discount such contingent payment back to time t.

So then you start with CDS Spread S_0, from the time of previous payment to time s (default).

That is the amount accrued on expectation.

Accrual = 𝑆_{0} Δ (𝑡_{𝑛 - 1}, 𝑠)

Then the expected present value of premium accrued due to a default in $[𝑠, 𝑠 + 𝑑 𝑠]$ in the nth premium period is…

The amount is

Accrual 𝑃 (𝑡, 𝑠) [- 𝑑 𝑄 (𝑡, 𝑠)]

Note that default can happen any time during the coupon period. So the value of accrual is integrated.

𝑆_{0} \int_{𝑡_{𝑛 - 1}}^{𝑡_{𝑛}} Δ (𝑡_{𝑛 - 1}, 𝑠) 𝑃 (𝑡, 𝑠) [- 𝑑 𝑄 (𝑡, 𝑠)]

This is a quantity such that expected accrual in case of default between t_n-1 and s. You can sum this over the life of the CDS. Then you will have a sum of such integrals, over each payment period.

This is the exact formula:

Sum over all the premium payment periods to calculate the expected present value of the premium accrued

𝑃 𝐴 = 𝑆_{0} \sum_{𝑛 = 1}^{𝑁} \int_{𝑡_{𝑛 - 1}}^{𝑡_{𝑛}} Δ (𝑡_{𝑛 - 1}, 𝑠) 𝑃 (𝑡, 𝑠) [- 𝑑 𝑄 (𝑡, 𝑠)]

The present value of the premium leg therefore becomes

𝑃 𝑉_{premium} = 𝑆_{0} [\sum_{𝑛 = 1}^{𝑁} Δ (𝑡_{𝑛 - 1}, 𝑡_{𝑛}) 𝑃 (𝑡, 𝑡_{𝑛}) 𝑄 (𝑡, 𝑡_{𝑛}) + 𝑃 𝐴 / 𝑆_{0}]

Approximation of Protection Leg

This is an exact expression. But there’s a problem with integration. You need the survival probability at each point.

You can do an approximation,

by valuing the function at the end of the interval and taking an average.

Important assumption: the payment happens at the end, when default occurs.

𝑃 𝐴 / 𝑆_{0} \approx \frac{1}{2} Δ (𝑡_{𝑛 - 1}, 𝑡_{𝑛}) 𝑃 (𝑡, 𝑡_{𝑛}) [𝑄 (𝑡, 𝑡_{𝑛 - 1}) - 𝑄 (𝑡, 𝑡_{𝑛})]

Basically, the change in survival probability represents the probability of default in that interval.

Sometimes, it’s difficult to have information about the survival probability. If we look at bonds, we make the assumption that every year has the same probability of default. In terms of the formula, it’s exact, but the approximation is more useful.

Here we assume that on average, default will happen halfway.

Valuation of Protection Leg (O’Kane 6.6)

The protection leg is a contingent payment of par minus recovery on the face value of credit following a credit event.

We have an uncertain quantity which is paid at default.

We have an assumption that we have a certain recovery rate, but we don’t know exactly how much can be recovered (depends on the restructuring process).

The price of a security which pays an uncertain amount (φ(τ)) at time of default:

\tilde{𝐷} (𝑡, 𝑇) = 𝔼 [𝑒^{- \int_{𝑡}^{𝜏} 𝑟 (𝑠) 𝑑 𝑠} \cdot 𝜙 (𝜏) \cdot 𝕀_{𝜏 \leq 𝑇}]

This is a more general expression, but in general in the problems, we assume

𝔼 [𝜙 (𝜏)] = 1 - 𝑅

If you recover nothing, the payment is the full principal. So we assume this is a fixed number, depending on the type of loan. This is independent of interest rates and default time.

Then the present value of the protection leg is..

𝑃 𝑉_{protection} = (1 - 𝑅) \int_{𝑡}^{𝑇} 𝑃 (𝑡, 𝑠) [- 𝑑 𝑄 (𝑡, 𝑠)]

You have one payment but you don’t know when you’ll get it. So you take the expectation.

The infinitesimal here is the probability of default for each infinitesimal, then discount it.

This is the general formula, which is complex to use.

Approximation of Protection Leg

One simpler approach is to discretize the time between t and T into k intervals:

𝜖 = \frac{𝑇 - 𝑡}{𝑘}

Then we can turn this integral into a sum over discrete intervals.

𝑃 𝑉_{protection} = (1 - 𝑅) \sum_{𝑘 = 1}^{𝐾} 𝑃 (𝑡, 𝑘 𝜖) [𝑄 (𝑡, (𝑘 - 1) 𝜖) - 𝑄 (𝑡, 𝑘 𝜖)]

Another approximation:

We know that P(t, T) is a monotonically decreasing function of T, so we can define bounds

lower:

𝐿 = (1 - 𝑅) \sum_{𝑘 = 1}^{𝐾} 𝑃 (𝑡, 𝑡_{𝑘}) [𝑄 (𝑡, 𝑡_{𝑘 - 1}) - 𝑄 (𝑡, 𝑡_{𝑘})]

Basically, here we have a difference in survival probabilities, where one is discounted. The problem is that the cash flows have to be discounted, but right now it’s discounted at the end of the interval of kε. The discount depends on when the default happens, which can be at any time during the interval.

𝑈 = (1 - 𝑅) \sum_{𝑘 = 1}^{𝐾} 𝑃 (𝑡, 𝑡_{𝑘}) [𝑄 (𝑡, 𝑡_{𝑘 - 1}) - 𝑄 (𝑡, 𝑡_{𝑘})]

The lower bound is for large value of T, upper bound is for smaller value

Then we take an average to get:

𝑃 𝑉_{protection} \approx \frac{1}{2} (𝐿 + 𝑈)

And then for a fair value of a CDS, based on no-arbitrage argument

𝑃 𝑉_{premium} = 𝑃 𝑉_{protection}

SPREAD FORMULA:

Then we can finally solve for the spread $𝑆_{0}$

𝑆_{0} = \frac{(1 - 𝑅) \sum_{𝑘 = 1}^{𝐾} [𝑃 (𝑡, 𝑡_{𝑘}) + 𝑃 (𝑡, 𝑡_{𝑘 - 1})] [𝑄 (𝑡, 𝑡_{𝑘 - 1}) - 𝑄 (𝑡, 𝑡_{𝑘})]}{\sum_{𝑛 = 1}^{𝑁} Δ (𝑡_{𝑛 - 1}, 𝑡_{𝑛}) 𝑃 (𝑡, 𝑡_{𝑛}) [𝑄 (𝑡, 𝑡_{𝑛 - 1}) + 𝑄 (𝑡, 𝑡_{𝑛})]}

Therefore, we need the full discount curve, the survival curve, this information.

Example

S = spread

We assume that payment occurs at the end of the year. We have the survival probability that the name survives, then we have an expected payment which is discounted to the present. Let’s do one calculation:

If we have a static probability of default, probability of default in second period is Q*(1-Q), probability of survival in second period is Q^2.

Expected payment in 3rd year is Q^3 * S.

PV of expected payment is Q^3 * S * e^{-r * T}

The PV of accrual payment in event of default:

The more general formula is on the average of default happens halfway, so our time of default is 1.5, 2.5, etc.

If during third year, the probability of default (from table) is 0.0192.

Then the coupon is halved, and the accrual payment is 0.5S. If it’s every month, then you divide that by 12 as well.

Expected accrual payment at t = 2.5 iii

With the CDS rate on the market, you can make an inference about the implied probability of default.

If mid market spread for a 5 year CDS is 100bps per year, then the conditional default probability is 1.61%.

Implied Hazard Rates from CDS Spreads

For the problem of estimation of default probabilities and corresponding hazard rates, we will used the so-called JPMorgan model. This is the credit curve, the “term structure” of probabilities of default.

Review of Hazard Rates

τ = time to default

CDF of probability of default is F(t) = P(τ ≤ t)

For the purpose of this, we will take survival probabilities, which are 1 - F(t)

The hazard rate is either h or λ by convention:

The survival probability is equal to the

𝑆 (𝑡) = 𝑒^{- \int_{0}^{𝑡} ℎ (𝑢) 𝑑 𝑢}

If the hazard rate is constant, you have e-ht function. But this doesn’t have to be constant. Typically, we will assume it’s piecewise constant, depends on how we estimate it.

If S(t) is differentiable, then

ℎ (𝑡) = - \frac{𝑑}{𝑑 𝑡} ln 𝑆 (𝑡) = \frac{𝐹^{'} (𝑡)}{1 - 𝐹 (𝑡)}

This is the relationship between the hazard rate, survival, and default probabilities.

If you have a risky ZCB with zero recovery that pays $1 at T:

\tilde{𝑃} (0, 𝑇) = 𝔼 [𝑃 (0, 𝑇) * 𝕀_{𝜏 > 𝑇}]

where

𝑃 (0, 𝑇) = 𝑒^{- \int_{0}^{𝑇} 𝑟 (𝑢) 𝑑 𝑢}

Then,

survival probabilities are

𝑆 (𝑡) = \frac{\tilde{𝑃} (0, 𝑇)}{𝑃 (0, 𝑇)}

We can also note that the risky ZCB value must be less than the riskless ZCB..

average of hazard rates will be time 0

Sometimes it’s easier to calculate an average hazard rate for some maturity, then afterwards compute the piecewise constant hazard rate.

We can determine h_i from successive survival probabilities.

We have the survival probability, but to express in terms of hazard rates.

If you assume hazard rates are known

## Calibration and SDE Parameter Estimation - URL: https://sharifhsn.dev/blog/computational-methods-week-10/ - Structured data: https://sharifhsn.dev/api/posts/computational-methods-week-10.json - Description: Parameter estimation for stochastic differential equations. investments by the West after dictator Ceaușescu failed to invade Czech republic he stopped imports and increased… - Date: 2025-04-01 - Exact published timestamp: 2025-04-01 - Topics: Computational Methods, Calibration, Maximum Likelihood, Stochastic Differential Equations, Parameter Estimation - Categories: Computational Methods - Source: Computational Methods in Quantitative Finance - Source URL: None - Content source format: typst

Plan for Today

Parameter estimation for stochastic differential equations.

Romania

investments by the West after dictator Ceaușescu failed to invade Czech republic

he stopped imports and increased exports

romania makes the best shoes (dress shoes)

they exported with the best leather, shit leather was for romanian shoes

black market, people stole best shoes off line

they had tons of money, but nothing to buy

we had alcohol, that’s it

85 is some stupid stuff (1984)

Calibration

You observe some data. Most of the time, you observe the data in time. That is traditional for our domain, and many others.

We say okay, I need to understand the dynamic of this data. In order to do this, we introduce randomness, because the randomness allows you to create data which is different. There is no way in hell you can predict the next data. I can try to predict the trend, and predict the variability around that trend. In our area of finance, it turns out that the stuff is so random, we have to go to these SDEs to understand anything.

In that stochastic equation we have two terms, drift and diffusion. Drift is trend, diffusion is variability.

Once I hypothesize a model, the next step is to find the parameters. For BS, estimate $𝜎$ and $𝜇$ . You can calibrate, or estimate from path of process (more complicated).

Calibration, you look at a derivative. A function of an underlying.

We need financial derivatives. We have seen derivatives like calls and puts. We hypothesize a model, where you come up with a formula for the derivative prices. For example:

𝐶 (𝑆, 𝐾, 𝑇, 𝑟, parameters:, 𝜃_{1}, 𝜃_{2}, 𝜃_{3})

Let’s take the Ornsteinn-Uhlbeck model: This is a mean reverting process.

𝑆_{𝑡} = 𝜃_{1} (𝜃_{2} - 𝑆_{𝑡}) 𝑑 𝑡 + 𝜃_{3} 𝑑 𝑊_{𝑡}

This is called the mean reverting Ornstein-Uhlenbeck model. How do you estimate this?

Let’s hypothesize an option on this, with all the known parameters known.

Given this, I’m going to solve a differential equation and get some sort of formula for the solution.

Once you have that, you’re done.

Then you’re basically done, because you have now a massive complex optimization problem.

Let’s say you observe $𝐶_{1}, 𝐶_{2}, \dots 𝐶_{𝑘}$ market option prices.

Then you calibrate the model to these option prices.

So you take

min_{𝜃_{1}, 𝜃_{2}, 𝜃_{3}} \sum_{𝑖 = 1}^{𝑘} (𝐶 (𝑆, 𝐾_{𝑖}, 𝑇_{𝑖}, 𝑟, 𝜃_{1}, 𝜃_{2}, 𝜃_{3}) - 𝐶_{𝑖})^{2}

(keeping in mind which are subscript i, different for each call option)

You can change this. Let’s say you have twenty options, but you’re more interested in fitting out of the money options, you can weight each of them. Let’s say you want the four options like this, then you weight one fifth of each of those options, and one fifth of the rest.

‘

Now in practice, for instance, we talked in the beginning of class about local vol. That’s a calibration method, we calibrate models to the observed values. In practice, if you’re doing real quantitative analysis, you do have these complicate model which have a combination of fitting/calibration, and another method (estimation).

Estimation

This is a very classical statistical problem.

We have the SDE:

𝑑 𝑋_{𝑡} = 𝑓 (𝑋_{𝑡}, 𝜃) 𝑑 𝑡 + 𝑔 (𝑋_{𝑡}, 𝜃) 𝑑 𝑊_{𝑡}

Then the O-U model looks like

𝑓 (𝑥, 𝜃) = 𝜃_{1} (𝜃_{2} - 𝑥)

𝑔 (𝑥, 𝜃) = 𝜃_{3}

Then for CIR,

𝑓 (𝜃) = 𝜃_{1} (𝜃_{2} - 𝑥)

𝑔 (𝑥) = 𝜃_{3} \sqrt{𝑥}

I want to estimate parameters based on data.

We assume that we have observations

𝑥_{1}, 𝑥_{2}, \dots 𝑥_{𝑛}

at times

𝑡_{1}, \dots, 𝑡_{𝑛}

They could be same distance but it could be any times you want to observe this process.

So how do you estimate the vector of parameters.

In theory, you do this by Maximum Likelihood Estimation (MLE). There’s also method of moments, and Bayesian methods

Bayesian requires specific distributions, kind of complicated.

Method of moments is kind of like calibration to observed moments.

MLE is theoretically the most powerful one.

How does it work?

It’s really simple:

It works with the density of these observations.

If you’re observing the values of the process at $𝑡_{1}, \dots 𝑡_{𝑛}$

You can form the joint density: $𝑓 (𝑥_{1}, 𝑥_{2} \dots 𝑥_{𝑛} | 𝜃)$

This result depends on the vector of parameters, where θ is given.

The method of maximum likelihood says that I’m actually observing x_1, x_n. The probability of observing those things should be the highest in those numbers.

So let’s form a function

𝐿 (𝜃) = 𝑓 (𝑥_{1}, \dots 𝑥_{𝑛} | 𝜃)

In this expression, I’m going to plug in al lof the observations I know, and make this function of a function of the parameters, which I don’t know.

Typically you take the log which is called the score function.

log 𝐿 (𝜃) = 𝑙 (𝜃)

Computer scientists came up with it, they sell better than mathematicians.

If we know the joint density, then we can solve this problem. The problem is that we don’t know the joint density. It’s a really complicated expression.

How do we do this?

The first trick: we need

𝑓 (𝑥_{1}, \dots 𝑥_{𝑛} | 𝜃) = 𝑓 (𝑥_{𝑛} | 𝑥_{1} \dots 𝑥_{𝑛 - 1}, 𝜃) 𝑓 (𝑥_{1}, \dots 𝑥_{𝑛 - 1} | 𝜃)

Joint divided by marginal.

You can continue this up until the very first one.

= 𝑓 (𝑥_{𝑛} | 𝑥_{𝑛 - 1}, 𝑥_{𝑛 - 2} \dots 𝑥_{1}, 𝜃) 𝑓 (𝑥_{𝑛 - 1} | 𝑥_{𝑛 - 2} 𝑥_{𝑛 - 1} \dots 𝑥_{1} 𝜃) \dots 𝑓 (𝑥_{2} | 𝑥_{1}, 𝜃) 𝑓 (𝑥_{1} | 𝜃)

Remember there should be theta all over the place.

These are all functions on one variable.

These are all solutions to my stochastic process.

Since x solves an SDE, then x is Markov.

Which means that it’s the same as

𝑥_{𝑛} | 𝑥_{𝑛 - 1}, 𝜃

All I need is the previous value. So this is the generalized function.

𝑥_{𝑛 - 1} | 𝑥_{𝑛 - 2}, 𝜃

etc.

So then the likelihood function is the product

𝐿 (𝜃) = \prod_{𝑖 = 2}^{𝑛} 𝑓 (𝑥_{𝑖} | 𝑥_{𝑖 - 1}, 𝜃) 𝑓 (𝑥_{1} | 𝜃)

This is actually not the same function. There’s another thing to be aware of. This is a homogeneous process.

Distribution does not depend on the time you are collecting it, only depends on the difference of the times.

If you’re looking at a non-homogeneous process, the distribution of today, tomorrow, and the next day are not the same as a year from today, next day from that, next day from that.

Stationary is always the same, homogeneous

A non homogeneous process would have a distribution of $𝑋_{1}, 𝑋_{2}, 𝑋_{3}$ vs $𝑋_{100}, 𝑋_{200}, 𝑋_{300}$ , this is always going to be very different.

$𝑋_{101}, 𝑋_{201}, 𝑋_{301}$ . For a non homogeneous process, these are also different. But for a homogeneous process, this will be the same as long as the time interval is the same, ANY TIME INTERVAL.

By the way, this is a joint distribution.

So let’s write this distribution,

If you’re modeling returns, you get

𝑅_{𝑡} = log 𝑆_{𝑡}

and

𝑅_{𝑡 + Δ 𝑡} - 𝑅_{𝑡} = (𝜇 - \frac{𝜎^{2}}{2}) Δ 𝑡 + 𝜎 Δ 𝑊_{𝑡}

And

log 𝑆_{𝑡 + Δ 𝑡} = log 𝑆_{𝑡} + (𝜇 - \frac{𝜎^{2}}{2}) Δ 𝑡 + 𝜎 Δ 𝑊_{𝑡}

The only randomness comes from Brownian motion, which has N(0, Δt)

I know the distribution of this thing, so

log 𝑆_{𝑡 + Δ 𝑡} | 𝑆_{𝑡} \sim 𝑁 (log 𝑆_{𝑡} + (𝜇 - \frac{𝜎^{2}}{2} Δ 𝑡, 𝜎^{2} Δ 𝑡)

This is a constant that I add to it, so the only thing that happens is the mean changes when I add to the normal.

Now that I know this, I can write down the density.

obviously, the logarithm of this thing is normal.

Which means S is e to this thing.

It’s just easier to write in the context of the logarithm

𝑓 (log 𝑆_{𝑡 + Δ 𝑡} | log 𝑆_{𝑡}, 𝜃) = \frac{1}{2 𝜋 𝜎^{2} Δ 𝑡} 𝑒^{- \frac{𝑋 - log 𝑆_{𝑡} - (𝜇 - \frac{𝜎^{2}}{2}) Δ 𝑡)^{2}}{2 𝜎^{2} Δ 𝑡}}

This is just the PDE of the normal

And notice that there’s no t, just Δt. So the only thing that matters is S_t and Δt.

That’s what homogeneous means.

What we know is:

The Feller Process

This is any Ito process, any SDE, which is written like

𝑑 𝑋_{𝑡} = 𝑓 (𝑋_{𝑡}, 𝜃) 𝑑 𝑡 + 𝑔 (𝑋_{𝑡}, 𝜃) 𝑑 𝑊_{𝑡}

Any process where’s the coefficients here don’t depend on time, only the stochastic process and the parameter.

The big deal is, any Feller process is homogeneous.

It turns out that as long as the coefficients don’t depend on time, the solutions don’t depend on time.

I make all these parentheses because it’s kind of crucial for us.

It’s not enough for us to have an equal distance between times, you need to have a homogeneous process. If you have daily, then they’re different, 100 different functions for 100 different days. Now I just have one function, which I plug in at different points in time.

The following is NON-homogeneous.

𝑑 𝑋_{𝑡} = 𝑓 (𝑋_{𝑡}, 𝑡, 𝜃) 𝑑 𝑡 + 𝑔 (𝑋_{𝑡}, 𝑡, 𝜃) 𝑑 𝑊_{𝑡}

I had a friend that was Russian and pronounce it gomogeneous.

That normal density is the derivation as well.

If we come back to it, it’s the same as

𝑝 (𝑦, 𝑥, Δ 𝑡 | 𝜃)

This is the transition probability from $𝑋 = log 𝑆_{𝑡}$ to $𝑌 = log 𝑆_{𝑡 + Δ 𝑡}$

Notice I don’t care what Δt is, it could be a different one.

= \frac{1}{\sqrt{2 𝜋 𝜎 Δ 𝑡}} 𝑒^{- \frac{(𝑦 - 𝑥 - 𝜈 Δ 𝑡)^{2}}{not done}}

This does work for GBM

We can write down explicitly what it is.

This function si solved, I write down the transition, I write the likelihood function, I take the logarithm.

The joint distribution is a product. The logarithm is a sum of these functions, it’s just easier to deal with sums.

In general, for other models than GBM, it is impossible to write an exact formula.

So what do you do?

A guy from Princeton in (1999, 2000) named Ait Sahalia got famous for doing this.

He came up with the idea of Pseudo MLE, also the Approximate Likelihood Method.

There is a package that does this, which does not quote this.

https://www.princeton.edu/~yacine/mle.pdf

Reminder: there is a transition from X to Y in times Δt which depends on θ.

We will replace p(y, x, Δt | θ) which we call pθ with a density hθ which depends maybe on some other parameters.

We choose hθ in the following way:

has to be simple

a constant is too simple, so it has to be related to this pθ, so:

hθ has to have the same moments as pθ

This is a transition distribution in Y.

The moments mean expected value of powers. We have this mgf. We know if two variables have same mgf, they have the same distribution. You can have two variables with several moments equal, but not the same distribution. This property only holds for ALL moments.

So this wouldn’t be possible. I’m going to pick some hθ so we have a few moments the same, maybe 1, 2, 3.

The method in Ait Sahalia, we will discretize SDE using Euler method.

We will approximate the integral with the LHS value times the increment, same as quadrature.

We will cover Monte Carlo next week.

𝑋_{𝑡 + Δ 𝑡} - 𝑋_{𝑡} = \int_{𝑡}^{𝑡 + Δ 𝑡} 𝑓 (𝑋_{𝑡}) 𝑑 𝑡 + \int_{𝑡}^{𝑡 + Δ 𝑡} 𝑔 (𝑋_{𝑡}) 𝑑 𝑊_{𝑡}

We’re skipping θ for laziness, but exists in every function.

This is the same as

= 𝑓 (𝑋_{𝑡}) Δ 𝑡 + 𝑔 (𝑋_{𝑡}) Δ 𝑊_{𝑡}

if we approximate using quadrature.

We know the distribution of Brownian motion as N(0, Δt)

Euler method says, that given X_t, I know the distribution!

𝑋_{𝑡 + Δ 𝑡} | 𝑋_{𝑡} = 𝑋_{𝑡} + 𝑓 (𝑋_{𝑡}) Δ 𝑡 + 𝑔 (𝑋_{𝑡}) Δ 𝑊_{𝑡}

And its distribution is

𝑋_{𝑡 + Δ 𝑡} | 𝑋_{𝑡} \sim 𝑁 (𝑋_{𝑡} + 𝑓 (𝑋_{𝑡}) Δ 𝑡, 𝑔^{2} (𝑋_{𝑡}) Δ 𝑡)

This is not the distribution, because we have an approximation on the integral with quadrature.

Then we can approximate pθ as

ℎ (Δ 𝑡, 𝑥, 𝑦) = \frac{1}{\sqrt{2 𝜋 𝑔^{2} (𝑥) Δ 𝑡}} 𝑒^{- \frac{- (𝑦 - 𝑥 - 𝑓 (𝑥) Δ 𝑡)^{2}}{2 𝑔^{2} (𝑥) Δ 𝑡}}

And if you give me f and g, and the parameters, I can technically write this down, very easily.

It is simply multiplied like before. The x and y will be replaced by observations.

How do you estimate analytically. You take the logarithm, which takes everything out and simplifies it.

You get a bunch of the gs and the fs squared, but it’s just sums. If this would be linear, it would be simply applying an average. But nobody said this is linear. f and g can be complicated, so you would solve them with a nonlinear optimizer.

Sim Diff Procedure

First, let’s see two things.

The O-U model: How do you solve it? Because it is solvable.

Reminder:

𝑑 𝑋_{𝑡} = 𝛼 (𝜇 - 𝑋_{𝑡}) 𝑑 𝑡 + 𝜎 𝑑 𝑊_{𝑡}

This is just notation for the stochastic integral, blah blah.

The thing about this process, I can write it like

𝑑 𝑋_{𝑡} + 𝛼 𝑋_{𝑡} 𝑑 𝑡 = 𝛼 𝜇 𝑑 𝑡 + 𝜎 𝑑 𝑊_{𝑡}

By the way, if there is no mean reverting here, it’s just called O-U. With mean reverting, it’s called mean reverting O-U. The alpha mu dt thing is gone in that case.

This is the real trick, which comes from differential equations.

If you look at the LHS, you should remember something. You remember integrating factors, the way to solve first order diffeqs.

And in fact that is a first order.

What do you do with integrating factor? I don’t remember, I remember it existed, but it’s complicated, with some weird formula.

But I do remember that if I take the derivative of $𝑒^{𝛼 𝑡}$ , then I’ll get $𝛼 𝑒^{𝛼 𝑡}$

That’s a powerful feature, to make a constant appear out of nowhere.

And we can also see another constant that appeared out of nowhere. So if we multiply everything here by that, you kind of get the derivative of the product to appear.

𝑒^{𝛼 𝑡} 𝑑 𝑋_{𝑡} + 𝑋_{𝑡} 𝛼 𝑒^{𝛼 𝑡} 𝑑 𝑡 = 𝜇 𝛼 𝑒^{𝛼 𝑡} 𝑑 𝑡 + 𝜎 𝑒^{𝛼 𝑡} 𝑑 𝑊_{𝑡}

Now this is a stochastics process, so it’s a little different from the regular product rule.

This process becomes

𝑑 (𝑒^{𝛼 𝑡} 𝑋_{𝑡})

I claim that. But how do you show this? What is the differential of two stochastic processes. That’s the Ito product rule. Which is the regular product rule plus the quadratic variation between the two terms.

When you take the quadratic variation between stochastic process and a deterministic process, that ends up being zero. So actually we can get rid of the quadratic variation part, so it ends up being the same as the regular product rule.

= 𝜇 𝑑 (𝑒^{𝛼 𝑡}) + 𝜎 𝑒^{𝛼 𝑡} 𝑑 𝑊_{𝑡}

The next step is to integrate, from 0 to t.

𝑒^{𝛼 𝑡} 𝑋_{𝑡} - 𝑋_{0} = 𝜇 (𝑒^{𝛼 𝑡} - 1) + \int_{0}^{𝑡} 𝜎 𝑒^{𝛼 𝑠} 𝑑 𝑊_{𝑠}

Notice this is an explicit solution.

𝑋_{𝑡} = 𝑒^{- 𝛼 𝑡} 𝑋_{0} + 𝜇 (1 - 𝑒^{- 𝛼 𝑡}) + \int_{0}^{𝑡} 𝜎 𝑒^{𝛼 (𝑠 - 𝑡)} 𝑑 𝑊_{𝑠}

Just rearranging to get this here. We can’t really touch the dW stuff

And this is random, because it has a stochastic integral.

Another trick.

Even for this simple expression, it is complicated to measure μ, α, or σ, because the stochastic integral is present. You can do the numerical thing and hope for the best.

I learned this trick when I was in Romania. If you have this equation

𝑋_{𝑡} - 𝑋_{0} = \int_{0}^{𝑡} 𝛼 (𝜇 - 𝑋_{𝑠}) 𝑑 𝑠 + \int_{0}^{𝑡} 𝑔 (𝑋_{𝑠}) 𝑑 𝑊_{𝑠}

I have some horrible expression like this, which I write as g, as complicated as you want.

So what is the expected value of this? I get scared, because it’s so ugly.

Because I have to know the pdf, and integrate x * pdf, which is horrible.

But actually, there’s a simple way to do this.

It’s all based on the fact that stochastic integrals are martingales. Being martingales, they have the same expectation at any moment in time. And the process is equal to 0 at 0, so that whole thing ends up becoming 0. The trick is to apply expectation everywhere.

𝔼 [𝑋_{𝑡}] - 𝔼 [𝑋_{0}] = 𝔼 [\int_{0}^{𝑡} 𝛼 (𝜇 - 𝑋_{𝑠}) 𝑑 𝑠] + 0

The next step is to realize is that this is two integrals, which will commute as long as the thing inside is finite. And it has to be finite.

= \int_{0}^{𝑡} 𝛼 (𝜇 - 𝔼 [𝑋_{𝑠}]) 𝑑 𝑠

And now this is an expectation, which is a number, it just depends on the time. We’ll define a function for this.

𝔼 [𝑋_{𝑡}] = 𝑢 (𝑡)

So we’re really solving this equation:

𝑢 (𝑡) - 𝑢 (0) = \int_{0}^{𝑡} 𝛼 (𝜇 - 𝑢 (𝑠)) 𝑑 𝑠

But this is now a deterministic equation which you can solve as diffeq using Calculus III.

You take the derivatives, which makes it simpler, because it mimics what we just did.

𝑑 𝑢 (𝑡) = 𝛼 (𝜇 - 𝑢 (𝑡)) 𝑑 𝑡

This the same equation we did a moment ago

𝑑 𝑢 (𝑡) + 𝛼 𝑢 (𝑡) 𝑑 𝑡 = 𝛼 𝜇 𝑑 𝑡

And this is a first order diffeq, where we can use the same e thing.

𝑑 (𝑒^{𝛼 𝑡} 𝑢 (𝑡)) = 𝛼 𝑢 𝑒^{𝛼 𝑡} 𝑑 𝑡

And now RHS integrates really easily, same as before

𝑒^{𝛼 𝑡} 𝑢 (𝑡) - 𝑢 (0) = 𝜇 (𝑒^{𝛼 𝑡} - 1)

𝑢 (𝑡) = 𝑢 (0) 𝑒^{- 𝛼 𝑡} + 𝜇 (1 - 𝑒^{- 𝛼 𝑡})

And actually we didn’t need to do all of this, because on the previous slide, we had the solution.

If we applied expectation directly to the Ito product rule we did, it would give the same formula, which is just more generalized.

𝔼 [𝑋_{𝑡}] = 𝔼 [𝑋_{0}] 𝑒^{- 𝛼 𝑡} + 𝜇 (1 - 𝑒^{- 𝛼 𝑡})

And you can observe the long term behavior, that it goes to μ.

That’s the theory. How do we estimate the stuff?

Code

I wrote this before coming to class, which is why I was late. This is part of a package called Sim.DiffProc, which you need to install. For some nonstandard packages, you need Rtools, looks like you don’t need it.

Now they have a very nice

Sim.DiffProc is not that good.

Simulation of Diffusion Processes.

They have a nice documentation for it.

They can do more things than this.

Here:

What You Should Remember

MLE Method

Approximation of Joint distribution with little pieces

Feller processes and what they are

This is stuff that actually shows your advantage over students from competing programs. They don’t do anything like this.

## CDS and Credit Markets - URL: https://sharifhsn.dev/blog/cds-and-credit-markets/ - Structured data: https://sharifhsn.dev/api/posts/cds-and-credit-markets.json - Description: The syllabus records “Introduction to Credit: Credit Default Swaps; relationship to Bonds; Credit as a traded asset class,” taught by Yee Han. The personal course export contains… - Date: 2025-03-31 - Exact published timestamp: 2025-03-31 - Topics: FX, Credit Risk, CDS, Credit Markets - Categories: FX - Source: FE-635 \| Risk Engineering - Source URL: None - Content source format: typst

Week 9 — 31 March 2025

The syllabus records “Introduction to Credit: Credit Default Swaps; relationship to Bonds; Credit as a traded asset class,” taught by Yee Han.

Source boundary

The personal course export contains this topic on the syllabus, but no worked CDS or bond-credit lecture notes. The earlier prose about premium and protection legs was a generated explanation, not a transcription of the available notes, so it has been removed. The class topic remains documented without inventing details.

## The Cost of AI-Generated Images - URL: https://sharifhsn.dev/blog/linkedin-2025-03-28-the-cost-of-ai-generated-images/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2025-03-28-the-cost-of-ai-generated-images.json - Description: New AI images stun the world 🤯 but are they worth the cost? 🤔 AI images have flooded social media seemingly overnight, with the specific trend of “Ghiblization” (transforming… - Date: 2025-03-28 - Exact published timestamp: 2025-03-28T13:00:08.066Z - Topics: AI, Creative Work - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/feed/update/urn:li:activity:7311371515738898434/ - Content source format: typst

New AI images stun the world 🤯 but are they worth the cost? 🤔

AI images have flooded social media seemingly overnight, with the specific trend of “Ghiblization” (transforming images into the style of animation studio Studio Ghibli) becoming a popular meme. These images from ChatGPT 4o are much higher quality than previous AI images.

That’s not a coincidence. 4o uses a different method of image generation than the old model, DALL-E. DALL-E was based around diffusion.

This method models the noising of an image through a stochastic process, which has a closed-form sampling method. Then, it uses Anderson’s reverse diffusion to denoise. The score function is estimated by the UNet neural network which is trained on real images that are noised.

The denoising process starts with pure Gaussian noise, then denoises at each step. It uses the attention mechanism just like LLMs do to recognize when a patch of noise looks like something in the prompt, and denoises it to fit the desired image. After only a few dozen steps, the image is denoised.

This process has benefits and drawbacks. It’s very computationally efficient, as it renders the whole image at once in a parallelizable fashion. However, this independence between batches causes a lack of spatial reasoning, which is why diffusion images often have messed up hands or text.

4o uses autoregression instead. You can think of this as similar to what ChatGPT does, where each part of the image is generated based on previous parts of the image, left to right and top to bottom.

This takes care of the spatial reasoning element, since each part of the image develops from earlier parts. But it requires a full forward pass through the transformer stack for each new image token, attending to all previous tokens. This makes the time complexity of the process O(n^3), with no opportunity for parallelization!

This is why access to 4o’s image generation capabilities are being limited, and Sam Altman has said “our GPUs are melting”. Much hay has been made about the energy consumption of AI, and I generally think that these worries are overblown compared to other uses of energy. But autoregressive image generation seems to push that limit hard.

What do you think? Are the higher quality images worth it?

## Hazard Rates and Credit Default Swaps - URL: https://sharifhsn.dev/blog/advanced-derivatives-week-09/ - Structured data: https://sharifhsn.dev/api/posts/advanced-derivatives-week-09.json - Description: Λ , the hazard rate, is convenient to work with. If we integrate over it, we can get the survival probability 𝑉 ( 𝑡 ) at time 𝑡 . We can also get the probability of default 𝑄 ( 𝑡… - Date: 2025-03-27 - Exact published timestamp: 2025-03-27 - Topics: Credit, Hazard Rates, Recovery Rates, Credit Default Swaps - Categories: Credit - Source: Advanced Derivatives - Source URL: None - Content source format: typst

Hazard Rate

$Λ$ , the hazard rate, is convenient to work with.

If we integrate over it, we can get the survival probability $𝑉 (𝑡)$ at time $𝑡$ .

We can also get the probability of default $𝑄 (𝑡)$ by doing $1 -$ survival probability.

Recovery Rate

If we have default, we will lose all future coupons/payoffs.

Typically we can recover something from the bond, depending on the structure of the debt. Bonds get paid before equity.

The recovery rate is defined as the price of the bond immediately after default as a percentage of FV.

Recovery rates DECREASE as default rates INCREASE. Recovery is not known in advance but it is estimated in a certain range based on historical data.

We can get the implied probability of default for a bond using some qualities.

the bond price
CDS spreads (will discuss)

CDS is basically an insurance instrument

You can also get historical data to construct hazard rates

Merton’s model also (will discuss)

Bond Prices

Obviously, there is a relationship between default and bond price. We can approximate default intensity over life of bond as

\frac{𝑠}{1 - 𝑅}

where $𝑠$ is the spread of the bond’s yield over the risk-free rate and $𝑅$ is the recovery rate.

We can look at a little more exact calculation.

Lecture 7, Part I, slide 13 (PDF p. 13) states the assumptions directly: a five-year corporate bond pays a 6% annual coupon semiannually; its continuously compounded yield is 7%, while a similar risk-free bond yields 5%. The risk-free bond is priced at 104.09 and the corporate bond at 95.34, giving an expected default loss of 8.75. The slide assumes a default probability of $𝑄$ per year and defaults halfway through each year, immediately before a coupon payment.

Slide 14 (PDF p. 14) evaluates a $40 recovery amount for each possible default time. It lists the risk-free value, loss given default, discount factor, and present value of expected loss:

Time (years)	Default probability	Recovery amount	Risk-free value	Loss given default	Discount factor	PV of expected loss
0.5	Q	40	106.73	66.73	0.9753	65.08 Q
1.5	Q	40	105.97	65.97	0.9277	61.20 Q
2.5	Q	40	105.17	65.17	0.8825	57.52 Q
3.5	Q	40	104.34	64.34	0.8395	54.01 Q
4.5	Q	40	103.46	63.46	0.7985	50.67 Q
Total						288.48 Q

Slide 15 (PDF p. 15) sets $288.48 𝑄 = 8.75$ and obtains $𝑄 = 3.03 %$ . It then notes that the analysis can allow defaults to occur more frequently and that several bonds provide more parameters for describing the default-probability distribution.

When we do pricing of a CDS, we will look more detail in this issue, where if defaults can happen at any time, we integrate over the domain.

The goal is to calculate such probability of default.

We have a coupon payment of $3. So the YTM is

You can estimate risk-neutral default rate each year.

It’s a bootstrap process. You start with lower maturity bonds, then get higher maturity bonds

Credit Default Swaps

Excess of n-bond yields of corporate bonds must equal CDS spread, otherwise there is arbitrage where you can either earn over the risk-free rate or borrow at less than the risk-free rate.

The CDS bond basis is the spread minus the excess bond yield.

From the arbitrage argument, the bond basis should be 0.

Historically, CDS bond basis $> 0$ .

One leg is the protection buyer, the premium leg,

and the other is the

When you value the premium leg,

The other element is the survival probability.

The second component is that in case of default, you have to pay the accrual

## Midterm Cheat Sheet - URL: https://sharifhsn.dev/blog/computational-methods-midterm-cheat-sheet/ - Structured data: https://sharifhsn.dev/api/posts/computational-methods-midterm-cheat-sheet.json - Description: Handwritten midterm reference for quadrature and Black-Scholes notation; several formulas are unclear in the source render. - Date: 2025-03-25 - Exact published timestamp: 2025-03-25 - Topics: Computational Methods, Midterm, Quadrature, Numerical Integration, Midpoint Rule, Simpson's Rule, Black-Scholes, Partial Differential Equations, Free Boundary - Categories: Computational Methods - Source: Handwritten CheatSheet.xopp; Drive-modified 2025-03-29. Approximate date within the Computational Methods in Quantitative Finance Docs history cluster (2025-01-21 through 2025-05-15). - Source URL: None - Content source format: typst

The handwritten page is headed “Computational Methods in Quantitative Finance Midterm Cheat Sheet.”

Quadrature / numerical integration

Readable labels: “Quadrature,” “numerical integration,” “Midpoint Rule,” and “Simpson’s Rule.” The line beginning “for Taylor:” and the formulas under the rule labels are unclear in the source.

Black-Scholes

The page marks “PDE” and “Free Boundary.” The associated PDE, Greek symbols, and boundary equations are unclear in the source.

The finite-difference grid and method derivations are covered in weeks 5 and 6. This supplement keeps the readable quadrature and section labels without repeating those derivations.

## GARCH Risk Statistics - URL: https://sharifhsn.dev/blog/garch-risk-statistics/ - Structured data: https://sharifhsn.dev/api/posts/garch-risk-statistics.json - Description: The syllabus records “Risk Management: GARCH — Statistics.” The personal course export contains the schedule and the workbook implementation notes for the Black–Scholes/FX… - Date: 2025-03-24 - Exact published timestamp: 2025-03-24 - Topics: FX, Risk Engineering, GARCH, Volatility Estimation - Categories: FX - Source: FE-635 \| Risk Engineering - Source URL: None - Content source format: typst

Week 8 — 24 March 2025

The syllabus records “Risk Management: GARCH — Statistics.” The personal course export contains the schedule and the workbook implementation notes for the Black–Scholes/FX material, but it does not contain a worked GARCH lecture or recurrence.

Source boundary

This entry is deliberately a source pointer: the GARCH lecture is listed in the syllabus, but the detailed class notes are not present in the exported document. I have removed the former generated GARCH explanation rather than attributing textbook material to the class.

## LIBOR Market Model and Hazard Rates - URL: https://sharifhsn.dev/blog/advanced-derivatives-week-08/ - Structured data: https://sharifhsn.dev/api/posts/advanced-derivatives-week-08.json - Description: The LIBOR Market Model (or Brace-Gatarek-Musiela Model ) is a model constructed in terms of the forward rates underlying caplet prices. HJM was defined as a process for the… - Date: 2025-03-20 - Exact published timestamp: 2025-03-20 - Topics: Fixed Income, LIBOR Market Model, Interest Rate Models, Hazard Rates - Categories: Fixed Income - Source: Advanced Derivatives - Source URL: None - Content source format: typst

LMM

The LIBOR Market Model (or Brace-Gatarek-Musiela Model)

is a model constructed in terms of the forward rates underlying caplet prices.

HJM was defined as a process for the forwards.

Basically the benefit was that it could model the entire term structure, as long as you input all the volatilities for the term structure.

But the problem is the calibration.

Notation

$𝑡_{𝑘}$ kth reset date

So $𝑡_{0} = 0$ , $𝑡_{1} = 0.25$ , $𝑡_{2} = 0.5$ , …

$𝐹_{𝑘}$ forward rate between $𝑘$ and $𝑘 + 1$

$𝑚 (𝑡)$ index for next reset date at time $𝑡$ .

So if $𝑡 = 0.4$ , then $𝑚 (𝑡) = 0.5$ .

$𝛿_{𝑘} = 𝑡_{𝑘 + 1} - 𝑡_{𝑘}$

So basically you can imagine that you have your forward curve,

$𝜉$ : $𝜉 (𝑡)$ = volatility of $𝐹_{𝑘} (𝑡)$ at time $𝑡$

Assume that we have only one factor

This factor is the forward risk neutral process with respect to $𝑃 (𝑡, 𝑡_{𝑘 + 1})$

Then, our process

𝑑 𝐹_{𝑘} (𝑡) = 𝜉_{𝑘} (𝑡) 𝐹_{𝑘} (𝑡) 𝑑 𝑊

The change is driven by the volatility, existing forward rate, and Brownian motion.

Applying some change of numeraire, skipping some steps, you get

the rolling forward rate process (in notes)

Price of a ZCB at time t_i which is discounted, the ratio of that to t_i+1

\frac{𝑃 (𝑡, 𝑡_{𝑖})}{𝑃 (𝑡, 𝑡_{𝑖 + 1})} = 1 + 𝛿_{𝑖} 𝐹_{𝑖} (𝑡)

If you apply the natural log, you get

ln 𝑃 (𝑡, 𝑡_{𝑖}) - ln 𝑃 (𝑡, 𝑡_{𝑖 + 1}) = ln [1 + 𝛿_{𝑖} 𝐹_{𝑖} (𝑡)]

This is the relationship between the ZCB and the forward.

Then if you apply Ito’s lemma… and equate coefficient of dW

something long…

Then you can take the substitution of the forward process.

\frac{𝑑 𝐹_{𝑘} (𝑡)}{𝐹_{𝑘} (𝑡)} = \sum_{𝑖 = 𝑚 (𝑡)}^{𝑘} \frac{𝛿_{𝑖} 𝐹_{𝑖} (𝑡) 𝜉_{𝑖} (𝑡) 𝜉_{𝑘} (𝑡)}{1 + 𝛿_{𝑖} 𝐹_{𝑖} (𝑡)} 𝑑 𝑡 + 𝜉_{𝑘} (𝑡) 𝑑 𝑊

This is the process followed by the forward rate between $𝑡_{𝑘}$ and $𝑡_{𝑘 + 1}$ , in a risk-neutral world.

In the limiting case, when this interval becomes smaller, this converges to HJM

The idea is being able to calibrate this model.

It can be simplified with:

Simplified Model

Assume that $𝜉_{𝑘} (𝑡)$ functions only on the number of whole accrued periods between the next date and $𝑡_{𝑘}$ .

Define $Λ_{𝑖}$ as the value of $𝜉_{𝑘} (𝑡)$ (volatility of the forward) when there are $𝑖$ such accrued periods.

Then the volatility of the forward can be redefined as

$𝜉_{𝑘} (𝑡) = Λ_{𝑘 - 𝑚 (𝑡)}$ which is a step function

And such values like $Λ_{𝑖}$ can be estimated from the volatilities used to value caplets in Black’s model.

Recall that to value a caplet, we have

𝐿 𝛿_{𝑘} 𝑃 (0, 𝑡_{𝑘 + 1}) [𝐹_{𝑘} 𝑁 (𝑑_{1}) - 𝑅_{𝑘} 𝑁 (𝑑_{2})]

blah blah blah $𝑑_{1}$ and $𝑑_{2}$

If we equate the variances between Black’s model and this, we get

𝜎_{𝑘}^{2} 𝑡_{𝑘} = \sum_{𝑖 = 1}^{𝑘} Λ_{𝑘 - 𝑖}^{2} 𝛿_{𝑖 - 1}

Lecture 5, slide 11 (PDF p. 11), gives the calibration example in its own words: “If Black volatilities for the first three caplets are 24%, 22%, and 20%, then $Λ_{0} = 24.00 %$ , $Λ_{1} = 19.80 %$ , and $Λ_{2} = 15.23 %$ .”

Implementation

How would I use this model to price a bond. If you observe the volatility, then you can price a bond under this?

This would be done via Monte Carlo simulation

\frac{𝑑 𝐹_{𝑘} (𝑡)}{𝐹_{𝑘} (𝑡)} = \sum_{𝑖 = 𝑚 (𝑡)}^{𝑘} \frac{𝛿_{𝑖} 𝐹_{𝑖} (𝑡) Λ_{𝑖 - 𝑚 (𝑡)} Λ_{𝑘 - 𝑚 (𝑡)}}{1 + 𝛿_{𝑖} 𝐹_{𝑖} (𝑡)} 𝑑 𝑡 + Λ_{𝑘 - 𝑚 (𝑡)} 𝑑 𝑊

Then, by Ito’s lemma,

𝑑 ln 𝐹_{𝑘} (𝑡) = [\sum_{𝑖 = 𝑚 (𝑡)}^{𝑘} \frac{𝛿_{𝑖} 𝐹_{𝑖} (𝑡) Λ_{𝑖 - 𝑚 (𝑡)} Λ_{𝑘 - 𝑚 (𝑡)}}{1 + 𝛿_{𝑖} 𝐹_{𝑖} (𝑡)}] 𝑑 𝑡 + same thing as above

Then we can approximate the drift by approximating $𝐹_{𝑖} (𝑡)$ and $𝑡$

Then,

𝐹_{𝑘 (𝑡_{𝑗 + 1})} = 𝐹_{𝑘 (𝑡)} exp ((\sum_{𝑖 = 𝑗 + 1}^{𝑘} \frac{𝛿_{𝑖} 𝐹_{𝑖 (𝑡_{𝑗})} Λ_{𝑖 - 𝑗 - 1}}{1 + 𝛿_{𝑖} 𝐹_{𝑖 (𝑡_{𝑗})}} - \frac{Λ_{𝑘 - 𝑗 - 1}^{2}}{2}) 𝛿_{𝑗} + Λ_{𝑘 - 𝑗 - 1} 𝜖 \sqrt{𝑑_{𝑗}})

Lecture 5, slide 8 (PDF p. 8), defines $𝑡_{𝑘}$ as the kth reset date, $𝐹_{𝑘 (𝑡)}$ as the forward rate between $𝑡_{𝑘}$ and $𝑡_{𝑘 + 1}$ , $𝑚 (𝑡)$ as the index for the next reset date at time $𝑡$ , and $𝛿_{𝑘} = 𝑡_{𝑘 + 1} - 𝑡_{𝑘}$ .

ipynb file is provided

Each approximation of the drift the forward rate remains constant.

If we define the rolling forward in the risk neutral world, it allows us to discount bond from one date to the next one.

In terms of simulation, what’s happening in the code is that if you assume that we want to simulate a zero curve with N accrued periods.

On each trial, we start with the forward rate at time 0, which is calculated from the initial zero curve.

$𝐹_{0} (0), 𝐹_{1} (0), \dots, 𝐹_{𝑁 - 1} (0)$

Then we can use an approximation formula (described above) to calculate $𝐹_{1} (𝑡_{1})$ , $𝐹_{2} (𝑡_{1})$ …

Limitations of the theoretical model

Credit Risk

LMM is a little complicated, this is just an introduction to the topic.

Credit risk is the probability that a debtor will default on their debts. We have always assumption that the cash flow is 100% likely i.e. with treasury bonds.

For single-name derivatives, we will look at CDS and CDOs. Then we will look at portfolios. We are interested in expected loss, with a discrete number of names in the portfolio, to estimate the losses in order to price the portfolio correctly.

Rating agencies will rate the credit risk of bonds.

S&P says AAA, AA, A, BBB, BB, B, CCC, CC, C

Moody’s has Aaa, Aa, A, Baa, Ba, B, Caa, Ca and C

Bonds with ratings of BBB and above are investment grade

These are measured as bands of probabilities.

Question about Diversification

By diversifying your portfolio into multiple assets, you decrease the credit risk. Then you can securitize the portfolio by splitting it into tranches, and then get the CDO.

And then payment in the bonds

You can have nonlinear dependence between variables, and tail dependence between variables which is conditional. Then the probability of default, what is the likelihood of this house being on fire if this other house is on fire? If the answer is not 0, then you have tail dependence.

We will look at different models that measure this.

Hazard Rates

Also known as default intensity, this is the probability of default for a certain time period conditional on no earlier default.

Unconditional default probability is from 0.

We are given a table that has cumulative default rates.

In order to calculate the probability that A defaults in the first year, you take the actual probability. Then if it’s the second year, we’re concerned with the cdf(2) - cdf(1).

Then the probability of survival is 1 - this

This is all unconditional default probability.

Then the conditional probabilities are known as default intensities or hazard rates

The unconditional probability of default within a certain default vs the survival probability.

The survival probability is V(t)

𝜆 (𝑡) Δ 𝑡 = \frac{𝑉 (𝑡) - 𝑉 (𝑡 + Δ 𝑡)}{𝑉 (𝑡)}

Then you get the ODE

\frac{𝑑 𝑉 (𝑡)}{𝑑 𝑡} = - 𝜆 (𝑡) 𝑉 (𝑡)

𝑉 (𝑡) = 𝑒^{- \int_{0}^{𝑡} 𝜆 (𝑡) 𝑑 𝑡}

Then the survival probability can be found by integrating over the hazard rate. So if you have a specification of the hazard rate, you can find the cumulative survival probability.

Then we can also discuss the cumulative default probability.

We can use the CDS to determine the piecewise constant hazard rates and then construct the survival probabilities.

So this is a calibration in which we can price the risk of default through CDS.

First we will look at bonds.

## Stochastic Volatility, Jumps, and Fourier Transforms - URL: https://sharifhsn.dev/blog/computational-methods-week-09/ - Structured data: https://sharifhsn.dev/api/posts/computational-methods-week-09.json - Description: Theoretical transformations, very technical How do you solve a Heston model using Fourier transforms? The very first stochastic volatility model introduced is the Hull-White model. - Date: 2025-03-18 - Exact published timestamp: 2025-03-18 - Topics: Computational Methods, Stochastic Volatility, Jump Processes, Poisson Process, Fourier Transform - Categories: Computational Methods - Source: Computational Methods in Quantitative Finance - Source URL: None - Content source format: typst

Theoretical transformations, very technical

Stochastic Volatility Models

How do you solve a Heston model using Fourier transforms?

Hull-White

The very first stochastic volatility model introduced is the Hull-White model.

𝑑 𝑆_{𝑡} = 𝑟 𝑆_{𝑡} 𝑑 𝑡 + \sqrt{𝑦_{𝑡}} 𝑆_{𝑡} 𝑑 𝑊_{𝑡}

𝑑 𝑌_{𝑡} = 𝜇_{𝑌} 𝑌_{𝑡} 𝑑 𝑡 + 𝜎_{𝑌} 𝑌_{𝑡} 𝑑 𝑍_{𝑡}

Has no solution:(

If the Brownian motions are uncorrelated, then there is no leverage effect, because that effect is a correlation between volatility and returns. But just because the Brownian motions are uncorrelated, doesn’t mean the processes are uncorrelated, they incorporate each other.

Leverage Effect

The leverage effect is the perceived correlation between returns and volatility, and news and returns. The idea is that positive news causes the stock to go up, which makes return go up, and vice versa for negative news.

However, the two effects are not the same. If there is good news the stock goes up, if there’s bad news, the stock goes down much more proportionally. This is called the leverage effect, which is a negative correlation between volatility and returns. Basically, the news increases vol, and as vol increases, returns go down.

You need to have stochastic volatility to have correlation between stochastic and deterministic processes.

I’ll mention two more stochastic volatility models.

SABR

We discussed SABR earlier, so this is a reminder. It looks like

𝑑 𝑆_{𝑡} = 𝑟 𝑆_{𝑡} 𝑑 𝑡 + 𝜎_{𝑡} 𝑆_{𝑡}^{𝛽} 𝑑 𝑊_{𝑡}

This is the most general form of the process, most that you see in practice have $𝑟 = 0$ because they are created by physicists who don’t like complicated models.

The $𝜎$ is the stochastic process, which has the process

𝑑 𝜎_{𝑡} = 𝛼 𝜎_{𝑡} 𝑑 𝑍_{𝑡}

The $𝑊$ and $𝑍$ are two Brownian motions which can be correlated with $𝜌$ .

This is also called the stochastic $𝛼$ - $𝛽$ - $𝜌$ model (SABR) for the three parameters and the stochastic volatility.

The authors of this made it so that once you calculate $𝜎$ , you can plug it into Black-Scholes and reuse all your old code.

Constant elasticity of variance (CEV)

Pioneered by Peter Carr (friend of the show) and Madan.

𝑑 𝑆_{𝑡} = 𝑟 𝑆_{𝑡} 𝑑 𝑡 + 𝜎 𝑆_{𝑡}^{\frac{𝛽}{2}} 𝑑 𝑊_{𝑡}

If you think about this in terms of stochastic models. This $𝛽$ is strictly less than $2$ , and $𝜎$ is greater than $0$ . You can write this as

𝜎 𝑆_{𝑡} 𝑆_{𝑡}^{\frac{𝛽 - 2}{2}}

Which means that the last term is the actual stochastic volatility.

That’s literally driven by the price evolution. This is inversely proportional to the value of the stock.

If you Itô this with $log 𝑆_{𝑡}$ , you get

𝑑 𝑋_{𝑡} = (𝑟 - \frac{𝜎^{2} 𝑆_{𝑡}^{\frac{𝛽 - 2}{2}}}{2}) 𝑑 𝑡 + 𝜎 𝑆_{𝑡}^{\frac{𝛽 - 2}{2}} 𝑑 𝑊_{𝑡}

Then the second term is the vol.

So the variance should be the vol squared.

𝕍 = 𝜎^{2} 𝑆_{𝑡}^{𝛽 - 2}

If you compute the derivative the change in the variance with respect to stock.

\frac{𝜕 𝕍}{𝜕 𝑆} = 𝜎^{2} (𝛽 - 2) 𝑆_{𝑡}^{𝛽 - 3}

You can rewrite this as

𝜎^{2} 𝑆_{𝑡}^{𝛽 - 2} \frac{𝛽 - 2}{𝑆_{𝑡}}

The point of doing this is to see that these left two terms are the exact same as the variance.

\frac{𝜕 𝕍}{𝜕 𝑆} = 𝕍 \frac{𝛽 - 2}{𝑆_{𝑡}}

Which is the same thing as saying

\frac{𝜕 𝕍}{𝕍} = (𝛽 - 2) \frac{𝜕 𝑆}{𝑆}

The change in variance is proportional to the change of stock.

The variance move elastically, proportional to the way the stock moves.

And they change inversely, because $𝛽$ is less than $2$ .

And actually, if $𝛽 = 2$ , then you get GBM in the formula.

𝑑 𝑆_{𝑡} = 𝑟 𝑆_{𝑡} 𝑑 𝑡 + 𝜎 𝑆_{𝑡} 𝑑 𝑊_{𝑡}

Which makes sense because that assumes constant volatility, where

\frac{𝜕 𝕍}{𝕍} = 0

When $𝛽 = 1$ , you get CIR

𝑑 𝑆_{𝑡} = 𝑟 𝑆_{𝑡} 𝑑 𝑡 + 𝜎 \sqrt{𝑆_{𝑡}} 𝑑 𝑊_{𝑡}

Changes in the stock become actually 1:1 with the stock in this case.

Jump Processes

Lonon is an expert on jump processes.

Poisson Process

Basically,

𝑁 \in {0, 1, \dots}

where at time t,

𝑁_{𝑡} \sim 𝑃 𝑜 𝑖 𝑠 𝑠 𝑜 𝑛 (𝜆 𝑡)

$𝜆$ quantifies the expected number of values for when $𝑡 = 1$ .

That’s how you scale it.

The question is, how do you simulate this?

There are two ways to do this:

Probability and Stochastic Processes (Florescu) has a lot of information on this.

I suggest you pirate my book because the publishers are thieves.

Method 1:

𝑋_{1}, 𝑋_{2}, 𝑋_{3}, \dots, 𝑋_{𝑛}

are iid $Exp (1 / 𝜆)$

Basically the expected amount over lifetime is $1 / 𝜆$ .

Then we let

𝑇_{1} = 𝑋_{1}, 𝑇_{2} = 𝑋_{1} + 𝑋_{2}, 𝑇_{3} = 𝑋_{1} + 𝑋_{2} + 𝑋_{3}, \dots

What I’m doing here is defining the event times for the Poisson process.

At some point t, if you were to plot the process, then you would get a bunch of jumps up to t.

Then

𝑁_{𝑡} = max_{𝑛} {𝑇_{𝑛} \leq 𝑡}

The only question is if it’s included or not, so le’ts be careful.

Actually, it’s

𝑁_{𝑡} = inf_{𝑛} {𝑇_{𝑛} > 𝑡}

You can understand how to create this!

You simply generate these exponentials from the distributions, and you know the times from the sum.

You give me the process, and the time t.

The jump is of value 1, so it always jumps by 1.

There is a marked Poisson process or compound Poisson process, where instead of moving it by 1, you move it by a random variable, and then you sum those. Once you understand this it’s very simple, you just generate a series of random variables and assign them to each T.

How is this useful?

In two lectures, we will learn about Monte Carlo simulations. These typically don’t have jumps. But if you add jumps to them, you get a jump process.

You will get your $𝑇_{1}$ , $𝑇_{2}$ , and $𝑇_{3}$ , etc.

You will be going up and down by random quantities at those times.

These jumps:

Trump announces tariffs at each time T, and it goes up and down based on what country he tariffs.

There is some process which is not Jumpy, looks more like a Brownian motion. To introduce the jumps, you simply shift the value by the value of the random variable.

The exponential distribution starts at $1 / 𝜆$ at $𝑡 = 0$ and then goes down, for $Exp (1 / 𝜆)$ .

Most of the times, you get a small value.

Secret: if you play video games, and play whatever discrete events that happen in time. If you play a gacha game and put coins in there to get the good Pokemon. You keep getting a crappy Pokemon, and suddenly they give you a good one. But now you say it stops because you have it. The time between Pokemons is a random variable. If you generate the Poisson process for this, and if you look at the path, and you think about it logically, if it happens 5 per hour, the jumps should happen evenly. But this is not how it really looks. If you look at the pdf, you’re much more likely to have smaller intervals than larger intervals.

Method 2 is based on the following two results.

If we look at interval $[0, 𝑡)$ , the number of events is distributed as $Poisson (𝜆 𝑡)$ .

Let’s say you have events that happen once a day. Trump issues things once per day. (I’m a Republican so don’t pick on me)

If you want to generate how many things this guy says this week, you make a random variable with parameter $𝑡 = 7$ , $𝜆 = 1$ .

This Poisson random variable.

Given there are $𝑁$ events in the interval $[0, 𝑡)$ that I’m generating, the times of the events (and this is proven in the book), are the order statistic from $𝑁$ uniform $[0, 𝑡]$ random variables.

That sounds fancy, but basically it says…

An order statistic. If you have $𝑛$ random variables iid, and you take them $𝑋_{1}, 𝑋_{2}, \dots, 𝑋_{𝑛}$ ,

the order statistics ordered like

𝑋_{(1)} \leq 𝑋_{(2)} \leq \dots 𝑋_{(𝑛)}

and this is just an ordered list. So the first order statistic is the smallest number. The order statistics have a distribution which depends on the original distribution, and it’s actually quite simple to work with them.

Then generate random variable

𝑁 \sim Poisson (𝜆 𝑡)

Then generate random

𝑁 \sim Uniform [0, 𝑡]

Let’s say Poisosn happesnt ob e 10.

I generate a variable

Generate the uniform, look at the numbers, then list them smallest to largest.

Let’s say it’s 1.1. It took Trump 1.1 days to say the first stupid things. Then you have 3, so it took him another day to say something else.

These are the times, then the magnitudes come from them.

rpois(1, 7)

for example, gives you 6

N=rpois(1,7)

Now you generate seven uniforms

runif(N, 0, 7)

This gives you the times, which are unsorted.

Now you sort them

sort(runif(N,0, 7))

Poisson gives us the number of events, and uniform gives us the actual times of the events.

Transformation Methods

Laplace Transform

This is very familiar to probabilists. This is also called the moment generating function.

If you define X as a random variable with pdf f(x)

Then we define the mgf

𝑀_{𝑋} : ℝ \to [0, \infty)

𝑀_{𝑋} (𝑡) = 𝔼 [𝑒^{𝑡 𝑋}]

If X has a pdf, then this is also defined as

= \int_{- \infty}^{\infty} 𝑒^{𝑡 𝑥} 𝑓 (𝑥) 𝑑 𝑥

This was invented by Laplace, but he invented it in physics, for functions that were positive support, from 0 to infinity. So it’s a little different. The mgf is called so because if you take the derivative of the function with respect to t, you get the moments. (You need to prove that the derivative commutes with the integral, not that difficult).

𝑀^{'} (𝑡) = \frac{𝑑 𝑀_{𝑋} (𝑡)}{𝑑 𝑡} = \int_{- \infty}^{\infty} 𝑥 𝑒^{𝑡 𝑥} 𝑓 (𝑥) 𝑑 𝑥 = 𝔼 [𝑋 𝑒^{𝑡 𝑥}]

And then

𝑀^{'} (0) = 𝔼 [𝑋]

The P&SP book will cover this in more detail.

The Laplace transform of f: (0, infinity) to R, looks like

𝐿 𝑓 (𝑡) = \hat{𝑓} (𝑡) = \int_{0}^{\infty} 𝑒^{𝑡 𝑥} 𝑓 (𝑥) 𝑑 𝑥

The only difference is that it’s positive.

You can always write mgf as sum of two Laplace transforms.

The Laplace transform has this nice inversion theorem. Moving past all the details, you should remember it as:

If f has Laplace transform Lf, then

𝑓 (𝑥) = \frac{1}{2 𝜋 𝑖} lim_{𝑇 \to \infty} \int_{𝐶 - 𝑖 𝑇}^{𝐶 + 𝑖 𝑇} 𝑒^{𝑡 𝑥} 𝐿 𝑓 (𝑡) 𝑑 𝑡

This limit exists only for certain Lf(t), and even if it exists it’s ugly. So undergrads will look at the table of Laplace transforms. Table of Laplace Transforms

Stanford uses t and s, Florescu uses x and t.

This is useful because if you take the derivative of the function, and apply the Laplace transform, it becomes a polynomial. It becomes $𝑡 𝐹 (𝑡) - 𝑓 (0)$ .

If you take the nth derivative, you get a bunch of derivatives evaluated at 0. It makes it easier to solve. It’s very useful for solving diffeqs.

In practice, this thing has two problems. The specific doesn’t exist. With small exceptions, the equation is too complicated to get the value of the function that corresponds to it.

If you’re interested in applying this and want to work with Laplace transform, use Mathematica which Dragos buys for Stevens

Fourier Transform

There is an equivalent to this in probability, which is the characteristic function, which is more complicated than the transform.

What is the Fourier transform? For f(x):

𝑓 : [0, \infty) \to ℝ

𝐹 (𝑡) = \int_{- \infty, \infty} 𝑒^{- 𝑖 𝑡 𝑥} 𝑓 (𝑥) 𝑑 𝑥

The only difference is the introduction of the i thing. In general, you can apply Euler’s identity for

𝑒^{𝑖 𝑎} = cos 𝑎 + 𝑖 sin 𝑎

So this becomes

𝐹 (𝑡) = \int_{- \infty, \infty} cos (𝑡 𝑥) 𝑓 (𝑥) 𝑑 𝑥 - 𝑖 \int_{- \infty, \infty} sin (𝑡 𝑥) 𝑓 (𝑥) 𝑑 𝑥

You can calculate this using real integrals. Laplace is actually harder than this.

What is the connection with the characteristic function?

𝜑_{𝑥} (𝑡) = 𝔼 [𝑒^{𝑖 𝑡 𝑋}] = \int_{- \infty}^{\infty} 𝑒^{𝑖 𝑡 𝑥} 𝑓 (𝑥) 𝑑 𝑥

The only difference is that the characteristic doesn’t have a minus, which is not a big deal.

The minus doesn’t mean anything, really.

What is the connection between the characteristic and the moment? We can calculate moments from characteristic function.

𝜑_{𝑥}^{'} (𝑡) \frac{𝑑}{𝑑 𝑡} 𝜑_{𝑋} (𝑡)

You have to prove this works, and then take the complex function, which is not that hard.

= 𝔼 [(𝑖 𝑋) 𝑒^{𝑖 𝑡 𝑋}]

It’s kind of like you go inside and take the derivative as normal.

But if you take this

𝜑_{𝑥}^{'} (0) = 𝑖 𝔼 [𝑋]

This becomes slightly more complicated, because you get the powers

𝜑_{𝑥}^{″} (0) = 𝑖^{2} 𝔼 [𝑋^{2}]

And this continues in general.

What is the advantage, why do we do this Fourier transform and not stick to the Laplace transform?

This Fourier transform always exists. And there is also an inverse Fourier transform. It is basically the same idea. You get a diffeq and a simpler equation, and then the inverse gives you the solution.

There is something more that exists here! That is discrete Fourier transform. That is the big deal.

Generally speaking, you will run into the same problem. You get this horrible expression from the Fourier transform, and you can’t get the pdf from it. Engineers have invented this approximation. You express the original function in terms of cosines and sines, and then you know the transforms and inverse transforms from there.

Table of Fourier Transform Pairs

I had an argument a long time ago…

The whole point is, when you do a Fourier transform, you go into the frequency domain of your function. If your function is a sinus, you get one value. If you have a combination of sinuses, then you get a multitude of frequencies. The whole point of the DFT is that you express the function through the bases of sinuses.

Discrete Fourier transform

There’s math here, but nobody actually uses it.

What we do instead, is that we call a package and say this is my function, calculate the DFT, put it into an equation, solve it, calculate IFT, and then just do it that way.

Applications of Fourier Transform

Peter Carr was a guy at Bloomberg, and him and Ionut argued.

He said he was solving stochastic vol formula analytically, Ionut says this is not possible. Carr admits that it’s just very fast.

This is used to solve the Heston model.

𝑑 𝑆_{𝑡} = 𝑆_{𝑡} (𝑟 𝑑 𝑡 + \sqrt{𝑉_{𝑡}} 𝑑 𝑊_{𝑡})

𝑑 𝑉_{𝑡} = 𝐾 (𝜃 - 𝑉_{𝑡}) 𝑑 𝑡 + 𝜎 \sqrt{𝑉_{𝑡}} 𝑑 𝑍_{𝑡}

$𝑊$ and $𝑍$ can be correlated with $𝜌$ .

Original Heston model is uncorrelated, there is an extension by Wiggins which is correlated.

The solution is not that complicated.

There is also something called the Feller condition.

In order for this Heston model to be nicely behaved (homogeneous), you should have

2 𝐾 𝜃 > 𝜎^{2}

The problem is, given this model for my stochastic process, which is actually quite realistic, we want to find the price of an option.

t is time now

S is stock price

K is strike price

T is time of maturity

r is risk free rate

NEW: V is the value of this variance process

𝐶 (𝑡, 𝑆, 𝐾, 𝑇 - 𝑡, 𝑟, 𝑉)

If you consider the other points observable, then this is a function of C(t, S, V)

We used to solve things as

𝑡 \in [0, 𝑇]

𝑆 \in (0, 𝑖 𝑛 𝑓 𝑡 𝑦)

𝕍 \in (0, \infty)

Now we solve this equation with respect to three parameters.

Call option =

𝔼^{𝑄} [𝑒^{- 𝑟 (𝑇 - 𝑡)} (𝑆_{𝑇} - 𝐾)_{+} | {ℱ︀}_{𝑡}]

This is the general formula.

You can write this as, by splitting into two different situations, when in the money and out of the money

= 𝔼^{𝑄} [𝑒^{- 𝑟 (𝑇 - 𝑡)} (𝑆_{𝑇} - 𝐾)_{+} 𝕀_{{𝑆_{𝑇} > 𝐾}}

Where we disregard the out of the money part because it’s worthless.

= 𝔼^{𝑄} [𝑒^{- 𝑟 (𝑇 - 𝑡)} 𝑆_{𝑇} 𝕀_{{𝑆_{𝑇} - 𝐾}} | {ℱ︀}_{𝑡}] - 𝐾 𝔼^{𝑄} [𝑒^{- 𝑟 (𝑇 - 𝑡)} 𝕀_{{𝑆_{𝑇} > 𝐾}}]

Then we can take out the e term because it’s just a number,

= 𝑒^{- 𝑟 (𝑇 - 𝑡)} 𝔼 [𝑆_{𝑇} 𝕀_{{𝑆_{𝑇} - 𝐾}} | {ℱ︀}_{𝑡}] + 𝐾 𝑒^{- 𝑟 (𝑇 - 𝑡)} \dots

Then this first term will be considered $𝑃_{1} (𝑡, 𝑆, 𝑉)$ , and the second term is $𝑃_{2} (𝑡, 𝑆, 𝑉)$ .

What he said is if you notice the very first property of the Fourier transform is that it is linear.

So both $𝑃_{1}$ and $𝑃_{2}$ must solve the Heston PDE.

Then let’s apply the Fourier transform to the original Heston PDE.

This is more complicated than just going from t to x as before, because we have three variables. So we do:

\hat{𝑓} (𝜙, 𝑥, 𝑣)

So he postulated that

\hat{𝑓} (𝜙, 𝑥, 𝑣) = 𝑒^{𝐶 (𝜏, 𝜙) + 𝐷 (𝜏, 𝜙) 𝑣 + 𝑖 𝜙 𝑥}

This has some theoretical reasoning, but he says that once you plug it in, there is a solution.

Next Week

Estimating parameters, optimizations.

## Short-Rate Calibration and HJM - URL: https://sharifhsn.dev/blog/advanced-derivatives-week-07/ - Structured data: https://sharifhsn.dev/api/posts/advanced-derivatives-week-07.json - Description: We’ll look specifically at Vasicek and CIR. Then we will make some remarks on the Hull-White two factor model, then some more complex model. The code for this in a package in R.… - Date: 2025-03-13 - Exact published timestamp: 2025-03-13 - Topics: Fixed Income, Calibration, Vasicek, Cox-Ingersoll-Ross, Hull-White, HJM - Categories: Fixed Income - Source: Advanced Derivatives - Source URL: None - Content source format: typst

Calibration

We’ll look specifically at Vasicek and CIR. Then we will make some remarks on the Hull-White two factor model, then some more complex model.

The code for this in a package in R.

Vasicek

First we will simulate, then look at parameters.

𝑑 𝑟_{𝑡} = 𝑎 (𝑏 - 𝑟_{𝑡}) 𝑑 𝑡 + 𝜎 𝑑 𝑊_{𝑡}

We will use the Euler discretization scheme.

𝑟_{𝑡 + Δ 𝑡} - 𝑟_{𝑡} = 𝑎 (𝑏 - 𝑟_{𝑡}) Δ + 𝜖_{𝑡 + Δ}

Then we will make a Binomial tree that uses such discretization.

where

𝜖_{𝑡 + Δ} \sim 𝑁 (0, 𝜎^{2} Δ)

Then in general,

𝑟_{𝑖 Δ} = 𝑟_{(𝑖 - 1) Δ} + 𝑎 (𝑏 - 𝑟_{𝑡}) Δ + 𝜖_{𝑡 + Δ}

Cox-Ingersoll-Ross

𝑑 𝑟_{𝑡} = 𝑎 (𝑏 - 𝑟_{𝑡}) 𝑑 𝑡 + 𝜎 \sqrt{𝑟_{𝑡}} 𝑑 𝑊_{𝑡}

Then do the same thing.

Two-Factor

Then the two factor Vasicek model can be defined with correlated factors.

There’s a short rate and long rate factor.

One such method of calibration is the maximum likelihood estimator (MLE)

For Vasicek with real world data, consider

sample short rates $𝑟_{0}, 𝑟_{Δ}, 𝑟_{2 Δ}$ ,

for Vasicek $𝑓 (𝑟_{𝑡 + 𝑠} | 𝑟_{𝑡})$ normal density

with $𝔼 [𝑟_{𝑡 + 𝑠} | 𝑟_{𝑡}] = 𝑏 + (𝑟_{𝑡} - 𝑏) 𝑒^{- 𝑎 𝑠}$

and

𝕍 [𝑟_{𝑡 + 𝑠} | 𝑟_{𝑡}] = \frac{𝜎^{2}}{2 𝑎} (1 - 𝑒^{- 2 𝑎 𝑠})

Then the MLE method determines this for the distribution.

It’s a function of parameters given the

Typically you take the ln of the likelihood and determine the maximum given the function.

𝛼^{*} = (1 - 𝑒^{- 2 𝑎 Δ}) 𝑏

𝛽^{*} = 𝑒^{- 𝑎 Δ}

𝜎^{*} = \sqrt{\frac{𝜎^{2}}{2 𝑎} (1 - 𝑒^{- 2 𝑎 Δ})}

𝔼 [𝑟_{𝑖 Δ} ∣ 𝑟_{(𝑖 - 1) Δ}] = 𝛼^{*} + 𝛽^{*} 𝑟_{(𝑖 - 1) Δ}

𝕍 [𝑟_{𝑖 Δ} ∣ 𝑟_{(𝑖 - 1) Δ}] = (𝜎^{*})^{2}

Then we take the natural log of the likelihood for the normal density plugging in these parameters.

MLE estimates

𝑎 = - \frac{ln ({\hat{𝛽}}^{*})}{Δ}

𝑏 = \frac{{\hat{𝑎}}^{*}}{1 - {\hat{𝛽}}^{*}}

Risk-Neutral Calibration

Assume that we have ZCB prices/discount factors (same thing).

The price of a ZCB under the Vasicek model is

𝑃 (𝑡, 𝑇) = 𝑒^{𝐴 (𝑡, 𝑇) - 𝐵 (𝑡, 𝑇) 𝑟}

where

𝐵 (𝑡, 𝑇) = \frac{1}{𝑎^{*}} (1 - 𝑒^{- 𝑎^{*} (𝑇 - 𝑡)})

𝐴 (𝑡, 𝑇) = (𝐵 (𝑡, 𝑇) - (𝑇 - 𝑡)) (𝑏^{*} - \frac{𝜎^{2}}{2 (𝑎^{*})^{2}}) \dots

for the calibration, you would want to minimize for n bonds

\sum_{𝑖 = 1}^{𝑛} (𝑃_{Vasicek} - 𝑃_{market})^{2}

You might use a nonlinear optimizer.

Hull White Two Factor

𝑑 𝑟 = [𝜃 (𝑡) + 𝑢 - 𝑎 𝑟] 𝑑 𝑡 + 𝜎_{1} 𝑑 𝑊_{1}

𝑑 𝑢 = - 𝑏 𝑢 + 𝜎^{2} 𝑑 𝑊_{2}

where we have another process u, and two volatilities/brownian motions.

u in this context is a random mean reversion level.

The previous model fits the term structure at time 0 and defines such mean reverting level, which is subject to some randomness. We have some historical data for this.

You also have ρ the correlation between dW_1 and dW_2, and the rest are constants.

Spot rate volatility structure for Hull-White two-factor model.

𝜎_{𝑅} (𝑡, 𝑇) = \frac{1}{𝑇 - 𝑡} \sqrt{[𝐵 (𝑡, 𝑇)^{2} 𝜎_{1}]^{2} + [𝐶 (𝑡, 𝑇) 𝜎_{2}]^{2} + 𝑤 𝜌 𝜎_{1} 𝜎_{2} 𝐵 (𝑡, 𝑇) 𝐶 (𝑡, 𝑇)}

Here is an example:

Price a 5 year zero coupon bond after one year i.e. four years remaining to maturity.

It has the following parameters:

t = 1, T = 5, r(1) = 0.05, u(1) = 0.01.

In order to calculate, all the implementation is in the PDF.

First calculate

B(1, 5)

B(0, 5)

B(0, 1)

C(1, 5)

C(0, 5)

C(0, 1)

ln A(1, 5)

P(1, 5)

Then you can construct the term structure will fit at time 0, and the evolution of this structure.

Then in this case, we have an analytical solution for σ = 0.0110

10 year and 15 year interest rates will have a lot of correlations, you can graph correlations against interest rate times and get the term structure.

Limitations of One-Factor or Two-Factor Models

Most involve only one factor of uncertainty.

The models do not give the freedom in choosing the volatility structure.

Looking at a simple example of rates, we can see that we have different volatilities for maturities, and these tend to be correlated.

Based on these considerations, we need a more general approach in specifying the volatility environment and allow multiple factors

This is the HJM Model (Heath Jarrow, and Marton Model)

In the previous short rate models, we had a single source of randomness, Brownian motion, and the instantaneous correlation between forward rates and maturities is equal to 1.

We will consider some notation.

P(t, T) is the price of ZCB of $1 at time t maturing at T.

Ω will be the vector of past and present values of interest rates and bond prices at time t that are relevant for determining bond price volatility at that time, basically describes the information set.

Then

ν(t, T, Ω) this is nu, the volatility of P(t, T)

The process for P(t, T) looks like

𝑑 𝑃 (𝑡, 𝑇) = 𝑟 (𝑡) 𝑃 (𝑡, 𝑇) 𝑑 𝑡 + 𝜈 (𝑡, 𝑇, Ω_{𝑡}) 𝑃 (𝑡, 𝑇) 𝑑 𝑊 (𝑡)

Then we have some characteristics at maturity.

The forward rate is determined by the derivative with respect to the ln change in ZCB over time.

The dynamics of the forward curve (ZCB) are only dependent on the volatility. Then we can say this is a risk-neutral process for the forward f that depends only on ν.

Then we have a process for the instantaneous forward dF which is only determined by the volatility structure. So if we have ν, then we have the term structure

There is a link between the dirft and the standard deviation of the instantaneous forward rate.

But the extension to several factors with the market is expressed int erms of the instantaneous forward rate, which is not directly observable, and it’s difficult to calibrate. And also the process for the short rate in HJM model is non-Markovian.

What is the connection between the HJM model and the previous short rate models?

We look just at the single factor model.

If you integrate the instantaneous forward rate, this is the actual short rate.

And generally you can see how the terms depend on each part of the ν expression. That’s why r is non-Markovian, a non-recombining tree.

## American Options and Cubic Splines - URL: https://sharifhsn.dev/blog/computational-methods-week-07/ - Structured data: https://sharifhsn.dev/api/posts/computational-methods-week-07.json - Description: Cubic Splines, you solve something that is like a tridiagonal system Artificial Neural Network multilayer perceptron… the whole point of this is that you can exercise this at any… - Date: 2025-03-11 - Exact published timestamp: 2025-03-11 - Topics: Computational Methods, American Options, Linear Complementarity, Successive Over-Relaxation, Cubic Splines - Categories: Computational Methods - Source: Computational Methods in Quantitative Finance - Source URL: None - Content source format: typst

Three New Topics

Cubic Splines, you solve something that is like a tridiagonal system

Artificial Neural Network multilayer perceptron…

Copulas (useful for $680$ )

American Option Valuation

page 192 of the QF book

the whole point of this is that you can exercise this at any time.

So when do we exercise, at time $𝜏$ ?

𝜏 \in [0, 𝑇]

There is a short, bad theorem that will help us.

If the underlying process does not pay dividends, and is continuous, then the price of an American call and a European call are identical.

You can prove this (the proof in the book is incorrect):

Clearly,

𝐶_{𝐴} (𝑆, 𝑡) \geq 𝐶_{𝐸} (𝑆, 𝑡)

American call for price S and time t, and European, because the American has extra on top of the call. And this is also true for the Put.

𝐶_{𝐴} (𝑆, 𝑡) \geq (𝑆_{𝑡} - 𝐾)_{+}

𝑃_{𝐴} (𝑆, 𝑡) \geq (𝐾 - 𝑆_{𝑡})_{+}

You can argue this from a no-arbitrage argument, because if this were not true, you could instantly buy the option and exercise it and make money.

In general, no-arbitrage arguments go by assuming that there is a strict inequality, then showing how you can buy low and sell high and make money instantly.

Furthermore,

𝐶_{𝐴} (𝑆, 𝑡) \geq 𝑆_{𝑡} - 𝐾 𝑒^{- 𝑟 (𝑇 - 𝑡)}

Why? We can do another non-arbitrage argument. You can put some money in a bank and get that K term risk-free. If we take the opposite of this, we can buy the cheap thing and sell the expensive thing. You buy the call option and the K ZCB (which becomes negative when you move it to the other side).

First you short sell 1 share of stock ( $𝑆_{𝑡}$ )

Then you borrow $𝐾 𝑒^{- 𝑟 𝑡}$ and put it in a bank

Then you buy 1 call

Because of this inequality, I will receive $𝑆_{𝑡}$ , and take my portion of the money I get and after taking away the call premium and the $𝐾 𝑒^{- 𝑟 𝑡}$ , I’m left with a positive quantity.

Everything with a minus is a liability, everything with a plus I have.

At time $𝑇$ , my $𝑆$ becomes $𝑆_{𝑇}$ , which I have to give back because I short sold it. Then I will receive $𝐾$ and it will make up the balance

Note that Kert is less than K, because e is raised to a negative exponent. And therefore

𝐶_{𝐴} (𝑆, 𝑡) \geq 𝑆_{𝑡} - 𝐾 𝑒^{- 𝑟 (𝑇 - 𝑡)} \geq 𝑆_{𝑡} - 𝐾

Because the value is always greater than the early payoff, then you should never exercise it early.

The derivation in the book is wrong.

This doesn’t work for puts.

This result will also hold for continuous dividends, not discrete dividends.

Free Boundary

To understand the American option problem, we have to understand the free boundary problem. In the differential equation described, you have a system of equations you have to tie down with boundaries. There are two types of boundaries, Neumann in terms of the actual function, Dirichlet which is expressed in the derivative. This is one boundary on the top and one boundary at the bottom (and the terminal condition). These are tied down to these curves.

Now a free boundary problem (which the American problem is) has another condition. This condition is not in a specific location. When are you going to exercise? When the expected future value of my option is going to be equal to or less than the value when I exercise now. If I exercise now and make more money, I should do that. But you don’t know what value of S and t will give you this. The problem is that you have to find this boundary.

In the book, it refers to the physics problem. A lot of the applications of these math problems come from physics. In our case, let’s say I price an American Put. The payoff is at

(𝐾 - 𝑆_{𝑇})_{+}

You can prove that because the form of the function is monotonic. This is the final payoff, so it decreases as S increases, and then stays the same, so it is not. Then the value of the put option is also monotonic.

There exists a price for the stock

𝑆_{𝑓} (𝑡)

At any time t, this stock price exists, the “frontier” price (another name for boundary)

we exercise if

𝑆_{𝑡} < 𝑆_{𝑓} (𝑡)

and we hold if

𝑆_{𝑡} > 𝑆_{𝑓} (𝑡)

What is the point of this? It’s a put option, so you make money if the stock price goes down. If the stock price is high, you make no money, and you want to hold. However, if the price is too low, maybe it will bounce up, so I should probably exercise.

If you look instantaneously, like a fraction of a second right before maturity, and you should definitely exercise if the option is in the money. So there’s a region that has no exercise and a region where you do exercise.

There’s a theorem:

\frac{𝜕 𝑃_{𝐴}}{𝜕 𝑆} (𝑆_{𝑓}, 𝑡) = - 1

The derivative right at the frontier is $- 1$ . If you’re exercising at the frontier, the value you get is $𝐾 - 𝑆_{𝑓}$ , so you have take the derivative with respect to $𝑆_{𝑓}$ , it becomes $- 1$ . There are a couple more steps in the actual proof because you need to say that it’s continuous and so on but this is the basic idea.

𝑆_{𝑡} \leq 𝑆_{𝑓}

, then we exercise and get immediately

(𝐾 - 𝑆_{𝑡})_{+}

Then the free boundary problem tells us:

𝑃 (𝑆_{𝐹}, 𝑡) = (𝐾 - 𝑆_{𝑓})_{+}

and

\frac{𝜕 𝑃_{𝐴}}{𝜕 𝑆} (𝑆_{𝑓}, 𝑡) = - 1

As opposed to the other boundary problem, where it’s at fixed $𝑇$ , and you don’t know what $𝑆_{𝑓}$ is.

Linear Complementarity Problem (LCP)

(\frac{𝜕 𝑃}{𝜕 𝑡} + \frac{1}{2} 𝜎^{2} 𝑆^{2} \frac{𝜕^{2} 𝑃}{𝜕 𝑆^{2}} + 𝑟 𝑆 \frac{𝜕 𝑃}{𝜕 𝑆} - 𝑟 𝑃) (𝑃_{𝐴} (𝑆, 𝑡) - (𝐾 - 𝑆)_{+}) = 0

For the first parentheses, this is geq 0, and the second parenthesis is also geq 0 . On each side of the space, one of them is 0. It’s the same as the tree. You went to a point, and if that point is more worth it to exercise, you would store the value that comes from the tree. In American, you’re never going to solve the early exercise.

Go back to your code for the American put and make the notes where you early exercise in red, and where you don’t exercise in black. In the lower part of the tree, you always exercise, and in the higher part you never do.

Left paren is you cannot exercise, and right paren is where you do. If you don’t exercise, the fair value is the european value.

If this stock value happens to be falling, then you exercise and right paren takes precedence.

We set up the problem so that it takes over when you exercise.

And then we also have the free boundary problems mentioned earlier.

Basically, we solve the LCP using finite difference.

In this, there are a bunch of steps that are reducing the problem, mentioned in the book like logarithm transformation, time transformation, and then you get equation 7.4.1

When you look at the finite difference method, you are looking backwards from the $𝑏_{𝑖}$ column at the end, and then $𝑢_{𝑖 + 1}$

(however we are going forward not backward)

And then you solve it with $𝐴 𝑈_{𝑖 + 1} = 𝑏_{𝑖}$

But that’s for European. With American we get inequalities

𝐴 𝑈_{𝑖 + 1} \geq 𝑏_{𝑖}

𝑈_{𝑖 + 1} \geq 𝑔_{𝑖 + 1}

The book will mention this g condition, describes the no-arbitrage condition.

To solve this system, we have to use what we learned last time about solutions.

We can use Jacobi or Gauss-Seidel. Gauss-Seidel is more computationally efficient because you only need one vector. Sometimes Jacobi is faster because Gauss-Seidel goes further, so it might go further in the wrong direction.

Successive Over Relaxation

Recently, 70 years ago, they invented Successive Over Relaxation (SOR) which is used a lot in machine learning.

We don’t have much ML in the core of financial engineering, so we should be putting some methodology.

Bishop had a million ML methods and didn’t explain how.

𝑈^{(𝑘)} = 𝑈^{(𝑘 - 1)} + (𝑈^{(𝑘)} - 𝑈^{(𝑘 - 1)}

This illustrates an innovation from an old thing to a new thing. This particular example does nothing. Let’s do this instead. Instead of moving all the way to $𝑈^{𝑘}$ , let’s move a little bit with $𝜔$

𝑈^{(𝑘)} = 𝑈^{(𝑘 - 1)} + 𝜔 (𝑈^{(𝑘)} - 𝑈^{(𝑘 - 1)})

If you make it less than 1, than you’re moving a fraction or otherwise you’re moving too much according to Gauss-Seidel.

Somehow that doesn’t make sense because we don’t have U^(k) yet, so how do we do this?

We’ll store the whole Jacobi expression into a single variable $𝑦_{𝑗}$

Then you calculate

𝑈_{𝑗}^{(𝑘)} = 𝑈_{𝑗}^{(𝑘 - 1)} + 𝜔 (𝑦_{𝑗} - 𝑢_{𝑗}^{(𝑘 - 1)})

So basically instead of moving all the way with $𝑦_{𝑗}$ , you preserve it a little bit, and you decide how much you move based on $𝜔$ . You can also change $𝜔$ at every step, but this is not generally done. This is the SOR method. This is used to solve American options.

You don’t need to know the excruciating details, but you should know the big ideas.

Cubic Splines

These were developed to approximate functions

Let’s say we observe $𝑓 : (𝑎, 𝑏) \to ℝ$

You don’t know the value of the function, maybe it’s really complicated, you want to approximate it.

in regression, you have a bunch of points, and you fit one line.

Let’s say our points are all over the place, and a line does not really fit.

You could group all the points as some kind of average, between the x and y coordinates, so you get multiple centers of mass for each region.

Then you want the curve to go through those points. The initial point distribution is totally irrelevant.

It doesn’t have to be a function. What if I have something that goes in circles, or has behavior that doesn’t go in circles. This is the computer science extension which uses B-splines, and conceptually there is no difference.

This is the principle. I do an endpoint approximation. We want to approximate f with piecewise polynomials, different polynomials on each segment.

We have $𝑛 + 1$ knots, these known points. It’s kind of like an anchor point to tie down your curve.

In our process, we take

𝑡_{𝑖} = 𝑎 + \frac{𝑏 - 𝑎}{𝑛} 𝑖

That fractional thing is our $Δ 𝑡$ .

They don’t have to be like this. In general, we can have $𝑡_{0}, 𝑡_{1}, \dots, 𝑡_{𝑛}$ and it will work the same way.

What is the condition? I want to make my curve smooth. I’m going to pick for each interval

[𝑡_{𝑖}, 𝑡_{𝑖 + 1}]

we have

𝑓 (𝑡) = 𝑃_{𝑖} (𝑡)

polynomial

What are the conditions on this polynomial?

Let’s understand what happens at

𝑡_{𝑖 - 1}, 𝑡_{𝑖}, 𝑡_{𝑖 + 1}

Well, one condition is that the polynomials have to connect, so

𝑃_{𝑖 - 1} (𝑡_{𝑖}) = 𝑃_{𝑖} (𝑡_{𝑖}) = 𝑓 (𝑡_{𝑖})

And this becomes two equations for each i. That becomes 2n conditions.

Then I have to stitch them.

I don’t actually know what the derivative of the polynomial is. If I did know, I could impose the extra conditions that has another 2n conditions, for a total of 4n. But this is only if you know the derivative is, which would make this a lot easier.

You need to pick a cubic spline.

The realistic condition:

\frac{𝜕 𝑃_{0}}{𝜕 𝑡} (𝑡_{𝑖}) = \frac{𝜕 𝑃_{1}}{𝜕 𝑡} (𝑡_{1})

You have $2 𝑛 - 2$ equations

So how do you interpolate?

You could take a linear polynomial

𝑃_{𝑖} (𝑡) = 𝑎_{𝑖} + 𝑏_{𝑖} 𝑡

This will clearly not work. You have $2 𝑛$ unknowns and $4 𝑛 - 2$ equations.

The same is true for quadratic.

Cubic gives us

𝑃_{𝑖} (𝑡) = 𝑎_{𝑖} + 𝑏_{𝑖} + 𝑐_{𝑖} 𝑡^{2} + 𝑑_{𝑖} 𝑡^{3}

Now we have $4 𝑛$ unknowns and $4 𝑛 - 2$ equations. So how do we fix this?

There are $𝑛 + 1$ points.

There’s a very clear order unless you somehow define it. And that determines how you stitch the functions. There’s an infinite number of solutions by fixing the two missing equations.

In traditinoal splines that come from statistics, the way it works is that there’s two possibilities.

There can be a free boundary problem where the

𝑆_{0}^{″} 𝑡 (0) = 𝑆_{𝑛} 𝑛^{″} 𝑡 (𝑛) = 0

When you say the second derivative is 0, that’s the maximum of the first derivative. It starts with the largest slope possible,

Then the second one is the clamped boundary

𝑆_{0}^{'} (𝑡_{0}) = 𝑓 (𝑡_{0})

𝑆_{𝑛}^{'} (𝑡_{𝑛}) = 𝑓^{'} (𝑡_{𝑛})

Because you don’t know the derivative, you constrain the curve to have a certain fixed slope.

You constrain the first one to start at a certain angle.

Then how do you solve this? It’s in terms of the functions and their derivatives.

I plug in these points and gets an equations in $𝑎_{𝑖}$ , $𝑏_{𝑖}$ , $𝑐_{𝑖}$ , and $𝑑_{𝑖}$ , and then the next parameters.

Basically I have four end parameters, and since these equations are relative to each other, it becomes a sparse system of equations.

It’s not tridiagonal, but it does have a system.

Then you solve it with Jacobi, Gauss-Seidel, SOR, etc. when you solve it in R.

These are very useful for smoothing approximations of curve. I have used them in my lecture earlier in the pricing the implied vol surface paper. This whole paper is inspired by two dimensional cubic splines. Basically, the Derman people created a bullshit local volatility problem. I thought about it and I realized, if you give me the option prices, and get the IV, and then fit a curve through these things. R had just created the 2D approximation and just used this. It’s basically the same as just described, but you do it with a line instead of points, but I don’t remember how and it’s pretty complicated.

There are many many applications to all of these numerical methods.

I learned about this realized volatility surface from these people, reading about this other method of approximation, and putting them together. That’s why it’s very important to learn approximation methods.

There’s many people that start the task, cut government waste, and then you go in there and have no clue what you’re doing. Method, knowing what to do is the most important thing.

Either you get videos from FOX that are BINGQILIN

## FX Volatility Smiles and Uncertainty - URL: https://sharifhsn.dev/blog/fx-volatility-smiles-and-uncertainty/ - Structured data: https://sharifhsn.dev/api/posts/fx-volatility-smiles-and-uncertainty.json - Description: The syllabus records “Black-Scholes shortcomings: smile effects.” The source export contains the volatility-surface helper below and explicitly labels it “Vol Surface Functions —… - Date: 2025-03-10 - Exact published timestamp: 2025-03-10 - Topics: FX, Volatility, Smiles - Categories: FX - Source: FE-635 \| Risk Engineering - Source URL: None - Content source format: typst

Week 7 — 10 March 2025

The syllabus records “Black-Scholes shortcomings: smile effects.” The source export contains the volatility-surface helper below and explicitly labels it “Vol Surface Functions — Not to be completed by Student.” I am preserving that status rather than presenting it as a finished calibration method.

Risk-reversal and fly interpolation

' ------------------- ' ------------------- ' ------------------- ' ------------------- ' -------------------
' ------------------- ' ------------------- ' ------------------- ' ------------------- ' -------------------
' ------------------- ' ------------------- ' ------------------- ' ------------------- ' -------------------
' ------------------- Vol Surface Functions - Not to be completed by Student ---------------------
Function RRFlyQuadraticFit(lowStrike As Double, _
                    atmFStrike As Double, _
                    highStrike As Double, _
                    ATMFVol As Double, _
                    RRvol As Double, _
                    FlyVol As Double, _
                    XpointATMF As Double, _
                    XPoint As Double)
    Dim VollowStrike As Double
    Dim VolATF As Double
    Dim VolHighStrike As Double
    Dim lowStrikeRelative As Double
    Dim highStrikeRelative As Double
    Dim XPointRelative As Double
    Dim zero As Double
    'If a crazy strike
    If Abs(XPoint) > atmFStrike * 1000 Or Abs(XPoint) < atmFStrike / 1000 Then
     RRFlyQuadraticFit = 0#
    'If not a crazy strike
    Else
     'Make everything relative to ATMF
     lowStrikeRelative = lowStrike - atmFStrike
     highStrikeRelative = highStrike - atmFStrike
     XPointRelative = XPoint - XpointATMF
     zero = 0#
     ' Low Strike vol
      VollowStrike = ATMFVol - RRvol / 2 + FlyVol
      VolATF = ATMFVol
       VolHighStrike = ATMFVol + RRvol / 2 + FlyVol
        RRFlyQuadraticFit = QuadraticFit(lowStrikeRelative, _
                                         VollowStrike, _
                                         zero, _
                                         VolATF, _
                                         highStrikeRelative, _
                                         VolHighStrike, _
                                         XPointRelative)
    End If
End Function
Function QuadraticFit(X1 As Double, _
                    Y1 As Double, _
                    X2 As Double, _
                    Y2 As Double, _
                    X3 As Double, _
                    Y3 As Double, _
                    XPoint As Double)
    Dim Acoef As Double
    Dim Bcoef As Double
    Dim Ccoef As Double
    Acoef = ((Y2 - Y1) * (X1 - X3) + _
         (Y3 - Y1) * (X2 - X1)) / _
        ((X1 - X3) * (X2 ^ 2 - X1 ^ 2) + _
         (X2 - X1) * (X3 ^ 2 - X1 ^ 2))
    Bcoef = ((Y2 - Y1) - Acoef * (X2 ^ 2 - X1 ^ 2)) _
        / (X2 - X1)
    Ccoef = Y1 - Acoef * X1 ^ 2 - Bcoef * X1
    QuadraticFit = Acoef * XPoint ^ 2 + Bcoef * XPoint + Ccoef
End Function
'------------------- AUXILIARY FUNCTIONS --------------------

Auxiliary normal terms

' D1 calculation
Function dOne(S, X, T, r, v)
    dOne = (Log(S / X) + (0.5 * v ^ 2) * T) / (v * (Sqr(T)))
End Function
' N(d1) calculation
Function NdOne(S, X, T, r, v)
    NdOne = Exp(-(dOne(S, X, T, r, v) ^ 2) / 2) / (Sqr(2 * Application.WorksheetFunction.Pi()))
End Function
' D2 calculation
Function dTwo(S, X, T, r, v)
    dTwo = dOne(S, X, T, r, v) - v * Sqr(T)
End Function
 ' N(D2) calculation
Function NdTwo(S, X, T, r, v)

The source gives an implementation sketch for interpolating a quadratic through low-strike, at-the-money-forward, and high-strike volatilities. It does not include a worked smile calibration or a claim that the helper was completed.

## Hull–White Interest Rate Trees - URL: https://sharifhsn.dev/blog/advanced-derivatives-week-06/ - Structured data: https://sharifhsn.dev/api/posts/advanced-derivatives-week-06.json - Description: This will be focused on numerical methods, next week we will have more examples. What is the difference between interest rate risk and stock price risk? One of the most important… - Date: 2025-03-06 - Exact published timestamp: 2025-03-06 - Topics: Fixed Income, Hull-White Model, Trinomial Trees, Interest Rate Models - Categories: Fixed Income - Source: Advanced Derivatives - Source URL: None - Content source format: typst

Numerical construction of the trinomial tree for the Hull-White model

This will be focused on numerical methods, next week we will have more examples.

What is the difference between interest rate risk and stock price risk?

One of the most important differences is that the discount rate will vary from node to node. We will look at an example:

We’ll assume that the payoff of a derivative is $100 (𝑟 - 0.11)_{+}$

We also know the probability of going up, staying and going down as 0.25, 0.5, 0.25

Figure 1: Hand-drawn Hull-White trinomial tree from the source notes.

Hull-White uses an alternative branching process. There is the typical (up med down) then (UP up med), (med down DOWN).

𝑑 𝑟 = [𝜃 (𝑡) - 𝑎 𝑟] 𝑑 𝑡 + 𝜎 𝑑 𝑧

So what is the procedure?

First we have to assume $𝜃 (𝑡) = 0$

And the start is $𝑟 (0) = 0$

The construction of this trinomial tree should match the expectation and variance of the process? Basically we run the simulation based on the parameters.

Then we determine $𝜃 (𝑡)$ so it matches the initial term structure.

Let’s look at the stages:

Figure 2: Handwritten tree-construction steps from the Week 6 source notes.

Then we set $Δ 𝑅 = 𝜎 \sqrt{3 Δ 𝑡}$

This particular equality is chosen for numerical convergence.

We need to determine which branching method applies at each node: (regular, up, down).

Let each node be $(𝑖, 𝑗)$ where $𝑡 = 𝑖 \cdot Δ 𝑡$ and $𝑅 * = 𝑗 Δ 𝑅$

So this is the position in the state space.

As for indices,

i is related to time, it is positive. j can be positive or negative because Hull-White allows negative interest rates.

switching the branching scheme is dependent on the index j

when a > 0

switch from branching regular to down for sufficiently large positive j.

switch from branching regular to up for sufficiently large negative j.

Let $𝑗_{max}$ and $𝑗_{min}$ for the value where you switch

Hull and White (posted on Canvas) show that the probabilities are always positive if $𝑗_{max}$ is equal to the smallest integer > $\frac{0.184}{𝑎 Δ 𝑡}$

𝑗_{min} = - 𝑗_{max}

The point where we change branching is both dependent on the speed of mean reversion and the length of the time step.

Therefore we can say that the probabilities $𝑝_{𝑢}$ , $𝑝_{𝑚}$ , and $𝑝_{𝑑}$ are chosen to match expected change and variance of change in R*. We will solve for these unknowns with equations. What are those equations?

That depends on the branching strategy:

regular:

𝑝_{𝑢} Δ 𝑅 - 𝑝_{𝑑} Δ 𝑅 = - 𝑎 𝑗 Δ 𝑅 Δ 𝑡

𝑝_{𝑢} Δ 𝑅^{2} + 𝑝_{𝑑} Δ 𝑅^{2} = 𝜎^{2} Δ 𝑡 + 𝑎^{2} 𝑗^{2}

𝑝_{𝑢} + 𝑝_{𝑚} + 𝑝_{𝑑} > 1

So this is your system of equations.

And by setting $Δ 𝑅 = 𝜎 \sqrt{3 Δ 𝑡}$ , our system becomes

𝑝_{𝑢} = \frac{1}{6} + \frac{1}{2} (𝑎^{2} 𝑗^{2} Δ 𝑡^{2} - 𝑎 𝑗 Δ 𝑡)

𝑝_{𝑚} = \frac{2}{3} - 𝑎^{2} 𝑗^{2} Δ 𝑡^{2}

𝑝_{𝑑} = \frac{1}{6} + \frac{1}{2} (𝑎^{2} 𝑗^{2} Δ 𝑡^{2} + 𝑎 𝑗 Δ 𝑡)

Note that these are all dependent on index j. We will look at an example of this.

Similarly, for up

𝑝_{𝑢} = \frac{1}{6} + \frac{1}{2} (𝑎^{2} 𝑗^{2} Δ 𝑡^{2} + 𝑎 𝑗 Δ 𝑡)

… I didn’t write them down in time.

And for down

𝑝_{𝑢} = \frac{7}{6} + \frac{1}{2} (𝑎^{2} 𝑗^{2} Δ 𝑡^{2} - 3 𝑎 𝑗 Δ 𝑡)

𝑝_{𝑚} = - \frac{1}{3} - 𝑎^{2} 𝑗^{2} Δ 𝑡^{2} + 2 𝑎 𝑗 Δ 𝑡

𝑝_{𝑑} = \frac{1}{6} + \frac{1}{2} (𝑎^{2} 𝑗^{2} Δ 𝑡^{2} - 𝑎 𝑗 Δ 𝑡)

Let’s take values of a zero curve

0.5

1.5

2.5

3.43

3.824

4.183

4.512

4.812

5.086

Building the first tree, we set our vertical spacing to the special numerical value $𝜎 \sqrt{3 Δ 𝑡}$

Then what is the value of j in our example?

𝜎 = 0.01

Then after we construct the tree, we have to displace the nodes to match the original term structure.

STEP 2

Convert this R* tree into a tree for R, where we displace the nodes so that we match the initial term structure.

Our displacement function is

𝛼 (𝑡) = 𝑅 (𝑡) - 𝑅 * (𝑡)

These can be calculated from the $𝜃 (𝑡)$ expression. Recall that

𝜃 (𝑡) = 𝐹_{𝑡} (0, 𝑡) + 𝑎 𝐹 (0, 𝑡) + \frac{𝜎^{2}}{2 𝑎} (1 - 𝑒^{- 2 𝑎 𝑡})

Note

𝑑 𝑅 = [𝜃 (𝑡) - 𝑎 𝑅] 𝑑 𝑡 + 𝜎 𝑑 𝑊

𝑑 𝑅 * = - 𝑎 𝑅 * 𝑑 𝑡 + 𝜎 𝑑 𝑊

So therefore if we substitute in,

𝑑 𝛼 = [𝜃 (𝑡) - 𝑎 𝛼 (𝑡)] 𝑑 𝑡

Then we can rearrange and integrate to get

𝛼 (𝑡) = 𝐹 (0, 𝑡) + \frac{𝜎^{2}}{2 𝑎^{2}} (1 - 𝑒^{- 𝑎 𝑡})^{2}

for infinitesimally small values.

So this is kind of a continuous function. In this case, we need finite Δt to match the term structure exactly. So it’s a little more simple

We define $𝛼_{𝑖}$ , which corresponds to each time step on the tree,

𝛼 (𝑖 Δ 𝑡) = 𝑅 (𝑖 Δ 𝑡) - 𝑅 * (𝑖 Δ 𝑡)

Here we’re interested in the discrete points on the tree.

Then the overall procedure here is that we are given some rates on the zero curve, but you can have multiple rates on the time step. So you have to price, and look at the value given by the market, and the value given by the tree.

Then we can define the present value of the security $𝑄_{𝑖, 𝑗}$ that pays $1 if node (i, j) is reached and $0 otherwise, so we can add up the probabilities and discount them. And we can calibrate α and Q to match the initial term structure.

So how do we start?

At 0, by definition, we pay $1

𝑄_{0, 0} = 1

And the displacement of the initial node is chosen to give the right price for a ZCB maturing at Δt. Since our zero rate is 3.824 for a one-year rate, we can set α_0 to be that. In an exercise, if you were to take Δ to be 0.5 it would be that rate.

And then $𝛼_{1}$ should be chosen in such a way to be a ZCB that matures in two years.

Then next we calculate $𝑄_{1, 1}, 𝑄_{1, 0}, 𝑄_{1, - 1}$ . i.e., at time i=1, there are three possibilities for j.

We use the probabilities, for $𝑄_{1, 1}$ it’s 0.1667. And then the discount factor is the value of R, in this case 3.824%.

we get the three values 0.1604, 0.6417, and 0.1604. And then we can use it to calculate $𝛼_{1}$ which is chosen to match the right price of a ZCB maturing at 2Δt.

We have the rate from the market, so we can price the bond with that rate. Such bond at 2Δt, you will discount with the rate applicable per branch, which is 1.732%. What we will see in this displaced tree, will be this original value 1.732% discounted.

𝑒^{- (𝛼_{1} + 0.01732) \times 1}

Then for node C, the rate 0 will be ignored so we just use α_1

Then having the price of the bond at nodes B, C, and D, we can calculate the price of the node.

𝐴 = 𝑄_{1, 1} 𝑒^{- (𝛼_{1} + 0.01732)} + 𝑄_{1, 0} 𝑒^{- 𝛼_{1}} + \dots

From the initial term structure, we can get the ZCB for two years as the “market price” = 0.9137.

And then we substitute in the Q values to get the total formula for the adjustment $𝛼_{1}$

ln [\frac{𝑄_{1, 1} 𝑒^{𝑅_{1, 1}} + 𝑄_{1, 0} + \dots}{P(0, 2)}]

The formal approach is to propagate state prices one time step at a time. If $𝑄_{𝑖, 𝑗}$ is the present value of receiving 1 at node $(𝑖, 𝑗)$ , then the next layer’s state price is the sum of discounted contributions from every node that can reach it:

$𝑄_{𝑖 + 1, 𝑘} = \sum_{𝑗 \in 𝑃_{𝑖 + 1, 𝑘}} 𝑄_{𝑖, 𝑗} 𝑝_{𝑖, 𝑗, 𝑘} 𝑒^{- 𝑅_{𝑖, 𝑗} Δ 𝑡}$ ,

where $𝑃_{𝑖 + 1, 𝑘}$ is the set of predecessor nodes and $𝑝_{𝑖, 𝑗, 𝑘}$ is the probability of moving from $(𝑖, 𝑗)$ to $(𝑖 + 1, 𝑘)$ . At each step, choose the displacement $𝛼_{𝑖}$ so the tree’s zero-coupon price matches the observed discount factor:

$𝑃 (0, (𝑖 + 1) Δ 𝑡) = \sum_{𝑘} 𝑄_{𝑖 + 1, 𝑘}$ .

Starting from $𝑄_{0, 0} = 1$ , this gives the iterative procedure: calibrate $𝛼_{𝑖}$ , propagate the state prices with the transition probabilities, and repeat for the next time step.

Recombining binomial tree

## Crank–Nicolson and the Heston Model - URL: https://sharifhsn.dev/blog/computational-methods-week-06/ - Structured data: https://sharifhsn.dev/api/posts/computational-methods-week-06.json - Description: Don’t use AI, you won’t learn anything. And it’s not really intelligent, it just takes things from the Internet. There are two different general finite difference schemes, which… - Date: 2025-03-04 - Exact published timestamp: 2025-03-04 - Topics: Computational Methods, Crank-Nicolson, Linear Systems, Heston Model, Finite Differences - Categories: Computational Methods - Source: Computational Methods in Quantitative Finance - Source URL: None - Content source format: typst

AI Rant

Don’t use AI, you won’t learn anything.

And it’s not really intelligent, it just takes things from the Internet.

Crank-Nicholson Finite Difference Methods

7.4 in the book.

There are two different general finite difference schemes, which are about which points you use in the grid.

In the explicit scheme, you go from many points to one point, so you can find out each point from the points in the boundary, kind of in a trinomial tree way.

In the implicit scheme, you go from one point to three points, which allows you to solve the systems, and find all the points altogether.

Crank-Nicholson is a variation of the implicit finite difference that is better than both, because it is more complicated.

We have the points. Remember that $𝑖$ is $Δ 𝑡$ , and $𝑗$ is $Δ 𝑥$ . We need three points for each derivative.

We will actually use the average of the points to calculate the x-type derivative.

This is the definition of ad derivative in a numerical sense:

\frac{𝜕 𝑢}{𝜕 𝑡} = \frac{𝑢 (𝑖 + 1, 𝑗) - 𝑢 (𝑖, 𝑗)}{Δ 𝑡}

For this one, we use the top yellow minus the bottom yellow. It’s a bit tricky because it’ saveraged, the point is to use both at the same time.

Then you divide by the interval between them.

\frac{𝜕 𝑢}{𝜕 𝑥} = \frac{\frac{1}{2} (𝑢 (𝑖, 𝑗 + 1) + 𝑢 (𝑖 + 1, 𝑗 + 1)) - \frac{1}{2} (𝑢 (𝑖, 𝑗 - 1) + 𝑢 (𝑖 + 1, 𝑗 - 1))}{2 Δ 𝑥}

Then the second derivative is horrendous. It works by using the two second derivatives and reducing to the value of the topmost plus the value of the bottom most minus twice the middle

\frac{𝜕^{2} 𝑢}{𝜕 𝑥^{2}} = \frac{\frac{1}{2} (𝑢 (𝑖, 𝑗 + 1) + 𝑢 (𝑖 + 1, 𝑗 + 1)) - (𝑢 (𝑖, 𝑗) + 𝑢 (𝑖 + 1, 𝑗)) + \frac{1}{2} (𝑢 (𝑖, 𝑗 - 1) + 𝑢 (𝑖 + 1, 𝑗 - 1))}{Δ 𝑥^{2}}

We have all these derivatives in the equation, but we will also use the u. We will use u_i for the last term, which is just u(t, x) for easier usage. I’m usinthe mid points, the yellow points, all over the place.

If I plug them into the equation, what do I get?

\frac{𝜕 𝑢}{𝜕 𝑡} + 𝜈 \frac{𝜕 𝑢}{𝜕 𝑥} + \frac{1}{2} 𝜎^{2} \frac{𝜕^{2} 𝑢}{𝜕 𝑥^{2}} - 𝑟 𝑢 = 0

where $𝜈 = 𝑟 - \frac{𝜎^{2}}{2}$

Then, plugging in our finite differences, it’s

\frac{𝑢 (𝑖 + 1, 𝑗) - 𝑢 (𝑖, 𝑗)}{Δ 𝑡} + 𝜈 \frac{𝑢 (𝑖, 𝑗 + 1) + 𝑢 (𝑖 + 1, 𝑗 + 1) - 𝑢 (𝑖, 𝑗 - 1) - 𝑢 (𝑖 + 1, 𝑗 - 1)}{4 Δ 𝑥} + \frac{1}{2} 𝜎^{2} \frac{𝑢 (𝑖, 𝑗 + 1) + 𝑢 (𝑖 + 1, 𝑗 + 1) - 2 (𝑢 (𝑖, 𝑗) + 𝑢 (𝑖 + 1, 𝑗)) + 𝑢 (𝑖, 𝑗 - 1) + 𝑢 (𝑖 + 1, 𝑗 - 1)}{2 Δ 𝑥^{2}} - 𝑟 \frac{𝑢_{𝑖 𝑗} - 𝑢_{𝑖 + 1, 𝑗}}{2} = 0

If this doesn’t work, please follow up because it might not work

I will isolate all terms that have terms u(i) because those are unknown, I already know i + 1. So I should factor out these three terms here to get our coefficients

𝑢 (𝑖, 𝑗 + 1) (\frac{𝜈}{4 Δ 𝑥} + \frac{𝜎^{2}}{4 Δ 𝑥^{2}}) - 𝑢 (𝑖, 𝑗) (\frac{1}{Δ 𝑡} + \frac{𝜎^{2}}{2 Δ 𝑥^{2}} + \frac{𝑟}{2}) + 𝑢 (𝑖, 𝑗 - 1) (- \frac{𝜈}{4 Δ 𝑥} + \frac{𝜎^{2}}{4 Δ 𝑥^{2}})

The point of being careful about this is that this is what gives us our A, B, and C. So it becomes Ax + By + Cz =…

And now this is the nightmare you have to move to the other side, and make sure to change the sign.

= - \frac{1}{Δ 𝑡} 𝑢 (𝑖 + 1, 𝑗) - \frac{𝜈}{4 Δ 𝑥} (𝑢 (𝑖 + 1, 𝑗 + 1) - 𝑢 (𝑖 + 1, 𝑗 - 1) - \frac{𝜎^{2}}{4 Δ 𝑥^{2}} (𝑢 (𝑖 + 1, 𝑗 + 1) - 2 𝑢 (𝑖 + 1, 𝑗) + 𝑢 (𝑖 + 1, 𝑗 - 1)) - \frac{𝑟}{2} 𝑢_{𝑖 + 1, 𝑗}

Now I have everything I need to do the pseudocode.

Pseudocode

Initialize $Δ 𝑥$ values.

Then I get the payoffs on the right most boundary based on $(𝑒^{𝑥 + 𝑛 Δ 𝑥} - 𝐾)_{+}$ or whatever.

That is my values for u(n, N), u(n, N - 1)…u(n, -N)

These equations supposedly work from top to bottom. Because I can’t go to infinity. I need to add two boundary numbers. And all of these numbers should recursively go backwards.

Then you will end up with the tridiagonal system with A B C, 0 A B C, 0 0 A B C, etc.

Then you will also need the $𝜆_{𝑢}$ and $𝜆 - 𝐿$ to finish the vector.

The system has the very same form as the implicit finite difference method. The only difference is that the last b vector is solved in a complex way.

Stability

How is this method stable with floating point errors?

You are dividing by A + B, which as long as it’s not tiny, it won’t explode.

Methods to Solve Linear Systems of Equations

Imagine you have to solve Ax = b.

Jacobi says, let’s do a trick. (This only works for nonzero diagonal systems).

𝐴 = [\begin{matrix} 𝑎_{11} & 𝑎_{12} \\ 𝑎_{21} & 𝑎_{22} \end{matrix}]

The diagonal cannot be zero.

You will take the lower triangle of the matrix L, the diagonal D, and the upper triangle U. This allows me to do

(𝐿 + 𝐷 + 𝑈) 𝑥 = 𝑏

You would normally take the inverse to get

𝑥 = 𝐴^{- 1} 𝑏

But Jacobi was a finance guy. When you try to invert 2x2 it’s okay, 3x3 is painful, and 4x4 is unbelievable. So before they had calculators, he wanted to invert something easier.

𝐷 𝑥 = 𝑏 - (𝐿 + 𝑈) 𝑥

Basically only inverting the part I know how to invert easily, since inverse diagonal is just the inverse of all its elements.

𝑥 = \frac{1}{𝐷} (𝑏 - (𝐿 + 𝑈) 𝑥)

That might seem useless, but then we can do this:

Let’s start with some random x^0, maybe all zeroes, then plug this into the equation, and take the following recurrence:

𝑋^{(} 𝑘) = 𝐷^{- 1} (𝑏 - (𝐿 + 𝑈) 𝑋^{(𝑘 - 1)})

This looks ugly in matrix form, but it’s more easily expressed as:

𝑎_{11} 𝑥_{1} = 𝑏_{1} - 𝑎_{12} 𝑥_{2} - \dots - 𝑎_{1 𝑛} 𝑥_{𝑛}

𝑥_{2} = \frac{1}{𝑎_{22}} (𝑏_{2} - 𝑎_{21} 𝑥_{1} - 𝑎_{23} 𝑥_{3} - \dots - 𝑎_{2 𝑛} 𝑥_{𝑛})

So you just move everything to the other side.

We’ll take $𝑥^{(0)}$ to be all zeroes.

Then you take the result in the next x, and recur it, and Jacobi proved that it will converge, AS LONG AS the diagonal elements are nonzero because notice that we have to divide by them.

There’s another generalization of this that will be useful when we talk about American options pricing.

Gauss-Seidel

This is the PPO of the Jacobi, where it’s a very small change from the Jacobi.

𝑥_{1}^{𝑘} = \frac{1}{𝑎_{11}} 𝑥_{2}^{(𝑘 - 1)} + 𝑎_{13} 𝑥_{3}^{(𝑘 - 1)} \dots + 𝑎_{1 𝑛} 𝑥_{𝑛}^{(𝑘 - 1)})

But if you notice in x_2, we already calculated the previous recurrence. And this does two steps in 1, and it has faster convergence.

This is a very common idea called the fixed point method. Whenever you solve equations, you look for things that are fixed. If you calculate a derivative, the reason is because it is the maximum. However, when you get close to the maximum, and you get bigger, then you get stuck. The gradient method goes up and up until you can’t go anymore. You need to see this pattern everywhere, because it’s very important.

The human brain is stupid.

Finite Difference for Heston

This is pretty tough, so we won’t get all the details.

There are some convergence issues with the explicit finite difference method. You need the conditions that

$Δ 𝑥 \geq 𝜎 \sqrt{3 Δ 𝑡}$ and $𝑁 = 𝑛$ .

Also, the order of convergence is $𝑂 (Δ 𝑥^{2} + Δ 𝑡)$ . So you want $Δ 𝑥$ to be as small as possible, to make convergence faster, so effectively the optimal is that it equals this.

We will pick an n, and then the following values fall out of it:

Δ 𝑡 = 𝑇 / 𝑛, Δ 𝑥 = 𝜎 \sqrt{3 Δ 𝑡}, 𝑁 = 𝑛

There’s nothing to choose.

But you have some more freedom for IFD (Implicit Finite Difference) and Crank Nicholson.

We have the same order of convergence

𝑂 (Δ 𝑥^{2} + Δ 𝑡)

but this is unconditionally convergent, so it doesn’t matter what values you pick, it doesn’t have this condition about being greater than whatever.

Let’s say that you want to be within epsilon of the true number.

So how do I make sure that I’m within? Technically you can’t, because big O has constants associated with it, but we might take

Δ 𝑥^{2} + Δ 𝑡 = 𝜖

Because there is no relationship between them, you can pick anything for either. Let’s say

$Δ 𝑥^{2} = \frac{𝜖}{2}$ and the same for $Δ 𝑡 = \frac{𝜖}{2}$ .

This gives you a known $Δ 𝑡$ which will give you n.

But $Δ 𝑥$ is not fixed.

But how far should I go? (N).

Because this process is Black-Scholes, we know that the spread at time T is lognormal, so $𝑋_{𝑇} \sim 𝑁 (𝑥_{0}, 𝜎^{2} 𝑇)$

So we need N such that $𝑁 Δ 𝑋 > 3.5 𝑠 𝑖 𝑔 𝑚 𝑎 \sqrt{𝑇}$

We know that most values are within 3 standard deviations, I do 3.5 just in case, but generally it doesn’t matter.

For the Crank-Nicholson, we have

𝑂 (Δ 𝑥^{2} + \frac{Δ 𝑡^{2}}{2})

So we handle it in kind of a similar way to IFD. How you split it is your choice.

Last thing is about Calculating Greeks:

These finite difference methods are useful for calculating these derivatives.

You are approximating derivatives, and the Greeks are derivatives.

Gamma, theta, vega, are instilled in the u-values.

Delta x is not the stock price, so you actually need to take e^x_0

For Delta:

\frac{𝑢_{0, 1} - 𝑢_{0, - 1}}{𝑒^{𝑥_{0} + Δ 𝑥} - 𝑒^{𝑥_{0} - Δ 𝑥}}

And similar for Gamma, we will need all three points, in the same way as the other second derivative was described.

The solution to the PDE is the same as the price of the option here, because that’s how we express the option. So we use the finite difference to numerically solve the PDE.

The best way to solve a PDE is analytically, then you’re done. But, I can count on my hand how many PDEs you can actually solve, just three. So in general, if you write anything (and the heat equation is one where there is a solution) it’s complex and you have to approximate it.

There is no formula for the Heston model.

It involves a characteristic and inverse characteristic function, which is the probabilistic version of a Fourier transform. And there are not that many known analytical solutions for the Fourier Transform.

So you do the discrete Fourier transform and the inverse discrete Fourier Transform.

That’s how the solution for the Heston model is expressed.

The ultimate way to solve for the Fourier is to do this finite difference method for the Heston model.

In order to solve the Heston model, we need to use a bunch of parameters. r is the reglar risk free interest rate.

𝑑 𝑆_{𝑡} = (𝑟 - 𝑞) 𝑆_{𝑡} 𝑑 𝑡 + \sqrt{𝑦_{𝑡}} 𝑆_{𝑡} 𝑑 𝑊_{𝑡}

This is the same as Blck-Scholes but $𝜎$ is replaced by this other stochastic process $𝑦_{𝑡}$ .

And the parameters are

𝑑 𝑌_{𝑡} = 𝜅 (\bar{𝑦} - 𝑦_{𝑡}) 𝑑 𝑡 + 𝜎 \sqrt{𝑦_{𝑡}} 𝑑 𝑊_{𝑡}^{2}

where the two Brownian motions are correlated with $𝜌$

This is a Cox Ingersoll Ross process (or Ornstein Uhlbeck)

Let’s consider Black-Scholes: one reason it’s nice is because you can do this logarithm transformation like this:

𝑑 𝑆_{𝑡} = 𝑟 𝑆_{𝑡} 𝑑 𝑡 + 𝜎 𝑆_{𝑇} 𝑑 𝑊_{𝑡}

𝑑 𝑅_{𝑡} = (𝑟 - \frac{𝜎^{2}}{2}) 𝑑 𝑡 + 𝜎 𝑑 𝑊_{𝑡}

So you might use this for interest rates, but actually it’s bad. So you usually use this Vasicek (or Ornstein Uhlerbeck) model:

𝑑 𝑅_{𝑡} = 𝜅 (𝑟 - 𝑅_{𝑡}) 𝑑 𝑡 + 𝜎 𝑑 𝑊_{𝑡}

where r is the long term interest rate what the Fed does an model the short term interest rate. The problem is that it can be negative.

So this was created in 1978 and it was rejected immediately. Then in 2008, interest rates went negative and now Vasicek is what people actually use. And this can actually be solved. If you’re taking a qualifying exam you should know how to do this.

Then you have the Cox-Ingersoll-Ross model where you take the mean reverting feature and then add a square root term

𝑑 𝑅_{𝑡} = 𝜅 (𝑟 - 𝑅_{𝑡}) 𝑑 𝑡 + 𝜎 \sqrt{𝑅_{𝑡}} 𝑑 𝑊_{𝑡}

That R_t makes the variance dependent ont he value of the proess. So when it goes close to 0, the volatility goes to 0 and the mean reverting term moves the process more to the mean. And this is the process used everywhere.

And you can see that this is the same dynamic for the Heston model.

However, unlike the CIR, this cannot be solved, does not have analytical solution.

One way to solve this is with transformations, which we will do after the midterm.

Another way is to come up with the PDE and solve the PDE with our other solutions.

Writing the PDE is not easy, you have to make a lot of no-arbitrage arguments.

Heston Option Price

v is actually the variance, it is the value for the Y process.

𝑢 (𝑡, 𝑆, 𝑣)

You can derive the equation, but I will write it down like I’m God.

q is the dividend thing. It’s a general thing for if you have a continuously paid dividend.

$𝜃$ is the $\bar{𝑦}$ in the heston model.

\frac{𝜕 𝑢}{𝜕 𝑡} = \frac{1}{2} 𝑣 𝑆^{2} \frac{𝜕^{2} 𝑢}{𝜕 𝑆^{2}} + 𝜌 𝜎 𝑣 𝑆 \frac{𝜕^{2} 𝑢}{𝜕 𝑆 𝜕 𝑣} + \frac{1}{2} 𝜎^{2} 𝑣 \frac{𝜕^{2} 𝑢}{𝜕 𝑣^{2}} - 𝑟 𝑢 + (𝑟 - 𝑞) 𝑆 \frac{𝜕 𝑢}{𝜕 𝑆} + 𝜅 (𝜃 - 𝑣) \frac{𝜕 𝑢}{𝜕 𝑣}

In order to get this nice equation, we also take a change of variable where $𝑡 = 𝑇 - 𝜏$ so this goes from a terminal problem to an initial problem. So I actually know the solution

𝑢 (0, 𝑆, 𝑣) = (𝑆 - 𝐾)_{+}

The reason why it’s like this is because in the solvers you have, they are designed to work as an initial value problem.

For all these terms, you need to substitute them and do the derivative stuff.

Everything else we’ve seen, but the mixed derivative is a little weird. It’s in the book, though. The bigger deal is that these coefficients depend on the location you are. They depend on where you are, they depend on S and v.

So in the equations, every time the coefficients are dependent on the grid. When you move back (or forward in this case) you have to keep track of your grid points and change your coefficients based on where you are.

There are two approaches:

Fixed Grid:

You will have three dimensional grid where you have to keep track of stock, time, AND variance. This is on page 190-191 of the book.

You read the corresponding points, you plug them into the equation, and the solution is solved in the same way.

The scheme here is an explicit scheme. The implicit scheme is absolutely horrendous for this.

Smart Grid:

Instead of using a rectangular grid, you will put your points in non-equally spaced space. The coefficients will become constant if you do it in this way. The coefficients are easy to solve.

Remember This

There are two different types of finite difference,

EXPLICIT AND IMPLICIT

and implicit has some variations.

It’s about discretizing the derivatives. You have to look at every equation you discretize, which points you bring into the equations, because the equation is a relationship between these points.

It’s all about the points you pick. If your boundary is not a straight line, say a curve, then you want to form a relationship that has three points you know, a point you don’t know.

That’s the explicit scheme. Which is easier, but it has some drawbacks and some limitations, as described relating to convergence.

Implicit gives you a system of equations, which will be linear, and eventually you will solve it.

Research

I wrote 3 or 4 papers on this.

All of the equations, if you look, it’s like

\frac{𝜕 𝑢}{𝜕 𝑡} = 𝐿 𝑢

where the operator L contains all this second derivatives, mixed derivatives, whatever.

We created a way to solve this, in a way that is parallel to the Jacobi idea.

You start with some u0 and plug it in here, take the derivative of the function, and then you get the derivative, which you can replug and recur.

Their methodology does both: it moves through the tree and recurs at the same time. It’s slow, but it works.

## Finite-Difference Sketches - URL: https://sharifhsn.dev/blog/computational-methods-finite-difference-sketches/ - Structured data: https://sharifhsn.dev/api/posts/computational-methods-finite-difference-sketches.json - Description: A handwritten finite-difference sketch with A/B/C labels; surrounding equation notation is unclear in the source render. - Date: 2025-03-04 - Exact published timestamp: 2025-03-04 - Topics: Computational Methods, Finite Differences, Grid Sketch, Numerical Methods - Categories: Computational Methods - Source: Handwritten FiniteDifference.xopp and matching autosave, byte-for-byte identical by SHA-256; Drive-modified 2025-03-04. Approximate date within the Computational Methods in Quantitative Finance Docs history cluster (2025-01-21 through 2025-05-15). - Source URL: None - Content source format: typst

The week 5 and 6 notes contain the broader finite-difference grid and explicit/implicit, Crank–Nicolson, and Heston derivations. This page is kept as a brief sketch supplement rather than repeating those derivations.

The hand-drawn diagram has the readable labels A, B, and C beside several tall, curved strokes. Other point labels and the nearby coefficient and equation notation are unclear in the source.

## FX Discrete Hedging and Greeks - URL: https://sharifhsn.dev/blog/fx-discrete-hedging-and-greeks/ - Structured data: https://sharifhsn.dev/api/posts/fx-discrete-hedging-and-greeks.json - Description: The syllabus labels this class “Black-Scholes shortcomings: discrete hedging; volatility uncertainty.” The source workbook expresses the risk measures as functions. It does not… - Date: 2025-03-03 - Exact published timestamp: 2025-03-03 - Topics: FX, Greeks, Discrete Hedging - Categories: FX - Source: FE-635 \| Risk Engineering - Source URL: None - Content source format: typst

Week 6 — 3 March 2025

The syllabus labels this class “Black-Scholes shortcomings: discrete hedging; volatility uncertainty.” The source workbook expresses the risk measures as functions. It does not contain a separate prose lecture summary, so the functions are kept here as the primary note.

$Δ$ , $𝜃$ , $Γ$ , vega, and $𝜌$

The functions below are transcribed from the workbook. BSDailyTheta uses a one-day change; BSVega uses a 0.01 volatility bump; the two $𝜌$ functions use a one-basis-point domestic or foreign-rate bump.

' Calculate the FX Delta of the option
Function BSDelta(ContractType As String, _
                 Forward As Double, _
                 Strike As Double, _
                 Days As Double, _
                 DiscRate As Double, _
                 ExtRate As Double, _
                 vol As Double)
    T = Days / 365
    S = Forward
    X = Strike
    r = DiscRate
    v = vol
    OptionType = ContractType
    If OptionType = "F" Or OptionType = "M" Then
        OptionDelta = Exp(-ExtRate * T)
    ElseIf OptionType = "D" Then
        OptionDelta = 0
     ElseIf OptionType = "C" Then
        OptionDelta = Exp(-ExtRate * T) * Application.NormSDist(dOne(S, X, T, r, v))
    ElseIf OptionType = "P" Then
        OptionDelta = Exp(-ExtRate * T) * (Application.NormSDist(dOne(S, X, T, r, v)) - 1)
    End If
     BSDelta = OptionDelta
End Function
Function BSDailyTheta(ContractType As String, _
                 Forward As Double, _
                 Strike As Double, _
                 Days As Double, _
                 DiscRate As Double, _
                 ExtRate As Double, _
                 vol As Double)
    T = Days / 365
    S = Forward
    X = Strike
    r = DiscRate
    rFor = ExtRate
    v = vol
    OptionType = ContractType
    ' Making Sure Strike is OK for MXN Deposit
    If OptionType = "M" Then
        X = 0.0000000001
    End If
    If OptionType = "F" Or OptionType = "M" Then
        Theta = BSPrice(ContractType, Forward * Exp(-(rFor - r) * (-1) / 365), Strike, Days - 1, DiscRate, ExtRate, vol)
        Theta2 = BSPrice(ContractType, Forward, Strike, Days, DiscRate, ExtRate, vol)
        Theta = Theta - Theta2
    ElseIf OptionType = "D" Then
        Theta = r * Exp(-r * T) * 1 / 365
    ElseIf OptionType = "C" Then
        Theta = BSPrice(ContractType, Forward * Exp(-(rFor - r) * (-1) / 365), Strike, Days - 1, DiscRate, ExtRate, vol)
        Theta2 = BSPrice(ContractType, Forward, Strike, Days, DiscRate, ExtRate, vol)
        Theta = Theta - Theta2
    ElseIf OptionType = "P" Then
        Theta = BSPrice(ContractType, Forward * Exp(-(rFor - r) * (-1) / 365), Strike, Days - 1, DiscRate, ExtRate, vol)
        Theta2 = BSPrice(ContractType, Forward, Strike, Days, DiscRate, ExtRate, vol)
        Theta = Theta - Theta2
    End If
  BSDailyTheta = Theta
End Function
' Calculate the Gamma of the contract
Function BSGamma(ContractType As String, _
                 Forward As Double, _
                 Strike As Double, _
                 Days As Double, _
                 DiscRate As Double, _
                 ExtRate As Double, _
                 vol As Double)
    T = Days / 365
    S = Forward
    X = Strike
    r = DiscRate
    v = vol
    OptionType = ContractType
    If OptionType = "D" Or OptionType = "M" Then
         Gamma = 0
    ElseIf OptionType = "F" Then
    Gamma = 0
    ElseIf OptionType = "C" Then
        Gamma = Exp(-(2 * ExtRate - DiscRate) * T) * NdOne(S, X, T, r, v) / (S * (v * Sqr(T)))
    ElseIf OptionType = "P" Then
        Gamma = Exp(-(2 * ExtRate - DiscRate) * T) * NdOne(S, X, T, r, v) / (S * (v * Sqr(T)))
    End If
  BSGamma = Gamma
End Function
' Function to calculate the Vega
Function BSVega(ContractType As String, _
                 Forward As Double, _
                 Strike As Double, _
                 Days As Double, _
                 DiscRate As Double, _
                 ExtRate As Double, _
                 vol As Double)
    T = Days / 365
    S = Forward
    X = Strike
    r = DiscRate
    v = vol
    OptionType = ContractType
    If OptionType = "F" Or OptionType = "D" Or OptionType = "M" Then
         Vega = 0
    ElseIf OptionType = "C" Then
        VegaBase = BSPrice(ContractType, Forward, Strike, Days, DiscRate, ExtRate, vol)
        VegaBump = BSPrice(ContractType, Forward, Strike, Days, DiscRate, ExtRate, vol + 0.01)
        Vega = VegaBump - VegaBase
    ElseIf OptionType = "P" Then
        Vega = 0
        VegaBase = BSPrice(ContractType, Forward, Strike, Days, DiscRate, ExtRate, vol)
        VegaBump = BSPrice(ContractType, Forward, Strike, Days, DiscRate, ExtRate, vol + 0.01)
        Vega = VegaBump - VegaBase
    End If
  BSVega = Vega
End Function
'Function that calculates the sensitivity to 1bps bump on Domestic interest rates (usually USD rates)
Function BSDailyRhoUSD(ContractType As String, _
                 Forward As Double, _
                 Strike As Double, _
                 Days As Double, _
                 DiscRate As Double, _
                 ExtRate As Double, _
                 vol As Double)
    T = Days / 365
    S = Forward
    X = Strike
    r = DiscRate
    rFor = ExtRate
    v = vol
    OptionType = ContractType
    ' Making Sure Strike is OK for MXN Deposit
    If OptionType = "M" Then
        X = 0.0000000001
    End If
    If OptionType = "F" Or OptionType = "M" Then
       Rho = BSPrice(ContractType, Forward * Exp(0.0001 * T), Strike, Days, DiscRate + 0.0001, ExtRate, vol)
       Rho2 = BSPrice(ContractType, Forward, Strike, Days, DiscRate, ExtRate, vol)
       Rho = Rho - Rho2
    ElseIf OptionType = "D" Then
       Rho = -T * Exp(-r * T) * 0.0001
    ElseIf OptionType = "C" Then
       Rho = BSPrice(ContractType, Forward * Exp(0.0001 * T), Strike, Days, DiscRate + 0.0001, ExtRate, vol)
       Rho2 = BSPrice(ContractType, Forward, Strike, Days, DiscRate, ExtRate, vol)
       Rho = Rho - Rho2
    ElseIf OptionType = "P" Then
       Rho = BSPrice(ContractType, Forward * Exp(0.0001 * T), Strike, Days, DiscRate + 0.0001, ExtRate, vol)
       Rho2 = BSPrice(ContractType, Forward, Strike, Days, DiscRate, ExtRate, vol)
       Rho = Rho - Rho2
'----------------------------------------------
    End If
  BSDailyRhoUSD = Rho
End Function
'Function that calculates the sensitivity to 1bps bump on External interest rates (usually MXN rates)
Function BSDailyRhoExt(ContractType As String, _
                 Forward As Double, _
                 Strike As Double, _
                 Days As Double, _
                 DiscRate As Double, _
                 ExtRate As Double, _
                 vol As Double)
    T = Days / 365
    S = Forward
    X = Strike
    r = DiscRate
    rFor = ExtRate
    v = vol
    OptionType = ContractType
    ' Making Sure Strike is OK for MXN Deposit
    If OptionType = "M" Then
        X = 0.0000000001
    End If
    If OptionType = "F" Or OptionType = "M" Then
       Rho = BSPrice(ContractType, Forward * Exp(-0.0001 * T), Strike, Days, DiscRate, ExtRate, vol)
       Rho2 = BSPrice(ContractType, Forward, Strike, Days, DiscRate, ExtRate, vol)
       Rho = Rho - Rho2
    ElseIf OptionType = "D" Then
       Rho = BSPrice(ContractType, Forward * Exp(-0.0001 * T), Strike, Days, DiscRate, ExtRate, vol)
       Rho2 = BSPrice(ContractType, Forward, Strike, Days, DiscRate, ExtRate, vol)
       Rho = Rho - Rho2
    ElseIf OptionType = "C" Then
       Rho = BSPrice(ContractType, Forward * Exp(-0.0001 * T), Strike, Days, DiscRate, ExtRate, vol)
       Rho2 = BSPrice(ContractType, Forward, Strike, Days, DiscRate, ExtRate, vol)
       Rho = Rho - Rho2
    ElseIf OptionType = "P" Then
       Rho = BSPrice(ContractType, Forward * Exp(-0.0001 * T), Strike, Days, DiscRate, ExtRate, vol)
       Rho2 = BSPrice(ContractType, Forward, Strike, Days, DiscRate, ExtRate, vol)
       Rho = Rho - Rho2
    End If
  BSDailyRhoExt = Rho
End Function

The implementation makes the discrete-hedging convention concrete: the “Greek” is a finite bump around the workbook’s forward, strike, rates, volatility, and 365-day maturity convention.

## Constant Maturity Swaps and Short-Rate Models - URL: https://sharifhsn.dev/blog/advanced-derivatives-week-05/ - Structured data: https://sharifhsn.dev/api/posts/advanced-derivatives-week-05.json - Description: Timing and quanto adjustments, we also looked at nonstandard swaps, LIBOR for LES? Another kind of nonstandard swap is… The definition of a CMS is an interest rate swap where the… - Date: 2025-02-27 - Exact published timestamp: 2025-02-27 - Topics: Fixed Income, Constant Maturity Swaps, Short Rate Models, Vasicek, Cox-Ingersoll-Ross, Hull-White - Categories: Fixed Income - Source: Advanced Derivatives - Source URL: None - Content source format: typst

Continuing Adjustments

Timing and quanto adjustments, we also looked at nonstandard swaps, LIBOR for LES? Another kind of nonstandard swap is…

Constant Maturity Swaps (CMS)

The definition of a CMS is

an interest rate swap where the floating rate equals the swap rate for another reference swap for a certain life

If you want to write an example, here you can say

floating payment on a CMS can be made every six months at rate equal to 5-year swap rate.

The frequency of this CMS rate must be smaller than the reference rate. Usually the swap rate is equal to a previously observed swap rate.

Furthermore, if we assume that we have reset dates $𝑡_{0}, 𝑡_{1}, 𝑡_{2} \dots$

And payment dates: $𝑡_{1}, 𝑡_{2}, 𝑡_{3} \dots$

We will also consider $𝐿$ = notional principal

What is going to be the floating payment? At $𝑡_{𝑖 + 1}$ , this would be the

swap rate $𝑠_{𝑖}$ applied to the principal for a period $𝜏_{𝑖}$

𝜏_{𝑖} 𝐿 𝑠_{𝑖}

where $𝜏_{𝑖} = 𝑡_{𝑖 + 1} - 𝑡_{𝑖}$

Now we have to do some adjustments. The convexity adjustment is, the frequency of these coupons is going to be the same as the bonds.

We will also have to do a timing adjustment, a combination of these two. If you apply these to another currency, then we will also have a quanto adjustment.

Let’s say $𝑦_{𝑖}$ is the forward value of the swap rate.

By using a convexity/timing adjustment.

We can say the realized swap rate is assumed:

𝑦_{𝑖} - \frac{1}{2} 𝑦_{𝑖}^{2} 𝜎_{𝑦_{𝑖}}^{2} 𝑡_{𝑖} \frac{𝐺_{𝑖}^{″} (𝑦)}{𝐺_{𝑖}^{'} (𝑦)} - \frac{𝑦_{𝑖} 𝜏_{𝑖} 𝐹_{𝑖} 𝜌_{𝑖} 𝜎_{𝑦, 𝑖} 𝜎_{𝐹, 𝑖} 𝑡_{𝑖}}{1 + {\bar{𝑇}}_{𝑖} 𝜏_{𝑖}}

where

$𝜎_{𝑦, 𝑖}$ is the volatility of the forward swap rate

$𝐹_{𝑖}$ is the current forward interest rate between $𝑡_{𝑖}$ and $𝑡_{𝑖 + 1}$

$𝜎_{𝐹, 𝑖}$ is the volatility of the forward rate

$𝐺_{𝑖} (𝑥)$ this is basically the price of a bond which pays at time i, the same amount as the reference swap.

The forward rate will be implied by the “Kaplan?” prices.

$𝜌_{𝑖}$ the correlation can be estimated from historical data.

Example

6 year CMS swap (life of swap is 6 years)

5 year swap rate is received, fixed rate of 5% is paid on notional $100M

exchange of payments is semiannual, on both 5-year underlying swap and on CMS swap.

The exchange rate on the payment date is determined from the swap rate on the previous payment date.

The term structure is flat at 5% per annum with semiannual compounding. (that’s where the timing adjustment comes in)

The volatility of the options on five-year swaps have 15% IV ( $𝜎_{𝑦, 𝑖}$ ) and in this simplified problem we’ll say caplets with 6 month tenor have a 20% IV. The correlation $𝜌_{𝑖}$ between each cap rate and each swap rate is 0.7

The fixed rate $𝑦_{𝑖}$ is 0.05. This is semiannual payment, so $𝜏_{𝑖} = 0.5$ . Given volatility $𝜎_{𝐹, 𝑖} = 0.20$ . Then the forward rate is also $𝐹_{𝑖} = 0.05$ from the term structure, and the correlation given is $𝜌_{𝑖} = 0.7$ .

Then what is the relevant function G in this case?

We have the reference rate which is a 5 year swap rate, which has a semiannual payment.

We can express this as a bond, so a sum of coupons. The 2.5M is the coupon payment (halved

𝐺_{𝑖} (𝑥) = \sum_{𝑖 = 1}^{10} \frac{2.5}{1 + \frac{𝑥}{2})^{𝑖}} + \frac{100}{(1 + \frac{𝑥}{2})^{10}}

Then the derivative is

𝐺_{𝑖}^{'} (𝑦_{𝑖}) = - 437.603, 𝐺_{𝑖}^{″} (𝑦_{𝑖}) = 2261.23

Then the total convexity/timing adjustment is, $0.0001197 𝑡_{𝑖}$

We’re going to have different values, so for example the 5 year swap rate t = 4 should be assumed to be 5.0479% instead of 5%.

Therefore the net cash flow at t = 4.5 should be 0.5 * 0.000479 * 100M.

Then we can calculate for all these cash payments, the value of the CMS

Models of the Short Rate

There are different specifications of such models. The models presented so far in our previous discussions make the assumption that the probability distribution of variables (interest rates, bond prices, …) are log normally distributed. That allows us to use Black’s model.

There are some issues with this kind of approach.

These models do not provide a description on how the interest rates evolve in time, just the underlying distribution.
Cannot be used to value American style options.

In this lecture, we will discuss the “term structure model”. Such interest rate model will describe the evolution of all zero-coupon interest rates, a little more general.

We will focus on some classical models.

The behavior of the “short-rate” or instantaneous short rate. This rate is considered in a risk-neutral world. Basically, that means in a very short period of time, between t and $𝑡 + Δ 𝑡$ , investors earn $𝑟 (𝑡) Δ 𝑡$ , the risk-free interest rate.

For example, the dollar money market account is a security that is worth $1 at time 0 and earns the instantaneous risk-free rate r at any given time. This r may be stochastic.

Furthermore, if g equals the money market account, then the process $𝑑 𝑔 = 𝑟 𝑔 𝑑 𝑡$ . Here we can see that the r the risk-free rate can be stochastic, so this will be the drift of g being stochastic, with volatility being 0.

We saw last lecture that

$\frac{𝑓}{𝑔}$ is a martingale in a world where the market price of risk is 0.

Therefore the expected value is

𝑓_{0} = 𝑔_{0} \hat{𝔼} [\frac{𝑓_{𝑇}}{𝑔_{𝑇}}]

Then if you take g as the money market account, then by definition

𝑔_{0} = 1

And for $𝑔_{𝑇}$ ? You might expect the e^rt general, but because this is stochastic, we express this as

𝑔_{𝑇} = 𝑒^{\int_{0}^{𝑇} 𝑟 𝑑 𝑡}

Therefore, the expectation of f at T must be discounted

𝑓_{0} = 𝔼 [𝑓_{𝑇} 𝑒^{- \int_{0}^{𝑇} 𝑟 𝑑 𝑡}]

You could also express this as

𝑓_{0} = \hat{𝐸} [𝑒^{- \bar{𝑟} 𝑇} 𝑓_{𝑇}]

where $\bar{𝑟}$ is the average value of r.

Interest Rate Derivative Valuation

A regular way to value an interest rate derivative, typically these will follow these steps:

Simulate short-term interest rate paths. (in some cases we will have closed-form solutions, but generally we simulate)
Calculate expected payoff on these paths.
Discount at average value of short rate on the sampled path.

In general, the value at time t of an interest rate derivative is based on providing a payoff $𝑓_{𝑇}$ at time T

\hat{𝔼} [𝑓_{𝑇} 𝑒^{- \bar{𝑟} (𝑇 - 𝑡)}]

Basically you calculate the value of such derivatives between t and T.

Furthermore, if you refer back to $𝑃 (𝑡, 𝑇)$ for the price function at time t of a ZCB with payoff of $1 at time T, we can say that

𝑃 (𝑡, 𝑇) = \hat{𝔼} [𝑒^{- \bar{𝑟} (𝑇 - 𝑡)}]

Now we will make connection to the interest rate.

Generally, if R(t, T) is continuously compounded interest rate at time t, for a term (T - t), then we can say

𝑃 (𝑡, 𝑇) = 𝑒^{- 𝑅 (𝑡, 𝑇) (𝑇 - 𝑡)}

then we can rearrange as

𝑅 (𝑡, 𝑇) = - \frac{1}{𝑇 - 𝑡} ln 𝑃 (𝑡, 𝑇)

Then finally we will arrive at the relationship

𝑅 (𝑡, 𝑇) = - \frac{1}{𝑇 - 𝑡} ln \hat{𝔼} [𝑒^{- \bar{𝑟} (𝑇 - 𝑡)}]

This equation is what enables the calculation of the term structure of the interest rate at any given time.

Based on the description of the risk-free rate, which may be stochastic, can help you describe the entire curve.

All the models we discuss will be models of this short rate, there are some models of the instantaneous forward, but generally they will be instantaneously spot. These classical models give descriptions of stochastic processes that the short rate will have this particular expression.

And this is the zero curve

Equilibrium Models

Rendleman & Bartter, Vasicek, Cox Ingersoll & Ross

These models start with some assumptions about the economic variables, then the variable process from the short rate r. This will explore what the process for r implies about the bond prices and option prices. This model might be one factor, or two factor. We will look now just at one factor models.

In a one factor equilibrium model, the process for r involves only one source of uncertainty.

𝑑 𝑟 = 𝑚 (𝑟) 𝑑 𝑡 + 𝑠 (𝑟) 𝑑 𝑊

This general specification, m(r) is the instantaneous drift, and s(r) is the instantaneous volatility (stdev).

For these models, m(r) and s(r) are assumed to be a function of r, but independent of time.

This is different from the no-arbitrage models.

Some of these models, historically, have been proposed. Some of them are not necessarily very good (lol).

Rendleman and Bartter

𝑑 𝑟 = 𝜇 𝑟 𝑑 𝑡 + 𝜎 𝑟 𝑑 𝑊

Where $𝜇$ and $𝜎$ are constant, it basically means that r follows GBM.

But this is bad because r doesn’t follow GBM.

These interest rates have some mean-reverting behavior, so that has to be included somehow. When r is high, it will tend to have negative drift, and when it’s low, it will tend to have positive drift.

Economic argument, high r will cause the economy to slow down, less borrowing, and vice versa. These models can also be calibrated, will be discussed next time.

Vasicek

𝑑 𝑟 = 𝑎 (𝑏 - 𝑟) 𝑡 𝑑 + 𝜎 𝑑 𝑊

where a, b, σ are constants. Here we can see the mean reverting component. In the long-run, the value, when r is small, it will be a positive number and increase, and vice versa. a is the speed of mean reversion, b is the mean.

This model has an analytical solution. Vasicek shows that if

𝑃 (𝑡, 𝑇) = \hat{𝔼} [𝑒^{- \bar{𝑟} (𝑇 - 𝑡)}]

then the price will be a function

𝑃 (𝑡, 𝑇) = 𝐴 (𝑡, 𝑇) 𝑒^{- 𝐵 (𝑡, 𝑇) \cdot 𝑟 (𝑡)}

This is a general form for the analytical solution to the Vasicek model, where

𝐵 (𝑡, 𝑇) = \frac{1 - 𝑒^{- 𝑎 (𝑇 - 𝑡)}}{𝑎}

𝐴 (𝑡, 𝑇) = exp {\frac{[𝐵 (𝑡, 𝑇) - 𝑇 + 𝑡] (𝑎^{2} 𝑏 - \frac{𝜎^{2}}{2})}{𝑎^{2}} - \frac{𝜎^{2} 𝐵^{2} (𝑡, 𝑇)}{4 𝑎}}

Gives you more flexibility, it’s an alternative.

There’s one more alternative…

Cox-Ingersoll-Ross

𝑑 𝑟 = 𝑎 (𝑏 - 𝑟) 𝑑 𝑡 + 𝜎 \sqrt{𝑟} 𝑑 𝑊

Again a, b, σ constants.

What is the advantage of this model? You can look at the evolution of term structure over time. It’s observed that as the interest rate changes, the volatility also changes. Therefore the model should include this behavior.

The price function for CIR has the same general expression as Vasicek:

𝑃 (𝑡, 𝑇) = 𝐴 (𝑡, 𝑇) 𝑒^{- 𝐵 (𝑡, 𝑇) 𝑟 (𝑡)}

The functions are just slightly different. We will discuss the similarities and properties of these models.

𝐵 (𝑡, 𝑇) = \frac{2 (𝑒^{𝛾 (𝑇 - 𝑡)} - 1)}{(𝛾 + 𝑎) (𝑒^{𝛾 (𝑇 - 𝑡)} - 1) + 2 𝛾}

𝐴 (𝑡, 𝑇) = {[\frac{2 𝛾 𝑒^{(𝑎 + 𝛾) (𝑇 - 𝑡) / 2}}{(𝛾 + 𝑎) (𝑒^{𝛾 (𝑇 - 𝑡)} - 1) + 2 𝛾}]}^{2 𝑎 𝑏 / 𝜎^{2}}

where

𝛾 = \sqrt{𝑎^{2} + 2 𝛾^{2}}

Basically, it has an analytical solution.

Properties of CIR vs Vasicek

A(t, T) and B(t, T) are different, but

P(t, T) is fundamentally the same general expression.

\frac{𝜕 𝑃 (𝑡, 𝑇)}{𝑟 (𝑡)} = - 𝐵 (𝑡, 𝑇) \cdot 𝑃 (𝑡, 𝑇)

This function B can be used therefore as an alternative to duration.

From the rate expression,

𝑅 (𝑡, 𝑇) = - \frac{1}{𝑇 - 𝑡} ln 𝑃 (𝑡, 𝑇)

The zero rate at time t for a period T - t is, if you take the natural log of this general AB expression, because it’s a product, you will get

𝑅 (𝑡, 𝑇) = - \frac{1}{𝑇 - 𝑡} ln 𝐴 (𝑡, 𝑇) + \frac{1}{𝑇 - 𝑡} 𝐵 (𝑡, 𝑇) 𝑟 (𝑡)

Here in this case, the entire term structure can be determined as a function of r(t) if a, b, σ are known.

Here we define the modified duration of a bond Q.

\hat{𝐷} = 𝐵 (𝑡, 𝑇)

The shape is going to be a function of r(t), and dependent on t.

The duration of such bond

\frac{Δ 𝑄}{𝑄} = - 𝐷 Δ 𝑦

If you consider that the yield of a bond.

Alternatively, the duration by Vasicek & CIR is

\frac{Δ 𝑄}{𝑄} = - \hat{𝐷} Δ 𝑟

\frac{𝜕 𝑄}{𝜕 𝑟} = - \hat{𝐷} 𝑄

Example

Let’s consider the zero-coupon bond lasting four years.

Duration s = 4, so 10bps parallel shift in the term structure should lead to decrease of 0.4% in the bond price, this is the sensitivity of the model. If we use Vasicek’s model, with a = 0.1, then

\hat{𝐷} = 𝐵 (0, 4) = \frac{1 - 𝑒^{- 0.1 \cdot 4}}{0.1} = 3.29

The duration is a function of the mean-reversion rate.

That means the short rate is 0.329%.

Can you calibrate a using the duration? Hmmmm.

Consider Q a portfolio of ZCBs (a coupon-bearing bond).

𝑃 (𝑡, 𝑇_{𝑖}) (1 \leq 𝑖 \leq 𝑚)

$𝑐_{𝑖}$ is the principal of the $𝑖^{th}$ bond.

Then the duration of our portfolio is

\hat{𝐷} = - \frac{1}{𝑄} \frac{𝜕 𝑄}{𝜕 𝑟} = - \frac{1}{𝑄} \sum_{𝑖 - 1}^{𝑚} \frac{𝜕 𝑃 (𝑡, 𝑇_{𝑖})}{𝜕 𝑟} 𝑐_{𝑖} = \sum_{𝑖 = 1}^{𝑚} \frac{𝑐_{𝑖} 𝑃 (𝑡, 𝑇_{𝑖})}{𝑄} {\hat{𝐷}}_{𝑖}

Where we have $\hat{𝐷}$ for a coupon bearing bond being the weighted average of the duration of the underlying ZCBs.

Vasicek/CIR Pricing Example

Suppose we are given a = 0.1, b = 0.1. The initial short rate is 10%. The initial stdev of short rate change in a short time Δt is $\sqrt{Δ 𝑡}$

Then we will get one point in the term structure, but we can use code to get the entire term structure.

No-Arbitrage Model

These models are designed to be consistent with today’s term structure.

Even if it’s used as parameters, it may not be exact.

Equilibrium models generate today’s term structure.

No-arbitrage models use term structure (the observed rates in the market) as an input.

In equilibrium models, drift is not a function of time.

In no-arbitrage models, drift is a function of time.

We will look briefly at the Ho-Lee and Hull-White model.

Ho-Lee Model

𝑑 𝑟 = 𝜃 (𝑡) 𝑑 𝑡 + 𝜎 𝑑 𝑊

$𝜎$ is the instantaneous standard deviation of the short rate.

These parameters are defined so they fit the initial term structure.

$𝜃 (𝑡)$ can be calculated analytically. For the instantaneous forward rate $𝐹_{𝑡} = \frac{𝜕 𝐹}{𝜕 𝑡}$ .

𝜃 (𝑡) = 𝐹_{𝑡} (0, 𝑡) + 𝜎^{2} 𝑡

As an approximation, you can take

𝜎 (𝑡) \approx 𝐹_{𝑡} (0, 𝑡)

This is something kind of similar to what we observe before. The drift is no longer a constant, it is chosen to match the initial term structure.

However, the price of a zero coupon bond.

𝑃 (𝑡, 𝑇) = 𝐴 (𝑡, 𝑇) 𝑒^{- 𝑟 (𝑡) (𝑇 - 𝑡)}

has the analytical solution (skipping derivations)

ln 𝐴 (𝑡, 𝑇) = ln \frac{𝑃 (0, 𝑇)}{𝑃 (0, 𝑡)} + (𝑇 - 𝑡) 𝐹 (0, 𝑡) - \frac{1}{2} 𝜎^{2} 𝑡 (𝑇 - 𝑡)^{2}

The advantage of this over previous models is that this will match our term structure. But it’s deficient in some components so it’s not so popular.

Assumes that all forward rates have the same stdev

Hull-White Model

𝑑 𝑟 = [𝜃 (𝑡) - 𝑎 𝑟] 𝑑 𝑡 + 𝜎 𝑑 𝑊

𝑑 𝑟 = 𝑎 [\frac{𝜃 (𝑡)}{𝑎} - 𝑟] 𝑑 𝑡 + 𝜎 𝑑 𝑊

a, σ are constant

This is the golden goose one factor model. It can be viewed as an extension of the Vasicek model that fits the term structure, with a time-dependent reversion level.

Or we can say that it’s similar to Ho-Lee with time-dependent reversion level.

𝜃 (𝑡) = 𝐹_{𝑡} (0, 𝑡) + 𝑎 𝐹 (0, 𝑡) + \frac{𝜎^{2}}{2 𝑎} (1 - 𝑒^{- 2 𝑎 𝑡})

That last term is small, so we can approximate it.

We can say that r follows the slope of the initial instantaneous forward rate curve. Basically this is the partial derivative.

Then we can calculate the bond prices,

𝑃 (𝑡, 𝑇) = 𝐴 (𝑡, 𝑇) 𝑒^{- 𝐵 (𝑡, 𝑇) 𝑟 (𝑡)}

𝐵 (𝑡, 𝑇) = \frac{1 - 𝑒^{- 𝑎 (𝑇 - 𝑡)}}{𝑎}

ln 𝐴 (𝑡, 𝑇) = ln \frac{𝑃 (0, 𝑇)}{𝑃 (0, 𝑡)} + 𝐵 (𝑡, 𝑇) 𝐹 (0, 𝑡) - \frac{1}{4 𝑎^{3}} 𝛾^{2} (𝑒^{- 𝑎 𝑇} - 𝑒^{- 𝑎 𝑡})^{2} (𝑒^{2 𝑎 𝑡} - 1)

Some other models

Black-Derman-Toy

𝑑 ln (𝑟) = [𝜃 (𝑡) - 𝑎 (𝑡) ln (𝑟)] 𝑑 𝑡 + 𝜎 (𝑡) 𝑑 𝑊

where $𝑎 (𝑡) = - \frac{𝜎^{'} (𝑡)}{𝜎 (𝑡)}$

Then volatility is related to the speed of mean reversion.

Black-Karasinski model is more general,

a(t) and σ(t) are determined independently.

It has the advantage that interest rates cannot be negative, and the future value is lognormal. However, it’s very difficult to model analytically.

These are simple one-factor models as well. The only thing is that they are very difficult to calibrate, especially the more general they get.

Bond Options

We can also have closed form solution for the price of a bond option for Vasicek and Hull-White models:

𝐿 𝑃 (0, 𝑠) 𝑁 (ℎ) - 𝐾 𝑃 (0, 𝑇) 𝑁 (𝐻 - 𝜎_{𝑃})

where h and σ are VERY COMPLICATED.

The idea is that you have a closed form solution and can solve using analytical methods.

Assignment 1 (BTW)

https://github.com/pola-rs/polars/issues/5255#issuecomment-1857124471

See this why you can’t have float linspace. Workaround is to multiply by 10 to an integer, floor, cast to int, do an arange, then recast to float and divide by the same magnitude.

## Differential Equations and PDE Methods - URL: https://sharifhsn.dev/blog/computational-methods-week-05/ - Structured data: https://sharifhsn.dev/api/posts/computational-methods-week-05.json - Description: Finite difference methods to approximate PDEs this week Next week we will continue this and do a little bit more complicated stuff. Homework is due right before spring break. - Date: 2025-02-25 - Exact published timestamp: 2025-02-25 - Topics: Computational Methods, Differential Equations, PDEs, Finite Differences, Black-Scholes - Categories: Computational Methods - Source: Computational Methods in Quantitative Finance - Source URL: None - Content source format: typst

Plan for the next weeks

Finite difference methods to approximate PDEs this week

Next week we will continue this and do a little bit more complicated stuff.

Homework is due right before spring break.

Exam will be any four hours during the weekend, questions will be a little different.

Differential Equations

The first thing is the ODEs, the ordinary differential equations.

There is an entire class about this in every program, as long as you do some kind of engineering science you need to take Calc 3, which covers these.

The reason they’re called ordinary is because they’re about finding a function $𝑓 (𝑥)$ .

The solution of the equation would be a function, expressed as derivatives of this function.

A fake example would be $2 𝑓^{″} (𝑥) + 3 𝑥 𝑓^{'} (𝑥) + 𝑥^{2} + 2 = 0$

subject to $𝑓 (0) = 3$

just to see how it looks.

Normally you spend a lot of time solving first-order, the most complicated one. You learn how to solve equations with constant coefficients, and things that can be reduced by transformations to having constant coefficients. Then you make a polynomial that solves the PDE.

PDEs

This is Calc 4, partial differential equations. You would only do this if your area requires it.

Not necessarily all of us have done this.

I thought this was the hardest, most horrible class. My professor purposely failed people (especially young women) so they would pay him for tutoring.

These will usually be a function of more than one variable, like

𝑓 (𝑡, 𝑥)

The second variable is usually time, but it doesn’t have to be. x can be multi-dimensional also. Let’s say it’s a line. A practical application would be the heat equation. You have a metal rod and you apply heat to one end, and you look at the distribution of heat over time, f will measure the heat, t will measure time, x will measure location. You could also have a metal plate, and then x is multidimensional.

[Embedded figure omitted from the text export.]

So $𝑓 (𝑡, 𝑥)$ is what we’re trying to find.

We’re solving in the time $𝑡 \in (0, \infty)$

Most of the time, these problems have the initial condition.

The equations that we will deal with will have a terminal boundary condition, ending at T. You know what the option value is at T.

Some equations will start at 0, but you can change between boundary condition 0 and T with change of variables. If you do the $𝜏 = 𝑇 - 𝑡$ , then it will go from terminal to initial.

The x we’re talking about is $𝑥 \in ℝ$ , aka one dimensional, but in general $𝑥 \in ℝ^{𝑛}$ .

NYHOPS

Here is a practical example of this. I worked a former Stevens professor some years ago on NYHOPS, New York Harbor Ovserving and Predicting System. Something solves a bunch of PDEs to do forecasts of things like salinity, etc. for the next 48 hours, on a revolving 6 hour period. Actually they just calculate salinity, and it turns out that water speed is driven by salinity.

So how does he do it?

He uses some other equation that is used for viscosity of water, but we don’t need to worry about it.

Because this $𝑓 (𝑡, 𝑥)$ has two variables, the equations will involve multiple derivatives, joint derivatives.

LEt’s say we have a Linear PDE, which we will define as $𝑢 (𝑡, 𝑥)$ . We will use u for our own purposes. This will only have second order derivatives

𝑎 \frac{𝜕^{2} 𝑢}{𝜕 𝑡^{2}} + 𝑏 \frac{𝜕^{2} 𝑢}{𝜕 𝑡 𝜕 𝑥} + 𝑐 \frac{𝜕^{2} 𝑢}{𝜕 𝑥^{2}} + 𝑑 \frac{𝜕 𝑢}{𝜕 𝑡} + 𝑒 \frac{𝜕 𝑢}{𝜕 𝑥} + 𝑓 𝑢 + 𝑗

This is the most general form of a linear PDE. Although they don’t have to be constant, for our definitions they need to be constant. The behavior of our solution, it turns out, is only governed by the second order derivatives. So we only need to worry about a, b, and c.

We will make a polynomial in α and β, where these represent the derivative with respect to t and x, respectively. Then,

𝑃 (𝛼, 𝛽) = 𝑎 𝛼^{2} + 𝑏 𝛼 𝛽 + 𝑐 𝛽^{2} + 𝑑 𝛼 + 𝑒 𝛽 + 𝑐 𝑡

And again we only worry about a, b, and c.

To reiterate: The nature of the PDE is determined by the properties of these 2nd order terms.

Specifically, we can look at these terms

$𝑎 𝛼^{2} + 𝑏 𝛼 𝛽 + 𝑐 𝛽^{2}$ and divide by $𝛽^{2}$ . THis gives us

𝑎 (\frac{𝛼}{𝛽})^{2} + 𝑏 \frac{𝛼}{𝛽} + 𝑐

And you can see this is a quadratic polynomial. It’s very well-studied and easy to solve. To solve for the determinant, it is

Δ = 𝑏^{2} - 4 𝑎 𝑐

We have different behavior based on this Δ.

There are three types of major equations.

Δ < 0

The equation has no real solutions, only complex conjugate solutions. These lead specifically to elliptic equations. One example of this that you would have done if you covered PDEs is the Laplace equation.

The Laplace equation looks like

\frac{𝜕^{2} 𝑢}{𝜕 𝑡^{2}} + \frac{𝜕^{2} 𝑢}{𝜕 𝑥^{2}} = 0

This is the simplest equation that is elliptic. It appears in thermodynamics. If you’re dealing with gases, behavior in physics, you will see this.

Δ > 0

This is called a hyperbolic equation. These terms (by the way) come from the behavior of the solution.

When you solve a PDE like this, if you don’t have boundary conditions, you get a function which conducts a field. You can look at the derivative of a function and it will tell you where the function goes. You have to tie it down to get one particular surface. The magnetic field is one such function.

The minimum is (this is trivial)

\frac{𝜕^{2} 𝑢}{𝜕 𝑡^{2}} - \frac{𝑝 𝑎 𝑟 𝑡 𝑖 𝑎 𝑙^{2} 𝑢}{𝜕 𝑥^{2}} = 0

And if you pay attention to where a and c are, you can see where the positive and negative come form.

This is called the wave equation, the jumprope equation which vibrates, useful for earthquakes. The way they do it is by solving an equation of this time, (although with more terms), and the constants are determined by the structure of the earth.

Δ = 0

This is called the parabolic PDE. The diffusion equation, if you’ve heard of diffusion tensor imaging, or something like that, uses these kinds of equations. You can see the image of the brain, but it’s not even correct. The brain is made of grey matter, the neurons, and the white matter, which are the axons that connect them. But nobody can cut open your brain to look at the structure. But the white matter decomposes immediately. Nobody knows how they are connected, so you have to put them in the MRI and see how the water molecules move. These are very intrusive, so you can’t do it for too long, then you have to make connections. You can’t see the cables, you can only see the regions. So someone says to take the PDE from the 1970s where they consider the axons are pipes that they heat up.

The heat equation is the most classical example:

\frac{𝜕 𝑢}{𝜕 𝑡} - \frac{𝜕^{2} 𝑢}{𝜕 𝑥^{2}} = 0

You have two terms that are missing, that’s the only way to get this equation. If you do this with parameters, then you get a square, which will become transformed into first-order aka not interesting. The only way to get second order you have to keep only one term. And in fact you can flip the t and x here and it’s not a big deal.

All finance of any kind uses this equation.

Because this is basically what you get when you have Markov processes. And in general diffusion of particles involves this.

Boundary Conditions

In general, PDEs are solved for $𝑡 \in (0, \infty) \times 𝑥 \in ℝ$ . So how does this space look? [Embedded figure omitted from the text export.]

But in finance, we are bounded by T:

If I don’t specify the boundary condition, the solution is floating, it can be infinite number of curves. I have to tie it down with the boundary T. And there’s another boundary at x = 0. You tie it down from three different places, t > 0, t < T, and x > 0. And I don’t care about anything other than when t = 0, because that’s where I am now.

[Embedded figure omitted from the text export.]

In a nutshell, to solve this, we create a domain, and we make a grid on this domain. In order for this grid to fit, we have to make squares. They solve the equation on each tiny square. Then they propagate the solution (I will show you how soon). This is nothing special, it is general and it applies to any PDE whatsoever.

Methodology

First: what is the PDE?

Let’s call $𝑉 (𝑡, 𝑆)$ for the value of an option at time t with asset price S.

We know that if the asset follows risk-neutral geometric Brownian motion, then V solves the following PDE:

\frac{𝜕 𝑉}{𝜕 𝑡} + 𝑟 𝑆 \frac{𝜕 𝑉}{𝜕 𝑆} + \frac{1}{2} 𝜎^{2} 𝑆^{2} \frac{𝜕^{2} 𝑉}{𝜕 𝑆^{2}} - 𝑟 𝑉 = 0

And this is the Black-Scholes PDE (technically, it’s Merton’s because he’s the one that did the PDE.

However this does not look like the heat equation, because the coefficients are not constant.

But you can make it constant through the use of transformation: $𝑆 = 𝑒^{𝑥}$ , which is the same as $𝑥 = ln 𝑆$

You can make the t go to T to solve the initial value as well, but this is a different way.

Then we can call

𝑉 (𝑆, 𝑡) = 𝑉 (𝑒^{𝑥}, 𝑡) = 𝑢 (𝑥, 𝑡)

\frac{𝜕 𝑉}{𝜕 𝑡} (𝑡, 𝑠) = \frac{𝜕 𝑢}{𝜕 𝑡} (𝑡, 𝑥)

Since t doesn’t do anything, this is easy.

But what is

$\frac{𝜕 𝑉}{𝜕 𝑆}$ in terms of x? I have to substitute two derivatives.

But you can do the chain rule to solve for dvdS

\frac{𝜕 𝑢}{𝜕 𝑡} + (𝑟 - \frac{𝜎^{2}}{2}) \frac{𝜕 𝑢}{𝜕 𝑥} + \frac{1}{2} 𝜎^{2} \frac{𝜕^{2} 𝑢}{𝜕 𝑥^{2}} - 𝑟 𝑢 = 0

And we’ll call r - σ^2/2 = μ for convenience.

Because this is constant coefficients, you can actually make this into a heat equation which is easy to solve. Merton solves it in a horrible way which is not like that. And you can reduce it into Black-Scholes.

But if there is no analytical solution, how do I approximate this solution? For Black-Scholes, we already have the analytical solution. We learn it as an example here, and once we learn the principle we can apply the methods for problems where there are only numerical solutions. NYHOPS does not have an analytical solution.

There’s two different ways of solving this. There is the explicit and implicit way. They are both finite difference methods. There is another way which is more appropriate to NYHOPS, but we will not cover that.

Explicit Finite Difference

First, we take our domain.

𝑡 \in [0, 𝑇)

x is different because we did this logarithm.

𝑥 = ln 𝑆 = (- \infty, \infty)

We need to discretize this domain. A square grid is the easiest, simplest way to do it. Most PDEs are solved in this way.

We substitute the derivative with the finite difference at each point on the grid. Then we can solve based on where the asset price is. In the explicit method, we use the boundary, and you move from there to the direction you want to go. And you do so explicitly (Will explain soon what that means). Depending on how many terms you have, you need t terms to get one term, in this case 3.

[Embedded figure omitted from the text export.]

Implicit

The difference in this equation, is that you won’t be able to explain each point in terms of one. You have to solve equations in general. Explicit describes the points explicitly in terms of previous points. For implicit, the value is expressed by the value implied by ALL the values on the previous points.

Differences

Generally, explicit is easier, but it may not converge. Implicit always converges.

BTW, NYHOPS works with curves and boundaries, which is called the finite element method. When you propagate these, you get inconsistencies, so if you have to keep propagating until you get convergence. You solve for the vertices of the cube. Now you have nine points available.

How

This is the actual math for doing this.

Discretize the Domain

We need to discretize t and x.

Δ 𝑡 = \frac{𝑇}{𝑛}

very simple.

Δx can be anything, with smaller being better. Is there an ideal relationship between Δx and Δt? There is, but we worry about that later.

𝑡 = (0, Δ 𝑡, 2 Δ 𝑡, \dots, 𝑛 Δ 𝑡)

𝑥 = (- 𝑁 Δ 𝑥, (- 𝑁 + 1) Δ 𝑋, \dots, 0, Δ 𝑥, \dots, 𝑁 Δ 𝑥)

So we have

2N + 1 points in x, and n + 1 points in t.

t thing doesn’t matter, x is crucial.

And we will notate the value of the function u as

𝑡_{𝑖} = 𝑖 Δ 𝑡

𝑥_{𝑖} = 𝑗 Δ 𝑥 - 𝑁

𝑢 (𝑡_{𝑖}, 𝑥_{𝑗}) = 𝑢_{𝑖 𝑗}

We need three derivatives

Explicit

This one is easy. The big difference is in how I make my derivatives.

They are calculated directly from these three points:

[Embedded figure omitted from the text export.]

Here are our equations:

\frac{𝜕 𝑢}{𝜕 𝑡} = \frac{𝑢_{𝑖 + 1, 𝑗} - 𝑢_{𝑖 𝑗}}{Δ 𝑡}

\frac{𝜕 𝑢}{𝜕 𝑥} = \frac{𝑢_{𝑖 + 1, 𝑗 + 1} - 𝑢_{𝑖 + 1, 𝑗 - 1}}{2 Δ 𝑥}

\frac{𝜕^{2} 𝑢}{𝜕 𝑥^{2}} = \frac{𝑢_{𝑖 + 1, 𝑗 + 1} - 2 𝑢_{𝑖 + 1, 𝑗} + 𝑢_{𝑖 + 1, 𝑗 - 1}}{Δ 𝑥^{2}}

To get it, you need those three points.

All of these things are now substituted in our PDE.

\frac{𝜕 𝑢}{𝜕 𝑡} + 𝜇 \frac{𝜕 𝑢}{𝜕 𝑥} + \frac{1}{2} 𝜎^{2} \frac{𝜕^{2} 𝑢}{𝜕 𝑥^{2}} - 𝑟 𝑢 = 0

Just the points with some constants like μ and σ.

Now you can determine those constants in terms of the other terms:

𝑢_{𝑖 𝑗} = Δ 𝑡 (\frac{𝜎^{2}}{2 Δ 𝑥^{2}} + \frac{𝜇}{2 Δ 𝑥}) 𝜇_{𝑖 + 1, 𝑗 + 1} + (a term) 𝑢_{𝑖 + 1, 𝑗} + another term 𝑢_{𝑖 + 1, 𝑗 - 1}

If we take those coefficient terms to be $𝑃_{𝑢}$ , $𝑃_{𝑚}$ , and $𝑃_{𝑑}$ , respectively, the probability of going up, staying the same, or going down, then it turns into a trinomial.

And it’s almost the same as the formula for the trinomial tree! The only difference is that explicit finite difference has $\frac{1}{1 + 𝑟 Δ 𝑡}$ inside the formulas, whereas the trinomial uses $\frac{1}{𝑒^{𝑟 Δ 𝑡}}$ aka discrete vs continuous. And of course as N increases we approach continuous.

Stability and Convergence

In order for this thing to converge, we must have $Δ 𝑥 \geq 𝜎 \sqrt{3 Δ 𝑡}$ which is the same condition as the trinomial tree. You can make Δx small, but not too small. Conceptually, Δx tells me how many points I have to solve for, and Δt tells me the number of steps I have to go through the tree. Because we have 3 to 1, we must have N > n. We’re working around the grid, and N tells me how many points I have. And you can see this visually, that you will get stuck if you have too many Δx compared to Δt: [Embedded figure omitted from the text export.]

If I pick $Δ 𝑥 = 𝜎 \sqrt{3 Δ 𝑡}$ .

If you calculate Δ you need two points in the origin, Γ you need three points. Sometimes that is useful.

Implicit Scheme

Fundamentally, if you understand the explicit, it’s similar.

Keep in mind the points. We’ll be using FOUR POINTS for the three derivatives.

[Embedded figure omitted from the text export.]

It’s hard to remember formulas. But it’s easier to remember how to derive them.

Now I’m going to plug them into the PDE, and then solve it. Remember that these coefficients are constants. Then you will get an equation relating these four points

𝐴 𝑢_{𝑖, 𝑗 + 1} + 𝐵 𝑢_{𝑖, 𝑗} + 𝐶 𝑢_{𝑖, 𝑗 - 1} = 𝑢_{𝑖 + 1, 𝑗}

The NUMBER ONE MOST IMPORTANT THING IS:

A, B, C are the same for all i,j. This is normal because the coefficients of the PDE don’t depend on the location or time, no t or x in it. That makes it simpler to solve!

Nonetheless, this is still an implicit solution.

A, B, C are not probabilities anymore.

We previously generalized American options by calculating the expected value of the future value of the option given that you are at that point. So you could take that point and compare what happens at exercise and store it. However, this doesn’t work, so it loses the interpretation of expected value.

There will be an exercise about this.

We have three unknowns and one equation, so we have to solve a lot of equations. We have 2N + 1 points. Because we are writing one equation for each set of three points, we will be lacking two equations, so we have 2N - 2 equations. So the system built here CANNOT be solved.

Therefore we need to come up with two more equations.

The two extra equations are coming from boundary conditions. Remember, N is supposed to be very large. And our underlying is the log of the stock, these boundaries relate to very high and very low stock value. It depends, because for a call, when a stock rises in value, the call becomes very valuable, for a put, it becomes worthless.

For Put, when $𝑆 ↑ \infty$ , then value is small. Under our property of log, it can’t be 0, although obviously we think it is.

\frac{𝜕 𝑉}{𝜕 𝑆} = 0

because it’s not going to change at all.

And then when $𝑆 ↓ 0$

\frac{𝜕 𝑉}{𝜕 𝑆} = - 1

This comes from the fact that K is a constant and derivative of S by itself is 1.

If we consider the topmost $𝑖, 𝑁$ . The difference

\frac{𝑢_{𝑖, 𝑁} - 𝑢_{𝑖, 𝑁 - 1}}{𝑒^{𝑁 Δ 𝑥 + 𝑥_{0}} - 𝑒^{(𝑁 - 1) Δ 𝑥 + 𝑥_{0}}} = 0

We should not be using u, but the corresponding Vs. We need to use the unknowns that we have, not other unknowns. So we can’t use dV/dS, we have to use the corresponding du/dS.

And we can get an extra equation from

𝑢_{𝑖, 𝑁} - 𝑢_{𝑖, 𝑁 - 1} = 0

For the bottom put, it’s

\frac{𝑢_{𝑖, - 𝑁} - 𝑢_{𝑖, - 𝑁 + 1}}{𝑒^{(- 𝑁 + 1) Δ 𝑥 + 𝑥_{0}} - 𝑒^{- 𝑁 Δ 𝑥 + 𝑥_{0}}}

and that becomes

𝑢_{𝑖, - 𝑁} - 𝑢_{𝑖, - 𝑁 + 1} = 𝜆_{𝐷}

where λ is some number.

𝑢_{𝑖, 𝑁} - 𝑢_{𝑖, 𝑁 - 1} = 𝜆_{𝑈}

𝑢_{𝑖, - 𝑁} - 𝑢_{𝑖, - 𝑁 + 1} = 𝜆_{𝐷}

And these are two extra equations that we can use to solve.

But this matrix is HUGE, 2N+1 by 2N+1, will cripple your computer. We deal with imbeciles that solve by exhaustive search, that’s how LLMs work. If you take these matrix 201x201 and ask R to solve it, it’s very fast. Then 1000x1000 is a little slower, but it can still do it. However, you can solve this with your brain, by derivation.

It is three formulas, which are easy to implement!

Since basically only the diagonal is populated, we are wasting our time doing typical matrix solving.

The idea is kinda cool. We know the matrix is solvable, because it’s invertible. We know that because the determinant exists and is not 0, easy to show.

There is a general method called the Jacobian, where you come up with some random numbers and plug them in. But there are two other methods to solve, the 5th grade ones: substitution and elimination.

And you can solve it in this way.

The first equation is

𝑢_{𝑁} 𝑢_{𝑁 - 1}

You can express the latter in terms of the former..

Then you can express $𝑢_{𝑁 - 2}$ in the same way, and keep bootstrapping to make it all functions of $𝑢_{𝑁}$ .

Then you go to the end and get

$𝑢_{𝑁 - 1}$ and $𝑢_{- 𝑁}$ and get $𝑢_{𝑁}$ from there, and then get everything else from there.

This is called in the book solving a tridiagonal system. Because you have three main diagonals, this is possible.

Technically you can solve this in general.

And this is in the book too.

We will assume matrix A [Embedded figure omitted from the text export.]

You’ll see that this works with the numbers being all different, although it’s easier in our case with the numbers being the same.

To solve this, we go through the motion:

𝑎_{11} 𝑥_{1} + 𝑎_{12} 𝑥_{2} = 𝑦_{1}

𝑥_{1} = \frac{1}{𝑎_{11}} 𝑦_{1} - \frac{𝑎_{12}}{𝑎_{11}} 𝑥_{2}

Let’s look at this structure. We have a number - another number times x_2. We’ll call those numbers C_1 and D_1.

𝑥_{1} = 𝐶_{1} + 𝐷_{1} 𝑥_{2}

Then

𝑥_{2} = \frac{𝑦_{2} - 𝑎_{21} 𝐶_{1}}{𝑎_{21} 𝐷_{1} + 𝑎_{22}} - \frac{𝑎_{23}}{𝑎_{21} 𝐷_{1} + 𝑎_{22}} 𝑥_{3}

And you can see it’s the same type of expression as x_1.

I came up with this on my own, the general idea of the derivation.

In general, each step i follows

𝑥_{𝑖} = 𝐶_{𝑖} + 𝐷_{𝑖} 𝑥_{𝑖 + 1}

where

𝐶_{𝑖} = \frac{𝑦_{𝑖} - 𝑎_{𝑖, 𝑖 - 1} 𝐶_{𝑖 - 1}}{𝑎_{𝑖, 𝑖 - 1} 𝐷_{𝑖 - 1} + 𝑎_{𝑖, 𝑖}}

and

𝐷_{𝑖} = - \frac{𝑎_{𝑖, 𝑖 + 1}}{𝑎_{𝑖, 𝑖 - 1} 𝐷_{𝑖 - 1} + 𝑎_{𝑖 𝑖}}

In general, $𝑦_{𝑖}$ is

𝑎_{𝑖 + 1, 𝑖} 𝑥_{𝑖} + 𝑎_{𝑖 + 1, 𝑖 + 1} 𝑥_{𝑖 + 1} + 𝑎_{𝑖 + 1, 𝑖 + 2} 𝑥_{𝑖 + 2} = 𝑦_{𝑖 + 1}

we can substitute the same C and D terms as we had done in 2, and turns out to be the exact same terms.

That means it’s really easy to program!

The only thing is, the end gets treated a little differently.

The last equation:

𝑎_{𝑛, 𝑛 - 1} 𝑥_{𝑛 - 1} + 𝑎_{𝑛 𝑛} 𝑥_{𝑛} = 𝑦_{𝑛}

We take that x value, and we say it’s equal to

𝑥_{𝑛 - 1} = 𝐶_{𝑛 - 1} + 𝐷_{𝑛 - 1} 𝑥_{𝑛}

It makes it easy to express in the future, because then you can move them all to the end. Once you plug that in backwards, you’re going to get x_n, so

𝑥_{𝑛} = \frac{𝑦_{𝑛} - 𝑎_{𝑛, 𝑛 - 1} 𝐶_{𝑛 - 1}}{𝑎_{𝑛, 𝑛 - 1} 𝐷_{𝑛 - 1} + 𝑎_{𝑛 𝑛}}

AND THEN I’M DONE!

## Black–Scholes Pricing for FX - URL: https://sharifhsn.dev/blog/fx-black-scholes-pricing/ - Structured data: https://sharifhsn.dev/api/posts/fx-black-scholes-pricing.json - Description: The syllabus calls this class “Introduction to Black-Scholes. BS usage for a practitioner. First and Second order risks. Xgamma. Intuitive awareness.” The source notes are the… - Date: 2025-02-24 - Exact published timestamp: 2025-02-24 - Topics: FX, Black–Scholes, Option Pricing - Categories: FX - Source: FE-635 \| Risk Engineering - Source URL: None - Content source format: typst

Week 5 — 24 February 2025

The syllabus calls this class “Introduction to Black-Scholes. BS usage for a practitioner. First and Second order risks. Xgamma. Intuitive awareness.” The source notes are the VBA workbook below, so the article preserves the implementation rather than replacing it with a summary.

Notes on inputs

(T = Days / 365)
(S = Forward) (the workbook expects forward FX)
(X = Strike)
(r = DiscRate)
(v = vol)
(ExtRate) is the foreign-currency rate.
All returned prices are in USD.

BSPrice

This is a cleaned transcription of the source workbook function. The branches cover calls, puts, forwards, and the two zero-coupon deposits recorded in the notes.

Function BSPrice(ContractType As String, _
                 Forward As Double, _
                 Strike As Double, _
                 Days As Double, _
                 DiscRate As Double, _
                 ExtRate As Double, _
                 vol As Double)
    T = Days / 365
    S = Forward
    X = Strike
    r = DiscRate
    v = vol
    OptionType = ContractType
    ' Call Option
    If OptionType = "C" Then
        BSPrice = Exp(-r * T) * (S * Application.NormSDist(dOne(S, X, T, r, v)) - X * Application.NormSDist(dTwo(S, X, T, r, v)))
    ' Put Option
    ElseIf OptionType = "P" Then
        BSPrice = Exp(-r * T) * (X * Application.NormSDist(-dTwo(S, X, T, r, v)) - S * Application.NormSDist(-dOne(S, X, T, r, v)))
    ' Fwd FX
    ElseIf OptionType = "F" Then
        BSPrice = Exp(-r * T) * (S - X)
    ' Deposit in USD (ZC in USD)
    ElseIf OptionType = "D" Then
        BSPrice = Exp(-r * T)
    ' Deposit in MXN (ZCB in MXN)
    ElseIf OptionType = "M" Then
        BSPrice = Exp(-r * T) * S
    End If
End Function

## Interest Rate Adjustments and Currency Swaps - URL: https://sharifhsn.dev/blog/advanced-derivatives-week-04/ - Structured data: https://sharifhsn.dev/api/posts/advanced-derivatives-week-04.json - Description: We may have to make some adjustments when we value interest rate derivatives. These are convexity, timing, and quant adjustments. We will see when it is appropriate to make these… - Date: 2025-02-20 - Exact published timestamp: 2025-02-20 - Topics: Fixed Income, Interest Rate Derivatives, Convexity Adjustment, Timing Adjustment, Change of Numeraire, Currency Swaps - Categories: Fixed Income - Source: Advanced Derivatives - Source URL: None - Content source format: typst

Adjustments

We may have to make some adjustments when we value interest rate derivatives. These are convexity, timing, and quant adjustments. We will see when it is appropriate to make these adjustments.

A general two-step procedure for valuing a European style derivative

First, calculate the expected payoff by assuming that the expected value of each underlying variable equals its forward value.

Then, we discount it based on the interest rate.

For non-standard interest rate derivatives, we need to modify the first step with adjustments to the forward value

If you look at the forward yields and forward prices, this kind of convexity adjustment may be necessary to make when the derivative is dependent on the yield of the bond. Basically, we define the forward yield on a bond as being the yield calculated from the forward bond price.

There is a non-linear relationship between bond yields and bond prices.

It follows that when the forward bond price equals the expected future bond price, the forward yield does not necessarily equal the expected future yield, you cannot have both.

Suppose:

$𝐵_{𝑇}$ = price of a bond at $𝑇$

$𝑦_{𝑇}$ = yield of a bond at $𝑇$

We have a relationship here that we can write as

$𝐵_{𝑇} = 𝐺 (𝑦_{𝑇})$

where G is a nonlinear function

$𝐹_{0}$ = forward bond price at time 0 for a transaction maturing at time $𝑇$

$𝑦_{0}$ = forward yield at time 0

We have a relationship where $𝐹_{0} = 𝐺 (𝑦_{0})$

Figure 1: Hand-drawn bond price and bond yield curves from the source notes.

If we make our forward equivalent to the expected future bond price, then

𝐹_{𝑇} = 𝔼 [𝐵_{𝑇}] = 𝐵_{2}

Then our yield is

𝔼 [𝑦_{𝑇}] = \frac{1}{3} \sum_{𝑖 = 1}^{3} 𝑦_{𝑖}

Such value is going to be greater than $𝑦_{2}$ .

Convexity Adjustment

Since we are incorporating convexity, we will adjust the expectation of the yield

Suppose that the payoff from a derivative at time T depends on the bond yield observed at T.

Define

$𝑦_{0}$ = forward bond yield for a contract maturing at $𝑇$

$𝑦_{𝑇}$ = bond yield at $𝑇$

$𝐵_{𝑇}$ = price of bond at $𝑇$

$𝜎_{𝑦}$ = volatility of forward bond yield

We have the relationship previously expressed that

$𝐵_{𝑇} = 𝐺 (𝑦_{𝑇})$

One thing we can do is expand $𝐺 (𝑦_{𝑇})$ in a Taylor series about $𝑦_{0}$ .

Here we can write

$𝐵_{𝑇} \approx 𝐺 (𝑦_{0}) + (𝑦_{𝑇} - 𝑦_{0}) 𝐺^{'} (𝑦_{0}) + \frac{1}{2} (𝑦_{𝑇} - 𝑦_{0})^{2} 𝐺^{″} (𝑦_{0})$

I can take the expectation of both expressions, left and right, in a world that is forward risk neutral with respect to a zero-coupon bond, maturing at T.

𝔼_{𝑇} [𝐵_{𝑇}] = 𝐺 (𝑦_{0}) + 𝔼_{𝑇} [𝑦_{𝑇} - 𝑦_{0}] 𝐺^{'} (𝑦_{0}) + \frac{1}{2} 𝔼_{𝑇} [(𝑦_{𝑇} - 𝑦_{0})^{2}] 𝐺^{″} (𝑦_{0})

What is $𝐺 (𝑦_{0})$ ? $𝐹_{0}$ , the forward bond price.

We would say the forward is the expected value. But this assumption does not work for the yield. Let us assume anyway.

$𝐹_{0} = 𝐸_{𝑇} [𝐵_{𝑇}]$

Now let’s crack this apart.

We can see that $𝑦_{𝑇} - 𝑦_{0}$ is a constant.

Also, $𝐸_{𝑇} [(𝑦_{𝑇} - 𝑦_{0})^{2}] = 𝜎_{𝑦}^{2} 𝑦_{0}^{2} 𝑇$ . Not sure why… maybe something to do with quadratic variation? I will post the proof online, using Itô’s lemma.

𝔼_{𝑇} [𝑦_{𝑇}] = 𝑦_{0} - \frac{1}{2} 𝑦_{0}^{2} 𝜎_{𝑦}^{2} 𝑇 \frac{𝐺^{″} (𝑦_{0})}{𝐺^{'} (𝑦_{0})}

This ½ term is considered the “convexity adjustment”.

It’s the difference between the expected bond yield and the forward bond yield at t=0.

Let’s assume that we have a GBM, and that we have a numeraire at time T.

Suppose that the growth rate of forward bond yield is denoted by $𝛼$ and its volatility $𝜎_{𝑦}$ . For example, we can use Itô’s lemma to calculate the process for the forward bond price.

𝑑 𝑦 = 𝛼 𝑦 𝑑 𝑡 + 𝜎_{𝑦} 𝑦 𝑑 𝑊

From Itô’s lemma we get

𝑑 [𝐺 (𝑦)] = [𝐺^{'} (𝑦) 𝑑 𝑦 + \frac{1}{2} 𝐺^{″} (𝑦) 𝜎_{𝑦}^{2} 𝑦^{2}] 𝑑 𝑡 + 𝐺^{'} (𝑦) 𝜎_{𝑦} 𝑦 𝑑 𝑊

Given that the expected growth rate of G(y) is zero,

We can find that

𝐺^{'} (𝑦) 𝛼 𝑦 + \frac{1}{2} 𝐺^{″} (𝑦) 𝜎_{𝑦}^{2} 𝑦^{2} = 0

𝛼 = - \frac{1}{2} \frac{𝐺^{″} (𝑦)}{𝐺^{'} (𝑦)} 𝜎_{𝑦}^{2} 𝑦

which results in our final convexity adjustment.

Applications for Interest Rate Derivatives

Consider an instrument that provides a cash flow at time T equal to the interest rate between T and t* applied to principal L.

The cash flow at time T is going to be equal to the principal and rate and tau

𝐿 𝑅_{𝑇} 𝜏

where $𝜏$ is $𝑇^{*} - 𝑇$ .

In order to apply this convexity adjustment, we have the relationship between price and yield. For a ZCB, G(y) is

𝐺 (𝑦) = \frac{1}{1 + 𝑦 𝜏}

From the previous equations, the expected value of such a rate is

𝔼_{𝑇} [𝑅_{𝑇}] = 𝑅_{0} - \frac{1}{2} 𝑅_{0} 𝜎_{𝑅}^{2} 𝑇 \frac{𝐺^{″} (𝑅_{0})}{𝐺^{'} (𝑅_{0})}

or, if you take the derivative of such expectations, you get

𝔼_{𝑇} [𝑅_{𝑇}] = 𝑅_{0} + \frac{𝑅_{0} 𝜎_{𝑅}^{2} 𝜏 𝑇}{1 + 𝑅_{0} 𝜏}

where $𝑅_{0}$ is the forward rate applicable between $𝑇$ and $𝑇^{*}$ .

Then the present value of the instrument is

𝑃 (0, 𝑇) 𝐿 𝜏 [𝑅_{0} + \frac{𝑅_{0} 𝜎_{𝑅}^{2} 𝜏 𝑇}{1 + 𝑅_{0} 𝜏}]

Numeric Example of Convexity Adjustment

An instrument:

payoff in 3 years equal to 1 year zero coupon rate multiplied by $1000

vol is 20%, yield curve is flat at 10%, annual compounding,

convexity adjustment is 10.9 bps.

Value of instrument is 75.95

If we have $𝑇 = 3$ , $𝑇^{*} = 4$ , $𝜏 = 1$ , $𝑅_{0} = 0.10$ , $𝜎_{𝑅} = 0.20$

Value of derivative is

1000 \times 1 \times [0.10 + \frac{{0.10}^{2} \cdot {0.20}^{2} \cdot 1 \cdot 3}{1 + 0.10 - 1}]

And you also have to discount the payoff, the yield curve is flat so it ends up being

\frac{1}{{1.10}^{3}}

Convexity Adjustment for Swap Rate

The adjustment is for the life of the swap. The derivative is on the swap observed at T, the life is from T to T + τ. What is different here? Just the reference instrument.

A direct application

Swap rate is approximated as the yield on a 12% bond.

𝐺 (𝑦) = \frac{0.12}{(1 + 𝑦)} + \frac{0.12}{(1 + 𝑦)^{2}} + \frac{1.12}{(1 + 𝑦)^{3}}

𝐺^{'} (𝑦) = - \frac{0.12}{(1 + 𝑦)^{2}} - \frac{0.24}{(1 + 𝑦)^{3}} - \frac{3.36}{(1 + 𝑦)^{4}}

Furthermore,

𝐺^{″} (𝑦) = \frac{0.24}{(1 + 𝑦)^{3}} + \frac{0.72}{(1 + 𝑦)^{4}} + \frac{13.44}{(1 + 𝑦)^{5}}

In this case, the forward yield $𝑦_{0} = 0.12$ .

G’(y_0) = −2.4018; G’‘(y_0) = 8.2546.

Then plug it into the convexity adjustment equation:

will get you 0.1236, 12.36%

Therefore the value of the instrument is

The basic procedure is to find the nonlinear function G, then take its derivatives, then plug the parameters in to this equation.

Timing Adjustment

Consider a case where a market variable v is observed at time T and its value is used to calculate the payoff that occurs later at time T*.

Define

v_T = value of v at time T

𝔼_{𝑇} [𝑣_{𝑇}]

= the expected value of $𝑣_{𝑇}$ in a world that is forward risk neutral (FRN) with respect to $𝑃 (𝑡, 𝑇)$

𝔼_{𝑇^{*}} [𝑣_{𝑇}]

= the expected value of $𝑣_{𝑇}$ in a world that is forward risk neutral (FRN) with respect to $𝑃 (𝑡, 𝑇^{*})$

𝑤 = \frac{𝑃 (𝑡, 𝑇^{*})}{𝑃 (𝑡, 𝑇)}

the ratio of the ZCB at different times, equal to the forward price of ZCB from $𝑇$ to $𝑇^{*}$ .

$𝜎_{𝑣}$ = volatility of $𝑣$

$𝜎_{𝑤}$ = volatility of $𝑤$

$𝜌_{𝑣 𝑤}$ = correlation between $𝑣$ and $𝑤$

So then what is this expectation at T*?

𝔼_{𝑇 *} [𝑣_{𝑇}]

Change of Numeraire

One way to look at this is by discussing the change of numeraire, what is the impact of that, if we’re following the market variable?

Assume that the variable is the price of a traded security f in a world where the market price of risk is $𝜆_{𝑖}$ .

𝑑 𝑓 = [𝑟 + \sum_{𝑖 = 1}^{𝑛} 𝜆_{𝑖} 𝜎_{𝑓, 𝑖}] 𝑓 𝑑 𝑡 + \sum_{𝑖 = 1}^{𝑛} 𝜎_{𝑓, 𝑖} 𝑑 𝑊_{𝑖}

When market price of risk is $𝜆_{𝑖}^{*}$

𝑑 𝑓 = [𝑟 + \sum_{𝑖 = 1}^{𝑛} 𝜆_{𝑖}^{*} 𝜎_{𝑓, 𝑖}] 𝑓 𝑑 𝑡 + \sum_{𝑖 = 1}^{𝑛} 𝜎_{𝑓, 𝑖} 𝑑 𝑊_{𝑖}

Then what is the effect of moving from the first world to the second world (star world)? The expected growth rate of price of any traded security f.

\sum_{𝑖 = 1}^{𝑛} (𝜆_{𝑖}^{*} - 𝜆_{𝑖}) 𝜎_{𝑓, 𝑖}

Define

𝑤 = \frac{ℎ}{𝑔}

Quantos

Derivatives where the payoff is defined using variables measured in one currency, and paid in another currency. Essentially this represents the exchange rate. It’s very similar to the timing adjustment.

No adjustment is necessary for vanilla swap, cap, or swaption.

Skipped Nonstandard, rest of stuff…

Currency Swaps

A swap for LIBOR in two currencies is theoretically worth 0.

However, this will have a spread in the exchange typically.

So we need to adjust for this.

There are also more complex swaps. They cannot be valued by assuming that forward rates will be realized.

LIBOR-in-arrears swap,

constant maturity swaps (CMS/CMT)

LIBOR-in-arrears

Rate is observed and paid at time T, not T*. Convexity adjustment to each forward rate underlying the swap.

This one is dependent on time when you have resets.

Figure 2: Handwritten sketch from the Week 4 source notes.

## PBMs Under Fire: What CVS's Response Leaves Out - URL: https://sharifhsn.dev/blog/linkedin-2025-02-18-pbms-under-fire-what-cvs-s-response-leaves-out/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2025-02-18-pbms-under-fire-what-cvs-s-response-leaves-out.json - Description: Trump has taken aim at pharmacy benefit managers, and the response from the industry seems to be… duck and cover. Prem Shah, executive VP at CVS, responded to skepticism on… - Date: 2025-02-18 - Exact published timestamp: 2025-02-18T16:47:24.726Z - Topics: Healthcare, Public Policy - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/feed/update/urn:li:activity:7297657973143666688/ - Content source format: typst

Trump has taken aim at pharmacy benefit managers, and the response from the industry seems to be… duck and cover. Prem Shah, executive VP at CVS, responded to skepticism on #Bloomberg today.

Pharmacy benefit managers (PBMs) act as a middleman between purchasers of prescription medicine, like health insurers. They negotiate directly with drug manufacturers and maintain drug formularies (the list of medicines covered by insurance). Theoretically, they’re able to consolidate these services for a large number of purchasers and save money that way.

That’s certainly what Prem Shah says. He repeated the claim found on the industry website that PBMs are responsible for over $1000 in savings per person per year. [1]

But that is decidedly at odds with the recent FTC report on PBMs, which found that they marked up cancer and HIV-related drugs by thousands of percent, resulting in $7.3 billion in additional revenue for the “Big 3” (CVS, ESI, and OptumRx). A previous report found that the percentage of dispensing revenue for specialty drugs like these going to PBMs increased from 54% to 68% between 2016 and 2023. [2]

Shah says that he hopes he can have a dialogue with the Trump administration regarding the value that PBMs provide. Time will tell how successful his efforts will be.

What do you think? Should these middlemen continue to exist?

[1] https://lnkd.in/eUfh9PPN [2] https://lnkd.in/eyNTjfq7

## Trinomial Trees, Dividends, and Greeks - URL: https://sharifhsn.dev/blog/computational-methods-week-04/ - Structured data: https://sharifhsn.dev/api/posts/computational-methods-week-04.json - Description: I posted an assignment, this assignment is due this week on Sunday. Let’s just talk about the homework. This covers material from first two lectures. This lecture and previous… - Date: 2025-02-18 - Exact published timestamp: 2025-02-18 - Topics: Computational Methods, Trinomial Trees, Dividends, Greeks, Finite Differences - Categories: Computational Methods - Source: Computational Methods in Quantitative Finance - Source URL: None - Content source format: typst

Homework

I posted an assignment, this assignment is due this week on Sunday. Let’s just talk about the homework.

This covers material from first two lectures.

This lecture and previous lecture (trees) will be HW 2.

Q: “Implement the Newton -whatever method, do we have to implement them alL?” A: You can implement all three for bonus

Q: Put, vs call option? A: Use both

Q: When you see equity data in the first part, what should we get? Tick-level data, just the price?

A: You need price of the underlying. You don’t need for daily or anything, the most important option derivative is $Δ$ and $Γ$ . Basically, you can calculate the price of the option if you know the price of the stock. You really needed to read the price of the stock, just get that price at the moment you download the option data.

Q: Do we have to include code in the appendix? A: You don’t need an appendix, but if you use Python, you should include the py file or Jupyter notebook, some way of checking that?

Q: How many strike prices do we need? A: You should use whatever makes sense to you. Normally, let’s use VIX as an example. That’s calculated from options on S&P 500. They literally take all of the options, because there’s tons of them, until they get to the two that have had zero trades in the last five minutes or so. That’s the rules. They basically get to options that haven’t been traded in a long time, so their prices are not current. Look at the most traded options, and don’t include options that don’t have a lot of volume. Just go enough in the tails to make sense for the data. What I want you to see in the option data and detail in the submission is the volatility smirk. I want to see some kind of decrease with strike price, and what happens when you look at next month volatility vs three months, then plot them all and see what you observe. Then it’s possible that the day you downloaded the data you don’t observe that, and that’s okay! The point is to explain what you get from the data.

Forgot to Discuss Last Time

When we calculate this option data, we said well, you basically step down in the tree, and then at every step you calculate the discounted expected value based on the nodes further on in the tree.

In Romania we have to do a thesis after Bachelor’s degree. In 1995, at the time, something just appeared. Before you type with a typewriter or write nicely with a pen. Computers had appeared, so Windows 3.1 is what we were using. It had a Word document processor. I was saving the files on the floppy disk. I was using the school computer. After I wrote half of my thesis, the file just crashed. So I couldn’t really open it. I was writing it with a pen, and I was typing it, so it was horrible, and I was about to quit school. And then a friend of his said that he was working at some company and said “I don’t know anything”. He was in charge of the Bucharest water system and routing teams to fix water when it breaks. You may not realize, but there’s a lot of emergencies that happen in a big city. He said that because I’m not doing anything else, I’ll type your thesis. And because of this experience, I have to save things. You guys are spoiled, everything is saved online

Combinatorial Formula

There’s not really a big deal about this.

We discussed trees and how they are constructed. The tree represents the volution of the stock.

[Embedded diagram omitted from the text export.]

You have to go down at the first step, then go all the way up. Or go down at the second step, then go up the rest. Or go down at third step after happening.

You can think about it in a combinatorial way, where it’s

(\binom{(3}{1)})

All of these paths have the same probability.

The probability that you end up in this particular node over here will be

(\binom{(3}{1) * 𝑝_{𝑢}^{2} * 𝑝_{𝑑}^{1}})

And this is exactly the binomial probability.

In general, if you look at n steps, at the top value, you know what the value is. Let’s say it’s an additive tree.

[Embedded diagram omitted from the text export.]

Because of this, you can calculate the payoff just by using the top price

(\binom{𝜑 (𝑒^{𝑥_{0} + 𝑛 Δ 𝑥_{𝑢}}) \times (𝑛}{0) 𝑝_{𝑢}^{𝑛}})

This is the probability of the one node.

My payoff is

𝜑 (𝑆_{𝑇})

, but it’s

𝔼 [𝜑 (𝑆_{𝑇}) | {ℱ︀}_{0}]

Conditioned on time 0.

So my expectation is payoff times probability.

(\binom{= \sum_{𝑖 = 0}^{𝑛} 𝜑 (𝑒^{𝑥_{0} + 𝑖 Δ 𝑥_{𝑢}}) \times (𝑛}{𝑖) 𝑝_{𝑢}^{𝑖} 𝑝_{𝑑}^{𝑛 - 𝑖}})

You are creating discrete paths for your process, which matches, in the limit, as n approaches infinity, the path of the general process.

That’s the combinatorics formula.

This formula doesn’t work if you price American options because they’re path-dependent. So I don’t know if I will end up with that final payoff, we have to calculate every step of the way.

Greeks

$𝛿$

𝛿 = \frac{𝜕 𝐶}{𝜕 𝑆}

The partial derivative of the call option with respect to the stock price, its sensitivity.

You’re supposed to calculate the change in the option price when the current price changes.

What most of the interviews will say, is to do the tree.

[Embedded diagram omitted from the text export.]

You would take the distance, the difference in actual values. When you step back in the tree, you get the call value, and you will actually calculate all these values. You can then approximate this $Δ$ as

\frac{𝑐 (𝑆_{0} 𝑢) - 𝑐 (𝑆_{0} 𝑑)}{𝑆_{0} 𝑢 - 𝑆_{0} 𝑑}

This is not a very good approximation because you’re not calculating the $Δ$ now, you’re calculating it in the future.

In the interview, you will construct the three step tree, and you’ll have the number and differences, and very easy to calculate.

This is not correct though. It’s supposed to be some value with very little difference in the stock.

So you should calculate 2 trees!

One starts from $𝑆_{0}$ , and one starts from $𝑆_{0} + Δ 𝑆$ , and $𝑆_{0} + Δ 𝑆$ . This will allow you to calculate the first and second derivative. You need the third point for the $Γ$ (second derivative).

Understand the idea! Don’t just apply what you read. Why am I doing it this way? There is no mystery to this, like math, it has to be logical. If it’s not logical, then it’s wrong.

This is a method that is very useful if you write it on a piece of paper. If you have a computer with you, a call using a tree takes a fraction of a second. So why wouldn’t I do two in a fraction of a second? It’s pretty easy to do the actual calculation, but this was not possible many years ago.

All the other derivatives are the same idea. All you do is vary the underlying derivative slightly, and hold everything constant.

Some are them are irrelevant, but you can!

That’s the one thing to mention from last class.

Trinomial Tree

It’s just the same thing as the binomial tree. There is literally no difference.

First of all, I want you to remember how we did the binomial tree.

𝔼 [Δ 𝑅^{disc}] = 𝔼 [Δ 𝑅^{cont}]

And same thing for variance.

And we did some other stuff to make sure that the probability distribution is correct. So we will use the same exact idea. Remember that from the continuous process $𝑑 𝑅_{𝑡}$ is the logarithm of the stock price.

𝑑 𝑅_{𝑡} = (𝑟 - \frac{𝜎^{2}}{2}) 𝑑 𝑡 + 𝜎 𝑑 𝑊_{𝑡}

𝔼 [Δ 𝑅_{𝑡}] = (𝑟 - \frac{𝜎^{2}}{2}) Δ 𝑡

𝕍 [Δ 𝑅_{𝑡}] = 𝜎^{2} Δ 𝑡

𝔼 [Δ 𝑅_{𝑡}^{2}] = 𝜎^{2} Δ 𝑡 + (𝑟 - \frac{𝜎^{2}}{2})^{2} Δ 𝑡^{2}

How do we do this?

Two conditions:

Has to converge
Has to be recombining

Same number of nodes, two above, and two below. $2 𝑛 + 1$ is still a linear increase.

[Embedded diagram omitted from the text export.]

That would be convenient.

If I did like this

[Embedded diagram omitted from the text export.]

It would be O(n^2), really bad.

For the recombining one, it would be

It turns out the condition you need to have is

Δ 𝑅_{𝑢} + Δ 𝑅_{𝑑} = 2 Δ 𝑅_{𝑚}

You can solve the whole thing with this. The problem is that you’re going to get a huge system. Six unknowns, and three equations. No matter what, we have three equations. One with the expectation, one with the variance, and probabilities must equal to 1.

The six unknowns is the three ps and three Rs.

This will have a huge number of solutions that will give us a trinomial tree. We won’t do that.

We’re going to say that this is too complicate,d so we will make on etree.

We will take $Δ 𝑅_{𝑢} = Δ 𝑅$ , $Δ 𝑅_{𝑑} = - Δ 𝑅$ , $Δ 𝑅_{𝑚} = 0$ .

[Embedded diagram omitted from the text export.]

Now we have four unknowns, and three equations, so still undetermined. But what’s going to happen is we will write the equations.

Calhoun Sidebar

some ex-students want to form a business, get help from students.

Δ 𝑅 𝑝_{𝑢} - Δ 𝑅 𝑝_{𝑑} = (𝑟 - \frac{𝜎^{2}}{2}) Δ 𝑡

And same thing for

Δ 𝑅^{2} 𝑝_{𝑢} + Δ 𝑅^{2} 𝑝_{𝑑} = 𝜎^{2} Δ 𝑡 + (𝑟 - \frac{𝜎^{2}}{2})^{2} Δ 𝑡^{2}

(here is supposed to be $- Δ 𝑅$ , but squared is he same thing)

𝑝_{𝑢} + 𝑝_{𝑚} + 𝑝_{𝑑} = 1

I have a choice of how large my tree should step. Technically, you can solve for everything(not really)

If you look at the probabilities, you can calculate expected value in terms of the parameters. Once you have the formula,

𝑝_{𝑢} = \frac{1}{2} (\frac{𝜎^{2} Δ 𝑡^{2} + (𝑟 - \frac{𝜎^{2}}{2})^{2} Δ 𝑡}{Δ 𝑅^{2}} + \frac{(𝑟 - \frac{𝜎^{2}}{2}) Δ 𝑡}{Δ 𝑅})

This probability has to be a number between 0 and 1. And you can guarantee this with a certain condition. In order for this to be true, we need a sufficient condition that

Δ 𝑅 \geq 𝜎 \sqrt{3 Δ 𝑡}

This is not easy to get, and it is sufficient, not necessary, so you can get tighter guarantees.

It’s important to understand what’s happening here fundamentally.

[Embedded diagram omitted from the text export.]

The more steps I add, that’s what $Δ 𝑡$ is. But I can’t make the increments too small, it must have some kind of spread. When you’re guaranteeing a minimum of spread, it captures the distribution, without that you don’t get the distribution. This formula is calculated numerically, but this is the intuition.

Order of Convergence

Order of convergence of the trinomial tree is $𝑂 (Δ 𝑅^{2} + Δ 𝑡)$ . What does that mean? As $Δ 𝑡$ goes to 0, $Δ 𝑅$ also has to go to 0 at the same rate. If I take $Δ 𝑡$ as 0.01, then I need $Δ 𝑅$ to be greater than 0.1 based on that formula above. If I have $Δ 𝑅 = 0.5$ , to square it 0.25, it will get too close. So I actually have to take them in the same convergence. And the best $Δ 𝑅$ uses this formula, so it’s just EQUAL to that square root. This makes sure that the tree has the fastest convergence.

Now the tree is unique, you calculate $Δ 𝑅$ , you get the probabilities, and it’s a unique tree that satisfies these assumptions. Professor, you didn’t explain to us what big O is.

This means that the call option value from the tree, minus the true value (unknown), the absolute value of that should be approximately $𝐶 (Δ 𝑅^{2} + Δ 𝑡)$ , where $𝐶$ is a constant.

The constant screws me up, nobody knows what it is. Even if you make your tree within one cents, if there’s a large constant it’s worthless. You can’t get exact numbers.

This is for the trinomial tree. If we do this same thing for binomial, then

$| Option (binomial) - True Value | \approx 𝑂 (Δ 𝑅^{2} + Δ 𝑡)$

So you wonder, why should I use a trinomial tree, which is simpler, gives me the same order of convergence? The answer is that there is no reason, other than path-dependent options.

If you are path dependent, then it’s sometimes better, because you get 3^n instead of 2^n, more paths to evaluate.

Exercise

How many total nodes are in a trinomial tree with n steps?

Dividends

Most stocks give you discrete dividends. This quarter gave us profit, that we will distribute to our shareholders. We get 20 cents from a $300 share of stock. Typically what happens is that it’s discrete cash, it gets automatically in your account, of this particular date of this particular year, every shareholder.

Mathematically, it’s very difficult. We will explain how to deal with cash dividends.

The simplest dividend is continuously paid dividends.

SPY Sidebar

(SPY is an ETF, which is done on multiple stocks, which each pay dividends. There are two ETFs, State Street and Vanguard, both of which track the same index. Because they own the stock, they will receive dividends. If they kept the dividends, they would have to pay taxes, which is very complicated. So what they do is distribute the dividends to the ETF shareholders. They calculate. Remember, the weights depend. The way ETF works, is that I own a basket of stocks. The basket is based on (there’s different types, but) in this case it’s market cap. You look to the top 500 assets which are in S&P 500. Then you look at the largest, say AAPL. The largest component accounts for say 2% of the entirety of my stocks. Market capitalization value for AAPL is $2.5B. The total value of my assets is $50B. If you divide, it gives you 2%. That is my proportion of shares that I will hold in my basket. The reason why they do this is because it’s very easy to calculate the return of the ETF by summing the returns of the components. Now what happens is, obviously the cap value of the company changes, every day, from $200 to $201. If you keep everything the same, the market capitalization value for AAPL grew. Proportionally then, the S&P 500 portfolio should hold more of AAPL. Obviously that’s not possible (maybe possible with new methodology), but not possible with old. Every day you have to change shares in the big basket. They do this in the middle of the quarter. When the imbecile decided to buy Twitter and call it X. THAT DUDE IS FROM SOUTH AFRICA, A WHITE DUDE FROM SOUTH AFRICA, WHAT CAN HE BE? THAT’S KIND OF RIDICULOUS, RIGHT? When that guy bought Twitter, he bought every share available in the market. The company is automatically withdrawn, it’s a private company one. It was listed on the S&P 500 and put another company, and then recalculate all of the shares. The whole thing is really simple. Every instrument on the equity market is trivial, they make it sound complicated. What’s a “basis point”? It’s a goddamn percentage.)

They take all the dividends, and scale them in such a way that it’s distributed to the owners of the SPY. It’s distributed when they do the rebalancing. When you open an account with any company (Fidelity, Ameritrade, eTrade) you give them your money, but they don’t put your money in your account. THey put it in some fund, which is their own managed fund. And actually, if you open an account, Fidelity ETF that they keep your money in, returns more money than your bank account. So you should buy Fidelity ETF instead of a bank account. Only difference is that if Fidelity goes under, you’re out of luck. Bank accounts get $250K FDIC coverage.

Continuing…

The value of the stock depreciates.

𝑑 𝑆_{𝑡} = (𝑟 - 𝛿) 𝑆_{𝑡} 𝑑 𝑡 + 𝜎 𝑆_{𝑡} 𝑑 𝑊_{𝑡}

The only difference between this and normal is that we have $𝛿$ , which is a known quantity.

We do the same basic constructions with $𝑟$ replaced by $𝑟 - 𝛿$ . This is not a big deal. Your function will take an input $𝑟$ , just replace that with $𝑟 - 𝛿$ .

Known proportional dividend

Let’s say $\hat{𝛿}$ is a proportion.

Elon Rant

Let’s say TSLA, because the owner is an imbecile. I never liked this guy. I played games since I got my first computer when I was 12. They made me think. One of my favorite games is called Path of Exile, which I played since it started 15 years ago. This idiot put out a video where he claims that he’s rank #5 in the hardcore version of the game. You don’t understand, I played this game for three years before I understood what was going on. It’s really complicated, it’s 10 years old and every 3 months the devs add something new. There’s so much stuff you can do, in ten different ways. Not only that, but you have to literally play it nonstop to be in the top rank. The company is called Grinding Gear game, they created this game after Diablo. Blizzard, the company that issued it, refused to support it, so these three people in NZ made a Diablo-like game. This imbecile puts out a video showing how he managed to be #5 on the hardcore ladder. Now hardcore means that if you die in the game one time, your character gets deleted. It’s horrible, basically, because if you make one mistake you die, and I’m going to break my computer. This guy basically opens the game and has a tab that says Elon’s Maps. This **** person has obviously paid another dude to play for him, from India or Romania or whatever, and that guy slaved for this guy, for money, clearly, for three weeks nonstop to become the top. And this guy just claims to be that. Why? What’s the point? And obviously people like me who have played for the longest time know that he has no idea what he’s doing. It’s pretty obvious, so what’s the point of doing this. You want to pretend you’re the smartest. I also like chess, I’ve been playing it for a long time. This dude said he was top two in his chess club. He always says top two, because if he said top, someone would say I remember the top.

Continuing

The company value is the total sum of the assets. When you are becoming a public company, you are selling the shares, and if you add the shares together, that’s the total number of assets. (You may also have debt, a liability), but otherwise that’s the value of your company. When you pay someone money (dividends), the value of your company decreases

This dividend is paid at time $𝜏$ . The value immediately becomes

𝑆_{𝜏} - \hat{𝛿} 𝑆_{𝜏}

Now what happens here, and why it’s a special case, is because this is

𝑆_{𝜏} (1 - \hat{𝛿})

And if you remember the multiplicative tree, it goes from

[Embedded diagram omitted from the text export.]

So you can just price this directly.

Then you look at the dividend payment at time $𝜏$ between 0 and $𝑇$ , and construct the tree. Then you drop at time $𝜏$ by this value, $𝑆$ to

𝑆 (1 - \hat{𝛿})

, and then keep computing the tree.

Now you have a tree that’s kind of broken, like someone hit it with a bat.

[Embedded diagram omitted from the text export.]

What about American options? In order to do this, you do the same thing as for the other tree, but you look what’s the value coming from the tree after τ, and compare it with before τ. If you have an option, you are not paid the dividend. There are going to be a bunch of places where you have to exercise. What’s happening here? I forgot to mention this, and this is a mistake in the book, too. Look at the value

(𝑝_{𝑢} 𝐶_{𝑢} + 𝑝_{𝑑} 𝐶_{𝑑}) 𝑒^{- 𝑟 Δ 𝑡}

When you look back at the Combinatorial Formula, you need to multiply by the discount factor.

But when you exercise, this is the expected value of the option in the future. When you compare here, you’re going to take the maximum of this value, and as if you had exercised right before τ, which is

(𝑆_{𝜏} - 𝐾)_{+}

If the dividend is sufficiently large, the optimal time to exercise is right before the dividend payment. Except that, this is a fake example, there isn’t such a thing.

Discretely paid cash dividends

The most complicated, the most common case.

We’ll say D is a quantity of cash that is constant regardless of the price of the stock. We’ll assume it’s known. You can’t do anything if the dividend is not known.

If you use the same idea, and you look at some time $𝑡_{𝑖}$ , and $𝑡_{𝑖 + 1}$ , and $𝜏$ is in between. The problem with this is that I have an $𝑆_{𝑢}$ and an $𝑆_{𝑑}$ . Let’s say I pay the dividend. The value of the stock in between these two is going to drop by $𝐷$ units. If we do like we did before, and drop $𝑆_{𝑢}$ by $𝐷$ and $𝑆_{𝑑}$ by $𝐷$ , and look at the next node, we have the lost the recombination property.

[Embedded diagram omitted from the text export.]

We have this tree that recombines all nice until time τ, then it becomes unmanageable. That’s the problem here. How do we fix this? The trick is kinda interesting. It’s a pretty complicated construction which is in the book. The idea is that we’re going to create a new tree. This new tree will follow

\tilde{𝑆_{𝑡}} = {\begin{cases} 𝑆_{𝑡} & 𝑡 > 𝜏 \\ 𝑆_{𝑡} - 𝐷 𝑒^{𝑟 (𝜏 - 𝑡)} & 𝑡 \leq 𝜏 \end{cases}

Normally, you drop every time $𝜏$ by the dividend. Now I drop every node below before the dividend, discounted by the time back. What’s happening here is you make all these trees before τ, nonre-combining. Then starting from that point, every point is recombining. This is better because it’s easier to handle exponential Big O at the beginning, where the numbers are smaller. The second thing that’s really smart about this is that if you have observed real markets, this is how real markets operate. Let’s say I own AAPL stock, and AAPl says that stock is $100 today, and then they say due to our quarterly earnings we will have 10¢ dividend per share on March 15, then the price will go down by exactly that amount. This is the present value of the dividend. But in the real world, there is no real price, there is a bid and ask spread. If you’re participating in HFT, every single winning team was running a market making strategy, from providing liquidity to the market.

If you’re doing a long-dated option with dividends, you can actually treat it as continuously paying.

Everything that we learn in class is not IRL.

Tree for Deterministic Time Varying Volatility

The stock price doesn’t follow geometric Brownian motion as Black-Scholes assumes. This is well-known.

We take $𝑅_{𝑡}$ , under the observed price.

𝑑 𝑅_{𝑡} = 𝜇 - \frac{𝜎^{2}}{2}) 𝑑 𝑡 + 𝜎 𝑑 𝑊_{𝑡}

We can take some daily time intervals, $𝑡_{1}, 𝑡_{2}, \dots, 𝑡_{𝑛}$ .

Then we take

log 𝑆_{𝑡} - log 𝑆_{𝑡 - Δ 𝑡}

Which is the literal continuously compounded return, which should be equal to this quantity in the formula.

That quantity is very simple, because the stochastic part is normal, then you add a constant.

Then your returns, say

𝑟_{𝑖} = log 𝑆_{𝑖} / 𝑠_{𝑖 - 1}

$𝑟_{𝑖}, \dots, 𝑟_{𝑛}$

These should be normal, you can test with a QQ plot. But it’s not normal, it is leptokurtic, it has fat tails.

You can try a different variability. The simplest thing to do is $𝜎 \sqrt{𝑡}$ . You will look at two different data points. It will still be normal, but we will make the variance

𝜎^{2} (𝑡 + Δ 𝑡) - 𝜎^{2} (𝑡)

What’s happening here is that some of the observations are coming from one variability, and others are coming from another. If this is possible, then it’s possible to be leptokurtic, combining $𝑁 (0, 1)$ and $𝑁 (0, 2)$ . That is one way in which you can make this leptokurtic distribution appear.

Having said that, the thing about this is the resulting process is not truly random. What happens is, and there’s a formula of this, if you price something from 0 to T, and you have the $𝜎$ which is a function of time, it is exactly the same as if you are pricing under GBM with a variability equal to the average (integral) of the function, this proof is in Shreve.

My Research

Quadrinomial tree. Tree with four nodes.

Somehow, somebody has estimated $𝜎 (𝑡)$ and $𝑟 (𝑡)$ , risk-free interest rate changes in time, and standard deviations changes in time. Take $Δ 𝑡 = 𝑇 / 𝑛$ . At times $𝑖 Δ 𝑡$ , we take the volatility values

𝜎_{𝑖} = 𝜎 (𝑖 Δ 𝑡)

and

𝑟_{𝑖} = 𝑟 (𝑖 Δ 𝑡)

and we will also define

𝜇_{𝑖} = 𝑟_{𝑖} - \frac{𝜎_{𝑖}^{2}}{2}

We take fixed $Δ 𝑋_{𝑢}$ and $Δ 𝑋_{𝑑}$ . We will do an additive tree. They are fixed for every $𝑡$ so that the tree is recombining. When I solve my tree, initially, for binomial and trinomial, on the right hand side, we calculate the value coming from the continuous process, which was the $(𝑟 - 𝜎^{2})$ thing, same for every $Δ 𝑡$ . Now it’s different because the $𝑟$ and $𝜎$ are changing, and at that particular time I have to match a different value.

So now the intervals have to be changing, because I don’t want my X to change.

[Embedded diagram omitted from the text export.]

You can see that the probabilities are different. And actually at each step we will assume that the probabilities are the same within the stpe. If you don’t ,then that’s stochastic volatility. For convenience, we will define

𝑝_{𝑢}^{𝑖} = 𝑝^{𝑖}, 𝑝_{𝑑}^{𝑖} = 1 - 𝑝^{𝑖}

I also don’t want to deal with $Δ 𝑋$ like this, so it will be the same $Δ 𝑋$ and $- Δ 𝑥$ .

So now we have our equations

𝑝_{𝑖} Δ 𝑥 - (1 - 𝑝_{𝑖}) Δ 𝑋

On the right hand side, it’s supposed to be the integral of an interval, and technically you can pick any point, but the left point will give us the stochastic process.

𝑝_{𝑖} Δ 𝑋 - (1 - 𝑝_{𝑖}) Δ 𝑋 = 𝜇_{𝑖} Δ 𝑡

𝑝_{𝑖} Δ 𝑋^{2} + (1 - 𝑝_{𝑖}) Δ 𝑋^{2} = 𝜎_{𝑖}^{2} Δ 𝑡 + 𝜇_{𝑖}^{2} Δ 𝑡^{2}

The system has a ton of unknowns, because it’s for every i. So there are actually 2n such unknowns.

How do we solve it? I’m going to do a thing. We’re going to express

𝑝_{𝑖} = \frac{1}{2} + \frac{𝜇_{𝑖} Δ 𝑡}{2 Δ 𝑋}

You can extract the $Δ 𝑋$ directly to get

Δ 𝑋^{2} = 𝜎_{𝑖}^{2} Δ 𝑡 + 𝜇_{𝑖}^{2} Δ 𝑡^{2}

We have a problem. You can see here that the $𝜇$ here changes at every $𝑖$ , but $Δ 𝑋$ can’t change. What you do, is you change this, this model doesn’t work.

The next step is to make every step change, and then take an average to get $Δ 𝑋$

Δ 𝑡_{𝑖}

Then

\overset{̅}{Δ 𝑡} = 𝑇 / 𝑛

and

Δ 𝑋 = \sqrt{\bar{𝜎^{2}} \overset{̅}{Δ 𝑡} + \bar{𝜇^{2}} {\overset{̅}{Δ 𝑡}}^{2}}

The point of these bars is the average.

\bar{𝜎^{2}} = \frac{1}{𝑛} \sum_{𝑖 = 1}^{𝑛} 𝜎_{𝑖}^{2}

etc.

You take these averages and put them here. The constructions can get complicated. But if you take the binomial tree with these values, you get a pretty good approximation for the moving volatility and the moving interest rate.

It’s given to you that $𝜎 (𝑡) = 0.4 𝑡 + sin (𝑡 - 𝜋)$ , or whatever. It’s deterministic.

The most complicated part

The problem is that none of these trees solve the Heston model (e.g.)

Heston:

\frac{𝑑 𝑆_{𝑡}}{𝑆_{𝑡}} = 𝑟 𝑑 𝑡 + \sqrt{𝑦_{𝑡}} 𝑑 𝑊_{𝑡}

where

𝑑 𝑦_{𝑡} = 𝛼 (\bar{𝑦} - 𝑦_{𝑡}) 𝑑 𝑡 + 𝜎 \sqrt{𝑦_{𝑡}} 𝑑 𝑍_{𝑡}

First of all, the construction I have applies to these types of stochastic processes. There are two conditions in the original paper. It can be any general function, for this construction to work, by the way.

The stock price must look like this, where it’s $𝑑 𝑆_{𝑡} / 𝑆_{𝑡}$ , so when you apply logarithm it becomes explicit, just $𝑦_{𝑡}$ .

The Brownian motions must also be independent. I had one student that worked on this, and I’m pretty sure it’s solvable, but he quit to go work at Millennium.

The question is, how do you construct a tree for this process?

It’s similar. The tree has to be recombining, and match the stochastic process. How it goes, the spread… Remember the condition $Δ 𝑡 > 3 𝜎 \sqrt{𝑡}$ . It’s commensurate with σ. The spread is captured by $\sqrt{𝑦_{𝑡}}$ . The problem is the variability, like we saw in the deterministic case, it changes. But unlike the deterministic case where we know, there’s 52 weeks in a year, we can plug 1/52 into this bar bar expression. But the problem is that $𝑍_{𝑡}$ is random, so we can’t get the complete thing. We can only get the distribution, and we can’t do the tree because we can’t match it.

So the idea that I had, was the following. I’m going to forget about dY_t, and work with the distribution of y_t, and at any moment in time t. I’m going to work at the distribution now. I can observe price, but I can’t observe volatility. Implied vol is a surface. You can get these values which will give you some stupid wrong distribution, but you can think of those values as being a histogram. Then I can simplify it and say it takes values y_1, y_2, and y_3.

[Embedded diagram omitted from the text export.]

The idea is to have a two-dimensional tree, which R on one axis and y on another.

From tis one point, I can form multiple different trees. Each point is a $𝜎$ . The R is a vertex, and I’m moving to four different points. I could move to a different four points for each $𝜎$ . For each one of these sets of four points, I can move to a different four on a different plane. It’s kind of like a huge pyramid. There’s a two dimensional pyramid, where you have values with a lot of facets.

Obviously this is impossible to calculate because it’s non-recombining.

Instead I will take a slice of $𝜎$ values and do a Monte Carlo, some $𝜎_{1}$ at $𝑡_{1}$ , $𝜎_{2}$ at $𝑡_{2}$ kinda thing. At each sample, I take a different tree. It’s kinda like a combination of Monte Carlo and binomial tree. I spent a year figuring out that binomial tree works, then sent it to a university and realized that there was a mistake and got 0 = 0. Then in 2015, they put it online. The first paper I published is wrong.

This is very efficient for pricing.

Section 6.11 in the book if you’re interested.

## Bond Options, Caps, Floors, and Swaptions - URL: https://sharifhsn.dev/blog/advanced-derivatives-week-03/ - Structured data: https://sharifhsn.dev/api/posts/advanced-derivatives-week-03.json - Description: We’re going to look at bond options, caps and floors, and everything about swaptions. Depending on time, we may go forward. Basically these are instruments whose payoff is… - Date: 2025-02-13 - Exact published timestamp: 2025-02-13 - Topics: Fixed Income, Interest Rate Derivatives, Bond Options, Black Model, Caps and Floors, Swaptions - Categories: Fixed Income - Source: Advanced Derivatives - Source URL: None - Content source format: typst

Interest Rate Derivatives

We’re going to look at bond options, caps and floors, and everything about swaptions. Depending on time, we may go forward.

Basically these are instruments whose payoff is dependent on the level of the interest rates. Generally, interest rates are more difficult to value than equity, or fixed derivatives.

One reason is that the behavior of one individual interest rate is more complicated than that of a stock price. For example, there is some mean reversion behavior. Another reason is that the valuation of many products needs to develop that describes the behavior of the entire zero coupon yield curve. Another reason is that the volatility is different at different maturities on this curve. Typically, it’s not a good assumption that volatility is the same for all maturities. Another characteristic is that interest rates are used for discounting the payoff and also for defining the payoff.

Bond Options

First of all, we’re going to discuss bond options.

A bond option is an option to buy or sell a particular bond by a particular date for a particular price. Typically, these bond options are traded over the counter, and they are typically embedded in bonds when they’re issued, to make them more attractive.

For example, callable bonds contain provisions that allow the issuing firm to buy back the bond at a pre-determined price at a certain time in the future. Usually they cannot be called for a few years, and the value of the call option is reflected in the yields on the bond.

Another example is the puttable bond, which contains provisions that allow the holder to demand early redemption at a pre-determined price in the future.

Obviously, such embedded put option will increase the value of the bond, these bonds tend to have a lower yield than a bond without such embedded option.

Other examples of instruments with embedded options: loans and deposits. The 5-year fixed-rate deposit issued by some institutions (banks) can be redeemed without penalty at any time, this is an American put option. You can see this deposit instrument as a bond.

Another example might be a bank loan with 5% per annum quote good for the next 2 months. By having this, you have the right to exercise over the next 2 months.

Basically there are many OTC bond options, and some embedded options are European. For this particular lecture, we will assume that the forward bond price has a constant volatility $𝜎_{𝐵}$ . This will allow us to use Black’s model.

There are two options for pricing interest rate options. One is to use a variant of Black’s model. Another is to use a no-arbitrage yield curve based model. Here we’re going to look at pricing such option using Black’s model. Such model will assume that the value of an interest rate, a bond price, or some other variable at time T in the future will have a log normal distribution.

Black’s Model Revisited

We will revisit Black’s model.

Consider a European call on an asset with strike K and maturity T. Define P(t, T) as the price at time t of a zero-coupon bond expiring at time T paying $1.

Sidebar: A martingale is a zero-drift stochastic process. In general, we can say that a variable $𝜃$ follows a martingale if $𝑑 𝜃 = 𝜎 𝑑 𝑊$ where $𝑑 𝑊$ is the Wiener process. In this case, $𝜎$ may be stochastic.

Then you have the martingale property defined as

𝔼 [𝜃_{𝑇}] = 𝜃_{0}

We can obtain an equivalent martingale result. Assume that $𝑓$ and $𝑔$ are prices of traded securities. We will assume that these prices are dependent on a single source of uncertainty. Furthermore, let’s define the ratio $𝜙 = 𝑓 / 𝑔$ . In this case, we’re going to call $𝑔$ the numeraire. $𝜙$ will be the relative price of $𝑓$ with respect to $𝑔$ , the numeraire. We are expressing $𝑓$ in terms of units of $𝑔$ . That’s why the security price of $𝑔$ is the numeraire.

Let’s assume also that we have volatilities for f and g. Assume the volatilities $𝜎_{𝑓}$ and $𝜎_{𝑔}$ in a world where the market price of risk is $𝜎_{𝑔}$ . We can see that the market price of risk is the volatility of g, then the ratio f/g is a martingale for all security prices.

Here we’re going to try to review this result.

We can consider the process f, and take df as

𝑑 𝑓 = (𝑟 + 𝜎_{𝑔} 𝜎_{𝑓}) 𝑑 𝑡 + 𝜎_{𝑓} 𝑓 𝑑 𝑊

In general, the process followed by the derivative $𝑓$ can be rewritten as $𝑑 𝑓 = 𝜇 𝑓 𝑑 𝑡 + 𝜎 𝑓 𝑑 𝑊$ , aka GBM. The value of $𝜇$ depends on risk preferences. If the market price of risk is 0, then you have $𝑑 𝑓 = 𝑟 𝑓 𝑑 𝑡 + 𝜎 𝑓 𝑑 𝑊$ .

By taking $𝜇 = 𝑟 + 𝜆 𝜎$ , where $𝜆$ is the market price of risk, or $𝜎_{𝑔}$ . Then you will end up with the same relationship as before.

𝑑 𝑓 = (𝑟 + 𝜎_{𝑔} 𝜎_{𝑓}) 𝑓 𝑑 𝑡 + 𝜎_{𝑓} 𝑓 𝑑 𝑊

𝑑 𝑔 = (𝑟 + 𝜎_{𝑔}^{2}) 𝑔 𝑑 𝑡 + 𝜎_{𝑔} 𝑔 𝑑 𝑊

Basically this derivation describes the process f/g. The market price of risk will make this process a martingale.

We can apply Itô’s lemma here (not showing work…) for ln f.

𝑑 ln 𝑓 = (𝑟 + 𝜎_{𝑔} 𝜎_{𝑓}) - \frac{𝜎_{𝑓}^{2}}{2}) 𝑑 𝑡 + 𝜎_{𝑓} 𝑑 𝑊

𝑑 ln 𝑔 = (𝑟 + \frac{𝜎_{𝑔}^{2}}{2}) 𝑑 𝑡 + 𝜎_{𝑔} 𝑑 𝑊

If we have these two processes, we can get the differences as

𝑑 (ln 𝑓 - ln 𝑔) = (𝜎_{𝑔} 𝜎_{𝑓} - \frac{𝜎_{𝑓}^{2}}{2} - \frac{𝜎_{𝑔}^{2}}{2}) 𝑑 𝑡 + (𝜎_{𝑓} - 𝜎_{𝑔}) 𝑑 𝑊

Therefore, if you combine these terms, and get

𝑑 (ln \frac{𝑓}{𝑔}) = - \frac{(𝜎_{𝑓} - 𝜎_{𝑔})^{2}}{2} 𝑑 𝑡 + (𝜎_{𝑓} - 𝜎_{𝑔}) 𝑑 𝑊

Here again we can use Itô’s lemma to get the process for f/g.

𝑑 (\frac{𝑓}{𝑔}) = (𝜎_{𝑓} - 𝜎_{𝑔}) \frac{𝑓}{𝑔} 𝑑 𝑊

Based on our previous definition/specification, f/g is a martingale.

Therefore it follows that

\frac{𝑓_{0}}{𝑔_{0}} = 𝔼_{𝑔} [\frac{𝑓_{𝑇}}{𝑔_{𝑡}}]

and

𝑓_{0} = 𝑔_{0} 𝔼_{𝑔} [\frac{𝑓_{𝑇}}{𝑔_{𝑇}}]

Now we can go back to the zero coupon bond price.

Let $𝐸_{𝑇}$ = expectation in a world that is forward risk neutral with respect $𝑇 (𝑡, 𝑇)$ .

What can we say about some of these values? $𝑔$ is the numeraire, so $𝑔_{𝑇} = 𝑝 (𝑇, 𝑇) = 1$ , the price of a ZCB at maturity.

$𝑔_{0} = 𝑝 (0, 𝑇)$

Then

𝑓_{0} = 𝑝 (0, 𝑇) 𝔼_{𝑇} [𝑓_{𝑇}]

Furthermore, for a European call option, with strike $𝐾$ and maturity $𝑇$ , the price of such option will be given by $𝑐$ .

𝑐 = 𝑝 (0, 𝑇) 𝔼_{𝑇} [(𝑆_{𝑇} - 𝐾)_{+}]

where $𝑆_{𝑇}$ is the asset price at time $𝑇$ .

Furthermore, let’s define F_0 and F_T as the forward price of an asset at times 0 and T.

Therefore the bond option price is

𝑐 = 𝑝 (0, 𝑇) [𝐹_{𝐵} 𝑁 (𝑑_{1}) - 𝐾 𝑁 (𝑑_{2})]

𝑝 = 𝑝 (0, 𝑇) [𝐾 𝑁 (- 𝑑_{2}) - 𝐹_{𝐵} 𝑁 (- 𝑑_{1})]

where

𝑑_{1} = \frac{ln (\frac{𝐹_{𝐵}}{𝐾}) + 𝜎_{𝐵}^{2} \frac{𝑇}{2}}{𝜎_{𝐵} \sqrt{𝑇}}

𝑑_{2} = 𝑑_{1} - 𝜎_{𝐵} \sqrt{𝑇}

The characteristic of this pricing model is that the bond price and the strike price should be the cash prices, not the quoted prices. Let’s look at an example.

Bond Options Example (Black’s)

Let’s consider a 10-month European call option on a 9.75 years bond with a face value of $1000. What we’re saying is when the option matures, the bond will still have 8 years and 11 months. The current bond price is $960, the strike is $1000, the 10-month risk-free interest rate is 10% per annum, the volatility of F_B for T=10 months is 9% per annum. The bond pays a coupon of 10% per year, semiannual payments. Therefore the coupon payments is $50, with the payments expected at 3 months and 9 months. Here the accrued interest at $25. Let’s suppose the risk-free interest rates are different for 3 months and 9 months, 9% and 9.5% per annum. This is a direct application of Black’s model.

In order to solve this, we can just apply Black’s model.

The value we need to calculate is the forward bond price $𝐹_{𝐵}$ , the expected value of the bond at some time in the future.

At some time in the future, you might have some coupons that are no longer considered in the calculations. Therefore the forward bond price is

𝐹_{𝐵} = \frac{𝐵_{0} - 𝐼}{𝑝 (0, 𝑇)}

where B_0 is bond price at time 0.

I is present value of coupons in (0, T)

In the next 10 months, we are expected to have two coupons. One in 3 months, and one in 9 months.

The current cash price is given

B_0 = 960

Then we can calculate the forward

𝐹_{𝐵} = \frac{𝐵_{0} - 𝐼}{𝑝 (0, 𝑡)}

small t in our case.

𝐼 = 50 𝑒^{- 0.25 \times 0.09} + 𝑡 0 𝑒^{- 0.75 \times 0.095} = 95.45

We are using the time of discounting (3 months and 9 months), the coupon payment (50), and the interest rate for each coupon (0.09 and 0.095).

Forward bond price therefore is

𝐹_{𝐵} = (960 - 95.45) 𝑒^{0.10 \times \frac{10}{12}} = 939.68

Since the 10 month interest rate is 10%

Let’s say it’s under specified. The strike price is 1000, but this problem does not specify if the strike price is the cash price to be paid for the bond or the quoted price. For the payment, if you have the quoted price, the value of the bond is going to be quoted price + accrual (dirty price). We can investigate these two cases.

If the strike price is the cash price that would be paid for the bond on exercise.

F_B = 939.68; K = 1000; $𝑝 (0, 𝑇) = 𝑒^{0.10 \times \frac{10}{12}} = 0.92$ ; $𝜎 = 0.2$

By applying the formula for the Black’s model for the call (not writing it all down)

Should be $9.49.

If the strike price is the quoted price.

We have one month accrual interest that must be added to K, because the most recent payment was at 9 months, and the expiration is at 10 months. Therefore K = 1000 + 100 * 1/12 = 1008.33

The rest of the characteristics are the same.

c = $7.97

Standard Deviation of ln B

Stdev will rise after 0, then fall.

Forward Bond and Forward Yield

The volatilities quoted for the bond options are many times the yield volatilities and the price volatilities. We can use the duration concept. Let’s suppose if D is the modified duration of the forward bond price at option maturity, then the relationship in the change of the forward price and the change in forward yield is given by

\frac{Δ 𝐹_{𝐵}}{𝐹_{𝐵}} \approx - 𝐷 Δ 𝑦_{𝐹}

Yield Vols vs Price Vols (Equation 28.4, page 652)

Caps and Floors

These are interest rate caps and floors. Basically, these instruments, you can think of them as insurance, protection against the increase or decrease in interest rate. They’re going to cap such an interest rate.

First of all, let’s consider a floating rate note where the interest rate is reset periodically to a floating rate. We have such reset periods. The time between the resets is known as the tenor. Let’s look a little bit at the mechanics.

Assume that the tenor is equal to 3 months.

The interest rate on the note for the first 3 months is equal to the initial rate for 3 months, as observed at t = 0
The interest rate on the note for the next 3 months is set equal to the 3-month floating rate prevailing in the market.

The interest rate cap is designed to provide insurance against interest rate on the floating rat rising above a certain level (“cap rate”).

For example, assume a principal of $10 million, tenor is 3 months, life of cap is 5 years, cap rate is 4%.

This is an important characteristic for the caps, because the payments are made quarterly, so the cap rate is expressed with quarterly compounding.

Furthermore, let’s assume on a particular reset date, we have the 3-month floating rate as 5%. In this case, a floating rate note would require a payment of the period (0.25) times the rate (0.05) times the principal (10M) = $125K.

However, with the 3-month rate capped at 4%, it resolves to $100K.

The cap provides protection, a payoff of such difference.

If you have such cap, then the cap provides a payoff of $25,000, 3 months later

So what is the value of the cap?

At each reset date that is happening during the life of the cap, the floating rate is observed.

If the floating rate is $\leq 4 %$ , which is smaller than the cap rate, then there is no payoff from the cap. Furthermore, if it’s $> 4 %$ , then the payoff is the excess rate applied to the principal.

In this case, we have 19 reset dates (5 years, $4$ tenors per year $- 1$ initial tenor), at times 0.25, 0.5, …, 4.75 years. There are 19 payoffs, which occur a tenor after, so at times 0.5, 0.75, …, 5 years.

Portfolio of Interest Rate Options

The cap can be expressed as a portfolio of call options on a reference floating rate (LIBOR maybe?).

Consider a cap with a total life $𝑇$ , principal $𝐿$ , and cap rate $𝑅_{𝐾}$ .

The reset dates are $𝑡_{1}, 𝑡_{2}, \dots, 𝑡_{𝑛}$ and $𝑡_{𝑛 + 1} = 𝑇$ .

We will define $𝑅_{𝑘}$ as the floating interest rate for a period between $𝑡_{𝑘}$ and $𝑡_{𝑘 + 1}$ observed at time $𝑡_{𝑘}$ . The small $𝑘$ can take values $1 \leq 𝑘 \leq 𝑛$ .

The cap leads to a payoff at time $𝑡_{𝑘 + 1}$ , based on the tenor. The payoff depends on the realization of the floating interest rate.

𝛿_{𝑘} 𝐿 (𝑅_{𝑘} - 𝑅_{𝐾})_{+}

This is applied to the principal. It’s annual, quarterly compounding, so this particular payoff will be applied for the period corresponding to the tenor, which we can mark with the $𝛿$ .

𝛿_{𝑘} = 𝑡_{𝑘 + 1} - 𝑡_{𝑘}

You may assume that tenors are equal, even if that’s not necessarily the case.

We can also consider it as a portfolio of puttable bond options, where the ZCB has payoff occurring at time $𝑡_{𝑘}$ .

The previous payoff at time $𝑡_{𝑘 + 1}$ is equivalent to

\frac{𝛿_{𝑘} 𝐿}{1 + 𝑅_{𝑘} 𝑆_{𝑘}} (𝑅_{𝑘} - 𝑅_{𝐾})_{+}

We can use algebra to determine that this is equivalent to

{(𝐿 - \frac{𝐿 (1 + 𝑅_{𝐾} 𝛿_{𝑘})}{1 + 𝑅_{𝑘} 𝛿_{𝑘}})}_{+}

Therefore the denominator is equivalent to the value at time $𝑡_{𝑘}$ of a ZCB that pays $𝐿 (1 + 𝑅_{𝑘} 𝛿_{𝑘})$ at time $𝑡_{𝑘 + 1}$ .

The floor is just the reverse of this, using put options, and $𝑅_{𝐾} - 𝑅_{𝑘}$ .

We can then view each call option in the portfolio that represents a cap as a caplet. We assume that the interest rate underlying each caplet is lognormal. Then we can use Black’s model.

A collar is an instrument that guarantees that the interest rate on the underlying floating rate always lies between two levels. You can achieve this by taking a long position in a cap and a short position in a floor. In terms of valuation…

Collar Example

The value of a caplet is

𝐿 𝛿_{𝑘} 𝑝 (0, 𝑡_{𝑘 + 1}) [𝐹_{𝑘} 𝑁 (𝑑_{1}) - 𝑅_{𝑘} 𝑁 (𝑑_{2})]

using the standard d1 and d2 for option valuation.

Where $𝐹_{𝑘}$ is the forward interest rate for a period between $𝑡_{𝑘}$ and $𝑡_{𝑘 + 1}$

In addition, we have to take some assumptions or make some estimates about the volatility of the forward interest rates

$𝜎_{𝑘}$ = volatility of the forward interest rate

Here you value all the caplets in order to determine the value of a cap.

Consider a contract that caps the LIBOR interest rate on $10M at 8% per annum (quarterly compounding) for 3 months starting in 1 year. This is a caplet, and could be an element of a cap.

LIBOR/swap zero curve is flat at 7% per annum,

volatility of the rate is 20% per annum

continuously compounded zero rate for all maturities is 6.9395%

Let’s consider our inputs

𝑡_{𝑘} = 1, 𝑡_{𝑘 + 1} = 1.25, 𝛿_{𝑘} = 0.25, 𝐹_{𝑘} = 0.07, 𝑅_{𝑘} = 0.08, 𝐿 = 10 𝑀, 𝜎 = 0.2

We need to calculate the discount factor and all the other values

𝑝 (0, 𝑡_{𝑘 + 1}) = 𝑝 (0, 1.25) = 𝑒^{- 0.069395 \times 1.25} = 0.9169

Furthermore, to calculate for our options

𝑑_{1} = - 0.5677

𝑑_{2} = - 0.7677

Then we can use the option valuation Black’s model to get the final value

Assumptions

We must use spot volatilities, where the vol is different for each caplet,

or one volatility, where it’s flat and the same for each cap.

Swaptions

The swaption or swap option gives the holder the right to enter into an interest rate swap in the future.

There are two kinds, the right to pay a fixed rate and receive LIBOR, or the right to receive fixed rate and pay LIBOR. (LIBOR = floating)

We have a single option on the swap rate with repeated payoffs.

This is a series of cash flows

\frac{𝐿}{𝑚} (𝑆_{𝑇} - 𝑆_{𝑘})_{+}

where

$𝐿$ = principal

$𝑚$ = frequency per year

$𝑆_{𝐾}$ = swap rate

$𝑆_{𝑇}$ = swap rate at time $𝑇$

The value of the swaption on the swap rate with repeated payoffs where the holder has right to pay $𝑆_{𝑘}$ .

Each of these payoffs will be discounted to the present.

\sum_{𝑖 = 1}^{𝑚 𝑛} \frac{𝐿}{𝑚} 𝑝 (0, 𝑇_{𝑖}) [𝑆_{0} 𝑁 (𝑑_{1}) - 𝑆_{𝐾} 𝑁 (𝑑_{2})]

$𝑆_{0}$ is the swap rate at time 0.

This is a natural extension of Black’s model, just with more discount factors.

This represents discount factors for mn payoffs ⬇️

\sum_{𝑖 = 1}^{𝑚 𝑛} 𝑝 (0, 𝑇_{𝑖})

We can define $𝐴$ as the value of a contract that pays $1 / 𝑚$ at times $𝑇_{𝑖}$ ( $1 \leq 𝑖 \leq 𝑚 𝑛$ )

With this notation, the value of the swaption will become

𝐿 𝐴 [𝑆_{0} 𝑁 (𝑑_{1}) - 𝑆_{𝐾} 𝑁 (𝑑_{2})]

Swaption Example

Suppose that the LIBOR yield curve is flat at 6% per annum with continuous compounding. Consider a swaption that gives the holder the right to pay 6.2% in a 3-year swap starting in 5 years. The volatility of the forward swap rate is 20%. The payments are semiannually and the principal is $100 million.

Let’s set our variables

For A, this is the sum of the discounted swaps. The yield curve is flat, so it’s always 6%. We are semiannual, so we start six months after 5 years when the swap starts.

𝐴 = \frac{1}{2} (𝑒^{- 0.06 \times 5.5} + 𝑒^{- 0.06 \times 6} + \dots + 𝑒^{- 0.06 \times 8} = 8

We also need $𝑆_{0}$ , the forward swap rate.

[…skips some steps, forgot to pay attention)

𝑅_{𝑐} = 𝑚 ln (1 + \frac{𝑅_{𝑚}}{𝑚})

Therefore

$𝑆_{0} = 0.0609$

$𝑆_{𝐾} = 0.062$

$𝑇 = 5$

$𝜎 = 0.2$

100 𝑀 \cdot 2.0035 [0.0609 𝑁 (0.1836) - 0.063 𝑁 (- 0.2636)] = $ 2.07 𝑀

Sensitivity

All these derivative have a delta, the DV01, the impact of a 1bps parallel shift in the zero curve.

Next Lecture

We will be doing adjustments, like convexity, time, and quant adjustments

Caps protect you, if the volatility rate is too high, you might have risk for a longer period.

The swap which is fixed for floating, you can enter in the future. Once you enter the swap,

## Tree Approximation and American Options - URL: https://sharifhsn.dev/blog/computational-methods-week-03/ - Structured data: https://sharifhsn.dev/api/posts/computational-methods-week-03.json - Description: We’re not going to use the arbitrage method, we are going to use the forest methods. You all understand the stochastic process. Fundamentally, it starts from a point 𝑆 0 , and… - Date: 2025-02-11 - Exact published timestamp: 2025-02-11 - Topics: Computational Methods, Binomial Trees, Option Pricing, American Options, Barrier Options - Categories: Computational Methods - Source: Computational Methods in Quantitative Finance - Source URL: None - Content source format: typst

Textbook - MF3

Tree Approximation

We’re not going to use the arbitrage method, we are going to use the forest methods.

Stochastic Process

You all understand the stochastic process.

Fundamentally, it starts from a point $𝑆_{0}$ , and has these continuous paths that are very weird, because they’re not differentiable. If you blow up the curve, it should smooth out, but actually it stays just as jagged. Therefore we draw an approximation where we draw lines on tiny intervals to approximate the Brownian motion.

The thing is, this is a diffusion process. You kind of know the distribution of the path, the $𝑑 𝑆_{𝑡}$ follows $𝜇 𝑆_{𝑡} 𝑑 𝑡 + 𝜎 𝑆_{𝑡} 𝑑 𝑊_{𝑡}$ .

This is an exponential of a normal, it’s called lognormal. This distribution has a known shape. A whole bunch of paths will end up at the end.

Discretization

The tree approximation says that the idea that this is continuous is bullshit. So let me do something else. Instead of something that goes all over the place, I’ll start at $𝑆_{0}$ and approximate the path by discrete intervals. At each step, the distribution should approximate the log-normal thing. It’s the same at each slice.

How is this working if the tree is discrete? But as the $Δ 𝑡$ shrinks, you will have more and more points, you take the interval to be smaller. Each point will have a certain probability, which will reflect the target distribution.

Kolmogorov

In probability theory, there’s a big theorem Kolmogorov.

These continuous time processes, as long as you have two, that match in their discrete time distribution. And if the distribution is the same, jointly, you can say it’s the same process.

And that’s why this worked. That’s the idea of any tree construction. You want to make it so that it matches each distribution.

There are two different requirements. It has to match this distribution, and it has to be recombining.

Recombining tree will go from one point to two points, the binomial tree. If you don’t recombine, you go from 4 to 8 to 16, n steps will create $2^{𝑛}$ paths, each distinct. If you do 64 steps, then you use all the bits in your computer. That means your computer will crash. 64 steps is very little, so the alternative is…

Recombining! You need to store the nodes, so it will grow at a polynomial rate, $1 + 2 + 3 + 4 + \dots$ , which is $𝑛 (𝑛 + 1) / 2 \to 𝑂 (𝑛^{2})$ , this is much better than $2^{𝑛}$ .

The number of paths are the same, you just use the same points.

Even though I basically have the number of paths, and I can store them in a computer, I still don’t do anything if I’m pricing path dependent options. There’s a problem: what if this isn’t a Markov process? If it doesn’t just depend where I am now, I have to remember where I was before, and that costs 2^n.

So this only works when you’re trying to approximate a Markov process, where you can assume that all information is encoded in the value at the end.

Sidebar: This is a difference between stochastic processes and LLMs. LLMs are just the most horrible things, they do exponential stuff, because the language is finite, not like numbers.

There’s a discussion in the textbook: Florescu’s work is a general diffusion approximation method (6.9)

The Traditional Tree

We’ve done the u and d before. But is that the only tree? It’s just a single tree. I’m not going to do that tree, because that tree is complicated. There are two trees you can make. There is the process

𝑑 𝑆_{𝑡} = 𝜇 𝑆_{𝑡} 𝑑 𝑡 + 𝜎 𝑆_{𝑡} 𝑑 𝑊_{𝑡}

Everything we’re doing is the geometric Brownian motion, we are not approximating any other stochastic processes. You could, but they think we’re stupid, so they only teach you one single tree of one type. Trees are fascinating. You can do whatever you want, but you need to understand how they work in order to use them.

What is the difference between additive tree and multiplicative tree?

𝑑 𝑆_{𝑡} = 𝜇 𝑆_{𝑡} 𝑑 𝑡 + 𝜎 𝑆_{𝑡} 𝑑 𝑊_{𝑡}

But then you need to do Girsanov to use risk-neutral measure

𝑑 𝑆_{𝑡} = 𝑟 𝑆_{𝑡} 𝑑 𝑡 + 𝜎 𝑆_{𝑡} 𝑑 𝑊_{𝑡}^{𝑄}

But this stochastic process is multiplicative. If you take an approximation and get

𝑆_{𝑡 + Δ 𝑡} - 𝑆_{𝑡} = \int_{𝑡}^{𝑡 + Δ 𝑡} 𝑟 𝑆_{𝑡} 𝑑 𝑡 + \int_{𝑡}^{𝑡 + Δ 𝑡} 𝜎 𝑆_{𝑡} 𝑑 𝑊_{𝑡}

This is a Riemann integral and stochastic integral. We can approximate this by taking the left point, the value in t.

If you do the homework, if you use a quadrature rule, we’re using the rectangle here. Here we’re using the stochastic integral approximation, because we know that the expectation of the increment of Brownian motion is 0.

𝑆_{𝑡 + Δ 𝑡} = 𝑆_{𝑡} (1 + 𝑟 Δ 𝑡 + 𝜎 Δ 𝑊_{𝑡})

The resulting tree that you use to approximate $𝑆_{𝑡}$ is called multiplicative. That’s why we took S and multiplied it by u or by d. That value is supposed to be that thing in parentheses, or something that converges to it.

What if I don’t care about this, and I want to approximate in the most efficient way possible? If you take $𝑅_{𝑡}$ to be the logarithm of $𝑆_{𝑡}$ , and apply Itô to $𝑅_{𝑡}$ , you get

𝑑 𝑅_{𝑡} = (𝑟 - \frac{𝜎^{2}}{2}) 𝑑 𝑡 + 𝜎 𝑑 𝑊_{𝑡}^{𝑄}

And then if you discretize,

𝑅_{𝑡 + Δ 𝑡} = 𝑅_{𝑡} + (𝑟 - \frac{𝜎^{2}}{2}) Δ 𝑡 + 𝜎 Δ 𝑊_{𝑡}

And this is direct, not an approximation. The next R_t is the previous one plus something.

What is the difference?

In the multiplicative case, you go from some S value with some $𝑝_{𝑢}$ and $𝑝_{𝑑}$ to $𝑆 𝑢$ and $𝑆 𝑑$ . But, this is equivalent to $log 𝑆$ to $log 𝑆 + log 𝑢$ or $log 𝑆 + log 𝑑$ . These two trees are completely equivalent.

This is possible because logarithm is a monotonically increasing function, so it can be a one to one transformation.

We’re going to do additive because we’re not imbeciles, much easier than multiplicative.

The two conditions:

The tree needs to be recombining. I can do that as long as I am dealing with a Markov process. That’s why we learn that GBM is a Markov process in 610, because it’s very valuable.
We need that calculated approximating derivative value needs to converge to the value calculated using the continuous process. I don’t really care about the continuous process. I’m trying to approximate the option price/premium. As long as it converges to the true value, then I’m fine.

Convergence

There is a value at time t, a random variable. I quantify the value by looking at the probability I reach every single point, where there’s a probability at each point, a discrete distribution, which should converge to the continuous version.

$𝑅_{𝑡}^{𝑑}$ is the random variable obtained using a discrete tree.

Let $𝑅_{𝑡}$ be the random variable using the continuous time process.

A stochastic process (by the way), there are two things. It depends on the path, and also time. It’s technically a function of two things. For each path, you have times, so you can slice the stochastic process by either the path or the time. What I’m talking about here, it’s at some fixed t, with some random variable coming from the tree and from the stochastic process. We want our tree to converge to the stochastic process.

The payoff should

𝔼 [𝑒^{- 𝑟 (𝑇 - 𝑡)} 𝜑 (𝑅_{𝑇}^{𝑑}) | {ℱ︀}_{𝑡}]

(discount factor doesn’t really matter)

This value should converge to

𝔼 [𝑒^{- 𝑟 (𝑇 - 𝑡)} 𝜑 (𝑅_{𝑇}) | {ℱ︀}_{𝑡}]

In the theory of random variables, we have tons of convergences. Any convergence will work.

The weaker type of convergence is convergence in distribution. It says that you have a series of random variables that converge in distribution to a target as long as the distribution of the random variables converges to the distribution of the target. With this tree, that’s what we are aiming to do.

In most trees, you take Δt to go to 0.

We are going to be doing the additive tree.

From $𝑅_{0}$ , we can go only one. Either you’ll go to $𝑅_{0} + Δ 𝑅_{𝑢}$ or $𝑅_{0} + Δ 𝑅_{𝑑}$ , with some probability $𝑝_{𝑢}$ and $𝑝_{𝑑}$ .

Everything is done in terms of Δt.

What about the next step?

Technically, I can use a different probability. But we have a very nice GBM, with a normal distribution in Brownian motion. As long as we have the same size of the interval, they have the same distribution, and we can use the same probability.

𝑅_{0} + 2 Δ 𝑅_{𝑢}, 𝑅_{0} + Δ 𝑅_{𝑢} + Δ 𝑅_{𝑑}, 𝑅_{0} + Δ 𝑅_{𝑢} + Δ 𝑅_{𝑑}, 𝑅_{0} + 2 Δ 𝑅_{𝑑}

Then you notice that the two middle values can recombine. You don’t need to have any relationship between $Δ 𝑅_{𝑢}$ and $Δ 𝑅_{𝑑}$ , and it doesn’t matter.

The other condition is that it has to converge to the continuous distribution. We are working with the additive case, which is very easy. So we know that

𝑅_{𝑡 + Δ 𝑡} - 𝑅_{𝑡} \sim 𝑁 ((𝑟 - \frac{𝜎^{2}}{2}) Δ 𝑡, 𝜎^{2} Δ 𝑡)

The tree with this R^d_t, is an increment with only two values.

It’s a very simple Bernoulli distribution, with a probability to $Δ 𝑅_{𝑢}$ and $Δ 𝑅_{𝑑}$ .

The normal is characterized by mean and variance. So therefore we will advance a general theorem, which was proved in terms of Central Limit Theorem.

There’s a French guy who proved it with binomial distribution. You have a number of trials, and a number of successes, and he looked at average number of successes. As n goes to infinity, the distribution of the average minus p and divided by blah blah converges to the standard normal. And that’s what started the whole Central Limit Theorem, and it was extended to different distributions by Laplace.

If you decrease the time interval and you look at the probabilities, they will be binomial probabilities, exactly the same. But there’s a fundamental reason why this is happening.

Florescu has proven a more general convergence theorem for trees, discusses infinitesimal generator.

I’m going to look at the moments of R_t^d and set them equal for discrete and continuous, to get our probabilities.

How do we solve for these four unknowns?

Mean

Δ 𝑅_{𝑢} 𝑝_{𝑢} + Δ 𝑅_{𝑑} 𝑝_{𝑑} = (𝑟 - \frac{𝜎^{2}}{2}) Δ 𝑡

Variance

Pretty ugly, so we won’t equate, we’ll just equate the second moment. We’ll use the property that

𝔼 [𝑅_{𝑡}^{2}] = 𝕍 [𝑅_{𝑡}] + (𝔼 [𝑅_{𝑡}])^{2}

Therefore

Probabilities

𝑝_{𝑢} + 𝑝_{𝑑} = 1

What else?

DeMoivre proved binomial converges to normal. The normal is characterized completely by the mean and variance, so this is all we need. We don’t need to look for anything else.

And actually any number of four parameters solved will give us the tree. There’s an infinite number of trees that solve this.

In practice…

This was actually initiated and proven by finance guys, who all came up with it from complicated arguments.

They solve it for one specific case.

If you take ΔR_u = -ΔR_d = ΔR, where you go up and down by the same quantity. Now you will have in this system, three equations with three unknowns, which creates a unique solution.

This is called the Cox-Ingersoll-Ross tree, very famous, the first.

You can also take p_u = p_d = ½. The probabilities are constant, but the values may be different. This is the Trigeorgis tree.

For CIR, expressed multiplicatively, you get

$𝑆_{𝑡} 𝑒^{Δ 𝑅}$ , and $𝑆_{𝑡} 𝑒^{- Δ 𝑅}$

And to solve it, there’s no mystery. It’s just a system.

$𝜎$ and $𝑟$ are given to you, by implied volatility and the interest rate. $Δ 𝑡$ is your choice, typically expressed by $𝑇 / 𝑛$ .

Sidebar: Technically, some trees are better than others. The one with symmetric values is faster to converge, but in general they have the same order of approximation.

It depends, however, on what you’re approximating.

Let’s approximate the European option. This is simple because the payoff only depends on the terminal value. Given this, you have that the price of the premium is

𝔼 [𝑒^{- 𝑟 𝑇} 𝜑 (𝑆_{𝑇}) | {ℱ︀}_{0}]

There is a simple derivation that makes sense to me, and this is why the tree functions.

\= \\mathbb{E}\[e^{-r(T-\\Delta t)} \\mathbb{E}\[e^{-r\\Delta t} \\varphi(S\_T)|\\mathcal{F}\_{T-\\Delta t}\]|\\

This uses the tower property of conditional expectations. As long as you discount back by something bigger, then you can get it back. Now you can see that this is conditioned on the penultimate step. And you can continue to nest these expectations so that each expectation is conditioned on each step.

What this tells you is basically, you can look at the final value $𝜑 (𝑆_{𝑇})$ and come back to the previous step, calculate the expectation there, and go on.

Then I can calculate the European call if I get $𝑆_{𝑇}$ from $𝑅_{𝑇}$ .

The discount factor is actually irrelevant because either you do it at first, or you do it every step, or at the end, doesn’t matter.

You will need to research and construct this algorithm.

Algorithm

inputs:

r: interest rate
σ: volatility
S_0: initial stock price
T: time to expiration
K: strike price
n: number of steps

Then decide what tree you’re going to use, CIR or Trigeorgis

𝑅_{0} = log 𝑆_{0}

Δ 𝑡 = 𝑇 / 𝑛

Then you calculate ΔR. The formula is in the book, based on the tree. Same for p_u and p_d.

What makes us special?

I solve problems that no one else can solve, I know math and coding. Nonstandard pricing comes in, it’s the Stevens student that says I can do it.

Then you calculate the terminal values at the end. There are n + 1 points after n steps. You have to create a vector of values $𝑅 [0] = 𝑅_{0} + 𝑛 Δ 𝑅$ , and $𝑅 [1] = 𝑅 [0] - 2 Δ 𝑅$

An algorithm is not something you just wake up and do, you have to think about it on paper, then code it. That’s why we have immigrants.

You’re working with vectors, that have n + 1 values, if you step back and take the two values and

There’s no book with this information.

Sidebar: Nash Equilibrium

If everyone in the room has one objective, and they want to maximize their profit, they all have to interact. The shares are being handed out, you get a mini-market, where they exchange assets to do this. This guy Nash proved that if we all have an objective, there is only one way in which the wealth is distributed that maximizes everyone’s objectives. The theory doesn’t work because it assumes that everyone has the same objective all the time, which is not understanding human beings. If the Nash equilibrium existed, there would be no trading, nobody would desire to have more.

American Options

Other than some horrible equations, you have to use trees. Monte Carlo doesn’t work, because you would like to exercise at the maximum value, but you don’t know that at the time, only when you get to the end.

If you are in the tree, you will compare with the expected future value. The discount factor needs to be done in the tree here because you will need to know what the discounted expected value is. You do it from the point and you calculate the value. You have two values, and you take whichever one is higher. This is a very useful way of thinking, related to reinforcement learning. This is based on the idea that you take the best action now which maximizes my future expected reward.

Physics is baby shit, with stupid matrix algebra. Finance is way harder.

You can also use trees to approximate Heston or SABR.

There’s one thing to mention.

Call vs Put

There’s an argument in the book about put call parity.

The American call and the European call are the same exact number. This is a mathematical thing because of the stochastic process we do, which is continuous.

There was no place in the tree when you stepped back when it was optimal to exercise.

But in the put, there are places in which it is appropriate to exercise, which creates a larger number than European put option.

Dividends are also a different question, or if the stochastic process jumps, which creates value for the American call.

Pretty much every path-dependent option can be done through this trick. The fundamental thing is the expected value of the future, conditioned on the particular node you’re in, which is what you’re storing.

HW1 is not about trees, HW2 is about trees.

I’m going to ask you to price a barrier option.

Barrier Option

There are two types, the IN and the OUT type.

IN option (can be call or put), is worthless unless the stock price hits the barrier. At that point, the option activates. An UP and IN call option, means the barrier is up from the starting price, IN means that once the barrier is reached, then the option is activated, and it’s a call option, which has its own strike price.

The OUT option starts as a regular option, which becomes worthless if the barrier is hit.

Let’s say you go the Bank of America and say you want to buy options. Instead of having a fixed rate on my house, I want to buy variable rates. Then BofA looks at instruments created by a quant, which gives the rate. That’s an option that’s provided by a specialist.

Options are issued by the owner of a stock. You can write covered options if you own the stock. You can write an option if your broker allows it, or if you won the call. You can provide naked calls, where you don’t have the actual share. I have 100 shares up for bidding. I could, using the money I bought from an option, buy another option, and close my contract no matter what.

Let’s say I hold onto my option until expiry, and receive my money, and then my share is gone. The broker will settle the contract.

The other thing you should look at is, if you look at this data, you should explain the value of the options. There’s something called open interest (openint) and volume. Volume is the number of contracts being transacted, typically per trading day. Open interest is how many contracts are actually open. The number of contracts open is usually very small compared to transacted. The same contract is being bought and sold, 50 issuances traded a thousand times.

This is relevant because it applies here. Basically, sometimes you don’t want to have the contract becomes active. There’s no risk of the option cannot be exercised. Barrier options are typically used for fixed income instruments. The barrier technically here depends on the underlying, which for fixed income is the rate, which is floating. For that, you want to have protection because you don’t want to pay more than e.g. 5%, but you don’t want to activate that unless the thing goes crazy. That’s when you put the barrier in place.

$𝑆_{0}$ the contract is either going to be a call, with $𝑆_{𝑇} - 𝐾$ or a put with $𝐾 - 𝑆_{𝑇}$

This is not very relevant because the option will become active no matter what.

But the barrier matters.

DOWN, UP, IN, and OUT are the four parameters, with four possibilities.

There is a relationship between this.

Let’s say the barrier $= 𝐵$ . Let’s talk about a CALL UP and IN. This option only becomes valuable when the barrier is hit from below. YOu can model this with an indicator

𝕀_{{𝑆_{𝑡} > 𝐵}} (𝑆_{𝑇} - 𝐾)_{+}

What about UP and OUT? If this is happening, then I get zero, otherwise you get the payoff

𝕀_{{𝑆_{𝑡} \leq 𝐵}} (𝑆_{𝑇} - 𝐾)_{+}

But these two sets are complementary to each other. If you add this using probability theory, you will see that these sum up to 1.

So if you take a call that’s up and in, and a call that’s up and out, you will get a regular call.

That means that you don’t have to calculate both of the barrier options, you only need to calculate the regular call and the one that you figure this out.

The reason this matters is because the OUT is way easier to price than the IN. The reason is because we are using a tree. If you look at a path in the tree, this is complicated because you have to keep track of the paths. The difficulty is with the point which is under the barrier, but it could be reached by crossing the barrier. If it’s IN, then it becomes valuable. So you have to look at all the paths that cross the barrier, so complicated.

But with the OUT, if it passes the barrier, then you set to zero, and don’t count them at all.

## FX Parity and Forward Trading - URL: https://sharifhsn.dev/blog/fx-parity-and-forward-trading/ - Structured data: https://sharifhsn.dev/api/posts/fx-parity-and-forward-trading.json - Description: The syllabus records this class as “Stochastic Processes — Intro to FX: FX parity; Spot and Forward trading; FX conventions.” The source document does not contain a prose… - Date: 2025-02-10 - Exact published timestamp: 2025-02-10 - Topics: FX, FX Parity, Forward Contracts - Categories: FX - Source: FE-635 \| Risk Engineering - Source URL: None - Content source format: typst

Week 3 — 10 February 2025

The syllabus records this class as “Stochastic Processes — Intro to FX: FX parity; Spot and Forward trading; FX conventions.” The source document does not contain a prose derivation of covered interest parity, so I am keeping the implementation notes and the topic record rather than filling that gap with a generated explanation.

Notes on inputs

The workbook notes record the conventions used by the later pricing functions:

Days-to-time conversion uses a 365-day year.
1. is the forward FX rate, not spot FX.
Contract types are (C) (call), (P) (put), (F) (forward), (D) (deposit in local currency, usually USD), and (M) (deposit in the foreign currency, usually MXN).
(DiscRate) is the discounting-currency rate; (ExtRate) is the foreign-currency rate.
Prices are returned in USD. Notionals are entered in MXN except for a USD deposit ((D)).

The quote direction and domestic/foreign labels therefore need to be fixed before comparing a spot, forward, or parity calculation.

Source boundary

The personal course export contains the syllabus and these workbook conventions, but no worked parity example. This article intentionally marks that source boundary instead of presenting an LLM-generated derivation as class notes.

## Yield Curves and Bootstrapping - URL: https://sharifhsn.dev/blog/advanced-derivatives-week-02/ - Structured data: https://sharifhsn.dev/api/posts/advanced-derivatives-week-02.json - Description: We started to look at interest rates. In the first part we’re going to discuss various interpolation methods, functions that are useful for modeling the yield curve. That’s… - Date: 2025-02-06 - Exact published timestamp: 2025-02-06 - Topics: Fixed Income, Yield Curves, Interpolation, Bootstrapping, Nelson-Siegel, Smith-Wilson - Categories: Fixed Income - Source: Advanced Derivatives - Source URL: None - Content source format: typst

From Last Time

We started to look at interest rates. In the first part we’re going to discuss various interpolation methods, functions that are useful for modeling the yield curve.

That’s useful because many times you might have to discount or use forward rates or discount factors for maturities that are not available. For this reason you have to do some interpolation.

Yield Curve Interpolation Methods

What do we want to achieve?

In general, the interpolation problem can be expressed in the following format:

given some data as a function of time $𝑡_{1}, 𝑡_{2}, \dots, 𝑡_{𝑛}$ and $𝑥_{1}, 𝑥_{2}, \dots, 𝑥_{𝑛}$ known
construct a continuous function $𝑥 (𝑡)$ that satisfies $𝑥 (𝑡_{𝑖}) = 𝑥_{𝑖}$ for $𝑖 = 1, 2, \dots, 𝑛$

What does it mean that the function is continuous? It means that at all points, the limit on left and right is the same, and that it passes through all points.

Criteria for choosing the Interpolation Method

Smoothness of the forward curve is desirable

What does this mean? The derivative should be continuous i.e. it should be differentiable. If the model has smooth curves, it should be more accurate because there’s no reason to have rough changes in curvature.

How local is the interpolation method?

If an input changes, does the interpolation function change locally or globally? Simple interpolation, linear interpolation have a local impact. Typically, each input just changes

It’s good to have interpolation that impacts globally, but maybe not with very high sensitivity.

Are the forwards not only continuous, but also stable?

The degree of stability is the maximum basis point change in the forward curve given a basis point change in move in input (???? look this up)

Linear Methods

Let’s graph t against r

For $𝑡_{𝑖 - 1} < 𝑡 < 𝑡_{𝑖}$ (this is the general case for local interpolation

𝑟 (𝑡) = \frac{𝑡 - 𝑡_{𝑖 - 1}}{𝑡_{𝑖} - 𝑡_{𝑖 - 1}} \cdot 𝑟_{𝑖} + \frac{𝑡_{𝑖} - 𝑡}{𝑡_{𝑖} - 𝑡_{𝑖 - 1}} \cdot 𝑟_{𝑖 - 1}

This is a simple rate from drawing a straight line between two points and picking the rate that corresponds to time

What is important for this class is to have consistent curves, to prevent arbitrage.

If this is the rate, how do we get the forward?

The price of a zero coupon bond is the continuously discounted forward rate

𝑃 (0, 𝑡_{1}, 𝑡_{2}) = 𝑒^{- 𝑓 (0, 𝑡_{1}, 𝑡_{2}) \cdot (𝑡_{2} - 𝑡_{1})}

Then the forward rate is just the natural log of this scaled to time

𝑓 (0, 𝑡_{1}, 𝑡_{2}) = - \frac{ln 𝑃 (0, 𝑡_{1}, 𝑡_{2})}{𝑡_{2} - 𝑡_{1}}

Zero coupon bonds can also be determined in this way.

𝑓 (0, 𝑡_{1}, 𝑡_{2}) = - \frac{ln 𝑃 (0, 𝑡_{2}) - ln 𝑃 (0, 𝑡_{1})}{𝑡_{2} - 𝑡_{1}}

In the limit, we can use the derivative

𝑓 (𝑡) = - \frac{𝑑}{𝑑 𝑡} ln 𝑃 (𝑡) - \frac{𝑑}{𝑑 𝑡} [𝑟 (𝑡) \cdot 𝑡]

So this is the relationship between the forward and the zero rate r(t) Now we can determine a consistent forward. Therefore f(t) is the derivative rate of this product

𝑓 (𝑡) = 𝑟 (𝑡) + 𝑡 𝑟^{'} (𝑡)

So if you take such derivatives, you can substitute the formula for the instantaneous forward.

𝑓 (𝑡) = \frac{(2 𝑡 - 𝑡_{𝑖 - 1}) 𝑟_{𝑖} + (𝑡_{𝑖} - 2 𝑡) 𝑟_{𝑖 - 1}}{𝑡_{𝑖} - 𝑡_{𝑖 - 1}}

This is linear, so you can do a function which is linear in log rates if you prefer

Linear in log rates

ln (𝑟 (𝑡)) = \frac{𝑡 - 𝑡_{𝑖 - 1}}{𝑡_{𝑖} - 𝑡_{𝑖 - 1}} ln (𝑟_{𝑖}) + \frac{𝑡 - 𝑡_{𝑖 - 1}}{𝑡_{𝑖} - 𝑡_{𝑖 - 1}} \cdot ln (𝑟_{𝑖 - 1})

So this is kind of a linear interpolation between two points, but in general these curves are not linear, they have some curvature

Exponential Interpolation

The discount factors are

𝑑 (𝑡_{1}) = 𝑒^{- 𝑟_{1} (𝑡_{1} - 𝑡_{0})}

𝑑 (𝑡_{2}) = 𝑒^{- 𝑟_{2} (𝑡_{2} - 𝑡_{0})}

This is exponential interpolation for the discount curve, which will give us

𝑟_{1} = - \frac{ln 𝑑 (𝑡_{1})}{𝑡_{1} - 𝑡_{0}}

𝑟_{2} = - \frac{ln 𝑑 (𝑡_{2})}{𝑡_{2} - 𝑡_{0}}

Then we define

𝑟 (𝑡_{𝑎}) = 𝜆 𝑟 (𝑡_{1}) + (1 - 𝜆) 𝑟 (𝑡_{2})

Now we’re going to calculate the discount factor

𝑑 (𝑡_{𝑎}) = 𝑒^{- 𝑟 (𝑡_{𝑎}) (𝑡_{𝑎} - 𝑡_{0})} = 𝑒^{- (𝜆 𝑟 (𝑡_{1}) + (1 - 𝜆) 𝑟 (𝑡_{2})) (𝑡_{𝑎} - 𝑡_{0})}

= 𝑒^{- 𝜆 𝑟 (𝑡_{1}) (𝑡_{𝑎} - 𝑡_{0})} \cdot 𝑒^{- (1 - 𝜆) 𝑟 (𝑡_{2}) (𝑡_{𝑎} - 𝑡_{0})}

Cubic Spline Interpolation

Given n zero rates of n distinct maturities,

we will denote jth maturity and zero rate pair, which we will denote as

$(𝑡_{𝑗}, 𝑅_{𝑗})$ where $𝑗 = 1, 2, \dots, 𝑛$

The idea is to use such polynomials

How many can we create? n-1 cubic polynomials.

These polynomials will have different coefficients

𝑅 (0, 𝑡) = {\begin{cases} 𝛽_{1, 0} + 𝛽_{1, 1} 𝑡 + 𝛽_{1, 2} 𝑡^{2} + 𝛽_{1, 3} 𝑡^{3} & 𝑡 \in [𝑡_{1}, 𝑡_{2}] \\ 𝛽_{2, 0} + 𝛽_{2, 1} 𝑡 + 𝛽_{2, 2} 𝑡^{2} + 𝛽_{2, 3} 𝑡^{3} & 𝑡 \in [𝑡_{2}, 𝑡_{3}] \\ 𝛽_{𝑛 - 1, 0} + 𝛽_{𝑛 - 1, 1} 𝑡 + 𝛽_{𝑛 - 1, 2} 𝑡^{2} + 𝛽_{𝑛 - 1, 3} 𝑡^{3} & 𝑡 \in [𝑡_{𝑛 - 1}, 𝑡_{𝑛}] \end{cases}

That’s the system, but how do we solve it?

What might be the problem with this?

Right now we have four parameters, and it’s 4 * (n-1). 4n-4 unknowns, n-1 determinants, so this is an undetermined system. We don’t have information in this system.

We can add some more information. The curve is continuous, so we can say that each part should be equal. Skipping some steps, we can say that

If R(0, t) is continuous and twice differentiable in t,

then n - 1 equations simplifies to (in a more compact form)

observed at time 0, maturity t

𝑅 (0, 𝑡) = 𝑎 + 𝑏 (𝑡 - 𝑡_{1}) + 𝑐 (𝑡 - 𝑡_{1})^{2} + \sum_{𝑘 = 1}^{𝑛 - 1} 𝑑_{𝑘} (𝑡 - 𝑡_{𝑘})_{+}^{3}

This is dependent on the value of t, obviously.

Let’s say you have a value of t in between t_2 and t_3.

\sum_{𝑘 = 1}^{𝑛 - 1} 𝑑_{𝑘} (𝑡 - 𝑡_{𝑘})_{+}^{3} = 𝑑_{1} (𝑡 - 𝑡_{1})_{+}^{3} + 𝑑_{2} (𝑡 - 𝑡_{2})_{+}^{3} + 𝑑_{3} (𝑡 - 𝑡_{3})_{+}^{3} + \dots

The first two terms will have positive values, and the rest will be 0

Now we’ve restricted our unknowns to a, b, c, and the discount factor $𝑑_{𝑞}, 𝑑_{2}, \dots 𝑑_{𝑛 - 1}$ which is n + 2 unknowns with n discrete rates. This is still not fully determined.

You may see different results in different implementations based on the next steps, in Python or R.

We need two more conditions. It depends on the problem. Typically, there are conditions that make sense.

The first derivative at the ends of the intervals, we can make the assumption that the convexity is 0 right at the end of the intervals.

lim_{𝑡 \to 𝑡_{1}} 𝑅^{″} (0, 𝑡) = 0

This counts as two conditions because it’s left and right.

Then you have n + 2 linear equations and n + 2 unknowns. Now we are fully determined. The solution is very easy to obtain 😏

Let’s take the vector

R_1 is going to correspond to t_1

Then you have to take the second order derivative, so what do you get for c? First derivative is 2c

In another way, you could write this as

[\begin{matrix} 𝑅_{1} \\ 𝑅_{2} \\ ⋮ \\ 𝑅_{𝑛} \\ 0 \\ 0 \end{matrix}] = 𝐴 \cdot [\begin{matrix} 𝑎 𝑏 𝑐 𝑑_{1} ⋮ 𝑑_{𝑛 - 3} 𝑑_{𝑛 - 2} 𝑑_{𝑛 - 1} \end{matrix}]

Which we can solve by inverting the matrix

𝐴^{- 1} \cdot [\begin{matrix} 𝑅_{1} \\ 𝑅_{2} \\ ⋮ \\ 𝑅_{𝑛} \\ 0 \\ 0 \end{matrix}] = 𝐴^{- 1} \cdot 𝐴 \cdot [\begin{matrix} 𝑎 𝑏 𝑐 𝑑_{1} ⋮ 𝑑_{𝑛 - 3} 𝑑_{𝑛 - 2} 𝑑_{𝑛 - 1} \end{matrix}]

And then

[\begin{matrix} 𝑎 \\ 𝑏 \\ 𝑐 \\ 𝑑_{1} \\ ⋮ \\ 𝑑_{𝑛 - 3} \\ 𝑑_{𝑛 - 2} \\ 𝑑_{𝑛 - 1} \end{matrix}] = 𝐴^{- 1} \cdot [\begin{matrix} 𝑅_{1} \\ 𝑅_{2} \\ ⋮ \\ 𝑅_{𝑛} \\ 0 \\ 0 \end{matrix}]

We got the values in A by taking the second order derivative, and the limit approaching t_1

Explanation of derivatives

𝑅 (0, 𝑡) = 𝑎 + 𝑏 (𝑡 - 𝑡_{1}) + 𝑐 (𝑡 - 𝑡_{1})^{2} + \sum_{𝑘 = 1}^{𝑛 - 1} 𝑑_{𝑘} (𝑡 - 𝑡_{𝑘})_{+}^{3}

𝑅^{'} (0, 𝑡) = 𝑏 + 2 𝑐 (𝑡 - 𝑡_{1}) + \sum_{𝑘 = 1}^{𝑛 - 1} 3 𝑑_{𝑘} (𝑡 - 𝑡_{𝑘})_{+}^{2}

𝑅^{″} (0, 𝑡) = 2 𝑐 + \sum_{𝑘 = 1}^{𝑛 - 1} 6 𝑑_{𝑘} (𝑡 - 𝑡_{𝑘})_{+}

Then the limit will show this collapse

lim_{𝑡 \to 𝑡_{1}} 𝑅^{″} (0, 𝑡) = 2 𝑐

lim_{𝑡 \to 𝑡_{𝑛}} 𝑅^{″} (0, 𝑡) = 2 𝑐 + 6 𝑑_{1} (𝑡_{𝑛} - 𝑡_{1}) + 6 𝑑_{2} (𝑡_{𝑛} - 𝑡_{2}) \dots

But the matrix is a more elegant form.

This particular function will pass through all of these points, and it will be a beautiful curve.

Functional Form of Yield Curve

In general, if coupon bonds are available, then use the functional form

We’re going to look at three popular models:

Nelson-Siegel Model

This is no longer an interpolation function, it’s more of a fitting function.

We will write the specification and the characteristics for this function, and understand the procedure

𝑅 (0, 𝑡) = 𝛽_{0} + 𝛽_{1} (\frac{1 - 𝑒^{- 𝜆 𝑡}}{𝜆 𝑡}) + 𝛽_{2} (\frac{1 - 𝑒^{- 𝜆 𝑡}}{𝜆 𝑡} - 𝑒^{- 𝜆 𝑡})

where we have $𝛽_{0}, 𝛽_{1}, 𝛽_{2}$ constants and $𝜆 > 0$ to specification. This is finding four parameters.

This particular function has some characteristics, we can look at these.

lim_{𝑡 \to \infty} 𝑅 (0, 𝑡) = 𝛽_{0}

If you take a longer maturity, it goes to a constant. So $𝛽_{0}$ should be associated with the long-term interest rate, like the level of it. Before that, it has some curvature. What about at 0?

Well, you would end up a 0/0, so you can apply L’Hopital’s rule.

lim_{𝑡 \to 0} = 𝛽_{0} + lim_{𝑡 \to 0} 𝑒^{- 𝜆 𝑡} + 𝛽_{2} lim_{𝑡 \to 0} (𝑒^{- 𝜆 𝑡} - 1)

= 𝛽_{0} + 𝛽_{1}

This is the spot interest rate, at t = 0.

Level Slope Curvature represent the three betas.

We have a numerical example just to see

Example

Assume that there are N coupon bonds.

For i = 1, 2, \ldots N, the ith bond has the following specification

B_i = cash(dirty) price of bond

m_i = remaining number of coupon payments

These bonds have different maturities, and they may have different specifications in the coupon payments (frequency and amount)

t_i^j = coupon payment date for j = 1, \ldots m_i

t_i^1 = date of next coupon payment

t_i^{m_i} = maturity date and date of last coupon

𝛿_{𝑖} = 𝑡_{𝑖}^{𝑗 + 1} - 𝑡_{𝑖}^{𝑗}

P_i = principal

C_i (amount of each coupon) = $𝛿_{𝑖} 𝑃_{𝑖} \cdot 𝐶 𝑜 𝑢 𝑝 𝑜 𝑛 𝑅 𝑎 𝑡 𝑒$

B_i = Market Quote (Clean Price) + $𝐶_{𝑖} \frac{𝛿_{𝑖} - 𝑡_{𝑖}^{1}}{𝛿_{𝑖}}$ (Accrual of Interest)

Theoretical Price:

\hat{𝐵_{𝑖}} = \sum_{𝑗 = 1}^{𝑚_{𝑖}} 𝑒^{- 𝑅 (0, 𝑡_{𝑖}^{𝑗}) \cdot 𝑡_{𝑖}^{𝑗}} \cdot 𝐶_{𝑖} + (𝑇 ℎ 𝑒 𝑛 𝐷 𝑖 𝑠 𝑐 𝑜 𝑢 𝑛 𝑡 𝑃 𝑟 𝑖 𝑛 𝑐 𝑖 𝑝 𝑎 𝑙) 𝑒^{- 𝑅 (0, 𝑡_{𝑖}^{𝑚_{𝑖}}) \cdot 𝑡_{𝑖}^{𝑚_{𝑖}}} \cdot 𝑃_{𝑖}

Basically you can calculate this for all bonds.

If you want to write this as an optimization problem, we want to find

$𝛽_{0}, 𝛽_{1}, 𝛽_{2}, 𝜆$ that minimizes

\sum_{𝑖 = 1}^{𝑁} 𝑤_{𝑖} (\hat{𝐵_{𝑖}} - 𝐵_{𝑖})^{2}

that satisfies

{𝛽_{0} > 0 𝛽_{0} + 𝛽_{1} > 0 𝜆 > 0

This w_i item is the weight assigned to bond i, if you want to equally weight them then it’s 1.

Choices for w_i:

w_i = $\frac{1}{𝑡_{𝑖}^{𝑚_{𝑖}}}$ This one is a function of maturity, so longer maturity will have smaller weights. This will fit better on shorter maturity than longer maturity.
w_i = $- \frac{1}{𝐵_{𝑖}} \cdot \frac{𝜕 𝐵_{𝑖}}{𝜕 𝑌 𝑇 𝑀}$ This one is dependent on the duration of the bond

Siegel-Svensson

Adds a second hump term (2 possible maximaminima

Let R(0, t) be the zero rate for maturity t, then

𝑅 (0, 𝑡) = 𝛽_{0} + 𝛽_{1} (\frac{1 - 𝑒^{- 𝜆_{1} 𝑡}}{𝜆_{1} 𝑡}) + 𝛽_{2} (\frac{1 - 𝑒^{- 𝜆_{1} 𝑡}}{𝜆_{1} 𝑡} - 𝑒^{- 𝜆_{1} 𝑡}) + 𝛽_{3} (\frac{1 - 𝑒^{- 𝜆_{2} 𝑡}}{𝜆_{2} 𝑡} - 𝑒^{- 𝜆_{2} 𝑡})

We have some constraints:

{𝛽_{0} > 0 𝛽_{0} + 𝛽_{1} > 0 𝜆_{1}, 𝜆_{2} > 0

The optimization problem for the previous problem was for price of the bonds. But otherwise we can write it as

We have the factor function, then the fitting function, then the objective function. That’s how you get Nelson-Siegel.

For the optimization methods, there are some characteristics,

constrOptim.nl is what he uses

estimates

Smith-Wilson

Generates a smoothed interpolated and extrapolated term structures that fit the spot market rates.

has simplicity, works for low number of market data points but performance

This going to work like an interpolation, but it also extrapolates for longer maturities.

Parameters:

UFR: ultra long-term forward rate

LLP: last liquid point, where the zero coupon market support ends (e.g. 20 years)

CP: Convergence

Assignment

This homework is due in two weeks. The first problem is similar to what we did last time, given some cash flows from different bonds, fill in the curve.

Determine the forward curve, bump it, and then if you change it, what’s the change? If you have 10 years, you have 10 such cases, bump for each year.

## Black–Scholes, Calibration, and Implied Volatility - URL: https://sharifhsn.dev/blog/computational-methods-week-02/ - Structured data: https://sharifhsn.dev/api/posts/computational-methods-week-02.json - Description: Hanlon 2 is occupied by the system administrator, Zheng Xing. He will move his class so that we can occupy there. They will need to remote desktop into the lab. I should have… - Date: 2025-02-04 - Exact published timestamp: 2025-02-04 - Topics: Computational Methods, Black-Scholes, Calibration, Implied Volatility, Stochastic Volatility - Categories: Computational Methods - Source: Computational Methods in Quantitative Finance - Source URL: None - Content source format: typst

Bad News

Hanlon 2 is occupied by the system administrator, Zheng Xing. He will move his class so that we can occupy there. They will need to remote desktop into the lab.

I should have come here early to help him 🙁

There’s two modes, Zoom Mode, and Bypass Zoom. If you bypass, you can’t use microphone or camera.

Doesn’t really matter, next week will be different.

Plan

We’ll look at some stochastic models. Then we will look at finding roots, and finding volatility, and so on.

Black-Scholes Model

Traditional model:

𝑑 𝑆_{𝑡} = 𝜇 𝑆_{𝑡} 𝑑 𝑡 + 𝜎 𝑆_{𝑡} 𝑑 𝑊_{𝑡}

The thing about this equation, is that it is under the objective probability measure.

You can see that there is a μ parameter that characterizes the stock.

Last time we talked about MLE. That is concerned with estimating the parameters, such as μ and σ.

This can only be done when you observe the actual stochastic process prices. It’s an estimation that you do under this objective probability measure.

However, when you calculate option prices, you go to the equivalent measure Q which is risk-neutral. The equation is slightly different.

𝑑 𝑆_{𝑡} = 𝑟 𝑆_{𝑡} 𝑑 𝑡 + 𝜎 𝑆_{𝑡} 𝑑 𝑊_{𝑡}^{𝑄}

We can observe this by doing the Girsanov transformation.

So what’s the difference here?

The option prices are always under Q.

Calibration

There is a different method to estimate parameters, called calibration.

We’re going to mention this idea here, and then expand later. Estimation and calibration are different. Estimation is a very standard method in statistics. The condition of estimation is that you observe the option prices. Calibration is a term invented in finance, which is only applicable here. It deals with this particular situation where you observe option prices $𝑐_{1}, 𝑐_{2}, \dots 𝑐_{𝑛}$ . These are derivative prices, not the underlying prices, derived under Q. If you’re estimating parameters for this, you’re going to be estimating r and σ. We’re lucky because under Black-Scholes because σ is the same. But we can’t estimate μ at all, we need stock prices to do that, because μ vanishes.

The method is, we would set theoretical prices “equal” to observed prices, and then we solve for parameters. And this is called calibrating parameters. In traditional statistics, you assume that this is a number. But when you do this methodology, and you estimate (e.g.) implied volatility, you would see how it changes based on different option prices. But then you say that σ is different, so how do you adjust for that? When you calibrate to real data, then you get it. The problem is that the model is bullshit.

Every time you use derivatives, you are estimating under Q.

The reason we are using BS and implied volatility is because there is a formula for this. You should know the formula. C the option price is a function of both S and t, the observed stock price and the time. However each price depends on other stuff, like K the strike price, r the interest rate, and T the time to maturity, and σ the volatility. It looks like there are six things in this equation, but it actually only depends on time to maturity, so there are five things.

𝐶 (𝑆, 𝑇 - 𝑡, 𝐾, 𝑟, 𝜎) = 𝑆 𝑁 (𝑑_{1}) - 𝐾 𝑒^{- 𝑟 (𝑇 - 𝑡)} 𝑁 (𝑑_{2})

Where

𝑑_{1} = \frac{log (𝑆 / 𝑘) + (𝑟 + 𝜎^{2} / 2) (𝑇 - 𝑡)}{𝜎 \sqrt{𝑇 - 𝑡}}

𝑑_{2} = \frac{log (𝑆 / 𝑘) + (𝑟 - 𝜎^{2} / 2) (𝑇 - 𝑡)}{𝜎 \sqrt{𝑇 - 𝑡}}

And N is the normal cdf. If you know K and you divide by K everywhere, you can get the same equation in terms of moneyness, which is determined by S/k. Moneyness is used in practice, because, for example, if you calculate at the

money option for AAPL, where it’s $600 at the money, then the following options are $605, then $610. But for AMD, it’s $10 at the money, then $10.5, then $11. By using moneyness, you scale all of these in the same way.

S is observed, K is observed, r is not really observed but you keep it fixed. You just go to the Federal Reserve and pick whatever risk-free interest rate, just keep it consistent, 3M or 3Y, whatever.

Time to maturity is a small value. Everything here is expressed in years. You need to keep this consistent: everything is in years. Unit conversions are a big problem.

This will give you yearly volatility. This is something well understood. S&P 500 has volatility 0.2, IBM 0.6. This is relatively stable. You could express everything in terms of days, but then you would get a scaled daily volatility which is very small. Because of this normality assumption.

You can get returns here by doing a transformation and applying Ito. To be clear, this is under Q.

𝑅_{𝑡} = log 𝑆_{𝑡}

𝑑 𝑅_{𝑡} = (𝑟 - \frac{𝜎^{2}}{2}) 𝑑 𝑡 + 𝜎 𝑑 𝑊_{𝑡}^{𝑄}

You always eliminate S this way when you take the logarithm.

Look at the form of this! If you discretize this process, you get

𝑅_{𝑡 + Δ 𝑡} - 𝑅_{𝑡} = \int_{𝑡}^{𝑡 + Δ 𝑡} (𝑟 - \frac{𝜎^{2}}{2}) 𝑑 𝑡 + 𝜎 (𝑊_{𝑡 + Δ 𝑡} - 𝑊_{𝑡})

You can see the increment of the Brownian motion, which is normal.

This is why the return is very nice, because this is the increment.

Then you get the continuously compounded return, which is normal and also independent.

This is very nice to work with, and it only works for the Brownian motion.

Continuously Compounded Return

Typically if you calculate return over a year, let’s say you have an asset that changes over time. That has the following return:

𝑟_{𝑡} = \frac{𝑆_{𝑡 + Δ 𝑡} - 𝑆_{𝑡}}{𝑆_{𝑡}}

This is the simple return, and tells you what happens at the end. But what happened in the middle? When you make the increments smaller, you eventually get

𝑅_{𝑡} = log (\frac{𝑆_{𝑡 + Δ 𝑡}}{𝑆_{𝑡}})

Every moment in time, I’m going to earn more. This is a mathematical construct and doesn’t exist. But it’s convenient, because it gives us the mean plus variance of the normal.

Approximating PDEs

If V is a European type option, you can only exercise at maturity. V(T) is then only a function of S(T). It only depends on the stock price at maturity. You can have European call or put, as long as the value only depends on the stock price at maturity. We can show that V solves a particular type of education, which is called the parabolic PDE.

\frac{𝜕 𝑉}{𝜕 𝑡} + \frac{1}{2} 𝜎^{2} 𝑆^{2} \frac{𝜕^{2} 𝑉}{𝜕 𝑆^{2}} + 𝑟 𝑆 \frac{𝜕 𝑉}{𝜕 𝑆} - 𝑟 𝑉 = 0

This is the PDE, but in order to solve this, you need boundary conditions to solve this. The proper solution is a surface, with an infinite number of solutions. We know that

𝑆 \in [0, \infty]

𝑡 \in [0, 𝑇]

Let’s consider the boundary at T, the time of payoff.

𝑉 (𝑆, 𝑇) = 𝑓 (𝑆_{𝑇})

We are not necessarily concerned with the function itself, it could be K - S, we’re not worried about that.

In the theory of PDEs, there are boundary conditions expressed in the function itself, or of derivatives of the function. there are also Dirichlet conditions.

We can’t put a condition at t = 0, because that’s what we’re looking for, the price at t = 0.

We can put a condition on the boundary at S = 0, and when S converges to ∞.

These are specific, and depend on the option you are pricing. If you are looking at a call option, going to infinity becomes S - K. which becomes 1 if you derive with respect to S. Going to 0 becomes 0 because

We will talk about this later. But these conditions are important!

These boundaries involved are Dirichlet boundary condition

Greeks

The Greeks involve the derivatives of the parameters. We have five Greeks.

K is fixed so it doesn’t have a derivative.

\frac{𝜕 𝑉}{𝜕 𝑇} = 𝜃

When time changes, how does the option change?

\frac{𝜕 𝑉}{𝜕 𝑟} = 𝜌

The interest rate changes for time to time, but it’s not that big of a deal.

\frac{𝜕 𝑉}{𝜕 𝜎} = 𝑉 𝑒 𝑔 𝑎

By this theory, σ is supposed to be constant. So this doesn’t really exist, it’s kind of a joke.

\frac{𝜕 𝑉}{𝜕 𝑆} = 𝛿

\frac{𝜕^{2} 𝑉}{𝜕 𝑆^{2}} = 𝛾

These are the big kahunas.

Delta is the sensitivity to changes in stock price.

Gamma measures the concavity or convexity, wherever you are on the curve.

These are used in hedging.

Job Market Tangent

Fortunately, finance is very competitive. So it’s very hard to find a job based on your specialty, becoming a quant. JP Morgan has 300 quants in total, throughout its whole company, out of tens of thousands. 200 of them are in England. It’s very hard to become a quant.

Those are real quants that do this shit. The good news is that this program doesn’t only prepare you to become a quant. It prepares you to become a quant in any company. JP Morgan has 300 real quants, but it has data scientists, quantitative analysts, that do other related things.

Heston Model

I’ve been making fun of this stochastic process BS, what is the alternative? Why do we keep learning about IV if it doesn’t really work?

Two reasons. 1. Most people in finance are imbeciles, so BS is the only thing they can understand. Ten years ago IV wasn’t on Yahoo Finance. 2. It’s a self-fulfilling prophecy, because everyone uses it. I’m going to buy Trump stock because he’s elected, so it goes up. Now there’s lots of people buying, but nobody selling. But the guy selling will see that the model says that they’re overpaying, so he will sell.

The Heston model is not useful for vanilla option prices. If you happen to work as a quant, you won’t work with idiots trying to buy Trump stock. You are going to deal with another quant who works at Barclays who says I have $500 million cash flow, and I want to ensure that the rate I’m getting is 5%, so I’m going to do a swap. Give me the price of a swap. This is a completely nontraditional thing, some instrument that you need to price. In order to understand this non-standard situation, you need to understand the simple model.

𝑑 𝑆_{𝑡} = 𝑟 𝑆_{𝑡} 𝑑 𝑡 + \sqrt{𝑉_{𝑡}} 𝑆_{𝑡} 𝑑 𝑊_{𝑡}^{1}

𝑑 𝑉_{𝑡} = 𝐾 (𝜃 - 𝑉_{𝑡}) 𝑑 𝑡 + 𝜎 \sqrt{𝑉_{𝑡}} 𝑑 𝑊_{𝑡}^{2}

These two Brownian motions are correlated by ρ

𝜌 𝑑 𝑡 = 𝔼 [𝑑 𝑊_{𝑡}^{1} 𝑑 𝑊_{𝑡}^{2}]

You can write a Heston PDE and solve it with the methodology in the book.

Let’s just understand this structure.

Note that S_t is linear, so we can get rid of it and make it explicit in terms of the process V_t and the Brownian motion. We can do the Ito to the logarithm of S_t.

The Brownian motion part can be negative. You can approximate the increment of V_t by the dt, because the increment is normally distributed.

Mean Reversion

θ - V_t will push towards θ because it reverts to it. The increment will be negative if V_t is above θ, and positive if V_t is below θ. And the mean of this process is actually θ, which we can calculate. In practice, K is called the speed of mean reversion.

If the speed of mean reversion determines how much it goes back and forth across the mean. Keep in mind that this is stochastic and not guaranteed, but this is the average trajectory.

SABR - Stochastic α β ρ

This is one of the first stochastic volatility models to be created, and was used in the insurance industry.

𝑑 𝑆_{𝑡} = 𝛼_{𝑡} 𝑆_{𝑡}^{𝛽} 𝑑 𝑊_{𝑡}

𝑑 𝛼_{𝑡} + 𝜈 𝑑_{𝑡} 𝑑 𝑍_{𝑡}

ρ correlation

This doesn’t have a solution, but it’s very popular.

Assume that the stochastic process follows the SABR model. THen you have an option price, but with no formula. But you do have a formula for the IV of that option price. So what you do is calculate that IV, and then plug it into BS. The reason that it was powerful because interest rate models had used BS already in their code, so they would need to change all their code to use something else. They were all spaghetti code idiots.

There used to be a guy at JP Morgan who was 70 years old and kept working. The reason was because he wrote code in COBOL 30 years ago and was running $5 billion instruments valued every day based on this model. Jamie Diamond commanded that they rewrite everything in Python, so he’s gone now.

Cox-Ingersoll-Ross

𝑑 𝑉_{𝑡} = 𝛼 (\bar{𝑉} - 𝑉_{𝑡}) 𝑑 𝑡 + 𝜎 \sqrt{𝑉_{𝑡}} 𝑑 𝑊_{𝑡}

I like α for speed of mean reversion and $\bar{𝑉}$ for my mean reversion.

The square root thing is called the vol of vol parameter.

You can use this for stock prices directly.

Before they went back to Vasicek, they used this for interest rates, or anything dealing with them like swaps, caps, floors, etc.

The question is, what’s the expected value and squared?

We can solve this very neatly by expressing this as an integral.

𝑉_{𝑡} - 𝑉_{0} = \int_{0}^{𝑡} 𝛼 (\bar{𝑉} - 𝑉_{𝑠}) 𝑑 𝑠 + \int_{0}^{𝑡} 𝜎 \sqrt{𝑉_{𝑠}} 𝑑 𝑊_{𝑠}

𝔼 [𝑉_{𝑡}] - 𝔼 [𝑉_{0}] = 𝔼 [\int_{0}^{𝑡} 𝛼 (\bar{𝑉} - 𝑉_{𝑡}) 𝑑 𝑡] + 𝔼 [\int_{0}^{𝑡} 𝜎 \sqrt{𝑉_{𝑡}} 𝑑 𝑊_{𝑡}]

The expectation of the right side is 0 because it’s a Brownian motion.

For the left side, as long as what is under the Riemann integral is finite, the integral order doesn’t matter. So we can flip them, and we want to know the expectation.

We will define a deterministic function here based on the time t.

𝔼 [𝑉_{𝑡}] = 𝑣 (𝑡)

Now we can determine

𝑣 (𝑡) - 𝑣 (0) = \int_{0}^{𝑡} 𝛼 (\bar{𝑉} - 𝑣 (𝑠)) 𝑑 𝑠

This is a Riemann integral that we can directly solve.

Then take the derivative.

𝑣^{'} (𝑡) = 𝛼 (\bar{𝑉} - 𝑣 (𝑡))

𝑑 𝑣 = 𝛼 \bar{𝑣} 𝑑 𝑡 - 𝛼 𝑣 (𝑡) 𝑑 𝑡

α appears out of nowhere if we derive

𝑒^{𝛼 𝑡}

so we will actually multiply this to make it appear

𝑒^{𝛼 𝑡} 𝑑 𝑉 + 𝛼 𝑒^{𝛼 𝑡} 𝑑 𝑡 𝑉 = 𝛼 𝑒^{𝛼 𝑡} \bar{𝑉} 𝑑 𝑡

Now we can integrate to give us the solution.

I can’t see the solution from where I’m sitting, check the textbook for this

The trick is literally useful when you can get some nice mean here in terms of prices. If you can say that, if the stochastic part was deterministic, then it would be easy to solve.

V^2 is left as exercise.

Tips: Apply Itô, then use the same trick to get rid of the stochastic part.

Implied Volatility

Implied volatility fits somewhere in between the bid and ask price for options.

Mathematicians solve this problem with a formula to calculate option price, such as BS.

In that model, σ is the only thing we don’t know. In order to solve this, we will set the option price formula to the average of the bid and ask (mid) and then solve for σ. You can rearrange as the difference between the theoretical price and the mid is 0. Then you can say you’re finding the root of a function.

𝑓 (𝜎) = 𝐶 (𝜎) - 𝑚 𝑖 𝑑 = 0

This value is called the implied volatility.

There is an obvious problem with this. This is for K_1 and t_1. What if I have a different strike price, or a different time to maturity? Then I’m going to have a different value. This is despite the fact that volatility is calculated on the underlying stock price and therefore should be constant.

This is called the volatility smile, or really the volatility smirk because the upwards trajectory is only in one direction. If you graph moneyness against implied volatility, you get the smirk. This only goes for one time. For T_2 larger than T_1, the smile is less clear, it’s shallower. How do you calculate implied vol? You might want to take an average of these points. Or you could take an average of a certain range.

Let’s say you have the data for option prices. Let’s do this

min \sum_{𝑖 = 1}^{𝑛} (𝐶 (𝑆, 𝑇_{𝑖}, 𝐾_{𝑖}, 𝑟, 𝜎) - 𝐶^{𝑖})^{2} \cdot 𝑊_{𝑖}

Where you weight each price and you minimize the volatility. This is a fix, because the model is crap.

There’s something better.

Local Volatility

This was pioneered by a bunch of math guys at Bloomberg. Dupire, Derman (director of FE at NYU, now retired).

They said, let’s fit everything. We’ll look at these values and create the local vol surface. It is a bad model, but it’s what people use. The industry uses a combination of local vol and stochastic vol. The professor’s students who work use this.

Local vol is not that complicated. Remember the equation for stock price from BS?

𝑑 𝑆_{𝑡} = 𝑆_{𝑡} (𝑟 - \frac{𝜎^{2}}{2}) 𝑑 𝑡 + 𝜎 𝑆_{𝑡} 𝑑 𝑊_{𝑡}

What these guys said, is that the dt term is crap, we’ll say it’s 0. The σ is clearly not constant, but I don’t want it to be a stochastic process. Then the natural thing to do is replace σ with σ(t) which is a function of time. Now this is a failure because nothing changes.

Now they have tons of data, and this is used for interest rates. Interest rates have two important times, the tenor and the maturity. Tenor is how much time remains. So they need a way to fit based on this information.

Let’s do σ(t, S_t). Where do we get this function from?

In the dV PDE, we have σ. So let’s extract σ.

\\sigma(T, K)^2 = 2\\frac{\\tfrac{\\partial C}{\\partial t} + (r_T  - 2T)K\\tfrac{\\partial K}{Z}

This is pretty much the same question in terms of strike price and time.

For each particular option price, I can estimate these derivatives. I’m going to market, r_t, and Q_t re known.

Then you can construct a vol surface from this, which you can find in Bloomberg.

The whole thing is bullshit. I wrote a paper about this. You’re fixing the problem by creating another problem. These derivatives are a big problem. How do you solve for the derivatives? You use finite differences (we will mention later when talking about approximating PDEs). If you have a function f, the derivative df/dx is approximately f(x+Δx) - f(x). As long as you take Δx goes to 0, this will converge to the derivative. This is called the finite difference, which you can express in multiple ways. This is first-order difference, and then you can use it substitute the derivatives. To get the second derivative, you do the first difference of the first difference.

You can see that this is the strike price minus a little bit. But the problem is that it DOES NOT EXIST. In practice, strike prices are very far apart from each other. They justify this for use in interest rate models, in which the interest rate increments are very small. But then with interest rates, the change in time is VERY FAR, so each of them will fail.

Paper

This stuff is not new, in 1998. Black-Derman-Toy? Model. Florescu heard about it and got mad, and wrote a paper to critique this.

Personal reminder: look at this on Canvas.

ChatGPT

Gradient descent is “Brute force”, so optimization actually SUCKS, brute force is the best thing.

Bisection

If I have f(x), how do I find x_0 such that f(x_0) = 0. This is finding the root of a function, which is a famous problem.

Bisection is very stupid and very simple, and it only works for our problem.

It’s designed on the fact that if you have a root on an interval, then you see that

f(a) * f(b+a/2) < 0, and if it’s true, then you can bisect this interval and look for the root, repeating like binary search.

None of this requires the BS formula, you just need a way to get the option value for a set of parameters. You could use an approximation method like a tree or finite difference. But because you are doing this a lot, you would like to have an analytical solution to be faster.

A couple of issues with this:

Interval $[𝑎, 𝑏]$ needs to contain the root. There cannot be two roots. The option price is increasing in terms of $𝜎$ , and this is monotonically increasing, so there is only one root. The root is always positive, we know that IV is always positive, so you can always use $𝑎 = 0$ . Idiots tell you IV is always less than 1 because it’s a percentage. But the IV depends on the actual Call option price. If the price is way weird, then it’s not. The bid might not have been traded in a while, and the ask price moved with the stock price, so the spread is now huge. So now the average is way off. If you consistently get the root to be $𝑎$ or $𝑏$ for certain options, it is usually a problem with your data. IV is usually not huge, although it is possible when stock price moves a lot but option prices don’t. If you use $[0, 1]$ , you won’t capture that. The fix is simple, to use $[0, 10]$ , or $[0, 4]$ . In two steps, you’re at $[0, 1]$ .
This only works for R → R. You need to map into R to do comparisons of greater or less than, like Euclidean distance.

Newton Method

𝑥_{1} - 𝑥_{0} = \frac{𝑓 (𝑥_{1}) - 𝑓 (𝑥_{0})}{𝑓^{'} (𝑥_{0})}

My goal is to find x where f(x) = 0, aka the root.

Let’s rearrange

𝑥_{1} = 𝑥_{0} - \frac{𝑓 (𝑥_{0}}{𝑓^{'} (𝑥_{0})}

where we delete f(x_1)

Then I can keep doing this until the two points are the “same” (within ε)

This is the recurrence formula.

It works for R_n generalized, but there is a problem.

If you start too far away, you’re going to go in a weird direction.

Trump

I saw a conference from Trump on Tuesday when he brought a CEO from Morocco, and the Sam Altman guy. He thinks that because it creates words, it’s alive.

Trump said we need to be leaders in AI. “I am prepared to give you $500 billion” The next day Chinese do it cheaply.

Next class, approximate stochastic processes using trees.

First homework is due… will be determined tomorrow. We will have 2-3 weeks to do it.

Old Paper

Introduction

We are concerned with estimating expected future volatility. One approach is with historical data. Another is to use market option prices and derive volatility from option valuation formulas.

Doing this derivation requires some assumptions based on which formula you’re using. Black-Scholes requires asset prices to follow geometric Brownian motion and constant volatility. Since volatility is constant, then this disagrees with the real-life volatility smile.

In order to account for this, we might think to change volatility based on the stock price. This typically means some kind of stochastic volatility model, which is not easy to solve for. In a special case, if the volatility is deterministic, then we can use the Black-Scholes PDE to solve option prices.

DK, D, and R have developed “local volatility” which does the following:

“Their methods attempt to fit a cross section of option prices and deduce the future behavior of volatility as anticipated by market participants. Rather than give a formula or a structural form for the volatility function, they search for a binomial or trinomial lattice that achieves an exact cross-sectional fit of reported option prices.”

In this way it’s similar to implied volatility, but it attempts to capture the volatility surface.

## Bond Pricing, Duration, and DV01 - URL: https://sharifhsn.dev/blog/advanced-derivatives-week-01/ - Structured data: https://sharifhsn.dev/api/posts/advanced-derivatives-week-01.json - Description: Take-home exam format, 24/48 hours, lots of writing code, during finals week. If a company wants money, it has options. It can issue stock, get a bank loan, or issue a bond. What… - Date: 2025-01-30 - Exact published timestamp: 2025-01-30 - Topics: Fixed Income, Bonds, Yield Curves, Bond Pricing, DV01, Duration, Convexity - Categories: Fixed Income - Source: Advanced Derivatives - Source URL: None - Content source format: typst

Final Exam

Take-home exam format, 24/48 hours, lots of writing code, during finals week.

Fixed Income Instruments

If a company wants money, it has options. It can issue stock, get a bank loan, or issue a bond.

What is the cost of borrowing money?

What determines this cost?

One such characteristic is that it can be determined by the market. If more people want to lend money at a certain maturity, it can lower the interest rate, with money being the commodity.

This rate is determined by the central bank, which has influence on short-term borrowing. Some governments can determine the cost of borrowing.

When a particular entity wants to borrow money, it depends on rating score. There’s a probability they don’t pay back, that’s risk.

Inflation also plays a role in long-term expectations.

Money and Bond Markets

The main factor in the variability is the interest rates. The bonds and rates have an inverse relationship.

You can see every fixed income instruments as a stream of known (or rather contingent) cash flows.

The interest rate is not a single interest rate for a particular maturity.

If you look at Bloomberg, you will have many different rates that are associated with the Treasury, individual corporate entities, and so forth.

The money market is traded with banks and corporations. Cash flow delivery can be done up to 13 months. Can be done in cash, short-term securities. Deposits, T-bills, CDs, commercial paper, OR repo agreements. There’s a contract to repurchase a security sometime in the future, which will correspond to a rate.

There’s an interbank and eurocurrency markets. Banks are required to keep a percentage of deposits as reserve with their local Federal Reserve bank. Basically the banks will borrow reserves from each other overnight to avoid falling below this threshold. The effective federal funds rate is determined by the market, and influenced by the Federal Reserve, the open market operations.

Also, there’s a reference rate, the SOFR (Secured OVernight FInancing Rate), this is a broad measure of the cost of borrowing cash overnight, which is collateralized by treasury securities. For a long time LIBOR was taken as a reference rate, SOFR has replaced it. SOFR is an evaluated median of transaction medium, repo data collected by the Bank of New York.

The methodology is a little more complicated but this is a 50th percentile of volume. LIBOR was the rate at which banks were willing to loan to each other at. They didn’t necessarily have to have the transaction, they just had to post a quote. SOFR is data-driven.

Many fixed income securities like government bonds, agency debt, municipal debt, etc.

Creditworthiness

Some interest rates we might consider to be risk-free, like treasuries or US government agency securities.

There are also low-risk instruments, floating coupon bonds indexed by floating rate, swaps, futures, forward rate agreements. Here the idea is, by looking at the market, you’ll be able to find many such rates. So the yield curve you’ll find is not necessarily unique, here’s one for treasuries, corporate bonds, etc.

Here we’re going to look at the construction of such yield curves.

Bootstrapping Yield Curve

Definitions

A coupon paying bond is a contract that pays a fixed coupon at future times. Typically this is given in terms of the annual rate, like 5% per annum. Such coupon is based on the principal amount of the bond. This is paid with a certain frequency. This coupon bond has the cash flow of the reimbursement of the notional value of the bond.

Example

Let’s say we have a bond that pays 5% annual coupon with $𝑇 = 5$ years until maturity.

Let’s assume that the principal $𝑃 = 100$ . The cash flow diagram is very simple, with cash flows of $5 at each year after year 0, and then the max cash flow at the end.

The zero coupon bond for a maturity $𝑡$ guarantees the payment of one unit at maturity. This is important because we use this extensively. The price of a zero coupon bond paying $1 at maturity $𝑡$ is called the discount factor, $𝑑 (𝑡)$ or $𝑃 (𝑡, 𝑇)$ , the price of a bond maturing at $𝑇$ at time $𝑡$ .

In general, the price of zero coupon is less than 1. As the maturity of the contract increases, its value decreases. If you take

\frac{𝜕 𝑃 (𝑡, 𝑇)}{𝜕 𝑇} < 0

A par coupon bond is worth par, 100% of the notional. This is $𝑐 (𝑡)$ .

A forward rate agreement is a commitment to lend money at a specified future rate for a specified period time. The rate of a forward loan is called forward rate. The forward rate is $𝑓 (𝑡)$ from $𝑡 - 1$ to $𝑡$ .

A floating rate note is a contract ensuring payment of a floating rate at future dates and pays a last cash flow reimbursing the notional at T.

A swap pays fixed for floating to exchange periodic interest payment.

Bond Pricing

If you just observe the prices of the bonds, it doesn’t contain enough information to know value. You also have to compare the return holding the bond. One measure of the performance is the Yield to Maturity. This is the rate of return on a bond if it is held to maturity.

This is expressed as an annual rate. It’s important to understand the frequency of the rate to compare fairly. Based on current price we can compute a discount rate such that the PV of the future bond cash flows match the current price.

The constant discount rate represents the yield of the bond.

Example

Consider a coupon bond with 4 years to maturity at 10% annual coupon rate. Assume current bond price is $90. What is YTM? By definition, price is $𝑃 𝑉 = 90$ . Typically you take the notional in these to be $100. Therefore the price is

90 = \frac{10}{1 + 𝑦} + \frac{10}{(1 + 𝑦)^{2}} + \frac{10}{(1 + 𝑦)^{3}} + \frac{10}{(1 + 𝑦)^{4}} + \frac{100}{(1 + 𝑦)^{4}}

Solve for $𝑦 = 0.1338 = 13.38 %$ .

This would be the yield of this bond. If the bond price is equal to the principal then what is the yield?

100 = \frac{10}{1 + 𝑦} . . .

Then $𝑦 = 0.10 = 10 %$ . So then the yield is the same as the coupon rate when the bond is par.

Clean/Dirty

Clean price, also known as quoted price, is the price of the bond excluding interest that has accrued since issue or the most recent coupon payment. So the price of the bond is supposed to include this accrual of the value of that bond. However, it is not quoted with this accrual, so the clean price is quoted. Otherwise you would have to adjust rates continuously.

Dirty price, also known as cash price, is the price of a bond which includes the accrued interest. Let’s look at some calculations

Example

Consider buying a 3-year 12% annual coupon bond. (30/360 day count convention), 360 days and 30 days in a month, within one month from the first coupon. What is the YTM?

The coupon value is $𝑐$ . Then the final cash flow is $100 + 𝑐$ .

Let’s say you buy at time $𝑡$ , which is before $𝑡_{1}$ , one month before the first coupon payment. There is some accrued interest at time $𝑡$ . The formula is the same, with the cash flows discounted to time $𝑡$ . What is the accrual in this case? It’s of 11 months, which by our convention is $330 / 360 \times 𝑐$ . The coupon is 12% of the principal 100, therefore 12, so $330 / 360 \times 12 = 11$ .

We can say as a general formula that

𝐷 𝑖 𝑟 𝑡 𝑦 𝑃 𝑟 𝑖 𝑐 𝑒 = 𝐶 𝑙 𝑒 𝑎 𝑛 𝑃 𝑟 𝑖 𝑐 𝑒 + 𝐴 𝑐 𝑐 𝑟 𝑢 𝑒 𝑑 𝐼 𝑛 𝑡 𝑒 𝑟 𝑒 𝑠 𝑡

100 + 11 = \frac{12}{(1 + 𝑦)^{\frac{30}{360}}} + \frac{12}{(1 + 𝑦)^{\frac{390}{360}}} + \frac{112}{(1 + 𝑦)^{\frac{750}{360}}}

Using a solver, you can find that $𝑦 = 11.97 %$ .

The quotations would have to adjust continuously. Dirty prices can be reported but they are not typically quoted.

Sensitivity to Yield

Consider a fixed income instrument with price $𝑃$ and yield $𝑌$ .

We will define DV01 as the dollar value of one basis point.

𝐷 𝑉 01 = - \frac{Δ 𝑃}{10, 000 \cdot Δ 𝑦}

Negative means that as one increases, the other decreases.

We also have duration, which is

𝐷 = - \frac{1}{𝑃} \cdot \frac{Δ 𝑃}{Δ 𝑦}

and convexity

𝐶 = \frac{1}{𝑃} \cdot \frac{Δ^{2} 𝑃}{Δ 𝑦^{2}}

This is written as a difference, but you can generalize this as a derivative.

You can approximate the rate of change of the price of a bond as being dependent on the duration and convexity with a formula combining them.

Example 1

Consider a 1-year zero coupon bond with annual yield $𝑦$ . What is the relationship between $𝑃$ and $𝑦$ ?

𝑃 = \frac{1}{1 + 𝑦}

Then using our formula for DV01, we get

𝐷 𝑉 01 = - \frac{1}{10, 000} \frac{𝑑 𝑃}{𝑑 𝑌}

We can use the quotient rule (or chain rule!) to take the derivative, which results in

= \frac{1}{10, 000 (1 + 𝑦)^{2}}

## The Cost of Transit Delays - URL: https://sharifhsn.dev/blog/linkedin-2025-01-28-the-cost-of-transit-delays/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2025-01-28-the-cost-of-transit-delays.json - Description: As a frequent user of public transit, I know how impactful delays are both on my productivity and my mental health. Being stuck in Secaucus Junction because of some nonsense… - Date: 2025-01-28 - Exact published timestamp: 2025-01-28T14:46:42.541Z - Topics: Transportation, Public Policy - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/feed/update/urn:li:activity:7290017452040544257/ - Content source format: typst

As a frequent user of public transit, I know how impactful delays are both on my productivity and my mental health. Being stuck in Secaucus Junction because of some nonsense happening in NYC is a feeling all too familiar for me.

Researchers from Stevens Institute of Technology have quantified the impact of disinformation in the media by using AI to model disruption scenarios from the social media alerts for the PATH train. They used clustering techniques like K-means and BERTopic to classify different disruption scenarios, then used Monte Carlo simulation to generate potential train flows given these scenarios, calculating lost time and rerouting costs.

They discovered that even mild disruptions could lead to stations being closed as much as 11% of the time, if timed at certain periods like major sporting events. I’ve seen how the waiting area fills up with disgruntled Knicks fans after a Hawks game, and it’s a ticking time bomb for a targeted attack.

In our modern, social media-driven world, the potential for information propagation causing mayhem is more significant than ever before. It is incumbent on us all to remain calm when such disruptions happen, knowing that the cause may not be mere negligence, but active maliciousness by an attacker.

Read more here: https://lnkd.in/epaDBfH5

The underlying study: https://lnkd.in/ehDt9RNv

## Stochastic Processes and Computational Pricing - URL: https://sharifhsn.dev/blog/computational-methods-week-01/ - Structured data: https://sharifhsn.dev/api/posts/computational-methods-week-01.json - Description: I’m out on vacation this week, so I watched the recording A stochastic processes is merely a collection of indexed random variables. The indexing of these variables confers a… - Date: 2025-01-28 - Exact published timestamp: 2025-01-28 - Topics: Computational Methods, Stochastic Processes, Numerical Methods, Option Pricing - Categories: Computational Methods - Source: Computational Methods in Quantitative Finance - Source URL: None - Content source format: typst

I’m out on vacation this week, so I watched the recording

Textbook

Stochastic Processes

A stochastic processes is merely a collection of indexed random variables. The indexing of these variables confers a structure onto the rvs which gives them some properties.

The indexing structure can be a set or an interval, which would imply a discrete or continuous stochastic process, respectively.

Syllabus

Late assignments are not accepted under any circumstances, at least 24 hours in advance must be warned.

Attendance is mandatory for in-person, not for online.

Old book with pseudocode with formula.

which is a lot more

Main things we cover:

Monte carlo approximation

Finite difference

Trees

Black-Scholes PDE solution we study in 610

Most models do not have a formula though

So how do we estimate the value then? That’s what this class is about

There are two fundamental ways

Approximate the process. Let’s say I know the path for sure. I can calculate the value by using the payoff formula.

I can generate millions of paths and then average them. That is very slow.

Or I could look at the probability of each path, and then average them? That’s trees.

Or you could solve the PDE. You can get complicated PDEs, but they always have a very similar structure.

## Rust 1.84: Three Infrastructure Changes - URL: https://sharifhsn.dev/blog/linkedin-2025-01-10-rust-1-84-three-infrastructure-changes/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2025-01-10-rust-1-84-three-infrastructure-changes.json - Description: A new version of the Rust language has been released: Rust 1.84.0! Although there aren’t any new language features, this release kickstarts three big infrastructure improvements… - Date: 2025-01-10 - Exact published timestamp: 2025-01-10T13:25:09.372Z - Topics: Rust, Developer Tools - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/feed/update/urn:li:activity:7283473947021459456/ - Content source format: typst

A new version of the Rust language has been released: Rust 1.84.0!

Although there aren’t any new language features, this release kickstarts three big infrastructure improvements for Rust that have been a long time coming: minimum supported Rust version (MSRV), a new trait solver, and pointer provenance.

MSRV is a community-driven standard for Rust libraries that seek to be compatible with users that require older versions of Rust. For a while now, Rust has allowed libraries to declare an MSRV, but it doesn’t do anything other than inform the user. If, for example, a library bumped the MSRV up when adding a new feature, the user would be forced to manually specify an older version of the library, which is hacky and difficult to maintain. Now, Rust’s package manager Cargo will detect the user’s MSRV and only download versions of libraries compatible with it.

Rust’s powerful type system uses type theory to evaluate the correctness of the code’s type specifications. As new features have been requested from Rust over time, the system that solves these types has bolted on the required features as they are implemented, resulting in a hacky system that’s difficult to extend. The new from-scratch trait solver is simpler and is a necessary prerequisite to some highly-requested features, like specialization.

In systems programming languages like C, users will often need to work with raw pointers to memory i.e. integers. Rust uses references by default, which are pointers to memory that are tied to the object that created them, and therefore contain information relating to that object, like its mutability. This information is called provenance. References are what allows Rust to declare itself “memory-safe”. However, it is sometimes necessary to construct pointers out of thin air, instead of references to an object, for example when working in an embedded environment. Previously, only raw pointers could be used for this, resulting in an unwieldy, buggy, C-like interface for users. Now, pointers can contain provenance, which gives them a safer API.

Nothing flashy, but all great QOL improvements!

https://lnkd.in/eiBrWqPd

## Exotic Options - URL: https://sharifhsn.dev/blog/exotic-options/ - Structured data: https://sharifhsn.dev/api/posts/exotic-options.json - Description: This is the last lecture we have for this class. Next week will be a review session. Exam will be online . December 21, We’re going to be pricing exotic options. We are focusing… - Date: 2024-12-12 - Exact published timestamp: 2024-12-12 - Topics: Stochastic Calculus, Exotic Options, Barrier Options - Categories: Stochastic Calculus - Source: FE-610 \| Stochastic Calculus - Source URL: None - Content source format: typst

Lecture Notes

Exotic Options

This is the last lecture we have for this class. Next week will be a review session.

Exam will be online. December 21,

We’re going to be pricing exotic options. We are focusing on two types.

Barrier Options

Barrier options can be in vs out.

Let’s say we have a stock process that starts at $𝑆_{0}$ and evolves, where we generate a path via Brownian motion. In the past, we have only been concerned with the strike price, which is the line that determines whether an option will be exercised. We’re going to institute a new variable $𝐵$ . The relationship of the barrier is to the stock price is governed by up vs down. If the barrier is lower, it’s down. And whether it’s a knock in instrument, then your option starts worthless. It will remain worthless unless your stock touches the barrier. Then it could potentially be exercised at time of maturity. But it’s worthless until it’s out. That’s Frankestien’s monster that comes alive.

Down and out would be a normal person and an electric fence. You walk into it and you die.

Lookback Options

Lookback options look at the largest price in the lifetime subtracted by the smallest price in the lifetime. This is always going to be exercised, although it might not be worth anything. In a sense, there is no optionality present, although the value is the same as that of an option.

We’re going to need to do a little bit more background math.

We’re going to be really interested in the maximum our stock price achieves, for both barrier and lookback options.

For a BM W(t), define $𝑀 (𝑡) = {max}_{0 \leq 𝑢 𝑙 𝑒 𝑞 𝑡} 𝑊 (𝑢)$

The joint distribution of a Brownian motion and its maximum is

$𝑓_{𝑀 (𝑡), 𝑊 (𝑡)} (𝑚, 𝑤) = \frac{2 (𝑍 𝑚 - 𝑤)}{𝑡 \sqrt{2 𝜋 𝑡}} 𝑒^{- \frac{1}{2 𝑡} (2 𝑚 - 𝑤)^{2}}$

We have a lot of tools in our toolbox which allows us to get the individual distribution of the maximum.

It’s not going to be done by integrating through W.

I’m going to define two new processes.

We need to come up with these distribituions

As a point of consideration:

If I wanted to look at the maximum value of my stock price, knowing that my stock price grows via

$𝑆 (𝑡) = 𝑆 (0) exp {(𝛼 - \frac{𝜎^{2}}{2}) 𝑡 + 𝜎 𝑊 (𝑡)}$

achieves its maximum for

${max}_{0 \leq 𝜇 \leq 𝑡}$ This may not be the same as when the Brownian motion is largest. So we need more work! How to get the maximum of a Brownian motion with dirft. And that’s the point of creating this $\hat{𝑊}$ which incorporates the drift term.

The risk-neutral probability measure

$\hat{𝑊} = \tilde{𝑊} (𝑡) + 𝛼 𝑡$

That would be thai Brownian motion with drift

I could also write this as

$= \tilde{𝑊} (𝑡) + \int_{0}^{𝑡} 𝛼 𝑑 𝑢$

If I was to create a process $\tilde{𝑇 ℎ 𝑒 𝑡 𝑎} (𝑢) = 𝛼$ , wouldn’t this now be

$= \tilde{𝑊} (𝑡) + \int_{0}^{𝑡} \tilde{Θ} (𝑢) 𝑑 𝑢$

And that is the Girsanov process!

If I define my Radon Nikodym derivative

$\tilde{𝑍} (𝑡) = exp {- \int_{0}^{𝑡} \tilde{Θ} (𝑢) 𝑑 \tilde{𝑊} (𝑢) - \frac{1}{2} \int_{0}^{𝑡} {\tilde{Θ}}^{2} (𝑢) 𝑑 𝑢}$

I would get

$= 𝑒^{- 𝛼 \tilde{𝑊} (𝑡) - \frac{1}{2} 𝛼^{2} 𝑡}$

Using $\hat{𝑊}$ here

The source note switches notation to $\hat{𝑊}$ here without recording the complete expression.

Under $\hat{ℙ}$ , $\hat{𝑊} (𝑡)$ is a Brownian motion.

So I know that they have the joint distribution with a closed form.

So how do we go from risk-neutral distribution to normal?

Lemma 5.2.1 says

$\tilde{𝔼} [𝑌] = 𝔼 [𝑌 𝑍 (𝑡)]$

This also implies

$\hat{𝔼} [𝑌] = \tilde{𝔼} [𝑌 \tilde{𝑍} (𝑡)]$

Then under my risk-neutral probability measure,

something something lemma 5.2.5

Markovs,

martingales,

geometric Brownian motion,

it’s all in the lookback option.

The only time you get speed and accuracy is with the closed-form solution.

Final homework is due in a few days.

Make sure you look over all of the material.

You can go to the review session at the ENGAGE on Monday night.

Whatever questions you have are answered on Monday and Thursday.

## Options and Volatility - URL: https://sharifhsn.dev/blog/options-and-volatility/ - Structured data: https://sharifhsn.dev/api/posts/options-and-volatility.json - Description: The straddle strategy is owning a call and put option with the same characteristics. If I end up long a call option and short a put, it ends up being a forward contract. I can… - Date: 2024-12-12 - Exact published timestamp: 2024-12-12 - Topics: Risk Management, Options, Volatility - Categories: Risk Management - Source: FE-535 \| Risk Management - Source URL: None - Content source format: typst

Lecture Notes

Options

The straddle strategy is owning a call and put option with the same characteristics.

If I end up long a call option and short a put, it ends up being a forward contract.

I can cook with stocks and bonds to replicate a put option.

paul wilmott

Some famous trader who spoke about Black Scholes

https://web.archive.org/web/20080724100130/http://www.wilmott.com/blogs/paul/index.cfm/2008/4/29/Science-in-Finance-IX-In-defence-of-Black-Scholes-and-Merton

Exam

Replicate Black-Scholes model in Excel.

Simulate to verify.

Will be on exam!

Expect to be able to price a bond

discounted payoffs at each time step

Does this also have a closed-form solution? Open question: verify.

you might get coupons, which I can put in a savings account to accumulate interest

the bond can get more expensive if it has appetizers like this

it also reduces duration risk (look this up)

In futures hedging on equities, you can invest in treasury bonds in order to make your portfolio $𝛽$ -neutral

And the same idea will lead you to having a bond portfolio duration neutral

Another item to check.

extraction of duration from real bond price data.

Process is in Exam II Review.

The Black Scholes model tells you what a

The intuition behind implied volatility:

black-scholes relies on historical volatility

but a risk manager might use forward looking information in the volatility

this comes from the implied volatility

VIX

VIX measures implied volatility in the market. It is the fear gauge.

Let’s look at calls and puts

If we graph their prices as a function of strike price, they form this interesting graph

𝑔 𝑟 𝑎 𝑝 ℎ

And then we get this area under the curve, and that is the typical calculation method for VIX.

But what happens when the market crashes

People are paying higher prices, insurance. Pushes the put price up which actually increases

start with the black-scholes model

Numerical exercise over the winter (Open question: complete it).

## Default Fields in Rust Structs - URL: https://sharifhsn.dev/blog/linkedin-2024-12-08-default-fields-in-rust-structs/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2024-12-08-default-fields-in-rust-structs.json - Description: A recent RFC (request for comment) for the Rust language has divided the community: default fields in structs. In order to implement a default version of a struct in Rust, one… - Date: 2024-12-08 - Exact published timestamp: 2024-12-08T16:50:39.228Z - Topics: Rust, Programming Languages - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/feed/update/urn:li:activity:7271566862621020161/ - Content source format: typst

A recent RFC (request for comment) for the Rust language has divided the community: default fields in structs.

In order to implement a default version of a struct in Rust, one has to implement the Default trait on it. The easiest way to do this is to use the derive macro for Default, which requires every single one of the structs fields to implement Default.

This works fine for structs composed of simple elements that already implement Default, like integers and strings, but for more complex objects, you will need to either implement Default through an impl block for each of the struct’s unimplemented fields, or have an impl block for the whole struct.

Both of these workarounds have disadvantages. Implementing Default for every struct field type you design is tedious, especially if they are rarely used in a Default context. Additionally, it is often possible that a struct field type needs different Default implementations in different parent struct contexts, rendering this workaround impossible. The other solution, writing a bespoke impl Default block for every parent struct, is prone to code duplication, a common source of errors.

The simple solution is default field types. These allow users to specify the default initializer for a struct field type tersely within a parent struct. You can see in this attached image how much easier this is.

The controversy arises from the fact that Rust already has ways of dealing with this problem. In addition to the workarounds I mentioned earlier, the “pure” way of doing this would involve instantiating a new struct field type for every different Default implementation. In the example I’ve shown here, instead of being an i128, the age field would have a special type PetAge(i128) which would have a separate impl Default. However, this creates a lot of boilerplate and bloats code size, which is always a code smell.

Additionally, the RFC has the limitation that the in-struct Default initializer must be const, a limitation that impl Default does not face. This is important because some of the most common use-cases of this in third-party derive macros that provide this functionality are for non-const initialization, particularly String initialization.

You can read more discussion about this on the RFC pull request page: https://lnkd.in/eTZ8FPd7

What do you think of this upcoming feature? Is it useful, or just needless complexity?

## Hedging Bonds - URL: https://sharifhsn.dev/blog/bond-hedging/ - Structured data: https://sharifhsn.dev/api/posts/bond-hedging.json - Description: Notes added to lab 4 and discussion which delves into the analytic part establishing the connection between the two. Futures are less flexible because they are standardized, and… - Date: 2024-12-05 - Exact published timestamp: 2024-12-05 - Topics: Risk Management, Bonds, Hedging - Categories: Risk Management - Source: FE-535 \| Risk Management - Source URL: None - Content source format: typst

Lecture Notes

Lab 4

Notes added to lab 4 and discussion which delves into the analytic part establishing the connection between the two.

Futures/Forwards

Futures are less flexible because they are standardized, and you have to announce the price constantly because you’re on the open market.

The forward contract you sleep for a year, wake up, and they honor an agreement.

Hedging Bonds

The $𝛽$ on bonds is the duration.

If we zoom in very far, we can see the earth is flat and get the slope (derivative).

As rates go up, bond prices go down.

## Rust 1.83 Brings More `const` - URL: https://sharifhsn.dev/blog/linkedin-2024-11-28-rust-1-83-brings-more-const/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2024-11-28-rust-1-83-brings-more-const.json - Description: A happy Thanksgiving to all! Today, in addition to my loved ones, I’m grateful for a brand-new version of Rust, now with more const ! const evaluation is one of the most… - Date: 2024-11-28 - Exact published timestamp: 2024-11-28T16:16:30.611Z - Topics: Rust, Programming Languages - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/feed/update/urn:li:activity:7267934391442583554/ - Content source format: typst

A happy Thanksgiving to all! Today, in addition to my loved ones, I’m grateful for a brand-new version of Rust, now with more const! const evaluation is one of the most important steps in making Rust more ergonomic in memory-constrained environments, such as embedded computing or the Linux kernel, as it allows for more code to be evaluated before execution.

For example, let’s say that a program requires the result of some complex mathematical expression to run. Computing this at runtime would add significant overhead and could violate memory constraints. By placing this computation in a const context (either a const expression or a const function), the computation is guaranteed to run at time of compilation, with the result stored in the program to be used immediately upon execution.

This new release allows users to take mutable references in a const context, greatly expanding what kind of expressions are allowed and improving the flexibility of const code. As a result, many common functions in the standard library are now stabilized in the const context, including many operations on floating point numbers. I’m looking forward to how this new feature will enable better patterns in embedded Rust!

https://lnkd.in/eC3x_U52

## Poisson Processes - URL: https://sharifhsn.dev/blog/poisson-processes/ - Structured data: https://sharifhsn.dev/api/posts/poisson-processes.json - Description: in-class final will be decided by Thanksgiving All of the stuff we learned about stochastic calculus has been organized around the Ito process, which is continuous. We’re going… - Date: 2024-11-28 - Exact published timestamp: 2024-11-28 - Topics: Stochastic Calculus, Poisson Processes, Jump Diffusion - Categories: Stochastic Calculus - Source: FE-610 \| Stochastic Calculus - Source URL: None - Content source format: typst

Lecture Notes

in-class final will be decided by Thanksgiving

Poisson Processes

All of the stuff we learned about stochastic calculus has been organized around the Ito process, which is continuous. We’re going to learn about how we can apply these ideas to other processes.

What exactly is a Poisson process?

If $𝜏$ is a random variable with this pdf

$𝑓_{𝜏} (𝑡) = {𝜆 𝑒^{- 𝜆 𝑡} 𝑡 \geq 0 0 𝑡 < 0$

This is the exponential process with the cdf

This is a memoryless random variable. Think of a light bulb. If the lifetime of a lightbulb is memoryless, I’m saying that if I screw in a light bulb today, and the probability that the lightbulb would last for more than a year is the same. All we have to know is that it hasn’t died yet. How we can prove this is true is with cdfs?

If I know that

$ℙ (𝜏 \leq 𝑡) = 𝑞 - 𝑒^{- 𝜆 𝑡}$

Then

$ℙ (𝜏 > 𝑡) = 𝑒^{- 𝜆 𝑡}$

by definition. Then what is the probability that

$ℙ (𝜏 > 𝑡 + 𝑠 | 𝜏 > 𝑠)$

to answer that, we would simply use conditional probabilities. This would be the same as

$= \frac{ℙ ((𝜏 > 𝑡 + 𝑠) \cap (𝜏 > 𝑠))}{ℙ (𝜏 > 𝑠)}$

But because this is a strict subset on top,

$= \frac{ℙ (𝜏 > 𝑡 + 𝑠)}{ℙ (𝜏 > 𝑠)}$

$= \frac{𝑒^{- 𝜆 (𝑡 + 𝑠)}}{𝑒^{- 𝜆 𝑠}}$

$= 𝑒^{- 𝜆 𝑡}$

If I’m trying to model a stock process that has discontinuitied, it would be really inconvenient if I had to go back and find a jump to understand my process. We want the Markov property, so the process needs to be Markov no matter what happened in the past: including a jump. Therefore, the process should be totally memoryless, so there is no information encoded in it that we can condition into.

We’re going to use our process to model the time in between events.

For our list of jump times $𝜏$ , we can define

This counting process is memoryless because it’s built off of exponential random variables which are memoryless.

It’s a counting process, so we’re only dealing with nonnegative numbers.

We need something that can go to infinity because there could b infinite.

We’re going to have independent increments just like Brownian motion. All of these are stationary and independent.

For these increments: $𝔼 [𝑁 (𝑡) - 𝑁 (𝑠)] = 𝜆 (𝑡 - 𝑠)$

$𝕍 [𝑁 (𝑡) - 𝑁 (𝑠)] = 𝜆 (𝑡 - 𝑠)$

Starting with this idea that

$ℙ (𝑁 (𝑡) - 𝑁 (𝑠) = 𝑘) = \frac{{(𝜆 (𝑡 - 𝑠))}^{𝑘} \times 𝑒^{- 𝜆 (𝑡 - 𝑠)}}{𝑘!}$

If I was tasked with determining what is the expected value, what would I do? The same for any discrete random variable:

$𝔼 [𝑁 (𝑡) - 𝑁 (𝑠)] = \sum_{𝑘 = 0}^{\infty} 𝑘 ℙ (𝑁 (𝑡) - 𝑁 (𝑠) = 𝑘) = \sum_{𝑘 = 0}^{\infty} 𝑘 \frac{{(𝜆 (𝑡 - 𝑠))}^{𝑘} \times 𝑒^{- 𝜆 (𝑡 - 𝑠)}}{𝑘!}$

I can potentially cancel a k, but that would be dividing by zero.

But this is 0 when $𝑘 = 0$ , so we can just remove that term.

$= \sum_{𝑘 = 1}^{\infty} 𝑘 \frac{{(𝜆 (𝑡 - 𝑠))}^{𝑘} \times 𝑒^{- 𝜆 (𝑡 - 𝑠)}}{𝑘!}$

So now I can kill that k

$= \sum_{𝑘 = 1}^{\infty} \frac{{(𝜆 (𝑡 - 𝑠))}^{𝑘} \times 𝑒^{- 𝜆 (𝑡 - 𝑠)}}{(𝑘 - 1)!}$

We can do a reindex and say

$𝑙 = 𝑘 - 1$

$= \sum_{𝑙 = 0}^{\infty} \frac{{(𝜆 (𝑡 - 𝑠))}^{𝑙 + 1} \times 𝑒^{- 𝜆 (𝑡 - 𝑠)}}{𝑙!}$

This is really close to fitting the power series

$𝑒^{𝑥} = \sum_{𝑘 = 0}^{\infty} \frac{𝑥^{𝑘}}{𝑘!}$

What would keep from taking advantage of that here? I can factor out the e part that’s a constant. Now that I have one two many ls, I can take one out.

$= 𝜆 (𝑡 - 𝑠) \times 𝑒^{- 𝜆 (𝑡 - 𝑠)} \times \sum_{𝑙 = 0}^{\infty} \frac{{(𝜆 (𝑡 - 𝑠))}^{𝑙}}{𝑙!}$

Then by the power series form, this is

$= 𝜆 (𝑡 - 𝑠) \times 𝑒^{- 𝜆 (𝑡 - 𝑠)} \times 𝑒^{𝜆 (𝑡 - 𝑠)} = 𝜆 (𝑡 - 𝑠)$

Therefore this is proven.

How could I get the variance?

Let’s build on our knowledge of the expectation:

$𝔼 [{(𝑁 (𝑡) - 𝑁 (𝑠))}^{2}] = \sum_{𝑘 = 0}^{\infty} \frac{𝑘^{2} {(𝜆 (𝑡 - 𝑠))}^{𝑘} \times 𝑒^{- 𝜆 (𝑡 - 𝑠)}}{𝑘!}$

If it’s squared, then we can use the same steps to take out zero through reindexing

$= \sum_{𝑙 = 0}^{\infty} \frac{(𝑙 + 1) {(𝜆 (𝑡 - 𝑠))}^{𝑙 + 1} \times 𝑒^{- 𝜆 (𝑡 - 𝑠)}}{𝑙!}$

So I’m going to factor out one of those $𝜆 (𝑡 - 𝑠)$ and distribute $𝑙 + 1$

$= 𝜆 (𝑡 - 𝑠) \times (\sum_{𝑙 = 0}^{\infty} \frac{𝑙 {(𝜆 (𝑡 - 𝑠))}^{𝑙} \times 𝑒^{- 𝜆 (𝑡 - 𝑠)}}{𝑙!} + \sum_{𝑙 = 0}^{\infty} \frac{{(𝜆 (𝑡 - 𝑠))}^{𝑙} \times 𝑒^{- 𝜆 (𝑡 - 𝑠)}}{𝑙!})$

The first term is just the expectation. The second is interesting. It’s the sum of the probabilities of all of the outcomes of the Poisson distribution, so it must be 1 based on the definition of a density.

$= 𝜆^{2} {(𝑡 - 𝑠)}^{2} + 𝜆 (𝑡 - 𝑠)$

What would be the moment generating function with dummy variable $𝑢$ ?

$𝛾_{𝑁 (𝑡)} (𝑢) = 𝔼 [𝑒^{𝑢 𝑁 (𝑡)}]$

To calculate this expected value, this is the summation

$= \sum_{𝑘 = 0}^{\infty} \frac{𝑒^{𝑢 𝑘} \times {(𝜆 𝑡)}^{𝑘} \times 𝑒^{- 𝜆 𝑡}}{𝑘!}$

We’re going to repeat some of the tricks from earlier.

Martingale Property

Poisson being a counting process, this is not a martingale. It only increases. But what we can do is create a new process

$𝑀 (𝑡) = 𝑁 (𝑡) - 𝜆 𝑡$

This creates a compensated Poisson process

In order to prove this a martingale, we will see

$𝔼 [𝑀 (𝑡) | ℱ︀ (𝑠)] = 𝔼 [𝑁 (𝑡) | ℱ︀ (𝑠)] - 𝜆 𝑡$

$= 𝔼 [𝑁 (𝑡) - 𝑁 (𝑠) | ℱ︀ (𝑠)] + 𝑁 (𝑠) - 𝜆 𝑡$

But because of the memoryless nature, we can take out what’s independent.

$= 𝑁 (𝑠) - 𝜆 𝑠 = 𝑀 (𝑠)$

Compound Process

We can’t just say the process could go up by 1. That’s not interesting to us. What we do here is make a compound Poisson process which uses $𝑁 (𝑡)$ to determine how many events have happened until $𝑡$ . Then it draws the sizes of those events from another distribution that we are going to say are iid called ${𝑌_{𝑖}}$ with expectation $𝛽$ .

$𝑄 (𝑡) = \sum_{𝑖 = 1}^{𝑁 (𝑡)} 𝑌_{𝑖}$

We have a result called Wald’s Equation, which states that

$𝔼 [\sum_{𝑖 = 1}^{𝑛 (𝑇)} 𝑌_{𝑖}] = 𝔼 [𝑁 (𝑡)] 𝔼 [𝑌_{𝑖}]$

Let’s say I have lootboxes in my video game. The number of boxes I get is also random. But the number of boxes I get is independent of how much is in each individual box. Those are independent distributions. If I expect 5 items in a box, and on average I expect to get 10 boxes, then I expect to get 50 items. That’s what this equation is saying. We can also show this by iterated conditioning, but this is easier

Then the stock price will evolve like so

$𝑆 (𝑡) = 𝑆 (0) exp {(𝑟 - 𝜆 𝑘 - \frac{𝜎^{2}}{2}) 𝑡 + 𝜎 𝑊 (𝑡)}$

Then the log returns would be normally distributed if GBM was reality.

You could do a Q-Q plot (quantile quantile) You figure out what point of data in your plot would be in which quantile, and then do that for another data.

We have leptokurtosis in our real weights which make this not a straight line (heavy tails).

There are strategies to deal with this:

Regime Switching: You assume that your stock follows four or five, or even more geometric Brownian motion. There are some periods where you’re in low or high volatility, and then you combine those, which explains some of leptokurtosis
Stochastic Volatility: This is a different stock model like the Heston stock market model, which is that instead of $𝜎 (𝑡)$ being multiplied by the stock price, it’s multiplied by a different stochastic process with a different Brownian motion
Fractional Brownian Motion: Brownian motion which doesn’t have independent increments.

Thomas Lonon’s PhD was on jump diffusion, so he wants to talk about that!

Jump Diffusion

Using our compound Poisson process, we can created a compensated compound process

$𝑄 (𝑡) - 𝛽 𝜆 (𝑡)$

And this is a martingale.

We will require that every process has independent increments for now.

We’re now going to define the stochastic integral

where $𝑋$ has jumps and is not a

$𝑋 (𝑡) = 𝑋 (0) + 𝐼 (𝑡) + 𝑅 (𝑡) + 𝐽 (𝑡)$

where $𝐽 (𝑡)$ is an adapted, right-continuous process with $𝐽 (0) = 0$ and $𝐽 (𝑡) = {lim}_{𝑠 ↓ 𝑡} 𝐽 (𝑠)$ .

Also known as cadlag, continuous from the right, and limit exists from the left, from the French. This is very important to adapted processes.

$\int_{0}^{𝑡} 𝜙 (𝑠) 𝑑 𝑋 (𝑠) = \int_{0}^{𝑡} 𝜙 (𝑠) Γ (𝑠) 𝑑 𝑊 (𝑠) + \int_{0}^{𝑡} 𝜙 (𝑠) Θ (𝑠) 𝑑 𝑠 + \sum_{0 < 𝑠 \leq 𝑡} 𝜙 (𝑠) Δ 𝐽 (𝑠)$

Well how do you calculate this $Δ 𝐽 (𝑠)$ ?

It’s $𝐽 (𝑠) - 𝐽 (𝑠_{-})$

This will give the magnitude of the jump at time $𝑠$ , and 0 everywhere else.

We will still need the square-integrability to hold true, in order for this stochastic integral to be a martingale, even though it has jumps.

Quadratic Variation

How do we prove quadratic variation? We go back to norms of partitions.

$[𝑋, 𝑋] (𝑡) = {lim}_{∥ Π ∥ \to 0} \sum_{𝑗 = 0}^{𝑛 - 1} (𝑋 (𝑡_{𝑗 + 1}) - 𝑋 (𝑡_{𝑗}))^{2}$

And this ends up being 0, because the uncountably infinite number of zeros from the continuous part of the process destroys the countable number of nonzero terms in the jump part of the process.

Ito formula for a Jump Process

$𝑓 (𝑋 (𝑡)) = 𝑓 (0, 𝑋 (0)) + \int_{0}^{𝑡} 𝑓^{'} (𝑋 (𝑠)) 𝑑 𝑋^{𝑐} (𝑠) + \frac{1}{2} \int_{0}^{𝑡} 𝑓^{″} (𝑋 (𝑠)) 𝑑 𝑋^{𝑐} (𝑠) 𝑑 𝑋^{𝑐} (𝑠) + \sum_{0 < 𝑠 \leq 𝑇} [𝑓 (𝑋 (𝑠)) - 𝑓 (𝑋 (𝑠_{-}))]$

This is the same as the Ito decomposition for the continuous aprt of the process plus the jumps.

We’re going to prove that our Brownian motion and our Poisson process are independent, with the Ito formula.

Let $𝑢_{1}$ and $𝑢_{2}$ be our dummy variables:

$𝑌 (𝑡) = 𝑒^{𝑢_{1} 𝑊 (𝑡) + 𝑢_{2} 𝑁 (𝑡) - \frac{1}{2} 𝑢_{1}^{2} 𝑡 - 𝜆 𝑡 (𝑒^{𝑢_{2}} - 1)}$

The continuous part is time and the Brownian motion, $𝑡$ and $𝑊 (𝑡)$ . The jump part is the jumps, given by $𝑁 (𝑡)$ .

We can prove that these are independent by showing that the expectation of their joint moment generating function is the same as the product of their individual moment generating functions.

Two Dimensional Ito Formula

For two dimensional jump processes, you have the integrals for both.

We can get an Ito product rule for Ito processes.

$𝑋_{1} (𝑡) 𝑋_{2} (𝑡) = 𝑋_{1} (0) 𝑋_{2} (0) + \int_{0}^{𝑡} 𝑋_{2} (𝑠_{-}) 𝑑 𝑋_{1} (𝑠) + \int_{0}^{𝑡} 𝑋_{1} (𝑠_{-}) 𝑑 𝑋_{2} (𝑠) + jump cross-terms$

We need this to prove the Doleans-Dade exponential

$𝑍^{𝑋} (𝑡) = 𝑒^{𝑋^{𝑐} (𝑡) - \frac{1}{2} [𝑋^{𝑐}, 𝑋^{𝑐}] (𝑡)} \prod_{0 < 𝑠 \leq 𝑡} (1 + Δ 𝑋 (𝑠))$

We claim that this process is the solution to the SDE

$𝑑 𝑍^{𝑋} (𝑡) = 𝑍^{𝑋} (𝑡 -) 𝑑 𝑋 (𝑡)$

With the initial condition

We will need this result to establish our risk-neutral probability measure.

How do we get the differential for a jump process?

We can change the intensity of a Poisson process through the risk-neutral probability measure $\tilde{ℙ}$ The uniqueness of the risk-neutral measure comes into doubt when you’re looking for a unique solution. You don’t have a complete market, so you calibrate the market to the existing market.

This is in contrast to GBM which always has a unique solution.

Diffusion

Let’s look again at these parameters:

$𝑆 (𝑡) = 𝑆 (0) exp {(𝑟 - 𝜆 𝑘 - \frac{𝜎^{2}}{2}) 𝑡 + 𝜎 𝑊 (𝑡)}$

Theorem 11.5.1

For

$𝐻 (𝑡) = log 𝑆 (𝑡)$

$𝑆 (𝑡) = 𝑆 (0) + \int_{0}^{𝑡} 𝑆 (𝑢) 𝑑 𝐻^{𝑐} (𝑢) + \frac{1}{2} \int_{0}^{𝑡} 𝑆 (𝑢) (𝑑 𝐻^{𝑐} (𝑢))^{2} + \sum_{0 < 𝑢 \leq 𝑡} (𝑆 (𝑢) - 𝑆 (𝑢_{-}))$

We know some of these things

$= 𝑆 (0) + \int_{0}^{𝑡} 𝑆 (𝑢) (𝑟 - 𝜆 𝑘 - \frac{𝜎^{2}}{2}) 𝑑 𝑢 + \int_{0}^{𝑡} 𝑆 (𝑢) 𝜎 𝑑 \tilde{𝑊} (𝑢) + \frac{1}{2} \int_{0}^{𝑡} 𝑆 (𝑢) 𝜎^{2} 𝑑 𝑢$

And the rest of this stuff…

$+ \sum_{0 < 𝑢 \leq 𝑡} (𝑆 (𝑢) - 𝑆 (𝑢_{-}))$

We can say that the $𝜎^{2}$ portions cancel out, so for a jump at time $𝑢$ ,

$𝑆 (𝑢) = 𝑆 (0) 𝑒^{(𝑟 - 𝜆 𝑘 - \frac{𝜎^{2}}{2}) 𝑢 + 𝜎 𝑊 (𝑢) + \sum_{𝑖 = 0}^{𝑁 (𝑢_{-})} 𝑋_{𝑖} + 𝑋_{𝑁 (𝑢)}}$

Then for the jump increment,

$𝑆 (𝑢) - 𝑆 (𝑢_{-}) = 𝑆 (𝑢_{-}) (𝑒^{𝑋_{𝑁 (𝑢)}} - 1) Δ 𝑁 (𝑢)$

Let’s look at the discount process

$𝐷 (𝑡) = 𝑒^{- 𝑟 𝑡}$ $𝐷 (𝑡) 𝑆 (𝑡) = 𝑆 (0) + \int_{0}^{𝑡} - 𝜆 𝑘 𝐷 (𝑢) 𝑆 (𝑢) 𝑑 𝑢 + \int_{0}^{𝑡} 𝜎 𝑆 (𝑢) 𝐷 (𝑢) 𝑑 \tilde{𝑊} (𝑢) + \sum_{0 < 𝑢 \leq 𝑡} 𝐷 (𝑢) 𝑆 (𝑢_{-}) (𝑒^{𝑋_{𝑁 (𝑢)}} - 1) Δ 𝑁 (𝑢)$

## Linear Hedging - URL: https://sharifhsn.dev/blog/linear-hedging/ - Structured data: https://sharifhsn.dev/api/posts/linear-hedging.json - Description: Let’s say I’m trading futures. What is the log return of this? The futures contract at time 1 is worth 𝑆 0 𝑒 𝑟 , then at time 2 is 𝑆 0 𝑒 2 𝑟 , etc. If you look at log returns,… - Date: 2024-11-21 - Exact published timestamp: 2024-11-21 - Topics: Risk Management, Hedging, Linear Risk - Categories: Risk Management - Source: FE-535 \| Risk Management - Source URL: None - Content source format: typst

Linear Risk Management

Let’s say I’m trading futures. What is the log return of this?

The futures contract at time 1 is worth $𝑆_{0} 𝑒^{𝑟}$ , then at time 2 is $𝑆_{0} 𝑒^{2 𝑟}$ , etc.

If you look at log returns, then it’s simply a linear function of the underlying returns.

I can use a futures contract to hedge my $𝛽$ (systematic risk)

Foreign Exchange Rates

If I have a 1 year change Q, you can calculate this as

Everything will get more expensive when Trump goes into office, making the dollar weaker relative to the euro. Tariffs will have an inflationary effect.

Unitary Hedge

If you’re going to trade futures on Japanese yen, each future is ¥12.5 million, which is about $80,000.

We have a risky asset called JPY, and 125 million worth. We have promised to get this payment. If someone pays me today, I will get this inventory. But they’re going to give me ¥125 million in 7 months, so I’m not sure what will happen. If we think about this under no-arbitrage pricing, or GBM, we could end up anywhere from the best case to the worst case. In the best case, JPY will become stronger, so the yen you receive is worth more. The worst case is the opposite.

International companies are getting paid in the currency for each place they are located in, so they are sensitive to currencies. How do you hedge this?

Sell a futures contract at the time you receive the payment for the export. That way you can lock in the exchange rate right now.

However, there’s a potential loss if the yen doesn’t weaken. You could have made more money, but you chose to lock in at a weaker price. This is basis risk.

To calculate the PnL for basis risk, you need Q as the amount of yen transacted and S and F as spot and future rates:

$𝑄 𝑆_{𝑇} - 𝑄 𝑆_{0} = 𝑄 \times (𝑆_{2} - 𝑆_{1})$

If you plug these into the equations for the forward rates, you get

$𝑆_{0} [𝑒^{Θ 𝑡} - 1] - 𝑆_{0} [𝑒^{Θ (𝑇 - 𝑡)} - 1]$

You can think of the hedge position as something called “trading the basis”, the difference between the underlying and the futures is the basis, and then the difference is between those two times.

Week 12

Lecture Notes

Federal Reserve

His friend worked for the OCC at the Fed, they make a lot of money when they need to fix problems. He says, Trump will keep Fed more employed. Results from politics take a long time to show up.

Unitary Hedge (continued)

We say that we’re going to get a delivery of ¥125 million in a few months. And if we map out the change in exchange rate between now and then, there becomes a risk in our GBM that the exchange rate weakens and the yen loses value. The value at risk is $- 𝑆_{0} + 𝑆_{𝑡}$ .

Why does technical analysis work? It’s because everyone uses it.

Hedging is trying to eliminate uncertainty. We need to compare apples to apples, so we want the underlying and hedge vehicle to be 100% correlated with equal volatilities. The underlying goes like

Optimal Hedge

The best?

Let’s say for each million dollars on stocks, bonds, whatever, how do I come up with the decision to hedge apples with oranges? Optimal means that it minimizes the objective function.

A unit hedge has the quantities transacted as being identical. But in general, how do we decide to transact.

The example is jet fuel and heating oil. These will actually move together, and this is lab 4. Let’s imagine you’re an airline company that wants to manage liability for jet fuel and minimize the cost to pay.

Now we’re adding a futures contract. Let $Δ 𝑆$ be the change in dollar value of inventory, and $Δ 𝐹$ be the change in dollar value of the futures contract. The change in value of portfolio is

$Δ 𝑉 = Δ 𝑆 + 𝑁 \times Δ 𝐹$

What is the optimal N?

We can either think about $𝑉$ as the whole value of the portfolio, and also the change in the value of the portfolio. I’m buying today, selling tomorrow, what is the difference? That’s my unhedged $Δ 𝑆$ . For $Δ 𝐹$ , we have to know how much we’re going to sell today and buy tomorrow to balance these out. Let’s calculate variance: $𝕍 [Δ 𝑉] = 𝕍 [Δ 𝑆] + 𝑁^{2} 𝕍 [Δ 𝐹] + 2 𝑁 Cov (Δ 𝑆, Δ 𝐹)$

Now how do you optimize this function?

We actually want this equate this to 0.

How would we determine the effectiveness of the hedge? By a decrease in portfolio volatility.

$𝜎_{Δ 𝑉}^{2} (𝑁^{*}) = 𝜎_{Δ 𝑆}^{2} - \frac{𝜎_{𝑆, 𝐹}^{2}}{𝜎_{Δ 𝐹}^{2}}$

What happens in a regression?

volatility of y is

The optimal hedge ratio is

$𝑁^{*} = - 𝛽_{𝑠, 𝑓} \frac{𝑄 \times 𝑠}{𝑄_{𝑓} \times 𝑓}$

in the long position, and negative for the short position.

$𝑄$ is the inventory being hedged, in this case ¥125M. $𝑄_{𝑓}$ is how many contracts you’re actually getting, in this case 12.5M.

So what is $𝛽$ actually? It’s estimated from the price time series on the underlying and the futures. We’re trying to explain the movement of the underlying as a result of the futures contract, and we will regress one on the other. That $𝛽$ is the coefficient of the futures contracts in question.

If you hedge a liability, then you go long, because you want the upside.

For an asset you short, for a liability you long.

A famous example will be used for lab 4.

Working with real data, all you have to do is figure out the $𝛽$ between the two (regression) and then hedge the liability (go long).

Amal says that Srinath says that in 610

## Stochastic Differential Equations - URL: https://sharifhsn.dev/blog/stochastic-differential-equations/ - Structured data: https://sharifhsn.dev/api/posts/stochastic-differential-equations.json - Description: Came late because my brother couldn’t get up 🙁 where 𝛽 is the drift and 𝛾 is the diffusion. Diffeqs always need a boundary condition, given by - Date: 2024-11-21 - Exact published timestamp: 2024-11-21 - Topics: Stochastic Calculus, SDEs, Diffusions - Categories: Stochastic Calculus - Source: FE-610 \| Stochastic Calculus - Source URL: None - Content source format: typst

Lecture Notes

Came late because my brother couldn’t get up 🙁

Stochastic Differential Equations

A SDE is of the form

$𝑑 𝑋 (𝑢) = 𝛽 (𝑢, 𝑋 (𝑢)) 𝑑 𝑢 + 𝛾 (𝑢, 𝑋 (𝑢)) 𝑑 𝑊 (𝑢)$

where $𝛽$ is the drift and $𝛾$ is the diffusion.

Diffeqs always need a boundary condition, given by

$𝑋 (𝑡) = 𝑥, 𝑡 \geq 0, 𝑥 \in ℝ$

In order to solve, we want to find

Most of the time, these don’t have closed-form solutions, so you have to solve numerically with Euler’s method.

We can express SDE as PDE using Feynman-Kac, if our solution

This is the end of the book of the final.

Over the next two weeks, we will look at what we can do based on what we’ve been learning.

Next week, we will introduce the idea of non-continuous stock processes i.e. jumps.

## Multidimensional Market Models - URL: https://sharifhsn.dev/blog/multidimensional-market-models/ - Structured data: https://sharifhsn.dev/api/posts/multidimensional-market-models.json - Description: The Girsanov states that The norm of a process is the Euclidean Norm: ∥ Θ ( 𝑡 ) ∥ = ∑ 𝑗 = 1 𝑑 Θ 𝑗 2 ( 𝑡 ) And so the square of the norm is just that. - Date: 2024-11-14 - Exact published timestamp: 2024-11-14 - Topics: Stochastic Calculus, Market Models, Girsanov - Categories: Stochastic Calculus - Source: FE-610 \| Stochastic Calculus - Source URL: None - Content source format: typst

Week 10

Lecture Notes

Midterm Review

Multi-Dimensional Girsanov

The Girsanov states that

$𝑍 (𝑡) = exp (- \int_{0}^{𝑡} Θ (𝑢) 𝑑 𝑊 (𝑢))$

The norm of a process is the Euclidean Norm: $∥ Θ (𝑡) ∥ = \sqrt{\sum_{𝑗 = 1}^{𝑑} Θ_{𝑗}^{2} (𝑡)}$

And so the square of the norm is just that.

I have my Brownian motion

$\tilde{𝑊} (𝑡) = 𝑊 (𝑡) + \int_{0}^{𝑡} Θ (𝑢) 𝑑 𝑢$

And this is true for each dimension of a Brownian motion.

Once I’m in this fixed space, let’s take a martingale $𝑀 (𝑡)$ . There is an adapted dimensional process $Γ (𝑢)$ such that

$𝑀 (𝑡) = 𝑀 (0) + \int_{0}^{𝑡} Γ (𝑢) 𝑑 𝑊 (𝑢)$

nonrandom initial constant plus Ito integral

There has to be vector under $\tilde{ℙ}$ that satisfies this.

We already proved that any Ito integral with respect to a martingale is a martingale. Importantly, we can see that this is an Ito integral because we are evaluating from the left point, not mid or right point.

Multidimensional Market Model

This is a representation of this model. We are going to assume we have $𝑚$ stocks, and $𝑑$ different sources of noise. We’re not saying which one we have more of. The valuation of a golf club would be based on commodities like wood and iron, which are two different sources of noise. We’re going to set it up in such a way that every source of noise could, but doesn’t have to, impact your stock.

The value of Nintendo doesn’t care about fluctuations in the wood commodity. The coefficient of that source of noise would be 0.

The drift $𝛼$ is applied to the stock, and then we will create the cov matrix based on the sources of noise which is applied to the stock.

$𝑑 𝑆_{𝑖} (𝑡) = 𝛼_{𝑖} (𝑡) 𝑆_{𝑖} (𝑡) 𝑑 𝑡 + 𝑆_{𝑖} (𝑡) \sum_{𝑗 = 1}^{𝑑} 𝜎_{𝑖 𝑗} (𝑡) 𝑑 𝑊_{𝑗} (𝑡), 𝑖 = 1, \dots, 𝑚$ Are these Brownian motions independent? We can find their covariance by

$Cov (𝐵_{𝑖}, 𝐵_{𝑘}) = 𝔼 [𝐵_{𝑖} 𝐵_{𝑘}] - 𝔼 [𝐵_{𝑖}] 𝔼 [𝐵_{𝑘}]$

We know that these are Brownian motions, so their expectation is 0.

$= 𝔼 [𝐵_{𝑖} 𝐵_{𝑘}]$

Since we don’t know how they’re related, the best way to approach this is via Ito decomposition.

$𝑑 (𝐵_{𝑖} 𝐵_{𝑘}) = 𝐵_{𝑖} 𝑑 𝐵_{𝑖} + 𝐵_{𝑘} 𝑑 𝐵_{𝑖} + 𝑑 𝐵_{𝑖} 𝑑 𝐵_{𝑘}$

Definition of risk-neutral: $\tilde{ℙ}$ and $ℙ$ have to be equivalent.

And every single discounted stock price $𝐷 (𝑡) 𝑆_{𝑖} (𝑡)$ must be a martingale.

When you rearrange this, you get

$𝛼_{𝑖} - 𝑅 = \sum_{𝑗 = 1}^{𝑑} 𝜎_{𝑖 𝑗} Θ_{𝑗}$

And this is a system of $𝑚$ equations with $𝑑$ unknowns (dimensions).

Fundamental Theorem

The expectation of an indicator is the same as the probability measure of the function being indicated.

Because prices are unique, you can only have a unique risk-neutral probability measure on them.

## Risk-Neutral Valuation and Interest Rate Parity - URL: https://sharifhsn.dev/blog/risk-neutral-valuation-and-parity/ - Structured data: https://sharifhsn.dev/api/posts/risk-neutral-valuation-and-parity.json - Description: We’re going to continue derivatives, uninteresting because I’m taking 620. Then we will start linear risk, more interesting. A financial engineer is someone who doesn’t know… - Date: 2024-11-14 - Exact published timestamp: 2024-11-14 - Topics: Risk Management, Risk-Neutral Valuation, Interest Rate Parity - Categories: Risk Management - Source: FE-535 \| Risk Management - Source URL: None - Content source format: typst

Lecture Notes

We’re going to continue derivatives, uninteresting because I’m taking 620. Then we will start linear risk, more interesting.

Risk Neutral Valuation

A financial engineer is someone who doesn’t know finance or engineering.

I can bundle two products together and have an outcome, so the price of the combination should be the outcome. A derivative has some underlying value. The forward contract has the value of the stock price with the risk-free price increase of the interest rate.

The futures of EUR/USD tell us what people think the exchange rate will be at that time. The expected value.

We need to ensure that there is no arbitrage in this process.

The intuition of risk-neutral is that you will replace the $𝜇$ in the GBM with the risk-free rate, because that’s the expected return. This keeps things consistent with no-arbitrage pricing.

Why do dividends matter when it comes to the forward contract?

You can buy the asset today and start getting dividends, or you can get the forward contract that will give you the asset in the future.

The higher the dividend, the more I’m missing. So if the yield of my dividend is q, then the future price should be lowered to

$𝐹_{𝑡} (𝑇) = 𝑆_{𝑡} 𝑒^{(𝑟 - 𝑞) 𝜏}$

What do we do if the forward price is high? Remember, that’s how you acquire the asset through the contract, and then the $𝑆_{𝑡}$ formula is how you acquire it through the underlying. You want to short the contract and buy the underlying.

Example Forward Dividend

We have

$𝑆_{0} = 990$

$𝐹 = 1000$

$𝑞 = 0.02$

$𝑟 = 0.04$

$𝜏 = 0.25$

The fair price for $𝐹$ is $995, so there is a $5 mispricing per contract.

Interest Rate Parity

When you’re thinking of buying a stock that pays dividend, the same argument is used for foreign exchange. If I’m buying euros, the exchange rate should take into account the risk-free interest rate for euros, which is different from dollars.

Let’s say I pay $106 to get €100. There is a foreign interest rate that the price will increase at, different from the USD risk-free interest rate.

This is a similar formula as for dividends

$𝐹_{𝑡} (𝜏) = 𝑆_{𝑡} 𝑒^{(𝑟 - 𝑟^{*}) 𝜏}$

where r* is the foreign interest rate.

Let’s say you need to deliver ₤1 million six months from now, when the GBP/USD rate is 1.308, and USD risk-free rate 2%. You might think that the forward price should be

$𝐹_{0} (0.5) = 1.308 𝑒^{0.02 \times 0.5} \approx 1.321$ .

But actually this forward contract is higher than the fair price because it doesn’t take into account the foreign interest rate. If this was the true price, you would have arbitrage.

In order to exploit this, I would borrow $1.308 today and buy €

If the market becomes stronger, than there’s inflation, which will cause the interest rates to go up.

Futures

## Joint and Conditional Distributions - URL: https://sharifhsn.dev/blog/joint-and-conditional-distributions/ - Structured data: https://sharifhsn.dev/api/posts/joint-and-conditional-distributions.json - Description: Keep in mind that you can’t separate 𝜖 out from 𝑌 because they are not necessarily independent. We can use the fact that 𝜖 is complementary to divide up what 𝑋 means. But notice… - Date: 2024-11-11 - Exact published timestamp: 2024-11-11 - Topics: Probability Theory, Conditional Distributions, Covariance - Categories: Probability Theory - Source: FE-540 \| Probability Theory - Source URL: None - Content source format: typst

Lecture Notes

(came late…)

Example 7.8

(a)

Keep in mind that you can’t separate $𝜖$ out from $𝑌$ because they are not necessarily independent.

We can use the fact that $𝜖$ is complementary to divide up what $𝑋$ means.

$ℙ (𝑋 \leq 𝑥) ℙ (𝜖 = 1) + ℙ (- 𝑋 \leq 𝑥) ℙ (𝜖 = - 1)$

But notice that those two X densities are symmetric! Geometrically, they mirror each other the normal density. And so we can say that these two are equal.

$= ℙ (𝑋 \leq 𝑥)$

And there we go! That’s the same as X, so we can say that

$𝑌 = 𝑁 (0, 1)$

What about the joint distribution?

We can create a similar negative contrast.

Open question: work through the joint distribution exercise.

(c)

Since both X and Y are normal, the expectation is 0.

$Cov (𝑋, 𝑌) = 𝔼 [𝑋 𝑌] - 𝔼 [𝑋] 𝔼 [𝑌] = 𝔼 [𝑋, 𝑌]$

We can make this easier by splitting this into two situations.

$= 𝔼 [𝑋 \cdot 𝜖 𝑋] = 𝔼 [𝜖] 𝔼 [𝑋^{2}] = 0$

The conclusion is that “X and Y are uncorrelated” because their covariance is equal to 0.

(d)

But they are not independent! This is a great counterexample to the false intuition that variables that are uncorrelated are necessarily independent. That’s why it’s not enough to calculate the covariance to determine independence.

$𝔼 [𝑋^{2} 𝑌^{2}] = 𝔼 [𝑋^{2} 𝜖^{2} 𝑋^{2}] = 𝔼 [𝜖^{2} 𝑋^{4}] = 𝔼 [𝜖^{2}] 𝔼 [𝑋^{4}] = 1 \times 3 = 3$

$𝔼 [𝑋^{2}] = 1$

$𝔼 [𝑌^{2}] = 𝔼 [𝜖^{2} 𝑋^{2}] = 𝔼 [𝜖^{2}] 𝔼 [𝑋^{2}] = 1$

And that indicates

$𝔼 [𝑋^{2} 𝑌^{2}] \neq 𝔼 [𝑋^{2}] 𝔼 [𝑌^{2}]$

And that is a proof by contradiction!

Example 7.12

(a)

In order to prove that this is a density, we need to prove that that the area under its function (integral) is equal to 1 over the entire domain.

We need to do a double integral here because the domain is across two variables.

The result becomes the integral over the exponential distribution which is known to be 1.

(b)

We are going to take some basic properties. What is the marginal distribution for each variable?

$𝑓_{𝑌} (𝑦) = 𝑎^{2} 𝑦 𝑒^{- 𝑎 𝑦}$

And we can recognize they take the form

$𝑓_{𝑌} (𝑦) = 𝜆^{𝑘} 𝑒^{- 𝜆 𝑦} / 𝑘!$

Example 7.13

(a)

This type of problem is a way to flip around already understood concepts. We still need the same formula for density, but now we’re actually solving for something instead it proving something we already know.

We are taking a sum here because $ℙ (𝑋 = 𝑥)$ is the notation used for discrete random variables.

Conditional Distribution: The Discrete Case

We have the marginal probability of which we are already familiar. $𝑝_{𝑋} (𝑥) = ℙ (𝑋 = 𝑥) = \sum_{𝑦} ℙ (𝑋 = 𝑥, 𝑌 = 𝑦) = \sum_{𝑦} 𝑝 (𝑥, 𝑦)$

From now on we will use the following notation: $𝑝_{𝑋 ∣ 𝑌} (𝑥 ∣ 𝑦)$

which is the conditioning random variable.

$𝑝_{𝑋 ∣ 𝑌} (𝑥 ∣ 𝑦) = ℙ (𝑋 = 𝑥 ∣ 𝑌 = 𝑦) = \frac{ℙ (𝑋 = 𝑥, 𝑌 = 𝑦)}{ℙ (𝑌 = 𝑦)} = \frac{𝑝 (𝑥, 𝑦)}{𝑝_{𝑌} (𝑦)}$

Conditional CDF

$𝐹_{𝑋 ∣ 𝑌} (𝑥 ∣ 𝑦) = ℙ (𝑋 \leq 𝑥 ∣ 𝑌 = 𝑦) = \sum_{𝑡 \leq 𝑥} ℙ (𝑋 = 𝑡 ∣ 𝑌 = 𝑦)$

$= \sum_{𝑡 \leq 𝑥} 𝑝_{𝑋 ∣ 𝑌} (𝑡 ∣ 𝑦)$

We can use all of these expressions, which are the same. They might be useful in different contexts, though.

8.16

Conditional Expectation

$𝔼 [𝑋 ∣ 𝑌 = 𝑦] = \sum_{𝑥} 𝑥 ℙ (𝑋 = 𝑥 ∣ 𝑌 = 𝑦) = \sum_{𝑥} 𝑥 𝑝_{𝑋 ∣ 𝑌} (𝑥 ∣ 𝑦)$

And this expectation generalizes across any function of X.

This is the same result of linear expectation.

8.19

The joint density is not actually easy to determine here.

What is the probability that (Y = 3)? It’s a geometric random variable, based on our knowledge of the “trials until you have success” property of geometric random variables.

$\frac{(𝑥 - 1)^{2}}{𝑥^{2}}$

Let’s see the probability for when (Y = 3). We will take the probability of every value of (x).

We actually have a restriction on our domain here.

There’s a common identity we can use for the geometric distribution:

$𝑔 (𝑥) = \sum_{𝑛 = 0}^{\infty} 𝑥^{𝑛} = \frac{1}{1 - 𝑥}$

$𝑔^{'} (𝑥) = \sum_{𝑛 = 1}^{\infty} 𝑛 𝑥^{𝑛 - 1} = \frac{1}{(1 - 𝑥)^{2}}$

$𝑥 𝑔^{'} (𝑥) = \sum_{𝑛 = 0}^{\infty} 𝑛 𝑥^{𝑛} = \frac{𝑥}{(1 - 𝑥)^{2}}$

## Forward Contracts - URL: https://sharifhsn.dev/blog/forward-contracts/ - Structured data: https://sharifhsn.dev/api/posts/forward-contracts.json - Description: On problem 2, either you optimize for risk-free or risky assets. The question becomes unconstrained optimization. Either we have the same thing as Exam 1, released on 13 or 19?… - Date: 2024-11-07 - Exact published timestamp: 2024-11-07 - Topics: Risk Management, Forward Contracts, Futures - Categories: Risk Management - Source: FE-535 \| Risk Management - Source URL: None - Content source format: typst

Lecture Notes

Project

On problem 2, either you optimize for risk-free or risky assets. The question becomes unconstrained optimization.

Final

Either we have the same thing as Exam 1, released on 13 or 19? Bonus question on Project 2, but requires reading. Stochastic Interest Rate Models (Vasicek)

Yields are going up, because people believe that Trump will create inflationary pressure.

Last video in Modules is worth watching!

Forward Contract

How do you price a forward contract? The underlying asset is unknown. Both parties have some views on how this process will evolve, but the price must be fair.

He has to pay $𝐹_{𝑡} (𝑇)$ .

And he can sell it for the value of the underlying $𝑆_{𝑡}$ .

We can simulate each time, where one party ins and one loses. How do you determine the actual price.

By expectation, they should both be better off.

$𝔼 [𝑉_{𝐴}] = 𝔼 [𝑉_{𝐵}]$

Week 11

Lecture Notes

We’re going to continue derivatives, uninteresting because I’m taking 620. Then we will start linear risk, more interesting.

Risk Neutral Valuation

A financial engineer is someone who doesn’t know finance or engineering.

The futures of EUR/USD tell us what people think the exchange rate will be at that time. The expected value.

We need to ensure that there is no arbitrage in this process.

The intuition of risk-neutral is that you will replace the $𝜇$ in the GBM with the risk-free rate, because that’s the expected return. This keeps things consistent with no-arbitrage pricing.

Why do dividends matter when it comes to the forward contract?

You can buy the asset today and start getting dividends, or you can get the forward contract that will give you the asset in the future.

The higher the dividend, the more I’m missing. So if the yield of my dividend is q, then the future price should be lowered to

$𝐹_{𝑡} (𝑇) = 𝑆_{𝑡} 𝑒^{(𝑟 - 𝑞) 𝜏}$

Example Forward Dividend

We have

$𝑆_{0} = 990$

$𝐹 = 1000$

$𝑞 = 0.02$

$𝑟 = 0.04$

$𝜏 = 0.25$

The fair price for $𝐹$ is $995, so there is a $5 mispricing per contract.

Interest Rate Parity

Let’s say I pay $106 to get €100. There is a foreign interest rate that the price will increase at, different from the USD risk-free interest rate.

This is a similar formula as for dividends

$𝐹_{𝑡} (𝜏) = 𝑆_{𝑡} 𝑒^{(𝑟 - 𝑟^{*}) 𝜏}$

where r* is the foreign interest rate.

Let’s say you need to deliver ₤1 million six months from now, when the GBP/USD rate is 1.308, and USD risk-free rate 2%. You might think that the forward price should be

$𝐹_{0} (0.5) = 1.308 𝑒^{0.02 \times 0.5} \approx 1.321$ .

But actually this forward contract is higher than the fair price because it doesn’t take into account the foreign interest rate. If this was the true price, you would have arbitrage.

In order to exploit this, I would borrow $1.308 today and buy €

If the market becomes stronger, than there’s inflation, which will cause the interest rates to go up.

Futures

Standardized, traded on exchanges.

They are marked to market, which means that every day someone is going to bother you to ask you what the price is.

You can also buy these on margin, so the trade will clear. An initial margin must be posted before you initiative a position. The clearinghouses will ensure that if the person I’m trading with disappears, they will find someone else to take his place and give me my money.

There’s a joke in Hull, for trading cattle. There was a new intern who didn’t close the position and ended up double short.

## Risk-Neutral Measures - URL: https://sharifhsn.dev/blog/risk-neutral-measures/ - Structured data: https://sharifhsn.dev/api/posts/risk-neutral-measures.json - Description: Let’s go into the world of sports gambling. Hypothetically, we’re going to have two teams competing, Team A and Team B. Objectively, Team A is the better team. Let’s say in the… - Date: 2024-11-07 - Exact published timestamp: 2024-11-07 - Topics: Stochastic Calculus, Risk-Neutral Pricing, Girsanov - Categories: Stochastic Calculus - Source: FE-610 \| Stochastic Calculus - Source URL: None - Content source format: typst

Lecture Notes

Motivation - Sports Betting

Let’s go into the world of sports gambling. Hypothetically, we’re going to have two teams competing, Team A and Team B. Objectively, Team A is the better team. Let’s say in the real world odds, they are 4:1, meaning that Team A is 4x as likely to win as Team B. That means that

$ℙ (Team A wins) = \frac{4}{5}$

$ℙ (Team B wins) = \frac{1}{5}$

If I use these odds for betting, let’s say I have a $10 stake on Team A. If Team A wins, I will receive my original stake + a quarter of the original stake, or $12.50. If I bet on Team B, then if Team B wins, I will receive the original stake + four times the original stake, or $50.

A bookie is collecting bets. He rounds up 40 people to bet on the game. Half bet on Team A, and half bet on Team B. Assuming that the stake is $10, they have collected $400.

If Team A wins, then the $400 collected will be paid out to the 20 individuals for the value they should get

$400 - 20 \times $ 12.50 = $ 150$

The bookie will make $150 if A wins. If B wins, then

$400 - 20 \times $ 50 = - $ 600$ The bookie will lose $650 if B wins.

Let this amount be $𝑋$ . What is $𝔼 [𝑋]$ ?

$𝔼 [𝑋] = \frac{4}{5} ($ 150) + \frac{1}{5} (- $ 600)$

$= 0$ This would be what would happen if the bookie was doing this out of the goodness of their heart. What if they took 1% of the winnings (still generous)? That’s calculated out of the winnings, so 50 cents for A winning, and $8 for B winning.

This is in expectation. What if B wins four games in a row? You’re out $4000, minus this small amount.

The bookie is much more likely to notice that A and B are equally bet. They don’t care about who wins, just how people are betting. People are betting evenly on the two teams. So if the bookie gives 1:1 odds instead of 4:1, the new numbers are:

if A wins, the bettor receives $10 + $10(1/1) = $20. And the same for B.

If A wins,

$400 - 20 (20) = 0$

And the same for B, so expected value is 0. But if we take a little off the top, which is calculated off the difference between total received and original stake for each person and then scaled to the number of people.

The preferred approach for bookies is to do this because this way, you’ll always make money.

This is the motivation for risk-neutral pricing.

Risk-Neutral Measure

We are going to define a new $\tilde{ℙ}$ which is a probability measure defined as

$\tilde{ℙ} (𝐴) = \int_{𝐴} 𝑍 (𝜔) 𝑑 ℙ (𝜔)$

on the original probability space, where $𝑍$ is a nonnegative random variable.

In order for this to be a valid probability measure, we have to define

$\tilde{ℙ} (Ω) = \int_{Ω} 𝑍 (𝜔) 𝑑 ℙ (𝜔) = 𝔼 [𝑍] = 1$

I could have two probability measures on a fair coin and unfair coin.

For every random variable $𝑍$ that satisfies these properties, we can use this to define a new probability measure on the same event space ( $Ω$ ) and sample space ( $ℱ︀$ ).

The expectation under the new probability measure is the same as multiplying with Z

$\tilde{𝔼} [𝑋] = 𝔼 [𝑋 𝑍]$

Equivalent Measures

What’s of particular importance is that we are going to have equivalent probability measures.

When you create this random variable Z, you can create any probability measure. But it might not be equivalent. Let’s draw a distinction between identical and equivalent probability measures.

If for all $𝜔 \in Ω$

$ℙ (𝜔) = \tilde{ℙ} (𝜔)$

then these are identical probability measures.

If for any set $𝐴 \in ℱ︀$ such that $ℙ (𝐴) = 0 \to \tilde{ℙ} (𝐴) = 0$

and if for any set $𝐴 \in ℱ︀$ , $\tilde{ℙ} (𝐴) = 0 \to ℙ (𝐴) = 0$

Equivalence means that they agree on what can’t happen, they agree on what has measure 0. We care about this when we start pricing instruments. Under real world probability measurements, it’s impossible for a stock to triple in five days, we can’t have that happen in our risk-neutral artificial world.

We can then define

$𝑍 = \frac{𝑑 \tilde{ℙ}}{𝑑 ℙ}$

Using this Radon-Nikodym derivative, we can confirm these results from earlier. For $𝑋$ nonnegative RV, then

$\tilde{𝔼} [𝑋] = \int_{Ω} 𝑋 (𝜔) 𝑑 \tilde{ℙ} (𝜔)$

This is the more theoretical construct where $𝜔$ could be anything. Because we’re not going to just get numerical values, it’s hard to write a numerical representation of that space. IF we’re dealing with the real number line, this becomes

$\int_{- \infty}^{\infty} 𝑋 𝑓_{𝑋} (𝑥) 𝑑 𝑥$

To deal with this expected value, I will multiply it by

$\frac{𝑑 ℙ (𝜔)}{𝑑 ℙ (𝜔)}$

which becomes

$= \int_{Ω} 𝑋 (𝜔) 𝑍 (𝜔) 𝑑 ℙ (𝜔)$

which simplifies to

$= 𝔼 [𝑋 𝑍]$

which we are trying to prove.

Adapted Stochastic Process

$𝑍 (𝑡) = 𝔼 [𝑍 | ℱ︀ (𝑡)]$

This is an adapted stochastic process because it depends on information at time t.

What is

$𝑍 (0) = ?$

It would just be 1 because there’s nothing to condition on so it’s just the expectation.

What about if it’s at time T, maturity? Then Z is measurable with respect to all that information, which is Z, the Radon-Nikodym derivative.

So $𝑍 (𝑡)$ is a stochastic process that begins at 1 and goes to the Radon-Nikodym derivative.

Is it a martingale? We can use iterated conditioning to figure out that yes, it is.

What if we add another RV into the mix? $𝔼 [𝑌 𝑍 (𝑡)] = 𝔼 [𝑌 𝔼 [𝑍 | ℱ︀ (𝑡)] | ℱ︀ (0)]$

What do I know about Y? It’s measurable with respect to $ℱ︀ (𝑡)$ , so we can put what’s known back in, in the same way we can take it out.

$= 𝔼 [𝔼 [𝑌 𝑍 | ℱ︀ (𝑡)] | ℱ︀ (0)]$

Iterated conditioning shows that this is

$= 𝔼 [𝑌 𝑍]$

Side note: why are we doing this? In math, you have theorems, corollaries and lemmas. Theorems are widely used, versatile and powerful results. Corollaries are results proved using just one theorem. Lemmas are little disposable proofs that are used to prove one thing that you will use somewhere later. Sometimes these lemmas become famous, like Ito’s lemma. We are building towards Girsanov.

The partial averaging theorem states

$\int_{𝐴} 𝔼 [𝑋 | 𝒢︀] (𝜔) 𝑑 ℙ (𝜔) = \int_{𝐴} 𝑋 (𝜔) 𝑑 ℙ (𝜔)$

For the problem we’re looking at, this will become

$\int_{𝐴} 𝑌 (𝜔) 𝑑 \tilde{ℙ} (𝜔) = \int_{𝐴} \tilde{𝔼} [𝑌 | ℱ︀ (𝑡)] (𝜔) 𝑑 \tilde{ℙ} (𝜔)$

What this theorem is telling me that I want to prove that the expectation of Y conditioned on information up to s is all that.

If I can confirm this is true, then I can prove this lemma.

If I’m integrating the left hand expression, it should be the same as the right hand expression.

We’re going to disassemble the toaster and then fix the broken wire inside, then reassemble the toaster.

Rewrite

$\int_{𝐴} \frac{1}{𝑍 (𝑠)} 𝔼 [𝑌 𝑍 (𝑡) | ℱ︀ (𝑠)] 𝑑 \tilde{ℙ} (𝜔)$

$\int_{Ω} \frac{1}{𝑍 (𝑠)} 𝔼 [𝑌 𝑍 (𝑡) | ℱ︀ (𝑠)] 𝕀_{{𝐴}} 𝑑 \tilde{ℙ} (𝜔)$

And that’s the definition of the expected value, and we can bring indicator inside because it’s nonrandom

$= \tilde{𝔼} [\frac{1}{𝑍 (𝑠)} 𝔼 [𝕀_{{𝐴}} 𝑌 𝑍 (𝑡) | ℱ︀ (𝑠)]]$

We can use the lemma about the transformation of expectation into a different expectation using Z.

$\tilde{𝔼} [𝑋] = 𝔼 [𝑋 𝑍 (𝑠)]$

To get that

$= 𝔼 [𝔼 [𝕀_{{𝐴}} 𝑌 𝑍 (𝑡) | ℱ︀ (𝑠)]]$

We did all that to take out the tilde. So we can use iterated conditioning to get

$= 𝔼 [𝕀_{{𝐴}} 𝑌 𝑍 (𝑡)]$

$= \tilde{𝔼} [𝕀_{{𝐴}} 𝑌]$ Which we can work backwards to prove the original.

Let $𝑋 (𝑡)$ an Ito process.

$𝑋 (𝑡) = 𝑋 (0) + \int_{0}^{𝑡} Δ (𝑢) 𝑑 𝑊 (𝑢) + \int_{0}^{𝑡} Θ (𝑢) 𝑑 𝑢$

Let’s discuss the conditions for when an Ito process is a martingale.

$𝔼 [𝑋 (𝑡) | ℱ︀ (𝑠)] = 𝔼 [𝑋 (0) + \int_{0}^{𝑡} Δ (𝑢) 𝑑 𝑊 (𝑢) + \int_{0}^{𝑡} Θ (𝑢) 𝑑 𝑢 | ℱ︀ (𝑠)]$

We know that an Ito integral with respect to a martingale is a martingale. I know that the expected value of it will be 0. But the conditioned expectation up until time s will be that integral. We have the result for the stochastic part, but not the Riemann part.

$= 𝑋 (0) + \int_{0}^{𝑠} Δ (𝑢) 𝑑 𝑊 (𝑢) + 𝔼 [\int_{0}^{𝑡} Θ (𝑢) 𝑑 𝑢 | ℱ︀ (𝑠)]$

Let’s rewrite that as a combination of integrals

$= 𝑋 (0) + \int_{0}^{𝑠} Δ (𝑢) 𝑑 𝑊 (𝑢) + 𝔼 [\int_{0}^{𝑠} Θ (𝑢) 𝑑 𝑢 + \int_{𝑠}^{𝑡} Θ (𝑢) 𝑑 𝑢 | ℱ︀ (𝑠)]$

And I know that the s part is known.

$= 𝑋 (0) + \int_{0}^{𝑠} Δ (𝑢) 𝑑 𝑊 (𝑢) + \int_{0}^{𝑠} Θ (𝑢) 𝑑 𝑢 + 𝔼 [\int_{𝑠}^{𝑡} Θ (𝑢) 𝑑 𝑢 | ℱ︀ (𝑠)]$

which we can finally rewrite, using our original, as

$= 𝑋 (𝑠) + 𝔼 [\int_{𝑠}^{𝑡} Θ (𝑢) 𝑑 𝑢 | ℱ︀ (𝑠)]$

I can’t adjust this in any way. But we want to know whether it’s a martingale. So in order for us to conclude this, then that last term has to be 0. Are there restrictions that we can place on $Θ (𝑢)$ to make sure that this is a martingale regardless of s and t? It has to always be 0. So we can get rid of it.

$𝑋 (𝑡) = 𝑋 (0) + \int_{0}^{𝑡} Δ (𝑢) 𝑑 𝑊 (𝑢)$

and

$𝑑 𝑋 (𝑡) = Δ (𝑡) 𝑑 𝑊 (𝑡)$ We stumbled over another way to get a martingale. If we take the differential of our process and there’s dt terms, then it can’t be a martingale because $Θ (𝑡)$ is not 0.

Girsanov

The Big One.

What this theorem states, is that if I start with a Brownian motion, and some adapted process $Θ (𝑡)$ that is square integrable. Using any, we can define a new Z(t)

$𝑍 (𝑡) = exp (- \int_{0}^{𝑡} Θ (𝑢) 𝑑 𝑊 (𝑢) - \frac{1}{2} \int_{0}^{𝑡} Θ^{2} (𝑢) 𝑑 𝑢)$

And define a new stochastic process which is the Brownian motion plus the integral of the process

$\tilde{𝑊} (𝑡) = 𝑊 (𝑡) + \int_{0}^{𝑡} Θ (𝑢) 𝑑 𝑢$

Then that stochastic process is a Brownian motion.

And this is all for the goal of finding risk-neutral probability measures.

First, let’s prove that the expected value of Z is 1.

$𝑍 (𝑡) = exp (- \int_{0}^{𝑡} Θ (𝑢) 𝑑 𝑊 (𝑢) - \frac{1}{2} \int_{0}^{𝑡} Θ^{2} (𝑢) 𝑑 𝑢)$

Let’s find the differential to see if it’s a martingale.

$𝑑 𝑍 (𝑡) = 𝑍 (𝑡) 𝑑 𝑌 (𝑡) + \frac{1}{2} 𝑍 (𝑡) (𝑑 𝑌 (𝑡))^{2}$

$= - 𝑍 (𝑡) Θ (𝑡) 𝑑 𝑊 (𝑡) - \frac{1}{2} 𝑍 (𝑡) Θ^{2} (𝑡) 𝑑 𝑡 + \frac{1}{2} 𝑍 (𝑡) Θ^{2} (𝑡) 𝑑 𝑡$

$= - 𝑍 (𝑡) Θ (𝑡) 𝑑 𝑊 (𝑡)$

Because all of the dt combinations cancel out, this is a martingale.

We will use the Lévy theorem to prove that $\tilde{𝑊} (𝑡)$ is a Brownian motion. It needs to be a martingale, continuous paths, starts at 0, and quadratic variation is t.

The only one of these should change because of a probability measure. So we only need to show that Brownian motion is a martingale.

Now we can finally use the partial averaging proof

$\tilde{𝔼} [\tilde{𝑊} (𝑡) | ℱ︀ (𝑠)] = \frac{1}{𝑍 (𝑠)} 𝔼 [\tilde{𝑊} (𝑡) 𝑍 (𝑡) | ℱ︀ (𝑠)]$

This is the direct result of that lemma. The only thing standing in our way to finishing this proof is establishing that this is a martingale under $ℙ$ .

Is $\tilde{𝑊} (𝑡) 𝑍 (𝑡)$ a $ℙ$ -martingale? What’s the differential? By the product rule, it’s

$𝑑 (\tilde{𝑊} (𝑡) 𝑍 (𝑡)) = \tilde{𝑊} (𝑡) 𝑑 𝑍 (𝑡) + 𝑍 (𝑡) 𝑑 \tilde{𝑊} (𝑡) + 𝑑 \tilde{𝑊} (𝑡) 𝑑 𝑍 (𝑡)$

Well we already proved this result for the various differentials, so it’s

$= - \tilde{𝑊} (𝑡) Θ (𝑡) 𝑍 (𝑡) 𝑑 𝑊 (𝑡) + 𝑍 (𝑡) 𝑑 𝑊 (𝑡) + 𝑍 (𝑡) Θ (𝑡) 𝑑 𝑡 + (𝑑 𝑊 (𝑡) + Θ (𝑡) 𝑑 𝑡) (- Θ (𝑡) 𝑍 (𝑡) 𝑑 𝑊 (𝑡))$

That last term gets multiplied out, dt * anything is 0, becomes the negative to that third term, so we only get dW(t), so it’s a martingale.

This completes our proof.

Discount Process

## A Rainbow on Election Day - URL: https://sharifhsn.dev/blog/linkedin-2024-11-06-a-rainbow-on-election-day/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2024-11-06-a-rainbow-on-election-day.json - Description: A brief pause from election-day politics to notice an unexpected rainbow. - Date: 2024-11-06 - Exact published timestamp: 2024-11-06T19:56:37.382Z - Topics: Campus Life, Personal - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/posts/sharif-haason_i-was-walking-on-campus-this-morning-preoccupied-activity-7260017251611815936-2p_J - Content source format: typst

I was walking on campus this morning, preoccupied with thoughts of the election, when I noticed a sprinkler on some flowers next to me. The water caught the light in just the right way to create a huge rainbow. The beauty of it stunned me, and I stood for a minute just staring at it. I reflected on how the small things in life are what’s important. There will be plenty of time to debate and discuss the impact of this election, but that rainbow will only exist for a brief period.

If your thoughts, like mine, are stormy with politics today, take a moment and look for a rainbow near you.

## Random Vectors and Independence - URL: https://sharifhsn.dev/blog/random-vectors-and-independence/ - Structured data: https://sharifhsn.dev/api/posts/random-vectors-and-independence.json - Description: Beginning chapter 7, vectors of random variables. We are concerned with a random vector The cdf of this vector is We can also say that if we have a random vector (X, Y) then - Date: 2024-11-04 - Exact published timestamp: 2024-11-04 - Topics: Probability Theory, Random Vectors, Independence - Categories: Probability Theory - Source: FE-540 \| Probability Theory - Source URL: None - Content source format: typst

Week 10

Lecture Notes

Random Vectors

Beginning chapter 7, vectors of random variables. We are concerned with a random vector

$𝑋 = (𝑋_{1}, \dots, 𝑋_{𝑁})$

The cdf of this vector is

$𝐹_{𝑋} (𝑥_{1}, \dots, 𝑥_{𝑁}) = ℙ (𝑋_{1} \leq 𝑥_{1}, \dots 𝑋_{𝑁} \leq 𝑥_{𝑛})$

We can also say that if we have a random vector (X, Y) then

If you compute joint cdf, we can get marginal distribution (distribution for a subset, maybe one element) very easily

$𝐹_{𝑋} (𝑥) = {lim}_{𝑦 \to \infty} 𝐹_{(𝑋, 𝑌)} (𝑥, 𝑦)$

and equivalently

$ℙ (𝑋 \leq 𝑥) = {lim}_{𝑦 \to \infty} ℙ (𝑋 \leq 𝑥, 𝑌 \leq 𝑦)$

Remark 7.7 says components of vectors are independent

These results hold for every random vector.

Discrete Random Vectors

Now we will talk about discrete random vectors.

The formal definition 7.8 says that the rv is discrete when the image set $𝑋 (Ω)$ is a finite or countable subset of $ℝ^{𝑛}$ , the product space between n different real number lines.

The pmf for a discrete random vector is given by the basic formula

$ℙ (𝑋_{1} = 𝑥_{1}, \dots 𝑋_{𝑁} = 𝑥_{𝑁})$

The density dictates that the sum must be 1.

$\sum_{𝑥 \in 𝑋 (Ω)} ℙ (𝑋 = 𝑥) = 1$

Then the cdf is

$𝐹_{𝑋} (𝑥_{1}, \dots, 𝑥_{𝑛}) = \sum_{𝑢_{1} \leq 𝑥_{1}} \dots \sum_{𝑢_{𝑁} \leq 𝑥_{𝑁}} ℙ (𝑋_{1} = 𝑢_{1}, \dots, 𝑋_{𝑁} = 𝑢_{𝑁})$

In order to intuitively understand why the marginal distribution calculation works for joint distribution, imagine a table of values, which has the probability for each event in the sample space. The rows are of the first random variable’s sample space, and the columns are for the second. Then, in order to calculate the probability of an event happening for the second random variable, you need to total all the probabilities from each row in the table.

Independence

Example 7.4

We have (X, Y) random vector. where

$𝑋 (Ω) = 𝑌 (Ω) = {0, 1}$

and

$ℙ (𝑋 = 0, 𝑌 = 0) = (1 - 𝑝)^{2}$

$ℙ (𝑋 = 1, 𝑌 = 1) = 𝑝^{2}$

$ℙ (𝑋 = 0, 𝑌 = 1) = ℙ (𝑋 = 1, 𝑌 = 0) = 𝑝 (1 - 𝑝)$

How do we prove independence?

We need to prove

$ℙ (𝐴 \cap 𝐵) = ℙ (𝐴) ℙ (𝐵)$

or for joint distributions,

$ℙ (𝑋 = 𝑥, 𝑌 = 𝑦) = ℙ (𝑋 = 𝑥) ℙ (𝑌 = 𝑦)$

We can use the marginal distribution calculation for this.

$ℙ (𝑋 = 0) = ℙ (𝑋 = 0, 𝑌 = 0) + ℙ (𝑋 = 0, 𝑌 = 1) = (1 - 𝑝)^{2} + 𝑝 (1 - 𝑝) = 1 - 𝑝$

$ℙ (𝑋 = 1) = 𝑝^{2} + 𝑝 (1 - 𝑝) = 𝑝$

$ℙ (𝑌 = 0) = 1 - 𝑝$

$ℙ (𝑌 = 1) = 𝑝$

And then now you can verify for each combination of different X and Y that the independence equality holds. Left as exercise for reader 😅

Continuous Random Vectors

We can consider any function $𝑓 : ℝ^{𝑛} \to ℝ$ a density if it’s positive and

$\int_{ℝ^{𝑛}} 𝑓 (𝑥) 𝑑 𝑥 = \int_{ℝ_{1}} \dots \int_{ℝ_{𝑁}} 𝑓 (𝑥_{1}, \dots, 𝑥_{𝑛}) 𝑑 𝑥_{1}, \dots 𝑑 𝑥_{𝑛} = 1$

We can say that the random vector has this density given by f if

$ℙ ((𝑋_{1}, \dots 𝑋_{𝑛}) \in 𝐷) = \int_{𝐷} 𝑓 (𝑥_{1}, \dots, 𝑥_{𝑛}) 𝑑 𝑥_{1} \dots 𝑑 𝑥_{𝑛}$

Where D is the product space. And in fact, for the cdf, this D is the Borel set

For the marginal density, instead of a summation of all the others, you actually do the integral of all the others

$𝑓_{𝑥_{𝑖}} (𝑥_{𝑖}) = \int_{ℝ^{𝑛 - 1}} 𝑓 (𝑥_{1}, \dots, 𝑥_{𝑛}) 𝑑 𝑥_{1} \dots 𝑑 𝑥_{𝑖 - 1} 𝑑 𝑥_{𝑖 + 1} \dots 𝑑 𝑥_{𝑛}$

Uniformly Distributed Vector

The joint density function is given by

$𝑓 (𝑥, 𝑦) = \frac{1}{(𝑏 - 𝑎) (𝑑 - 𝑐)} 𝕀_{(𝑎, 𝑏]} (𝑥) 𝕀_{(𝑐, 𝑑]} (𝑦)$

Two Dimensional Normal

The mapping which is given by

$𝑓 (𝑥, 𝑦) = \frac{1}{2 𝜋} 𝑒^{- \frac{𝑥^{2} + 𝑦^{2}}{2}}$

This is for both variables being (N(0, 1)).

If we want to prove that a function is a density, we need to prove only positivity and equivalence to one on the integral over the interval.

Change of Variable

Let’s take a random vector X with density $𝑓_{𝑋}$ , and $𝑌 = 𝜑 (𝑋)$ where this function is a multivariate function that transforms X. The density of Y is, in the most general case,

$𝑓_{𝑌} (𝑦) = 𝑓_{𝑋} (𝜑^{- 1} (𝑦)) | det (𝐽 (𝜑^{- 1} (𝑦))) |$

J the Jacobian is a matrix of partials, from

$\frac{𝜕 𝜑_{1}}{𝜕 𝑥_{1}} \dots \frac{𝜕 𝜑_{1}}{𝜕 𝑥_{𝑛}}$

going left to right

You’re kind of stacking these multivariate functions on top of each other:

$𝜑 = (𝜑_{1} (𝑥_{1}, \dots, 𝑥_{𝑛}), \dots, 𝜑_{𝑛} (𝑥_{1}, \dots, 𝑥_{𝑛}))$

To get the inverse of $𝜑$ , you treat it as a system

${𝑢 = 𝑥 𝑦 𝑣 = \frac{𝑥}{𝑦} \to 𝑢 𝑣 = 𝑥^{2} \to 𝑥 = \sqrt{𝑢 𝑣}, 𝑦 = \sqrt{\frac{𝑢}{𝑣}}$

$𝜑^{- 1} (𝑢, 𝑣) = (\sqrt{𝑢 𝑣}, \sqrt{\frac{𝑢}{𝑣}})$

Then you can get the Jacobian from the vector

$\frac{𝜕}{𝜕 𝑢} \sqrt{𝑢 𝑣} \frac{𝜕}{𝜕 𝑣} \sqrt{𝑢 𝑣} \frac{𝜕}{𝜕 𝑥} \sqrt{\frac{𝑢}{𝑣}} \frac{𝜕}{𝜕 𝑣} \sqrt{\frac{𝑢}{𝑣}}$

Then calculate each partial, then take the determinant with a calculator.

Convolution

The convolution will allow you to find the density of the sum of random variables.

$(𝑓 * 𝑔) (𝑥) = \int_{ℝ} 𝑓 (𝑦) 𝑔 (𝑥 - 𝑦) 𝑑 𝑦 = \int_{ℝ} 𝑔 (𝑧) 𝑓 (𝑥 - 𝑧) 𝑑 𝑧$

Take the convolution of the two density functions to get the result.

Examples

If you’re determining pairwise independence, you can use the procedure described earlier.

But for mutual independence, you will need to add all them together

## Derivatives and Clearinghouses - URL: https://sharifhsn.dev/blog/derivatives-and-clearinghouses/ - Structured data: https://sharifhsn.dev/api/posts/derivatives-and-clearinghouses.json - Description: The Treynor ratio is intended to capture systematic risk, but doesn’t tell you anything about portfolio returns. Sharpe ratio doesn’t tell you anything about systematic risk. R^2… - Date: 2024-10-31 - Exact published timestamp: 2024-10-31 - Topics: Risk Management, Derivatives, Clearinghouses - Categories: Risk Management - Source: FE-535 \| Risk Management - Source URL: None - Content source format: typst

Lecture Notes

Exam Review

The Treynor ratio is intended to capture systematic risk, but doesn’t tell you anything about portfolio returns. Sharpe ratio doesn’t tell you anything about systematic risk.

R^2 tells you much of the variance is explained by the model.

The variance has a systematic component and error component.

$𝕍 [𝑅] = 𝛽^{2} 𝜎_{𝑚}^{2} + 𝜎_{𝜖}^{2}$

Systematic volatility is the $𝛽 𝜎_{𝑚}$

Derivatives

There’s Q finance, which is risk-neutral, that will give you martingale theories and all that. In our class we are considering P finance based off physical probabilities, real market stuff. When you get into stochastic models and pricing, then you have to make all these assumptions under Q finance.

What is a derivative? A derivative is an instrument whose value is derived by another underlying instrument. Let’s say we have the ETF SPY. We can take options on SPY, and that option is a derivative.

What would a farmer like to do? He makes bourbon, so he needs to hedge corn. He uses derivatives to do that.

OTC markets are actually much larger than exchanges.

The farmer would like to buy corn. He wants to secure $1 million in corn.

How do you buy S&P? You can buy SPY, but you can also buy a futures contract. The notional is 50x the index, which is $250,000. If you want to trade these contracts, you need a lot of capital.

What’s the difference between bonds/equities and derivatives? Let’s say you have a publicly owned company. You can raise money by issuing a bond, and gaining assets. Bonds/equities are a way of raising money for a company.

Derivatives are an agreement between two parties, with a winner and a loser. So it needs to be a fair game.

Flash boys by Michael Lewis, HFT trading.

Exchanges organize trades, match buyers and sellers together. The farmer would like to sell bourbon for a certain price, but if the second party disappears, that’s a disappointment. The exchange monitors both parties and makes sure they’re both healthy, and have the capital to fulfill their obligations.

OTC market is a network of traders, financial institutions that negotiate contracts on an individual basis.

There was more oversight in OTC after the financial crisis. It used to be a bilateral agreement with no middleman, but it’s becoming more similar to exchanges. Forwards are the same as futures, just OTC vs exchanges. You can have FX futures, depends on underlying

The true value of OTC derivatives contracts is about $22 trillion, only 3% of notional. If they are liquid and traded on a day to day basis, then I can see the mark to market price

Clearinghouses and Robinhood

Clearinghouses are the plumbing of the financial system. If you make a trade, let’s say you use Robinhood, they hold the money. This is when VaR is useful, because they put some money on margin, to compensate for the value at risk of the money being traded.

Robinhood got put in front of Congress because they weren’t allowing GME traders to trade.

They couldn’t allow traders to buy more GME, so if you can’t buy you can only sell.

It wasn’t a conspiracy. Robinhood got a margin call. They were telling Robinhood that their assets were way too risky. Because it was all in one asset, that was very volatile, it was risky. So they were margin called for $3 billion.

There was probably model risk that caused this, which caused reputational risk.

Could GME happen again? Another faculty, in literature, was asking about GME, but it’s too late.

Reddit coordinated in a way that hedge funds couldn’t.

Central Counterparties

CCPs provide more flexibility, they’re clearinghouses for OTC.

We have a financial network.

Project

We have three assets here, not two, and that’s the difficulty. What we do is have a closed form solution for this, based on minimizing mean variance.

I take the most and least efficient fund. When I have a risk-free rate, the dynamics change. My pension, some is risk-free, some is market.

## Stochastic Calculus Review - URL: https://sharifhsn.dev/blog/stochastic-calculus-review/ - Structured data: https://sharifhsn.dev/api/posts/stochastic-calculus-review.json - Description: Recognize that this is some function that looks like 𝑓 ( 𝑎 , 𝑏 ) , like where 𝑎 is a deterministic component and 𝑏 is a stochastic component. Then solve your partials - Date: 2024-10-31 - Exact published timestamp: 2024-10-31 - Topics: Stochastic Calculus, Review, Pricing - Categories: Stochastic Calculus - Source: FE-610 \| Stochastic Calculus - Source URL: None - Content source format: typst

Content Review

Problem Set 5

$𝑌 (𝑡) = 𝑒^{𝑡 𝑊 (𝑡)}$

Find the differential

$𝑑 𝑌 (𝑡)$

Recognize that this is some function that looks like $𝑓 (𝑎, 𝑏)$ , like

$𝑓 (𝑎, 𝑏) = 𝑒^{𝑎 𝑏}$

where $𝑎$ is a deterministic component and $𝑏$ is a stochastic component.

Then solve your partials

$𝑓_{𝑎} = 𝑏 𝑒^{𝑎 𝑏}$ $𝑓_{𝑏} = 𝑎 𝑒^{𝑎 𝑏}$

$𝑓_{𝑏 𝑏} = 𝑎^{2} 𝑒^{𝑎 𝑏}$

then using the Ito formula, this is

$𝑑 𝑌 (𝑡) = 𝑊 (𝑡) 𝑒^{𝑡 𝑊 (𝑡)} 𝑑 𝑡 + 𝑡 𝑒^{𝑡 𝑊 (𝑡)} 𝑑 𝑊 (𝑡) + \frac{1}{2} 𝑡^{2} 𝑒^{𝑡 𝑊 (𝑡)} 𝑑 𝑡$

Question on why we’re able to get away with this? $𝔼 [𝑒^{𝜎 𝑊 (𝑡)}] = 𝑒^{\frac{1}{2} 𝜎^{2} 𝑡}$

This is because of a result, confirmed in the result, that

$𝑋 \sim 𝑁 (𝑎, 𝑏^{2})$ implies $𝔼 [𝑒^{𝑢 𝑋}]$

Which is the moment generating function with respect to the dummy variable $𝑢$ , is

$= 𝑒^{𝑎 𝑢 + \frac{1}{2} 𝑏^{2} 𝑢^{2}}$

We’re using this off-label. So if we plug it into this we can do this. What if $𝜎$ is adapted to time? Can we use this result? Not necessarily. Being an adapted process means that you’re measurable with respect to filtration at that time. But if $Δ 𝑡$ is a random process, then this is a totally different problem.

$\frac{𝑑 𝑆 (𝑡)}{𝑆 (𝑡)} = 𝜇 𝑑 𝑡 + 𝜎 𝑑 𝑊 (𝑡)$

Can you take the integral from both sides? You can, but it causes some issues.

$\int_{0}^{𝑡} \frac{𝑑 𝑆 (𝑢)}{𝑆 (𝑢)} = 𝛼 𝑡 + 𝜎 𝑊 (𝑡)$

Doesn’t really give you a closed form approximation.

From the derivation of Black Scholes

Expectation of the discount factor, $𝑆_{𝑡} - 𝐾$ , then you have an indicator function.

Indicator is transformed, where does that came from

$𝔼 [𝑒^{- 𝑟 𝑇} (𝑆 (𝑇) - 𝐾) 𝕀_{{𝑑 * < 𝑊 (𝑇)}}]$

We can substitute

What is random? $𝑊 (𝑡)$ . What it’s distribution? normal over real line.

Then take the infinite integral.

Question from sample midterm: $𝐼 (𝑡) = \int_{0}^{𝑡} Δ (𝑢) 𝑑 𝑊 (𝑢)$

Find $𝔼 [𝐼^{4} (𝑡)]$

And $Δ (𝑢)$ is $3 𝑢$

There are a couple ways to do this! There’s the way with the hint, and other ways. Any time you’re doing an expected value of a non-well-defined process, like an Ito process raised to a power, look at the Ito decomposition instead of the process as a whole.

$𝑑 (𝐼^{4} (𝑡)) = 4 𝐼^{3} (𝑡) 𝑑 𝐼 (𝑡) + 6 𝐼^{2} (𝑡) (𝑑 𝐼 (𝑡))^{2}$

And then

$𝑑 𝐼 (𝑡) = 3 𝑡 𝑑 𝑊 (𝑡)$

based on what we already know…

$(𝑑 𝐼 (𝑡))^{2} = 9 𝑡^{2} 𝑑 𝑡$ based on what we know

$𝐼^{4} (𝑡) = \int_{0}^{𝑡} 4 𝐼^{3} (𝑢) Δ (𝑢) 𝑑 𝑊 (𝑢) + \int_{0}^{𝑡} 6 𝐼^{2} (𝑢) (9 𝑢^{2}) 𝑑 𝑢$

When I take the expectation, I can see that an Ito integral with respect to a martingale is a martingale, so the first term is gone.

$𝔼 [𝐼^{4} (𝑡)] = \int_{0}^{𝑡} 54 𝑢^{2} 𝔼 [𝐼^{2} (𝑢)] 𝑑 𝑢$

Could you be more clever about this?

This is an Ito integral of a deterministic integrand, so it has to be normally distributed. The fourth moment of any normal variable is three times the variance.

This would give you the answer. But if it was like in the homework problem where you want to do $𝐼^{6}$ , you would have to do it in the Ito decomposition way.

An Ito decomposition results in A nonrandom initial constant, a Riemann integral $𝑑 𝑢$ , and an Ito integral $𝑑 𝑊 (𝑡)$ .

Understand that this formula is

Generalized geometric Brownian motion.

If you recognize that if you have a process that can be rewritten as e^Ito process, do we need to do our partials? If you can reliably get the correct answer without doing that, you can, but it’s better to do the partials.

Itos’ formula:

$𝑓 (𝑡, 𝑋 (𝑡)) = 𝑓 (0, 𝑋 (0)) + \int_{0}^{𝑡} 𝑓_{𝑎} (𝑢, 𝑋 (𝑢)) 𝑑 𝑢 + \int_{0}^{𝑡} 𝑓_{𝑏} (𝑢, 𝑋 (𝑢)) 𝑑 𝑋 (𝑢) + \frac{1}{2} \int_{0}^{𝑡} 𝑓_{𝑏 𝑏} (𝑢, 𝑋 (𝑢)) (𝑑 𝑋 (𝑢))^{2}$

$𝑋 (𝑡)$ is an Ito process

$𝑢$ is your index

Brownian motion is definitely an Ito process. In order to be an Ito process, you need to be able to be expressed as the nonrandom constant + Riemann + stochastic.

Quadratic variation of a process

$𝑋 (𝑡) = 3 - \int_{0}^{𝑡} 𝑊 (𝑢) 𝑑 𝑢 + \int_{0}^{𝑡} 𝑢 𝑑 𝑊 (𝑢)$

Quickest way to find this is recognizing that this is an Ito process.

Quadratic variation of an Ito process is

$\int_{0}^{𝑡} Δ^{2} (𝑢) 𝑑 𝑢$ REMEMBER THIS

How do we get the expected value of an Ito integral squared? The Ito isometry.

If I have the expression

If I tell you $𝑢 = 2$ and $𝑡 = 4$ ,

$\int_{0}^{𝑡} 𝑊 (𝑡) 𝑢^{2} 𝑑 𝑊 (𝑢)$

THAT IS NOT AN ADAPTED PROCESS.

Week 9

General Probability Theory

$Ω$ is the sample space of all possible outcomes $𝜔$ .

A $𝜎$ -algebra is the set of all sets generated from a starting set of outcomes, their complements, and the unions of all complements. Conceptually, it represents an addressable subset in a potentially uncountably infinite sample space. The $𝜎$ -algebra is represented by $ℱ︀$ .

The probability measure $ℙ$ distributes values of probability on every set in the $𝜎$ -algebra $ℱ︀$ , which must add up to 1 across $Ω$ .

The Borel set is the $𝜎$ -algebra generated by the set of all closed intervals on the real number line. Conceptually, it represents an addressable subset of real numbers.

A random variable is a mapping from outcomes $𝜔$ to a Borel subset. This could be a single number or an interval.

The distribution measure of a random variable $𝑋$ is the probability measure that $𝑋$ is in each Borel subset. This is computed by finding the preimage of a Borel subset, which is an event in the sample space, and finding the probability measure of that event.

Two different random variables could have the same distribution measure, and the same random variable could have two different distribution measures under two different probability measures.

The expectation of a random variable is the average result of the random variable. This can be computed in many ways. For a discrete random variable, you can directly compute the result of each outcome mapped through the random variable, multiply them by the probability of the outcome, and sum them. For a continuous random variable, you must use an integral.

$𝔼 [𝑋] = \int_{Ω} 𝑋 (𝜔) 𝑑 ℙ (𝜔)$

(Definition 1.3.3)

Expectations have the property of linearity.

Information and Conditioning

We can model the information we have regarding an uncertain outcome with $𝜎$ -algebras.

If we know which set of outcomes that the outcome must belong to based on the current information, we can generate a $𝜎$ -algebra from that set of outcomes.

A filtration is a sequence of such $𝜎$ -algebras that represent the information available over time until time $𝑡$ .

Filtrations always increase in information over time.

The filtration at time $𝑡$ represents all the information at time $𝑡$ , and it is used for conditioning. We can say that a random variable is measurable with respect to a filtration if the information in the filtration can be used to determine the value of the random variable.

For any two sets, independence is generally defined by

$ℙ (𝐴 \cap 𝐵) = ℙ (𝐴) \cdot ℙ (𝐵)$

Two $𝜎$ -algebras are independent if this is true for all sets in each $𝜎$ -algebra.

Two random variables are independent if the $𝜎$ -algebras generated from them are independent.

Any functions on independent random variables will also result in independent random variables.

Not gonna bother with joint densities and crap.

A conditional expectation allows us to use information from a filtration to estimate the expectation of a random variable without it being completely measurable. In shorthand, we might use a random variable as a condition to indicate that the filtration is the $𝜎$ -algebra generated by the random variable.

Conditional expectations have the following properties (Theorem 2.3.2)

Linearity
Taking out what is known: if a random variable $𝑋$ is measurable with respect to the condition, it can be taken out. $𝔼 [𝑋 𝑌 | 𝒢︀] = 𝑋 𝔼 [𝑌 | 𝒢︀]$
Iterated conditioning: if we have a $𝜎$ -algebra $𝐻$ with less information, we can subsume the $𝜎$ -algebra $𝐺$ with more information into it. $𝔼 [𝔼 [𝑋 | 𝒢︀] | ℋ︀] = 𝔼 [𝑋 | ℋ︀]$
Independence: if a random variable $𝑋$ is independent of the condition, the condition is irrelevant. $𝔼 [𝑋 | 𝒢︀] = 𝔼 [𝑋]$

A stochastic process is a collection of random variables indexed at time $𝑡$ .

An adapted stochastic process is a stochastic process where each random variable is measurable by the filtration at time $𝑡$ . Conceptually, it is a process which only depends on information available at each time $𝑡$ .

A martingale is an adapted stochastic process which is expected to remain the same value at all times. Formally, (Definition 2.3.5)

$𝔼 [𝑀 (𝑡) | ℱ︀ (𝑠)] = 𝑀 (𝑠)$ for all $0 \leq 𝑠 \leq 𝑡 \leq 𝑇$

Submartingales and supermartingales observe this property for $\geq$ and $\leq$ , respectively.

A Markov is an adapted stochastic process where we can determine the expectation of the process using only the previous value. In other words, all of the information (filtration) up to time $𝑠$ is encoded in the value. Formally (2.3.6)

$𝔼 [𝑓 (𝑋 (𝑡)) | ℱ︀ (𝑠)] = 𝑔 (𝑋 (𝑠))$

Brownian Motion

The symmetric random walk is an adapted stochastic process which moves up and down by 1 by the flip of a coin at each time step. It has the property that each of its increments are independent. The expectation of the increment is 0, and its variance is $Δ 𝑡$ . The walk has the martingale property, which can be shown by splitting the current value into an increment, which is independent, and the previous value, which is measurable.

The scaled symmetric random walk approximates the Brownian motion.

$𝑊^{(𝑛)} (𝑡) = \frac{1}{\sqrt{𝑛}} 𝑀_{𝑛 𝑡}$

as $𝑛$ approaches $\infty$ .

I’m going to skip a lot of stuff relating to random walks, assuming that it won’t be relevant.

Brownian motion $𝑊 (𝑡)$ has the following properties:

Independent increments
Normally distributed increments
Expectation of increments is 0
Variance of increments is $Δ 𝑡$
Martingale

The first-order variation of a function is the definite integral from time 0 to $𝑇$ of the absolute value of its derivative. It can also be computed by partitioning time $𝑇$ into partition $Π$ . As $∥ Π ∥$ (the norm of $Π$ , aka size of its largest increment) approaches 0, we’re interested in summing the absolute value of each of these increments. Essentially we’re measuring how much change there is in the function.

The quadratic variation is computed in the same way, but with the increments being squared. Formally (3.4.1)

$[𝑓, 𝑓] (𝑇) = {lim}_{∥ Π ∥ \to 0} \sum_{𝑗 = 0}^{𝑛 - 1} [𝑓 (𝑡_{𝑗 + 1}) - 𝑓 (𝑡_{𝑗})]^{2}$

Typically, quadratic variation is 0 for any differentiable function. The quadratic variation of Brownian motion $[𝑊, 𝑊] (𝑇)$ is $𝑇$ . Conceptually, we can understand this as Brownian motion accumulating quadratic variation at rate one per unit time.

The differential of time $𝑑 𝑡$ multiplied by any other differential is 0. The differential of Brownian motion $𝑑 𝑊 (𝑡)$ multiplied by itself is $𝑑 𝑡$ . This relates to quadratic variation because

$𝑑 [𝑊, 𝑊] (𝑡) = 𝑑 𝑊 (𝑡) 𝑑 𝑊 (𝑡) = 𝑑 𝑡$

Geometric Brownian motion is an application of Brownian motion to stock prices. The process is described by

$𝑆 (𝑡) = 𝑆 (0) exp {𝜎 𝑊 (𝑡) + (𝛼 - \frac{1}{2} 𝜎^{2}) 𝑡}$

And log returns in this process are described by

$log \frac{𝑆 (𝑡_{𝑗 + 1})}{𝑆 (𝑡_{𝑗})} = 𝜎 (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗})) + (𝛼 - \frac{1}{2} 𝜎^{2}) (𝑡_{𝑗 + 1} - 𝑡_{𝑗})$

Skipping first passage time distribution, reflection principle, maximum to date

Stochastic Calculus

The purpose of stochastic calculus is to integrate an adapted stochastic process on a Brownian motion.

$\int_{0}^{𝑇} Δ (𝑡) 𝑑 𝑊 (𝑡)$ This $Δ$ describes our position in an asset at time $𝑡$ , which is dependent on the change in price. This integral is the value of the portfolio at time $𝑇$ . The change in price is a stochastic process based on Brownian motion, which means that we cannot integrate with a Riemann/Lebesgue integral.

In order to construct the Itô integral, we add up increments of $𝑊 (𝑡)$ and multiply them by the stock position $Δ (𝑡)$ at that time. The discrete analogy is that of $𝑊 (𝑡)$ as stock price and the time intervals being trading dates. The value of the portfolio is therefore the sum of the product of: the change in stock price each day, and the position held in stock on that day. Taking this discrete analogy continuously creates the integral.

The Itô integral is described as follows:

$𝐼 (𝑡) = \int_{0}^{𝑡} Δ (𝑢) 𝑑 𝑊 (𝑢)$

The Itô integral is a martingale. This is proved here.

Final Methods

## Student's t and Heavy-Tailed Distributions - URL: https://sharifhsn.dev/blog/student-t-and-heavy-tailed-distributions/ - Structured data: https://sharifhsn.dev/api/posts/student-t-and-heavy-tailed-distributions.json - Description: Definition 5.4.6 says that the pdf is then X is a Student t-distribution with degrees of freedom 𝑡 𝑛 . Proposition 5.48 will tell us all the moments, which are not easy to prove. - Date: 2024-10-28 - Exact published timestamp: 2024-10-28 - Topics: Probability Theory, Student t, Heavy Tails - Categories: Probability Theory - Source: FE-540 \| Probability Theory - Source URL: None - Content source format: typst

Student T-distribution

Definition 5.4.6 says that the pdf is

$𝑓 (𝑥) = \frac{Γ (\frac{(𝑛 + 1)}{2})}{\sqrt{𝜋 𝑛} Γ (\frac{𝑛}{2})} {(1 + \frac{𝑥^{2}}{𝑛})}^{- \frac{𝑛 + 1}{2}}$

then X is a Student t-distribution with degrees of freedom $𝑡_{𝑛}$ .

Proposition 5.48 will tell us all the moments, which are not easy to prove.

$𝔼 [𝑋] = 0$ if $𝑛 > 1$ , and

$𝕍 [𝑋] = \frac{𝑛}{𝑛 - 2}$ if $𝑛 > 2$

This distribution has a heavier tail than normal, but not as heavy as the next

Heavy Tailed Distribution

Definition 5.4.9 tells us the Pareto distribution which describes the power law in natural sciences. Income of people follows a heavy tail, there are always people earning a lot of money. Benford law erlates to cryptography, Zipf law relates to economics.

$𝑓 (𝑥) = \frac{𝑎 𝑏^{𝑎}}{𝑥^{𝑎 + 1}} 𝐼_{[𝑏, \infty]} (𝑥)$

I’m only looking at the value above b which is a positive quantity, it’s decaying at a power of a + 1. A larger a means lighter tails.

Proposition 5.50 says that if we integrate this, you get 1, so this is a density function.

Proposition 5.51 tells us the moments:

$𝔼 [𝑋] = \frac{𝑎 𝑏}{𝑎 - 1}$

$𝕍 [𝑋] = \frac{𝑎 𝑏^{2}}{(𝑎 - 1)^{2} (𝑎 - 2)}$

There is a link between the Pareto and exponential distribution. This is proposition 5.53.

If you have a exponential distribution

$𝑍 \sim Exp (𝑎)$ , then for a $𝑏 > 0$ , then

$𝑋 = 𝑏 \cdot 𝑒^{𝑍}$ is a Pareto distribution $𝑃 (𝑎, 𝑏)$ .

Theorem 5.10

Log-Normal Distribution

Definition 5.54. The Black-Scholes model follows this distribution.

It’s called log normal because the log of it is normal

$𝑓 (𝑥) = \frac{1}{𝑥 𝜎 \sqrt{2 𝜋}} 𝑒^{- \frac{(ln 𝑥 - 𝜇)^{2}}{2 𝜎^{2}}} 𝐼_{(0, \infty)} (𝑥)$

then $𝑋 \sim LogNormal (𝜇, 𝜎^{2})$

We can also describe the cdf of the distribution:

$𝐹_{𝑋} (𝑡) = ℙ (𝑋 \leq 𝑡) = Φ (\frac{ln 𝑡 - 𝜇}{𝜎})$

where $Φ$ is the cdf of the normal distribution.

Proposition 5.55 will tell us the density proof, will be on final

Proposition 5.56 tells us the moments:

$𝔼 [𝑋] = 𝑒^{𝜇 + \frac{1}{2} 𝜎^{2}}$

$𝕍 [𝑋] = (𝑒^{𝜎^{2}} - 1) 𝑒^{𝑡 𝜇 + 𝜎^{2}}$

Laplace Distribution

Definition 5.59

$𝑓 (𝑥) = \frac{𝜃}{2} 𝑒^{- 𝜃 | 𝑥 |}$

Something is different is that it’s the absolute value of $𝑥$ , and this applies to the whole real line.

Proposition 5.60 proves the density is 1

Proposition 5.61 tells us the moments

$𝔼 [𝑋] = 0$

the proof:

$𝔼 [𝑋] = \frac{𝜃}{2} \int_{- \infty}^{\infty} 𝑥 𝑒^{- 𝜃 | 𝑥 |} 𝑑 𝑥$

Because this is an odd function, we can say that the expectation must be 0.

$𝕍 [𝑋] = \frac{1}{𝜃}$

Double Exponential Distribution

5.4.10

$𝑓 (𝑥) = {𝑝 𝛼_{1} 𝑒^{- 𝛼_{1} 𝑥}, 𝑥 > 0 (1 - 𝑝) 𝛼_{2} 𝑒^{𝛼_{2} 𝑥}, 𝑥 \leq 0$

If x is positive, it’s exponential, if it’s negative, it follows a different exponential distribution

Examples

These come from the second textbook G, where homework is.

5.1

the pdf

$𝑓 (𝑥) = \frac{2}{𝜋 (1 + 𝑥^{2})} 𝕀_{[0, \infty)} (𝑥)$

show that it’s a density function and that the expected value does not exist

We need to solve for the integral over the range to be 1.

$\int_{0}^{\infty} \frac{2}{𝜋 (1 + 𝑥^{2})} 𝑑 𝑥 = \frac{2}{𝜋} \int_{0}^{\infty} \frac{1}{1 + 𝑥^{2}} 𝑑 𝑥$

And that’s just arctan

Then you get one by solving for arctan.

For expectation, the proof is done by u-substitution for x^2 + 1, and then you can show that the expectation is infinite, which means it doesn’t exist.

5.2

If $𝑋 \sim Unif [0, 1]$ , show that $𝑋^{2} \sim Beta (\frac{1}{2}, 1)$

We will use theorem 5.10 for solving for pdf in order to get $𝑋^{2}$ from the uniform. We can use $𝑥^{2}$ as our function because under the interval $[0, 1]$ it’s monotonic increasing.

We need to find the inverse of our function, and the derivative of our function. That’s

$ℎ^{- 1} (𝑦) = \sqrt{𝑦}$

$ℎ^{'} (𝑥) = 2 𝑥$

$\frac{1}{2 \sqrt{𝑦}}$

Now we have to prove that this is the density function for beta.

We can substitute the numbers given to us in the beta formula, and then use the gamma decomposition to give us actual numbers.

5.3

We want to show that $𝑌 = 𝑋^{2}$ has the density function The density function is the derivative of the cdf, so if you want to show the pdf

4.7

cdf of a discrete random variable

$𝐹 (𝑥) = {0 𝑥 < 0 𝑥 / 40 \leq 𝑥 < 1 1 / 21 \leq 𝑥 < 2 1 / 12 𝑥 + 1 / 22 \leq 𝑥 < 3 1 𝑥 \geq 3$

How do you compute the probabilities?

$ℙ (𝑋 < 2) = \frac{1}{2}$ We can directly compute this easily. What about

$ℙ (𝑋 = 2) = ℙ (𝑋 \leq 2) - ℙ (𝑋 < 2)$

This is because it’s a cdf and you’re getting rid of all that came before which is actually under a different function. Same for

$ℙ (1 \leq 𝑋 < 3) = ℙ (𝑋 < 3) - ℙ (𝑋 < 1)$

What about the greater than? We can use the complement.

$ℙ (𝑥 > \frac{3}{2}) = 1 - ℙ (𝑥 \leq \frac{3}{2})$

So if you’re doing equal, you get rid of everything up to that point. For a point in an interval this is actually 0 which makes sense because the probability of any individual point is usually 0.

$ℙ (𝑥 = \frac{5}{2}) = ℙ (𝑥 \leq \frac{𝑡}{2}) - ℙ (𝑋 < \frac{5}{2}) = 0$

4.8

If we flip a coin twice, and X is the number of tails, what is the cdf of X?

Let’s calculate

$𝐹 (𝑡) = ℙ (𝑋 \leq 𝑡) = {0 𝑡 < 0 ℙ (𝑥 = 0) 0 \leq 𝑡 < 1 ℙ (𝑥 = 0 \cup 𝑥 = 1) 1 \leq 𝑡 < 2 1 𝑡 \geq 2$

4.10

if we consider cdf of X

$𝐹 (𝑡) = {0 𝑡 < 0 \frac{1}{2} 𝑡^{2} 0 \leq 𝑡 < 1 𝑘 (4 𝑡 - 𝑡^{2}) 1 \leq 𝑡 < 2 1 𝑡 \geq 2$

In order to consider the independence of two sets

$𝐴 = {\frac{1}{2} \leq 𝑥 \leq \frac{3}{2}}$

$𝐵 = {1 \leq 𝑥}$

We want the probability of the intersection of those sets to be the same as the product of their probabilities. In this case, they are not, so they are not independent.

5.18

Deck of 52 cards. Draw the cards with replacement until an ace is drawn. Calculate the probability that at least 10 draws are needed to get the first ace.

If you see “until the first” it means, “until the first success event”, which means the geometric distribution.

This is the $𝑋 \sim 𝐺 𝑒 𝑜 (\frac{1}{13})$

Then we can get the pmf as

$ℙ (𝑋 = 𝑛) = {(\frac{12}{13})}^{𝑛 - 1} \frac{1}{13}$ How do you do at least ten draws?

$ℙ (𝑋 \geq 10) = \sum_{𝑛 = 10}^{\infty} {(\frac{12}{13})}^{𝑛 - 1} \frac{1}{13}$

$= \frac{1}{13} \sum_{𝑛 = 10}^{\infty}$

We will do a change of variable of $𝑘 = 𝑛 - 10$

$= \frac{1}{13} \sum_{𝑘 = 0}^{\infty} {(\frac{12}{13})}^{𝑘 + 9}$

$= \frac{1}{13} {(\frac{12}{13})}^{9} \sum_{𝑘 = 0}^{\infty} {(\frac{12}{13})}^{𝑘}$

VERY IMPORTANT FORMULA FOUND HERE:

$\sum_{𝑛 = 0}^{\infty} 𝑝^{𝑛} = \frac{1}{1 - 𝑝}$

We will use this to solve geometric distributions.

Now let’s look at continuous time

6.8

Typical example: we are given the density and are asked to solve for the moments.

$𝑓 (𝑥) = \frac{27}{490} (𝑒 𝑥^{2} - 2 𝑥), \frac{2}{3} < 𝑥 < 3$

$𝔼 [𝑋] = \int_{- \frac{2}{3}}^{3} 𝑥 \cdot \frac{27}{490} (3 𝑥^{2} - 2 𝑥) 𝑑 𝑥$

We can solve this integral which should give us the answer 283/120. Left as exercise to reader.

7.7

Check notes…

6.3

$𝑓 (𝑥) = \frac{2}{𝑥^{2}}, 1 < 𝑥 < 2$

Find the distribution and pdf of $𝑌 = 𝑋^{2}$

same as the uniform question.

There are many ways to solve this. Theorem 5.10 is one way to solve it. You can also start from the basic principle of how cumulative distribution works

$𝐺 (𝑡) = ℙ (𝑌 \leq 𝑡) = ℙ (𝑋^{2} \leq 𝑡) = ℙ (- \sqrt{𝑡} \leq 𝑥 \leq \sqrt{𝑡}) = ℙ (𝑥 \leq \sqrt{𝑡}) = ℙ (𝑥 \leq \sqrt{𝑡}) - ℙ (𝑋 \leq - \sqrt{𝑡}$

The last probability we canceled so we are only concerned with the middle probability,

$ℙ (1 \leq 𝑋 \leq \sqrt{𝑡})$

We have to integrate to get this

$= \int_{1}^{\sqrt{𝑡}} \frac{2}{𝑥^{2}} 𝑑 𝑥 = - \frac{2}{𝑥} |_{𝑥 = 1}^{𝑥 = \sqrt{𝑡}} = - \frac{2}{\sqrt{𝑡}} + 2$

Then we can find that $𝐺 (𝑡)$ is substituted this.

Then we can find the density function, which is known to be the derivative of the cdf from earlier:

$𝑔 (𝑡) = 𝐺^{'} (𝑡)$

That’s it from the examples.

## Remembering the Boarding-School History - URL: https://sharifhsn.dev/blog/linkedin-2024-10-27-remembering-the-boarding-school-history/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2024-10-27-remembering-the-boarding-school-history.json - Description: A reflection on President Biden's apology for federal Indian boarding schools. - Date: 2024-10-27 - Exact published timestamp: 2024-10-27T15:20:30.081Z - Topics: Education, Native American History - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/posts/sharif-haason_president-biden-apologizes-for-treatment-activity-7256323884658888704-1-gA - Content source format: typst

I was moved by this emotional announcement by President Biden, apologizing for an often forgotten aspect of oppression of Native Americans: the forced kidnapping of their children to boarding schools where they were stripped of their language and culture. This policy was only ended in 1978 with the Indian Child Welfare Act. It’s a common attitude of Americans that the mistreatment of Native Americans was something that happened a long time ago. Statements like these by our president are therefore important for educating the public and I urge you to listen to his speech and understand the magnitude of suffering that the history books have skipped over.

## Interest Rate Risk and Duration - URL: https://sharifhsn.dev/blog/interest-rate-risk-and-duration/ - Structured data: https://sharifhsn.dev/api/posts/interest-rate-risk-and-duration.json - Description: The Economic Value of Mean Squared Error: Evidence from Portfolio Selection What’s the intuition? Everyone talks about optimization, rational agents doing the best they can. A… - Date: 2024-10-24 - Exact published timestamp: 2024-10-24 - Topics: Risk Management, Interest Rate Risk, Duration - Categories: Risk Management - Source: FE-535 \| Risk Management - Source URL: None - Content source format: typst

Lecture Notes

Prof’s Paper

The Economic Value of Mean Squared Error: Evidence from Portfolio Selection

What’s the intuition? Everyone talks about optimization, rational agents doing the best they can. A colleague told the president of Stevens that it’s all cost benefit analysis. The question is how you measure that. We know we have model risk, you get $\hat{𝑤}$ never $𝑤^{*}$ , only an estimate. Expectation vs reality. We’ll never get to that target. There will always be some loss.

How do we mitigate that discrepancy. Instead of trying to shrink the y-axis, you actually work with the x-axis, which is where the MSE comes from. If you want to handle this model risk ex ante, before you start, you apply this to the portfolio selection problem.

We’ll eventually show that there’s a linear relationship between MSE and decision loss. What made this paper more interesting is that, everyone talks about ML, prediction, optimization. And there’s interesting literature on those issues.

On the website.

Bond

The book is very qualitative, and it talks about LTCM. It talks about the 10 days value at risk, which is too high. Chapter 14 Model Risk is a very interesting big picture chapter.

We know what compounding is.

Fixed Coupon Bond

A fixed coupon bond, e.g. corporate treasury bonds, will issue you coupons regularly over time.

You will pay today $- 𝑃$ at time 0. Then you will get appetizers, some coupon $𝑐$ . And then on the time of maturity, you’ll get another appetizer and then the major face value dinner $𝑐 + 𝐹$ .

The total cash flow is $𝑐 \cdot 𝑇 + 𝐹$

But this is not discounted. So this value is definitely less than $𝑃$ .

For our risk-free rate $𝑦$ , we get

$𝑃_{0} = 𝑐 \times 𝐹 \sum_{𝑡 = 1}^{𝑇} \frac{1}{(1 + 𝑦)^{𝑡}} + \frac{𝐹}{(1 + 𝑦)^{𝑇}}$ And the nice closed form solution

$𝑃_{0} = \frac{𝑐}{𝑦} \times 𝐹 [1 - \frac{1}{(1 + 𝑦)^{𝑇}}] + \frac{𝐹}{(1 + 𝑦)^{𝑇}}$

which comes from the fact that this is a geometric series.

Greeks help you understand the sensitivity of the solution.

We can get some analytical insights here because of the closed-form nature of Black-Scholes.

A bond is called selling at par, if the current price is equal to the face value.

If we think about the derivative with respect to y, it might not look so nice. But it actually ends up looking like a nice graph with a downward slope. And that’s because when interest rates go up, bond prices go down.

What’s actually being done to calculate this? You don’t want bond yields to go up, that means people are selling more bonds. When people buy more bonds, the yields go down, which is good.

You are inferring the yield from these prices. Before we took the yield for granted, but that’s not what you actually do.

The current yield is calculated by the annual coupon rate divided by the current market price. The yield to maturity

Campbell Harvey talks about how the inversion of the yield curve predicts a recession. Why does this occur? Let’s think about this from a risk management point of view.

This means that there’s more pressure on the prices in the long term.

During a recession, the central bank will have a soft landing by lowering interest rates. What the market is doing, they look at information. When they are holding treasury bonds with a longer maturity, you can think of it as a simple discounting factor.

If we have a bond that pays an annual coupon of 8%, and a YTM of 6%, then the price will actually decrease after a year. And this is because the appetizers are really good, so if you miss out on one then the bond will become less valuable.

We can think of an asset that pays a continuous rate over time which is reflected by GBM, which would give you the same idea.

LTCM

Great book about bonds. Liar’s Poker he describes the strategies when it comes to mortgages, prepayments. Mortgages were bailed out and a lot of trading strategies developed around that. And that helps you understand the price of these instruments.

The nice thing about fixed income

Long-Term Capital management (LTCM) was a hedge fund formed in the mid-1990s. All the main people like Scholes were a part of this. They had huge returns in their first year, then they got into international bond pricing, and then Russia defaulted which caused almost all of their bonds to fail. And the banks bailed them out. From When Genius Failed. Three main players, two academics Merton and Scholes. John Merriwether traditional trader as well. Classic case of FRM. We had premium bonds and discount bonds. We’ll find treasury bonds domestic and foreign with the same maturity, coupons, same characteristics. So their price should be the same. LTCM went long on the cheaper ones and short on the more expensive ones, to exploit this arbitrage. Russia defaulted and they went bust. Goldman, Chase, Lehman helped break it up, but the Fed stepped in and bailed them out.

The reason that there was a divergence between those bonds to begin with was th liquidity of the bonds to begin with.

Interest Rate Risk

All bonds are subject to interest rate risk. The major question is what happens if yields go up? They keep moving, which causes prices to go up and down. It comes down to monetary policy, where an investor or risk manager should proactively think about how it impacts your portfolio.

Think about it from the CAPM the single risk factor is the market. The same argument is here, that the systematic risk is the interest rate risk.

We’re at a 5% interest rate, what happens if the interest rate goes up? My bond price goes down. But I want to map those basis points into an actual risk measurement.

Let’s try to do a first-order approximation for this function. This goes to duration risk, or price sensitivity. The bond price is a function of yields. $𝑃_{0} = 𝑓 (𝑦)$

We want to know how the price will change when we get

$𝑓 (𝑦 + Δ 𝑦)$ Let me do a Taylor expansion. I start with the price at the initial point $𝑃_{0}$ , which is $𝑓 (𝑦_{0})$ . Then what else do I have? I have the first derivative with respect to that yield curve.

$𝑃_{1} = 𝑃_{0} + 𝑓^{'} (𝑦_{0}) \times (Δ 𝑦) + \frac{1}{2} 𝑓^{″} (𝑦_{0}) \times (Δ 𝑦)^{2}$ If we ignore that second derivative, just looking at first-order, we see

$𝑃_{1} - 𝑃_{0} = 𝑓^{'} (𝑦_{0}) Δ 𝑦$

Now we can find our change of price for a given change in yield. That’s the sensitivity, and therefore the exposure.

This derivative is called the duration.

$𝑓^{'} (𝑦_{0}) = - 𝐷 𝐷 = - 𝐷^{*} 𝑃_{0} = - \frac{𝐷}{1 + 𝑦} 𝑃_{0}$

These different Ds are different measures of duration. DD is the Dollar Duration, the notional amount. D* is the modified duration. That duration is a nonnegative measure, so we make it negative to express the inverse relationship between yield and bond price.

The analytical solution is quite difficult, you can look it up on your own, so we’ll just look at the numerical solution.

The Macaulay duration, which is D, is actually more intuitive. It can be computed as

$𝐷 = \sum_{𝑡 = 1}^{𝑇} 𝑤_{𝑡} 𝑡$

$𝑤_{𝑡} = \frac{𝐶 𝐹_{𝑡} / (1 + 𝑦)^{𝑡}}{𝑃_{0}}$

Let’s go back to the definition of the fixed coupon bonds. What’s the risk?

Each coupon gets its own weight. It takes me time to recover my investment. The weight at time t, I’m getting a cash flow one year from now. After taking into account the time value of money, how much money did I recover from that first cash flow. I want to think about this as, I put money away, how long does it take for me to receive money, and then how long to get back my full investment.

We can see that macaulay duration equals maturity. This is easy to see when we have zero coupon bonds. It doesn’t matter what the yield is, it’s just maturity.

The higher the yield, the lower duration is. There is less weight assigned to the distant future. If you live in an environment with much higher yield, you can reinvest that as compounding.

Frequency matters because you have more coupons. And if you get more coupons, that means you have a lower duration.

If you have no coupons, that means you have a very long duration.

You can have these sensitivities for options as well, these are the Greeks.

Second order sensitivity for bond price is convexity. Duration is expected value of time, convexity is the second moment of time.

Convexity has $𝑤_{𝑡}$ , which is scaled by the discount factor squared. It gives you an approximation for the true value of the price.

Portfolio Duration and Convexity

I want to meet certain duration targets in my portfolio.

The market is expecting the interest rate to drop by 25 bps. That means you want to increase your duration, which is your exposure to the yield. If you work with modified duration, it’s the same as a $𝛽$ , it’s a weight for each asset, in terms of exposure. Portfolio duration (modified duration) is just the average of the duration of each of your bonds, modified by their weights.

The way you solve the bond portfolio problem is by defining it as a matrix multiplication, where the matrix of the bond durations is inverted and multiplied by the intended portfolio duration, and the resultant vector should be the weights.

The least squares solution comes into the picture if you have multiple bonds, not just two. You can do a generalized inverse on an e.g. 2x4 matrix and then solve it that way.

## Multidimensional Itô Calculus - URL: https://sharifhsn.dev/blog/multidimensional-ito-calculus/ - Structured data: https://sharifhsn.dev/api/posts/multidimensional-ito-calculus.json - Description: We’re going to be talking about multivariable stochastic calculus. All of this material is fair game for the midterm. Next week is the midterm review session. Bring questions!… - Date: 2024-10-24 - Exact published timestamp: 2024-10-24 - Topics: Stochastic Calculus, Multidimensional Models, Correlation - Categories: Stochastic Calculus - Source: FE-610 \| Stochastic Calculus - Source URL: None - Content source format: typst

Week 7

Lecture Notes

We’re going to be talking about multivariable stochastic calculus. All of this material is fair game for the midterm. Next week is the midterm review session. Bring questions!

We’re mostly going to be doing applications and extensions today, which is not too bad.

How do we deal with more than one source of randomness? We need to have more than one Brownian motion. We have been talking about the money market and a risky asset, but there are multiple stocks in existence. We need multiple sources of noise. We can’t use the single source of noise, because Tropicana and Amtrak are impacted differently from e.g. the price of steel. How do we work with multiple sources of noise?

Let’s define what it means to have a d-dimensional Brownian motion. In order to consider a 10-dimensional Brownian motion as opposed to 10 Brownian motions, we need to have each process to be a Brownian motion, and these are all independent. We need to define the associated filtration on this space. Information should accumulate, and everything should be measurable with respect to that filtration, and each of those increments should be independent.

Because each one is a Brownian motion in its own right, the quadratic variation for each Brownian motion is t. I’m still going to have

$𝑑 𝑊_{𝑖} (𝑡) 𝑑 𝑊_{𝑖} (𝑡) = 𝑑 𝑡$

We still don’t know the cross-variation between two independent Brownian motions? $𝑑 𝑊_{𝑖} (𝑡) 𝑑 𝑊_{𝑗} (𝑡) = ?$

We will prove that this is 0.

Similar to what we did for proving the quadratic variation to be t, we’re going to do a mean squared convergence. We will create a process $𝐶_{Π}$ the sample cross-variation:

$𝐶_{Π} = \sum_{𝑘 = 0}^{𝑛 - 1} (𝑊_{𝑖} (𝑡_{𝑘 + 1}) - 𝑊_{𝑖} (𝑡_{𝑘})) (𝑊_{𝑗} (𝑡_{𝑘 + 1}) - 𝑊_{𝑗} (𝑡_{𝑘}))$

This would be the real cross variation if it went to 0. We are going to study the expectation and variance as the norm of the partition goes to 0.

If we can define these and figure these out, we can show that the cross variation will converge to the variable that has this mean and this variance.

How could I calculate the expected value? Let’s just use the definition

$𝔼 [𝐶_{Π}] = 𝔼 [\sum_{𝑘 = 0}^{𝑛 - 1} (𝑊_{𝑖} (𝑡_{𝑘 + 1}) - 𝑊_{𝑖} (𝑡_{𝑘})) (𝑊_{𝑗} (𝑡_{𝑘 + 1}) - 𝑊_{𝑗} (𝑡_{𝑘}))]$

I have linearity, so I can move the expectation inside the summation.

$= \sum_{𝑘 = 0}^{𝑛 - 1} 𝔼 [(𝑊_{𝑖} (𝑡_{𝑘 + 1}) - 𝑊_{𝑖} (𝑡_{𝑘})) (𝑊_{𝑗} (𝑡_{𝑘 + 1}) - 𝑊_{𝑗} (𝑡_{𝑘}))]$

Now we use independence. These two Brownian motions are independent from each other, so their increments are independent. Using this independence, we say

$= \sum_{𝑘 = 0}^{𝑛 - 1} 𝔼 [(𝑊_{𝑖} (𝑡_{𝑘 + 1}) - 𝑊_{𝑖} (𝑡_{𝑘}))] 𝔼 [(𝑊_{𝑗} (𝑡_{𝑘 + 1}) - 𝑊_{𝑗} (𝑡_{𝑘}))]$

Well what is that expected value? 0!

$= 0$

What is the limit as the norm of the partition goes to 0, of 0? 0!

Now let’s do variance $𝕍 [𝐶_{Π}]$

We could try the traditional expectation of squares method, but we would then have to work with $𝐶_{Π}^{2}$ . What is that?

$𝐶_{Π}^{2} = \sum_{𝑘 = 0}^{𝑛 - 1} (𝑊_{𝑖} (𝑡_{𝑘 + 1}) - 𝑊_{𝑖} (𝑡_{𝑘})) (𝑊_{𝑗} (𝑡_{𝑘 + 1}) - 𝑊_{𝑗} (𝑡_{𝑘})) \cdot \sum_{𝑘 = 0}^{𝑛 - 1} (𝑊_{𝑖} (𝑡_{𝑘 + 1}) - 𝑊_{𝑖} (𝑡_{𝑘})) (𝑊_{𝑗} (𝑡_{𝑘 + 1}) - 𝑊_{𝑗} (𝑡_{𝑘}))$

Which isn’t wrong, but is maybe more daunting than we need. Maybe there’s a better way we can tackle this. We’re going to take advantage of properties we know about variance. In particular, we have this result:

$𝕍 [𝛼 𝑋 + 𝛽 𝑌] = 𝛼^{2} 𝕍 [𝑋] + 2 𝛼 𝛽 𝑐 𝑜 𝑣 (𝑋, 𝑌) + 𝛽^{2} 𝕍 [𝑌]$

In order for this to be useful, we need these numbers to be independent. We have the property that future increments are independent of the filtration, so we will have the independence between the product of the increments.

Aside: The cross variation of a process with respect to itself is the same as quadratic variation.

We can then write

$𝕍 [𝐶_{Π}] = \sum_{𝑘 = 0}^{𝑛 - 1} 𝕍 [(𝑊_{𝑖} (𝑡_{𝑘 + 1}) - 𝑊_{𝑖} (𝑡_{𝑘})) (𝑊_{𝑗} (𝑡_{𝑘 + 1}) - 𝑊_{𝑗} (𝑡_{𝑘}))]$

because of this independence. Let’s examine the variance of one of these elements.

To get this variance, we can take the expectation of squares here.

$= 𝔼 [{Inc}_{𝑖}^{2} \cdot {Inc}_{𝑗}^{2}] - 𝔼 {[{Inc}_{𝑖} \cdot {Inc}_{𝑗}]}^{2}$

I know for any Brownian motion that the variance is length of time, because that’s quadratic variation. Therefore it’s just that value squared.

$= (𝑡_{𝑘 + 1} - 𝑡_{𝑘})^{2}$

For this normal random variable, the expected value is 0, so in this case the variance is the same as the variance squared. Now if we look at the limit of the norm of the partition going to 0 for the variance, we get

${lim}_{∥ Π ∥ \to 0} \sum_{𝑘 = 0}^{𝑛 - 1} (𝑡_{𝑘 + 1} - 𝑡_{𝑘})^{2}$

But that’s just the formula for the quadratic variation of time, which must be 0 because it’s a continuous differentiable function.

If it’s converging to mean 0 and variance 0, then that’s just 0.

We have proven that the cross variation is 0, and therefore the differential of the cross variation is also 0.

If you multiply a Brownian motion by itself, you get the variance not the mean. This result is only for independent Brownian motions.

Two Dimensional Ito Process

This leads us to our two dimensional Ito process. Big distinction between just two Brownian motions and the two-dimensional, because dimensional means independent.

We could write a standard Ito process as

$𝑋 (𝑡) = 𝑋 (0) + \int_{0}^{𝑡} Δ (𝑢) 𝑑 𝑊 (𝑢) + \int_{0}^{𝑡} Θ (𝑢) 𝑑 𝑢$

But now my $Δ$ and $𝑊$ are both vectors, and we take the dot product of these, which is a scalar vector multiplication

Why don’t we have a dot product with the $Θ$ s? Can you have multi-dimensional time? No. We don’t want multidimensional time, so we don’t have multidimensional $Θ$ . $Θ$ could depend on the process, but you wouldn’t decompose with reference to different aspects of movements in time.

What if we take the differential?

$𝑑 𝑋 (𝑡) = 𝜎_{1} (𝑡) 𝑑 𝑊_{1} (𝑡) + 𝜎_{2} (𝑡) 𝑑 𝑊_{2} (𝑡) + Θ (𝑡) 𝑑 𝑡$

Now what if we square it? Too long to write out, but it follows all the rules of multiplication. We proved earlier that $(𝑑 𝑊_{1} (𝑡))^{2}$ is dt, and the same for $𝑑 𝑊_{2}$ . We also know that the product of those is 0 because their cross variation is 0, and the product of dt and any of these dW is 0. So most of it is gone. The only thing that sticks around is

$= (𝜎_{1}^{2} (𝑡) + 𝜎_{2}^{2} (𝑡)) 𝑑 𝑡$ I can use this result. I know that this is a handy way to get the quadratic variation of an Ito process by integrating this differential.

$[𝑋, 𝑋] (𝑡) = \int_{0}^{𝑡} (𝜎_{1}^{2} (𝑢) + 𝜎_{2}^{2} (𝑢)) 𝑑 𝑢$

There’s a lot of different ways to do this. $Σ$ is just standing in here for any adapted stochastic process. We are going to define $𝑋 (𝑡)$ by its own Ito process. These multidimensional Ito processes have the nonrandom constant, d Ito integrals, and then the Riemann of time.

Let’s say we have two well-defined multidimensional Ito processes in this way. Defined as such, what is

$𝑑 𝑋 (𝑡) 𝑑 𝑌 (𝑡) = ?$

Let’s not write out all these terms. We know that dW multiplied by any differential except itself is 0, and dt multiplied by any differential is 0. So we get the terms

$= (𝜎_{𝑋, 1} + 𝜎_{𝑋, 2}) (𝜎_{𝑌, 1} + 𝜎_{𝑌, 2})$

This is all to build up to the concept of the two dimensional Ito formula. We are going to use the Taylor series expansion with three variables now. That would say that

$𝑓 (𝑡_{𝑘 + 1}, 𝑋_{𝑘 + 1}, 𝑌_{𝑘 + 1}) = 𝑓 (𝑡_{𝑘}, 𝑋_{𝑘}, 𝑌_{𝑘}) + 𝑓_{𝑡} (𝑡_{𝑘}, 𝑋_{𝑘}, 𝑌_{𝑘}) (𝑡_{𝑘 + 1} - 𝑡_{𝑘}) + 𝑓_{𝑋} (𝑡_{𝑘}, 𝑋_{𝑘}, 𝑌_{𝑘}) (𝑋_{𝑘 + 1} - 𝑋_{𝑘}) + 𝑓_{𝑌} (𝑡_{𝑘}, 𝑋_{𝑘}, 𝑌_{𝑘}) (𝑌_{𝑘 + 1} - 𝑌_{𝑘})$ And all the second order terms. But we don’t worry about those… we can use the same arguments from the past and move the first element to the left as a telescoping sum. When we sum all of these together and get the limits, we can turn all those increment terms into differentials. And then because the second order involves multiplying differentials, these all become 0. Not all of them though. The cross terms are not 0 based on our definition, so we still need to consider $𝑓_{𝑋 𝑋}$ , $𝑓_{𝑋 𝑌}$ , and $𝑓_{𝑌 𝑌}$ . But we can rage quit on third order because that definition includes dt, so we will eliminate those and get 0.

For future reference: $𝑋 = 𝑋 (𝑡)$ , $𝑌 = 𝑌 (𝑡)$ , to save my fingers.

$𝑑 𝑓 (𝑡, 𝑋, 𝑌) = 𝑓_{𝑡} (𝑡, 𝑋, 𝑌) 𝑑 𝑡 + 𝑓_{𝑋} (𝑡, 𝑋, 𝑌) 𝑑 𝑋 + 𝑓_{𝑦} (𝑡, 𝑋, 𝑌) 𝑑 𝑌 + \frac{1}{2} 𝑓_{𝑋 𝑋} (𝑡, 𝑋, 𝑌) 𝑑 𝑋^{2} + 𝑓_{𝑋 𝑌} (𝑡, 𝑋, 𝑌) 𝑑 𝑋 𝑑 𝑌 + \frac{1}{2} 𝑓_{𝑌 𝑌} (𝑡, 𝑋, 𝑌) 𝑑 𝑌^{2}$

These are all partials.

We have the product rule of Ito processes. And we still need to include the cross term. In normal calculus, this term is 0. $𝑑 (𝑋 𝑌) = 𝑋 𝑑 𝑌 + 𝑌 𝑑 𝑋 + 𝑑 𝑋 𝑑 𝑌$

Let’s do our partials to prove this. $𝑓 (𝑡, 𝑋, 𝑌) = 𝑋 𝑌$

$𝑓_{𝑡} = 0, 𝑓_{𝑋} = 𝑌, 𝑓_{𝑌} = 𝑋, 𝑓_{𝑋 𝑋} = 0, 𝑓_{𝑋 𝑌} = 1, 𝑓_{𝑌 𝑌} = 0$

Then using our formula, that’s

$= 0 𝑑 𝑡 + 𝑌 𝑑 𝑋 + 𝑋 𝑑 𝑌 + \frac{1}{2} (0) 𝑑 𝑋^{2} + 1 𝑑 𝑋 𝑑 𝑌 + \frac{1}{2} (0) 𝑑 𝑌^{2}$

$= 𝑋 𝑑 𝑌 + 𝑌 𝑑 𝑋 + 𝑑 𝑋 𝑑 𝑌$

Let’s find the differential of the discounted stock process

$𝑒^{- 𝑟 𝑡} 𝑆 (𝑡)$ I could do the whole stochastic mess we did before, or I could recognize that this is something where I can use the Ito product rule.

$𝑑 (𝑒^{- 𝑟 𝑡}) = 𝑒^{- 𝑟 𝑡} 𝑑 𝑆 + 𝑆 𝑑 (𝑒^{- 𝑟 𝑡}) + 𝑑 (𝑒^{- 𝑟 𝑡}) 𝑑 𝑆$

$𝑑 (𝑒^{- 𝑟 𝑡}) = - 𝑟 𝑒^{- 𝑟 𝑡} 𝑑 𝑡$

Warning: you need the dt, or it’s not a differential, it’s a derivative! A derivative is the rate of change, and the differential is the instantaneous change. The derivative does not exist because Brownian motion is not smooth anywhere, so we have to use the differential.

Any continuous process with t and W(t) is an Ito process. All of our results are only for continuous processes.

Lévy

This is a very handy result. If we have a martingale M relative to filtration $ℱ︀$ , the starting value is 0, it’s continuous, and the quadratic variation is t, then it is a Brownian motion. We don’t actually need the normally distributed part to prove this.

Quadratic variation and variance are different things! Quadratic variation is how much second order variation a process will have on one path, and it’s a random variable. The variance is how much you expect the process to differ from its mean on average, which is a single number.

We are going to prove that this is a Brownian motion. We will need all of these components. Because M is continuous, it must be a continuous process, which means we have an Ito decomposition, so M is an Ito process.

So we can now use the Ito formula to look at an arbitrary function f on M $𝑓 (𝑡, 𝑀 (𝑡)) = 𝑓 (0, 𝑀 (0)) + \int_{0}^{𝑡} 𝑓_{𝑎} (𝑢, 𝑀 (𝑢)) 𝑑 𝑢 + \int_{0}^{𝑡} 𝑓_{𝑏} (𝑢, 𝑀 (𝑢)) 𝑑 𝑀 (𝑢) + \frac{1}{2} \int_{0}^{𝑡} 𝑓_{𝑏 𝑏} (𝑢, 𝑀 (𝑢)) 𝑑 𝑀^{2} (𝑢)$

I know that M(0) is equal to 0 from rule 2, and we also know that $𝑑 𝑀^{2} (𝑢)$ is $𝑑 𝑢$ because of rule 4 (definition of quadratic variation).

$𝑓 (𝑡, 𝑀 (𝑡)) = 𝑓 (0, 0) + \int_{0}^{𝑡} (𝑓_{𝑎} (𝑢, 𝑀 (𝑢)) + \frac{1}{2} 𝑓_{𝑏 𝑏} (𝑢, 𝑀 (𝑢))) 𝑑 𝑢 + \int_{0}^{𝑡} 𝑓_{𝑏} (𝑢, 𝑀 (𝑢)) 𝑑 𝑀 (𝑢)$

This is a trick where we use a sledgehammer as a scalpel. We have established this true for any function f, so it has to be true for one particular function. That particular function is if

$𝑓 (𝑎, 𝑏) = 𝑒^{𝑢 𝑏 - \frac{1}{2} 𝑢^{2} 𝑎}$

with a nonrandom dummy variable $𝑢$ that we don’t know the value of. Let’s get our partials.

$𝑓_{𝑎} = - \frac{1}{2} 𝑢^{2} 𝑓 (𝑎, 𝑏)$

$𝑓_{𝑏} = 𝑢 𝑓 (𝑎, 𝑏)$ $𝑓_{𝑏 𝑏} = 𝑢^{2} 𝑓 (𝑎, 𝑏)$ If I plugged this in, I would get

$𝑓 (𝑡, 𝑀) = 𝑒^{𝑢 𝑀 - \frac{1}{2} 𝑢^{2} 𝑡}$ I have now that (skipping some steps)

$𝑒^{𝑢 𝑀 - \frac{1}{2} 𝑢^{2} 𝑡} = 1 + \int_{0}^{𝑡} 𝑢 𝑓 (𝑢, 𝑀) 𝑑 𝑀$

If I take the expected value, I can factor out the nonrandom component

$𝔼 [𝑒^{𝑢 𝑀}] 𝑒^{- \frac{1}{2} 𝑢^{2} 𝑡} = 1 + 𝔼 [\int_{0}^{𝑡} 𝑢 \cdot 𝑓 (𝑢, 𝑀) 𝑑 𝑀]$

Exercise 4.1 asked us to prove that an Ito integral with respect to a martingale is still a martingale, which was a further proof of the class proof on a Brownian motion. Therefore the expectation of the Ito integral, which is itself a martingale, must be 0, because that was its initial value.

$𝔼 [𝑒^{𝑢 𝑀}] = 𝑒^{\frac{1}{2} 𝑢^{2} 𝑡}$

And that is the moment generating function of a normal variable with mean 0 and variance t.

So you can prove that the martingale is normally distributed. Then you can go forward and say that it’s a Brownian motion (proof left as exercise).

Now let’s do a two dimensional Lévy.

The same four properties are important to us, with the tweak that the cross variation has to be 0.

We are considering each martingale in the set. We can use the same proof we did before to find the expectation and variance of a two dimensional Brownian motion to get the solution for every function f, and then generalize to

$𝑓 (𝑡, 𝑋, 𝑌) = 𝑒^{𝑢_{1} 𝑋 + 𝑢_{2} 𝑌 - \frac{1}{2} (𝑢_{1}^{2} + 𝑢_{2}^{2}) 𝑡}$

which results in

$𝔼 [𝑒^{𝑢_{1} 𝑀_{1} + 𝑢_{2} 𝑀_{2}}] = 𝑒^{\frac{1}{2} 𝑢_{1}^{2} 𝑡} 𝑒^{\frac{1}{2} 𝑢_{2}^{2} 𝑡}$

which is the joint moment generating function.

What we have expressed here is that the joint moment generating functions are the product of the individual moment generating functions, which means that the martingales are independent. That’s the only requirement to prove dimensionality, since we’ve already proved that they’re individually Brownian motions.

Correlated Stock Prices

Let’s say you have two stock prices and you don’t want them to be independent, since that’s how the real world works. The claim is that we can express differentials for the stock processes:

$𝑑 𝑆_{1} = 𝛼_{1} 𝑆_{1} 𝑑 𝑡 + 𝜎_{1} 𝑆_{1} 𝑑 𝑊_{1}$

$𝑑 𝑆_{2} = 𝛼_{2} 𝑆_{2} 𝑑 𝑡 + 𝜎_{2} 𝑆_{2} (𝜌 𝑑 𝑊_{1} + \sqrt{1 - 𝜌^{2}} 𝑑 𝑊_{2})$

We can all agree that $𝑆_{1}$ follows Brownian motion, so we know its closed form solution.

$𝑆_{2}$ is only a Brownian motion if that big term on the right is a Brownian motion.

Let’s think abstractly. We have a Brownian motion

$𝐵_{1} = 𝑊_{1}$

And another

$𝐵_{2} = 𝜌 𝑊_{1} + \sqrt{1 - 𝜌^{2}} 𝑊_{2}$

Lévy helps us prove this by giving us the four conditions to satisfy. We need the martingale property. So let’s see what the expectation is on the filtration.

$𝔼 [𝜌 𝑊_{1} (𝑡) + \sqrt{1 - 𝜌^{2}} 𝑊_{2} (𝑡) | ℱ︀ (𝑠)]$

We can separate through linearity.

$= 𝜌 𝔼 [𝑊_{1} (𝑡) | ℱ︀ (𝑠)] + \sqrt{1 - 𝜌^{2}} 𝔼 [𝑊_{2} (𝑡) | ℱ︀ (𝑠)]$

We already know that Brownian motion is a martingale, so we can use that information here.

$= 𝜌 𝑊_{1} (𝑠) + \sqrt{1 - 𝑝^{2}} 𝑊_{2} (𝑠)$

Which by definition is

$= 𝐵_{𝑤} (𝑠)$

It also fulfills the zero property if you substitute. What about continuity? It’s a linear combination of continuous processes, so it must be continuous.

Final: what is quadratic variation? We can solve that simply through $𝑑 𝐵^{2}$

$𝑑 𝐵_{2}^{2} = 𝜌^{2} 𝑑 𝑡 + (1 - 𝑝^{2}) 𝑑 𝑡$

$= 𝑑 𝑡$

So it must be a Brownian motion. Now pulling us back to our original stock process.

$𝑆_{1} = 𝑆_{1} (0) 𝑒^{(𝛼_{1} - \frac{𝜎_{1}^{2}}{2}) 𝑡 + 𝜎_{1} 𝐵_{1}}$

We have chosen our correlation coefficient to be between −1 and 1 so that we don’t deal with imaginary numbers.

If we take the cross variation of these stock processes.

These are two Brownian motions, but they have correlation. Their cross variation is not 0! But what about their covariance?

What is the correlation between our sources of noise? We need to know the covariance. We already know the variances of Brownian motion.

I’m going to have that

$𝔼 [𝐵_{1} | 𝐵_{2}] = \int_{0}^{𝑡} 𝐵_{1} 𝑑 𝐵_{2} + \int_{0}^{𝑡} 𝐵_{2} 𝑑 𝐵_{1} + \int_{0}^{𝑡} 𝑑 𝐵_{1} 𝑑 𝐵_{2}$

We have Ito integral with respect to martingale, so that becomes 0

$= 𝔼 [\int_{0}^{𝑡} 𝜌 𝑑 𝑢] = 𝜌 𝑡$

Therefore the covariance is

$\frac{Cov (𝐵_{1}, 𝐵_{2})}{\sqrt{𝕍 [𝐵_{1}]} \sqrt{𝕍 [𝐵_{2}]}} = 𝜌$

This is the Kolesky decomposition of these two correlated Brownian motion into independent Brownian motions.

What is the correlation of my two stocks given by these processes? We get this through the original equation that $𝑆_{1}$ follows GBM, and we know that structure

$𝔼 [𝑆_{1}] = 𝔼 [𝑆_{1} (0) 𝑒^{(𝛼_{1} - \frac{𝜎_{1}^{2}}{2}) 𝑡 + 𝜎_{1} 𝐵_{1}}]$

Skipping some steps,

$= 𝑆_{1} (0) 𝑒^{𝛼_{1} 𝑡}$

How do we get the variance of $𝑆_{1}$ ?

$𝕍 [𝑆_{1}] = 𝔼 [𝑆_{1}^{2}] - 𝔼 [𝑆_{1}]^{2}$

This would be

$= 𝑆_{1}^{2} (0) 𝑒^{2 𝛼_{1} 𝑡 + 𝜎_{1}^{2} 𝑡} - 𝑆_{1}^{2} (0) 𝑒^{2 𝛼 𝑡}$

We would then take the square root of this to get the standard deviation, so in order to plug this in we need to know the covariance between $𝑆_{1}$ and $𝑆_{2}$ , which is

$𝔼 [𝑆_{1} 𝑆_{2}] - 𝔼 [𝑆_{1}] 𝔼 [𝑆_{2}]$

I’ve created these through correlated Brownian motion, so I can’t just say they’re independent. I know that they’re Ito processes though! So I can decompose.

$𝑑 (𝑆_{1} 𝑆_{2}) = 𝑆_{1} 𝑑 𝑆_{2} + 𝑆_{2} 𝑑 𝑆_{1} + 𝑑 𝑆_{1} 𝑑 𝑆_{2}$

$= 𝑆_{1} 𝛼_{2} 𝑆_{2} 𝑑 𝐵_{2}$

## Gamma, Beta, and Chi-Squared Distributions - URL: https://sharifhsn.dev/blog/gamma-beta-and-chi-squared-distributions/ - Structured data: https://sharifhsn.dev/api/posts/gamma-beta-and-chi-squared-distributions.json - Description: Definition 5.36 says that X is a Chi squared distribution whose pdf is described by This is actually linked to the gamma distribution There is also a remark which says that the… - Date: 2024-10-21 - Exact published timestamp: 2024-10-21 - Topics: Probability Theory, Gamma Distribution, Beta Distribution - Categories: Probability Theory - Source: FE-540 \| Probability Theory - Source URL: None - Content source format: typst

Week 8

Lecture Notes

Chi-Squared

Definition 5.36 says that X is a Chi squared distribution whose pdf is described by

$𝑓 (𝑥) = \frac{1}{2^{𝑛 / 2} Γ (\frac{𝑛}{2})} 𝑥^{\frac{𝑛}{2} - 1} 𝑒^{- \frac{𝑥}{2}}$

where $𝑋 \sim 𝑋_{𝑛}^{2}$

This is actually linked to the gamma distribution

$𝑋 \sim 𝑋_{𝑛}^{2}$ , then $𝑋 \sim Γ (\frac{𝑛}{2}, \frac{1}{2})$

There is also a remark which says that the gamma distribution has an additive property. And if you inverse a gamma distribution, you get the inverse gamma distribution.

Beta Distribution

Described by 5.4.5

Gamma distribution is defined under the gamma function. We will need a two variable beta function. The intermediate function is defined as

$B (𝑎, 𝑏) ≔ \int_{0}^{1} 𝑥^{𝑎 - 1} (1 - 𝑥)^{𝑏 - 1} 𝑑 𝑥$

The integral is univariate. There’s an apparent symmetry. If we do a change of variable of $𝑦 = 1 - 𝑥$ then

$B (𝑏, 𝑎) = \int_{0}^{1} 𝑥^{𝑏 - 1} 𝑦^{𝑎 - 1} 𝑑 𝑥$

and

$= \int_{1}^{0} (1 - 𝑦)^{𝑏 - 1} 𝑦^{𝑎 - 1} (- 𝑑 𝑦)$

Which when you cancel out the negative, they end up being equal. And in fact it can be shown that

$B (𝑎, 𝑏) = B (𝑏, 𝑎)$

and a more useful result proposition 5.39, which is that

$B (𝑎, 𝑏) = \frac{Γ (𝑎) Γ (𝑏)}{Γ (𝑎 + 𝑏)}$

this will be very useful later. Proving this is very tedious calculus.

We can now introduce the beta distribution through definition 5.40. If $𝑥$ has the following pdf

$𝑓_{𝑎, 𝑏} (𝑥) = \frac{𝑥^{𝑎 - 1} (1 - 𝑥)^{𝑏 - 𝑎}}{B (𝑎, 𝑏)} 𝐼_{[0, 1]} (𝑥)$

then it’s in the beta distribution.

Proposition 5.41 says that the integral is 1.

Proposition 5.42 tells us the expected value and variance.

$𝔼 [𝑋] = \frac{𝑎}{𝑎 + 𝑏}$

$𝕍 [𝑋] = \frac{𝑎 𝑏}{(𝑎 + 𝑏)^{2} (𝑎 + 𝑏 + 1)}$ To prove this, we take the definition of the expectation

$𝔼 [𝑋] = \int_{0}^{1} 𝑥 𝑓_{𝑎, 𝑏} (𝑥) 𝑑 𝑥$

$= \frac{1}{B (𝑎, 𝑏)} \int_{0}^{1} 𝑥^{𝑎} (1 - 𝑥)^{𝑏 - 1} 𝑑 𝑥$

We can then adjust the $𝑎$ to make it agree with the beta function

$= \frac{B (𝑎 + 1, 𝑏)}{B (𝑎, 𝑏)}$

Now I can adjust it to be the gamma function instead.

$= \frac{Γ (𝑎 + 𝑏)}{Γ (𝑎) Γ (𝑏)} \cdot \frac{Γ (𝑎 + 1) Γ (𝑏)}{Γ (𝑎 + 𝑏 + 1)}$

We can cancel out some things, and then utilize the property of the gamma function that $Γ (𝑛 + 1) = 𝑛 Γ (𝑛)$

$\frac{𝑎}{𝑎 + 𝑏}$

Then for variance, skipping some steps, we can show that

$𝔼 [𝑋^{2}] = \frac{1}{B (𝑎, 𝑏)} \int_{0}^{1} 𝑥^{𝑎 + 1} (1 - 𝑥)^{𝑏 - 1} 𝑑 𝑥$

Using a similar method, we can rewrite this as

$= \frac{B (𝑎 + 2, 𝑏)}{B (𝑎, 𝑏)}$

Now we have from gamma

$= \frac{Γ (𝑎 + 2) Γ (𝑏)}{Γ (𝑎 + 𝑏 + 2)} \cdot \frac{Γ (𝑎 + 𝑏)}{Γ (𝑎) Γ (𝑏)}$

canceling and substituting with our values, it’s

$= \frac{(𝑎 + 1) 𝑎}{(𝑎 + 𝑏 + 1) (𝑎 + 𝑏)}$

Then to calculate variance, we get

$= \frac{(𝑎 + 1) 𝑎}{(𝑎 + 𝑏 + 1) (𝑎 + 𝑏)} - \frac{𝑎^{2}}{(𝑎 + 𝑏)^{2}}$

which will simplify into our answer

## Black–Scholes - URL: https://sharifhsn.dev/blog/black-scholes/ - Structured data: https://sharifhsn.dev/api/posts/black-scholes.json - Description: We will either lend or borrow from a bank at interest rate r. In that scenario, the amount of wealth that i have at maturity time T. My wealth would be the value of my shares and… - Date: 2024-10-17 - Exact published timestamp: 2024-10-17 - Topics: Stochastic Calculus, Black–Scholes, Option Pricing - Categories: Stochastic Calculus - Source: FE-610 \| Stochastic Calculus - Source URL: None - Content source format: typst

Week 6

Lecture Notes

Black-Scholes

We will either lend or borrow from a bank at interest rate r. In that scenario, the amount of wealth that i have at maturity time T. My wealth would be the value of my shares and the rest of my wealth increased by the interest rate.

I could say that the value of my portfolio at the next time increment can be expressed as the position I take in the stock at the original time, times the current value of the stock.

$𝑋 (𝑡_{𝑗}) = Δ (𝑡_{𝑗}) 𝑆 (𝑡_{𝑗}) + 𝑋 (𝑡_{𝑗}) - Δ (𝑡_{𝑗}) 𝑆 (𝑡_{𝑗})$

No time has actually elapsed here, so there is no interest rate applied. I have merely shifted my wealth into a different asset. The $Δ$ is the position in the asset, and $𝑋$ is the position in the money market i.e. wealth.

Then the value of my portfolio at time $𝑡_{𝑗 + 1}$ , (we’ll pun in a bit in the notation here) when I’m talking about my interest rate $𝑟$ , you’re earning interest over time $𝑇$ . Now we’re using it for a partition, and then we’ll annualize it.

When we move to $𝑡_{𝑗 + 1}$ , you have an interest rate.

$𝑋 (𝑡_{𝑗 + 1}) = Δ (𝑡_{𝑗}) 𝑆 (𝑡_{𝑗 + 1}) + (1 + 𝑟) (𝑋 (𝑡_{𝑗}) - Δ (𝑡_{𝑗}) 𝑆 (𝑡_{𝑗}))$

If we want to know how much our wealth is changing, that’s the value $𝑋 (𝑡_{𝑗 + 1}) - 𝑋 (𝑡_{𝑗})$ , which ends up being

$𝑋 (𝑡_{𝑗 + 1}) - 𝑋 (𝑡_{𝑗}) = Δ (𝑡_{𝑗}) (𝑆 (𝑡_{𝑗 + 1}) - 𝑆 (𝑡_{𝑗})) + (𝑋 (𝑡_{𝑗}) - Δ (𝑡_{𝑗}) 𝑆 (𝑡_{𝑗}) (1 + 𝑟 - 1)$

This gives us the intuition to understand how the process is actually evolving. If we are to look at infinitesimally smaller pieces of time, we would eventually have the differential

$𝑑 𝑋 (𝑡) = Δ (𝑡) 𝑑 𝑆 (𝑡) + 𝑟 (𝑋 (𝑡) - Δ (𝑡) 𝑆 (𝑡)) 𝑑 𝑡$

If I assume that my stock follows a geometric Brownian motion, which is the assumption in the Black-Scholes model, and I replace $𝑑 𝑆 (𝑡)$ with that, I get

$= Δ (𝑡) (𝛼 𝑆 (𝑡) 𝑑 𝑡 + 𝜎 𝑆 (𝑡) 𝑑 𝑊 (𝑡)) + 𝑟 (𝑋 (𝑡) - Δ (𝑡) 𝑆 (𝑡)) 𝑑 𝑡$

And once I plug in, I can group these terms together and get

$= 𝑟 𝑋 (𝑡) 𝑑 𝑡 + Δ (𝑡) 𝑆 (𝑡) (𝛼 - 𝑟) 𝑑 𝑡 + Δ (𝑡) 𝜎 𝑆 (𝑡) 𝑑 𝑊 (𝑡)$

just picking and choosing my terms. The reason I have written it in this way is to see three different dynamics on display in the change of my portfolio

$𝑟 𝑋 (𝑡) 𝑑 𝑡$ is the underlying return. If I had taken that money and put it in the bank, that’s how much I would return. Is there an advantage in investing in the stock instead of just using the money market? Then we have to have a risk premium. That’s given by $Δ (𝑡) 𝑆 (𝑡) (𝛼 - 𝑟) 𝑑 𝑡$ , which is the dynamic from choosing to invest in the risky asset instead of the riskless asset. Then we have noise. There is uncertainty that comes about in the value of our portfolio if we choose to invest in our asset instead of the money market. The more money in the asset, our volatility should go up. And that’s what we see in $Δ (𝑡) 𝜎 𝑆 (𝑡) 𝑑 𝑊 (𝑡)$ where this value is proportional to the value of $Δ (𝑡)$ .

Our task right now is to try and come up with a methodology to price an instrument. There should be some kind of relationship between these dynamics. I want to study not just how my processes change, but also my discounted process. When I compare money, I need to be comparing apples to apples. We need to look at dynamics of our stock and portfolio. If I wanted to define my discount stock process, I would multiply my stock process by the discount factor: $𝑒^{- 𝑟 𝑡} 𝑆 (𝑡)$

How would I go about getting the differential? We could rewrite this as an Ito process because we know GBM is an Ito process. We just have to know what our function $𝑓 (𝑎, 𝑏)$ is.

$𝑑 (𝑒^{- 𝑟 𝑡} 𝑆 (𝑡)) = 𝑑 𝑓 (𝑡, 𝑆 (𝑡))$

$𝑓 (𝑎, 𝑏) = 𝑒^{- 𝑟 𝑎} 𝑏$

$𝑓_{𝑎} = - 𝑟 𝑒^{- 𝑟 𝑎} 𝑏$

$𝑓_{𝑏} = 𝑒^{- 𝑟 𝑎}$

$𝑓_{𝑏 𝑏} = 0$ This is going to be

$= - 𝑟 𝑒^{- 𝑟 𝑡} 𝑆 (𝑡) 𝑑 𝑡 + 𝑒^{- 𝑟 𝑡} 𝑑 𝑆 (𝑡) + \frac{1}{2} (0) (𝑑 𝑆 (𝑡))^{2}$

This now gives me that

$= - 𝑟 𝑒^{- 𝑟 𝑡} 𝑆 (𝑡) 𝑑 𝑡 + 𝑒^{- 𝑟 𝑡} 𝛼 𝑆 (𝑡) 𝑑 𝑡 + 𝑒^{- 𝑟 𝑡} 𝜎 𝑆 (𝑡) 𝑑 𝑊 (𝑡)$

which I can group the $𝑑 𝑡$ terms together and get

$= 𝑒^{- 𝑟 𝑡} 𝑆 (𝑡) (𝛼 - 𝑟) 𝑑 𝑡 + 𝑒^{- 𝑟 𝑡} 𝜎 𝑆 (𝑡) 𝑑 𝑊 (𝑡)$

We can see here how the discounted stock process would evolve, as this is the differential of the discounted stock process. How would I go about figuring out the differential of the discounted portfolio process? We use the same Ito process method. $𝑑 (𝑒^{- 𝑟 𝑡} 𝑋 (𝑡)) = - 𝑟 𝑒^{- 𝑟 𝑡} 𝑋 (𝑡) 𝑑 𝑡 + 𝑒^{- 𝑟 𝑡} 𝑑 𝑋 (𝑡)$

Plugging in what we know, we know

$= - 𝑟 𝑒^{- 𝑟 𝑡} 𝑋 (𝑡) 𝑑 𝑡 + 𝑟 𝑒^{- 𝑟 𝑡} 𝑋 (𝑡) 𝑑 𝑡 + 𝑒^{- 𝑟 𝑡} Δ (𝑡) 𝑆 (𝑡) (𝛼 - 𝑟) 𝑑 𝑡 + 𝑒^{- 𝑟 𝑡} Δ (𝑡) 𝜎 𝑆 (𝑡) 𝑑 𝑊 (𝑡)$

And, if I combine terms, the first two will cancel.

$= 𝑒^{- 𝑟 𝑡} (Δ (𝑡) 𝑆 (𝑡) (𝛼 - 𝑟)) 𝑑 𝑡 + 𝑒^{- 𝑟 𝑡} Δ (𝑡) 𝜎 𝑆 (𝑡) 𝑑 𝑊 (𝑡)$

Which we will slightly modify by pulling out $Δ (𝑡)$

$= Δ (𝑡) (𝑒^{- 𝑟 𝑡} 𝑆 (𝑡) (𝛼 - 𝑟) 𝑑 𝑡 + 𝑒^{- 𝑟 𝑡} 𝜎 𝑆 (𝑡) 𝑑 𝑊 (𝑡))$

Which we can recognize as being the differential of the discounted stock price

$= Δ (𝑡) 𝑑 (𝑒^{- 𝑟 𝑡} 𝑆 (𝑡))$

We’re going to assume that there is some initial stock price and wealth here for the purposes of math.

This makes sense, because once we discount our portfolio, we’re ignoring wealth that would be accrued by earning money in the money market, so all the change will come from the changes in my discounted stock process. And the amount of change will be proportional to how many shares we own of the underlying asset.

Now we are going to try and do what is called a replicating portfolio. This portfolio will perfectly match the value of the derivative of a security. If I have portfolios A and B, if at maturity, I know that A will perfectly match the value of B, then I know they will always have to have had the same value. If at any point in time, A was different from B, then you would just long one and short the other. At maturity, when you know they will be the same, you would close and get arbitrage. That is a guaranteed way to game the system, which cannot happen. One of the assumptions we will always make is that arbitrage cannot exist in a healthy, functioning market. We will try to construct a portfolio of a riskless and risky asset that will perfectly replicate a call option. In order to have A match B, moving in lockstep, we have to know how they’re both going to change. I know now how a discounted portfolio will evolve. How will discounted security evolve?

Let $𝑐 (𝑡, 𝑆 (𝑡))$ represent the value of a derivative security. We will mostly talk about call options today, but we will be explicit about when the calculations require them to be a call option. In this case, $𝑐$ is any derivative. The first and second partial derivatives should exist and all that.

If we are to look at how it evolves, i.e. its differential, we would use the Ito formula. Here we will adopt the book’s notation where we use $𝑡$ and $𝑥$ instead of $𝑎$ and $𝑏$

$𝑑 𝑐 (𝑡, 𝑆 (𝑡)) = 𝑐_{𝑡} (𝑡, 𝑆 (𝑡)) 𝑑 𝑡 + 𝑐_{𝑥} (𝑡, 𝑆 (𝑡)) 𝑑 𝑆 (𝑡) + \frac{1}{2} 𝑐_{𝑥 𝑥} (𝑡, 𝑆 (𝑡)) (𝑑 𝑆 (𝑡))^{2}$

That’ll give me

$= 𝑐_{𝑡} (𝑡, 𝑆 (𝑡)) 𝑑 𝑡 + 𝑐_{𝑥} (𝑡, 𝑆 (𝑡)) 𝛼 𝑆 (𝑡) 𝑑 𝑡 + 𝑐_{𝑥} (𝑡, 𝑆 (𝑡)) 𝜎 𝑆 (𝑡) 𝑑 𝑊 (𝑡) + \frac{1}{2} 𝑐_{𝑥 𝑥} (𝑡, 𝑆 (𝑡)) 𝜎^{2} 𝑆^{2} (𝑡) 𝑑 𝑡$

Because that is the definition of $(𝑑 𝑆 (𝑡))^{2}$ . Again we will group $𝑑 𝑡$ and $𝑑 𝑊 (𝑡)$ together. This shows that keeping track of these is very important.

That gives us that

$= (𝑐_{𝑡} (𝑡, 𝑆 (𝑡)) + 𝛼 𝑆 (𝑡) 𝑐_{𝑥} (𝑡, 𝑆 (𝑡)) + \frac{1}{2} 𝜎^{2} 𝑆^{2} (𝑡) 𝑐_{𝑥 𝑥} (𝑡, 𝑆 (𝑡)) 𝑑 𝑡 + 𝑐_{𝑥} (𝑡, 𝑆 (𝑡)) 𝜎 𝑆 (𝑡) 𝑑 𝑊 (𝑡)$

This is how an arbitrary derivative security will evolve at time $𝑡$ . But, that’s not what we really want to study. We want to compare things to what is the evolution of discounted security. Automatically, when we have the Ito decomposition of this process, we can see that it’s an Ito process. So isn’t this just another example of us using Ito processes?

$𝑑 (𝑒^{- 𝑟 𝑡} 𝑐 (𝑡, 𝑆 (𝑡))) = - 𝑟 𝑒^{- 𝑟 𝑡} 𝑐 (𝑡, 𝑆 (𝑡)) 𝑑 𝑡 + 𝑒^{- 𝑟 𝑡} 𝑑 𝑐 (𝑡, 𝑆 (𝑡))$

But we just figured that out, so we can plug in. This whole expression becomes

$= 𝑒^{- 𝑟 𝑡} (- 𝑟 𝑐 (𝑡, 𝑆 (𝑡)) + 𝑐_{𝑡} (𝑡, 𝑆 (𝑡)) + 𝛼 𝑆 (𝑡) 𝑐_{𝑥} (𝑡, 𝑆 (𝑡)) + \frac{1}{2} 𝜎^{2} 𝑆^{2} (𝑡) 𝑐_{𝑥 𝑥} (𝑡, 𝑆 (𝑡)) 𝑑 𝑡 + 𝑒^{- 𝑟 𝑡} 𝜎 𝑆 (𝑡) 𝑑 𝑊 (𝑡)$ I now have two dynamics well expressed: the change in a portfolio and the change in a derivative security. I need a list of instructions. I need a way to match one portfolio to another portfolio. I can’t really change the value in my derivative security. Let’s say A is one share of a derivative security. B is my portfolio where I trade in the underlying, so I choose my $Δ (𝑡)$ where they perfectly match. To this end, I need to match the endpoint $𝑋 (𝑇) = 𝑐 (𝑇, 𝑆 (𝑇))$ and the change in the processes:

$𝑑 (𝑒^{- 𝑟 𝑡} 𝑋 (𝑡)) = 𝑑 (𝑒^{- 𝑟 𝑡} 𝑐 (𝑡, 𝑆 (𝑡)))$

Would it be enough for me to match a ball thrown if I raised it to the same height, or if I rolled it across the ground to reach the same point? No, I need to match both horizontal and vertical change for them to match. And those values are expressed by $𝑑 𝑡$ and $𝑑 𝑊 (𝑡)$ , respectively. For the latter term, we get

$𝑑 𝑊 (𝑡) : 𝑒^{- 𝑟 𝑡} Δ (𝑡) 𝑆 (𝑡) 𝜎 = 𝑒^{- 𝑟 𝑡} 𝜎 𝑆 (𝑡) 𝑐_{𝑥} (𝑡, 𝑆 (𝑡))$

And if I match those, I’m going to see that

$Δ (𝑡) = 𝑐_{𝑥} (𝑡, 𝑆 (𝑡))$ That is the choice for $Δ$ that will make these two portfolios match as far as changes with respect to Brownian motion. If I match the $𝑑 𝑡$ terms, then I’m going to have (we’ll assume $𝑐$ means $𝑐 (𝑡, 𝑆 (𝑡))$

$𝑑 𝑡 : 𝑒^{- 𝑟 𝑡} Δ (𝑡) 𝑆 (𝑡) (𝛼 - 𝑟) = 𝑒^{- 𝑟 𝑡} (- 𝑟 𝑐 + 𝑐_{𝑡} + 𝛼 𝑆 (𝑡) 𝑐_{𝑥} + \frac{1}{2} 𝜎^{2} 𝑆^{2} (𝑡) 𝑐_{𝑥 𝑥})$

We can see that we have a $𝛼 𝑐 𝑥$ on both sides, so if we take those out, we get

$0 = - 𝑟 𝑐 + 𝑐_{𝑡} + 𝑟 𝑆 (𝑡) 𝑐_{𝑥} + \frac{1}{2} 𝜎^{2} 𝑆^{2} (𝑡) 𝑐_{𝑥 𝑥}$ And this is 100% where the Black-Scholes PDE comes from. This should work for underlying asset, so I should be able to come up with some differential equation where

$0 = - 𝑟 𝑐 (𝑡, 𝑥) + 𝑐_{𝑡} (𝑡, 𝑥) + 𝑟 𝑋 𝑐_{𝑥} (𝑡, 𝑥) + \frac{1}{2} 𝜎^{2} 𝑋^{2} 𝑐_{𝑥 𝑥} (𝑡, 𝑥)$

This gives me a PDE I could theoretically solve to get the formula for my derivative security. Because these two match at maturity and are in lockstep at every other point in time, we get

$𝑋 (0) = 𝑐 (0, 𝑆 (0))$

Call Options

Now let’s start pricing an instrument. We have now moved from an arbitrary derivative to a call option. The PDE was just made works for any process that derives its value from an Ito process. Every PDE or DE has an infinite number of solutions. Boundary constraints give us finite solutions. What boundary constraints make sense for a call option. If we live in this world, a call option is the right, but not the obligation to buy an asset for a fixed price at maturity. $𝐾$ represents this price. We can look at the value of this derivative security at time $𝑇$ (or as it approaches maturity). If I know I can buy an asset at $𝐾$ dollars by owning this derivative security, let’s say $𝐾$ is 100, and current trading is 150. I would choose to buy this security right away, so I can buy the security at 100 and exercise it immediately. If the current trading was 80, I wouldn’t buy. Therefore it’s

${lim}_{𝑡 \to 𝑇} 𝑐 (𝑡, 𝑥) = (𝑋 - 𝐾)_{+}$

That only gives me one of my values. There are two other boundaries to consider. Could we look at evolutions with respect to variable $𝑥$ ? What am I actually saying here? The underlying stock has gone bankrupt. So why the hell would I ever want to buy that security? My call option is now worthless

${lim}_{𝑋 \to 0} 𝑐 (𝑡, 𝑥) = 0$

What about the limit as $𝑥$ goes in the other direction. Could there be a traded stock that’s worth 1000 per share. What about 30,000, 500,000? Nothing says that couldn’t happen. There is no upper bound on the value a stock could take. So the value of the derivative security is that price minus the strike price. The strike price must be irrelevant as the price increases, so it just becomes the price.

${lim}_{𝑥 \to \infty} 𝑐 (𝑡, 𝑥) \approx 𝑋$

Black-Scholes-Merton

The formula is

$𝑐 (𝑡, 𝑥) = 𝑥 𝑁 (𝑑_{+} (𝑇 - 𝑡, 𝑥)) - 𝐾 𝑒^{- 𝑟 (𝑇 - 𝑡)} 𝑁 (𝑑_{-} (𝑇 - 𝑡, 𝑥))$

But let’s derive it so we can understand the pieces. This is bare-bones and we’re going to make a few assumptions that we will explain further after the midterm. We just want intuition and a slight tweak. The slight tweak is that GBM will be rewritten as

$𝑆 (𝑡) = 𝑆 (0) exp (𝑟 - 𝜎^{2} / 2) 𝑡 + 𝜎 𝑊 (𝑡))$

$𝑑 𝑆 (𝑡) = 𝑟 𝑆 (𝑡) 𝑑 𝑡 + 𝜎 𝑆 (𝑡) 𝑑 𝑊 (𝑡)$ We will rigorously derive this later.

To get the fair price, we get the value when it is exercised, discount it today’s time, and then take its expected value.

Now we’ll say the value of our derivative at time 0 is

$𝑐 (0, 𝑆 (0)) = 𝔼 [𝑒^{- 𝑟 𝑇} 𝑐 (𝑇, 𝑆 (𝑇))]$ Being a call option, this becomes

$= 𝔼 [𝑒^{- 𝑟 𝑇} (𝑆 (𝑇) - 𝐾)_{+}]$

That + makes this a little trickier to integrate, which we will have to do because we want an expected value. To aid with that, we’ll ask the question: when do we exercise? When $𝑆 (𝑇) > 𝐾$ . And that’s the same as

$𝑆 (0) exp ((𝑟 - \frac{1}{2} 𝜎^{2}) 𝑡 + 𝜎 𝑊 (𝑡)) > 𝐾$

Everything there is known except for the source of randomness, which is the Brownian motion. We can get bounds for the Brownian motion because we know its distribution. So, solving this for Brownian motion by dividing and taking the log, we get

$(𝑟 - \frac{𝜎^{2}}{2}) 𝑇 + 𝜎 𝑊 (𝑇) > log (\frac{𝐾}{𝑆 (0)})$ This becomes

$𝑊 (𝑇) > \frac{log (\frac{𝐾}{𝑆 (0)}) - (𝑟 - \frac{𝜎^{2}}{2}) 𝑇}{𝜎}$

which we will indicate by the convenience variable $𝑑 *$

We can re-express this expectation as the indicator function based on this inequality, and then get the integral for it.

$= \int_{- \infty}^{\infty} 𝑒^{- 𝑟 𝑇} (𝑆 (0) 𝑒^{(𝑟 - 𝜎^{2} / 2) 𝑇 + 𝜎 𝑋} - 𝑘) 𝐼_{𝑋 > 𝑑 *} 𝑓_{𝑉 (𝑇)} (𝑥) 𝑑 𝑥$

This then becomes

$= \int_{𝑑 *}^{\infty} 𝑒^{- 𝑟 𝑡} (𝑆 (0) 𝑒^{(𝑟 - 𝜎^{2} / 2) 𝑇 + 𝜎 𝑥} - 𝐾) \frac{1}{\sqrt{2 𝜋 𝑡}} 𝑒^{- \frac{𝑥^{2}}{2 𝑇}} 𝑑 𝑥$

Now we just evaluate this integral, which is really two integrals, so let’s split it up.

$= \int_{𝑑 *}^{\infty} 𝑆 (0) 𝑒^{- 𝑟^{2} / 2 𝑇 + 𝜎 𝑋} \frac{1}{\sqrt{2 𝜋 𝑇}} 𝑒^{- \frac{𝑋^{2}}{2 𝑇}} 𝑑 𝑥 - 𝐾 𝑒^{- 𝑟 𝑇} \int_{𝑑 *}^{\infty} \frac{1}{\sqrt{2 𝜋 𝑇}} 𝑒^{- \frac{𝑋^{2}}{2 𝑇}} 𝑑 𝑥$

Looking at this second integral, this is the integral of mY PDF for Brownian motion.

We want to convert to a standard normal. Our mean is 0, and our variance is T. So this second integral ends up being the same as

$\int_{\frac{𝑑 *}{𝜎}}^{\infty} \frac{1}{\sqrt{2 𝜋}} 𝑒^{- 𝑧^{2} / 2} 𝑑 𝑧$

The normal distribution is symmetric around 0, so we can rewrite this in reverse as well

$\int_{- \infty}^{- \frac{𝑑 *}{𝜎}} \frac{1}{\sqrt{2 𝜋}} 𝑒^{- 𝑧^{2} / 2} 𝑑 𝑧$

which is the standard normal of something I didn’t have time to write down called $𝑑_{-}$ Now for the other integral. We have

$𝑆 (0) \int_{𝑑 *}^{\infty} 𝑒^{- 𝜎^{2} / 2 𝑇 + 𝜎 𝑥} \frac{1}{\sqrt{2 𝜋 𝑇}} 𝑒^{- \frac{𝑋^{2}}{2 𝑇}} 𝑑 𝑋$

which we can rearrange algebraically and take the standard normal of to get

$= 𝑆 (0) \int_{\frac{𝑑 * - 𝜎 𝑇}{\sqrt{𝑇}}}^{\infty} \frac{1}{\sqrt{2 𝜋}} 𝑒^{- 𝑧^{2} / 2} 𝑑 𝑧$

That lower term is the same as $𝑑_{-} + 𝜎 \sqrt{𝑇}$ . This is going to be the same as

$\frac{log (\frac{𝑆 (0)}{𝐾}) + (𝑟 - 𝜎^{2} / 2) 𝑇}{𝜎 \sqrt{𝑇}} + 𝜎 𝑇$

so we can combine these and get

$\frac{log (\frac{𝑆 (0)}{𝐾}) + (𝑟 + \frac{𝜎^{2}}{2}) 𝑇}{𝜎 \sqrt{𝑇}} = 𝑑_{+}$

Open question: the source note ends this derivation here.

This is set up to work at any point in time up to the time of maturity.

If I’m going to say that $𝑐 (𝑡, 𝑥) = 𝑥 𝑁 (𝑑_{+}) - 𝐾 𝑒^{- 𝑟 (𝑇 - 𝑡)} 𝑁 (𝑑_{-})$

where

$𝑑_{\pm} = \frac{log (\frac{𝑥}{𝑘}) + 𝑟 \pm \frac{𝜎^{2}}{2} (𝑇 - 𝑡)}{𝜎 \sqrt{𝑇 - 𝑡}}$

and this standard normal function is the CDF

What is the limit as t approaches T? What does matter is the first piece, the behavior of the logarithm. Since the denominator is going to 0, we might say that it’s going to infinity. But that depends on whether that log is positive or negative, which tells us the direction.

${lim}_{𝑡 \to 𝑇} 𝑐 (𝑡, 𝑥) \frac{log (\frac{𝑋}{𝑘})}{𝜎 \sqrt{𝑇 - 𝑡}} = {\infty 𝑥 \geq 𝐾 - \infty 𝑥 < 𝐾$

What can you tell me about the standard normal CDF evaluated at infinity? It’s 1, by definition. What is it evaluated at negative infinity? It’s 0.

${lim}_{𝑡 \to 𝑇} 𝑐 (𝑡, 𝑥) = {𝑥 - 𝐾 𝑥 \geq 𝐾 0 𝑥 < 𝐾$

What about when x approaches 0? Then the log goes to negative infinity

${lim}_{𝑥 \to 0} 𝑐 (𝑡, 𝑥) = 0$

What about x goes to infinity? We subtract our discounted K, which just becomes X.

${lim}_{𝑥 \to \infty} 𝑐 (𝑇, 𝑥) = 𝑥$

So this satisfies our boundary conditions.

If we consider our Black-Scholes

$𝑐 (𝑡, 𝑥) = 𝑥 𝑁 (𝑑_{+}) - 𝐾 𝑒^{- 𝑟 (𝑇 - 𝑡)} 𝑁 (𝑑_{-})$

What is the partial derivative? You might think it’s $𝑁 (𝑑_{+})$ , and it is, but not for the reasons you think. We will take $𝜏 = 𝑇 - 𝑡$ , and expand out the definition of the derivatives of the normal distribution.

$𝑐_{𝑥} = 𝑁 (𝑑_{+}) + 𝑥 \frac{1}{\sqrt{2 𝜋 𝜏}} 𝑒^{- 𝑑_{+}^{2} / (2 𝜏)} \frac{1}{𝑥 𝜎 \sqrt{𝜏}} - 𝐾 𝑒^{- 𝑟 𝜏} \frac{1}{\sqrt{2 𝜋 𝜏}} 𝑒^{- 𝑑_{-}^{2} / (2 𝜏)} \frac{1}{𝑥 𝜎 \sqrt{𝜏}}$

Which we can rearrange to

$= 𝑁 (𝑑_{+}) + \frac{1}{𝑥 𝜎 \sqrt{2 𝑝 𝑖}} (𝑥 𝑒^{- \frac{𝑑_{+}^{2}}{2}} - 𝐾 𝑒^{- 𝑟 𝜏} 𝑒^{- \frac{- 𝑑_{-}^{2}}{2}})$

Earlier, we had the result that $𝑑_{-} + 𝜎 \sqrt{𝜏} = 𝑑_{+}$

So we can derive that

$𝑑_{-}^{2} = 𝑑_{+}^{2} - 2 𝑑_{+} 𝜎 \sqrt{𝜏} + 𝜎^{2} 𝜏$

and

$\frac{𝑑_{-}^{2}}{2} = \frac{𝑑_{+}^{2}}{2} - 𝑑_{+} 𝜎 \sqrt{𝜏} + \frac{𝜎^{2} 𝜏}{2}$

We also know that

$𝑑_{+} 𝜎 \sqrt{𝜏} = log (\frac{𝑥}{𝐾}) + (𝑟 + \frac{𝜎^{2}}{2}) 𝜏$

So we can substitute this back in to get, knowing this, can’t be bothered to write it out but it shows that the end solution is still $𝑁 (𝑑_{+})$

This becomes important when we determine $𝑐_{𝑥 𝑥}$ $𝑐_{𝑥 𝑥} (𝑡, 𝑥) = 𝑁^{'} (𝑑_{+}) \frac{1}{𝑥 𝜎 \sqrt{𝜏}}$

proof is left as exercise to reader

and $𝑐_{𝑡}$ is

$𝑐_{𝑡} (𝑡, 𝑥) = - 𝑟 𝐾 𝑒^{- 𝑟 𝜏} 𝑁 (𝑑_{-}) - \frac{𝑥 𝜎}{2 \sqrt{𝜏}} 𝑁^{'} (𝑑_{+})$

If you take these results and plug them into Black-Scholes, you will get that it satisfies that formula.

We’ll do a slight rewrite of Black-Scholes to show that

$𝑟 𝑐 (𝑡, 𝑥) = 𝑐_{𝑡} (𝑡, 𝑥) + 𝑟 𝑥 𝑐_{𝑥} (𝑡, 𝑥) + \frac{1}{2} 𝑟^{2} 𝑥^{2} 𝑐_{𝑥 𝑥} (𝑡, 𝑥)$ $= - 𝑟 𝐾 𝑒^{- 𝑟 𝜏} 𝑁 (𝑑_{-}) - \frac{𝑥 𝜎}{2 \sqrt{𝜏}} 𝑁^{'} (𝑑_{+}) + 𝑟 𝑥 𝑁 (𝑑_{+}) + \frac{1}{2} 𝜎^{2} 𝑥^{2} \frac{1}{𝑥 𝜎 \sqrt{𝜏}}$

Which when we simplify out gives

$- 𝑟 𝐾 𝑒^{- 𝑟 𝜏} 𝑁 (𝑑_{-}) + 𝑟_{𝑥} 𝑁 (𝑑_{+})$

which is the same answer of Black-Scholes even though it’s derived from probability.

Put Option

We don’t need to start from scratch here.

With portfolio A, I buy one share of a forward. For our purposes, all calls, puts, and forwards share the same maturity $𝑇$ and the same strike $𝐾$ . What is going to be the value of this portfolio at maturity? $𝐴 (𝑇) = (𝑆 (𝑇) - 𝐾)$ What about portfolio B? I am going to buy one share of stock, and short a zero coupon bond (aka one payout) that is going to pay $𝐾$ at maturity. For this portfolio B, at time 0, we’re going to have the value of the asset we bought one share of, and we shorted the bond, so we’re going to owe somebody K dollars in the future, discounted to today:

$𝐵 (0) = 𝑆 (0) - 𝐾 𝑒^{- 𝑟 𝑇}$

What about at $𝑇$ ? $𝐵 (𝑇) = 𝑆 (𝑇) - 𝐾$

Which is the same as a forward. In order for the market to disallow arbitrage, these two portfolios must have always had the same price.

$𝐵 (𝑡) = 𝑠 (𝑡) - 𝐾 𝑒^{- 𝑟 𝜏}$

A forward can be valued in this manner, purely through arbitrage arguments.

Now we have portfolio $𝛾$ where we buy one call option and short one put option.

$𝛾 (0) = 𝑐 (0, 𝑆 (0)) - 𝑝 (0, 𝑆 (0))$

What is the value at maturity? $𝛾 (𝑇) = 𝑐 (𝑇, 𝑆 (𝑇)) - 𝑝 (𝑇, 𝑆 (𝑇))$ We know the formula for a call option, so we can substitute that in. What about the put? If the price is trading below your strike, then you buy it at the current price, and sell it at the strike price.

$= (𝑆 (𝑇) - 𝐾)_{+} - (𝐾 - 𝑆 (𝑇))_{+}$

But if you notice, only one of these can be positive. So it ends up always being

$= 𝑆 (𝑇) - 𝐾$ So the value of my forward is the same as that

$𝑓 (𝑡, 𝑆 (𝑡)) = 𝑐 (𝑡, 𝑆 (𝑡)) - 𝑝 (𝑡, 𝑆 (𝑡))$

This is called put-call parity. It requires no assumptions about the stock process, it is purely based on market dynamics.

We can use our Ito magic here

$𝑓 (𝑡, 𝑥) = 𝑐 (𝑡, 𝑥) - 𝑝 (𝑡, 𝑥)$

$𝑥 - 𝐾 𝑒^{- 𝑟 𝜏} = 𝑥 𝑁 (𝑑_{+}) - 𝐾 𝑒^{- 𝑟 𝜏} 𝑁 (𝑑_{-}) - 𝑝 (𝑡, 𝑥)$

One assumption is that the risk-free rate is constant. There’s a modified version with a stochastic risk-free rate.

We can therefore isolate the put option and find that

$𝑝 (𝑡, 𝑥) = 𝐾 𝑒^{- 𝑟 𝜏} (1 - 𝑁 (𝑑_{-})) - 𝑥 (1 - 𝑁 (𝑑_{+}))$

playing around with our normal distributions, we know that this is the same as

$= 𝐾 𝑒^{- 𝑟 𝜏} 𝑁 (- 𝑑_{-}) - 𝑥 𝑁 (- 𝑑_{+})$

which is the value of the put. We didn’t have to rederive everything because of the put-call parity.

Q&A

You can use the boundary and the PDE to find the value of a put option, or you can use probability arguments. We derived the generic derivative PDE and used boundary constraints, which we will need later. We didn’t actually use the PDE to get the formula for the call option, we just use it to confirm the option price.

There’s not necessarily a good motivation for the intuition between the d+ and d-, the d* is when the option is actually exercised, and the d+ and d- are modifications of that. This d* is arguably the boundary for Brownian motion for the option to be exercised, that’s the only real world application.

Because I have these nice boundary constraints on the strike price, it’s more common to price the put and then get the call from the put-call parity. Prof did this in this PhD dissertation.

There’s not always a real-world analogue in a mathematical construct, like for quadratic variation.

$𝑁_{𝐷 +}$ is the number you would use to create the replicating portfolio, but there’s no intuitive reason why it would do this.

Midterm will cover everything we’ve discussed in the first four chapters, including multidimensional stochastic differential equations.

## Probability and Regression Review - URL: https://sharifhsn.dev/blog/probability-and-regression-review/ - Structured data: https://sharifhsn.dev/api/posts/probability-and-regression-review.json - Description: We can actually do this from a portfolio perspective, it’s the same idea. We have some weight What’s the variance? 𝕍 [ 𝑍 ] = 𝑤 1 𝜎 1 2 + 𝑤 2 2 𝜎 2 2 + 2 𝑤 1 𝑤 2 𝜎 12 It’s the… - Date: 2024-10-17 - Exact published timestamp: 2024-10-17 - Topics: Risk Management, Probability, Regression - Categories: Risk Management - Source: FE-535 \| Risk Management - Source URL: None - Content source format: typst

Exam Review

Chapter 2: Fundamentals of Probability

For a

$𝑍 = 𝑎 𝑋 + 𝑏 𝑌$

then

$𝔼 [𝑍] = 𝑎 𝔼 [𝑋] + 𝑏 𝔼 [𝑌]$

and for variance,

$𝕍 [𝑍] = 𝑎^{2} 𝕍 [𝑋] + 𝑏^{2} 𝕍 [𝑌] + 2 𝑎 𝑏 Cov (𝑋, 𝑌)$

We can actually do this from a portfolio perspective, it’s the same idea. We have some weight

$𝑍 = 𝑤_{1} \cdot 𝑅_{1} + 𝑤_{2} 𝑅_{2}$

What’s the variance? $𝕍 [𝑍] = 𝑤_{1} 𝜎_{1}^{2} + 𝑤_{2}^{2} 𝜎_{2}^{2} + 2 𝑤_{1} 𝑤_{2} 𝜎_{12}$

It’s the same in this exercise where we substitute the weights for the coefficients and the random variables for X and Y.

Another way to look at this problem is through matrix algebra.

We have a vector v,

$\vec{𝑤} = [\begin{matrix} 𝑎 𝑏 \end{matrix}]$

We can rewrite $𝑍$ as

$𝑍 = {\vec{𝑤}}^{𝑇} \vec{𝑅}$

Then the expectation and variance becomes

$𝔼 [𝑍] = {\vec{𝑤}}^{𝑇} 𝔼 [𝑅]$

This is nice because if you have a covariance matrix you don’t have to look at all the pairwise. The question gives us that this X and Y have covariance 0.35. Be mindful of the difference between covariance and correlation.

The formula for covariance is that

$𝜎_{12} = 𝜌 𝜎_{1} 𝜎_{2}$

so that’s where correlation comes into it. That $𝜌$ is correlation.

We can then rewrite our problem into

$𝑤^{'} Σ 𝑤$

based on considering this as matrix.

If you $100 today, it can either go to $110 or $90. Up is 0.3, Down is 0.7. We’ll have the same increase and decrease amounts and probabilities at each step.

We’re actually interested in the value of

$𝔼 [𝑆_{2} | 𝑆_{0}]$

But in this case we have discrete outcomes, not like in GBM. If we have a random variable, it’s either going to be 120, 100, or 80. There is a probability assigned to each of these for a given $𝑆_{0}$ .

We know that the probability for 120 is 0.3 * 0.3 or 0.09, for 80 it’s 0.7 * 0.7 = 0.49. For 110 we have to consider both the possibility of 0.3 * 0.7 up and down and 0.7 * 0.3 down and up, which becomes 0.42. And we can show that these all add to 1.

So how do we actually calculate probability? We think about it simply as

$𝔼 [𝑋] = \sum_{\forall 𝑥} ℙ (𝑋 = 𝑥) \times 𝑥$

and

$𝔼 [𝑥^{2}] = \sum_{\forall 𝑥} ℙ (𝑋 = 𝑥) \times 𝑥^{2}$

And we can generalize this to any function $𝑔 (𝑋)$ .

Therefore the simple calculation is multiplying each probability of each outcome by its value and adding them together.

For variance we can use the same logic.

$𝕍 [𝑆_{2}] = 𝔼 [𝑆_{2}^{2}] - 𝔼 [𝑆_{2}]^{2}$

For the second one it’s simple, it must be $92^{2}$ . And for the expectation of squares we can use the $𝑔 (𝑋)$ property.

Then afterwards we take the square root to get the volatility.

What about ten steps ahead? It’s not as intuitive if we continue to use the tree model, there are too many possibilities.

We can actually solve this problem using binomial process.

One way you can think about it is success and no success. So, we could think about it as, we have a process

$𝐵_{2} \sim Binom (2, 0.3)$

And the question is of how many possibilities do we have, and what’s the probability of success? Let’s use some reasoning

When $𝐵_{2} = 0$ , we know that $𝑃_{2} = 80$ . So we can think about it as

$𝑃_{2} = 0 \cdot 20$

As this is a random variable which modifies our answer. And we get either 80, 100, or 120 depending on the value $𝐵_{2}$ . And we can in fact say that

$ℙ_{2} = 80 + 𝐵_{2} \cdot 20$ Therefore,

$𝔼 [𝑃_{2}] = 80 + 20 𝔼 [𝐵_{2}]$

$𝕍 [𝑃_{2}] = 0 + 20^{2} 𝕍 [𝐵_{2}]$

So now let’s try to understand the distribution of our random variable $𝐵_{2}$ .

We can actually leverage some common knowledge which tells us that the answer is based on the formulas.

If we have a general

$𝑋 \sim 𝑁 (𝜇, 𝜎^{2})$

and we want to know

$ℙ (𝑋 < 𝑎)$

We know this is the same thing as asking

$𝜙 (\frac{𝑎 - 𝜇}{𝜎})$

And then you can use the Excel formula

NORMDIST((a - mu)/ sigma)

Remember that in order to solve this, you can consult the table or Excel. There is a standardized reference.

5.

If we have stock returns iid, with annual vol of 30% what is daily VaR at 99% confidence level.

This is tricky because we have to convert between annual and daily. If we have iid variables, than the VaR can be written as

$𝑉 𝑎 𝑅 (𝑋, 𝛼) = 𝜎 𝑍_{1 - 𝛼}$

If we consider daily, we have

${VaR}_{𝑑} = 𝜎_{𝑑} \cdot 𝑍_{1 - 𝛼}$

and we are given that $𝛼$ is 0.01.

We are not given information about $𝜎_{𝑑}$ , but we know $𝜎_{𝐴} = 0.3$ . To some extent, we are looking at the aggregation. If we have log returns, the annual returns of A is the summation of these returns from days until 252. So

$𝜎_{𝐴}^{2} = 252 \cdot 𝜎_{𝑑}$

That means that actually

$𝜎_{𝑑} = \frac{1}{\sqrt{252}} \cdot 𝜎_{𝐴}$

We can plug this value in to get $𝜎_{𝑑}$ .

That’s how we move between periods. For the $𝑍$ , we just refer to the normal distribution table, which is the 99th percentile.

6.

In order to get the 10 day VaR, we use the formula

$𝑉 𝑎 𝑅_{𝑑} = \sqrt{𝑑} \cdot 𝑉 𝑎 𝑅$ .

7.

What does value at risk mean?

In 1 out of however many days (represented by confidence level), the portfolio will decline by the value at risk or more.

e.g. for 95% confidence level, this is 1 out of 20 days.

Regression Analysis

There’s a portfolio on Canvas, an Excel spreadsheet in the FE_535_Session_01_Portfolio_CAPM.xlsx. The file looks at the composition of TSLA returns. That’s what regression does.

8.

What is the correlation between these two assets? What is $𝛽$ , what does it stand for? It is the derivative of the regression, how will covariance change x?

$𝛽 = \frac{Cov (𝑋, 𝑌)}{𝕍 [𝑋]}$

If $𝛽$ is positive, that tells us correlation is positive, so we can consider them related. This is also

$\frac{𝜎_{𝑥 𝑦}}{𝜎_{𝑥}^{2}}$

We can play a little trick here to get

$\frac{𝜎_{𝑥 𝑦}}{𝜎_{𝑥} 𝜎_{𝑦}} \cdot \frac{𝜎_{𝑦}}{𝜎_{𝑥}}$

And that first expression is the correlation $𝜌_{𝑥 𝑦}$

So we have calculated

$𝛽 = 𝜌_{𝑥 𝑦} \frac{𝜎_{𝑦}}{𝜎_{𝑥}}$

Let’s take the square here.

$𝛽^{2} = 𝜌_{𝑥 𝑦}^{2} \frac{𝜎_{𝑦}^{2}}{𝜎_{𝑥}^{2}}$

I can write the volatility of this asset as systematic volatility.

$𝜎_{𝑦}^{2} = \frac{𝛽^{2} 𝜎_{𝑥}^{2}}{𝜌_{𝑥 𝑦}^{2}}$

There’s a proportion of total volatility that will be explained by this variable x.

If we have

$𝑅^{2} = \frac{𝛽^{2} 𝜎_{𝑥}^{2}}{𝜎_{𝑦}^{2}}$

this is the meaning of this statistical property.

We can actually explain the total volatility as the systematic volatility ( $𝛽$ ) as

$𝜎_{𝑦}^{2} = 𝛽^{2} 𝜎_{𝑥}^{2} + 𝜎_{𝑒}^{2}$

When you look at

## Continuous Random Variables and Transformations - URL: https://sharifhsn.dev/blog/continuous-random-variables-and-transformations/ - Structured data: https://sharifhsn.dev/api/posts/continuous-random-variables-and-transformations.json - Description: Recall for a discrete random variable. The moment formula is The moment for a continuous rv is the integral All the possibilities are the real line here. - Date: 2024-10-14 - Exact published timestamp: 2024-10-14 - Topics: Probability Theory, Continuous Distributions, Change of Variables - Categories: Probability Theory - Source: FE-540 \| Probability Theory - Source URL: None - Content source format: typst

Lecture Notes

Recall for a discrete random variable. The moment formula is

$𝔼 [𝑋] = \sum_{𝑥 \in 𝑋 (Ω)} 𝑥 ℙ (𝑋 = 𝑥)$

The moment for a continuous rv is the integral

$𝔼 [𝑋] = \int_{- \infty}^{\infty} 𝑥 𝑓 (𝑥) 𝑑 𝑥$

All the possibilities are the real line here.

Most of today’s class will deal with integrals. We are going to calculate a lot of these, just like a single variable calculus class. We will also learn multivariable calculus for multiple variables.

Generalization of expectation for any function is definition 5.9 (skip technical assumptions)

$𝔼 [ℎ (𝑋)] = \int_{- \infty}^{\infty} ℎ (𝑥) 𝑓 (𝑥) 𝑑 𝑥$

You would just compute the antiderivative of $ℎ (𝑥)$ and then evaluate the integral by subtracting the top from bottom. For an indicator variable, it is 0 to 1 not -infinity to infinity.

Now I want to go further, I want to know the full distribution, not just the expectation, of $ℎ (𝑋)$

This is given by theorem 5.10 which tells us how to determine the probability density function. The PDF of $𝑌 = ℎ (𝑋)$ is given by

$𝑓_{𝑌} (𝑦) = \frac{1}{ℎ^{'} (ℎ^{- 1} (𝑦))} 𝑓 (ℎ^{- 1}) (𝑦) 𝐼_{{𝑦 \in ℎ (𝐷)}}$

Here, $𝑓 (ℎ^{- 1})$ denotes the composition of f on inverse h. First I apply the inverse h, then I apply the f. Sometimes you use the notation $𝑓 \circ ℎ^{- 1}$

How do you prove this? We will sketch. We will make the assumption that $ℎ$ is non-decreasing.

$ℙ (𝑌 \leq 𝑦) = ℙ (ℎ (𝑋) \leq 𝑦) = ℙ (𝑋 \leq ℎ^{- 1} (𝑦))$

We can do the inverse because it’s non-decreasing.

This is true, so we can write this probability as the integral of the PDF of X

$= \int_{- \infty}^{ℎ^{- 1} (𝑦)} 𝑓 (𝑥) 𝑑 𝑥$

If I want to say the PDF of the random variable $𝑌$ , then it gets the $𝑌$ subscript; without the subscript, it’s for $𝑋$ .

Now we will do a change of variables

$𝑥 = ℎ^{- 1} (𝑧)$

We know that $𝑥$ is inside the integral, so we will write the equivalence that when it is in between those two values, it is equivalent to the event that $𝑧$ is in between $- \infty$ and $𝑦$ .

This is important because we will change the integration range:

$= \int_{- \infty}^{𝑦} 𝑓 (ℎ^{- 1} (𝑧)) 𝑑 ℎ^{- 1} (𝑧)$

You can move this part to the front

$= \int_{- \infty}^{𝑦} 𝑓 (ℎ^{- 1} (𝑧)) \cdot | (ℎ^{- 1})^{'} (𝑧) | 𝑑 𝑧$

We will see that

$𝑓_{𝑌} (𝑦) = \frac{𝑑}{𝑑 𝑦} ℙ (𝑌 \leq 𝑦)$

Now if we take the derivative of this integral, then we get

$\frac{𝑑}{𝑑 𝑦} \int_{- \infty}^{𝑦} 𝑓 (ℎ^{- 1} (𝑧)) | (ℎ^{- 1})^{'} (𝑧) | 𝑑 𝑧$

We get something like the fundamental theorem of calculus here, or part of it, or Leibniz rule. you have the following result: $\frac{𝑑}{𝑑 𝑥} (\int_{- \infty}^{𝑥} 𝑓 (𝑢) 𝑑 𝑢) = 𝑓 (𝑥)$

This identity is learned in calculus.

So, we can show that everything inside doesn’t depend on $𝑦$ , so we can just evaluate the integral with it.

$= 𝑓 (ℎ^{- 1} (𝑦)) | (ℎ^{- 1})^{'} (𝑦) |$

Then we will use this result, from a second theorem of calculus called the implicit function theorem. This is more real analysis, here’s a simple example.

$ℎ^{- 1} (ℎ (𝑦)) = 𝑦$

Now the idea is we want to do differentiation on both sides.

$\frac{𝑑}{𝑑 𝑦} (ℎ^{- 1} (ℎ (𝑦)) = \frac{𝑑}{𝑑 𝑦} (𝑦)$

Because we have this equality, the derivative preserves the equality (maybe not integration because of range). Applying the chain rule, we get

$(ℎ^{- 1})^{'} (ℎ (𝑦)) = \frac{1}{ℎ^{'} (𝑦)}$

And then we do a change of variable.

$𝑦 = ℎ^{- 1} (𝑢)$

y is just a dummy variable, so we can apply this here, to finally show that

$= 𝑓 (ℎ^{- 1} (𝑦)) | \frac{1}{ℎ^{'} (ℎ^{- 1} (𝑦))} |$ Interesting corollary 5.11 that if X has pdf f and we have a Y which is aX + b, then pdf of Y is

$𝑓_{𝑌} (𝑦) = \frac{1}{| 𝑎 |} 𝑓 (\frac{𝑦 - 𝑏}{𝑎})$

This is a special case when the $ℎ (𝑥)$ is $𝑎 𝑋 + 𝑏$ , can be shown by differentiation. All you need to find is $ℎ^{'} (𝑥)$ and $ℎ^{- 1} (𝑦)$

We can essentially think of $𝑌$ as a linear combination of a different random variable whose pdf is known, and then we can derive the pdf from the existing pdf f.

5.4.1

We are concerned with a uniform random variable defined continuously on $[𝑎, 𝑏]$ on $ℝ$

Definition 5.13 says that the pdf of this is

$𝑓 (𝑥) = \frac{1}{𝑏 - 𝑎}, 𝑥 \in [𝑎, 𝑏]$

We will use the notation to indicate uniform distribution

$𝑥 \sim 𝑈 [𝑎, 𝑏]$

Let’s do something non-trivial. Proposition 5.15 says that if we have $𝑋 \sim 𝑈 [𝑎, 𝑏]$ , then

$𝔼 [𝑋] = \frac{𝑎 + 𝑏}{2}$

$𝕍 [𝑋] = \frac{(𝑏 - 𝑎)^{2}}{12}$

How do we prove? We can create the integral based on the definition

We have from algebra that

$𝑏^{3} - 𝑎^{3} = (𝑏 - 𝑎) (𝑎^{2} + 𝑎 𝑏 + 𝑏^{2})$

which we can use to prove the variance.

We are generally solving for

$\frac{1}{𝑏 - 𝑎} \int_{𝑎}^{𝑏} 𝑓 (𝑥) 𝑑 𝑥$

and that’s how we solve for expectation, for $𝑓 (𝑥)$ being $𝑥$ or $𝑥^{2}$

We have to turn these types of problems into a pdf, that will be done on the midterm.

What is the cdf for a uniform variable?

By proposition 5.16, it’s

$𝐹_{𝑋} (𝑡) = {0 if 𝑡 < 𝑎 \frac{𝑡 - 𝑎}{𝑏 - 𝑎} if 𝑡 \in [𝑎, 𝑏] 1 if 𝑡 > 𝑏$

This is the same as $ℙ (𝑋 \leq 𝑡)$

We can accomplish this case conclusion by the integral

$= \int_{𝑎}^{𝑡} \frac{1}{𝑏 - 𝑎} 𝐼_{[𝑎, 𝑏]} (𝑢) 𝑑 𝑢$ Very simple proof for this.

Now let’s consider a simple example 5.8 where $𝑋 \sim 𝑈 [0, 1]$ , we will compute $ℙ (\frac{1}{4} \leq 𝑥 \leq \frac{3}{4})$

To solve, this we will solve as an integral

$= \int_{\frac{1}{4}}^{\frac{3}{4}} 𝑓 (𝑥) 𝑑 𝑥 = \int_{\frac{1}{4}}^{\frac{3}{4}} 1 𝑑 𝑥 = \frac{1}{2}$

This is one way to solve this, or you could separate it into cdf up to ¾ and ¼.

5.4.2

The next thing is the exponential distribution, definition 5.17. We introduce the auxiliary parameter $𝜆 > 0$ , then

$𝑓 (𝑥) = 𝜆 𝑒^{- 𝜆 𝑥}, 𝑥 > 0, 𝑋 \sim Exp (𝜆)$

That is what is meant when we are given a distribution. How do we verify that this result is a pdf? Proposition 5.18 says

$\int_{- \infty}^{\infty} 𝑓 (𝑥) 𝑑 𝑥$

The range of integration starts from 0

$= 𝜆 \int_{0}^{\infty} 𝑒^{- 𝜆 𝑥} 𝑑 𝑥$

Now we can do some calculus

$\frac{𝑑}{𝑑 𝑥} 𝑒^{- 𝜆 𝑥} = 𝑒^{- 𝜆 𝑥} (- 𝜆)$

$𝑒^{- 𝜆 𝑥} = \frac{\frac{𝑑}{𝑑 𝑥} 𝑒^{- 𝜆 𝑥}}{- 𝜆}$

$𝑒^{- 𝜆 𝑥} 𝑑 𝑥 = 𝑑 (\frac{𝑒^{- 𝜆 𝑥}}{- 𝜆})$

Now we can do integration by parts, which is a very important technique for this chapter that we will do over and over again.

If we are interested in integrating

$𝜆 𝑓 (𝑥) 𝑑 𝑔 (𝑥) = 𝑓 (𝑥) \cdot 𝑔 (𝑥) - 𝜆 𝑔 (𝑥) - \int 𝑔 (𝑥) 𝑑 𝑓 (𝑥)$

And this is the process of integration by parts.

(We don’t actually need integration by parts here, but we will need to do it for the moments)

Now we will show the moments by proposition 5.19

$𝔼 [𝑋] = \frac{1}{𝜆}$

$𝕍 [𝑋] = \frac{1}{𝜆^{2}}$

We will now try to evaluate this integral

$= 𝜆 \int_{0}^{\infty} 𝑥 𝑒^{- 𝜆 𝑥} 𝑑 𝑥$

And this is where we need integration by parts

$= 𝜆 \int_{0}^{\infty} 𝑥 𝑑 \frac{𝑒^{- 𝜆 𝑥}}{- 𝜆}$ First we take the product

$= 𝜆 (𝑥 \frac{𝑒^{- 𝜆 𝑥}}{- 𝜆} |_{𝑥 = 0}^{𝑥 = \infty} - \int_{0}^{\infty} \frac{𝑒^{- 𝜆 𝑥}}{- 𝜆} 𝑑 𝑥)$

Variance is not easy because you need to integrate twice by parts.

$𝔼 [𝑋^{2}] = \int_{0}^{\infty} 𝑥^{2} 𝜆 𝑒^{- 𝜆 𝑥} 𝑑 𝑥 = 𝜆 \int_{0}^{\infty} 𝑥^{2} 𝑒^{- 𝜆 𝑥} 𝑑 𝑥 = 𝜆 \int_{0}^{\infty} 𝑥^{2} 𝑑 (\frac{𝑒^{- 𝜆 𝑥}}{- 𝜆})$

We will try now to get rid of that denominator to get

$= - \int_{0}^{\infty} 𝑥^{2} 𝑑 𝑒^{- 𝜆 𝑥}$

Integration by parts:

$- 𝑥^{2} 𝑒^{- 𝜆 𝑥} |_{𝑥 = 0}^{𝑥 = \infty} - \int_{0}^{\infty} 𝑒^{- 𝜆 𝑥} 𝑑 𝑥^{2}$

$= - (0 - \int_{0}^{\infty} 𝑒^{- 𝜆 𝑥} 2 𝑥 𝑑 𝑥)$

$= 2 \int_{0}^{\infty} 𝑒^{- 𝜆 𝑥} 𝑥 𝑑 𝑥$

And we already have shown this result so it is

$= \frac{2}{𝜆^{2}}$

And then after subtracting by the expectation squared you get

$𝕍 [𝑋] = \frac{1}{𝜆^{2}}$

What is the cdf of the exponential random variable? Proposition 5.20 will tell us that

$𝐹_{𝑋} (𝑥) = 1 - 𝑒^{- 𝜆 𝑥}$

Here is the proof, which is just calculation

$𝐹_{𝑋} (𝑥) = \int_{0}^{𝑥} 𝑓 (𝑢) 𝑑 𝑢 = \int_{0}^{𝑥} 𝜆 𝑒^{- 𝜆 𝑢} 𝑑 𝑢$

The difference between a CDF and a statistical measure is that we’re measuring up to (x), not infinity, but the technique is the same.

$= 𝜆 \int_{0}^{𝑥} 𝑑 (\frac{𝑒^{- 𝜆 𝑢}}{- 𝜆}) = 𝜆 \frac{𝑒^{- 𝜆 𝑢}}{- 𝜆} |_{𝑢 = 0}^{𝑢 = 𝑥}$

There is a memoryless property of the exponential distribution.

$ℙ (𝑋 > 𝑡 + 𝑠 | 𝑋 > 𝑡) = ℙ (𝑋 > 𝑠), 𝑠, 𝑡 \geq 0$

I know that my exponential variable is bigger than t. There is some interval between t and t + s which is length s. I know my random variable $𝑋$ is bigger than t + s, given that it is bigger than t. But actually it doesn’t really matter what the probability is between t and t + s, there is no memory of the starting point. You only care about the probability after s. (We will not prove this)

Let’s take example 5.9, it will be easy to prove the cdf

$𝑓_{𝑋} (𝑡) = \frac{1}{100} 𝑒^{- \frac{𝑡}{100}}, 𝜆 = \frac{1}{100}$

Then we get expectation by the formula $\frac{1}{𝜆}$ How do we solve if we are given probability between an interval like

$ℙ (0 \leq 𝑋 \leq 50)$

We take the integral between these two points.

$= \int_{0}^{50} \frac{1}{100} 𝑒^{- \frac{𝑡}{100}} 𝑑 𝑡$

And we can actually do a change of variable here from

$𝑠 = \frac{𝑡}{100}$

Which will give us

$= \int_{0}^{\frac{1}{2}} 𝑒^{- 𝑠} 𝑑 𝑠 = \frac{𝑒^{- 𝑠}}{- 1} |_{𝑠 = 0}^{𝑠 = \frac{1}{2}} = \frac{𝑒^{- \frac{1}{2}} - 1}{- 1} = 1 - 𝑒^{- \frac{1}{2}}$

5.4.3

The normal distribution by definition 5.22 has the distribution function

$𝑓 (𝑥) = \frac{1}{\sqrt{2 𝜋 𝜎^{2}}} 𝑒^{- \frac{(𝑥 - 𝜇)^{2}}{2 𝜎^{2}}}, 𝑥 \in ℝ$

denoted by

$𝑋 \sim 𝑁 (𝜇, 𝜎^{2})$ We have to show that the normal distribution is a PDF, which is not very straightforward.

Proposition 5.24

Proposition 5.25 tells us that the expected value and variance are

$𝔼 [𝑋] = 𝜇$

$𝕍 [𝑋] = 𝜎^{2}$

which is very convenient for us. But we will need to formally verify this

We will copy this pdf into the integral

$𝔼 [𝑋] = \frac{1}{\sqrt{2 𝜋 𝜎^{2}}} \int_{- \infty}^{\infty} 𝑥 𝑒^{- \frac{(𝑥 - 𝜇)^{2}}{2 𝜎^{2}}} 𝑑 𝑥$

And we are going to do a change of variable

$𝑦 = 𝑥 - 𝜇$

We have the same integration region because they are infinite. We can replace x in this whole function.

$= \frac{1}{\sqrt{2 𝜋 𝜎^{2}}} \int_{- \infty}^{\infty} (𝑦 + 𝜇) 𝑒^{- \frac{𝑦^{2}}{2 𝜎^{2}}} 𝑑 𝑦$

We will separate the integral into two parts

$= \frac{1}{\sqrt{2 𝜋 𝜎^{2}}} \int_{- \infty}^{\infty} 𝑦 𝑒^{- \frac{𝑦^{2}}{2 𝜎^{2}}} 𝑑 𝑦 + 𝜇 \int_{- \infty}^{\infty} 𝑒^{- \frac{𝑦^{2}}{2 𝜎^{2}}} 𝑑 𝑦$

The first integrand is an odd function. This means that $𝑓 (𝑥) = - 𝑓 (- 𝑥)$ . Even functions are $𝑓 (𝑥) = 𝑓 (- 𝑥)$ . For any integral from -infinity to infinity, we can combine all the negatives and all the positives. And if it is symmetric around 0 with the opposite, which is the meaning of odd, then when they’re added it should give you 0.

Then we can recognize that for the second integrand that it must be equal to $𝜇$ by the fact that the normal distribution area is equal to 1 (assumed).

Now for the variance. By definition, because we know the expectation already, we get

$𝕍 [𝑋] = 𝔼 [(𝑋 - 𝜇)^{2}]$

which by the pdf is

$= \frac{1}{\sqrt{2 𝜋 𝜎^{2}}} \int_{- \infty}^{\infty} (𝑥 - 𝜇)^{2} 𝑒^{- \frac{(𝑥 - 𝜇)^{2}}{2 𝜎^{2}}} 𝑑 𝑥$

Then we can do a change of variable, and then solve via polar coordinates on page 138

$= \frac{1}{\sqrt{2 𝜋 𝜎^{2}}} \int_{- \infty}^{\infty} 𝑦^{2} 𝑒^{- \frac{𝑦^{2}}{2 𝜎^{2}}} 𝑑 𝑦$

Remark 5.2.7 says that if $𝑋 \sim 𝑁 (0, 1)$ , then $𝜇 + 𝜎 𝑋 \sim 𝑁 (𝜇, 𝜎^{2})$

The inverse operation

$𝑌 \sim 𝑁 (𝜇, 𝜎^{2}), \frac{𝑌 - 𝜇}{𝜎} \sim 𝑁 (0, 1)$

is known as standardization.

Proof: if you let

$𝑌 = 𝜇 + 𝜎 𝑋 = ℎ (𝑋)$

Using the corollary 5.11 for linear combination we can prove this.

Useful property of normal distribution. If we have two normal distributions $𝑋$ and $𝑌$ that are independent, then

$𝑋 + 𝑌 \sim 𝑁 (𝜇_{1} + 𝜇_{2}, 𝜎_{1}^{2} + 𝜎_{2}^{2})$ We have to do a convolution to solve this, involves integration between the two things.

5.4.14

Last one for today is the gamma distribution.

We need some preparation, in the gamma function.

$Γ (𝑥) = \int_{0}^{\infty} 𝑡^{𝑥 - 1} 𝑒^{- 𝑡} 𝑑 𝑡$

Proposition 5.30 says the properties of the gamma function. We have the property (a) that

$Γ (𝑥 + 1) = 𝑥 Γ (𝑥), 𝑥 > 0$

and (b) that

$Γ (𝑛 + 1) = 𝑛!$

and (c), which the source note does not prove,

$Γ (\frac{1}{2}) = \sqrt{𝜋}$

Proof of (a) is that

$= \int_{0}^{\infty} 𝑡^{𝑋} 𝑒^{- 𝑡} 𝑑 𝑡 = \int_{0}^{\infty} 𝑡^{𝑋} (𝑑 \frac{𝑒^{- 𝑡}}{- 1})$

$𝑡^{𝑥} \frac{𝑒^{- 𝑡}}{- 1} |_{𝑡 = 0}^{𝑡 = \infty} - \int_{0}^{\infty} \frac{𝑒^{- 𝑡}}{- 1} 𝑑 (𝑡^{𝑥})$

We can separately calculate $𝑑 𝑡^{𝑋} = 𝑥 𝑡^{𝑥 - 1} 𝑑 𝑡$ by the power rules.

$= \int_{0}^{\infty} 𝑒^{- 𝑡} 𝑥 𝑡^{𝑥 - 1} 𝑑 𝑡$

$𝑥 \int_{0}^{\infty} 𝑡^{𝑥 - 1} 𝑒^{- 𝑡} 𝑑 𝑡 = 𝑥 Γ (𝑥)$

The technique is to do the integration by parts and factor out certain quantities.

If we want to find out (b), then we can do induction once we find $Γ (2)$

$Γ (2) = \int_{0}^{\infty} 𝑡 𝑒^{- 𝑡} 𝑑 𝑡 = \int_{0}^{\infty} 𝑡 𝑑 \frac{𝑒^{- 𝑡}}{- 1} = 𝑡 \cdot \frac{𝑒^{- 𝑡}}{- 1} |_{𝑡 = 0}^{𝑡 = \infty} - \int \frac{𝑒^{- 𝑡}}{- 1} 𝑑 𝑡$

$= \int_{0}^{\infty} 𝑒^{- 𝑡} 𝑑 𝑡$

What about the pdf? Given by definition 5.31

$𝑓_{𝑎, 𝜆} (𝑥) = \frac{𝜆^{𝑎}}{Γ (𝑎)} 𝑒^{- 𝜆 𝑥} 𝑥^{𝑎 - 1}$

$𝑋 \sim Γ (𝑎, 𝜆)$

It has a lot of weird special cases.

Proposition 5.32 shows that this is a true density function. We will do integration from 0 to the infinity of this density

$\int_{0}^{\infty} \frac{𝜆^{𝑎}}{Γ (𝑎)} 𝑒^{- 𝜆 𝑥} 𝑥^{𝑎 - 1} 𝑑 𝑥$

$= \frac{𝜆^{𝑎}}{Γ (𝑎)} \int_{0}^{\infty} {(\frac{1}{𝜆} 𝑦)}^{𝑎 - 1} \cdot 𝑒^{- 𝑦} \frac{1}{𝜆} 𝑑 𝑦$

$= \frac{𝜆^{𝑎}}{Γ (𝑎)} \int_{0}^{\infty} \frac{1}{𝜆^{𝑎 - 1}} \cdot \frac{1}{𝜆} \cdot 𝑒^{- 𝑦} \frac{1}{𝜆} 𝑑 𝑦$

After we cancel out the $𝜆$

$= \frac{1}{Γ (𝑎)} \int_{0}^{\infty} 𝑦^{𝑎 - 1} 𝑒^{- 𝑦} 𝑑 𝑦$

which is just the definition of gamma

$= \frac{Γ (𝑎)}{Γ (𝑎)} = 1$

Proposition 5.33 says for E and V that

$𝔼 [𝑋] = \frac{𝑎}{𝜆}$

$𝕍 [𝑋] = \frac{𝑎}{𝜆^{2}}$

We know from the definition of expectation and combining similar terms that

$𝔼 [𝑋] = \int_{0}^{\infty} 𝑥 \cdot \frac{𝜆^{𝑎}}{Γ (𝑎)} 𝑥^{𝑎 - 1} 𝑒^{- 𝜆 𝑥} 𝑑 𝑥 = \frac{𝜆^{𝑎}}{Γ (𝑎)} \int_{0}^{\infty} 𝑥^{𝑎} 𝑒^{- 𝜆 𝑥} 𝑑 𝑥$

Let’s do a change of variable to $𝑦 = 𝜆 𝑥$ $= \frac{𝜆^{𝑎}}{Γ (𝑎)} \cdot \int_{0}^{\infty} {(\frac{𝑦}{𝜆})}^{𝑎} \cdot 𝑒^{- 𝑦} \cdot \frac{1}{𝜆} 𝑑 𝑦$

Taking this out, I get

$\frac{𝜆^{𝑎}}{Γ (𝑎)} \cdot \frac{1}{𝜆^{𝑎}} \frac{1}{𝜆} \int_{0}^{\infty} 𝑦^{𝑎} 𝑒^{- 𝑦} 𝑑 𝑦$

Here we can use the result from before of the definition of gamma to substitute the integral.

$= \frac{𝑎}{𝜆}$

To get variance, we have

$𝔼 [𝑋^{2}] = \frac{𝜆^{𝑎}}{Γ (𝑎)} \cdot \int_{0}^{\infty} 𝑥^{𝑎 + 1} 𝑒^{- 𝜆 𝑥} 𝑑 𝑥$

Change of variable…

$= \frac{𝜆^{𝑎}}{Γ (𝑎)} \cdot \int_{0}^{\infty} {(\frac{𝑦}{𝜆})}^{𝑎 + 1} 𝑒^{𝑦} 𝑑 𝑦$

$= \frac{𝑎 (𝑎 + 1)}{𝜆^{2}}$

$𝕍 [𝑋] = \frac{𝑎}{𝜆^{2}}$

I can sequentially use the value of a to get this result.

Finally, proposition 5.35 shows us that for $𝑋 \sim 𝑁 (0, 1)$ , then $𝑋^{2} \sim Γ (\frac{1}{2}, \frac{1}{2})$

Our proof is that

$𝐹_{𝑋^{2}} (𝑡) = ℙ (𝑋^{2} \leq 𝑡) = ℙ (- \sqrt{𝑡} \leq 𝑋 \leq \sqrt{𝑡})$

$= ℙ (𝑋 \leq \sqrt{𝑡}) - ℙ (𝑋 \leq - \sqrt{𝑡})$

x is a normal density, so we can write this as

now we’re going to differentiate on both sides

$𝑓_{𝑋^{2}} (𝑡) = \frac{𝑑}{𝑑 𝑡} 𝐹_{𝑋^{2}} (𝑡)$ . The source note starts the derivative calculation here; the remaining algebra is not recorded.

These two minus signs cancel out and you get both

And the two will cancel out and you can put this downstairs

Then you can rewrite this exactly as $Γ (\frac{1}{2}, \frac{1}{2})$

$𝑓_{\frac{1}{2}, \frac{1}{2}} (𝑥) = \frac{(\frac{1}{2})^{\frac{1}{2}}}{Γ (\frac{1}{2})} \cdot 𝑒^{- \frac{1}{2} 𝑥} 𝑥^{- \frac{1}{2}}$

And then we can know from our definition that

$\frac{1}{\sqrt{2} Γ (\frac{1}{2})} = \frac{1}{\sqrt{2 𝜋}}$

Using the progressive values of gamma makes it easier to do calculations with them.

## Bonds and the Time Value of Money - URL: https://sharifhsn.dev/blog/bonds-and-time-value-of-money/ - Structured data: https://sharifhsn.dev/api/posts/bonds-and-time-value-of-money.json - Description: We started with the foundation, risk-reward trade offs, portfolios. Then we moved to quantitative analysis on individual assets. Mostly we have focused on a single game, a single… - Date: 2024-10-10 - Exact published timestamp: 2024-10-10 - Topics: Risk Management, Bonds, Time Value of Money - Categories: Risk Management - Source: FE-535 \| Risk Management - Source URL: None - Content source format: typst

Bond Fundamentals

We started with the foundation, risk-reward trade offs, portfolios. Then we moved to quantitative analysis on individual assets. Mostly we have focused on a single game, a single price. But now we can think we have covered quantitative analysis. We are now going to the third book, the financial products and instruments. This would be for the non-finance people. We’re not going to flip any coins right now. In session 5 we will be flipping coins and doing finance, that’s derivatives.

Bonds give us a nice intuition, they have a lot of data. Risk is a function that goes down. Duration tells us the sensitivity of the instrument to a certain factor, in this case we will say interest raete risk. Then when we go to portfolios, we will study that.

Bonds will help us understand valuations as well.

Risk management starts with asset pricing. Seems to make sense. We can look at the sensitivity, quantify the risk. Let’s think about bonds. What are bonds subject to? Interest rates. If interest rates go up, bond prices go down. How would I price something? Think about what happened during the COVID days. Interest rates were near zero, so tech companies did well. They have cash flows coming in the future, and their assets are subject to some valuation. A simple way to think about it is the value of a payment in the future today, discount cash flow. Now we need to think, whatever will happen in the future, what is the value today? And that is the time value of money.

If you have a friend that wants to give you $100, it’s better to get it now than to get it five years from now.

We think about duration as having the $𝛽$ when it comes to bonds. We have a linear risk exposure, and it tells us how much risk there is.

What is a bond? It’s a debt. We have the borrower and the lender. The lender gives the money to the borrower, and the borrower returns the money plus interest. If there is a chance that your money will not come back, you will not lend your money to anybody, so interest rates compensate for that. Let’s assume that there’s no credit risk. I’ll lend money for a year. What if I didn’t lend that money? What else could I do with it? I could invest it in a risk-free investment and would get some interest. So why would I lend that money unless the return was at least that good?

Prices also go up over time. Think about consumption. I can get 100 apples for $100 at Stevens. One year from now, the apple will become $2. If someone gives me $100 in the future, I could only buy 50 apples. If we froze the apples, we were able to acquire more apples because we bought them now.

An entrepreneur is able to increase their equity by getting a million dollar loan and improve their business. The lender is happy to lend to them because the entrepreneur pays them back. It’s a win-win situation. Financial instruments can give you situations like this. In no arbitrage pricing situations, it’s a game, so what is the fair price? Stevens issued a bond of $200 million to build the towers, and helped raise capital. You can also have municipal bonds from Hoboken. Between these two parties of the borrower and the lender, you might have a one-time exchange, or you might have a coupon, a periodic payment. The face value of the bond, the principle, is the main course at the restaurant. The coupons are the appetizers. This keeps the lender happy because they just want the passive income from the bond.

I have a pricing equation which denotes

$𝑃_{0} = \sum_{𝑡 = 1}^{𝑇} 𝐶 𝐹_{𝑠, 𝑡} \times ℙ (𝑠_{𝑡}) \times \frac{1}{(1 + 𝑦)^{𝑡}}$

We will assume for simplicity right now that the probability of the return $ℙ (𝑠_{𝑡})$ is always 1.

If you have to pay back $100 for $50, that’s 100% interest.

Bonds are considered fixed-income, because they give based on contractual agreement. We get yields based on how many investors are buying them. Unlike a stochastic process, when you start getting close to time T, things get interesting. The value of the bond approaches the face value.

You can think about it as the yield $𝑦$ on an investment/bond as well. We also call this the IRR (Internal Rate of Return) or EAR (Effective Annual Rate). What is the return on my investment of the bonds?

$𝑃_{0} = \sum_{𝑡 = 1}^{𝑇} \frac{𝐶 𝐹_{𝑡}}{(1 + 𝑦)^{5}}$

People started trading today, and at the end there’s a closing price. By contract, we know the bond is a payment over 30 years, and bloomberg reports the yield on this investment. We get a small coupon on 2 year bonds, but we get a high r price and a lower yield. 30 years is much better.

These bonds will pay you yields on an annual (European) or semi-annual (American) basis. This becomes

$(1 + 𝑦) = {[1 + \frac{𝑦_{2}}{2}]}^{2}$

Let’s give an example. If we have the yield of 0.1 and an investment of 1, our investment will go from 1, to 1.1, to 1.21, to 1.31…

Now let’s go for semi-annual. We will multiply by 1.05 every six months, which means that every year we will multiply by 1.1025. This is a little more.

Now what if we do continuous, and exponential growth.

${lim}_{𝑑 \to \infty} {(1 + \frac{𝑦 (𝑑)}{𝑑})}^{𝑑} = 𝑒^{𝑦 (𝑐)}$ Before we talked about geometric Brownian motion, where we have a process that grows, a continuous differential time drift.

The question is what investment would give you the highest EAR. If they all have the same interest, then the answer is continuous compounding.

You can use the rule of 72 to determine how long it takes for an investment to double the investment. The number of years multiplied by the interest rate should be around 72.

## Itô Calculus - URL: https://sharifhsn.dev/blog/ito-calculus/ - Structured data: https://sharifhsn.dev/api/posts/ito-calculus.json - Description: This is seemingly one of the simplest integrals you could do. If we’re going to do this integral, how we’re going to do that is to create a process Δ 𝑛 ( 𝑢 ) → 𝑊 ( 𝑢 ) that… - Date: 2024-10-10 - Exact published timestamp: 2024-10-10 - Topics: Stochastic Calculus, Ito Lemma, Geometric Brownian Motion - Categories: Stochastic Calculus - Source: FE-610 \| Stochastic Calculus - Source URL: None - Content source format: typst

Week 5

Lecture Notes

Baby’s First Stochastic Integral

This is seemingly one of the simplest integrals you could do.

$\int_{0}^{𝑡} 𝑊 (𝑢) 𝑑 𝑊 (𝑢)$

If we’re going to do this integral, how we’re going to do that is to create a process $Δ_{𝑛} (𝑢) \to 𝑊 (𝑢)$ that converges so that this integral equals

$= {lim}_{𝑛 \to \infty} \int_{0}^{𝑡} Δ_{𝑛} (𝑢) 𝑑 𝑊 (𝑢)$

We essentially want this integral to converge to the integral of a Brownian motion.

We’re going to define

We’re going to assume wer’e cutting up our timespan $[0, 𝑡]$ into $𝑛$ equal sized pieces.

What we’re doing here is that if we have this path of a Brownian motion, we will create time partitions where we are cutting it up into equal spaces. And our process takes the value of the motion at time 0 until time 1, then resets to time 1, then time 2, so it turns it into a random walk. As $𝑛$ gets larger and larger, this converges to Brownian motion pointwise, no place for discontinuities to hide.

I can now express this integral as

The widths come from the change in Brownian motion.

We’ll denote $𝑊_{𝑗} = 𝑊 (\frac{𝑗 𝑡}{𝑛})$ for convenience, so this becomes

$= \sum_{𝑗 = 0}^{𝑛 - 1} 𝑊_{𝑗} (𝑊_{𝑗 + 1} - 𝑊_{𝑗})$

We’re going to look at the expansion of this squaredd

$\frac{1}{2} \sum_{𝑗 = 0}^{𝑛 - 1} (𝑊_{𝑗 + 1} - 𝑊_{𝑗})^{2} = \frac{1}{2} \sum_{𝑗 = 0}^{𝑛 - 1} 𝑊_{𝑗 + 1}^{2} - \sum_{𝑗 = 0}^{𝑛 - 1} 𝑊_{𝑗} 𝑊_{𝑗 + 1} + \frac{1}{2} \sum_{𝑗 = 0}^{𝑛 - 1} 𝑊_{𝑗}^{2}$

The 1/2 will become relevant later

Now, first thing I’m going to observe is that I could take this first term and rewrite it as

$\sum_{𝑘 = 1}^{𝑛} 𝑊_{𝑘}^{2}$

I’m doing a reindex and starting at $𝑘$ . What would this be when $𝑘 = 0$ ? 0. So this will be the same thing as

$= \sum_{𝑘 = 0}^{𝑛} 𝑊_{𝑘}^{2} = 𝑊_{𝑛}^{2} + \sum_{𝑗 = 0}^{𝑁 - 1} 𝑊_{𝑗}^{2}$

The expansion of this is

$= \frac{1}{2} 𝑊_{𝑛}^{2} + \sum_{𝑗 = 0}^{𝑛 - 1} (𝑊_{𝑗}^{2} - 𝑊_{𝑗} 𝑊_{𝑗 + 1})$

Well, this is

$= \frac{1}{2} 𝑊_{𝑛}^{2} + \sum_{𝑗 = 0}^{𝑛 - 1} 𝑊_{𝑗} (𝑊_{𝑗} - 𝑊_{𝑗 + 1})$

This term here is very close to where we left off in the original expansion. I can move this to the other side, which takes out the −1, which makes them correspond.

We have just proven that

$\sum_{𝑗 = 0}^{𝑛 - 1} 𝑊_{𝑗} (𝑊_{𝑗 + 1} - 𝑊_{𝑗}) = \frac{1}{2} 𝑊_{𝑛}^{2} - \frac{1}{2} \sum_{𝑗 = 0}^{𝑛 - 1} (𝑊_{𝑗 + 1} - 𝑊_{𝑗})^{2}$

After substituting that back in, I now have that

$\int_{0}^{𝑡} 𝑊 (𝑢) 𝑑 𝑊 (𝑢) = {lim}_{𝑛 \to \infty} (\frac{1}{2} 𝑊_{𝑛}^{2} - \frac{1}{2} \sum_{𝑗 = 0}^{𝑛 - 1} (𝑊_{𝑗 + 1} - 𝑊_{𝑗})^{2})$

What is $𝑊_{𝑛}$ ?

This doesn’t depend on $𝑛$ . If you look very carefully at the summation of the difference of squares, that’s quadratic variation, so $𝑡$

$= \frac{1}{2} 𝑊^{2} (𝑡) - \frac{𝑡}{2}$

The quadratic variation is the precise property which makes stochastic integral different

This is the correct answer if we use the left-hand endpoints i.e. if this is an adaptive process.

Right hand endpoint also

Stratonyvich integral if we use midpoint, so that’s intuitive. Why don’t we use it? It’s not adaptive/measurable, it requires you to evaluate the value of the expression halfway through the integral. That would be the equivalent of choosing your trading position at the middle of the day for the entire day. That’s impossible. So we have to use Ito integrals.

Differential Form

If we were dealing with regular calculus, and you asked me to find the differential of a function, I would look at it and use the chain rule. But Brownian motion isn’t differentiable.

$𝑑 𝑓 (𝑊 (𝑡)) = 𝑓^{'} (𝑊 (𝑡)) 𝑊^{'} (𝑡) 𝑑 𝑡 = 𝑓^{'} (𝑊 (𝑡)) 𝑑 𝑊 (𝑡)$

We’re still not going to have it correct because of quadratic variation. We actually have

$𝑑 𝑓 (𝑊 (𝑡)) = 𝑓^{'} (𝑊 (𝑡)) 𝑑 𝑊 (𝑡) + \frac{1}{2} 𝑓^{″} (𝑊 (𝑡)) 𝑑 𝑡$

If we look at the integral form and we look at both sides,

By the theorem of calculus, integral of differential should be difference between bounds. So we must prove this.

We’re going to use the Taylor series to come up with some expansions. This is going to be

$𝑓 (𝑥) = \sum_{𝑘 = 0}^{\infty} \frac{𝑓^{(𝑘)} (𝑎) (𝑥 - 𝑎)^{𝑘}}{𝑘!}$

writing out the first few terms, this is

$= 𝑓 (𝑎) + 𝑓^{'} (𝑎) (𝑥 - 1) + \frac{1}{2} 𝑓^{″} (𝑎) (𝑥 - 𝑎)^{2} + \frac{1}{6} 𝑓^{‴} (𝑎) (𝑥 - 𝑎)^{3} \dots$

So, I have that $𝑓 (𝑥) - 𝑓 (𝑎)$ equals all of this stuff

This is true for any function that we can do the Taylor approximation on. We’re going to keep this in mind and then I want to talk about telescoping sums. For a sequence ${𝑋_{𝑖}}_{𝑖 \in {1, \dots, 𝑛}}$ what can you tell me about this/

$\sum_{𝑖 = 0}^{𝑛 - 1} (𝑋_{𝑖 + 1} - 𝑋_{𝑖}) = 𝑋_{1} - 𝑋_{0} + 𝑋_{2} - 𝑋_{1} + 𝑋_{3} - 𝑋_{2} + \dots + 𝑋_{𝑛 - 1} - 𝑋_{𝑛 - 2} + 𝑋_{𝑛} - 𝑋_{𝑛 - 1}$

Well this is going to cancel out in between so we end up just getting the difference between the endpoints.

$= 𝑋_{𝑛} - 𝑋_{0}$

We now have most of our pieces.

I’m going to say that for a given partition $Π$ , I’m going to let $𝑥 = 𝑊 (𝑡_{𝑗 + 1})$ , $𝑎 = 𝑊 (𝑡_{𝑗})$ . I’m going to rewrite this Taylro series expansions with these variables. I now have

$𝑓 (𝑊 (𝑡_{𝑗 + 1})) - 𝑓 (𝑊 (𝑡_{𝑗})) = 𝑓^{'} (𝑊 (𝑡_{𝑗})) (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗})) + \frac{1}{2} 𝑓^{″} (𝑊 (𝑡_{𝑗})) (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗}))^{2} + \frac{1}{6} 𝑓^{‴} (𝑊 (𝑡_{𝑗})) (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗}))^{3} + \dots$

Then, I am now going to take (and remember, if I can do it to both sides I can do anything) I am going to add both of these sides up

$\sum_{𝑗 = 0}^{𝑛 - 1} (𝑓 (𝑊 (𝑡_{𝑗 + 1})) - 𝑓 (𝑊 (𝑡_{𝑗}))) = \sum_{𝑗 = 0}^{𝑛 - 1} (𝑓^{'} (𝑊 (𝑡_{𝑗})) (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗})) + \frac{1}{2} 𝑓^{″} (𝑊 (𝑡_{𝑗})) (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗}))^{2} + \frac{1}{6} 𝑓^{‴} (𝑊 (𝑡_{𝑗})) (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗}))^{3} + \dots)$

The left is just our telescoping summation. We have various summations on the other side.

$𝑓 (𝑊 (𝑡)) - 𝑓 (𝑊 (0)) = \sum_{𝑗 = 0}^{𝑛 - 1} 𝑓^{'} (𝑊 (𝑡_{𝑗})) (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗})) + \frac{1}{2} \sum_{𝑗 = 0}^{𝑛 - 1} 𝑓^{″} (𝑊 (𝑡_{𝑗})) (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗}))^{2} + \frac{1}{6} \sum_{𝑗 = 0}^{𝑛 - 1} 𝑓^{″} (𝑊 (𝑡_{𝑗})) (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗}))^{3} + \dots$

Because this is for an arbitrary partition, let’s look at the limit as the norm of partition goes to 0. Nothing happens on the left side, because there’s no index involved there. So we’ll leave it untouched.

$𝑓 (𝑊 (𝑡)) - 𝑓 (𝑊 (0)) = {lim}_{∥ Π ∥ \to 0} \sum_{𝑗 = 0}^{𝑛 - 1} 𝑓^{'} (𝑊 (𝑡_{𝑗})) (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗})) + \frac{1}{2} {lim}_{∥ Π ∥ \to 0} \sum_{𝑗 = 0}^{𝑛 - 1} 𝑓^{″} (𝑊 (𝑡_{𝑗})) (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗}))^{2} + {lim}_{∥ Π ∥ \to 0} \frac{1}{6} \sum_{𝑗 = 0}^{𝑛 - 1} 𝑓^{″} (𝑊 (𝑡_{𝑗})) (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗}))^{3} + \dots$

We’re writing this all down because we will need to refer back to this later! Now, let’s look and see how this limit affects this. As we’re dealing with the limit of the norm of this partition going to 0, this change in Brownian motion is going to converge to $𝑑 𝑊 (𝑢)$ the differential. That is the definition of the differential. That means that this term inside the exponent is just the differential. So we get the first term as $𝑑 𝑊 (𝑢)$ , then the second term as $𝑑 𝑢$ because that’s what we determined. And after we multiply it one more time, it will be 0, because that’s also proven. So we can throw out all terms after the second order. This gives us one Ito integral and one Riemann integral. What we have here is

$= \int_{0}^{𝑡} 𝑓^{'} (𝑊 (𝑢)) 𝑑 𝑊 (𝑢) + \frac{1}{2} \int_{0}^{𝑡} 𝑓^{″} (𝑊 (𝑢)) 𝑑 𝑢$ At this point, everybody should know that the 1/2 appears because of the Taylor expansion. Everybody makes the mistake and forgets it, so you need to be extra vigilant on homework and tests for this.

Ito Formula

That leads us to being able to prove Theorem 4.4.1 the Ito formula for Brownian motion:

Let $𝑓 (𝑡, 𝑥)$ be a function for which the partial derivatives etc.

$𝑓 (𝑇, 𝑊 (𝑇)) = 𝑓 (0, 𝑊 (0)) \dots$

I would strongly suggest that you use a different form of the Ito formula.

$𝑓 (𝑎, 𝑏)$ is some function such that $𝑓_{𝑎}$ , $𝑓_{𝑏}$ , and $𝑓_{𝑏 𝑏 𝑏 𝑏}$ exist.

$𝑓 (𝑡, 𝑊 (𝑡)) = 𝑓 (0, 𝑊 (0)) + \int_{0}^{𝑡} 𝑓_{𝑎} (𝑢, 𝑊 (𝑢)) 𝑑 𝑢 + \int_{0}^{𝑡} 𝑓_{𝑏} (𝑢, 𝑊 (𝑢)) 𝑑 𝑊 (𝑢) + \frac{1}{2} \int_{0}^{𝑡} 𝑓_{𝑏 𝑏} (𝑢, 𝑊 (𝑢)) 𝑑 [𝑊, 𝑊] 𝑑 𝑢$

This lets us remember where all the pieces are coming from. We can see that $𝑓$ is only considered with respect to these dummy variables.

Where does this formula come from? The two dimensional Taylor series expansion!

For some function, centered around the point $(𝑎_{0}, 𝑏_{0})$

$𝑓 (𝑎, 𝑏) - 𝑓 (𝑎_{0}, 𝑏_{0}) = 𝑓_{𝑎} (𝑎_{0}, 𝑏_{0}) (𝑎 - 𝑎_{0}) + 𝑓_{𝑏} (𝑎_{0}, 𝑏_{0}) (𝑏 - 𝑏_{0})$

These are all the first order derivatives. Then we get

$+ \frac{1}{2} 𝑓_{𝑎 𝑎} (𝑎_{0}, 𝑏_{0}) (𝑎 - 𝑎_{0})^{2} + \frac{1}{2} 𝑓_{𝑎 𝑏} (𝑎_{0}, 𝑏_{0}) (𝑎 - 𝑎_{0}) (𝑏 - 𝑏_{0}) + \frac{1}{2} 𝑓_{𝑏 𝑎} (𝑎_{0}, 𝑏_{0}) (𝑏 - 𝑏_{0}) (𝑎 - 𝑎_{0}) + \frac{1}{2} 𝑓_{𝑏 𝑏} (𝑎_{0}, 𝑏_{0}) (𝑏 - 𝑏_{0})^{2} + \frac{1}{6} + \dots$

into the second and third order and so on.

We will let $𝑎_{0} = 𝑡_{𝑗}$ , $𝑎 = 𝑡_{𝑗 + 1}$ , $𝑏_{0} = 𝑊 (𝑡_{𝑗})$ , $𝑏 = 𝑊 (𝑡_{𝑗 + 1})$

We’re going to do the same sum to telescoping sum trick here.

$𝑓 (𝑡, 𝑊 (𝑡)) - 𝑓 (0, 0) = \sum_{𝑗 = 0}^{𝑛 - 1} 𝑓_{𝑎} (𝑡_{𝑗}, 𝑊 (𝑡_{𝑗})) (𝑡_{𝑗 + 1} - 𝑡_{𝑗}) + \sum_{𝑗 = 0}^{𝑛 - 1} 𝑓_{𝑏} (𝑡_{𝑗}, 𝑊 (𝑡_{𝑗})) (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗})) + \frac{1}{2} \sum_{𝑗 = 0}^{𝑛 - 1} 𝑓_{𝑎 𝑎} (𝑡_{𝑗}, 𝑊 (𝑡_{𝑗})) (𝑡_{𝑗 + 1} - 𝑡_{𝑗})^{2} + \sum_{𝑗 = 0}^{𝑛 - 1} 𝑓_{𝑎 𝑏} (𝑡_{𝑗}, 𝑊 (𝑡_{𝑗})) (𝑡_{𝑗 + 1} - 𝑡_{𝑗}) (𝑊 (𝑡_{𝑗 + 1} - 𝑊 (𝑡_{𝑗})) + \frac{1}{2} \sum_{𝑗 = 0}^{𝑛 - 1} 𝑓_{𝑏 𝑏} (𝑡_{𝑗}, 𝑊 (𝑡_{𝑗})) (𝑡_{𝑗 + 1} - 𝑡_{𝑗}) (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗}))$

assume that fab and fba are the same

The differential notation makes the limit transparent: $𝑑 𝑢$ is first-order time, $𝑑 𝑊 (𝑢)$ is the Brownian increment, $𝑑 𝑢$ times any differential is 0, and $(𝑑 𝑊 (𝑢))^{2} = 𝑑 𝑢$ through quadratic variation. The only nonzero second-order term is therefore the quadratic-variation term. That is how we get the Itô formula.

Using this formula, we can now try and solve a simple exercise. The only stochastic part of the Ito formula is in the $𝑓_{𝑏}$ part. So let’s try and move everything to the other side of the expression.

$\int_{0}^{𝑡} 𝑓_{𝑏} (𝑢, 𝑊 (𝑢)) 𝑑 𝑊 (𝑢) = 𝑓 (𝑡, 𝑊 (𝑡)) - 𝑓 (0, 𝑊 (0)) - \int_{0}^{𝑡} 𝑓_{𝑎} (𝑢, 𝑊 (𝑢)) 𝑑 𝑢 - \frac{1}{2} \int_{0}^{𝑡} 𝑓_{𝑏 𝑏} (𝑢, 𝑊 (𝑢)) 𝑑 𝑢$

So, if I was given the problem, what is

$\int_{0}^{𝑡} 𝑊 (𝑢) 𝑑 𝑊 (𝑢)$

That seems to suggest that

$𝑓_{𝑏} (𝑢, 𝑊 (𝑢)) = 𝑊 (𝑢)$

What would be $𝑓_{𝑏} (𝑎, 𝑏)$ ? I don’t want to work with my functions with stochastic crap, I want to use well-defined dummy variables?

$𝑓_{𝑏} (𝑎, 𝑏) = 𝑏$

If I do this substitution, then this is what i results in.

Then what is

$𝑓_{𝑏 𝑏} (𝑎, 𝑏)$ ? 1.

I am going to do this the hardest possible way right now, to demonstrate how it’s done.

$𝑓 (𝑎, 𝑏) = \frac{𝑏^{2}}{2} + 𝑔 (𝑎) + 𝐶$

with $𝑔$ being some arbitrary function, which has to be true because of all these derivatives. NOTE: this $𝑓_{𝑏}$ is the notation for a partial derivative with respect to $𝑏$ .

We will prove that $𝑔 (𝑎)$ and $𝐶$ must be 0, so we don’t have to worry about them later.

$𝑓_{𝑎} (𝑎, 𝑏) = 𝑔^{'} (𝑎)$

With these choices, I’m going to be able to say that this

$\int_{0}^{𝑡} 𝑊 (𝑢) 𝑑 𝑊 (𝑢) = \frac{𝑊^{2} (𝑡)}{2} + 𝑔 (𝑡) + 𝐶 - \frac{𝑊^{2} (0)}{2} - 𝑔 (0) - 𝐶 - \int_{0}^{𝑡} 𝑔^{'} (𝑢) 𝑑 𝑢 - \frac{1}{2} \int_{0}^{𝑡} 𝑑 𝑢$

This is just plugging are previous values in the formula. We can cancel out the constant, so it will be 0 in the future.

What is the integral of $𝑔$ ? It would just be $𝑔 (𝑡) - 𝑔 (0)$ . So it didn’t matter at all what function we used, so we’ll just use 0 in the future. What’s left over is

$= \frac{𝑊^{2} (𝑡)}{2} - \frac{𝑡}{2}$

which is the same exact answer.

Another Baby

$\int_{0}^{𝑡} 𝑢 𝑑 𝑊 (𝑢)$

We know that the partial derivative $𝑓_{𝑏} (𝑢, 𝑊 (𝑢)) = 𝑢$ , so $𝑓_{𝑏} (𝑎, 𝑏) = 𝑎$ , $𝑓_{𝑏 𝑏} = 0$ .

So what is $𝑓$ ? $𝑓 = 𝑎 𝑏$ . We know that $𝑓_{𝑎} = 𝑏$ .

Professor recommendation: do everything with $𝑎$ and $𝑏$ , which is very handy. You don’t get messed up about the connections between $𝑡$ and $𝑊 (𝑡)$ . This becomes

$= 𝑡 𝑊 (𝑡) - \int_{0}^{𝑡} 𝑊 (𝑢) 𝑑 𝑢$

because everything else is 0. This is the fast way of doing it, just calculate the partial derivatives and solve through dummy variables until you can’t anymore.

Can you integrate Brownian motion? That value represents the area under Brownian motion. Knowing that every time Brownian motion is generated, it has a different value every time, you have to represent it as an integral, there’s no simplification because you can’t know more. Lonon will troll the students by putting Riemann integrals on exams and homework. Students always try to throw away all the other math when they learn stochastic calculus.

Ito Processes

One benefit of these stochastic integrals is that we can create the Ito process. Our only requirement is that our processes have to be adapted.

$𝑋 (𝑡) = 𝑋 (0) + \int_{0}^{𝑡} Δ (𝑢) 𝑑 𝑊 (𝑢) + \int_{0}^{𝑡} Θ (𝑢) 𝑑 𝑢$

We can show that the quadratic variation of this process is the same as the quadratic variation of the integral

$[𝑋, 𝑋] (𝑡) = \int_{0}^{𝑡} Δ^{2} (𝑢) 𝑑 𝑢$

Now let’s talk about the differential of a stochastic process.

$𝑑 𝑋 (𝑡) = {lim}_{𝛿 \to 0^{+}} (𝑋 (𝑡 + 𝛿) - 𝑋 (𝑡))$

$= {lim}_{𝛿 \to 0^{+}} (\int_{𝑡}^{𝑡 + 𝛿} Δ (𝑢) 𝑑 𝑊 (𝑢) + \int_{𝑡}^{𝑡 + 𝛿} Θ (𝑢) 𝑑 𝑢)$

When we’re talking about integrals over these tiny increments, it’s something different.

${lim}_{𝛿 \to 0^{+}} (Δ (𝑡) (𝑊 (𝑡 + 𝛿) - 𝑊 (𝑡)) + Θ (𝑡) (𝑡 + 𝛿 - 𝑡))$

These limits are the same. This means that when you take the limit, you get

$= Δ (𝑡) 𝑑 𝑊 (𝑡) + Θ (𝑡) 𝑑 𝑡$

We can now take the differential of any integral really quickly. This is a really simple trick.

What if we squared it?

$(𝑑 𝑋 (𝑡))^{2} = (Δ (𝑡) 𝑑 𝑊 (𝑡) + Θ (𝑡) 𝑑 𝑡)^{2}$

Let’s expand the square

$= Δ^{2} (𝑡) (𝑑 𝑊 (𝑡))^{2} + 2 Δ (𝑡) Θ (𝑡) 𝑑 𝑡 𝑑 𝑊 (𝑡) + Θ^{2} (𝑡) (𝑑 𝑡)^{2}$ Let’s simplify almost all of this. We know that all the dt squared and dtdW(t) are 0

$= Δ^{2} (𝑡) 𝑑 𝑡$

This is an approach that will work for a continuous process. An Ito process has to be a continuous process. It’s a nonrandom variable plus an Ito integral plus a Riemann integral. We can use this approach: Since I know that

$(𝑑 𝑋 (𝑡))^{2} = 𝑑 [𝑋, 𝑋] (𝑡)$

That means that

$\int_{0}^{𝑡} (𝑑 𝑋 (𝑢))^{2} = [𝑋, 𝑋] (𝑡)$

IF $𝑋$ IS CONTINUOUS.

This is the quick proof.

We can now define an Ito integral with respect to an Ito process.

$\int_{0}^{𝑡} Γ (𝑢) 𝑑 𝑋 (𝑢) = \int_{0}^{𝑡} Γ (𝑢) Δ (𝑢) 𝑑 𝑊 (𝑢) + \int_{0}^{𝑡} Γ (𝑢) Θ (𝑢) 𝑑 𝑢$

Your stock process would follow some stochastic process, and this reprseents a trading strategy

We can now find, where $Δ$ is a simple process that dictates how we choose stocks,

$\int_{0}^{𝑡} Δ (𝑤) 𝑑 𝑆 (𝑤)$

We can vary it based on the stock price. We can actually use this Ito formula to take differentials as well as integrals. With the expression $𝑓 (𝑡, 𝑊 (𝑡))$ , if we assume

$𝑑 𝑓 (𝑡, 𝑊 (𝑡)) = 𝑓_{𝑎} (𝑡, 𝑊 (𝑡)) 𝑑 𝑡 + 𝑓_{𝑏} (𝑡, 𝑊 (𝑡)) 𝑑 𝑊 (𝑡) + \frac{1}{2} 𝑓_{𝑏 𝑏} (𝑡, 𝑊 (𝑡)) (𝑑 𝑊 (𝑡))^{2}$

If I said this function was geometric Brownian motion

$𝑓 (𝑡, 𝑊 (𝑡)) = 𝑆 (0) 𝑒^{(𝛼 - \frac{𝜎^{2}}{2}) 𝑡 + 𝜎 𝑊 (𝑡)}$

For simplicity, we’ll replace with a and b

$𝑓 (𝑎, 𝑏) = 𝑆 (0) 𝑒^{(𝛼 - \frac{𝜎^{2}}{2}) 𝑎 + 𝜎 𝑏}$

Then what are our partials? $𝑓_{𝑎} = (𝛼 - \frac{𝜎^{2}}{2}) 𝑓 (𝑎, 𝑏)$

$𝑓_{𝑏} = 𝜎 𝑓 (𝑎, 𝑏)$

$𝑓_{𝑏 𝑏} = 𝜎^{2} 𝑓 (𝑎, 𝑏)$

Open question: look more into partial derivatives and why this is true.

This means that

$𝑑 𝑆 (𝑡) = (𝛼 - \frac{𝜎^{2}}{2}) 𝑆 (𝑡) 𝑑 𝑡 + 𝜎 𝑆 (𝑡) 𝑑 𝑊 (𝑡) + \frac{1}{2} 𝜎^{2} 𝑆 (𝑡) 𝑑 𝑡$

Those terms cancel, which means this whole thing is

$= 𝛼 𝑆 (𝑡) 𝑑 𝑡 + 𝜎 𝑆 (𝑡) 𝑑 𝑊 (𝑡)$

or, written slightly differently,

$\frac{𝑑 𝑆 (𝑡)}{𝑆 (𝑡)} = 𝛼 𝑑 𝑡 + 𝜎 𝑑 𝑊 (𝑡)$

Now we can see why we’re looking at GBM. When Black-Scholes were looking for a way to model the stock prices. They looked at this instantaneous drift $𝛼$ and the $𝜎$ that is the diffusion. The problem is that this is nonsense because you can have negative stocks in that model. So now they look at proportional changes in the stock processes, which is sort of like instantaneous return. This is what this dynamic is showing us. We have drift and diffusion.

That means that our original integral for choosing stocks is

$\int_{0}^{𝑡} Δ (𝑢) 𝑑 𝑆 (𝑢) = \int_{0}^{𝑡} Δ (𝑢) 𝛼 𝑆 (𝑢) 𝑑 𝑢 + \int_{0}^{𝑡} Δ (𝑢) 𝜎 (𝑢)$

Our process is coming from deterministic drift and random diffusion. That also finally gets us to theorem 4.4.6, which is the best version of this formula, the Ito formula for an Ito process.

The source starts the general Itô-process formula here but leaves it unfinished (Open question: finish the displayed formula from the slides).

The first argument corresponds to the first differential $𝑑 𝑢$ , and then the second one is done once than twice, which is what goes into the quadratic variation.

Why is this true? Let’s go back to the two dimensional Taylor series.

If we use $𝑋$ the Ito process instead of $𝑊$ Brownian motion, you end up with the same combinations. $𝑑 𝑢 𝑑 𝑋 (𝑢)$ . We can create a general rule that $𝑑 𝑡$ / $𝑑 𝑢$ multiplied by any other differential is 0. REMEMBER THIS.

The only difference is that the quadratic variation of the Ito process, which is $Δ^{2} (𝑢) 𝑑 𝑢$

This also can be expressed in differential form, which loses a few details.

Generalized GBM

You can also have generalized Geometric Brownian Motion, which is different from regular GBM. We can now say that $𝛼$ and $𝜎$ are no longer inputs, they are stochastic processes in themselves. They are still adapted, but they are stochastic.

IF I define

$𝑆 (𝑡) = 𝑆 (0) exp (\int_{0}^{𝑡} (𝛼 (𝑢) - \frac{𝜎^{2} (𝑢)}{2}) 𝑑 𝑢 + \int_{0}^{𝑡} 𝜎 (𝑢) 𝑑 𝑊 (𝑢))$

I can’t do the differential hre normally because I don’t have everything substitutable, we don’t have W(t) really. We can’t write this as a function of $𝑡$ and $𝑊 (𝑡)$

But I can define a new stochastic process

$𝑋 (𝑡) = \int_{0}^{𝑡} (𝛼 (𝑢) - \frac{𝜎^{2} (𝑢)}{2}) 𝑑 𝑢 + \int_{0}^{𝑡} 𝜎 (𝑢) 𝑑 𝑊 (𝑢)$

Can I write $𝑆 (𝑡)$ now as an Ito process?

$𝑆 (𝑡) = 𝑆 (0) 𝑒^{𝑋 (𝑡)} = 𝑓 (𝑡, 𝑋 (𝑡))$

My Ito formula is going to tell me that

$𝑑 𝑆 (𝑡) = 𝑓_{𝑎} (𝑡, 𝑋 (𝑡)) 𝑑 𝑡 + 𝑓_{𝑏} (𝑡, 𝑋 (𝑡)) 𝑑 𝑋 (𝑡) + \frac{1}{2} 𝑓_{𝑏 𝑏} (𝑡, 𝑋 (𝑡)) (𝑑 𝑋 (𝑡))^{2}$

This helps me know what I’m missing to find this differential.

$𝑓 (𝑎, 𝑏) = 𝑆 (0) 𝑒^{𝑏}$

Therefore, $𝑓_{𝑎} = 0$ , $𝑓_{𝑏} = 𝑓$ , $𝑓_{𝑏 𝑏} = 𝑓$

$𝑑 𝑋 (𝑡) = (𝛼 (𝑡) - \frac{𝜎^{2} (𝑡)}{2}) 𝑑 𝑡 + 𝜎 (𝑡) 𝑑 𝑊 (𝑡)$

$(𝑑 𝑋 (𝑡))^{2} = 𝜎^{2} (𝑡) 𝑑 𝑡$

This is because… I don’t know he didn’t explain 🙁

$𝑑 𝑆 (𝑡) = 0 𝑑 𝑡 + 𝑆 (𝑡) 𝑑 𝑋 (𝑡) + \frac{1}{2} 𝑆 (𝑡) (𝑑 𝑋 (𝑡))^{2}$

$= 𝑆 (𝑡) (𝛼 (𝑡) - \frac{𝜎^{2} (𝑡)}{2}) 𝑑 𝑡 + 𝑆 (𝑡) 𝜎 (𝑡) 𝑑 𝑊 (𝑡) + \frac{1}{2} 𝑆 (𝑡) 𝜎^{2} (𝑡) 𝑑 𝑡$ All of that is $𝑆 (𝑡) 𝑑 𝑋 (𝑡)$

We can cancel out the $𝜎^{2}$ portion and create

$= 𝑆 (𝑡) 𝛼 (𝑡) 𝑑 𝑡 + 𝑆 (𝑡) 𝜎 (𝑡) 𝑑 𝑊 (𝑡)$

This is the same as GBM, but we need to derive this by using the Ito process.

The Ito formula for Ito processes works for everything, which makes it the best one, 4.4.6.

An interesting result of theorem 4.4.9 is that

$𝐼 (𝑡) = \int_{0}^{𝑡} Δ (𝑠) 𝑑 𝑊 (𝑠)$

The random variable is normally distributed. Let’s prove that.

The key to this proof is that I’m going to first define

$𝑋 (𝑡) = 𝑢 𝐼 (𝑡) - \frac{𝑢^{2}}{2} \int_{0}^{𝑡} Δ^{2} (𝑠) 𝑑 𝑠$ .

What is $𝔼 [𝑒^{𝑋 (𝑡)}] = ?$

Let’s look at the differential, the Ito decomposition. This makes it much more clear if this is an Ito process.

$𝑑 (𝑒^{𝑋 (𝑡)}) = 𝑒^{𝑋 (𝑡)} 𝑑 𝑋 (𝑡) + \frac{1}{2} 𝑒^{𝑋 (𝑡)} (𝑑 𝑋 (𝑡))^{2}$

But what we don’t have well defined is $𝑑 𝑋 (𝑡)$ . But that shouldn’t be too bad.

$𝑑 𝑋 (𝑡) = 𝑢 𝑑 𝐼 (𝑡) - \frac{𝑢^{2}}{2} Δ^{2} (𝑡) 𝑑 𝑡$

I know what is given to us by 4.4.9

$= 𝑢 Δ (𝑡) 𝑑 𝑊 (𝑡) - \frac{𝑢^{2}}{2} Δ^{2} (𝑡) 𝑑 𝑡$

$𝑑 (𝑒^{𝑋 (𝑡)}) = 𝑒^{𝑋 (𝑡)} 𝑢 Δ (𝑡) 𝑑 𝑊 (𝑡) - 𝑒^{𝑋 (𝑡)} \frac{𝑢^{2}}{2} Δ^{2} (𝑡) 𝑑 𝑡 + \frac{1}{2} 𝑒^{𝑋 (𝑡)} 𝑢^{2} Δ^{2} (𝑡) 𝑑 𝑡$

REMEMBER THAT $(𝑑 𝑋 (𝑡))^{2}$ is $Δ^{2} 𝑑 𝑡$

Canceling out $𝑒$ terms, we get

$= 𝑒^{𝑋 (𝑡)} 𝑢 Δ (𝑡) 𝑑 𝑊 (𝑡)$

Integrating both sides, we get

$𝑒^{𝑋 (𝑡)} = 𝑒^{𝑋 (0)} + \int_{0}^{𝑡} 𝑒^{𝑋 (𝑠)} 𝑢 Δ (𝑠) 𝑑 𝑊 (𝑠)$

I can see an interesting thing here. This formula must be a martingale, because it’s a constant plus a martingale. All Ito integrals with respect to Brownian motion are martingales.

$= 1 + \int_{0}^{𝑡} 𝑒^{𝑋 (𝑠)} \dots$

The expected value is the same as when it’s conditioned on zero information, which is 1.

$𝔼 [exp (𝑢 𝐼 (𝑡) - \frac{𝑢^{2}}{2} \int_{0}^{𝑡} Δ^{2} (𝑠) 𝑑 𝑠)] = 1$

$Δ$ is explicitly nonrandom, so $𝐼 (𝑡)$ is the only stochastic part of it.

That means you actually have

$𝔼 [𝑒^{𝑢 𝐼 (𝑡)}] = exp \dots$

That’s how you get a moment generating function, for a normal random variable, for mean 0 and variance given by this integral.

Vasicek

We don’t only use stochastic processes for stock prices. There are many random things in finance, like interest rates. There are different models for this. This is given by this stochastic differential equation:

$𝑑 𝑅 (𝑡) - (𝛼 - 𝛽 𝑅 (𝑡)) 𝑑 𝑡 + 𝜎 𝑑 𝑊 (𝑡)$

It ahsa closed form solution, which we can prove.

This Vasicek model has a mean reverting component. We can see this by looking at the behavior of this process if the current value of interest rate $𝑅 (𝑡) < \frac{𝛼}{𝛽}$ , then $(𝛼 - 𝛽 𝑅 (𝑡)) > 0$ . If our interest rate falls below this level, our drift term becomes strictly positive. We’re going to have a general upward drift if this happens. But what if it goes below that? We get pulled back down. What happens if we hit it perfectly? We still have noise in our model from $𝜎 𝑑 𝑊 (𝑡)$ , so we will never have an equilibrium.

Closed form solution:

$𝑅 (𝑡) = 𝑅 (0) 𝑒^{- 𝛽 𝑡} + \frac{𝛼}{𝛽} (1 - 𝑒^{- 𝛽 𝑡}) + 𝜎 𝑒^{- 𝛽 𝑡} \int_{0}^{𝑡} 𝑒^{𝛽 𝑢} 𝑑 𝑊 (𝑢)$

We’ll pick the Ito process

$𝑋 (𝑡) = \int_{0}^{𝑡} 𝑒^{𝛽 𝑢} 𝑑 𝑊 (𝑢)$

Because this will give us

$𝑅 (𝑡) = 𝑓 (𝑡, 𝑋 (𝑡))$

We need to find $𝑑 𝑋 (𝑡)$

$𝑑 𝑋 (𝑡) = 𝑒^{𝛽 𝑡} 𝑑 𝑊 (𝑡)$ We’ll substitute with $𝑎 𝑎 𝑛 𝑑$ b$

$𝑓 (𝑎, 𝑏) = 𝑅 (0) 𝑒^{- 𝛽 𝑎} + \frac{𝛼}{𝛽} (1 - 𝑒^{- 𝛽 𝑎}) + 𝜎 𝑒^{- 𝛽 𝑎} 𝑏$

What’s our partials? $𝑓_{𝑎} = - 𝛽 𝑅 (0) 𝑒^{- 𝛽 𝑎} + 𝛼 𝑒^{- 𝛽 𝑎} - 𝛽 𝜎 𝑒^{- 𝛽 𝑎} 𝑏$

$𝑓_{𝑏} = 𝜎 𝑒^{- 𝛽 𝑎}$

$𝑓_{𝑏 𝑏} = 0$

If we run our Ito formula:

$𝑑 𝑅 (𝑡) = (- 𝛽 𝑅 (0) 𝑒^{- 𝛽 𝑡} + 𝛼 𝑒^{- 𝛽 𝑡} - 𝛽 𝜎 𝑒^{- 𝛽 𝑡} 𝑋 (𝑡)) 𝑑 𝑡 + 𝜎 𝑒^{- 𝛽 𝑡} 𝑑 𝑋 (𝑡) + \frac{1}{2} (0) (𝑑 𝑋 (𝑡))^{2}$

Let’s factor out this $- 𝛽$ and cancel out $𝑑 𝑋 (𝑡))$ when we factor it in.

$= - 𝛽 (𝑅 (0) 𝑒^{- 𝛽 𝑡} - \frac{𝛼}{𝛽} 𝑒^{- 𝛽 𝑡} + 𝜎 𝑒^{- 𝛽 𝑡} 𝑋 (𝑡)) 𝑑 𝑡 + 𝜎 𝑑 𝑊 (𝑡))$

We are very close to being $𝑅 (𝑡)$ . The only difference is $\frac{𝛼}{𝛽}$ So we’ll add and subtract it.

$= - 𝛽 (- \frac{𝛼}{𝛽} + 𝑅 (0) 𝑒^{- 𝛽 𝑡} + \frac{𝛼}{𝛽} - \frac{𝛼}{𝛽} 𝑒^{- 𝛽 𝑡} + 𝜎 𝑒^{- 𝛽 𝑡} 𝑋 (𝑡)) 𝑑 𝑡 + 𝜎 𝑑 𝑊 (𝑡)$

When we cancel out the denominator with $𝛽$ , we get

$= (𝛼 - 𝛽 𝑅 (𝑡)) 𝑑 𝑡 + 𝜎 𝑑 𝑊 (𝑡)$

This is very nice because we can get the expected value of $𝑅 (𝑡)$ more easily.

$𝔼 [𝑅 (𝑡)] = 𝔼 [𝑅 (0) 𝑒^{- 𝛽 𝑡} + \frac{𝛼}{𝛽} (1 - 𝑒^{- 𝛽 𝑡}) + 𝜎 𝑒^{- 𝛽 𝑡} \int_{0}^{𝑡} 𝑒^{𝛽 𝑡} 𝑑 𝑊 (𝑠)]$

Whenever you’re calculating expectations, you need to figure out what’s random, because the expectation of a constant is itself.

The only random thing is the Ito integral, which is a martingale, so we get

$= 𝑅 (0) 𝑒^{- 𝛽 𝑡} + \frac{𝛼}{𝛽} (1 - 𝑒^{- 𝛽 𝑡})$

What would be the long-term rate?

${lim}_{𝑡 \to \infty} 𝔼 [𝑅 (𝑡)] = \frac{𝛼}{𝛽}$

This term is called the long-term interest rate, what it’s mean reverting to.

This seems like a pretty interesting process. Ito process, closed form solution, but it’s got a big flaw. The only thing that’s random is the integral.

If you add a constant to a normal distribution, it’s still normal. We just agreed that this process is normally distributed, which means that there’s a nonzero probability that the interest rate could be negative. Depending on what you’re modeling, like risk-free interest rates in a functioning economy, this could be really bad.

Some very intelligent people saw this problem and fixed it.

Cox-Ingersoll-Ross

The CIR interest rate model adds in one thing: a square root. In order for the interest rate to go negative, it has to go to 0. This model causes noise to disappear as interest rates go to 0, so upward drift increases.

$𝑑 𝑅 (𝑡) = (𝛼 - 𝛽 𝑅 (𝑡)) 𝑑 𝑡 + 𝜎 \sqrt{𝑅 (𝑡)} 𝑑 𝑊 (𝑡)$

Unfortunately, there is no obvious closed-form solution. Fortunately, we’re built different. We would look at

$𝑑 (𝑒^{𝛽 𝑡} 𝑅 (𝑡)) = 𝑑 𝑓 (𝑡, 𝑅 (𝑡))$

where the dummy function is

$𝑓 (𝑎, 𝑏) = 𝑒^{𝛽 𝑎} 𝑏$

$𝑓_{𝑎} = 𝛽 𝑓$

$𝑓_{𝑏} = 𝑒^{𝛽 𝑎}$

$𝑓_{𝑏 𝑏} = 0$

So the differential becomes

$= 𝛽 𝑒^{𝛽 𝑡} 𝑅 (𝑡) 𝑑 𝑡 + 𝑒^{𝛽 𝑡} 𝑑 𝑅 (𝑡)$

$= 𝛽 𝑒^{𝛽 𝑡} 𝑅 (𝑡) 𝑑 𝑡 + 𝑒^{𝛽 𝑡} (𝛼 - 𝛽 𝑅 (𝑡)) 𝑑 𝑡 + 𝑒^{𝛽 𝑡} 𝜎 \sqrt{𝑅 (𝑡)} 𝑑 𝑊 (𝑡)$

We see what happens is that the terms get canceled because we chose this specific differential.

$= 𝛼 𝑒^{𝛽 𝑡} 𝑑 𝑡 + 𝜎 \sqrt{𝑅 (𝑡)} 𝑒^{𝛽 𝑡} 𝑑 𝑊 (𝑡)$

I can integrate both sides and get

$𝑒^{𝛽 𝑡} 𝑅 (𝑡) - 𝑒^{𝛽 (0)} 𝑅 (0) = \int_{0}^{𝑡} 𝛼 𝑒^{𝛽 𝑢} 𝑑 𝑢 + \int_{0}^{𝑡} 𝜎 \sqrt{𝑅 (𝑢)} 𝑒^{𝛽 𝑢} 𝑑 𝑊 (𝑢)$

I can kind of get $𝑅 (𝑡)$ by itself but not all the way because I can’t get rid of the square root. There is no fancy thing I can look at to make this thing behave. But I will still have

$𝑅 (𝑡) = 𝑅 (0) 𝑒^{- 𝛽 𝑡} + \frac{𝛼}{𝛽} (1 - 𝑒^{- 𝛽 𝑡}) + 𝑒^{- 𝛽 𝑡} \int_{0}^{𝑡} 𝜎 \sqrt{𝑅 (𝑡)} 𝑒^{𝛽 𝑢} 𝑑 𝑊 (𝑢)$

Which means I can get the expectation, which would not necessarily be given by the original expression.

If I tried doing that with the original, I depend on $𝑅 (𝑢)$ . But in my new form, the only $𝑅 (𝑢)$ is contained in the stochastic integral which we know has expectation 0.

## Minneapolis Housing and Zoning - URL: https://sharifhsn.dev/blog/linkedin-2024-10-08-minneapolis-housing-and-zoning/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2024-10-08-minneapolis-housing-and-zoning.json - Description: What Minneapolis changed in its zoning rules, and what happened to housing and rents. - Date: 2024-10-08 - Exact published timestamp: 2024-10-08T21:50:44.977Z - Topics: Housing, Urban Policy - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/posts/sharif-haason_minneapolis-land-use-reforms-offer-a-blueprint-activity-7249536724404432898-dpDs - Content source format: typst

Some of you likely tuned into the vice presidential debate last week. Tim Walz touted the success of housing policy in Minneapolis, so what did that actually look like? And how can those lessons be applied to the nationwide housing problem?

The key behind Minneapolis’s housing was the Minneapolis 2040 Plan, approved in 2019 and put into place starting in 2020. This was a comprehensive set of 100 policies designed towards the goals of making housing affordable, increasing jobs, and making sure that urban areas fulfill environmental and social considerations.

As all developers know, the primary obstacle to building housing is zoning. If a particular area is zoned to only allow single-family homes (SFHs), then the amount of development that can occur in that area is limited, as SFHs are large relative to their number of residents. There’s a fundamental density limitation which prevents SFHs from solving the housing problem in cities.

The 2040 Plan upzoned Minneapolis in two ways. First, it requires residential areas which were SFH zoned to allow the building of duplexes and triplexes. This policy had very little impact on actual development because the rules required these homes to have the same floor-to-area ratio (FAR) as SFHs to prevent them from being larger. This placed significant design constraints on duplexes/triplexes to the point that they weren’t worth developing.

A much more successful initiative was multifamily zoning on through streets, which the 2040 Plan calls “corridors”. These streets have bustled with business activity for ages, but their growth was stunted by their SFH zoning. The FAR requirements for corridor housing were relaxed, which allowed larger apartment buildings with 20-50 housing units to be constructed. This type of construction was the vast majority of housing that was permitted and built.

There’s one more deregulation that developers say has been instrumental here: the elimination of citywide parking minimums. Minneapolis, like many other cities, used to require apartments in some areas to have parking. This meant that developers had to build additional parking structures, which studies show adds about $50,000 in per-unit development costs. It’s no wonder growth was stunted with such a massive burden on developers.

What was the impact of this? Between 2017 and 2022, rents in Minneapolis stayed the same, when rents in Minnesota in general rose 14%. The levels of homelessness over the same period dropped 12% in Minneapolis and rose 14% in Minnesota. High rent and homelessness are the most pressing issues in the U.S. related to housing policy, so the plan can be considered a resounding success and a model for similar cities.

What do you think of these policies? Should they be implemented in your city?

Read the linked analysis: Minneapolis Land Use Reforms Offer a Blueprint for Housing Affordability.

## Stochastic Integrals - URL: https://sharifhsn.dev/blog/stochastic-integrals/ - Structured data: https://sharifhsn.dev/api/posts/stochastic-integrals.json - Description: We left off last week talking about first passage time. We said that the cdf was Last thing before Stochastic Calculus we should talk about maximum to date . This will be… - Date: 2024-10-03 - Exact published timestamp: 2024-10-03 - Topics: Stochastic Calculus, Stochastic Integrals, Ito Integral - Categories: Stochastic Calculus - Source: FE-610 \| Stochastic Calculus - Source URL: None - Content source format: typst

Week 4

Lecture Notes

We left off last week talking about first passage time. We said that the cdf was

Last thing before Stochastic Calculus we should talk about maximum to date. This will be beneficial in the latter half of the semester. This allows us to work with barrier options and other exotic financial instruments.

I define a process $𝑀 (𝑡) = {max}_{0 \leq 𝑢 \leq 𝑡} 𝑊 (𝑤_{0})$

This process takes the largest position of my Brownian motion. Is this Markov? In order for this to be the case, the only thing we need to calculate future valuations of my process is simply knowing the most recent value of my process. We could have a different Brownian motion that reaches the same peaks, but ends up being much lower at the end. In that green path, it would be much less likely for our maximum to date to change, because it’s much further from its current maximum than the blue path.

But we could create a multidimensional process that is Markov that incorporates both the Brownian motion and the maximum to date: $(𝑊 (𝑡), 𝑀 (𝑡))$ , which is 2-D Markov. Just knowing the maximum gives us no idea, we need both.

An interesting thing happens when we’re working with these maximums.

$𝑀 (𝑡) \geq 𝑚 \Leftrightarrow 𝜏_{𝑚} \leq 𝑡$ for $𝑚 > 0$

Why? Remember that $𝜏_{𝑚}$ is first passage time. I know that I’ve reached level $𝑚$ before time $𝑡$ , which means that the Brownian motion reached a maximum of at least $𝑚$ , which means the maximum at time $𝑡$ must be greater than or equal to that.

We can rewrite our reflection equality with the maximum to date with this property.

$ℙ (𝑀 (𝑡) \geq 𝑚, 𝑊 (𝑡) \leq 𝑤) = ℙ (𝑊 (𝑡) \geq 2 𝑚 - 𝑤)$

This is simply the reflection equality replacing one with the other.

If this is true, we can use this to get the joint distribution between Brownian motion and its maximum. Assume the joint distribution exists: $𝑓_{𝑀 (𝑡), 𝑊 (𝑡)} (𝑚, 𝑤)$

If this is my joint distribution with respect to these dummy variables, I could rewrite the probability measures in my reflection equality:

$\int_{- \infty}^{𝑤} \int_{𝑚}^{\infty} 𝑓_{𝑀 (𝑡), 𝑊 (𝑡)} (𝑥, 𝑦) 𝑑 𝑥 𝑑 𝑦 = \int_{2 𝑚 - 𝑤}^{\infty} \frac{1}{\sqrt{2 𝜋 𝑡}} 𝑒^{\frac{- 𝑧^{2}}{2 𝑡}} 𝑑 𝑧$

Now I can take the derivative of both sides with respect to $𝑚$

$- \int_{- \infty}^{𝑤} 𝑓_{𝑀 (𝑡), 𝑊 (𝑡)} (𝑚, 𝑦) 𝑑 𝑦 = - \frac{1}{\sqrt{2 𝜋 𝑡}} 𝑒^{\frac{- (2 𝑚 - 𝑤)^{2}}{2 𝑡}} (2)$

We can take the derivative now with respect to $𝑤$ (and get rid of negatives)

$𝑓_{𝑀 (𝑡), 𝑊 (𝑡)} (𝑚, 𝑤) = \frac{2}{\sqrt{2 𝜋 𝑡}} 𝑒^{\frac{- (2 𝑚 - 𝑤)^{2}}{𝑤 𝑡}} (\frac{- 2 (2 𝑚 - 𝑤) (- 1)}{2 𝑡})$

This ends up being

$= \frac{2 (2 𝑚 - 𝑤)}{𝑡 \sqrt{2 𝜋 𝑡}} 𝑒^{- (2 𝑚 - 𝑤)^{2} / (2 𝑡)}$

This is the joint density of $(𝑀 (𝑡), 𝑊 (𝑡)$

We also have the conditional distribution given that Brownian motion is normal, so we can take the joint over the marginal:

$𝑓_{𝑀 (𝑡) | 𝑊 (𝑡)} (𝑚 | 𝑤) = \frac{2 (𝑤 𝑚 - 𝑤)}{𝑡} 𝑒^{- \frac{(2 𝑚 - 𝑤)^{2}}{2 𝑡}}$

We will not worry about this too much until later. This process is mostly useful because it’s an example of an adaptive stochastic process which isn’t Markov, although most are.

Stochastic Calculus (Integrands)

Now it’s the big kahuna.

What does this mean: $\int_{0}^{𝑇} Δ (𝑡) 𝑑 𝑊 (𝑡) = ?$

In order for this to be understood, $Δ (𝑡)$ needs to be an adapted stochastic process.

Let’s go back to Riemann. The Riemann sum integrals mean you take some function where we chop up the x into little pieces. We take observations of the function at each $𝑥$ . What we’re going to do in this course is still this. It’s just that we’re also going to be working with a Brownian motion. Instead of getting a $Δ 𝑥$ , we could get our $Δ 𝑊 (𝑡)$ . You take the heights from the result of the adapted stochastic process, but the change comes from the stochastic process. However, every Brownian motion has a different result, it’s random. This means that the stochastic integral itself is a random variable. Why do we care? Let’s start with a specific, motivated example.

A simple process is one that given a partition, $Δ (𝑡)$ is a constant.

The random walk is a simple process.

Given a partition of $𝑥$ , you take one value that you hold at every partition piece which changes randomly.

If we put ourselves under the constraint that $Δ (𝑡)$ is a simple adapted process.

We’re going to construct a hypothetical. We have a stock. The value of the stock is given by Brownian motion $𝑊 (𝑡)$ . $Δ (𝑡)$ is the position I hold in the stock, with these rules: At the beginning of the day, I take the position of the stock and hold it for the rest of the day, until the next day. I have no idea what position I’m going to hold tomorrow.

Let’s start at day 1

$𝐼 (𝑡) = Δ (𝑡_{0}) (𝑊 (𝑡) - 𝑊 (𝑡_{0}))$

Hanging out here in Day 1 $𝑡_{1} \leq 𝑡 < 𝑡_{2}$ I’ve made

$𝐼 (𝑡) = Δ (𝑡_{1}) (𝑊 (𝑡) - 𝑊 (𝑡_{1})) + Δ (𝑡_{0}) (𝑊 (𝑡_{1}) - 𝑊 (𝑡_{0}))$

from the previous day as well.

Check on Day 3

$𝐼 (𝑡) = Δ (𝑡_{2}) (𝑊 (𝑡) - 𝑊 (𝑡_{2})) + Δ (𝑡_{1}) (𝑊 (𝑡) - 𝑊 (𝑡_{1})) + Δ (𝑡_{0}) (𝑊 (𝑡_{1}) - 𝑊 (𝑡_{0}))$

For Day K $𝑡_{𝑘} \leq 𝑡 < 𝑡_{𝑘 + 1}$ we have made since we started investing

$𝐼 (𝑡) = Δ (𝑡_{𝑘}) (𝑊 (𝑡) - 𝑊 (𝑡_{𝑘})) + \sum_{𝑖 = 0}^{𝑘 - 1} Δ (𝑡_{𝑖}) (𝑊 (𝑡_{𝑖 + 1}) - 𝑊 (𝑡_{𝑖}))$

This essentially is the size of my position which is multiplied by

I want to look at thai integral

$\int_{𝑡_{𝑗}}^{𝑡_{𝑗 + 1}} Δ (𝑢) 𝑑 𝑊 (𝑢)$

If I know that my $Δ$ is a simple process, it’s going to take a value at the beginning of my time increment and hold that value for the entire time.

$= \int_{𝑡_{𝑗}}^{𝑡_{𝑗 + 1}} Δ (𝑡_{𝑗}) 𝑑 𝑊 (𝑢)$

How does my $Δ (𝑡_{𝑗})$ change as I move? It doesn’t, so I can pull it out

$= Δ (𝑡_{𝑗}) \int_{𝑡_{𝑗}}^{𝑡_{𝑗 + 1}} 𝑑 𝑊 (𝑢)$

which by the fundamental theorem of calculus is

$= Δ (𝑡_{𝑗}) (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗}))$

Therefore, I can rewrite the sum in my return calculation as this integral.

If I make these substitutions, I have that

$𝐼 (𝑡) = \int_{𝑡_{𝐾}}^{𝑡} Δ (𝑢) 𝑑 𝑊 (𝑢) + \sum_{𝑖 = 0}^{𝑘 - 1} \int_{𝑡_{𝑖}}^{𝑡_{𝑖 + 1}} Δ (𝑢) 𝑑 𝑊 (𝑢)$

But obviously the summation of an integral is still an integral.

$= \int_{0}^{𝑡} Δ (𝑢) 𝑑 𝑊 (𝑢)$

What does it mean to integrate a process other than Brownian motion? We shall see.

We can wave our hands a bit about the trading of a stock. You could trade at the beginning of the day and at lunch, or every minute. Not the best idea but you could do that. This time works with a partition. This is now our Ito integral. What makes it an Ito integral is that our partition pieces are from $𝑡_{𝑘}$ to $𝑡_{𝑘 + 1}$ , where they include the left and take that value, and exclude the right. We are always taking the left Riemann integral. I can’t decide at lunch what price I bought the stock at in the morning. I can only make those choices based on the information I had at that time. That’s why adapted processes are a big deal and everything needs to be measurable. If you trade with unavailable information, that is insider trading, very illegal. So that’s why we have to use Ito integrals.

Interestingly, these Ito integrals are martingales.

Proof that Ito Integral is a Martingale

Prove that $𝐼 (𝑡)$ is a martingale.

Here’s our situation. We have an arbitrary partition of our time, for which there exists some $𝑘$ such that $𝑡$ belongs to that partition piece $𝑡 \in [𝑡_{𝑘}, 𝑡_{𝑘 + 1}]$ . To prove that this is a martingale, I need to prove for that $𝑠 \leq 𝑡$ ,

$𝔼 [𝐼 (𝑡) | ℱ︀ (𝑠)] = 𝐼 (𝑠)$

This has to work whether $𝑠$ is in the same time piece or in an earlier one, so there are two different cases to account for.

Case 1: $\exists 𝑙 < 𝑘 s.t. 𝑠 \in [𝑡_{𝑙}, 𝑡_{𝑙 + 1}]$

It could be in the one at the start or the one just before.

I’m going to be able to say that

$𝐼 (𝑡) = \sum_{𝑗}^{𝑙 - 1} Δ (𝑡_{𝑗}) (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗})) + Δ (𝑡_{𝑙}) (𝑊 (𝑡_{𝑙 + 1}) - 𝑊 (𝑡_{𝑙})) + \sum_{𝑗 = 𝑙 + 1}^{𝑘 - 1} Δ (𝑡_{𝑗}) (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗})) + Δ (𝑡_{𝑘}) (𝑊 (𝑡) - 𝑊 (𝑡_{𝑘}))$

All days before $𝑙$ , day $𝑙$ , and all days from $𝑙$ to $𝑘$ p to how far we are into day $𝑘$

Because this is a sum, we can split this up by linearity of expectations:

$𝔼 [𝐴 | ℱ︀ (𝑠)] = 𝔼 [\sum_{𝑗 = 0}^{𝑙 - 1} Δ (𝑡_{𝑗}) (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗})) | ℱ︀ (𝑠)]$

We will use the properties of expectations here. I would say at this point that linearity is already used, but let’s try something else. What is the latest time index referred to in this summation? It’s $𝑙$ . But here’s the thing. Because $𝑠 \in [𝑡_{𝑙}, 𝑡_{𝑙 + 1}]$ , then $𝑠 \geq 𝑡_{𝑙}$ . So everything is known, so we can take out everything and leave 1.

$= 𝐴$

Let’s try $𝐵$

$𝔼 [𝐵 | ℱ︀ (𝑠)] = 𝔼 [Δ (𝑡_{𝑙}) (𝑊 (𝑡_{𝑙 + 1}) - 𝑊 (𝑡_{𝑙}) | ℱ︀ (𝑠)]$

$= Δ (𝑡_{𝑙}) 𝔼 [𝑊 (𝑡_{𝑙 + 1}) - 𝑊 (𝑡_{𝑙}) | ℱ︀ (𝑠)]$

Let’s brute force this with linearity, and then do the elegant solution later

$= Δ (𝑡_{𝑙}) [𝔼 [𝑊 (𝑡_{𝑙 + 1}) | ℱ︀ (𝑠)] - 𝔼 [𝑊 (𝑡_{𝑙}) | ℱ︀ (𝑠)]]$

Generally in order to break up these problems we try to add a 0 to break it up into an independent piece and a measurable piece.

$𝔼 [𝑊 (𝑡_{𝑙 + 1}) - 𝑊 (𝑠) + 𝑊 (𝑠) | ℱ︀ (𝑠)]$

We know future increments are independent. Therefore, we can use linearity to split them, independence to drop the first condition, and measurability to take out what’s known from the second part.

We could have solved this with a single step if we remember that Brownian motion is a martingale.

Now let’s consider $𝐶$

$𝔼 [Δ (𝑡_{𝑗}) (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗})) | ℱ︀ (𝑠)]$

We know that the Brownian motion part is independent, but the $Δ$ isn’t. And none of this is measurable. We can use iterated conditioning, though. For $ℱ︀ (𝑠) \subseteq ℱ︀ (𝑡)$

$𝔼 [𝔼 [𝑋 | ℱ︀ (𝑡)] | ℱ︀ (𝑠)] = 𝔼 [𝑋 | ℱ︀ (𝑠)]$

I can do this by using the property of filtrations.

$= 𝔼 [𝔼 [Δ (𝑡_{𝑗}) (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗})) | ℱ︀ (𝑡_{𝑗})] | ℱ︀ (𝑠)]$

We have a hard time picking this time, because it’s true for a lot of them. We know that it’s time $𝑡$ or later. My adapted process is measurable with respect to time $𝑡_{𝑗}$ or later. So we can use $𝑡_{𝑗}$ . We can take out what’s known now.

$= 𝔼 [Δ (𝑡_{𝑗}) 𝔼 [𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗}) | ℱ︀ (𝑡_{𝑗})] | ℱ︀ (𝑠)]$

We know that the stochastic process is independent and that it’s a martingale, so we can get rid of that whole thing and call it 0.

$= 0$

So the conditional expectation of $𝐶$ is 0.

Finally, let’s consider $𝐷$ .

$𝔼 [𝐷 | ℱ︀ (𝑠)] = 𝔼 [Δ (𝑡_{𝑘}) (𝑊 (𝑡) - 𝑊 (𝑡_{𝑘})) | ℱ︀ (𝑠)]$

We can use the same exact proof for $𝐶$ that uses iterated conditioning for $𝑡_{𝑘}$ .

We have just proven that we can get rid of C and D, so our final expectation is

$𝔼 [𝐼 (𝑠)] = \sum_{𝑗 = 0}^{𝑙 - 1} Δ (𝑡_{𝑗}) (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗})) + Δ (𝑡_{𝑙}) (𝑊 (𝑠) - 𝑊 (𝑡_{𝑙}))$

We need to touch all of our spinning plates to understand why we’re doing. When $𝑠$ occurs on day $𝑙$ . If we think about this intuitively, this value is the same as $𝐼 (𝑠)$ because it is all the spinning plates going up to $𝑙 - 1$ plus the time into $𝑙$ that comes to $𝑠$ . That means

$= 𝐼 (𝑠)$

So it’s a martingale

Case 2 $𝑠 < 𝑡; 𝑠, 𝑡 \in [𝑡_{𝑘}, 𝑡_{𝑘 + 1})$

We are trying to figure out

$𝔼 [𝐼 (𝑡) | ℱ︀ (𝑠)] =$

Knowing that $𝑠$ and $𝑡$ occur on the same day, we don’t need all those pieces. So this is simply

$= 𝔼 [\sum_{𝑗 = 0}^{𝑘 - 1} Δ (𝑡_{𝑗}) (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗})) + Δ (𝑡_{𝑘}) (𝑊 (𝑡) - 𝑊 (𝑡_{𝑘})) | ℱ︀ (𝑠)]$

What is the relationship of $𝑠$ to time $𝑡_{𝑘}$ ? It’s greater than or equal to.

In this summation, all of these pieces are measurable because $𝑠$ is greater than $𝑘$ .

$= \sum_{𝑗 = 0}^{𝑘 - 1} Δ (𝑡_{𝑗}) (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗})) + 𝔼 [Δ (𝑡_{𝑘}) (𝑊 (𝑡) - 𝑊 (𝑡_{𝑘})) | ℱ︀ (𝑠)]$

By taking out what is known, we can get

$= \sum_{𝑗 = 0}^{𝑘 - 1} Δ (𝑡_{𝑗}) (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗})) + Δ (𝑡_{𝑘}) 𝔼 [𝑊 (𝑡) - 𝑊 (𝑡_{𝑘}) | ℱ︀ (𝑠)]$

This is not a future increment. The time is smack dab in the middle of the increment. In this case, you should add a 0, which allows you to take out what is known and show independence.

$= \sum_{𝑗 = 0}^{𝑘 - 1} Δ (𝑡_{𝑗}) (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗})) + Δ (𝑡_{𝑘}) (𝑊 (𝑠) - 𝑊 (𝑡_{𝑘}))$

which is just

$= 𝐼 (𝑠)$

Thus, we prove that the Ito integral is a martingale for all cases.

The fact that the Ito integral is a martingale is a lifesaver later int eh course. That means that $𝐼 (0) = 0$

Ito Isometry

We will now look at the expectation of $𝐼^{2}$

$𝔼 [𝐼^{2} (𝑡)] = 𝔼 [\int_{0}^{𝑡} Δ^{2} (𝑢) 𝑑 𝑢]$

This gives me the variance of the Ito integral. Let’s prove that this is the case.

Let’s do a slight modification of $𝐼 (𝑡)$ for the purpose of these proofs.

$𝐷_{𝑗} = {𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗}), for 𝑗 \in {0, 1, \dots 𝑘 - 1} 𝑊 (𝑡) - 𝑊 (𝑡_{𝑘}), for 𝑗 = 𝑘$

Which we do so that we can define

$𝐼 (𝑡) = \sum_{𝑗 = 0}^{𝑘} Δ (𝑡_{𝑗}) 𝐷_{𝑗}$

$𝐼^{2} (𝑡) = (\sum_{𝑗 = 0}^{𝑘} Δ (𝑡_{𝑗}) 𝐷_{𝑗}) (\sum_{𝑗 = 0}^{𝑘} Δ (𝑡_{𝑗}) 𝐷_{𝑗})$

I’m also going to observe at this point.

Let’s think a little about how matrices work. We could envision a matrix like this:

$= \sum_{𝑗 = 0}^{𝑘} Δ^{2} (𝑡_{𝑗}) 𝐷_{𝑗}^{2} + 2 \sum_{0 \leq 𝑖 < 𝑗 \leq 𝑘} Δ (𝑡_{𝑖}) Δ (𝑡_{𝑗}) 𝐷_{𝑖} 𝐷_{𝑗}$

Let’s think about filtrations

$ℱ︀ (𝑡_{𝑗}) : Δ (𝑡_{𝑖}) Δ (𝑡_{𝑗}) 𝐷_{𝑖}$ is measurable

$𝐷_{𝑗}$ is a future value and therefore is independent

We must have independence between random variables if one of them is measurable by a $𝜎$ -algebra and the other is independent of that $𝜎$ -algebra.

$𝔼 [𝐼^{𝑤} (𝑡)] = 𝔼 [\sum_{𝑗 = 0}^{𝑘} Δ^{2} (𝑡_{𝑗}) 𝐷_{𝑗}^{2}] + 2 𝔼 [\sum_{0 \leq 𝑖 < 𝑗 \leq 𝑘} Δ (𝑡_{𝑖}) Δ (𝑡_{𝑗}) 𝐷_{𝑖} 𝐷_{𝑗}]$

We can use linearity of expectations here

$= \sum_{𝑗 = 0}^{𝑘} 𝔼 [Δ^{2} (𝑡_{𝑗})] 𝔼 [𝐷_{𝑗}^{2}] + 2 \sum_{0 \leq 𝑖 < 𝑗 \leq 𝑘} 𝔼 [Δ (𝑡_{𝑖}) Δ (𝑡_{𝑗}) 𝐷_{𝑖}] 𝔼 [𝐷_{𝑗}]$

Since $𝐷_{𝑗}$ is Brownian motion, the expected value is 0 which makes that entire sum 0, and the expectation of $𝐷_{𝑗}^{2}$ is the variance, which is known to be $𝑡$

$= \sum_{𝑗 = 0}^{𝑘} 𝔼 [Δ^{2} (𝑡_{𝑗})] (𝑡_{𝑗 + 1} - 𝑡_{𝑗})$

Let’s take it the other way with linearity

$= 𝔼 [\sum_{𝑗 = 0}^{𝑘} Δ^{2} (𝑡_{𝑗}) (𝑡_{𝑗 + 1} - 𝑡_{𝑗})]$

What do you call it when you add up all the area of those rectangles? This is the definition of a Riemann integral! This is nearly

$= 𝔼 [\int_{0}^{𝑡} Δ^{2} (𝑢) 𝑑 𝑢]$

Which is the variance of any Ito integral.

All we know about $Δ$ is that it’s an adapted stochastic process, so we can’t simplify it further. But we will get one numeric value for the variance. We can further prove that this is the quadratic variation for an Ito integral.

$[𝐼, 𝐼] (𝑡) = \int_{0}^{𝑡} Δ^{2} (𝑢) 𝑑 𝑢$

This is similar to the isometry except that it’s not the expectation. This is because variation can be a random variable. Why do these look so similar then? Let’s prove it.

Let’s say I have a partition defined as

$Π = {𝑡_{0}, 𝑡_{1}, 𝑡_{2}, \dots, 𝑡_{𝑘 + 1}}$

$Π_{𝑖} = {𝑠_{0} = 𝑡_{𝑖}, 𝑠_{1}, 𝑠_{2}, 𝑠_{3}, \dots, 𝑠_{𝑛 + 1} = 𝑡_{𝑖 + 1}}$

Essentially, let’s create a new sub-partition for each partition piece.

$[𝐼, 𝐼] (𝑡) = {lim}_{∥ Π ∥ \to 0} \sum_{𝑗 = 0}^{𝑚 - 1} (𝐼 (𝑡_{𝑗 + 1}) - 𝐼 (𝑡_{𝑗}))^{2}$

I want to note that the difference in the sum, assuming that we are in a single time partition piece, means that

$= {lim}_{𝑖 = 0}^{𝑘 + 1} {lim}_{∥ Π ∥ \to 0} \sum_{𝑗 = 0}^{𝑛 - 1} (𝐼 (𝑠_{𝑗 + 1}) - 𝐼 (𝑠_{𝑗}))^{2}$

variation over entire partition is sum of variation over each sub-partition.

Let’s look just at the difference in the sum. If we expand this squared part and eliminate it, we can show the required cancellation (Open question: work through the algebra explicitly).

We know that $Δ (𝑡_{𝑖})$ is going to take one value and hold it because it is a simple process.

This limit is the very definition of the quadratic variation of Brownian motion over that time piece, so we can simplify it to time.

$= \sum_{𝑖 = 0}^{𝑘 - 1} Δ^{2} (𝑡_{𝑖}) (𝑡_{𝑖 + 1} - 𝑡_{𝑖})$

which because it’s a simple process is

$= \int_{0}^{𝑡} Δ^{2} (𝑢) 𝑑 𝑢$

Keep in mind that there are no expectations because we are looking for variation not variance

Square Integrability

Very important consideration! Our Ito integrals are only going to exist if this process $Δ (𝑡)$ is such that the variance is finite, since infinite variance is not a well-defined concept. This is known as the square-integrability condition.

Once that is satisfied, we can define $Δ_{𝑛} (𝑡)$ as a sequence of simple processes which together converge to a non-simple process $Δ (𝑡)$ in this way

${lim}_{𝑛 \to \infty} 𝔼 [\int_{0}^{𝑇} | Δ_{𝑛} (𝑡) - Δ (𝑡) |^{2} 𝑑 𝑡] = 0$

Then I can define my Ito integral as

$\int_{0}^{𝑡} Δ (𝑢) 𝑑 𝑊 (𝑢) = {lim}_{𝑛 \to \infty} \int_{0}^{𝑡} Δ_{𝑛} (𝑢) 𝑑 𝑊 (𝑢)$

If you can pull all this off, you get these properties:

Continuity. The integral is continuous. Brownian motion is continuous, and the integral over the short interval $[𝑠, 𝑡]$ goes to zero as $𝑡$ approaches $𝑠$ . Thus a discontinuity in the adapted integrand does not create a jump in the integral.
Adaptivity: every piece is measurable
Linearity: these are all just summations, so you can apply all linearity properties
Martingale (proven)
Ito isometry (proven)
Quadratic variation (proven)

The first stochastic integral:

$\int_{0}^{𝑡} 𝑊 (𝑢) 𝑑 𝑊 (𝑢)$

essentially

$\int 𝑥 𝑑 𝑥$

but it does something a little weird, not just the reverse power rule.

The result of this integral is entirely due to the fact that quadratic variation is nonzero.

Next week we will prove this.

## A CUPS Vulnerability Explained - URL: https://sharifhsn.dev/blog/linkedin-2024-09-27-cups-linux-vulnerability/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2024-09-27-cups-linux-vulnerability.json - Description: An overview of the 2024 CUPS printer-discovery vulnerability and its risk. - Date: 2024-09-27 - Exact published timestamp: 2024-09-27T11:29:52.298Z - Topics: Cybersecurity, Linux - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/posts/sharif-haason_attacking-unix-systems-via-cups-part-i-activity-7245394208822300672-9IQ7 - Content source format: typst

A 9.9 Linux CVE was discovered! Heartbleed was only 8.5, so you know that this is something to pay attention to. If you know any server administrators, text them some encouragement as they’ll surely be going through it at work today. Here’s some more detail on what the exploit is: ⬇️

CUPS is the Common Unix Printing System, a general service for printers that is used widely by Unix-like systems, including most distributions of Linux and macOS. The vulnerability was found in its printer discovery service cups-browsed. CUPS listens for packets at port 631 from any host. These packets trigger printer discovery, where CUPS tries to get the printer attributes from this arbitrary server. It can set its own name and the URL for the printing protocol to access, which could be an attacker-controlled host. There is also a printer attribute called FoomaticRIPCommandLine, which arbitrarily executes its string value when a print job is executed. This exists to provide flexibility for systems that need configuration before they can print… but it’s also a huge attack vector and has been the site of multiple previous CVEs.

In essence: any computer connected to the public Internet is vulnerable to an attacker adding a malicious printer to it. If a user executes a print job with the malicious printer, then the arbitrary code, which could be malware, is executed. Keep in mind that the attacker can name the printer whatever they like. How many of you would think twice before printing to “HP DeskJet”?

How is this mitigated? The easiest way to do this is to disable the cups-browsed service. Automatic printer discovery is more of a nice-to-have than a need, and with such a glaring security flaw, the correct decision is obvious. In general, considering how flawed CUPS is, you should avoid printing from Unix-like systems when possible.

Want to know more? The author of the CVE published an excellent article that goes over all of the details of the exploit: Attacking UNIX Systems via CUPS, Part I.

## Brownian Motion - URL: https://sharifhsn.dev/blog/brownian-motion/ - Structured data: https://sharifhsn.dev/api/posts/brownian-motion.json - Description: The most important properties of Brownian motion are that it is a martingale (the expected value of the process is 0) and it accumulates quadratic variation at a rate of one per… - Date: 2024-09-26 - Exact published timestamp: 2024-09-26 - Topics: Stochastic Calculus, Brownian Motion, Quadratic Variation - Categories: Stochastic Calculus - Source: FE-610 \| Stochastic Calculus - Source URL: None - Content source format: typst

Week 3

Reading - Brownian Motion (Shreve 3)

The most important properties of Brownian motion are that it is a martingale (the expected value of the process is 0) and it accumulates quadratic variation at a rate of one per unit time.

Scaled Random Walk

The symmetric random walk is given by a fair coin which goes up 1 on heads and down 1 on tails. The increments are independent from each other. The expected value is zero. Because the increments are independence, we can show that the symmetric random walk is a martingale.

Brownian Motion

A Brownian motion is a continuous function where the increments at time $𝑡$ are independent and each of the increments is normally distributed with expectation 0 (martingale) and variance $𝑡_{𝑖 + 1} - 𝑡_{𝑖}$ . The difference between this and the scaled walk is that the scaled walk is linear between each time step, but Brownian motion is never linear. You can think of it as an infinitely scaled walk.

The $𝜔$ in this case is the path of the Brownian motion.

We can say with respect to information, it accumulates, and the path at any time can be described by the current filtration, which is what sets the path must fall into by probability measure.

Lecture Notes

Where we left off last week, we were talking about the quadratic variation of Brownian motion where we could prove that it was $𝑡$ almost surely.

$[𝑊, 𝑊] (𝑡) = 𝑡$

What are the implications of this? We could think of this as a function of time. The quadratic variation grows as a function of time with value $𝑡$ . This Brownian motion is the limit of the scaled symmetric random walk. This process is jagged everywhere. If you were to zoom in on this process, it would appear just as jagged because it’s constantly moving up and down. If there’s no place where it’s smooth, there’s no place where it’s differentiable. Does that mean there’s no way to talk about derivatives? We will now only concern ourselves with differentials, not derivatives. Minor nit in English: differentiable actually means the derivative exists, not the differential.

Let’s say we have a stochastic process $𝑋 (𝑡)$ ? We will define the differential of this process $𝑑 𝑋 (𝑡)$

$𝑑 𝑋 (𝑡) = {lim}_{𝛿 \to 0^{+}} (𝑋 (𝑡 + 𝛿) - 𝑋 (𝑡))$

This is the instantaneous change, not the instantaneous rate of change (the derivative).

This will exist for any function, smooth or not.

We are defining the differential of Brownian motion as $𝑑 𝑊 (𝑡)$ . Let’s point out something that comes about from looking at these differentials. We established last week that the quadratic variation is just time. BUt let’s think about the differential of the quadratic variation of time. It is a random variable indexed by time i.e. a stochastic process, so we can take the differential of it.

$𝑑 [𝑊, 𝑊] (𝑡) = {lim}_{𝛿 \to 0^{+}} ([𝑊, 𝑊] (𝑡 + 𝛿) - [𝑊, 𝑊] (𝑡))$

I’m looking at this second order variation up until time $𝑡 + 𝛿$ , then time $𝑡$ , so I just end up with the change between $𝑡$ and $𝑡 + 𝛿$ .

$= {lim}_{𝛿 \to 0^{+}} (𝑡 + 𝛿 - 𝑡)$

$= {lim}_{𝛿 \to 0^{+}} (𝛿)$

You could call that 0, but it’s not really a 0 as far as differentials are concerned. This is change of time, i.e. $𝑑 𝑡$ .

This is sort of equivalent to saying that you have a partition consisting of just the points

$Π = {𝑡, 𝑡 + 𝛿}$

and if you want to do quadratic variation over this region, it would be

${lim}_{∥ Π ∥ \to 0} \sum_{𝑗 = 0}^{𝑛 - 1} (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗}))^{2}$

(formula for second order variation) but if $𝛿$ goes to 0, the norm must go to 0, right?

I could have just said it’s the limit

$= {lim}_{𝛿 \to 0^{+}} (𝑊 (𝑡 + 𝛿) - 𝑊 (𝑡))^{2}$

The endpoint minus the beginning point, squared.

But this is just the differential Brownian motion squared

$= (𝑑 𝑊 (𝑡))^{2}$

We have a consequence of quadratic variation of Brownian motion being $𝑡$ , we can express $(𝑑 𝑊 (𝑡))^{2} = 𝑑 𝑡$ . And this gives us a very necessary substitution when we are doing actual stochastic calculus. The reason for that is the nonzero quadratic variation of Brownian motion.

In addition to quadratic variation, we are potentially interested in cross variation, which is related to quadratic variation in the same way that covariance is related to variance. It’s defined as

for processes $𝑋 (𝑡)$ and $𝑌 (𝑡)$ , the cross variation is

$[𝑋, 𝑌] (𝑡) = {lim}_{| | Π | | \to 0} \sum_{𝑗 = 0}^{𝑛 - 1} (𝑋 (𝑡_{𝑗 + 1}) - 𝑋 (𝑡_{𝑗})) (𝑌 (𝑡_{𝑗 + 1}) - 𝑌 (𝑡_{𝑗}))$

If I let $𝑋 (𝑡))$ be Brownian motion $𝑊 (𝑡)$ and $𝑌 (𝑡)$ equal to time, I could get the cross variation of Brownian motion and time in this manner.

$= {lim}_{| | Π | | \to 0} \sum_{𝑗 = 0}^{𝑛 - 1} (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗})) (𝑡_{𝑗 + 1} - 𝑡_{𝑗})$

I’m going to pull out the biggest piece of my Brownian motion and say

$\leq {lim}_{| | Π | | \to 0} {max}_{0 \leq 𝑘 \leq 𝑛 - 1} | 𝑊 (𝑡_{𝑘 + 1}) - 𝑊 (𝑡_{𝑘}) | \sum_{𝑗 = 0}^{𝑛 - 1} (𝑡_{𝑗 + 1} - 𝑡_{𝑗})$

I could apply the limit to each of those pieces separately. But I don’t have to apply the limit because this sum is just $𝑡$ .

This limit comes directly from the fact that Brownian motion is continuous. There’s no place for discontinuity. Because of that, the limit must be 0.

So we can confirm that the cross variation between a continuous stochastic process, and a continuous differentiable stochastic process must be 0, because the differentiability leads it to being 0.

I know that this cross variation is less than equal to that expression. That gives me the upper bound on cross variation. I also know that

$[𝑊, 𝑡] (𝑡) \geq {lim}_{| | 𝑃 𝑖 | | \to 0} - {max}_{0 \leq 𝑘 \leq 𝑛 - 1} | 𝑊 (𝑡_{𝑘 + 1}) - 𝑊 (𝑡_{𝑘}) | \sum_{𝑗 = 0}^{𝑛 - 1} (𝑡_{𝑗 + 1} - 𝑡_{𝑗})$

I can make the same exact argument, which means it must be 0. We’re not actually using the norm of the partition here, we’re just looking for the biggest piece, which must be 0 because the change in Brownian motion goes to 0.

$𝑑 [𝑋, 𝑌] (𝑡) = 𝑑 𝑋 (𝑡) 𝑑 𝑌 (𝑡)$

But, since I know that the cross variation between time and Brownian motion is 0, I know that that value must be 0. Even more, because I know that the quadratic variation of any continuous differentiable process is 0, I know that the quadratic variation of $𝑑 𝑡$ is 0, so $(𝑑 𝑡)^{2}$ is 0.

Next slides will have hand waving:

Joint normal distribution

Technically, if you took a rigorous probability theory class, you learned about the joint normal distribution. Looking at normal variables that are not independent of each other. In that sense, we define the joint normal distribution through a vector of the means and a matrix of the covariances. We bring this up to stress the idea that if we have increments that are independent and normally distributed, the variables are jointly normally distributed. If we wanted to look at $𝐶 𝑜 𝑣 (𝑊 (𝑠), 𝑊 (𝑡))$ where $𝑠 < 𝑡$ :

$𝔼 [𝑊 (𝑠) 𝑊 (𝑡)] - 𝔼 [𝑊 (𝑠)] 𝔼 [𝑊 (𝑡)]$

We know that $𝑊 (𝑠)$ is the segment to time $𝑠$ , so the expectation must be 0 because all of these are 0. So it just becomes

$𝔼 [𝑊 (𝑠) 𝑊 (𝑡)$

They overlap, so they’re definitely not independent. How do we figure this out? Let’s add a zero.

$= 𝔼 [𝑊 (𝑠) 𝑊 (𝑡) - 𝑊^{2} (𝑠) + 𝑊^{2} (𝑠)]$

using linearity of expected values

$= 𝔼 [𝑊 (𝑠) 𝑊 (𝑡) - 𝑊^{2} (𝑠)] + 𝔼 [𝑊^{2} (𝑠)]$

we can write this as

$= 𝔼 [𝑊 (𝑠) (𝑊 (𝑡) - 𝑊 (𝑠))] + 𝔼 [𝑊^{2} (𝑠)]$

and write this as

$= 𝔼 [𝑊 (𝑠)] 𝔼 [𝑊 (𝑡) - 𝑊 (𝑠)] + 𝔼 [𝑊^{2} (𝑠)]$

because that expectation is 0, that means that we’re just looking at the second moment, which is the variance of s, which is just s

$= 𝑠$

It turns out that the covariance of $𝑊 (𝑠)$ and $𝑊 (𝑡)$ is just $𝑠$ , the smaller time.

If I were to create jointly normally distributed variables, with the covariance structure of just the minimums of the time and means equal to 0, this is an identical declaration of Brownian motion. You could also use joint moment generating functions (but this is too complicated for us to go over right now).

We will only be using definition 3.3.1 for Brownian motion. You MUST understand this by heart.

Filtration

Once we have a Brownian motion, now we can define a filtration on that probability space. It’s a collection of $𝜎$ -algebras, the previous sets are included in the next set. The Brownian motion $𝑊 (𝑡)$ must be $ℱ︀ (𝑡)$ -measurable, which means we can use the information at time $𝑡$ to know that Brownian motion, and the increment after $𝑊 (𝑡)$ is totally independent of the filtration. We can have a stochastic process $𝛿 (𝑡)$ that is adapted to the filtration if the variable is $ℱ︀ (𝑡)$ . All the stochastic processes we are considering are adapted. We can only consider past data because we live in the real world, so we will use information at that time.

Brownian motion is a martingale. The argument is the same for the symmetric random walk being a martingale.

It also turns out that Brownian motion will be Markov (consider this later).

Arguably, the most famous stock market model in the world is geometric Brownian motion. This is what the Black-Scholes model is based on.

$𝑆 (𝑡) = 𝑆 (0) 𝑒^{(𝛼 - \frac{1}{2} 𝜎^{2}) 𝑡 + 𝜎 𝑊 (𝑡)}$

What’s interesting is if we look at this formula, we can break down log returns in this manner.

You can think of it as having an initial stock price which is multiplied by an exponential function which gives you an incremental change. It could grow or shrink, but it never goes to 0. The $𝛼 - 𝜎^{2} / 2$ this is a drift. If it’s a positive term then the drift will increase. $𝜎 𝑊 (𝑡)$ is the uncertainty in the stock price, known as volatility. The $𝜎$ determines how much impact the noise has. If $𝜎$ is small, then the noise has a small impact. If it’s large, it has a large impact. This makes it very important, but it’s very difficult to observe. You could look at long-term behavior and get an idea of the drift, if we wanted to calibrate our alphas (?), but we can’t observe $𝜎$ . We can try through historical volatility, implied volatility, local volatility, and realized volatility.

Realized volatility says we observe the stock sufficiently often. Ideally as often as possible, hopefully tick data. If we have stock prices $𝑆 (𝑡_{0}), 𝑆 (𝑡_{1}), \dots 𝑆 (𝑡_{𝑚})$

I could create the log return of any of my stock prices:

$log (\frac{𝑆 (𝑡_{𝑗 + 1})}{𝑆 (𝑡_{𝑗})}) = (𝛼 - \frac{1}{2} 𝜎^{2}) (𝑡_{𝑗 + 1} - 𝑡_{𝑗}) + 𝜎 (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗}))$

Let’s say we have a vector of our log returns.

$\sum_{𝑗 = 0}^{𝑚 - 1} (log (\frac{𝑆 (𝑡_{𝑗 + 1})}{𝑆 (𝑡_{𝑗})}))^{2} = \sum_{𝑗 = 0}^{𝑚 - 1} [(𝛼 - \frac{1}{2} 𝜎^{2})^{2} (𝑡_{𝑗 + 1} - 𝑡_{𝑗})^{2} + 2 (𝛼 - \frac{1}{2} 𝜎^{2}) 𝜎 (𝑡_{𝑗 + 1} - 𝑡_{𝑗}) (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗})) + 𝜎^{2} (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗}))^{2}]$

Here’s where the statement of “often” comes into play. I could apply the summation separately, the drift becomes constant and it gets pulled out, and then you get the quadratic variation of time, which makes that first element 0

The second element is approximately the cross variation of Brownian motion and time, which is also approximately 0 (determined earlier), so you only end up with the last piece, which becomes the quadratic variation of Brownian motion, observed until time $𝑇$ which becomes $𝑇$ .

So, in essence, we get that the realized volatility is approximately equal to

$𝜎_{𝑅 𝑉} \approx \sqrt{\frac{1}{𝑇_{2} - 𝑇_{1}} \sum_{𝑗 = 0}^{𝑛 - 1} (log (\frac{𝑆 (𝑡_{𝑗 + 1})}{𝑆 (𝑡_{𝑗})}))^{2}}$

Implied volatility comes about from a later formula for pricing European options model.

We have a strike price, and all the stuff associated with option price, so we could find the volatility implied by the option price that exists.

Markov Property

One way we can think of the Markov property is that for an adapted stochastic process if the conditional probability distribution for future states dependent on everything that has happened up to that point of time is the same as for future states, just using the most recent number, that process is Markov.

We are able to prove that Brownian motion is Markov.

This might seem like a cheap, semantic argument, but he believes this argument has merit.

My problem is, for some function $𝑓$ , we’re going to look at

$𝔼 [𝑓 (𝑊 (𝑡)) | ℱ︀ (𝑠)]$

there’s very little we can say about martingales because we don’t know anything about $𝑓$ other than that it’s a function. We can’t talk about growth or decline or anything. Let’s think about how we could manipulate this.

In the past, you’ve seen the trick of adding the zero. That doesn’t work here because it’s wrapped in a function.

Let’s go to an “entirely different” problem now. Hypothetically, let’s say you have a dummy variable $𝑋$ , like from algebra, some number you don’t know, a placeholder, not a random variable.

$𝔼 [𝑓 (𝑊 (𝑡) - 𝑊 (𝑠) + 𝑥) | ℱ︀ (𝑠)]$

The only thing that is random is the increment of bBrownian motion. What do I know about the relation of Brownian motion to a filtration? It’s independent. IBecause this is independent, we can drop the condition:

$= 𝔼 [𝑓 (𝑊 (𝑡) - 𝑊 (𝑠) + 𝑥]$

even more obviously, the only random thing is the increment. It’s a normally distributed variable, using the definition of expectation it is

$= \int_{- \infty} \infty 𝑓 (𝑤 + 𝑥) \frac{1}{\sqrt{2 𝜋 (𝑡 - 𝑠)}} 𝑒^{\frac{- 𝑤^{2}}{2 (𝑡 - 𝑠)}} 𝑑 𝑤$

$= 𝑔 (𝑥)$

This is basically a function because $𝑤$ gets integrated out

For every function $𝑓$ , there is a $𝑔$ .

What would be $𝑔 (7)$ ?

$𝑓 (7) = 𝔼 [𝑓 (𝑊 (𝑡) - 𝑊 (𝑠) + 7) | ℱ︀ (𝑠)]$

This works with any dummy variable you put in.

What if you use $𝑊 (𝑠)$ ?

$𝑔 (𝑊 (𝑠)) = 𝔼 [𝑓 (𝑊 (𝑡)) | ℱ︀ (𝑠)]$

Markov is just saying that you have this function conditioned on the information $ℱ︀$ . This is a common type of proof in this class: we prove it for everything so it becomes true for one thing.

Transition Density

You can express this as a transition density. There is a way to manipulate this through that process instead of this.

Stopping Time

A stopping time $𝜏$ is a random variable that takes the value $[0, \infty]$ and satisfies that $𝔼 [𝜏 | ℱ︀ (𝑡)] = 𝑡$ , then $𝔼 [𝜏 | ℱ︀ (𝑢)] = 𝑡$ for $𝑢 > 𝑡$

In other words it’s a random variable that if there’s enough information that I’m stopping at time $𝑡$ , no future information can change that I’m stopping at time $𝑡$ .

An American option is the perfect example. You’re watching an instrument’s price, and you are tempted to exercise because the price is above the strike price. After, you see this amazing rally after you exercise, and you want to exercise now. There was enough information at the time you exercised that made you feel like you should have exercised. Nothing that happens after can change that fact.

Fact: A martingale stopped at a stopping time is a martingale.

Exponential Martingale

The process $𝑍 (𝑡)$ is

$𝑍 (𝑡) = 𝑒^{𝜎 𝑊 (𝑡) - \frac{1}{2} 𝜎^{2} 𝑡}$

I need to show that the expectation of the function will always be the same.

$𝔼 [𝑍 (𝑡) | ℱ︀ (𝑠)]$

$= 𝔼 [𝑒^{𝜎 𝑊 (𝑡) - \frac{1}{2} 𝜎^{2} 𝑡} | ℱ︀ (𝑠)]$

We are not going to have much luck with that zeroing out method, because this exponential function is not linear. It’s not like Brownian motion even, the increments may not be independent.

Let’s try multiplying by 1!

$= 𝔼 [𝑍 (𝑡) \times \frac{𝑍 (𝑠)}{𝑍 (𝑠)} | ℱ︀ (𝑠)]$

this is okay because $𝑍 (𝑠)$ can’t be 0. Never do that, only if you’re a physicist.

Rewrite as

$= 𝔼 [𝑍 (𝑠) \times \frac{𝑍 (𝑡)}{𝑍 (𝑠)} | ℱ︀ (𝑠)]$

$= 𝔼 [𝑒^{𝜎 (𝑊 (𝑡) - 𝑊 (𝑠)) - \frac{1}{2} 𝜎^{2} (𝑡 - 𝑠)} 𝑍 (𝑠) | ℱ︀ (𝑠)]$

Now that we have an increment of Brownian motion $𝑊 (𝑡) - 𝑊 (𝑠)$ , we can make statements about that. It’s independent of the filtration. Let’s pull the measurable piece out

$= 𝑍 (𝑠) 𝑒^{- \frac{1}{2} 𝜎^{2} (𝑡 - 𝑠)} 𝔼 [𝑒^{𝜎 (𝑊 (𝑡) - 𝑊 (𝑠))} | ℱ︀ (𝑠)$

Because of independence, we can disregard our condition:

$= 𝑍 (𝑠) 𝑒^{- \frac{1}{2} 𝜎^{2} (𝑡 - 𝑠)} 𝔼 [𝑒^{𝜎 (𝑊 (𝑡) - 𝑊 (𝑠))}$

THis is why we had the homework problem on the moment-generating function. This is effectively a moment-generating function. For any value like this, the expectation is $𝑍 (𝑠)$ , which makes it a martingale.

First Passage Time

There are some interesting behaviors to Brownian motion

A random variable that tells me the first time that my Brownian motion reaches a certain level $𝑚$ : $𝜏_{𝑚} = min {𝑡 \geq 0; 𝑊 (𝑡) = 𝑚}$

This variable is a stopping time. Whatever happens after that is not of concern to me. Knowing that a martingale that stopped is a martingale, we can talk about our exponential process introducing a martingale.

We know that

$𝔼 [𝑍 (𝜏_{𝑚})] = 𝔼 [𝑍 (𝜏_{𝑚}) | ℱ︀ (0)]$

which must be 0 because it’s conditioned on first information.

This is a martingale, so it’s for any amount of time. Let’s test that:

${lim}_{𝑡 \to \infty} 𝕖 [𝑍 (𝑡 \land 𝜏_{𝑚})] = 1$

${lim}_{𝑡 \to \infty} 𝔼 [𝑒^{𝜎 𝑊 (𝑡 \land 𝜏_{𝑚}) - \frac{1}{2} 𝜎^{2} (𝑡 \land 𝜏_{𝑚})}]$

limit is a linear operator, so you can swap the order and separate them to solve both

${lim}_{𝑡 \to \infty} 𝑒^{- \frac{1}{2} 𝜎^{2} (𝑡 \land 𝜏_{𝑚})}$

If we’re talking about the first time we reach a given level, our Brownian motion could go negative forever, nothing says it has to turn positive. The path of Brownian motion could never reach the level I need it to. The only way to express that this never happens is that my stopping time is infinity.

This adds a wrinkle to this expression because we need to express it as two different possibilities

We could also write this as

All of this was necessary to do the right limit.

There are two cases. If we consider finite $𝜏_{𝑚}$ , then we can disregard $𝑡$ and say that it’s equal to $𝑒^{𝜎 𝑚}$ .

My process will always be less than $𝑒^{𝜎} 𝑚$ if $𝜏_{𝑚}$ is infinite because instead we disregard $𝜏_{𝑚}$ and look at $𝑡$ , because that means the stopping point is never reached.

So we can say that the expected value of all of this is equal to 1.

Now this is true for any value of $𝜎$ . Usually we make the restriction that $𝜎$ is positive. Let’s consider

We know that the probability measure of this is 1 in the case when $𝜏_{𝑚} < \infty$ , which means that we always expect to reach these values in finite time. But the expected amount of time to reach any of these levels is infinite.

Now we will look at $𝛼 = \frac{1}{2} 𝜎^{2}$

This will allow us to express

$𝔼 [𝑒^{- 𝛼 𝜏_{𝑚}}] = 𝑒^{- | 𝑚 | \sqrt{2 𝛼}}$

This is the Laplace transform (we will not need to know this).

An interesting thing is, given this expression, and I took the derivative of both sides with respect to $𝛼$ , I would get

$𝔼 [𝜏_{𝑚} 𝑒^{- 𝛼 𝜏_{𝑚}}] = \frac{| 𝑚 | \sqrt{2}}{2} 𝛼^{- \frac{1}{2}} 𝑒^{- | 𝑚 | \sqrt{2 𝛼}}$

If we look at $𝛼$ going to 0, we get

${lim}_{𝛼 \to 0^{+}} 𝔼 [𝜏_{𝑚} 𝑒^{- 𝛼 𝜏_{𝑚}}] = {lim}_{𝛼 \to 0^{+}} \frac{| 𝑚 | \sqrt{2}}{2} 𝛼^{- \frac{1}{2}} 𝑒^{- | 𝑚 | \sqrt{2 𝛼}}$

$= \infty$

So even though $𝜏_{𝑚}$ is finite almost surely, its expectation goes to infinity.

At this point, we have our first passage time random variable $𝜏_{𝑚}$ and we know a couple things about it. It’s finite almost surely, expected value infinity, only takes positive value. I’d like to find its exact distribution through its pdf. We will use the reflection equality.

$ℙ (𝜏_{𝑚} \leq 𝑡, 𝑊 (𝑡) < 𝑤) = ℙ (𝑊 (𝑡) > 𝑍_{𝑚} - 𝑤)$

The probability that I reached level $𝑚$ before time $𝑡$ and my Brownian motion ended below level $𝑤$ is the same as the probability as it ended above $2 𝑚 - 𝑤$

Let’s let $𝑤 = 𝑚$ . This becomes

$ℙ (𝜏_{𝑚} \leq 𝑡, 𝑊 (𝑡) < 𝑚) = ℙ (𝑊 (𝑡) > 𝑚)$

The probability that I reach level $𝑚$ and $𝑡$ time and end up beneath $𝑚$ is the same that I end up above $𝑚$ .

In general,

$ℙ (𝐴 \cap 𝐵) + ℙ (𝐴 \cap 𝐵^{𝑐}) = ℙ (𝐴)$

I have that

$ℙ (𝜏_{𝑚} \leq 𝑡) = 2 ℙ (𝑊 (𝑡) > 𝑚)$

But we know that probability! Because it’s normally distributed.

$= 2 \int_{𝑚}^{\infty} \frac{1}{\sqrt{2 𝜋 𝑡}} 𝑒^{- 𝑥^{2} / (2 𝑡)} 𝑑 𝑥$

If we do this u substitution $𝑢 = \frac{𝑥}{\sqrt{𝑡}}$ , that’s

$= 2 \int_{\frac{𝑚}{\sqrt{𝑡}}}^{\infty} \frac{1}{\sqrt{2 𝜋}} 𝑒^{\frac{- 𝑢^{2}}{2}} 𝑑 𝑢$

But this is the cdf, we want the pdf

In order to get the derivative of an interval like so,

$\frac{𝑑}{𝑑 𝑥} \int_{𝑔 (𝑥)}^{ℎ (𝑥)} 𝑓 (𝑦) 𝑑 𝑦 = 𝑓 (ℎ (𝑥)) ℎ^{'} (𝑥) - 𝑓 (𝑔 (𝑥)) 𝑔^{'} (𝑥)$

so in this context, it becomes

$\frac{𝑑}{𝑑 𝑡} ℙ (𝜏_{𝑚} \leq 𝑡) = 0 + \frac{1}{\sqrt{2 𝜋}} 𝑒^{\frac{- 𝑚^{2}}{2 𝑡}} \frac{𝑚}{2} 𝑡^{- \frac{3}{2}}$

That is the resulting CDF/density relationship. Open question: retrieve the exact displayed equation from the slides.

## Brownian Motion and Geometric Growth - URL: https://sharifhsn.dev/blog/brownian-motion-risk-models/ - Structured data: https://sharifhsn.dev/api/posts/brownian-motion-risk-models.json - Description: Under market efficiency, financial prices should exhibit a random walk. Eugene Fama is the big guy behind market efficiency. He was interviewed by the Financial Times ( Open… - Date: 2024-09-26 - Exact published timestamp: 2024-09-26 - Topics: Risk Management, Brownian Motion, Geometric Brownian Motion - Categories: Risk Management - Source: FE-535 \| Risk Management - Source URL: None - Content source format: typst

Univariate Stochastic Processes

Under market efficiency, financial prices should exhibit a random walk.

Eugene Fama is the big guy behind market efficiency. He was interviewed by the Financial Times (Open question: read the article) and shared the Nobel prize with Robert Shiller (Rational Exuberance). Fama came up with the efficient market hypothesis, which tells us that if people are seeking profits and you start trading that, putting millions in, the price will go up and it will become expensive. Any time there’s insight or information, that gets priced. A random walk means that it’s unpredictable what the next result will be. As soon as information comes in, everybody starts trading, the price goes up, maybe negative information the next day, and then price goes down. We will say that these returns are not correlated. How are market participants able to process and price information? Some people say that efficient markets are not possible for behavioral reasons. In our class, we will keep it on the iid assumption to keep it simple.

Week 4

Lecture Notes

Univariate Stochastic Processes (cont.)

Under the efficient market hypothesis, every new piece of information changes the price, which creates a random walk. This means you only care about the previous point because the price should encode all of the information up to that point, which is why it’s a Markov process. If we think about price jumps:

$𝑃_{1} = 𝑃_{0} (1 + 𝑅_{1})$

$𝑃_{2} = 𝑃_{1} (1 + 𝑅_{2})$

So the movement/change in that price only depends on the return. Our argument is that these returns are independent. We have this iid assumption so that returns are uncorrelated.

Markov property (weak):

$ℙ (𝑆_{𝑇_{𝑎}} | 𝑆_{𝑇 - 1}, 𝑆_{𝑇 - 2} \dots) = ℙ (𝑆_{𝑇 = 𝑎} | 𝑆_{𝑇 = 1})$

The next flip of a coin is independent of the previous flip.

On average these jumps will be positives, so we have a certain level of growth or expected return $𝜇$ . Our volatility $𝜎$ is how much it goes differently from expected. We need to pay attention to our time period $𝑡$ . $𝜇$ basically says that if we do these stochastic processes a million times, on average it goes to a certain point. Variance asks how much the process will be dispersed. Time asks where we’re modeling. Dispersion is generally larger with increased time, which is why options pricing makes options more expensive over long periods of time.

We will look at three different distributions:

Brownian Motion

$𝑍_{𝑡}$ : This is a natural phenomenon. The best biologist is a chemist, the best chemist is a physicist, the best physicist is a mathematician. Also known as Weiner process.

It might go up and down, but it only goes one way.

This has three different properties.

We are reviewing probability and statistics, since we are looking at quantitative analysis.

We have a process that starts at time 0, with value starting at 0.

The very important property is that the process has a Gaussian distribution:

$𝑍_{𝑡} 𝑁 (0, 𝑡)$

Its variance depends on time (think quadratic variation).

Let’s think about simulations. We’re sampling from the normal distribution.

In Excel, we have the function =NORM.INV(RAND(), 0, 1) which generates the next number.

What is the value of $𝑍_{2}$ now that we have observed $𝑍_{1}$ i.e. $𝑍_{2} | 𝑍_{1}$ ? The increments are independent and also follow a normal distribution:

$Δ 𝑍 = 𝑍_{𝑡} - 𝑍_{𝑠} 𝑁 (0, 𝑡 - 𝑠)$

In order to generate $𝑍_{2}$ , I need to think about the value of $𝑍_{1}$ .

$𝑍_{2} | 𝑍_{1} = 𝑍_{1} + Δ 𝑍_{2}$

where $𝑍_{1}$ is a constant and $𝑍_{2}$ is a random variable.

We can generate this by simulation to find the properties of the distribution. Eventually you get

$𝑍_{\frac{𝑑}{100}} = 𝑍_{0} + \sum_{𝑖 = 0}^{𝑑 - 1} Δ 𝑍_{\frac{𝑖}{100}}$

We can generate this into the distribution if we assume returns are iid as variance of summations:

$𝑍_{\frac{𝑑}{100}} \sim 𝑁 (0, 𝑑 \cdot 0.01)$

There should be an Excel spreadsheet related to Brownian motion:

FE_535_Session_02_games_SBM_sim

If we look at the expectation of the values here, this is a simulation of the terminal point of the Brownian path. We created this over smaller increments. I want to validate that this is in fact normal $𝑁 (0, 1)$ . We have a “large” simulation so according to the LLN the average should converge to 0. This is a validation of the two moments if we see −0.006788 and 0.985446m, then it’s close enough. If we look in aggregate from a distributional point of view, how do these prices look?

In R, you use the functions rnorm for a randomly generated normal distribution, and dnorm for a true normal distribution. If we lay them on each other, we can see that they are very similar for larger numbers of rnorm.

This gives a good tool to understand pricing.

General Brownian Motion

When we talk about normal distribution, there’s only single probabilities within a standard. There are infinite tables that depend on $𝜇$ and $𝜎$ . When we model this, we add two degrees of freedom. One is how the process is changing over time. I start at point $𝑋_{0}$ , but what it actually implies is that we have some expected increase like in real stocks. That number depends on the $𝜇$ . We could imagine a casino where $𝜇$ is negative. The other question is how volatile this is, which is $𝜎$ . This is the generalized process:

$𝑋_{𝑡 + Δ 𝑡} - 𝑋_{𝑡} = Δ 𝑋_{𝑡} = 𝜇 Δ 𝑡 + 𝜎 Δ 𝑍_{𝑡}$

where $Δ 𝑋_{𝑡} \sim 𝑁 (𝜇 Δ 𝑡, 𝜎^{2} Δ 𝑡)$

If we simulate this process over multiple periods instead of just using $Δ 𝑡$ , we get this.

$𝑋_{𝑡} \sim 𝑁 (𝑋_{0} + 𝜇 \cdot 𝑡, 𝜎^{2} 𝑡)$

The problem with the previous BM is that prices could take negative values. But actually stock prices cannot go negative. If you have a limited liability company, the equity value cannot go negative. Whoever owns those stocks is not responsible for e.g. the debts of the company.

You can also think about it as a percentage change, which cannot go to 0 just by multiplication.

Whatever assets you invested in, I lose only what I put in. I don’t care that much if the company goes bankrupt, the only thing that happens is that I lose the money I invested.

If we take this process with the random variable in the same way as with regular Brownian motion, we can see

$𝑋_{𝑡} = 𝑋_{0} + \sum Δ 𝑋 = 𝑋_{0} + \sum 𝑖 = 1^{100} 𝜇 \cdot 0.01 + 𝜎 \cdot Δ 𝑍_{𝑖} = 𝑋_{0} + 𝜇 \cdot 0.01 \cdot 100 + 𝜎 \cdot \sum_{𝑖 = 1}^{100} Δ 𝑍_{𝑖}$

Why am I wasting your time with this? If we go back to the initial notation, this 0.01 is the $Δ 𝑡$ , $𝑁$ is 100, and our $𝑡$ is $Δ 𝑡 \cdot 𝑁$ . For calculating $𝑍_{𝑖}$ , we cancel out all the deltas in between and get $𝑍_{100} - 𝑍_{0}$ . By construction $𝑍_{0} = 0$ , so this is $𝑍_{100}$ .

Now we have a solution for the process.

If I know what $𝑍_{𝑡}$ is going to be, that matrix, then I know what $𝑋_{𝑡}$ is going to be. This is also related to geometric Brownian motion. If we’re able to create $𝑍_{𝑡}$ , we’re done.

𝐶 𝑢 𝑒 𝑚 𝑒 𝑐 𝑜 𝑢 𝑔 ℎ 𝑖 𝑛 𝑔 𝑓 𝑜 𝑟 𝑡 𝑒 𝑛 𝑚 𝑖 𝑛 𝑢 𝑡 𝑒 𝑠

Lab

We have our sheet X_general_bm.

The variance of half year through is 50%.

$𝕍 [𝑍_{𝑡}] = 𝑡$

$𝔼 [𝑋_{𝑡}] =$

The simulated value basically tells you, you can go on and pick up the time of 50 which is half year from now, then take the average of that, take the variance of that, that will give you a proxy for the moments you actually want to calculate. Then you need to do this for six months then 1 year.

We have reference points of True Value that simulated value should resemble.

Let’s take the solution that

$𝑆_{𝑇} = 𝑆_{0} 𝑒^{𝑅_{𝑡}}, 𝑅_{𝑇} \sim 𝑁 ((𝜇 - \frac{𝜎^{2}}{2}) 𝑇, 𝜎^{2} 𝑇)$

T always refers to years for simplicity.

We can all relate to annual numbers.

If T was equal to days, then the parameters should represent that.

We need to find the expectation of $𝑆_{𝑇}$

$𝑆_{0}$ is a constant

$𝔼 [𝑆_{𝑇}] = 𝑆_{0} 𝔼 [𝑒^{𝑅_{𝑇}}]$

How do we get there analytically? By simulation we got a hint that it is the correct answer.

Wikipedia tells us the expected value of the log normal distribution, which our returns are.

$𝑒^{𝜇_{𝑋} + \frac{𝜎_{𝑋}^{2}}{2}}$

If we plug it in, we get

$𝔼 [𝑆_{𝑇}] = 𝑆_{0} 𝑒^{𝜇_{𝑋} + \frac{𝜎_{𝑋}^{2}}{2}}$

then when we simplify after ignoring volatility we get

$= 𝑆_{0} 𝑒^{𝜇 𝑇}$

You can repeat this process for variance, where pulling out the constant has to square it

$𝕍 [𝑆_{𝑇}] = 𝑆_{0}^{2} 𝕍 [𝑒^{𝑅_{𝑇}}]$

plug in wikipedia log normal variance for returns $= 𝑆_{0}^{2} [𝑒^{𝜎_{𝑋}^{2}} - 1] [𝑒^{2 𝜇_{𝑋} + 𝜎_{𝑋}^{2}}]$

$= 𝑓 (𝑆_{0}, 𝜇, 𝜎, 𝑇)$

Geometric Brownian Motion

The most common process to simulate stock prices is the Geometric Brownian Motion (GBM).

$Δ 𝑆_{𝑡} = 𝑆_{𝑡} 𝜇 Δ 𝑡 + 𝑆_{𝑡} 𝜎 Δ 𝑍_{𝑡}$

By design we know by the motion itself that

$𝑑 𝑙 𝑜 𝑔 (𝑆_{0}) = 𝜇 𝑑 𝑡$

In the other example we aggregated. How do you aggregate continuously? The integral, which is

$log (𝑆_{𝑇}) - log (𝑆_{0}) = 𝜇 \cdot 𝑇$

$log (\frac{𝑆_{𝑇}}{𝑆_{0}}) = 𝜇 \cdot 𝑇$

$𝑆_{𝑇} = 𝑆_{0} 𝑒^{𝜇 \cdot 𝑇}$

This is an ordinary differential equation (only one dimension, time)

We have a constant process which evolves with respect to time.

The distribution would be skewed upwards because this cannot take negative values.

We need to use something called Ito’s Lemma, which gives us the nice solution that

$log (\frac{𝑆_{𝑡 + Δ 𝑡}}{𝑆_{𝑡}}) = (𝜇 - \frac{𝜎^{2}}{2}) Δ 𝑡 + 𝜎 Δ 𝑍_{𝑡}$

which is equation 4.6 from the textbook.

If you take the $𝜎$ out with nothing stochastic, you end up with the ODE solution.

The point of the solution is to simulate the price at time $𝑡 + Δ 𝑡$ , aka returns over time $𝑅_{𝑇}$ .

The log-normal distribution tells us that if a random variable is log-normally distributed, then the natural log of the random variable has the normal distribution.

Excel formula for variance of geometric Brownian motion: =C8^2*EXP(2*C4*C3)*(EXP(C5^2*C3)-1).

Proposed simulation code:

=B2*EXP(NORMINV(RAND(), (Inputs!$C$4-Inputs!$C$5/2)*Inputs!$C$3, Inputs!$C$5*SQRT(Inputs!$C$3)))

## How rustc_codegen_clr Connects Rust to .NET - URL: https://sharifhsn.dev/blog/linkedin-2024-09-23-rust-mir-to-dotnet/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2024-09-23-rust-mir-to-dotnet.json - Description: A short guide to Rust's compiler pipeline and a backend that targets .NET. - Date: 2024-09-23 - Exact published timestamp: 2024-09-23T11:47:41.887Z - Topics: Rust, Compilers - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/posts/sharif-haason_rust-panics-under-the-hood-and-implementing-activity-7243949143541391360-ia_9 - Content source format: typst

If you’re a fan of compilers, Rust, or .NET, then Michał Kostrubiec’s blog post about his project rustc_codegen_clr is a must-read. This is one of the most exciting and fast-moving projects in the Rust sphere, all driven by the hard work of one talented programmer. Here’s a quick summary of what the project is and why it’s so significant:

Rust, like many other modern languages, goes through several steps of compilation where it is compiled into different intermediate representations (IRs) before becoming an executable application. The current translation layer goes like so: Rust -> HIR -> MIR -> LLVM IR -> native code. The key step is between MIR and LLVM IR. This is when the Rust compiler frontend rustc shells out to a different compiler backend to perform the actual code generation, or “codegen”. The default is LLVM, the highly popular and versatile Apple-supported backend used by Swift and C/C++ (through clang). Using an existing backend allowed the Rust team to focus on frontend development. Instead of compiling efficient code to every possible CPU architecture, they were able to take advantage of LLVM’s existing infrastructure and compile to their IR.

Although this works great for the vast majority of use cases, there are some hiccups. LLVM doesn’t support every architecture that the standard C/C++ compiler gcc does, limiting Rust’s use in embedded environments. It’s also quite slow because Rust emits a lot of type information by design, which takes a long time to compile since LLVM wasn’t built to optimize that. That’s why efforts have been made to use other backends for codegen, like gcc [2], and there even exists now native Rust backends like Cranelift. [3]

rustc_codegen_clr [4] does something different. It translates MIR to the C# IR CIL, which allows .NET to compile Rust code. This has two main effects. .NET can use Rust libraries, which are comparable in performance to C libraries while being more robust. Rust can also use .NET libraries, which would enable Rust programs to access much more of the Windows platform, including the Windows App SDK. Rust native modern Windows GUI is just on the horizon!

Michał’s blog thoroughly examines the technical details behind his implementation and is worth the read for anyone interested in the project!

1 · 2 · 3 · 4

## Discrete Distributions and Moments - URL: https://sharifhsn.dev/blog/discrete-distributions-and-moments/ - Structured data: https://sharifhsn.dev/api/posts/discrete-distributions-and-moments.json - Description: We will be covering discrete random variables this week. A discrete random variable is one whose image 𝑋 ( Ω ) is a finite/countable subset of ℝ . First thing we introduce is the… - Date: 2024-09-23 - Exact published timestamp: 2024-09-23 - Topics: Probability Theory, Discrete Distributions, Moments - Categories: Probability Theory - Source: FE-540 \| Probability Theory - Source URL: None - Content source format: typst

Week 4

Lecture Notes

We will be covering discrete random variables this week.

A discrete random variable is one whose image $𝑋 (Ω)$ is a finite/countable subset of $ℝ$ .

First thing we introduce is the probability mass function (pmf).

It is basically the probability measure assigned to a single value: $𝑃_{𝑋} ({𝑥}) = 𝑃 (𝑋 = 𝑥)$

Because of the law of total probability, the probability of all outcomes must sum to 1:

$\sum_{𝑥 \in 𝑋 (Ω)} 𝑃 (𝑋 = 𝑥) = 1 .$

Next thing is the cumulative distribution function (cdf).

Usually use the capital F to denote this: $𝐹 (𝑡) = 𝑃 (𝑋 \leq 𝑡) = \sum_{𝑥 \in 𝑋 (Ω) \cap 𝑥 \leq 𝑡} 𝑃 (𝑋 = 𝑥)$

A simple example for a discrete random variable is that if you roll a die, the outcome of the die will give you a discrete random variable. If I’m interested in $𝐹 (3) = 𝑃 (𝑋 \leq 3) = 𝑃 (𝑋 = 1) + 𝑃 (𝑋 = 2) + 𝑃 (𝑋 = 3)$

Another example:

We denote $𝑋$ as the number of times we roll the die until we get the outcome 6. If we’re lucky and roll the die once to get a 6, what’s the probability? For a fair die it’s $\frac{1}{6}$ . For the second roll, $𝑃 (𝑋 = 2) = \frac{5}{6} \times \frac{1}{6} = \frac{5}{36}$ : the first roll is not 6 and the second roll is 6.

The outcome of the first roll does not affect the second roll.

More difficult one, $𝑃 (𝑋 = 3)$

$𝑃 (𝑋 = 3) = \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}$

You can prove by induction that the formula is

$𝑃 (𝑋 = 𝑛) = {(\frac{5}{6})}^{𝑛 - 1} \times \frac{1}{6}, 1 \leq 𝑛$

This is a simple example.

I want to say a property of the cdf:

If you have $𝑋 : Ω \to ℝ$ is a discrete r.v., the cdf $𝐹 (𝑡)$ is piecewise constant and has finite/countable jumps.

If you think about cdf, it’s a probability that lies from 0 to 1.

The graph of discrete cdf is exclusive coming from the right side, so the hole is on the bottom of the jump, not the top. We accumulate in a discrete ways, which makes it like a staircase. Each piece corresponds to a constant value, and there are finite jumps.

For simplicity, we will denote the following notations:

$𝑋 (Ω) = {𝑎_{1}, 𝑎_{2}, 𝑎_{3}, \dots}$

$𝑝_{𝑖} = 𝑃 (𝑋 = 𝑎_{𝑖}), 𝑖 = 1$

Moments

Moments are expectations, expectations are kind of like intervals in calculus. Here we are talking about discrete random variables, so our integration is basically a summation. In the next class we are going to talk about continuous, where the integral is an integral. Integrals is approximated by the Riemann sum, and there are many techniques for evaluating them. But right now we are sticking to summations.

Let’s define the expectation, by definition 4.7.

If you have a discrete r.v. $𝑋$ , we say $𝑋$ is integrable if

$\sum_{𝑥 \in 𝑋 (Ω)} | 𝑥 | \cdot 𝑃 (𝑋 = 𝑥) < \infty$

If you map this to what you learned in calculus, this is absolute integrability. You know that if you are absolutely integrable, you are regular integrable. Why absolute value? This $𝑥$ can take positive or negative values. If this holds, we can define expectation as

$𝔼 [𝑋] = \sum_{𝑥 \in 𝑋 (Ω)} 𝑥 𝑃 (𝑋 = 𝑥)$

In order to help you understand this from a discrete random variable notation, you can put this in a table format.

$𝑋 = {outcome probability 𝑎_{1} 𝑝_{1} 𝑎_{2} 𝑝_{2}$

This will be a handy tool when doing calculations.

This summation is purely rigorous.

We need the non-absolute value sum to be less, because the probability cannot be negative. I’m not just saying that the expectation is finite. I’m saying that the expectation is bounded by something finite. $𝑋 < \infty$ is very different from $𝑋 < 𝑌 < \infty$ . The second one means that the value is bounded away from infinity, where $𝑌$ is the ceiling. The first one means you can approach infinity. This is important for mathematical analysis.

Proposition 4.8 says a property of the expectation is that expectation is linear. If you learn functional analysis (calculus on functions), you understand that expectation is a linear operator. The statement is this: $𝕖 [𝑎 𝑋 + 𝑏 𝑌] = 𝑎 𝐸 [𝑋] + 𝑏 𝐸 [𝑌]$ where $𝑎$ and $𝑏$ are constants in $ℝ$ .

How do we prove this?

We want to first establish that $𝔼 [𝑎 𝑋] = 𝑎 𝔼 [𝑋]$

There are two cases. When $𝑎 = 0$ , this is trivial because $𝑋$ becomes 0. How do you prove for $𝑎 \neq 0$ ?

We will define $𝑍 = 𝑎 𝑋$ . From the definition of a discrete random variable, its image set must be a countable subset of $ℝ$ . Therefore, $𝑍$ is a discrete random variable. This implies that $𝑍 (Ω)$ is countable. This is important because anything countable can be represented as a sum. $𝜋$ is uncountable and cannot be represented by a summation.

$𝔼 [| 𝑍 |] = \sum_{𝑥 \in 𝑋 (Ω)} | 𝑎 𝑥 | 𝑃 (𝑍 = 𝑎 𝑥)$

by the simple definition of expectation.

$= \sum_{𝑥 \in 𝑋 (Ω)} | 𝑎 𝑥 | 𝑃 (𝑎 𝑋 = 𝑎 𝑥)$

I’m going to factor out the $𝑎$ , which we can do because we know $𝑎 \neq 0$

$= | 𝑎 | \sum_{𝑥 \in 𝑋 (Ω)} | 𝑥 | \cdot 𝑃 (𝑋 = 𝑥)$

Based on our definition 4.7, we can say that this is less than infinity.

Then we can formally write down $𝑍$ without absolute value, since it’s bounded finitely.

$𝔼 [𝑍] = \sum_{𝑥 \in 𝑋 (Ω)} 𝑎 𝑥 𝑃 (𝑍 = 𝑎 𝑥) = 𝑎 \sum_{𝑥 \in 𝑋 (Ω)} 𝑥 𝑃 (𝑋 = 𝑥) = 𝑎 \cdot 𝔼 [𝑋]$

This might seem silly, but we need to be rigorous in math when proving things.

Now we want to prove $𝔼 [𝑎 𝑋 + 𝑏 𝑌] = 𝔼 [𝑎 𝑋] + 𝔼 [𝑎 𝑌]$

The proof is too long, has too much rigor, uses some techniques of interchanging the summation signs.

Second part of proposition is that if $𝑋 \geq 0$ , then $𝔼 [𝑋] \geq 0$ . If the random variable is greater than 0, then the expectation is greater than 0. The simple proof is that

$𝔼 [𝑋] = \sum_{𝑥 \in 𝑋 (Ω)} 𝑥 𝑃 (𝑋 = 𝑥) \geq 0$

If $𝑋 > 𝑌$ , then $𝔼 [𝑋] \geq 𝔼 [𝑌]$ , where we can show that the difference is positive

Theorem 4.9 the transfer formula says that for $𝑋$ discrete r.v., where $𝜑 : ℝ \to ℝ$ is a measurable function, then $𝜑 (𝑋)$ is a r.v. and is integrable iff

$\sum_{𝑥 \in 𝑋 (Ω)} | 𝜑 (𝑥) | 𝑃 (𝑋 = 𝑥) < \infty$

In this case I can formally write down the expectation to be the sum

$𝔼 [𝜑 (𝑋)] = \sum_{𝑥 \in 𝑋 (Ω)} 𝜑 (𝑥) 𝑃 (𝑋 = 𝑥)$

Examples

Here are some examples:

If we have a discrete r.v.

$𝑋 = {1 0.2 2 0.3 3 0.5, 𝔼 [𝑋] = ?$

You multiply horizontally and add them all up.

Represented in the cases format,

$𝔼 [𝑋] = \sum_{𝑖 = 1}^{\infty} 𝑎_{𝑖} 𝑝_{𝑖}$

For a measurable function $𝜑$ ,

$𝔼 [𝜑 (𝑋)] = \sum_{𝑖 = 1}^{\infty} 𝜑 (𝑎_{𝑖}) 𝑝_{𝑖}$

You only change the outcome, not the probability.

In this example,

$𝔼 [𝑋] = 1 \times 0.2 + 2 \times 0.3 + 3 \times 0.5 = 2.3$ .

What if we take $𝔼 [𝑋^{2}]$ ?

The square is a measurable function, so we can consider $𝜑 (𝑋) = 𝑋^{2}$

$𝔼 [𝜑 (𝑋)] = 1^{2} \times 0.2 + 2^{2} \times 0.3 + 3^{2} \times 0.5 = 5.9$

We can go a little crazier and talk about $𝜑 = sin 𝑋$

In all our examples, all of our functions are measurable. In terms of mathematical analysis, you might be interested in functions that are not measurable. But we don’t have to worry about that in our class.

Variance

The variance of a r.v. is defined to be

$𝕍 (𝑋) = 𝔼 [(𝑋 - 𝔼 [𝑋])^{2}]$

There is an intuitive explanation for this definition. How do you measure the variability if numbers in a set? What you do is take the expected value of the numbers in the set.

Let’s say we have a sequence $(𝑥_{1}, 𝑥_{2}, \dots, 𝑥_{𝑛})$

$𝔼 [𝑋] = 𝑥_{1} \cdot \frac{1}{𝑛} + 𝑥_{2} \frac{1}{𝑛} + \dots + 𝑥_{𝑛} \frac{1}{𝑛}$

We want to measure the distance from each data point to the middle point. The distance of each variable to that middle point is what we are counting. We square them because the difference might be positive or negative, so it makes them all 0.

What happens if we don’t square it?

$𝔼 [𝑋 - 𝔼 [𝑋]] = 𝔼 [𝑋] - 𝔼 [𝑋] = 0$

If you do some simple arithmetics, you can see the simple binomial expansion as

$(𝑎 + 𝑏)^{2} = 𝑎^{2} + 2 𝑎 𝑏 + 𝑏^{2}$

and if we apply this to our variance, we get

$𝕍 [𝑋] = 𝔼 [𝑋^{2} - 2 𝑋 𝔼 [𝑋] + (𝔼 [𝑋])^{2}]$

expected value of an expected value is the same thing because it’s just a constant

$= 𝔼 [𝑋^{2}] - 2 𝔼 [𝑋] 𝔼 [𝑋] + (𝔼 [𝑋])^{2}$

$= 𝔼 [𝑋^{2}] - (𝔼 [𝑋])^{2} \geq 0$

This is Jensen’s inequality, which shows the convexity of variance.

There are two operations, we take square, then expectation, and for the other, we take expectation first, then square. And this difference must be greater than or equal to 0.

Eventually, everything becomes calculus.

Definition 4.12 is of standard deviation, or std dev.

$𝜎 (𝑋) = \sqrt{𝕍 (𝑋)} = \sqrt{𝔼 [𝑋^{2}] - (𝔼 [𝑋])^{2}}$

This identity is very important:

$𝔼 [𝑋^{2}] = 𝕍 (𝑋) + (𝔼 [𝑋])^{2}$

This is very convenient for our calculations. Let’s think about calculating the variance for the cases format. Assuming this format, we can represent

$𝕍 [𝑋] = 𝔼 [𝑋^{2}] - (𝔼 [𝑋])^{2}$

$= \sum_{𝑖 = 1}^{\infty} 𝑎_{𝑖}^{2} 𝑝_{𝑖} - {(\sum_{𝑖 = 1}^{\infty} 𝑎_{𝑖} 𝑝_{𝑖})}^{2}$

That’s it. This is the formula to use on homework. USEFUL

This has all been theoretical, abstract concepts. From now on we are going to talk about examples of discrete random variables.

We are going to go through several classes of random variables.

Discrete Uniform Distribution

The discrete uniform distribution has all the values being the same.

$𝑋 (Ω) = {𝑥_{1}, 𝑥_{2}, \dots, 𝑥_{𝑛}}$ and pmf is given by

$𝑝_{𝑖} = 𝑃 (𝑋 = 𝑥_{𝑖}) = \frac{1}{𝑛}, \forall 𝑖 \in {1, 2, \dots, 𝑛}$

we denote this as

$𝑋 \sim 𝐷 𝑈 (𝑛)$ If you roll a fair die, the outcome of a die is a discrete uniform distribution. What is our cdf?

$𝐹 (𝑡) = {0 if 𝑡 < 1 \frac{𝑖}{6} if 𝑖 \leq 𝑡 \leq 𝑖 + 1, 𝑖 = 1, 2, 3, 4, 5 1 if 𝑡 \geq 6$ Let’s look at expected value (in general, not just for fair die)

$𝔼 [𝑋] = \sum_{𝑖 = 1}^{𝑛} 𝑖 \cdot \frac{1}{𝑛}$

Let’s make this clear that the image set is $𝑋 (Ω) = {1, 2, 3, \dots, 𝑛}$

Then we can determine the high school mathematics by taking the variable outside.

$= \frac{1}{𝑛} (1 + 2 + 3 + \dots + 𝑛)$

$= \frac{1}{𝑛} \frac{𝑛 (𝑛 + 1)}{2}$

$= \frac{𝑛 + 1}{2}$

Now let’s look at the variance.

$𝔼 [𝑋^{2}] = \sum_{𝑖 = 1}^{𝑛} 𝑖^{2} \cdot \frac{1}{𝑛} = \frac{1}{𝑛} \sum_{𝑖 = 1}^{𝑛} 𝑖^{2}$

By the same summation property

$= \frac{1}{𝑛} (1^{2} + 2^{2} + \dots + 𝑛^{2}) = \frac{1}{𝑛} \frac{𝑛 (𝑛 + 1) (2 𝑛 + 1)}{6} = \frac{(𝑛 + 1) (2 𝑛 + 1)}{6}$

To get the full variance we have

$𝕍 [𝑋] = 𝔼 [𝑋^{2}] - (𝔼 [𝑋])^{2} = \frac{(𝑛 + 1) (2 𝑛 + 1)}{6} - \frac{(𝑛 + 1)^{2}}{4}$

$= \frac{2 (𝑛 + 1) (2 𝑛 + 1) - 3 (𝑛 + 1)^{2}}{12}$

$= \frac{1}{12} (4 𝑛^{2} + 6 𝑛 + 2 - 3 𝑛^{2} - 6 𝑛 - 3)$

$= \frac{1}{12} (𝑛^{2} - 1)$

Bernoulli Distribution

$𝑋 = {1 𝑝 01 - 𝑝$ We will denote $𝑋 Bernoulli (𝑝)$ .

The most obvious example is flipping a coin. Our expectation is very simple.

$𝕡 = \frac{1}{2}$

$𝔼 [𝑋] = 1 \times 𝑝 + 0 \times (1 - 𝑝) = 𝑝$

For variance, it is

$𝔼 [𝑋^{2}] = 1^{2} \times 𝑝 + 0^{2} \times (1 - 𝑝) = 𝑝$

$𝕍 [𝑋] = 𝔼 [𝑋^{2}] - (𝔼 [𝑋])^{2} = 𝑝 - 𝑝^{2} = 𝑝 (1 - 𝑝)$

Binomial Distribution

This can be thought of as the sum of independent Bernoulli random variables. We will consider our pmf as

$𝑃 (𝑋 = 𝑘) = {(\binom{𝑛}{𝑘}) 𝑝^{𝑘} (1 - 𝑝)^{𝑛 - 𝑘} if 𝑘 \in {0, 1, 2, \dots, 𝑛} 0 otherwise$

Denote $𝑋 \sim Binom (𝑛, 𝑝)$

$(\binom{𝑛}{𝑘})$ is the combinatorial number, selecting $𝑘$ items out of $𝑛$ without caring about order

The formula is

$(\binom{𝑛}{𝑘}) = \frac{𝑛!}{𝑘! (𝑛 - 𝑘)!}$

The factorial is

$𝑛! = 𝑛 (𝑛 - 1) (𝑛 - 2) \dots 2 \times 1$

There is a property of the factorial that

$\frac{𝑛!}{(𝑛 - 1)!} = 𝑛$

which comes from the definition of the factorial.

We will consider more generally that

$\frac{𝑛!}{(𝑛 - 𝑘)!} = 𝑛 (𝑛 - 1) (𝑛 - 2) \dots (𝑛 - 𝑘 + 1)$

This fact will become more important.

The binomial expansion formula is that

$(𝑎 + 𝑏)^{𝑛} = \sum_{𝑘 = 0}^{𝑛} (\binom{𝑛}{𝑘}) 𝑎^{𝑘} 𝑏^{𝑛 - 𝑘}$

We will use this in many places.

The expectation of the binomial distribution

$𝔼 [𝑋] = 𝑛 𝑝$

Let’s prove this.

$𝔼 [𝑋] = \sum_{𝑘 = 0}^{𝑛} 𝑘 𝑃 (𝑋 = 𝑘) = \sum_{𝑘 = 0}^{𝑛} 𝑘 \cdot (\binom{𝑛}{𝑘}) 𝑝^{𝑘} (1 - 𝑝)^{𝑛 - 𝑘}$

We will use the technique of factoring out what we want to prove, and then showing that the rest is equal to 1. This is a common and powerful technique. By definition of choose,

$= \sum_{𝑘 = 1}^{𝑛} 𝑘 \cdot \frac{𝑛!}{𝑘! (𝑛 - 𝑘)!} 𝑝^{𝑘} (1 - 𝑝)^{𝑛 - 𝑘}$

Now we can factor out $𝑛 𝑝$

$= 𝑛 𝑝 \sum_{𝑘 = 1}^{𝑛} 𝑘 \cdot \frac{(𝑛 - 1)!}{𝑘! (𝑛 - 𝑘)!} \cdot 𝑝^{𝑘 - 1} (1 - 𝑝)^{𝑛 - 𝑘}$

We can simplify out that $𝑘$ and rewrite our $𝑛 - 𝑘$ for future rearranging and rewrite our for future rearranging

$= 𝑛 𝑝 \sum_{𝑘 = 1}^{𝑛} \frac{(𝑛 - 1)!}{(𝑘 - 1)! ((𝑛 - 1) - (𝑘 - 1))!} \cdot 𝑝^{𝑘 - 1} (1 - 𝑝)^{(𝑛 - 1) - (𝑘 - 1)}$

Next technique will be the change of variables in summations. Wherever I see $𝑘 - 1$ , I will change it to a dummy variable.

$= 𝑛 𝑝 \sum_{𝑘^{'} = 0}^{𝑛 - 1} \frac{(𝑛 - 1)!}{(𝑘^{'})! (𝑛 - 1 - 𝑘^{'})!} 𝑝^{𝑘^{'}} (1 - 𝑝)^{(𝑛 - 1) - 𝑘^{'}}$

We’re almost there, now we can rewrite as a choose

$= 𝑛 𝑝 \sum_{𝑘^{'} = 0}^{𝑛 - 1} (\binom{𝑛 - 1}{𝑘^{'}}) 𝑝^{𝑘^{'}} (1 - 𝑝)^{(𝑛 - 1) - 𝑘^{'}}$

We can now use the binomial expansion formula here.

$= 𝑛 𝑝 (𝑝 + 1 - 𝑝)^{𝑛 - 1}$

$= 𝑛 𝑝$

For variance we will use a slightly different formulation.

We will consider $𝔼 [𝑋 (𝑋 - 1)]$ , which is equivalent to $𝔼 [𝑋^{2}] - 𝔼 [𝑋]$ , because I want to cancel the factorial in a certain way., which is equivalent to , because I want to cancel the factorial in a certain way.

$𝔼 [𝑋 (𝑋 - 1)] = \sum_{𝑘 = 0}^{𝑛} 𝑘 (𝑘 - 1) (\binom{𝑛}{𝑘}) 𝑝^{𝑘} (1 - 𝑝)^{𝑛 - 𝑘}$

I can safely start from 2 because of this minus 1 I’m using. Now let’s do some cancellations/substitutions.

$\sum_{𝑘 = 2}^{𝑛} 𝑘 (𝑘 - 1) \frac{𝑛!}{𝑘! (𝑛 - 𝑘)!} 𝑝^{𝑘} (1 - 𝑝)^{𝑛 - 𝑘}$

Now we can divide $𝑘$ into its factorial.

$\sum_{𝑘 = 2}^{𝑛} \frac{𝑛!}{(𝑘 - 2)! (𝑛 - 𝑘)!} 𝑝^{𝑘} (1 - 𝑝)^{𝑛 - 𝑘}$

Let me take out the thing I want to prove, and use the $𝑛 - 𝑘$ technique. technique.

$= 𝑛 (𝑛 - 1) 𝑝^{2} \sum_{𝑘 = 2}^{𝑛} \frac{(𝑛 - 2)!}{(𝑘 - 2)! ((𝑛 - 2) - (𝑘 - 2))!} 𝑝^{𝑘 - 2} (1 - 𝑝)^{(𝑛 - 2) - (𝑘 - 2)}$

Using the $𝑘^{'}$ technique for $𝑘 - 2$ .

$= 𝑛 (𝑛 - 1) 𝑝^{2} \sum_{𝑘^{'} = 0}^{𝑛 - 2} \frac{(𝑛 - 2)!}{𝑘^{'}! (𝑛 - 2 - 𝑘^{'})!} 𝑝^{𝑘^{'}} (1 - 𝑝)^{(𝑛 - 2) - 𝑘^{'}}$

By the binomial expansion formula this is

$= 𝑛 (𝑛 - 1) 𝑝^{2} (𝑝 + 1 - 𝑝)^{𝑛 - 2}$

$= 𝑛 (𝑛 - 1) 𝑝^{2}$

Plugging this back into variance, we can show that

$𝕍 (𝑋) = 𝔼 [𝑋 (𝑋 - 1)] + 𝔼 [𝑋] - (𝔼 [𝑋])^{2}$

$= 𝑛 (𝑛 - 1) 𝑝^{2} + 𝑛 𝑝 - 𝑛^{2} 𝑝^{2}$

$= 𝑛^{2} 𝑝^{2} - 𝑛 𝑝^{2} + 𝑛 𝑝 - 𝑛^{2} 𝑝^{2}$

$= 𝑛 𝑝 (1 - 𝑝)$

You have to cancel a few things, use a change of variable, and group the sum into the binomial expansion formula. Those are the main techniques.

Stochastic calculus is less tedious but more conceptually difficult.

We have proposition 4.22: if $𝑋 \in Bin (𝑛, 𝑝)$ , $𝑌 \sim Bin (𝑚, 𝑝)$ , and $𝑋$ and $𝑌$ are independent, then their sum is binomial:

$𝑋 + 𝑌 \sim Bin (𝑛 + 𝑚, 𝑝)$

Geometric Distribution

The distribution has no cap, can go to infinity.

$𝑃 (𝑋 = 𝑘) = (1 - 𝑝)^{𝑘 - 1} \cdot 𝑝, 𝑘 = 1, 2, 3, \dots$

We can verify that this is in fact a probability distribution. We want to prove that

$\sum_{𝑘 = 1}^{\infty} 𝑃 (𝑋 = 𝑘) \sum_{𝑘 = 1}^{\infty} (1 - 𝑝)^{𝑘 - 1} 𝑝 = 𝑝 \sum_{𝑘 = 1}^{\infty} (1 - 𝑝)^{𝑘 - 1}$

We can expand this summation to

$= 𝑝 (1 + (1 - 𝑝) + (1 - 𝑝)^{2} + \dots)$

Based on the geometric formula, we can show that

$= 𝑝 \frac{1}{1 - (1 - 𝑝)}$

$= 1$

We will show some lemmas here, some results.

The power series is

$\sum_{𝑛 = 0}^{\infty} 𝑥^{𝑛} = \frac{1}{1 - 𝑥}, \forall 0 < 𝑥 < 1$

This is calculus:

$\sum_{𝑛 = 1}^{\infty} 𝑛 𝑥^{𝑛 - 1} = \frac{𝑑}{𝑑 𝑥} (\frac{1}{1 - 𝑥}) = \frac{1}{(1 - 𝑥)^{2}}$

We can also show

$\sum_{𝑛 = 2}^{\infty} 𝑛 (𝑛 - 1) 𝑥^{𝑛 - 2} = \frac{0 - 1 \times 2 (1 - 𝑥) (- 1)}{(1 - 𝑥)^{4}} = \frac{2}{(1 - 𝑥)^{3}}$

We can start calculating the expected value now, this is from proposition 4.26

$𝔼 [𝑋] = \sum_{𝑛 = 0}^{\infty} 𝑃 (𝑋 > 𝑛)$

$𝑃 (𝑋 > 𝑛) = \sum_{𝑘 = 𝑛 + 1}^{\infty} 𝑃 (𝑋 = 𝑘) = \sum_{𝑘 = 𝑛 + 1}^{\infty} 𝑝 {(1 - 𝑝)}^{𝑘 - 1} = 𝑝 {(1 - 𝑝)}^{𝑛} \times \sum_{𝑘 = 𝑛 + 1}^{\infty} {(1 - 𝑝)}^{𝑘 - 𝑛 - 1}$

I want to make this formula into my power series. Let $𝑘^{'} = 𝑘 - 𝑛 - 1$ . Then $𝑃 (𝑋 > 𝑛) = 𝑝 {(1 - 𝑝)}^{𝑛} \times \sum_{𝑘^{'} = 0}^{\infty} {(1 - 𝑝)}^{𝑘^{'}} = 𝑝 {(1 - 𝑝)}^{𝑛} \times \frac{1}{1 - (1 - 𝑝)} = {(1 - 𝑝)}^{𝑛}$ .

$𝑃 (𝑋 > 𝑛) = 𝑝 {(1 - 𝑝)}^{𝑛} \times \sum_{𝑘^{'} = 0}^{\infty} {(1 - 𝑝)}^{𝑘^{'}} = 𝑝 {(1 - 𝑝)}^{𝑛} \times \frac{1}{1 - (1 - 𝑝)} = {(1 - 𝑝)}^{𝑛}$ .

$𝑝 {(1 - 𝑝)}^{𝑛} \times \frac{1}{1 - (1 - 𝑝)}$

$= {(1 - 𝑝)}^{𝑛}$

Then we have expected value as the infinite sum of this

$𝔼 [𝑋] = \sum_{𝑛 = 0}^{\infty} (1 - 𝑝)^{𝑛} = \frac{1}{1 - (1 - 𝑝)} = \frac{1}{𝑝}$

For variance, we will use the same intermediate value of $𝔼 [𝑋 (𝑋 - 1)]$

$𝔼 [𝑋 (𝑋 - 1)] = \sum_{𝑘 = 2}^{\infty} 𝑘 (𝑘 - 1) \cdot 𝑝 (1 - 𝑝)^{𝑘 - 1}$

Take out what we want

$= 𝑝 (1 - 𝑝) \sum_{𝑘 = 2}^{\infty} 𝑘 (𝑘 - 1) (1 - 𝑝)^{𝑘 - 2}$

This follows from our third power series lemma

$= 𝑝 (1 - 𝑝) \frac{2}{𝑝^{3}}$

$= \frac{2 (1 - 𝑝)}{𝑝^{2}}$

So for our variance, we get

$𝕍 [𝑋] = \frac{2 (1 - 𝑝)}{𝑝^{2}} + \frac{1}{𝑝} - \frac{1}{𝑝^{2}}$

$= \frac{2 - 2 𝑝 + 𝑝 - 1}{𝑝^{2}}$

$= \frac{1 - 𝑝}{𝑝^{2}}$

Poisson will be beginning of next class

## Martingales and Random Walks - URL: https://sharifhsn.dev/blog/martingales-and-random-walks/ - Structured data: https://sharifhsn.dev/api/posts/martingales-and-random-walks.json - Description: (continuing from last time) A stochastic process is a collection of random variables indexed by time. If you start out at time 0 and each time you pick heads you go one step,… - Date: 2024-09-19 - Exact published timestamp: 2024-09-19 - Topics: Stochastic Calculus, Martingales, Random Walks - Categories: Stochastic Calculus - Source: FE-610 \| Stochastic Calculus - Source URL: None - Content source format: typst

Week 2

Lecture Notes

Measurability

(continuing from last time)

A stochastic process is a collection of random variables indexed by time. If you start out at time 0 and each time you pick heads you go one step, it’s now a walk. You have no idea where you’re going to be in the future, but you know the general distribution and have an idea of where you’re likely to do.

We also have a filtration which is a collection of $𝜎$ -algebras indexed by time. If we have enough information in the filtration to know what set your outcome is in, you get an adapted stochastic process. Mathematically, these variables are considered $ℱ︀$ -measurable by the filtration $ℱ︀$ .

Independence

Two $𝜎$ -algebras are independent if for every element $𝐴$ and $𝐵$ in both $𝜎$ -algebras:

$ℙ (𝐴 \cap 𝐵) = ℙ (𝐴) \times ℙ (𝐵)$

This is a strong mathematical result, but it is not useful because most useful $𝜎$ -algebras are infinite.

BUt if you can show that the joint cdf is the same as the products of the marginal cdfs (WILL COVER LATER) it will do the same thing.

We can say that two random variables are independent if the $𝜎$ -algebras generated by them are

One useful way to do this is by determining that if one variable is measurable by a $𝜎$ -algebra that is independent of a $𝜎$ -algebra generated by another variable, they are independent.

Variance

Variance $𝕍 (𝑋)$ is defined by:

$𝑉 𝑎 𝑟 (𝑋) = 𝕍 (𝑋) = 𝔼 [(𝑋 - 𝔼 [𝑋])^{2}] = 𝔼 [𝑋^{2}] - (𝔼 [𝑋])^{2}$

This variable is necessarily nonnegative, although it may be infinite. If your variance is negative, it is for sure wrong. The standard deviation of $𝑋$ is the square root of the variance.

The reason that it’s squared is that it’s always positive, which means that movements above the mean matter as much as movements below the mean. The reason it’s squared and not absolute value is that if you have tiny differences from the mean, and you square them, they disappear, while bigger ones get bigger. It minimizes tiny differences and maximizes larger ones. This is why it’s called the measurement of central tendency, because it measures how clustered values are around the mean.

Covariance

We have a sample space and we take variables. You measure how much variables fluctuate with each other with respect to the inputs from the sample space.

$Cov (𝑋, 𝑌) = 𝔼 {(𝑋 - 𝔼 [𝑋]) (𝑌 - 𝔼 [𝑌])}$

$= \sum_{𝜔 \in Ω} (𝑋 (𝜔) - 𝔼 [𝑋]) (𝑌 (𝜔) - 𝔼 [𝑌]) ℙ (𝜔)$

You’re going through every entry and seeing how much $𝑋$ differs from its mean and $𝑌$ differs from its mean, weighting them with their probability measure. It’s how they are different from each other in the same scenario.

We can say that two variables are uncorrelated if their covariance is 0. Independent variables are uncorrelated, but uncorrelated variables may not be independent. The value of covariance doesn’t tell you how correlated variables are, only the correlation coefficient does:

$𝜌 (𝑋, 𝑌) = \frac{Cov (𝑋, 𝑌)}{\sqrt{𝕍 (𝑋) 𝕍 (𝑌)}}$

If this value is 0, then they are uncorrelated. This coefficient is always $[- 1, 1]$ . −1 means perfectly inverse, and 1 means perfectly direct. This only has to do with direction, not necessarily magnitude.

Let’s consider a probability space. $𝒢︀$ is a sub- $𝜎$ -algebra of $ℱ︀$ , and $𝑋$ random variable. The conditional expectation of $𝑋$ given $𝒢︀$ is a random variable that is measurable i.e. $𝔼 [𝑋 | 𝒢︀]$ is $𝒢︀$ -measurable and partial averaging

If i limited myself just to a set $𝐴$ and I averaged $𝑋$ over $𝐴$ , I should get the same result as if I took the average of $𝑋$ . Underneath this subset, if I average over the $𝑋$ terms.

Conditional expectations are random variables.

Technically, the condition in a conditional expectation is a $𝜎$ -algebra.

Important theorems:

Linearity of conditional expectations: $𝔼 [𝑐_{1} 𝑋 + 𝑐_{2} 𝑌 | 𝒢︀] = 𝑐_{1} 𝔼 [𝑋 | 𝒢︀] + 𝑐_{2} 𝔼 [𝑌 | 𝒢︀]$ Let’s consider a situation where $𝑋$ can take one of two values and $𝑌$ can take one of two values, both dependent on $𝑎, 𝑏, 𝑐, 𝑑$ . If $𝑌 = 3$ , $𝜔 \in {𝑎, 𝑐}$ , then $𝔼 [𝑋 | 𝑌 = 3] = \sum_{𝜔 \in Ω} 𝑋 (𝜔) ℙ (𝜔 | 𝑌 = 3)$ and this is the same as $\sum_{𝜔 \in Ω} 𝑋 (𝜔) \frac{ℙ (𝜔 \cap {𝑎, 𝑐})}{ℙ ({𝑎, 𝑐})}$ from Bayes theorem Knowing that Y=3, that means $\sum_{𝜔 \in {𝑎, 𝑐}} 𝑋 (𝜔) \frac{ℙ (𝜔)}{2 / 3}$ $= \frac{3}{2} (𝑋 (𝑎) ℙ (𝑎) + 𝑋 (𝑐) ℙ (𝑐))$
We can take something that is known outside of an expectation. $𝔼 [𝑋 𝑌 | 𝒢︀] = 𝑋 𝔼 [𝑌 | 𝒢︀]$
Iterated conditioning. If $ℋ︀ \subset 𝒢︀$ , (think of this as information), then we know less information in $ℋ︀$ than in $𝒢︀$ . If I create a new random variable around $𝑋$ when you condition it just on this less information set, you get: $𝔼 [𝔼 [𝑋 | 𝒢︀] | ℋ︀] = 𝔼 [𝑋 | ℋ︀]$ We use iterated conditioning to complicate a process before simplifying it.
Independence: If $𝑋$ is independent of $𝒢︀$ , having the condition doesn’t do anything.
Conditional Jensen’s Inequality - not important

Martingale Processes

Consider an adapted stochastic process $𝑀 (𝑡)$ , for $0 \leq 𝑡 \leq 𝑇$ where $𝑇$ is a fixed positive number. The process needs to be such that if I know all the information up to time $𝑠$ , my best estimate for where it will be in the future, it will be the same as it is now. On average, I expect it to stay the same. Even though it probably won’t, I expect it to. If you expect it to grow, it’s a submartingale, and if you expect it to shrink, it’s a supermartingale.

Markov

If you have a stochastic process where for every non-negative function $𝑓$ on the number line, there is another $𝑔$ where you can describe the expectation for a future $𝑓 (𝑋 (𝑡))$ given the information in time $𝑠$ with $𝑔 (𝑋 (𝑠))$ . How much information do we need to calculate the value of this process? We would love to have a stock process that is Markov. This aligns with the idea that all past information is encoded in the current price, so we only need the current price and some function on that to determine a future price.

Random Walks

We can define the outcomes of the tosses of a coin where heads is 1, and tails is −1. The process of flipping a coin and calculating the sum of the values is the symmetric random walk $𝑀$ . These increments are independent from each other, whether symmetric or asymmetric. Knowing that there were a bunch of heads in the past doesn’t give you any information about future flips. Our increment is defined by:

$𝑀_{𝑘_{𝑖 + 1}} - 𝑀_{𝑘_{𝑖}} = \sum_{𝑗 = 𝑘_{𝑖} + 1}^{𝑘_{𝑖 + 1}} 𝑋_{𝑗}$

This starts at plus 1 because… wasn’t listening 😓

If we want to calculate the expected value of this, we can use the linearity of expectation to bring the expectation into the random variable: $\sum_{𝑖 = 𝑘_{𝑖} + 1}^{𝑘_{𝑖 + 1}} 𝔼 [𝑋_{𝑖}]$

By adding these together, we can see that the expectation of the random variable is 0.

Is variance a linear operator? If it is, then $Var (𝛼 𝑋 + 𝛽 𝑌) = 𝛼 Var (𝑋) + 𝛽 Var (𝑌)$

Using our definition of variance, this is the same as

$𝔼 [(𝛼 𝑋 + 𝛽 𝑌)^{2}] - 𝔼 [𝛼 𝑋 + 𝛽 𝑌]^{2}$

$= 𝛼^{2} 𝔼 [𝑋^{2}] + 2 𝛼 𝛽 𝔼 [𝑋 𝑌] + 𝛽^{2} 𝔼 [𝑌^{2}] - 𝛼^{2} 𝔼 [𝑋]^{2} - 2 𝛼 𝛽 𝔼 [𝑋] 𝔼 [𝑌] - 𝛽^{2} 𝔼 [𝑌]^{2}$

and this ends up equalling, once you cancel everything out

$= 𝛼^{2} Var (𝑋) + 2 𝛼 𝛽 Cov (𝑋, 𝑌) + 𝛽^{2} Var (𝑌)$

Clearly, this is not a linear operator. Only in the scenario where the variables are independent and their coefficient is 1, i.e. they are independent and identically distributed (iid), do they act as if they are linear.

When we look at the random walk $𝑀$ , we can see that each increment of the random walk is independent, and the coefficient is 1. So in this case, the summation of the variances happens to equal the variance of the summations.

What about $𝔼 [𝑋_{𝑗}^{2}]$ ? Because the negative outcome is squared, it becomes 1, which means that the variance is 1, so that the summation of variances is $𝑙 - 𝑘$ . If you have a walk that lasts 5 steps or 500 steps, the expectation is the same, but the variance goes from 5 to 500.

We can prove this symmetric random walk is a Martingale.

We need to prove for $𝑘 < 𝑙$ that

$𝔼 [𝑀 (𝑙) | ℱ︀ (𝑘)] = 𝑀 (𝑘)$

If we add a zero, nothing should happen, right?

$= 𝔼 [𝑀 (𝑙) - 𝑀 (𝑘) + 𝑀 (𝑘) | ℱ︀ (𝑘)]$

Now we can use the linearity of expectations to take out the $𝑀 (𝑘)$

$𝔼 [𝑀 (𝑙) - 𝑀 (𝑘) | ℱ︀ (𝑘)] + 𝔼 [𝑀 (𝑘) | ℱ︀ (𝑘)]$

for the purposes of this, $ℱ︀ (𝑘)$ is the information we would know at time $𝑘$ , and $𝑀 (𝑘)$ is definitely known with respect to that information. We also know that the difference between $𝑀$ is the increment, and that the increment is independent of information, so we can ignore it:

$𝔼 [𝑀 (𝑙) - 𝑀 (𝑘)] + 𝑀 (𝑘)$

We already know that the expectation of the increment is 0, so

$𝑀 (𝑘) = 𝑀 (𝑘)$

First Order Variation

This is the total change of the process. The first order variation in a function is the change in the y-axis. We’ll consider $Π$ as the partition of the process, going from ${𝑡_{0}, 𝑡_{1}, \dots, 𝑡_{𝑛}}$ , where this is a sequence of ordered values that ends at $𝑡$ . Let’s get an operator called the norm which gives you the length of the biggest piece of the partition: $| | Π | |$

We want a brute force way to calculate this first order variation for any function. We’ll cut up our function into a bunch of different pieces, and then look at the value of the function at every point (the “realizations” of this process). You can get all the changes in y at the different values and add them all up. In order to get the true value, we would add more and more points, maybe an infinite number of points. We need to take the limit as the norm of the partition approaches 0. The mechanical way to do this is to take the norm and cut it in half at every step:

$𝐹 𝑉_{𝑇} (𝑓) = {lim}_{| | Π | | \to 0} \sum_{𝑗 = 0}^{𝑛 - 1} | 𝑓 (𝑡_{𝑗 + 1}) - 𝑓 (𝑡_{𝑗}) |$

For a continuous and differentiable function, we can use the mean value theorem which means that for every increment, there exists a number which exists within that increment where the derivative of that point is equal to the slope between the beginning and end of the increment. In this case, we can use this fact to simplify this to get the Riemann integral.

Quadratic Variation

$[𝑓, 𝑓] (𝑇) = {lim}_{| | Π | | \to 0} \sum_{𝑗 = 0}^{𝑛 - 1} (𝑓 (𝑡_{𝑗 + 1}) - 𝑓 (𝑡_{𝑗}))^{2}$

Does not have a nice real world analogy. It is that same change squared. This is uninteresting outside of stochastic calculus because the quadratic variation for a continuous, differentiable function is always 0.

We can create a handy inequality here. We can take out the norm like this and say

${lim}_{| | Π | | \to 0} | | Π | | \sum_{𝑗 = 0}^{𝑛 - 1} (𝑓^{'} (𝑡_{𝑗}^{*}))^{2} (𝑡_{𝑗 + 1} - 𝑡_{𝑗})$

is greater than or equal to the same but with the difference squared, because the norm is the largest piece.

And because this is a continuously shrinking limit that’s being added together, we can say that this is an integral of a derivative. But we also know that because the term of the norm is here, that goes to 0, which means that the entire thing is 0. And because that is greater than or equal to the quadratic variation formula, that must also always be zero (because it has to be positive).

We will deal with some situations where the first order variation will be infinite, and quadratic variation will give us a more interesting result.

The random walk evolves in a discontinuous, non-differentiable way. It also is random, so there are multiple different versions of the same function. Subdividing this function with first order variation gives you that the expectation is 0 most of the time. We can consider it as

$𝐹 𝑉_{4} (𝑀) = \sum_{𝑗 = 1}^{4} | 𝑋_{𝑗} |$

so you get the total change as being 4. Open question: explain this step more clearly.

For quadratic order variation, this is

$[𝑀, 𝑀] (4) = \sum_{𝑗 = 1}^{4} (𝑋_{𝑗})^{2} = 4$

This does not necessarily occur! If the step size changed to 2 and −1, this is no longer true. The quadratic order variation is only the same as first order variation when the random walk is symmetric.

Scaled Symmetric Random Walk

Brownian motion is an approximation for a process that constantly fluctuates. Our random walk jumps up and down simplistically. We need to start taking steps a lot more often, but the problem with that is it would significantly magnify our variation without changing much. We also need to decrease step size. We create more and more steps by subdividing the x-axis, and make them smaller. This gives us our scaled symmetric random walk. We will take $𝑛$ times as many steps, but the step size is decreased by $\sqrt{𝑛}$ :

$𝑊^{(𝑛)} (𝑡) = \frac{1}{\sqrt{𝑛}} \sum_{𝑗 = 1}^{𝑛 𝑡} 𝑋_{𝑗}$

We still have independence of increments in this walk. The expected value of a walk is still 0. We can use the same arguments to determine that this is a martingale. But if we look at the variance of the increments in the scaled symmetric random walk, it is now

$Var (\sum_{𝑗 = 𝑛 𝑠 + 1}^{𝑛 𝑡} \frac{1}{\sqrt{𝑛}} 𝑋_{𝑗})$

This presents a problem for our attempt to convert between summation of variances because there is now a coefficient. But because $𝛼 = 𝛽$ in this case, Open question: explain this conversion more clearly.

$= \sum_{𝑗 = 𝑛 𝑠 + 1}^{𝑛 𝑡} \frac{1}{𝑛} Var (𝑋_{𝑗})$

$= \frac{1}{𝑛} (𝑛 𝑡 - 𝑛 𝑠) = 𝑡 - 𝑠$

The reason it has to be $\sqrt{𝑛}$ and not $𝑛$ is because in this formula, the variance would be going to 0 as $𝑛$ goes to $\infty$ .

The quadratic variation becomes $𝑡$ , but the first order variation goes to $\infty$ .

Normal Distribution

We will prove that the scaled symmetric walk approaches the normal distribution. Any normal variable is defined by $𝜇$ and $𝜎^{2}$ . It also has a

$𝜑_{𝑊^{(𝑛)} (𝑡)} (𝑠) = 𝔼 [𝑒^{𝑠 𝑊^{(𝑛)} (𝑡)}]$

$= 𝔼 [𝑒^{\sum_{𝑗 = 1}^{𝑛 𝑡} \frac{𝑠}{\sqrt{𝑛}} 𝑋_{𝑗}}]$

$= 𝔼 [\prod_{𝑗 = 1}^{𝑛 𝑡} 𝑒^{\frac{𝑠}{\sqrt{𝑛}} 𝑋_{𝑗}}]$

This is the expected value of the product as a whole bunch of independent variables, which means that it’s the same as the product of the expectations:

$= \prod_{𝑗 = 1}^{𝑛 𝑡} 𝔼 [𝑒^{\frac{𝑠}{\sqrt{𝑛}} 𝑋_{𝑗}}]$

Important: understand what’s random and what’s not. $𝑋_{𝑗}$ only is random. Given the expectation of $𝑋_{𝑗}$ being known:

$= \prod_{𝑗 = 1}^{𝑛 𝑡} (\frac{1}{2} 𝑒^{\frac{𝑠}{\sqrt{𝑛}}} + \frac{1}{2} 𝑒^{- \frac{𝑠}{\sqrt{𝑛}}})$

this is the same as

$= {(\frac{1}{2} 𝑒^{𝑠 / \sqrt{𝑛}} + \frac{1}{2} 𝑒^{- 𝑠 / \sqrt{𝑛}})}^{𝑛 𝑡}$

This is the moment generating function

We are trying to prove that

${lim}_{𝑛 \to \infty} 𝜑_{𝑊^{(𝑛)} (𝑡)} (𝑠) = 𝑒^{\frac{1}{2} 𝑠^{2} 𝑡}$

because this means that the walk is normally distributed. We can take the natural log of both sides

${lim}_{𝑛 \to \infty} log (𝜑_{𝑊^{(𝑛)} (𝑡)} (𝑠)) = \frac{1}{2} 𝑠^{2} 𝑡$

Let’s make a convenience variable $𝑥$

$𝑥 = \frac{1}{\sqrt{𝑛}}$

We are satisfied that the limit of $𝑛$ going to $\infty$ is the same as $𝑥$ going to 0.

${lim}_{𝑥 \to 0^{+}} (\frac{𝑡}{𝑥^{2}} log (\frac{1}{2} 𝑒^{𝑠 𝑥} + \frac{1}{2} 𝑒^{- 𝑠 𝑥}))$

this results in a 0/0, which is undesirable. In order to get rid of this situation, we will apply L’Hopital’s rule.

$\overset{𝐿}{=} \frac{𝑠 𝑡}{2} {lim}_{𝑥 \to 0^{+}} (\frac{\frac{1}{2} 𝑒^{𝑠 𝑥} - \frac{1}{2} 𝑒^{- 𝑠 𝑥}}{𝑥})$

which is still indeterminate, so we will apply it once again, which will give us the normal distribution

Open question: copy the limiting argument from the video.

Brownian Motion

Suppose there is a continuous function $𝑊 (𝑡)$ which starts at $𝑊 (0) = 0$ and depends on $𝜔$ . We have increments that are independent and each of the increments are normally distributed. The expectation is 0, and the variation is $𝑡_{𝑖 + 1} - 𝑡_{𝑖}$ . The quadratic variation of Brownian motion is known to be $𝑡$ . Because this is a continuous, non-differentiable variable, its first order variation goes to infinity.

Let’s talk about mean squared convergence. We can define a random process that will converge to our quadratic variation, and if its square does as well, and the mean of that square converges as well, the expectation must as well.

We have a sample quadratic variation for a random partition

$𝑄_{Π} = \sum_{𝑗 = 0}^{𝑛 - 1} (𝑤 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗}))^{2}$

We will make more intricate partitions as we go on so that the limit approaches 0.

I want to show that

${lim}_{| | Π | | \to 0} 𝔼 [𝑄_{Π}] = 𝔼 [𝑇]$

where $𝑇$ is a constant. Because $𝑇$ is a constant, expectation is itself and variance is 0. We can then show that we want $𝑄_{Π}$ to approach both the variance and the value of $𝑇$ .

$𝔼 [𝑄_{Π}] = 𝔼 [\sum_{𝑗 = 0}^{𝑛 - 1} (𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗}))^{2}]$

We can use the linearity of expectation to get

$= \sum_{𝑗 = 0}^{𝑛 - 1} 𝔼 [(𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗}))^{2}]$

Using our previous definition of variance as

$𝔼 [𝑋^{2}] = Var (𝑋) + 𝔼 [𝑋]^{2}$

We know that the mean of the increment of Brownian motion is 0 and its variation is time $𝑡$ .

So this becomes

$\sum_{𝑗 = 0}^{𝑛 + 1} (𝑡_{𝑗 + 1} - 𝑡_{𝑗}) = 𝑡$

therefore,

${lim}_{| | Π | | \to 0} 𝔼 [𝑄_{Π}] = 𝑡$

because the limit of a constant is the same as the constant.

$Var ((𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗}))^{2}) = 𝔼 [(𝑊 (𝑡_{𝑗 + 1}) - 𝑊 (𝑡_{𝑗}))^{4}]$

Without standardization, the fourth moment is $3 𝜎^{4}$

The variance of an increment is the duration of time.

${lim}_{∥ Π ∥ \to 0} Var (𝑄_{Π}) = 2 {lim}_{∥ Π ∥ \to 0} \sum_{𝑗} (𝑡_{𝑗 + 1} - 𝑡_{𝑗})^{2} = 0$

We have derived the quadratic variation of time. 🤯

Time is a continuous, differentiable function, so it’s 0.

Overall: we have determined that the quadratic variation of Brownian motion is equal to the time almost surely, since any individual continuous differentiable function formed by Brownian motion has probability 0 of occurring.

## Monte Carlo Risk Simulation - URL: https://sharifhsn.dev/blog/monte-carlo-risk-simulation/ - Structured data: https://sharifhsn.dev/api/posts/monte-carlo-risk-simulation.json - Description: 50 bps cut! Markets exploded today. Information gets priced in a nonlinear way Today’s topic will be a little independent of what we’ve done, so it will be more into the… - Date: 2024-09-19 - Exact published timestamp: 2024-09-19 - Topics: Risk Management, Monte Carlo, Simulation - Categories: Risk Management - Source: FE-535 \| Risk Management - Source URL: None - Content source format: typst

Week 3

Lecture Notes

Fed Talk

50 bps cut! Markets exploded today. Information gets priced in a nonlinear way

Today’s topic will be a little independent of what we’ve done, so it will be more into the quantitative analysis, simulation, law of large numbers.

Lab Solutions

labs are only 10% of the grade, one point or two is not a big deal, don’t stress

if you look at the annual sharpe ratio, you have an average sharpe ratio which changes over time.

We have weekly returns, based on the end price divided by the previous one. Continuous allows us to use derivations, even if real life is discrete. You can either cut into pieces or eat whole pizza pie. Does it matter? Depends on the assumption we’re making. We can actually represent this as a summation of returns. When we’re trying to understand the expectation, we can use the linearity of expectations.

Annualization also allows you to expand data.

Limit is 15 pages for projects, keep it compact, only wants to see results

R has apply function for functional programming

Open question: Regress $𝜇_{𝑝}$ on $𝜎_{𝑝}$ , what’s going to be the $𝛼$ , the $𝛽$ ?

You have this market representative agent who picks the most efficient portfolio with a high Sharpe ratio and says you don’t need to do something else. It’s the same with picking a pensions, dictates how you allocate your money, higher risk equity if you’re younger.

Monte Carlo Methods

Assumes some quantitative analysis background.

Monte Carlo is a simulation method. Example: let’s flip this coin and see the next result. We don’t know for sure, but if we run it 100 times, it might be around 50/50 heads and tails. We run an R simulation to do this more and more times. As N number of trials increases, we get closer to 50% on average. This applies to any distribution, which helps us avoid solving complicated mathematical equations. It gives us a lot of good solutions. Let’s have $𝑋 𝑁 (𝜇, 𝜎^{2})$ normal distribution.

What’s the expectation?

$𝔼 [𝑥] = 𝜇$

Second moment? Variance

$𝔼 [𝑥^{2}] = 𝜎^{2} + 𝜇^{2}$

Third moment (skew) becomes more complicated

Mean of (x^2) ends up being 5 when (x) is uniform on ((1,2)).

We use this a lot when looking at options and derivatives

If you ask one person how many M&Ms we can put in a jar, it will probably be wrong. But if you ask 1000 people, it will be more right on average. “The wisdom of the crowd.” This is the Law of Large Numbers. You can figure out the values and moments you’re actually looking at. We will price some games as our motivation for finance.

Maybe it’s the same guy, which makes it biased

Bayesian is far away, long-term, this thing only gets one chance.

Monte Carlo (MC) methods allow risk managers to avoid complex analytical solutions and derive complex distributions, price complex instruments. The major part of managing risk is identifying it and understanding its sensitivity. The simulation will help us stress test. What about when things go bad? This will be useful.

Black-Scholes became popular because no one had laptops.

MC relies heavily on the model’s assumption. It needs to have the right distribution, parameters, pricing function. If you have a contract to get $5 if this happens, if you model risk issues, what’s the parameters?

Model risk is always a part of things

Calibration of parameters will be taken for granted.

By flipping a coin, you can either get a $1 heads or lose a $1 tails. There’s a premium to participate in this game, so how much would you pay? Depends if it’s a fair coin or not a fair coin.

You can think of it as $𝜋$ as the probability of heads and $1 - 𝜋$ as tails. Therefore, we can model this problem as

$𝜋 \times 1 + (1 - 𝜋) * - 1 - 𝑝 = 0$

where $𝑝$ is the premium to pay into the game.

To solve this we get $𝑝 = 2 𝜋 - 1$ .

In the case when $𝜋$ is $0.5$ in the fair case, then the premium is 0. If $𝜋 < 0.5$ , then you actually get paid to participate.

In Atlantic City, they give you a free bus ride in order to get you in. The odds are against you, so as long as you come in, they’re making money.

How would you price this?

The price of a security is

$𝑝 = \sum_{𝑠 = 1}^{𝑆} 𝐶 𝐹_{𝑠} \times ℙ (𝑠)$

where cash flows from state $𝑠$ and the probability of $𝑠^{𝑎}$ is probability.

The martingale is the fair game, where expected value is the same as the initial investment.

Lottery tickets are not a fair game. You pay a higher premium for the euphoria of imagining you win.

Let’s consider a complicated game. We have a fair die with equal probability of the results. We want to roll the die 3 times, with three results $𝑋_{1}$ , $𝑋_{2}$ , and $𝑋_{3}$ . We consider $𝑋_{max}$ , where if it’s 6, you get a dollar, otherwise you get nothing. What is the premium for this game? This is actually a binary or digital option.

We can solve this two ways. Analytically, and numerically.

We can combine each of these states into an indicator variable, which will tell us whether the maximum is 6 or not.

$𝔼 [𝐼_{𝑛}] = 𝑝$

$ℙ (𝑋_{max} \leq 𝑖) = (𝑖 / 6)^{3}$

We understand this as $1 - (5 / 6)^{3} = 0.4213$

Numerical (my own)

PS C:\Users\sharif> python
>>> import rand
Traceback (most recent call last):
ModuleNotFoundError: No module named 'rand'
>>> import random
>>> random.randint()
TypeError: Random.randint() missing 2 required positional arguments: 'a' and 'b'
>>> random.randint(1, 6)
2
>>> v = [[random.randint(1, 6), random.randint(1, 6), random.randint(1, 6)] for i in range(1, 1000000)]
>>> m = [int(max(c) == 6) for c in v]
>>> sum(m)/len(m)
0.42147742147742145
>>> 1 - (5/6)**3
0.42129629629629617

Numerical (Professor’s)

Done in Excel, which approximates the answer.

We can establish the weak law as such

${lim}_{𝑁 \to \infty} \frac{\sum_{𝑛 = 1}^{𝑁} 𝑋_{𝑛}}{𝑁} \to 𝜇$

Sports betting is a binary option.

There’s a certain probability something will happen, and you will get nothing otherwise. We can play the game under different conditions than the maximum of dice rolls.

If the current stock price is $100, the stock goes beyond $110 next month, you get paid $1, if not you get nothing and you lose the down payment $𝑝$ . What is the fair price of $𝑝$ . We can consider two regions, the winning and losing. If we have some model that will allow us to model the different scenarios, we can analyze numerically what might happen.

We need the return to be above 9.53% (because of interest rates it’s not 10%), so our option price is

$ℙ (𝑅_{1} > 9.53 %) .$

If we know that returns are normally distributed on $𝑅_{1} 𝑁 (0.02, {0.04}^{2})$ , then

$ℙ (𝑅_{1} > 0.0953) = 1 - 𝜙 (\frac{0.0953 - 0.02}{0.04}) = 1 - 0.9701 \approx 3 %$

based on the formula

$= 𝜙 (\frac{𝐾 - 𝜇}{𝜎})$

If you have high volatility, then there’s a higher probability of exercising the option, and vice versa. Nobody writes options for penny stocks.

European options have a closed form solution. American options have no analytical solution. Asian options also don’t (what are these? don’t know).

## Random Variables and CDFs - URL: https://sharifhsn.dev/blog/random-variables-and-cdfs/ - Structured data: https://sharifhsn.dev/api/posts/random-variables-and-cdfs.json - Description: We can consider a mapping between all of the sets in the 𝜎 -algebra of one measurable space ( Ω , ℱ︀ ) to another sample space Ω 1 , where the mapping is measurable each set has… - Date: 2024-09-16 - Exact published timestamp: 2024-09-16 - Topics: Probability Theory, Random Variables, CDFs - Categories: Probability Theory - Source: FE-540 \| Probability Theory - Source URL: None - Content source format: typst

Week 3

Reading - Random Variables Generalities (FT 3)

Definition

We can consider a mapping between all of the sets in the $𝜎$ -algebra of one measurable space $(Ω, ℱ︀)$ to another sample space $Ω_{1}$ , where the mapping is measurable each set has a mapping. A random variable is a measurable function $𝑋 : Ω \to ℝ$ where the target sample space is the Borel set. Random variables are capital letters and their values are lowercase letters.

The sum, product, or composition of measurable functions are measurable functions, so those are still random variables.

We consider the preimage of $𝑋$ to be $𝑋^{- 1} (𝐴)$ , which is not precisely the inverse function. It’s the set in the $𝜎$ -algebra in the input sample space which generated the set $𝐴$ .

Distribution

The distribution of a random variable is itself a random variable, denoted by $ℙ_{𝑋} = ℙ \circ 𝑋^{- 1} (ℝ)$ . This variable is defined on the Borel set as a probability on the measurable space. We can consider the probability space of the distribution to be $(ℝ, 𝔹 (ℝ), ℙ 𝑋)$ i.e. the sample space is all real numbers, the function mapping is the Borel subset (every interval and combination of such), and the distribution is the probability measure.

If we have a sequence of disjoint sets in the Borel subset of $ℝ$ , then the distribution of $𝑋$ ( $ℙ (𝑋)$ ) on the unions of those sequences is the same as the sum of the probability measure on each individual set.

We will consider the notation ${𝑋 \in 𝐴}$ to mean that there’s an outcome $𝜔$ in sample space $Ω$ $𝑋$ which maps to the set $𝐴$ in $ℬ︀ (ℝ)$

Cumulative Distribution Function

Formally, the cdf of a random variable is $𝐹 (𝑥) = ℙ ({𝜔 : 𝑋 (𝜔) \leq 𝑥})$ or the probability measure of the outcomes where the mapping to the real numbers is less than a specified number x.

This function has the properties that it is increasing, right continuous, and it approaches 0 and 1 from left and right, respectively.

The infinimum $inf$ is the largest element smaller than every element of a set (aka it’s the element just to the left of the set in the order) and the supremum $sup$ is the element to the right.

If we consider the variable.

For every cdf that exists, there exists a random variable that has that cdf.

A Borel function is a measurable function that maps from $ℝ$ into $ℝ$ .

Lecture Notes

CDF

$𝐹 : ℝ \to [0, 1]$ with $𝐹 (𝑥) ≔ 𝑃_{𝑋} ((- \infty, 𝑥]) = 𝑃_{𝑋} (𝐵)$ , where $𝐵 = (- \infty, 𝑥]$

These are equivalent ways of writing a cdf. This counts the accumulated probability less than or equal to x.

More interestingly, the cdf has these properties: increasing

The increasing property is fairly intuitive. Rigorously, you can show that for $𝑦 \geq 𝑥$ , $(- \infty, 𝑥] \subseteq (- \infty, 𝑦]$ . The interval from x to y must be 0 or greater, so then the cdf to y must be greater than the cdf to x.

You can also show that it’s right continuous, proof is beyond the scope, involves monotone class

${lim}_{ℎ ↘︎ 0} 𝐹 (𝑥 + ℎ) = 𝐹 (𝑥)$ for any x in the domain.

Let’s examine the first quadrant of an equation. We have $𝑥$ . If we imagine the difference between that and $𝑥 + ℎ$ , in a continuous function this is well-defined. But we might have a discontinuous function. This cannot be a cumulative distribution function. This is a complicated statement.

${lim}_{𝑥 \to - \infty} 𝐹 (𝑥) = 0, {lim}_{𝑥 \to \infty} 𝐹 (𝑥) = 1$

We can prove this by showing that

${lim}_{𝑥 \to - \infty} 𝐹 (𝑥) = 0 = {lim}_{𝑥 \to - \infty} 𝑃_{𝑋} ((- \infty, 𝑥]) = 𝑃_{𝑋} (\emptyset) = 0$

Why is this empty set? Because if you have the interval $(- \infty, \infty]$ , left open interval says I don’t have it, right says I do, because they conflict then there must be an empty set.

${lim}_{𝑥 \to \infty} 𝐹 (𝑥) = 1 = 𝑃_{𝑋} ((- \infty, \infty]) = 𝑃_{𝑋} (ℝ) = 1$

This just becomes the probability measure of the sample space which is definitionally 1.

Most of the focus in this course will be calculation and not proof. But if you understand the proof, it will guide you in calculations.

Lemma 3.11: If we have $𝐹 (𝑋)$ as the cdf, then

$ℙ (𝑋 \geq 𝑥) = 1 - ℙ (𝑋 \leq 𝑥)$

This is known as the survival function. If you survive over sixty years, then you get your retirement money. This assumes that $ℙ (𝑋 = 𝑥) = 0$ which is true if $𝑋$ is a continuous random variable.

Proof is ${𝑋 > 𝑥} \cup {𝑋 \leq 𝑥} = ℝ = Ω$

Important identity:

$ℙ (𝑥 < 𝑋 \leq 𝑦) = 𝐹 (𝑦) - 𝐹 (𝑥)$

This proof comes from the set inequality: ${𝑋 \leq 𝑥} \cup {𝑥 < 𝑋 \leq 𝑦} = {𝑋 \leq 𝑦}$

And then if we take the probability measure of all of these then we see that the right side is the cdf of y, left side is cdf of x disjoint with the other part so you can add them by earlier definition

Also known

$ℙ (𝑋 = 𝑥) = 𝐹 (𝑥) - 𝐹 (𝑥 -)$ this is us taking the left limit where

$𝐹 (𝑥 -) ≔ {lim}_{𝑦 ↗︎ 𝑥} 𝐹 (𝑦) = {lim}_{ℎ ↘︎ 0} 𝐹 (𝑥 - ℎ)$

consequences of this? If this $𝐹 (𝑥)$ is left-continuous, it means that the limit is equal, which means this whole thing is equal to 0. However, it’s not left-continuous, then $ℙ (𝑋 = 𝑥) > 0$ strictly larger than 0. This looks like when the dot is on the right and the open circle is on the left. This is allowed, and it happens when $𝑋$ is a discrete random variable. Binomial, Poisson, Geometric random variables are all discrete.

Numerical

Roll a die, probability that number is smaller than five. (strictly less than on an exam)

$ℙ (𝑋 < 5) = ℙ (𝑋 \leq 4) = 𝐹 (4) = ℙ ({𝑋 = 1} \cup {𝑋 = 2} \cup {𝑋 = 3} \cup {𝑋 = 4})$

$= ℙ (𝑋 = 1) + ℙ (𝑋 = 2) + ℙ (𝑋 = 3) + ℙ (𝑋 = 4)$

$= \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6}$

$= \frac{2}{3}$

alternatively, I could do the same with 5 and 6 and use the 1 - property. Two random variables are independent if

$ℙ (𝑋 \in 𝐴, 𝑌 \in 𝐵) = ℙ (𝑋 \in 𝐴) \times ℙ (𝑌 \in 𝐵)$

or put another way

$= ℙ ({𝑋 \in 𝐴} \cap {𝑌 \in 𝐵})$

Intuitively, this means that the outcome of $𝑋$ does not affect the outcome of $𝑌$ .

For any $𝑥, 𝑦 \in ℝ$ , where $𝑥, 𝑦$ are independent

$ℙ (𝑋 \leq 𝑥, 𝑌 \leq 𝑦) = ℙ (𝑋 \leq 𝑥) \times ℙ (𝑌 \leq 𝑦)$

are mutually independent if

$ℙ (⋂_{𝑖 = 1}^{𝑛} {𝑥_{𝑖} \in 𝐵_{𝑖}}) = \prod_{𝑖 = 1}^{𝑛} ℙ (𝑋_{𝑖} \in 𝐵_{𝑖})$

They are also pairwise independent in this case.

For an arbitrary family of random variables where we have some index set $𝐼$ for $𝑋_{𝑖}$ , they are mutually independent if every finite subfamily is independent. That means any finite subset of the indices will cause this to be the case. This is not very useful as a way to prove independence because there are so many cases to prove, this is just a definition.

Let’s say we roll a fair die twice and we have $𝑋$ represent the maximum of both rolls. The first question we’re interested in calculating the image of $Ω$ through $𝑋$ and the distribution of $𝑋$ .

$Ω = {𝜔 = (𝑗_{1}, 𝑗_{2}), 𝑗_{1}, 𝑗_{2} \in {1, 2, 3, 4, 5, 6}}$

total of 36 elements

the probability of observing any particular outcome is $\frac{1}{36}$ because it’s fair. This $𝑋$ is clearly a mapping from the sample space to the set of six numbers. If we calculate probabilities:

$ℙ (𝑋 = 1) = ℙ_{𝑋} (𝜔 = (1, 1)) = \frac{1}{36}$

$ℙ (𝑋 = 2) = ℙ_{𝑋} (𝜔 = (1, 2)) + ℙ_{𝑋} (𝜔 = (2, 1)) + ℙ_{𝑋} (𝜔 = (2, 2)) = \frac{3}{36}$

$ℙ (𝑋 = 3) = ℙ (⋃_{𝑘 = 1}^{3} {𝜔 = (3, 𝑘} \cup ⋃_{𝑘 = 1}^{2} {𝜔 = (𝑘, 3)}) = \frac{5}{36}$

This is 3 then 2 because you don’t want to double count the same number.

$ℙ (𝑋 = 4) = ℙ (⋃_{𝑘 = 1}^{4} {𝜔 = (4, 𝑘} \cup ⋃_{𝑘 = 1}^{3} {𝜔 = (𝑘, 4)}) = \frac{7}{36}$

$ℙ (𝑋 = 5) = ℙ (⋃_{𝑘 = 1}^{5} {𝜔 = (5, 𝑘} \cup ⋃_{𝑘 = 1}^{4} {𝜔 = (𝑘, 5)}) = \frac{9}{36}$

$ℙ (𝑋 = 6) = ℙ (⋃_{𝑘 = 1}^{6} {𝜔 = (6, 𝑘} \cup ⋃_{𝑘 = 1}^{5} {𝜔 = (𝑘, 6)}) = \frac{11}{36}$

Now let’s get the cdf $𝐹 (𝑋)$ . Let’s consider each instance of $𝐹 (𝑏) = ℙ (𝑋 \leq 𝑏)$

Obviously, this has to be 0 for less than 1 because that’s not an outcome. For the rest of them, you add up the probabilities (because they’re disjoint).

Another example.

You are given

$ℙ (0 \leq 𝑋 \leq 1) = \frac{8}{12}$

$ℙ (0 \leq 𝑋 \leq 2) = \frac{7}{12}$

$ℙ (0 \leq 𝑋 \leq 3) = \frac{10}{12}$

$ℙ (𝑋 = 3) = ℙ (𝑋 \geq 4)$

We want to know the probability for up to 3.

We can solve for $ℙ (𝑋 = 0) = \frac{3}{12}$

We also want to show this inequality $𝑋 (𝜔) \leq 𝑌 (𝜔), \forall 𝜔 \in Ω$

You want to show that $𝐹_{𝑋} (𝑎) \geq 𝐹_{𝑌} (𝑎), 𝑎 \in ℝ$

Hint: $𝐹 (𝑋 (𝑎) = ℙ (𝑋 \leq 𝑎) ? ℙ (𝑦 \leq 𝑎) = 𝐹_{𝑌} (𝑎)$

More interesting example 3.4

cdf

you have to find the conditions on a, b, c, d, that $𝐹$ satisfies the properties of a cdf.

We have the property

${lim}_{𝑥 \to \infty} 𝐹 (𝑥) = 1 = 𝑑$

Because that’s if $𝑡 \geq 3$ .

a gives you the same solution of 0.

## What Washing Machines Can Teach Us About Tariffs - URL: https://sharifhsn.dev/blog/linkedin-2024-09-12-washing-machine-tariffs/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2024-09-12-washing-machine-tariffs.json - Description: A look at the 2018 washing-machine tariffs, consumer costs, and U.S. manufacturing. - Date: 2024-09-12 - Exact published timestamp: 2024-09-12T11:54:43.644Z - Topics: Tariffs & Trade, Trade Policy - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/posts/sharif-haason_what-washing-machines-can-teach-us-about-activity-7239964645996863489-Bq7D - Content source format: typst

Tuesday’s presidential debate featured tariffs as a topic, but they were explained poorly. Here’s a deep dive into a real-life Trump tariff to understand what his plan might mean: the 2018 washing machine tariffs.

Let’s say you’re a major retailer like Walmart. You might buy most of the washing machines that you sell at your stores from Samsung Electronics, which is based in South Korea, because of their cost and product popularity. In 2018, Trump instituted a tariff on imported washing machines starting at 20% for the first 1.2 million machines and then 50% for additional imports. This means that the majority of your washing machines just became 1.2x as expensive for you to purchase.

Where does that extra cost go? Well, like any other operating cost, it is passed down to the consumer. A study from University of Chicago in 2019 estimated that prices of washing machines increased 12% as a result of the tariffs. Dryers, a complementary good, rose in price as well. This price rise had a total consumer cost of $1.5 billion per year. [1]

However, it’s not all bad news. The expense of imports caused both Samsung and LG Electronics to each build a factory in the United States in previously blighted areas, investing hundreds of millions of dollars and creating over 2000 new jobs in the process. The price shocks faded as well, and by 2020, the cost of washing machines had returned to its previous levels. The tariffs expired in 2023 and were not renewed by President Biden. [2]

The UChicago study also had some flaws. It only analyzed the four month period with the highest prices and generalized it into a per-annum statistic, meaning that the true cost was very likely to be lower than the number they gave. [2]

One of the proposed benefits of tariffs is that they strengthen incumbent domestic industries, in this case Whirlpool Corporation. But a comprehensive report by the US International Trade Commission found that these manufacturers continued to decline in sales, wages, and capacity, with the gains in domestic manufacturing being entirely concentrated in the new Samsung and LG factories. It seems that the tariffs benefited some American workers, but they didn’t help American businesses at all. [3]

What’s the overall takeaway from the washing machine tariffs? If you take UChicago’s numbers at face value, each new job represented around $800,000 in consumer costs. Although the numbers are inflated for the period they analyzed, there was continued increased cost after the study was published as well, which likely balances out the number as a representation for the total cost, but not a per annum one.

What do you think? Were the tariffs worth it?

1 · 2 · 3

The linked analysis: What washing machines can teach us about the cost of tariffs.

## Probability Spaces and Filtrations - URL: https://sharifhsn.dev/blog/probability-spaces-and-filtrations/ - Structured data: https://sharifhsn.dev/api/posts/probability-spaces-and-filtrations.json - Description: Let’s define an infinite probability space Ω ∞ . This sample space is uncountably infinite, which means that the probability of any particular outcome is zero. Therefore, in… - Date: 2024-09-12 - Exact published timestamp: 2024-09-12 - Topics: Stochastic Calculus, Filtrations, Measure Theory - Categories: Stochastic Calculus - Source: FE-610 \| Stochastic Calculus - Source URL: None - Content source format: typst

Week 1

Readings

General Probability Theory

Infinite Probability Spaces

Let’s define an infinite probability space $Ω_{\infty}$ . This sample space is uncountably infinite, which means that the probability of any particular outcome is zero. Therefore, in order to determine the probability of an event, we define the formula for the probability of certain simple events, then use probability measures to determine the probability of more complicated events.

$𝜎$ -algebras form the basis of these simple events:

𝐸 𝑥 𝑝 𝑙 𝑎 𝑛 𝑎 𝑡 𝑖 𝑜 𝑛 𝑜 𝑓 (𝜎) - 𝑎 𝑙 𝑔 𝑒 𝑏 𝑟 𝑎 𝑠 𝑓 𝑜 𝑢 𝑛 𝑑 𝑖 𝑛 𝐿 𝑒 𝑐 𝑡 𝑢 𝑟 𝑒 𝑁 𝑜 𝑡 𝑒 𝑠

Lebesgue Measures

The probability of picking a number in between ⅓ and ⅔ is ⅔ - ⅓ = ⅓. This is also true for open intervals because the probability of an individual point occurring is 0.

We can apply $𝜎$ -algebras to the Lebesgue measure by generating a $𝜎$ -algebra from the collection of all closed intervals in our interval, like:

$[0, \frac{1}{3}], [\frac{1}{2}, 1], [\frac{1}{4}, \frac{230}{243}]$

and so on. Open intervals can be defined as a union of a sequence of closed intervals, so these are included as well. The $𝜎$ -algebra generated by this process is the Borel $𝜎$ -algebra, specifically of the subset $[0, 1]$ , which is $ℬ︀ [0, 1]$ . The Borel $𝜎$ -algebra is a well-defined set of events (remember, an event is a set) where the outcome is a real number. In this context, under the rules of $𝜎$ -algebras, an event is an interval, a union of intervals, a complement of an interval, or any combination of these.

Coin Tosses

Let’s think about coin tosses. Through filtration (see notes), we can define the probability of every event that can be described as a finite number of coin tosses. We can determine the probability of every individual sequence in $Ω_{\infty}$ by showing that their probabilities converge to zero since they are fractions that multiply together infinitely.

We can use the rules of $𝜎$ -algebras and probability measures to determine the probability of each of these events, like how the probability measure of the union of disjoint events (remember, events are sets) is equivalent to the summation of the probability measures of each event.

There are some events that are not easily computed unless we define them in this respect. Consider

${lim}_{𝑛 \to \infty} \frac{𝐻_{𝑛} (𝜔_{1} \dots 𝜔_{𝑛})}{𝑛} = \frac{1}{2}$

which describes all sequences of tosses where the long-run probability of choosing heads is ½. This set must be in ${ℱ︀}_{\infty}$ so we can determine its probability measure. If we choose positive integers $𝑚$ and $𝑛$ , we can define

$𝐴_{𝑛, 𝑚} = {𝜔; | \frac{𝐻_{𝑛} (𝜔_{1} \dots 𝜔_{𝑛})}{𝑛} - \frac{1}{2} | \leq \frac{1}{𝑚}}$

where this set is in ${ℱ︀}_{𝑛}$ i.e. that step in the filtration. We can define this as a combination of sets if we recognize that for each threshold $𝑚$ , there is an $𝑁$ where for all $𝑛$ after $𝑁$ i.e. for all sufficiently long sequences after a certain point, the average of the sequence meets the threshold, following the definition of a limit. By considering intersections as “all of these are true” and unions as “at least one must be true”, we can define this as

$𝐴 = ⋂_{𝑚 = 1}^{\infty} ⋃_{𝑁 = 1}^{\infty} ⋂_{𝑛 = 𝑁}^{\infty} 𝐴_{𝑛, 𝑚}$

By using the law of large numbers (large enough sequences will converge to the true value), we can see that $ℙ (𝐴) = 1$ for probability of ½, i.e. that the probability of an outcome belonging to this set is 1

Note here that this is not the same thing as saying that every outcome occurs in this set. We can easily imagine outcomes that do not. However, any individual event has a probability of 0 of occurring, so all those exceptional events together have a probability of 0. The way we describe this technically is that the event corresponding to the set with probability measure of 1 occurs almost surely.

Random Variables and Distributions

Let’s define a random variable in the context of all this measure theory. We’ll generalize the random variable to a function $𝑋$ with the property that (in probability space $(Ω, ℱ︀, ℙ)$

${𝑋 \in 𝐵} = {𝜔 \in Ω; 𝑋 (𝜔) \in 𝐵}$

where $𝐵$ is a Borel subset of $ℝ$ , this subset of $Ω$ is in $𝜎$ -algebra $ℱ︀$ . Essentially, this defines $𝑋$ as a function mapping outcome $𝜔$ to a Borel subset i.e. an interval or combination of intervals on the real number line. This is the formal way to turn probability measures into actual numbers (see dice example from lecture notes).

We are not interested in $𝑋$ taking a specific value as that will typically be 0, so we will usually talk about the probability that $𝑋 \in 𝐵$ .

Stock Prices

Let’s consider the binomial asset pricing model where $𝑆_{0} (𝜔) = 4$ , $𝑢 = 2$ , and $𝑑 = 1 / 2$ , which is generalized as

i.e. the stock price doubles on heads and halves on tails. Let’s think of the probability of the stock price now as the distribution of mass. $𝑆_{0} = 4$ occurs with probability one, so this random variable has a unit of mass on 4. If we move two step forwards, $ℙ {𝑆_{2} = 16} = 𝑝^{2}$ , $ℙ {𝑆_{2} = 4} = 2 𝑝 𝑞$ and $ℙ {𝑆_{2} = 1} = 𝑞^{2}$ , where $𝑝$ and $𝑞$ are the probabilities of $𝑢$ and $𝑑$ , respectively. There are three lumps of mass distributed between 16, 4, and 1, each with the size of their probabilities. Now let’s think of spreading that mass continuously. The distribution of a random variable is about the amount of mass in a set. This distribution measure $𝜇_{𝑋}$ of random variable $𝑋$ is a probability measure $𝑃$ , but it is done on $𝐵$ rather than subsets of $Ω$ . In other words, it measures the probability of $𝑋$ being in a particular Borel subset.

Random variables have distributions, but they are not the same thing. The variable is the mapping between the outcome and the real number line i.e. a function, and the distribution is a measure of the probability that the variable is a certain value, or more generally, in a certain Borel subset. A random variable can have two different distributions and two different random variables can have the same distribution. Consider $ℙ$ as the uniform measure of $[0, 1]$ . We have two random variables $𝑋 (𝜔) = 𝜔$ and $𝑌 (𝜔) = 1 - 𝜔$ . The distribution measure of $𝑋$ is uniform, which means that all outcomes are equally likely. The distribution measure of $𝑌$ is also uniform, despite being different from $𝑋$ , they’re both $𝑏 - 𝑎$ .

Now instead let’s define $ℙ$ as

$ℙ [𝑎, 𝑏] = \int_{𝑎}^{𝑏} 2 𝜔 𝑑 𝜔 = 𝑏^{2} - 𝑎^{2}, 0 \leq 𝑎 \leq 𝑏 \leq 1$

This is still a probability measure on $[0, 1]$ because $ℙ [0, 1] = 1$ . If you take a Borel subset of this interval, you can calculate its probability measure through the properties of probability measures applied to each interval that is unioned or complemented.

The distribution measure is one way to record a distribution. The other is the cumulative distribution function $𝑐 𝑑 𝑓$

$𝐹 (𝑥) = ℙ {𝑋 \leq 𝑥}, 𝑥 \in ℝ$

The difference is that distribution measure is a direct mapping of the idea of probability measure onto the Borel subset, while the cdf is a distribution measure on the specific Borel subset $𝜇_{𝑋} (- \infty, 𝑥]$ . It can also be converted backwards as $𝜇_{𝑋} (𝑥, 𝑦] = 𝐹 (𝑦) - 𝐹 (𝑥)$

Through the rules of probability measures, when the distribution measure is determined for every interval $[𝑎, 𝑏]$ in $ℝ$ , then it is determined for every Borel subset of $ℝ$ .

We can also get more detail on the distribution in two special functions.

We can have a density function $𝑓 (𝑥)$

$𝜇_{𝑋} [𝑎, 𝑏] = ℙ {𝑎 \leq 𝑋 \leq 𝑏} = \int_{𝑎}^{𝑏} 𝑓 (𝑥) 𝑑 𝑥$

which describes the distribution through its integral.

The probability mass function tells us that a random variable $𝑋$ takes one of the values in a particular sequence with probability 1. You could imagine, for instance, if $𝑋 = 𝜔^{2}$ , there is a probability of 1 that $𝑋$ is one of the values in the sequence of squared numbers. Strictly, this is defined where $𝑝_{𝑖} = ℙ {𝑋 = 𝑥_{𝑖}}$ and $\sum_{𝑖} 𝑝_{𝑖} = 1$ .

The sequence of numbers $𝑥_{1}, 𝑥_{2}, \dots$ is a Borel subset.

Random variables can be described by a density, a probability mass function, both, or neither. This last one is very advanced.

Let’s define the standard normal density:

$𝜑 (𝑥) = \frac{1}{\sqrt{2 𝜋}} 𝑒^{- \frac{𝑥^{2}}{2}}$

and the cumulative normal distribution function as

$𝑁 (𝑥) = \int_{- \infty}^{𝑥} 𝜑 (𝜉) 𝑑 𝜉$

This function is strictly increasing and maps $ℝ$ to $(0, 1)$ and has a strictly increasing inverse function $𝑁^{- 1} (𝑦)$ . Let’s consider the uniformly distributed random variable $𝑌$ on probability space $(Ω, ℱ︀, ℙ)$ , and $𝑋 = 𝑁^{- 1} (𝑌)$ .

The distribution measure is the probability measure that $𝑋 (𝜔)$ is in the interval defined. The distribution measure of $𝑌 𝜔$ therefore is $(𝑁 (𝑎), 𝑁 (𝑏)$ , which means that the distribution measure is $𝑁 (𝑏) - 𝑁 (𝑎)$ because $𝑌$ is uniformly distributed, which is the same as

$\int_{𝑎}^{𝑏} 𝜑 (𝑥) 𝑑 𝑥$ .

This distribution measure $𝜇_{𝑋}$ is known as the standard normal distribution, and any random variable with this distribution is a standard normal random variable. This technique of generating a standard normal random variable is called the probability integral transform and it is essential to Monte Carlo simulation.

A simpler way to get a standard normal random variable is to take this integral from arbitrary $𝑎$ and $𝑏$ as a probability measure for the random variable $𝑋 (𝜔) = 𝜔$ . This method works for any distribution, but it is not useful when we have multiple random variables with their own distributions and dependencies. For those cases we need a single probability space.

Expectation

We are interested in computing an “average value” of $𝑋$ in a probability space. For a finite $Ω$ , this is trivially:

$𝔼 𝑋 = \sum_{𝜔 \in Ω} 𝑋 (𝜔) ℙ (𝜔)$

The same can be done for a countably infinite $Ω$ .

However, if $Ω$ is uncountably infinite, expectation must be defined as an integral.

First, let’s consider the Riemann integral. This involves partitioning the area under the function in ever smaller subintervals and adding them up, which gets closer and closer to the true integral. There is an upper Riemann sum based on the maximum of the areas and lower based on the minimum, which both converge to the Riemann integral.

The reason we can’t use the Riemann integral for expectation is that our x-axis is $Ω$ , which is not necessarily a subset of $ℝ$ . We instead must partition the y-axis.

We have defined this as a Lebesgue sum, which is the sum of

Lecture Notes

$𝜎$ -algebras

The set of possible outcomes is the sample space $Ω$ . If you flip a coin, the sample space is ${𝐻, 𝑇}$

We are interested in the probability of events, which might be more complicated than an outcome. A $𝜎$ -algebra is a set of elements that belong to $ℱ︀$ the powerset of $Ω$ . The empty set must belong to it, all complements of sets must belong there, and the unions of all the composite sets must also belong there.

The $𝜎$ -algebra MUST contain $\emptyset$ and $Ω$ , and this particular algebra is called the trivial $𝜎$ -algebra because it is the simplest possible one. This might be confusing because $𝜎$ can represent both the standard deviation and the algebra, so use your brain for context.

A fancier $𝜎$ -algebra could be generated from a set $𝐴$ contained in it, which would be $𝜎 (𝐵) = {\emptyset, 𝐴, Ω, 𝐴^{𝑐}}$

Trick question: if I have a set $𝐵$ , what is it? It’s actually the same just with $𝐵$

A different question is $𝜎 (𝐴, 𝐵)$ . This would start with generators ${\emptyset, Ω, 𝐴, 𝐵, 𝐴^{𝑐}, 𝐵^{𝑐}, 𝐴 \cup 𝐵, (𝐴 \cup 𝐵)^{𝑐}}$

The number of elements in $𝜎$ -algebra can be found as 2 to the power of the number of distinct regions. For disjoint $𝐴$ and $𝐵$ , this is 3.

What about joint $𝐴$ and $𝐵$ like a Venn diagram. There’s the area outside of $𝐴$ and $𝐵$ , the area just in $𝐴$ , the area just in $𝐵$ , and the area shared by $𝐴$ and $𝐵$ , so 4, which means 16 elements.

$𝜎 (𝐴, 𝐵) = {\emptyset, Ω, 𝐴, 𝐵, 𝐴^{𝑐}, 𝐵^{𝑐}, 𝐴 \cup 𝐵, (𝐴 \cup 𝐵)^{𝑐},$

$𝐴 \cup 𝐵^{𝑐}, (𝐴 \cup 𝐵^{𝑐})^{𝑐} = 𝐴^{𝑐} \cap 𝐵 = 𝐵 \ 𝐴, 𝐵 \cup 𝐴^{𝑐}, (𝐵 \cup 𝐴^{𝑐})^{𝑐} = 𝐴 \ 𝐵, 𝐴^{𝑐} \cup 𝐵^{𝑐}, (𝐴^{𝑐} \cup 𝐵^{𝑐})^{𝑐} = 𝐴 \cap 𝐵,$

$({(𝐵 \cup 𝐴^{𝑐})}^{𝑐}) \cup ({(𝐴 \cup 𝐵^{𝑐})}^{𝑐})}$

Open question: draw out a venn diagram of what each of these mean

Let’s say we choose some random person at Stevens. Event A is undergraduate students, Event B is graduate students. Some people are neither (professors) and some people are both (accelerated students). The $𝜎$ -algebra expresses the event space.

Probability Measure

The function mapping $ℱ︀$ maps all of the sets in our $𝜎$ -algebra to events in the event space, which we can use for the probability measure $ℙ$ . We have two axioms:

$ℙ (Ω) = 1$
the $ℙ$ of a sequence of disjoint sets is equal to the sum of the

The probability space is the combination of the sample space, function mapping, and probability measure: $(Ω, ℱ︀, ℙ)$

If $ℙ (𝐴) = 1$ , then the event $𝐴$ occurs almost surely. Doesn’t mean it can’t happen, but it has zero probability of occurring.

Example: If I were to pick a completely random real number between 0 and 1, the probability I would pick a rational number is 0, even though there clearly are rational numbers that exist.

A random variable is a function $𝑋$ that maps from the event space to all real numbers $ℝ$ . You’ve been doing this your whole life without knowing it. Roll a six-sided die. What’s the sample space? It’s not 1, 2, 3, 4, 5, 6, it’s a set of weird looking symbols that we map onto numbers. You could equally create a $𝑌$ which says 6, 7, 8, 9, 10, 11.

The Borel subsets are every possible subdivision of the number line. It’s essentially a way to describe the real number line through set theory.

Expectation

Regular expectation is the probability of each outcome added together. How can you be certain that you’re exhaustive? We have to define elements in our sample space, and we use set theory to derive this:

$𝔼 [𝑋] = \sum_{𝜔 \in Ω} 𝑋 (𝜔) ℙ (𝜔)$

This looks like regular expectation, but it can be extended to understand the expectation of a function on a random variable as well.

Assume we have a fair coin. $𝑋 (𝜔)$ is 1 if $𝜔$ is heads, 0 if tails. $𝑌 (𝜔)$ is 10 and −2 for the same.

What is $𝔼 (𝑋)$ ? $\sum_{𝜔 \in Ω} 𝑋 (𝜔) ℙ (𝜔) = 𝑋 (ℎ) ℙ (ℎ) + 𝑋 (𝑡) ℙ (𝑡) = 1 * 0.5 + 0 * 0.5 = 0.5$

Let’s think more advanced:

$𝔼 [𝑒^{𝑥}] = 𝔼 [𝑔 (𝑥)] = \sum_{𝜔 \in Ω} 𝑔 (𝑋)$

Expected value are linear operators, so $𝔼 [𝛼 𝑋 + 𝛽 𝑌] = 𝛼 𝔼 [𝑋] + 𝛽 𝔼 [𝑌]$ this is very useful for operations

Jensen’s inequality mainly communicates that Variance is never negative: if $𝜑$ is a convex real-valued function and if $𝔼 [𝑋] < \infty$ , then

$𝜑 (𝔼 [𝑋]) < = 𝔼 [𝜑 (𝑋)]$

Probability being greater than 1 and variance being negative are massive red flags that should stop you in your tracks.

Change of Measure

We can shift between probability spaces and get different probability measures like so:

$𝑍 (𝜔) = \frac{\tilde{ℙ} (𝜔)}{ℙ (𝜔)}$

this is random variable which is almost surely nonnegative with $𝔼 [𝑍] = 1$ .

new function:

$\tilde{ℙ} (𝐴) = \int_{𝐴} 𝑍 (𝜔) 𝑑 ℙ (𝜔)$

so $\tilde{ℙ}$ is a probability measure.

We could consider the same thing for $Ω$ as the set instead of $𝐴$ , but the integral of $𝑍$ with respect to the probability measure over the sample space is just the same thing as $𝔼 [𝑍]$ , which is axiomatically 1.

$𝑣 𝔼 [𝑋 𝑍] = \int_{Ω} 𝑋 (𝜔)$

We know from the definition of $𝑍$ as the ratio of probability measure that $𝑍 (𝜔) 𝑑 ℙ (𝜔) = 𝑑 \tilde{ℙ}$ which gives us our integral over $Ω$ and $𝔼 [𝑍]$

Equivalent Measure

Equivalent probability measures only need to agree on which sets in $ℱ︀$ have probability zero. Let’s say we’ve bent a coin so much that now landing on heads has a 10% chance. The event space is still heads and tails, but the probability measure is different because the probability of the heads event and the tail event are different. Yet, these are still equivalent because they agree that the only set with probability zero is $\emptyset$ .

Let’s say you’re drawing a card from a deck and checking the suit. If you remove some cards, there’s still an equivalent probability measure. But if you remove all the clubs, it is not equivalent.

Radon-Nikodým derivative Open question: add this from slides

Filtration

A filtration is a collection of $𝜎$ -algebras where each one contains all the previous ones, like Russian nesting dolls. The best way to think about them is information that accrues over time at each step.

If we have random variable $𝑋$ , the $𝜎$ -algebra generated by $𝑋$ is $𝜎 (𝑋)$ , which is the collection of all subsets of $Ω$ where $(𝑋 \in 𝐵)$ for $𝐵$ Borel subset Open question: look up this up more formally.

Flipping a coin:

$Ω = {𝜔_{1}, 𝜔_{2}, 𝜔_{3} : 𝜔_{𝑖} \in {ℎ, 𝑡}}$

Before we even flip a coin, what do we know we would know about the coin flip? It’s either heads or tails.

$ℱ︀ (1) = {\emptyset, Ω, 𝐻, 𝑇}$

What if we flip it twice? We have four distinct outcomes now, HH, HT, TH, and TT.

$ℱ︀ (2) = {\emptyset, Ω, 𝐻 𝐻, 𝐻 𝑇, 𝑇 𝐻, 𝑇 𝑇, 𝐻 𝐻^{𝑐}, 𝐻 𝑇^{𝑐}, 𝑇 𝐻^{𝑐}, 𝑇 𝑇^{𝑐}, 𝐻 𝐻 \cup 𝐻 𝑇, 𝐻 𝐻 \cup 𝑇 𝐻, 𝐻 𝐻 \cup 𝑇 𝑇, 𝐻 𝑇 \cup 𝑇 𝐻, 𝐻 𝑇 \cup 𝑇 𝑇, 𝑇 𝐻 \cup 𝑇 𝑇}$

What about three times? 😅

$ℱ︀ (3) = {\emptyset, Ω, \dots}$ (256 elements)

Measurability

If $𝒢︀$ is also a $𝜎$ -algebra, and every set in $𝜎 (𝑋)$ is in $𝒢︀$ , then $𝑋$ is said to be $𝒢︀$ -measurable. If we have

$Ω = {𝑎, 𝑏, 𝑐, 𝑑}$ ,

and $𝑋 (𝜔)$ 2 if a, 0, if b or c, −2 if d

and $𝑌 (𝜔)$ 13 if a or b, −2 if c or d

and $𝑍 (𝜔)$ −2 if a or d, 3 if b or c

What is $𝜎 (𝑍)$ ? In order to generate a $𝜎$ -algebra from a random variable, we look at the partitions in the function and add those into the generator:

$𝜎 (𝑍) = {\emptyset, Ω, {𝑎, 𝑑}, {𝑏, 𝑐}}$

We can also more easily define these sets as $𝛼$ , $𝛽$ , $𝛾$ , etc. to simplify writing down.

We can see that $𝑋$ is $𝑍$ -measurable. But what does that actually mean?

Let’s say I know $𝑋 = 2$ . I do know the values of $𝑌$ and $𝑍$ , because the outcome has to be $𝑎$ .

But if $𝑋$ is 0, I have no idea what $𝑌$ is. I do know what $𝑍$ is.

## Risk-Adjusted Performance - URL: https://sharifhsn.dev/blog/risk-adjusted-performance/ - Structured data: https://sharifhsn.dev/api/posts/risk-adjusted-performance.json - Description: Let’s say we want to invest in long-term U.S. government bonds. The monthly average return of this is 0.47%, half of the equity market, and 2.3% standard deviation, less than… - Date: 2024-09-12 - Exact published timestamp: 2024-09-12 - Topics: Risk Management, CAPM, Risk-Adjusted Performance - Categories: Risk Management - Source: FE-535 \| Risk Management - Source URL: None - Content source format: typst

Comparing Multiple Assets

Let’s say we want to invest in long-term U.S. government bonds. The monthly average return of this is 0.47%, half of the equity market, and 2.3% standard deviation, less than half of the equity market. An investor must decide how much of each of these they want to invest in. Their ratio of volatility to return is similar, so it’s a tough decision. Similarly, a bank must decide how much leverage they can assume based on their assets divided by their equity, and how volatile they want their assets to be.

Risk-Adjusted Performance Measurement

The simplest metric for this is the Sharpe ratio (SR), the ratio of the average rate of return $𝜇 (𝑅_{𝑃})$ in excess of the risk-free rate $𝑅_{𝐹}$ to the absolute risk:

$𝑆 𝑅 = \frac{[𝜇 (𝑅_{𝑃}) - 𝑅_{𝐹}]}{𝜎 (𝑅_{𝑃})}$

The Sharpe ratio uses absolute terms, and it can use the VAR instead of return volatility in the denominator. Graphically, when comparing volatility and expected return, the Sharpe ratio is the slope of the line starting from cash (0 volatility, risk-free return) to the point of the instrument.

For relative risk, the information ratio (IR) is the ratio in excess of the benchmark:

$𝐼 𝑅 = \frac{[𝜇 (𝑅_{𝑃}) - 𝜇 (𝑅_{𝐵})]}{𝜎 (𝑅_{𝑃} - 𝑅_{𝐵})}$

An illustrative example is that you could have a portfolio with $𝑅_{𝑓} = 3 %$ , $𝜇 (𝑅_{𝑃}) = - 6 %$ , and $𝜎 (𝑅_{𝑃}) = 25 %$ , which is negative and terrible. But if the benchmark has a return of −10% and TEV of 8%, then the IR is $[(- 6 %) - (- 10 %)] / 8 % = 0.50$ which is positive and therefore good. So even though the absolute performance is bad, the relative performance is good.

You can calculate TEV from $𝜎_{𝑃}$ and $𝜎_{𝐵}$ , in addition to correlation $𝜌$ :

$𝜔^{2} = 𝜎_{𝑃}^{2} - 2 𝜌 𝜎_{𝑃} 𝜎_{𝐵} + 𝜎_{𝐵}^{2}$

The IR is used to judge active management skills, with an IR of 0.50 being “good”.

One disadvantage is that TEV does not adjust for average returns (Open question: give a more illustrative example of this)

Mixing Assets

A portfolio can be divided between assets, giving weight to each asset, where the portfolio weights sum to 1:

$\sum_{𝑖 = 1}^{𝑁} 𝑤_{𝑖} = 1$

Let’s try and shift a portfolio of all bonds to all weights. On the graph of volatility to return, this moves the point on a line. This is called an asset allocation problem. The shape of the line depends on the correlation coefficient $𝜌$ , which measures how related the assets are and how much they move together. The line is curved, which means that diversifying your portfolio with riskier assets (equities) can give you the same volatility as the pure bond portfolio. Given $𝜌 = 0.13$ , a portfolio with 77% bonds and 23% stocks:

$𝜎_{𝑃}^{2} = {0.77}^{2} {8.1}^{2} + 2 \times 0.77 \times 0.23 (0.13 \times 8.1 \times 19.2) + {0.23}^{2} {19.2}^{2} = 65.56$

The volatility is the square root, which is 8.1%, which is the same volatility that bonds have, but a return of 6.9% compared to 5.6%

Efficient Portfolios

Risk-adjusted Performance

We refer to the Sharpe ratio

$𝑆 𝑅_{𝑖} = \frac{𝜇_{𝑖} - 𝑅_{𝐹}}{𝜎_{𝑖}}$

The interest rate is free money for 5%, so why should I bother taking any risk beyond the 5%?

intuitive weighting of portfolios:

$𝑤_{𝑖} = \frac{𝑆 𝑅_{𝑖}}{\sum_{𝑗 = 1}^{𝑁} 𝑆 𝑅_{𝑗}}$

you can see data from the yahoo api for public traded apis from quantstats.shinyapps.com/yahoo

Annualize daily returns by multiplying by $\sqrt{252}$ , average number of trading days in a year. Why?

basic return formula:

$𝑟_{𝑖} = \frac{𝑃_{1} - 𝑃_{0}}{𝑃_{0}} = \frac{Δ 𝑃_{0}}{𝑃_{0}}, log -return = log (\frac{𝑃_{1}}{𝑃_{0}})$

If you annualize, it becomes

$𝑟_{𝑖} = \frac{𝑃_{252} - 𝑃_{0}}{𝑃_{0}}, annual log-return = log (\frac{𝑃_{252}}{𝑃_{0}})$

You can show the return as

$log \frac{𝑃_{1}}{𝑃_{0}} + log \frac{𝑃_{2}}{𝑃_{1}} + \dots + log \frac{𝑃_{252}}{𝑃_{251}}$

$𝑅_{𝐴} = \sum_{𝑑 = 1}^{252} 𝑅_{𝑑}$

summation of expectations is the expectation of the summation

mean is $𝜇_{𝐴} = 𝔼 [𝑅_{𝐴}] = 252 \times 𝜇_{𝑑}$

deviation is $𝜎_{𝐴}^{2} = 𝑉 [𝑅_{𝐴}] = 252 \times 𝜎_{𝑑}^{2}$

summation of variances is not variances of summations unless variables are iid

$𝑆 𝑅_{𝐴} = \frac{𝜇_{𝐴}}{𝜎_{𝐴}} = \frac{𝜇_{𝑑} \times 252}{𝜎_{𝑑} \sqrt{252}} = 𝑆 𝑅_{𝑑} \sqrt{252}$

Portfolio construction next week

Risk-Adjusted Performance (cont.)

The annual return is $𝑅_{𝐴} = log (\frac{𝑃_{252}}{𝑃_{0}})$

We argue that the expectation of $𝑅_{𝐴}$ is simply the proof that it’s going to be $𝔼 [𝑅_{𝐴}] = 252 𝔼 [𝑅_{𝑑}]$

And that the variance does as well

$𝕍 [𝑅_{𝐴}] = 252 𝕍 [𝑅_{𝑑}]$

This one guy went to industry from academia. How come you moved? Can you give me a Sharpe ratio of 3. The market is about 0.5-0.6, 1 is good, 2 is great, 3 is exceptional.

How Bad Things Can Get?

That’s risk, in a nutshell.

You can imagine a stock going up in a straight line. You could also imagine the stock ending up higher, but bouncing around a lot. How do you express the volatility here? Let’s look at downside risk.

In a distribution, $𝜎$ tells you how wide the distribution is.

Portfolio Construction

The professor constructed his portfolio for his pension looking for the highest return, and ended up on the upside because he had large market caps. He tells us to only invest in mutual funds, the risk premium is very worth it.

We can give better weights to stocks that have a higher Sharpe ratio.

We want a theoretically justified approach to setting our objective.

We want to maximize our return for a given level of risk, or minimize risk for a given return.

The mean variance model gives us the optimal portfolio. We have a vector of weights which is $𝑤$ , with some rules that $𝑤^{'} 1 = 1$ and $𝑤^{'} 𝜇 = 𝑚$ . The weights all sum to 1, and weights of the vector of mean returns should hit the mean target $𝑚$ , minimizing risk relative to $Σ$ the covariance matrix of asset returns

$min 𝜎_{𝑝}^{2} = 𝑤^{'} Σ 𝑤$

Let’s say I’m targeting 15% annual. When you have only two assets, this is easy because you only have two unknowns. It’s always solvable. But how do we solve this with more assets?

You end up with a curve that is perfectly solvable.

Let’s look at $Σ$ more closely. This has the variance of all of the stocks on the diagonal, which measures the volatility of each asset. We are also concerned with diversifying our portfolio, so we also want to minimize covariance between assets. Each covariance $𝜎_{𝑎 𝑏} = 𝜌 𝜎_{𝑎} 𝜎_{𝑏}$ .

If we’re available to capture $𝜇$ and $𝜎$ , we can get the reward and risk.

There is a slight inverse correlation between bond volume and stock volume because people will put their money in bonds when the market sucks.

The optimal solution is given by the closed form solution:

$𝑤 = 𝑤_{0} + \frac{1}{𝐴} 𝐵 𝜇$

This says that I have two funds, a low-risk fund $𝑤_{0}$ also known as the global minimum variance portfolio (GMV) and a sophisticated fund that goes long and short in those stocks $𝐵 𝜇$ , also considered $𝑤_{1}$ . We can use the IEF (treasury bonds) and SPY (S&P) as these, respectively. It actually doesn’t minimize risk to put all of our money in IEF because there’s no diversification. To minimize co-risk, we should invest a little in SPY.

Let’s consider the parameter $𝐴$ . What happens if $𝐴$ goes to $\infty$ ? High $𝐴$ values means we are very risk-averse and we invest a lot in the low-risk fund. The value of $𝑤_{1}$ can be considered how leveraged or risky we are, if it’s very high then that will be the case.

We can repeat this experiment with different values of $𝐴$ , which will give us a different set of weights for the optimal portfolio construction. We can calculate the mean and variance of all these portfolios, and if we graph this we can see the mean-variance efficient frontier (MVEF). This set of optimal portfolios has the minimum variance and maximum return for a given level of acceptable risk.

Part of the lab is going through the Excel spreadsheet, you can also code, Majeed says do both.

You can also modify this for Value at Risk but you lose the tractable closed-form solution. There is also SIVAR which does not violate convexity (Open question: look this up).

You can also construct special portfolios like the GMV portfolio. Let’s assume that the covariances of the matrix are all 0. Then the inverse of the matrix is just the reciprocal of every variance.

The Sharpe ratio portfolio has weight proportional to SR, and it can be an efficient portfolio.

Capital Asset Pricing Model

There are three aspects to it. There’s $𝛼$ , which is the unique value of an asset. There’s $𝛽$ , which is market risk. And there is $𝜖$ , which represents residual errors.

$𝑅_{𝑖} = 𝛼 + 𝛽 𝑅_{𝑚} + 𝜖_{𝑖}$

The market is full of participants that are all making choices.

The CAPM is an equilibrium model between some assets and a limited supply. William Sharpe came up with it. There is a linear relationship of returns on stocks and exposure to market risk. Taking market risk should be compensated.

If I want to minimize risk, I’m going to choose the risk-free portfolio. Let’s say I have a risk-free asset. There’s technically no such thing, but short-term treasury bonds are the closest thing we can get. Long-term treasury bonds do not stay fixed, they change a lot actually. It is a theoretical fact that if you introduce the risk-free rate, the efficient frontier becomes a straight line instead of a curve.

$𝑤 = \frac{1}{𝐴} Σ^{- 1} [𝜇 - 𝑅_{𝐹} 1]$

You can show analytically with a portfolio, the new frontier is

$𝜇_{𝑝} = 𝑅_{𝐹} + 𝜎_{𝑃} 𝜃$

Open question: look into $𝜃$ for extra credit

This is called the Sharpe portfolio or the tangent portfolio because it’s tangent to the line of the Sharpe ratio. If we have the most efficient portfolio, the demand for the most risky assets will be determined by the efficient portfolio. All agents in the market have the same risk aversion and they will pick the same asset.

I had a friend tell me that he trades crypto and they do technical analysis, and it works because everyone else is using technical analysis.

This becomes the market portfolio because everyone’s using the same portfolio.

And in fact, this is how you derive the CAPM. $𝐴$ constitutes the market portfolio.

What is the covariance between this vector of returns and the market portfolio? $𝐶 𝑜 𝑣 [𝑅, 𝑅_{𝑀}] = Σ 𝑤^{*}$

The end result of CAPM states that the mean return on each risky asset is

$𝜇_{𝑖} - 𝑅_{𝐹} = 𝛽_{𝑖} (𝜇_{𝑝} - 𝑅_{𝐹})$

The risk premium $𝛽_{𝑖}$ is how much you are exposed to the market, which increases compensation.

$𝛽_{𝑖} = \frac{𝜎_{𝑖 𝑝}}{𝜎_{𝑝}^{2}}$

Example: TSLA has a $𝛽$ of 2. This means that if the market swings, TSLA swings twice as much.

Leveraged ETFs (SPXL) gives you 3 times leverage on the market. Experimentally, it’s about 3.11

How realistic is this model? In equilibrium, this model should represent the relationship. Let’s collect data. If the data follows the line, then the theory is correct. The model exists to explain reality. It says that the relationship is linear, so in order for the theory to be correct, reality must also be linear. We can also think of this more strictly using regression modeling when we are benchmarking.

We can price the return on an asset by

$𝑅_{𝑖, 𝑡} - 𝑅_{𝐹} = 𝛼_{𝑖} + 𝛽_{𝑖} (𝑅_{𝑀}, 𝑡 - 𝑅_{𝐹}) + 𝜖_{𝑖, 𝑡}$

which when you take the expectation on both sides gives the same formula as before, as $𝛼$ and $𝜖$ disappear.

Let’s think about variance in price.

$𝕍 [𝑅_{𝑖}] = 𝛽^{2} 𝕍 [𝑅_{𝑀}] + 𝕍 [𝜖_{𝑖}]$

$\sum_{𝑖 = 1}^{𝑁} 𝜎_{𝑖}^{2} = \sum_{𝑖 = 1}^{𝑁} 𝛽_{𝑖}^{2} 𝜎_{𝑀}^{2} + \sum_{𝑖 = 1}^{𝑁} 𝜎_{𝜖, 𝑖}^{2}$

Monthly volatility of SPY is 4%. TSLA has a lot of idiosyncratic risk at 20% (represented by $𝜎_{𝜖}$ ).

Portfolio volatility ( $𝜎_{𝑝}^{2}$ ) is what we are concerned with.

$𝜎_{𝑝}^{2} = \frac{𝜎_{𝑚}^{2}}{𝑁} \times avg (𝛽_{𝑝}) + \frac{1}{𝑁} \sum_{𝑖 = 1}^{𝑁} 𝜎_{𝜖}^{2}$

If our number of stocks goes to infinity, the $𝜖$ shrinks to 0, and we get rid of idiosyncratic risk, which is part of the wisdom of diversification.

Relative Performance Measures

Sharpe ratio doesn’t necessarily take into account systematic risk factor, but you can replace volatility in the formula with $𝜎_{𝑝}$ .

The Treynor ratio only considers systematic risk i.e. exposure to the market, because when you’re comparing two funds you don’t care about the actual market risk since it’s the same between them: $𝑇 𝑅_{𝑖} = \frac{𝜇_{𝑖} - 𝑅_{𝐹}}{𝛽_{𝑖}}$

SPXL is not actually efficient if you use systematic risk because of how exposed it is.

Jensen’s $𝛼$ is what matters for hedge funds. Open question: acquired podcast on rentech, nvidia

RenTech makes 60% annual returns.

You have to take leverage in order to benefit from the market, but then you become exposed. If the market tanks, you’re not going to be happy.

Decomposing the return of the portfolio.

$𝛼_{𝑝} = 𝜇_{𝑝} - 𝛽_{𝑝} 𝜇_{𝑀}$

The Jensen’s $𝛼$

Let’s consider excess returns $\tilde{𝑅} = 𝑅 - 𝑅_{𝑓}$ . The ratio is

$𝑆 𝑅 = \frac{𝔼 [\tilde{𝑅}]}{\sqrt{𝕍 (\tilde{𝑅})}}$

Information Ratio replaces the risk-free rate with market rate (or some other benchmark). The error in this case is called the tracking error. We care about this because we don’t want to be behind the market where everything is priced.

When you’re working with volatility, if you pull a constant from inside to outside, you have to square it because volatility is expectation squared minus square of expectation

Benchmarking is important from a risk manager perspective.

we have confirmation bias for the results we like

Sharpe ratio of 2, but we don’t know where it comes from

Risk Measurement Evaluation

We have known knowns, which are known positions. Losses can still occur from bad luck and portfolio decisions. Value at Risk will tell us if our bad luck is too bad which indicates a flawed model.

Known unknowns could be ignoring important risk factors, or nonstationarity. When the market is bad, correlations increase because everyone starts selling.

We can simulate reality via back-testing. We have in-sample (IN) and out-of-sample (OUT) from our sample. We use the IN to get the weights of each portfolio using our mean-variance formulas, and then we test the model on the OUT sample.

What about stress testing? How much will your assets be able to sustain losses. Portfolio 2 has a big short position in the energy sector. If energy companies recover in the OUT period, the stressed asset return has an increase in variable $𝑈_{𝑡}$ which increases volatility.

Open question: do we need higher-order co-moments? See the paper.

## NVIDIA, Cisco, and Market Euphoria - URL: https://sharifhsn.dev/blog/linkedin-2024-09-10-nvidia-and-the-cisco-bubble/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2024-09-10-nvidia-and-the-cisco-bubble.json - Description: Bill Smead compares current enthusiasm for AI with the dot-com era's Cisco boom. - Date: 2024-09-10 - Exact published timestamp: 2024-09-10T13:30:09.913Z - Topics: AI, Financial Markets - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/posts/sharif-haason_just-listened-to-bill-smead-on-bloomberg-activity-7239263887978627072-OgsH - Content source format: typst

Just listened to Bill Smead on Bloomberg talk about a bearish outlook on the current financial euphoria the market is experiencing with respect to artificial intelligence.

He compares the current wave of excitement over NVIDIA to the excitement over Cisco during the early days of the Internet. The legitimate development, which was the Internet, depended on the presence of the router, which skyrocketed the valuation of Cisco. On March 27, 2000, reported earnings pushed it to a $600 billion market cap. It has since grown to 15 times that size, but it never went back to that valuation. Smead thinks that the legitimate development in this case is artificial intelligence, and NVIDIA holds the same role in this situation as Cisco did.

With respect to when the NVIDIA bubble will burst, he uses Buffett’s analogy of Cinderella at the ball. When the clock strikes midnight, everything will turn into pumpkins and mice, but there’s no clocks around so she can’t know when it will happen. One thing can be sure, though: the burst will be vicious.

What do you think of Smead’s predictions? Do they hold weight or are they overly pessimistic?

## Probability Measures and Bayes' Rule - URL: https://sharifhsn.dev/blog/probability-measures-and-bayes-rule/ - Structured data: https://sharifhsn.dev/api/posts/probability-measures-and-bayes-rule.json - Description: We have already considered the measurable space ( Ω , ℱ︀ ) of the sample space and 𝜎 -algebra. We can now define the probability measure ℙ as a measure defined on events in the 𝜎… - Date: 2024-09-09 - Exact published timestamp: 2024-09-09 - Topics: Probability Theory, Conditional Probability, Bayes Rule - Categories: Probability Theory - Source: FE-540 \| Probability Theory - Source URL: None - Content source format: typst

Week 2

Reading

FT-2 | Probability Measure

We have already considered the measurable space $(Ω, ℱ︀)$ of the sample space and $𝜎$ -algebra. We can now define the probability measure $ℙ$ as a measure defined on events in the $𝜎$ -algebra $ℱ︀$ , taking values in the unit interval $[0, 1]$ . This measures the size of the events, which has two properties: the probability of $Ω$ is 1, which means that all outcomes are accounted for in the sample space, and that the measure is countably additive, which allows us to extend the probability of individual events to the entire $𝜎$ -algebra.

There are three ways of thinking on how the numerical values of probability manifest. Frequentism assigns values directly from data. Objectivism assigns values based on some model which is considered to describe aspects of the universe. Subjectivism characterizes probability as measuring beliefs about the world.

The definition of probability measure is as follows:

$ℙ (Ω) = 1$

and for any sequence of disjoint events in $ℱ︀$ :

$ℙ (⋃_{𝑛 = 1}^{\infty} 𝐴_{𝑛}) = \sum_{𝑛 = 1}^{\infty} ℙ (𝐴_{𝑛})$

There are some other basic properties that arise from these definitions:

When $𝐴 \subset 𝐵$

$ℙ (𝐵 ∖ 𝐴) = ℙ (𝐵) - ℙ (𝐴)$

$ℙ (𝐴) \leq ℙ (𝐵)$

$ℙ (𝐴^{𝑐}) = 1 - ℙ (𝐴)$

$ℙ (\emptyset) = 0$

For all $𝐴, 𝐵$ :

$ℙ (𝐴 \cup 𝐵) = ℙ (𝐴) + ℙ (𝐵) - 𝑃 (𝐴 \cap 𝐵)$

Poincaré’s formula:

$ℙ (𝐴_{1} \cup 𝐴_{2} \cup \dots \cup 𝐴_{𝑛}) = \sum_{𝐼 \subset {1, 2, \dots, 𝑛}} (- 1)^{| 𝐼 | + 1} ℙ (⋂_{𝑖 \in 𝐼} 𝐴_{𝑖})$

$= \sum_{𝑖 = 1}^{𝑛} ℙ (𝐴_{𝑖}) - \sum_{𝑖 < 𝑗 \leq 𝑛} ℙ (𝐴_{𝑖} \cap 𝐴_{𝑗})$

$+ \sum_{𝑖 < 𝑗 < 𝑘 \leq 𝑛} ℙ (𝐴_{𝑖} \cap 𝐴_{𝑗} \cap 𝐴_{𝑘}) - \dots + {(- 1)}^{𝑛} \times ℙ (𝐴_{1} \cap 𝐴_{2} \dots \cap 𝐴_{𝑛})$

This is better expressed as the idea that the probability measure of the unions of sets can be expressed as the alternating signed partial sum of all the different partial intersections of the sequence of sets. It is proved by induction over the previous property.

Finite subadditivity (Boole’s inequality):

$ℙ (⋃_{𝑖 = 1}^{𝑛} 𝐴_{𝑖}) \leq \sum_{𝑖 = 1}^{𝑛} ℙ (𝐴_{𝑖})$

To show this, we shall set: $𝐵_{1} = 𝐴_{1}, 𝐵_{2} = 𝐴_{2} \cap 𝐴_{1}^{𝑐}, \dots, 𝐵_{𝑛} = 𝐴_{𝑛} \cap 𝐴_{𝑛 - 1}^{𝑐} \cap \dots \cap 𝐴_{1}^{𝑐}$

You can think of this every successive element being disjoint with all the previous elements, which means

$𝐵_{𝑖} \cap 𝐵_{𝑗} = \emptyset$

and also

$⋃_{𝑖 = 1}^{𝑛} 𝐵_{𝑖} = ⋃_{𝑖 = 1}^{𝑛} 𝐴_{𝑖}$

because the union of all the disjoint parts of the sets will include all the overlap in the sets.

A partition of a set $Ω$ is a collection of disjoint sets ${Ω_{𝑖}}$ where the union recreates the original set: $⋃_{𝑖} Ω_{𝑖} = Ω$

We can think of the probability of an event as the intersections of the event with a partition of $Ω$ . This is the law of total probability. For $(𝐴_{𝑖})_{𝑖 \in 𝐼}$ a partition of $Ω$ and every event $𝐵 \in ℱ︀$ :

$ℙ (𝐵) = \sum_{𝑖 \in 𝐼} ℙ (𝐵 \cap 𝐴_{𝑖})$

Uniqueness of Probability Measures

Let’s consider the idea of a $𝜆$ -system $𝒰︀$ which is a family of events where:

$Ω \in 𝒰︀$

For any $𝐴, 𝐵 \in 𝒰︀$ and $𝐴 \subset 𝐵$ , $𝐵 ∖ 𝐴 \in 𝒰︀$

For a sequence of disjoint events $𝐴_{𝑛}$ , indexed by $𝑛 \in 𝒰︀$ :

$⋃_{𝑛} 𝐴_{𝑛} \in 𝒰︀$

You might notice that this is fairly similar to the definition of a $𝜎$ -algebra. In fact, a $𝜎$ -algebra is a $𝜆$ -system, and every $𝜆$ -system closed to finite intersection is a $𝜎$ -algebra.

Now let’s consider a collection of sets $ℳ︀ \subseteq 𝑃 (Ω)$ where $𝒰︀ (ℳ︀)$ is the smallest $𝜆$ -system which contains $ℳ︀$ : $𝒰︀ (ℳ︀) = 𝜎 (ℳ︀)$

Note that this is distinct from the previous definition because although the $𝜆$ -system is known to be a $𝜎$ -algebra, it is now also defined to be the same as one which is generated by the same collection of sets.

Now let’s consider some probability measures $ℙ_{1}$ and $ℙ_{2}$ on the $𝜎$ -algebra $ℱ︀$ generated from a collection of sets $ℳ︀$ . If $ℙ_{1}$ and $ℙ_{2}$ coincide on $ℳ︀$ , then they coincide on $ℱ︀$ . This means that for every event, the two probability measures assign the same probabilities, which includes all the different kinds of events generated by the $𝜎$ -algebra. This also applies to partitions of a sample space.

Monotone Class

This is a variant of a $𝜆$ -system. A monotone class $ℳ︀$ of subsets in $Ω$ is closed under the formation of monotone unions and intersections, i.e.

$𝐴_{1}, 𝐴_{2}, \dots \in ℳ︀$ and $𝐴_{𝑛} \subseteq 𝐴_{𝑛 + 1}, \cup_{𝑛} 𝐴_{𝑛} = 𝐴 \Rightarrow 𝐴 \in ℳ︀$

where the sets are increasing in size, containing the previous one,, and the union of all of them equals the last one.

$𝐴_{1}, 𝐴_{2}, \dots \in ℳ︀$ and $𝐴_{𝑛} \supseteq 𝐴_{𝑛 + 1} \Rightarrow \cap_{𝑛} 𝐴_{𝑛} \in ℳ︀$

or where the sets are decreasing in size, containing the next one, and the intersection of all of them equals the last one.

Open question: more stuff about monotone classes only if the professor talks about them.

Monotone Convergence Properties of Probability

How do we deal with infinity?

If $𝐴_{𝑛}, 𝐴 \in ℱ︀$ and $𝐴_{𝑛} ↑ 𝐴$ , which means that $𝐴$ is a monotonically increasing class, then $ℙ (𝐴_{𝑛}) ↑ ℙ (𝐴)$ and the same is true for $𝐴_{𝑛} ↓ 𝐴$ .

Under countable subadditivity, the probability measure of the union of sets is less than or equal to the sum of the probability measures of those sets. That makes sense intuitively because it would be equal if the sets were disjoint, but the sets might overlap, making it less. We can use the above properties to extend this idea to countably infinite sets.

A null set is a set with probability measure 0. $\emptyset$ is the obvious example, and in infinite sets there may be others like the probability measure of a singleton set in the Borel $𝜎$ -algebra.

Conditional Probability

“In some cases the probability of an event happening depends not just on the experiment itself but on other information as well. Conditional probability forms a framework in which this additional information can be incorporated.”

$ℙ (𝐴 | 𝐵) = \frac{ℙ (𝐴 \cap 𝐵)}{ℙ (𝐵)}$

Obviously, the probability measure of $𝐵$ cannot be 0, which makes sense because in order for something to be conditioned, the condition has to have happened, which is not possible if the probability is 0.

We can get the multiplicative rule from this formula.

$ℙ (𝐴 \cap 𝐵) = ℙ (𝐴 | 𝐵) ℙ (𝐵),$

$ℙ (𝐴 \cap 𝐵 \cap 𝐶) = ℙ (𝐴 | 𝐵 \cap 𝐶) ℙ (𝐵 | 𝐶) ℙ (𝐶)$ , etc.

Let’s take an example. We have Urn 1 which contains 2 black balls and 3 white balls, and Urn 2 which contains 1 of each. One urn is chosen at random and one ball is chosen randomly from that urn. If a black ball is chosen, what is the probability it came from Urn 1?

Let’s assign our variables.

$𝐴$ is the event of choosing Urn 1, and $𝐵$ is the event of choosing a black ball. We’re trying to find $𝑃 (𝐴 | 𝐵)$ , so we need $𝑃 (𝐴 \cap 𝐵)$ . There are four total events in our sample space: { Urn1Black, Urn1White, Urn2Black, Urn2White }. The probability of Urn1Black is $0.5 \times 0.4 = 0.2$ . The probability of choosing a black ball is $0.5$ in the case of Urn 2 and $0.4$ in the case of Urn 1, which averages to $0.45$ . So the overall probability is $0.2 / 0.45$ , which is $4 / 9$ .

The total probability formula given $𝐴_{𝑛}$ partition of $Ω$ is

$ℙ (𝐵) = \sum_{𝑖 = 1}^{𝑛} ℙ (𝐵 | 𝐴_{𝑖}) ℙ (𝐴_{𝑖})$

for any event $𝐵 \in ℱ︀$ .

If Stanley wants to draw a card from 20, where 8 are favorable to him, what is the probability of him drawing a favorable card if he draws first, second, or third?

First is easy. $8 / 20 = 0.4$

For second, there are two different possibilities. Either the first draw was a favorable or unfavorable card. Let’s use the formula above.

$𝐵$ is the event of Stanley drawing a favorable card. There are two $𝐴$ events in consideration, which are the first card being favorable and unfavorable. $ℙ (𝐴_{1}) = 2 / 5$ and $ℙ (𝐴_{2}) = 3 / 5$ . If a favorable card was drawn first, then the new probability is $7 / 19$ . Otherwise, it’s $8 / 19$ . So the complete formula becomes

$ℙ (𝐵) = 2 / 5 \times 7 / 19 + 3 / 5 \times 8 / 19 = 0.4$

It’s clear that the same situation would arise if he drew the third card.

Bayes’ formula is the following (like the others it’s based on a partition $𝐴$ and arbitrary set $𝐵 \in ℱ︀$ ):

$ℙ (𝐴_{𝑗} | 𝐵) = \frac{ℙ (𝐵 | 𝐴_{𝑗}) ℙ (𝐴_{𝑗})}{\sum_{𝑖 = 1}^{𝑛} ℙ (𝐵 | 𝐴_{𝑖}) ℙ (𝐴_{𝑖})}$

“This theorem can be interpreted as: All of a sudden we find out $𝐵$ happened. In the light of this new information there should be a simple way to update all the probabilities of the sets in the partition $(ℙ (𝐴_{𝑗} | 𝐵))$ . Indeed, the Bayes rule provides just that.”

So you take the existing $𝐴$ , and then condition it on $𝐵$ , and then the calculation cascades across the events in the sample space i.e. $𝐴$ .

Independence

Two events $𝐴$ and $𝐵$ are defined as independent if

$ℙ (𝐴 \cap 𝐵) = ℙ (𝐴) ℙ (𝐵)$

which can be extended to the idea of being mutually independent (or simply independent) for a set of events:

$ℙ (⋃_{𝑗 \in 𝐽} 𝐴_{𝑗}) = \prod_{𝑗 \in 𝐽} ℙ (𝐴_{𝑗})$

There’s also the idea of jointly independent (or pairwise independent):

$ℙ (𝐴_{𝑖} \cup 𝐴_{𝑗}) = ℙ (𝐴_{𝑖}) ℙ (𝐴_{𝑗})$

which does not necessarily imply independence.

For two $𝜎$ -algebras $𝐺$ and $𝐻$ , they are $ℙ$ -independent if

$ℙ (𝐺 \cup 𝐻) = ℙ (𝐺) ℙ (𝐻)$

and the same applies for mutual independence as with events.

We can also show that if all $𝜎$ -algebras in a family are each generated from a collection of independent sets, they are independent. This independence is also associative. There’s an additional property of “disassociativity of the independence”, where independent families included in bigger independent families remain independent.

$𝜎$ -algebras generated by events are related in their independence.

## Bessent on Tariffs and the Deficit - URL: https://sharifhsn.dev/blog/linkedin-2024-09-05-bessent-tariffs-and-deficits/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2024-09-05-bessent-tariffs-and-deficits.json - Description: A response to Scott Bessent's arguments about tariffs, inflation, and fiscal policy. - Date: 2024-09-05 - Exact published timestamp: 2024-09-05T11:41:31.361Z - Topics: Tariffs & Trade, Fiscal Policy - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/posts/sharif-haason_bloombergradio-activity-7237424607861878784-wDAo - Content source format: typst

Listened to Scott Bessent, potential Secretary of the Treasury for a future Trump administration, today on Bloomberg Radio talk about his opinions on the fiscal policy of the Trump-Vance campaign and I have some thoughts.

He claims that Trump’s tariffs won’t be inflationary because it’s a one-time administrative cost adjustment that will be priced into the market. That might be true in the long-term, but the economic shift of a 10% tariff for the United States would be seismic and might even cause a recession to occur. You can’t compare the effects of specific tariffs like those he instituted on steel to the widespread tariffs he plans to implement.

He also deflected from criticisms of Trump’s tax cuts as contributing to the deficit by focusing on the principle of how spending on government programs should be reduced. But this is irrelevant on a fiscal level. On the balance sheet, tax cuts show up the same as government spending. He also claims that tariffs will help pay for it, but as I’ve already noted, there would be massive shifts in the market that would be caused by the tariffs, making the net impact on the deficit hard to predict even if the balance sheet shows a reduction.

What do you think about Bessent’s claims?

## Risk Measurement and Portfolio Construction - URL: https://sharifhsn.dev/blog/risk-measurement-and-portfolio-construction/ - Structured data: https://sharifhsn.dev/api/posts/risk-measurement-and-portfolio-construction.json - Description: Financial risk management involves identifying, assessing, measuring, and managing financial risks. Some risks are measurable, others are not. Risk cannot always be avoided, but… - Date: 2024-09-05 - Exact published timestamp: 2024-09-05 - Topics: Risk Management, Portfolio Construction, Market Risk - Categories: Risk Management - Source: FE-535 \| Risk Management - Source URL: None - Content source format: typst

Week 1

Reading - Risk Management

Financial risk management involves identifying, assessing, measuring, and managing financial risks. Some risks are measurable, others are not. Risk cannot always be avoided, but it can be mitigated. In order to leverage risk, you must measure it. The first centralized risk management tool was value at risk (VAR), developed in the 1990s. These tools measure risk in the portfolio context, as well as measuring on a forward-looking basis.

Risk Measurement

Example

The issue for a portfolio is whether expected profit warrants the assumed risk. The first step is to understand the profit and losses in the portfolio. $Δ 𝑃$ is the PnL variable, measured in a risk currency, like the dollar. This is the product of the initial investment $𝑃$ and future rate of return $𝑅_{𝑃}$ . Rate of return is a random variable described by a probability density function, which is a distribution. For example, the return on the S&P 500 follows a normal distribution, though with slightly worse risk. There is a 3% cumulative probability of losing 10% or more in a month. In order to reduce risk, money can be allocated to cash to benefit from the riskless interest rate for a reduced return.

Absolute vs. Relative Risk

Risk can be measured relative to a benchmark, such as the S&P 500 index, or an investor’s present value of liabilities compared to future value. This can also be adjusted for inflation.

Absolute risk is a shortfall relative to the value of the investment, expressed in terms of standard deviation: $𝜎 (Δ 𝑃) = 𝜎 (Δ 𝑃 / 𝑝) \times 𝑃 = 𝜎 (𝑅_{𝑃}) \times 𝑃$

Relative risk is relative to a benchmark index $𝐵$ . In this case, risk is measured via deviation in the tracking error $𝑒 = 𝑅_{𝑃} - 𝑅_{𝐵}$ scaled to the principal:

$𝜎 (𝑒) 𝑃 = [𝜎 (𝑅_{𝑃} - 𝑅_{𝐵})] \times 𝑃 = 𝜔 \times 𝑃$

where $𝜔$ is tracking error volatility (TEV), literally just the standard deviation in the difference between the returns in the investment and returns in the benchmark.

Evaluation of Risk Measurement

There are two parts to measuring risk. The easy part is that the scale of dollar returns should be proportional to the initial investment, which is just a simple scaling based on the value of $Δ 𝑃$ . Constructing the distribution of future rates of returns is difficult. We might have strong historical data like the S&P 500, but historical data doesn’t always work. Gold was stable until 1967 and then fluctuated wildly in price afterwards. Loss doesn’t mean risk management failed, so you have to measure it in other ways.

Known Knowns

These are risks that are identified and measured. They are unlikely to cause losses. A portfolio with VAR at 99% confidence of 14.4% would be highly unlikely to have consecutive losses of 15% or more and indicative of a flawed model.

Known Unknowns

These are model weaknesses that are known but not measured, inaccurate measurements of risk factors like volatilities, and mapping errors from positions to exposures. This is called model risk, which can be evaluated with stress tests that shock variables beyond typical ranges. For example, in 2007, banks over relied on historical data and credit rating and accrued too much exposure on risky assets.

Liquidity risk describes the ability to liquidate a position, which is based on the inherent liquidity of the asset as well as the size of the position. Some assets are very illiquid, like real estate, and if they are too large they cannot be cashed out quickly without disrupting the market.

Unknown Unknowns

These are out of scope events that cannot be considered typically, like regulatory risks, counterparty risks, and some amount of liquidity risk. This is also known as Knightian uncertainty, which is immeasurable risk, with only the government being the end risk manager.

Risk Management Failures

A risk manager should identify risks, assess and monitor them, manage them, and communicate them. A failure in risk management is a failure in at least one of these.

Portfolio Construction

Comparing Multiple Assets

Risk-Adjusted Performance Measurement

The simplest metric for this is the Sharpe ratio (SR), the ratio of the average rate of return $𝜇 (𝑅_{𝑃})$ in excess of the risk-free rate $𝑅_{𝐹}$ to the absolute risk:

$𝑆 𝑅 = \frac{[𝜇 (𝑅_{𝑃}) - 𝑅_{𝐹}]}{𝜎 (𝑅_{𝑃})}$

For relative risk, the information ratio (IR) is the ratio in excess of the benchmark:

$𝐼 𝑅 = \frac{[𝜇 (𝑅_{𝑃}) - 𝜇 (𝑅_{𝐵})]}{𝜎 (𝑅_{𝑃} - 𝑅_{𝐵})}$

You can calculate TEV from $𝜎_{𝑃}$ and $𝜎_{𝐵}$ , in addition to correlation $𝜌$ :

$𝜔^{2} = 𝜎_{𝑃}^{2} - 2 𝜌 𝜎_{𝑃} 𝜎_{𝐵} + 𝜎_{𝐵}^{2}$

The IR is used to judge active management skills, with an IR of 0.50 being “good”.

One disadvantage is that TEV does not adjust for average returns (Open question: give a more illustrative example of this)

Mixing Assets

A portfolio can be divided between assets, giving weight to each asset, where the portfolio weights sum to 1:

$\sum_{𝑖 = 1}^{𝑁} 𝑤_{𝑖} = 1$

$𝜎_{𝑃}^{2} = {0.77}^{2} {8.1}^{2} + 2 \times 0.77 \times 0.23 (0.13 \times 8.1 \times 19.2) + {0.23}^{2} {19.2}^{2} = 65.56$

The volatility is the square root, which is 8.1%, which is the same volatility that bonds have, but a return of 6.9% compared to 5.6%

Efficient Portfolios

If you diversify across a large number of stocks like the S&P 500, it seems more difficult. Markowitz simplifies the problem. He assumes that assets follow a jointly normal distribution i.e. normal distribution extended to higher dimensions, which means that the distribution of portfolio returns can be summarized by mean and variance. You need to identify an efficient set, which is a locus of points that represent portfolio mixes with the best risk-return characteristics. For an expected return $𝜇_{𝑝}$ , the risk should be minimized: $𝑀 𝑖 𝑛_{𝑤} 𝜎_{𝑝}^{2}$ where the portfolio return is equal to a specified value $𝑘$ , with a different efficient set for each value. A closed-form solution exists for every efficient set if there are no short-sale restrictions on portfolio weights Open question: what does this mean?, which is the linear combination between the global minimum-variance portfolio with the lowest volatility and the portfolio with the highest Sharpe ratio. This can be used for value at risk as well as regular variance, which is valuable for fat-tailed distributions Open question: why?. But in the general case, no closed-form solution exists.

Asset Pricing Theories

Capital Asset Pricing Model (CAPM)

The capital asset pricing model (CAPM) was developed by William Sharpe in 1964. The covariance structure of stocks is simplified with a one-factor model. Run a regression with the following returns at time $𝑡$ : stock $𝑖$ - $𝑅_{𝑖, 𝑡}$ , risk-free - $𝑅_{𝐹, 𝑡}$ , market - $𝑅_{𝑀, 𝑡}$

$𝑅_{𝑖, 𝑡} - 𝑅_{𝐹, 𝑡} = 𝛼_{𝑖} + 𝛽_{𝑖} [𝑅_{𝑀, 𝑡} - 𝑅 𝐹, 𝑡] + 𝜖 𝑖, 𝑡$

Lecture Notes

Risk Management

risk: the possibility that something bad might happen

everybody deals with risk, we are dealing with it in a financial context

making a decision to buy headphones, you risk it breaking and you’re out $150, so you might buy something cheaper if you think you’re likely to lose it, keep backup headphones

in the case of bond price, you have a financial instrument subject to interest rates,

they’re about to cut interest rates

let’s say we have interest rate risk

we graph interest rate y against P

graph of yield of bonds at quantstats.shinyapps.io/FRED/

the interest rate has moved significantly, was raised to about 5% in 2022-2023

trying to price a bond over ten years, the interest rate changes significantly, presents risk

four steps:

identify

analyze

assess impact

manage

when identifying risks,

you could find correlations, maybe spurious correlations

pricing assesses the impact of risk

transferring risk: insurance, like credit default swaps

typology of risks:

cyber risk - hacking and stuff

Market Risk

interest rates are a sort of market risk. the price for the bonds is exposed for something systematic.

i have this inventory of assets. a market is a representation of the whole economy, this is what we think of the market right now, and there’s a value for that.

NVidia is a big company, part of the market. all these asset managers hold trillions in assets and diversify across different products. if i want to buy the market, it’s a combination of assets including NVidia. I think the economy will do well so I will buy the market and it will go up. If I buy an ETF exchange traded fund, I’m buying everything, including NVidia.

But the market is going down because the economy is not doing well, so even though NVidia has launched a new product, they will go down as well. Stock prices go up and down because of information. For example, this Tuesday there was a poor economic outlook so people began to sell shares, driving down market

$𝑟_{𝑖} = 𝛼 + 𝛽 𝑟_{𝑚} + 𝜖_{𝑖}$

The return of the stock is influenced by two noises. the $𝛽$ is the systematic general market risk. Ignoring $𝛼$ , we have $𝜖_{𝑖}$ , which is the idiosyncratic specific market risk for the specific asset, like how Elon Musk is unpredictable, will tweet and might make the stock go down

In order to reduce the impact of $𝜖_{𝑖}$ , you diversify your portfolio to add justin whang and others to mitigate the risk of any individual stock. you hold a cross section of stocks, and you analyze them under different characteristics and groups and get their risk factors so you can combine them to mitigate risk

people have views of where the market is going, and that’s where the market price comes from.

Credit Risk

A credit score is a way to monitor creditworthiness. The lender lends $100 to the borrower at 20% interest. The risk is that the borrower will not pay back the loan. That risk is priced into the interest rate.

Interest rate also comes from inflation, as money becomes less valuable over time. The possibility of default is the same thing. As an individual I can get loans based on my credit score, but if it gets lowered I will be downgraded which will decrease my ability to obtain credit, decreasing my value. Same happens to companies

COVID kills all these companies, but the Fed bail out junk bonds and created positive sentiment to increase prices

Liquidity Risk

What moves prices is supply and demand. If there’s too much supply, I can buy a small amount as an individual and nothing happens, but if a big institution buys a lot it will have an impact.

If you’re a small player, liquidity doesn’t matter. but if you’re a bank trying to unleash millions of shares it becomes a liquidity issue.

The fed fund rate is overnight borrowing between banks, and is a good measure of liquidity. This is the cost just to borrow and lend. When Lehman Brothers went down, there was a big spike in the interest rate. Open question: read Afonoso 2011 stressed not frozen which goes over this. liquidity was still provided but the cost increased

Operational Risk

The risk of loss resulting from inadequate or failed internal processes, people, and systems or from external events

Every business has an input and an output. Databricks??? LOL

Bourbon is made of barley/corn input, bourbon comes out as output.

external events could be a shock, maybe machines are too old, not enough R&D

one of the big meat farmers got hacked, cyber risk led to operations risk

LTCM long term capital management was a highly leveraged company that had a lot of model risk

ML is LSTM, predict categorizations based on operational risk data, come with sensitivity analysis

corporate governance scandals with enron

How much do you trust the model? that’s operational risk

Other Risks

Business risks, consumer demand, pricing decisions, managing product innovation, just making bad decisions as a CEO

Strategic risks involves making large decisions about the firms direction long-term, major investments of capital, HR, management reputation

Reputation risk is danger firm will suffer a sudden fall in market standing, volkswagen with emissions, boeing with broken-down planes

Process

The reason to take risks is to pursue a reward, food, shelter, or bitcoin.

Two questions: Is the risk commensurate with the reward? And could we lower the risk for the same reward?

Modern Portfolio Theory

People used to only think about the return, now people think about reward in terms of risk-taking. The guy was Markowitz who came up with the Mean-Variance model, shared the Nobel prize with Sharpe for CAPM and also shared it with Merton Miller (who is this?)

Risks that are measured are easier to manage

you try to do the best you can given the circumstances

Let’s say we have ten assets to construct a portfolio. Measuring risk, evaluating the potential gains and losses. You could completely diversify equally by doing a naive classification, and it’s actually quite difficult to beat on average

consider the annual mean return of various ETFs. you might choose XLK because it has the highest mean return, but you don’t know anything about the risk. XLK is technology, which did well between 2018-2022 because of cutting interest rates but is not doing well now

There is a certain volatility, fluctuation, anxiety associated with holding shares, even if on average you are outperforming the market. We are all risk averse, it’s all about tolerance.

Capturing the risk dimension in terms of volatility on x axis and returns on y axis. We make these heuristics to determine the efficiency of the asset.

This is data from rational risk reward tradeoff

Motivate the idea of choosing between stocks

Risk-adjusted Performance

We refer to the Sharpe ratio

$𝑆 𝑅_{𝑖} = \frac{𝜇_{𝑖} - 𝑅_{𝐹}}{𝜎_{𝑖}}$

The interest rate is free money for 5%, so why should I bother taking any risk beyond the 5%?

intuitive weighting of portfolios:

$𝑤_{𝑖} = \frac{𝑆 𝑅_{𝑖}}{\sum_{𝑗 = 1}^{𝑁} 𝑆 𝑅_{𝑗}}$

you can see data from the yahoo api for public traded apis from quantstats.shinyapps.com/yahoo

Annualize daily returns by multiplying by $\sqrt{252}$ , average number of trading days in a year. Why?

basic return formula:

$𝑟_{𝑖} = \frac{𝑃_{1} - 𝑃_{0}}{𝑃_{0}} = \frac{Δ 𝑃_{0}}{𝑃_{0}}, log -return = log (\frac{𝑃_{1}}{𝑃_{0}})$

If you annualize, it becomes

$𝑟_{𝑖} = \frac{𝑃_{252} - 𝑃_{0}}{𝑃_{0}}, annual log-return = log (\frac{𝑃_{252}}{𝑃_{0}})$

You can show the return as

$log \frac{𝑃_{1}}{𝑃_{0}} + log \frac{𝑃_{2}}{𝑃_{1}} + \dots + log \frac{𝑃_{252}}{𝑃_{251}}$

$𝑅_{𝐴} = \sum_{𝑑 = 1}^{252} 𝑅_{𝑑}$

summation of expectations is the expectation of the summation

mean is $𝜇_{𝐴} = 𝔼 [𝑅_{𝐴}] = 252 \times 𝜇_{𝑑}$

deviation is $𝜎_{𝐴}^{2} = 𝑉 [𝑅_{𝐴}] = 252 \times 𝜎_{𝑑}^{2}$

summation of variances is not variances of summations unless variables are iid

$𝑆 𝑅_{𝐴} = \frac{𝜇_{𝐴}}{𝜎_{𝐴}} = \frac{𝜇_{𝑑} \times 252}{𝜎_{𝑑} \sqrt{252}} = 𝑆 𝑅_{𝑑} \sqrt{252}$

Portfolio construction next week

## Probability Spaces - URL: https://sharifhsn.dev/blog/probability-spaces/ - Structured data: https://sharifhsn.dev/api/posts/probability-spaces.json - Description: The purpose of probability is to obtain answers about real-life phenomena which do not have a predetermined outcome. We need a mathematical model to describe this so that we can… - Date: 2024-09-02 - Exact published timestamp: 2024-09-02 - Topics: Probability Theory, Measure Theory, Sigma Algebras - Categories: Probability Theory - Source: FE-540 \| Probability Theory - Source URL: None - Content source format: typst

Week 1

Reading

FT-1 | Probability Space

The purpose of probability is to obtain answers about real-life phenomena which do not have a predetermined outcome. We need a mathematical model to describe this so that we can provide answers to these questions. To this end, we define a probability space.

This space is defined on the sample space of all possible outcomes of a random experiment, denoted as $Ω$ . Each element (outcome) of $Ω$ is denoted by $𝜔$ , and a collection of outcomes is called an event, aka a subset of $Ω$ . Probability requires us to understand the size of an event. Not all subsets of $Ω$ can be measured, so we will define a collection of subsets on which a measure can be defined, the $𝜎$ -algebra.

A set is described in braces like so:

${𝑥 | 𝑥 satisfy property 𝒫︀}$

The set which does not contain an element is the empty set $\emptyset$ .

To say that an element belongs to a set is that $𝜔 \in 𝐴$ , and for it not to belong is $𝜔 \notin 𝐴$ .

A union of sets $𝐴 \cup 𝐵$ has elements that are in either of them.

An intersection of sets $𝐴 \cap 𝐵$ has elements that are in both of them.

The complement $𝐴^{𝑐}$ of set $𝐴$ is the set of elements in $Ω$ but not in $𝐴$

The difference of two sets $𝐴 ∖ 𝐵$ is the set of elements in $𝐴$ but not $𝐵$

The symmetric difference of two sets is $𝐴 Δ 𝐵 = (𝐴 ∖ 𝐵) \cup (𝐵 ∖ 𝐴)$ , which is the reverse of $𝐴 \cap 𝐵$ if $𝐴$ and $𝐵$ share elements. In terms of a Venn diagram, it would be the left and right circle parts without the middle part.

If $𝐴 \cap 𝐵 = \emptyset$ , then these sets are considered disjoint or mutually exclusive.

There is a distributive law (DeMorgan’s law) for intersection over union:

$𝐴 \cap (𝐵 \cup 𝐶) = (𝐴 \cap 𝐵) \cup (𝐴 \cap 𝐶)$

and the reverse for unions.

Let’s consider all possible subsets of $Ω$ as $𝒫︀ (Ω)$ . An algebra on $Ω$ is a collection of sets in $𝒫︀ (Ω)$ which is closed under complementarity and finite union.

There are three basic properties of the algebra $𝒜︀$ :

$Ω \in 𝒜︀$
If $𝐴 \in 𝒜︀$ , then $𝐴^{𝑐} \in 𝒜︀$
If $𝐴, 𝐵 \in 𝒜︀$ , then $𝐴 \cup 𝐵 \in 𝒜︀$

These properties naturally extend themselves to other properties, like the empty set being in the algebra, intersections being in the algebra, and the union of all different sets being in the algebra.

A $𝜎$ -algebra is a kind of algebra which is closed under countable unions. Regular algebras can have uncountable unions. These kinds of unions arise in stochastic processes. In order to use $𝜎$ -algebras when describing stochastic processes, we will define filtrations that increase $𝜎$ -algebras over time. When we try to prove properties of $𝜎$ -algebras, we will instead prove properties about their generators, and then extend them to the $𝜎$ -algebra.

Let’s consider a collection $𝒞︀$ of sets in $Ω$ . The $𝜎$ -algebra of $𝒞︀$ defined as $𝜎 (𝒞︀)$ has the properties that it contains all sets in $𝒞︀$ and it is the smallest $𝜎$ -algebra which is a subset of any other $𝜎$ -algebra which contains $𝒞︀$ .

Here are some facts about $𝜎$ -algebras:

$𝒫︀ (Ω)$ is the largest possible $𝜎$ -algebra on $Ω$ .

If $𝒞︀$ is already a $𝜎$ -algebra, then the generated $𝜎$ -algebra is the same.

If $𝐶$ is $\emptyset$ or $Ω$ , $𝜎 (𝒞︀)$ is ${\emptyset, Ω}$ , the trivial $𝜎$ -algebra.

Any collection of sets which has $𝐶$ has subset has the same $𝜎$ -algebra as $𝐶$ .

There are a lot of elements of $𝜎$ -algebras, so we typically consider them in the context of generator $𝒞︀$ .

A measurable space is a pair $(Ω, ℱ︀)$ where $Ω$ is the sample space and $ℱ︀$ is a $𝜎$ -algebra on $Ω$ .

For two $𝜎$ -algebras ${ℱ︀}_{1}$ and ${ℱ︀}_{2}$ , ${ℱ︀}_{1} \cap {ℱ︀}_{2}$ is a $𝜎$ -algebra (and in fact this can be generalized to any countable intersection of $𝜎$ -algebras, but ${ℱ︀}_{1} \cup {ℱ︀}_{2}$ is not a $𝜎$ -algebra. You would have to define $𝜎 ({ℱ︀}_{1} \cup {ℱ︀}_{2})$ , better denoted as ${ℱ︀}_{1} \lor {ℱ︀}_{2}$ .

To picture the Borel $𝜎$ -algebra, we must first envision our $Ω$ as a topological space where geometry is defined. We will specifically consider $Ω = ℝ$ , where our sample space is the real number line. The Borel $𝜎$ -algebra $ℬ︀$ is generated by the class of open subsets of $Ω$ . Think of this as every subdivision of the number line. Almost every subset of $ℝ$ is in $ℬ︀$ , except for some very exotic sets.

As an example: if you partition $Ω$ i.e. you split it into disjoint sets, the different unions of all those sets is the same as the $𝜎$ -algebra generated by all the disjoint sets. This can be proven by the fact that the complement of each collection of sets is present as a union of different sets, since they encompass the whole sample space.

## What the Price Gouging Prevention Act Says - URL: https://sharifhsn.dev/blog/linkedin-2024-08-27-price-gouging-prevention-act/ - Structured data: https://sharifhsn.dev/api/posts/linkedin-2024-08-27-price-gouging-prevention-act.json - Description: A summary of the 2024 bill's price-gouging language, defenses, and penalties. - Date: 2024-08-27 - Exact published timestamp: 2024-08-27T18:32:34.267Z - Topics: Consumer Protection, Public Policy - Categories: None - Source: LinkedIn - Source URL: https://www.linkedin.com/posts/sharif-haason_monopoly-round-up-price-gouging-vs-price-activity-7234266560796786688-BlTq - Content source format: typst

The newsletter linked here says that Kamala “was not specific on what [price gouging] meant”. There’s a bill currently in the Senate called the Price Gouging Prevention Act of 2024, which lays out these details. Is it possible that this bill’s contents are what Kamala is referring to?

Here’s a quick summary:

The specific language used for price gouging is the following: “It shall be unlawful for a person to sell or offer for sale a good or service at a grossly excessive price, regardless of the person’s position in a supply chain or distribution network.”
Small businesses with less than $100,000,000 in gross revenue are exempted from this.
Businesses can make an affirmative defense to the accusation of price gouging by a preponderance of the evidence (i.e. more likely than not) that the increase in price was both not in their control and incurred in the process of procurement, acquisition, distribution, or provision of the good or service.

There are also more specific rules in the case of an exceptional market shock, defined as “any change or imminently threatened change in the market for a good or service resulting from [any] cause of an atypical disruption in such market; or any period of time during which the President has declared a major disaster or emergency”.

If businesses offer goods at an excessive price compared to what they did before the shock or competing sellers, they cannot have unfair leverage, defined as either being large (>$1,000,000 gross revenue/yr), being a critical trading partner, discriminating between equal trading partners, engaging in deceptive practices, or having a dominant position in any market i.e. they have no competition or greater than 30-40 percent of a relevant market.

The punishment for price gouging is civil penalties, which for businesses without unfair leverage is a small amount, at most $25,000. For businesses with unfair leverage, the penalty is 5% of their revenues for the previous year.

Bottom line:

The bill targets large businesses that sell goods or services at grossly excessive prices without the affirmative defense that the price increase was not within their control. In addition, during exceptional market shocks, the bill focuses on companies with unfair leverage raising prices relative to other companies, with harsher penalties for those companies as well. Unlike what some critics have said, this is not a price control. It is intended to prevent larger players from using their market share during exceptional market shocks to gouge consumers.

The linked discussion: Monopoly Round-Up: Price Gouging vs. Price Fixing vs. Price Controls.

## Floating Point Architecture - URL: https://sharifhsn.dev/blog/floating-point-architecture/ - Structured data: https://sharifhsn.dev/api/posts/floating-point-architecture.json - Description: Floating point numbers are complicated enough in the abstract. How do systems programmers and hardware designers cope with the complexity of floating point? In the general… - Date: 2022-05-31 - Exact published timestamp: 2022-05-31 - Topics: floating-point, Computer Architecture - Categories: Computer Architecture - Source: Archive - Source URL: None - Content source format: typst

Floating point numbers are complicated enough in the abstract. How do systems programmers and hardware designers cope with the complexity of floating point?

Floating Point Instructions

In the general computer architecture, CPUs function by executing a stream of instructions. The set of instructions is defined by the hardware model, such as x86 or ARM. Instructions are typically tiny operations that need to be done extremely often. Some examples of instructions are mov to move values in and out of memory, sub to subtract numbers in registers, etc. In order to speed up floating point operations, hardware can include floating point instructions instead of letting software handle it.

Most of the time, instruction sets only allow floating point operations in the same precision. However, there is a useful optimization that is not often implemented in hardware. A hypothetical instruction to multiply two single precision numbers that results in a double precision number would be very useful. In particular, it would enable for the use of iterative improvement algorithms that would otherwise have compounding catastrophic cancellations if the precision was not increased temporarily.

Subexpression Evaluation

The order of operations is very important for preserving invariants in floating point. For example, floating point numbers do not necessarily follow the associative property, so compilers that optimize out parentheses will exhibit undefined behavior.

This can also cause issues when numbers with different precisions are operated on. There are two solutions to this problem. One is the solution that languages like OCaml take where all values in an arithmetic expression must have the same type. This means that integers will never be implicitly cast to floating point when they are operated with them, because they are not allowed to be operated with them anyway. However, this can be an overly strict restriction on the types of programs that can be written.

Another solution is the one that C takes, which is to establish rules for subexpression evaluation while allowing mixed type expressions. K&R C requires that every operation be done in double precision, but this can cause incompatibility between expressions and a stored value that are in different precisions.

A better way to do subexpression evaluation is to set a tentative precision for each expression, then increase the precision as needed in the wider expression. However, this means that the precision of a sub-expression can change with the expression it is embedded in, which is potentially unexpected behavior for a programmer.

IEEE Conformation

Languages like C conform to the IEEE specification, but the ways in which they do it can cause issues in implementation. If the exact rounding operation is implemented in hardware, then all the language has to do is call the appropriate instruction.

However, there are some aspects of floating point that are harder to represent. Floating point has a certain state associated with it, with rounding mode, flags, trap handlers, etc. This state must be read and written to, and must be preserved across subroutines.

NaNs also cause issue. The reflexive property is typically taken as invariant since it is in integer math, but it can be disastrous with NaNs. One massive consequence of the inclusion of NaNs is that floating point numbers cannot have a total order. This is because NaNs are explicitly unordered with respect to other floating point numbers. This has cascading consequences for the implementation of comparison operators like <, >, and =.

“Optimization”

Compilers make many optimizations to programs in order to increase performance or reduce instruction count. These optimizations are usually made with certain invariants in mind so that the accuracy of the program is not affected. However, because floating point has different invariants than integer math, compilers can make mistakes when optimizing floating point math. For example,

float ε = 1;
do {
    ε = 0.5 * ε;
} while (ε + 1 > 1);

will estimate $𝜖$ . However, a compiler might notice that ε + 1 > 1 for integer math is equivalent to ε > 0 and make that optimization, without considering that the expression $𝜖 \oplus 1$ has a special meaning that is different for floating point numbers than a test for positivity.

In general, many algorithms with floating point will exhibit expressions that, upon first blush, seem to be redundant in integer math. Having consideration for the ways in which floating point math can change values is important.

Exceptions

Trap handlers are empowered to be able to access variables in programs. However, computers which have parallel arithmetic may not necessarily be able to easily identify which operation threw an exception. Trap handlers must also be able to identify programs.

The reordering of certain instructions can also cause issues. Compilers will often change the order of certain arithmetic instructions when they are exact so that the semantics don’t change. However, if the operation traps, then it is not as easy to identify the operation that trapped, since the operation that happened in parallel will modify the trapped arithmetic.

One solution to this is presubstitution where a user that knows that an exception can create their own handler for an exception and substitute the value themselves beforehand. However, because it goes against IEEE-754, it’s unlikely to be proliferated.

All this information comes from the landmark paper What Every Computer Scientist Should Know About Floating-Point Arithmetic

## IEEE-754 - URL: https://sharifhsn.dev/blog/ieee-754/ - Structured data: https://sharifhsn.dev/api/posts/ieee-754.json - Description: IEEE-754 is the standard for floating point computation around the world. Thus, any real-world discussion of floating point must include it. IEEE-754 floating point numbers are… - Date: 2022-05-31 - Exact published timestamp: 2022-05-31 - Topics: floating-point, Computer Architecture - Categories: Computer Architecture - Source: Archive - Source URL: None - Content source format: typst

IEEE-754 is the standard for floating point computation around the world. Thus, any real-world discussion of floating point must include it.

Format

IEEE-754 floating point numbers are always in base 2 binary; this makes sense as it minimizes wobble which scales with $𝛽$ . It also allows for a special optimization that can add an extra bit of precision without increasing the size of the number. To demonstrate that, we can look at the number $001010.0011$ . The normalized floating point representation of this is $1.0100011 \times 2^{3}$ . Notice how the digit in front of the decimal point is always 1. It can’t be 0, because then we could just shift the decimal point and throw away the zero as we did with the initial number. Therefore, IEEE-754 assumes that the first bit is always 1 and only encodes the bits after the decimal point, granting an extra 1. There is an obvious problem with this: the number 0 cannot be represented. In IEEE-754, if the significand and exponent are all 0, then the number is the special case of 0.

IEEE-754 allows for four possible precisions: single, double, single-extended, and double-extended:

The formats use these parameter values:

Parameter	Single	Single-Extended	Double	Double-Extended
$𝑝$	24	32	53	64
$𝑒_{𝑚 𝑎 𝑥}$	+127	+1023	+1023	+16383
$𝑒_{𝑚 𝑖 𝑛}$	−126	−1022	−1022	−16382
bits for exponent	8	11	11	15
total # of bits	32	43	64	79

The usefulness of extended precision is that it grants a large amount of guard digits for internal calculations, for example in a calculator. There are fast algorithms for common transcendental functions like $log$ , but most of them have a large amount of possible error. By using extended precision internally then rounding to regular precision on display, calculators can quickly compute transcendental functions for users accurately.

The exponent value is calculated using a bias so it can represent negative values. The initial unbiased exponent is subtracted by $2^{𝑒 - 1} - 1$ , e.g. 127 for single precision. The reason that the $𝑒_{𝑚} 𝑎 𝑥$ is always bigger than $𝑒_{𝑚} 𝑖 𝑛$ is to prevent overflow when the reciprocal of the smallest numbers are taken. In turn, the reciprocal of the largest numbers will create underflow. In general, underflow is preferable to overflow, for reasons that will be clear when we discuss overflow in more depth.

Operations between floating point numbers in IEEE-754 are always exactly rounded. Having a guard digit can reduce error, but it does not guarantee the same result as exact rounding, so it cannot be used. There are methods using two guard digits and a “sticky” bit that can be used for efficient exact rounding, however. Exact rounding is important so that the result of two operations can always be guaranteed to be the same, regardless of hardware. The operations defined under IEEE to be exactly rounded are $+$ , $-$ , $\times$ , $/$ , $\sqrt{\cdot}$ , $%$ , and conversion with integers. Conversion between binary and decimal is not necessarily exactly rounded, because the most efficient conversion algorithms are not exactly rounded. Transcendental functions are also not specified as exactly rounded because there is no efficient algorithm that works across all hardware.

`NaN` and $\infty$

If you’ve used a calculator before, chances are you have seen both of these terms before. But what do they mean in the context of floating point?

Not all bit patterns in IEEE-754 follow the rules that were laid out earlier for significands and exponents. Some aren’t even valid floating point numbers! There are special values that are possible:

Special exponent values represent the following:

Exponent	Fraction	Represents
$𝑒 = 𝑒_{𝑚 𝑖 𝑛} - 1$	$𝑓 = 0$	$\pm 0$
$𝑒 = 𝑒_{𝑚 𝑖 𝑛} - 1$	$𝑓 \neq 0$	$0 . 𝑓 \times 2^{𝑒_{𝑚 𝑖 𝑛}}$ (denormalized)
$𝑒_{𝑚 𝑖 𝑛} \leq 𝑒 \leq 𝑒_{𝑚 𝑎 𝑥}$	—	$1 . 𝑓 \times 2^{𝑒}$ (normalized)
$𝑒 = 𝑒_{𝑚 𝑎 𝑥} + 1$	$𝑓 = 0$	$\pm \infty$
$𝑒 = 𝑒_{𝑚 𝑎 𝑥} + 1$	$𝑓 \neq 0$	NaN

The middle row represents normalized numbers, which we are already familiar with. We will discuss each of the other special values.

As mentioned earlier, 0 is an exception where the significand and exponent are both 0. But wait, why can 0 be positive and negative? IEEE-754 also defines that $+ 0 = - 0$ so the distinction seems useless. The reason both are allowed is because it preserves the sign of infinity when they interact. It also allows for functions that are only defined for positive or negative numbers to be able to safely include 0 in them.

Denormalized numbers are numbers with 0 as the hidden bit instead of 1. The reason they need to exist and break these rules is because small numbers being subtracted can end up underflowing to 0. It breaks a common invariant in code that the difference between two unequal numbers is not 0, and can cause bugs. Denormalized numbers gradually underflow to 0 so this problem does not occur. It also significantly reduces relative error for very small numbers.

The idea of infinity is an important one to measure for floating point. Unlike with integer arithmetic, division by zero is not an immediate error which aborts the operation. Rather, overflow can be an expected result which is handled for by using infinity. If the infinity ends up in a denominator, then the computation can underflow to 0 which can be an expected result.

NaN stands for “Not a Number” and is perhaps the most unique special valueNaN does not represent any computable value, and it is “infectious”; any operation withNaN will result in an NaN and NaNs always compare to false, even with other NaNs. They are a result of computations like $0 / 0$ or $\sqrt{- 1}$ which are not well-defined in real numbers, and also cannot be represented by infinity. These are all the operations that will produce an NaN:

These operations produce NaN:

Operation	Production
+	$\infty + (- \infty)$
×	$0 \times \infty$
/	$0 / 0$ , $\infty / \infty$
%	$𝑥 % 0$ , $\infty % 𝑦$
$\sqrt{\cdot}$	$\sqrt{𝑥}$ for $𝑥 < 0$

Unlike with denormalized numbers, the value in the significand does not necessarily represent any specific values. Often, some information about the operation will be placed in the significand as a signal. If an operation is done between a real number and an NaN, it will not change the significand of the NaN.

Exceptions

NaN and $\infty$ allow floating point calculations to continue in the face of special circumstances without aborting immediately. However, this behavior is not always appropriate. Implementations of IEEE-754 typically have trap handlers that will handle exceptions generated by such circumstances. There are five classes of exception that can each be set by a status flag: overflow, underflow, division by zero, invalid operations, and inexact operations. Invalid operations are any that involve NaN except for those when one of the operands are already NaN.

Operations in IEEE-754 must be exact, so if an operation is performed that is inexact, it will raise an exception. However, this raises an issue. Inexact exceptions can happen extremely often, and telling the OS to summon the trap handler every time harms performance. In order to prevent this, a software flag is typically enabled upon inexact exception to mask off future exceptions until the flag is reset.

Although trap handlers can abort the algorithm, they can also exhibit other behavior which can aid algorithms in being more efficient and accurate. For example, IEEE-754 will wrap around overflowing numbers by dividing the computed result by $2^{𝛼}$ . $𝛼$ is 192 for single precision and 1536 for double precision. This is useful in partial products when the operation might overflow and underflow at several points. By allowing the products to continue without aborting, the final product might cancel out under/overflows and be in range without having to stop the OS at every point.

Rounding

We stated earlier that round to even is the best and most commonly accepted method of rounding. This is the default mode of IEEE-754. However, there are other acceptable rounding modes that can be enabled for certain operations. These are round to 0, to $+ \infty$ , and to $- \infty$ . These modes turn rounding into floor/ceiling operations, which are useful for computing intervals.

We can represent floating point results as an interval between two numbers where the first is rounded to $- \infty$ and the second is rounded to $+ \infty$ . The exact result is somewhere between those two numbers. This representation can be useful to get an idea of how wide your error is when calculating a result. If the result with single precision has a very wide interval, then you might redo the result in double precision and so on to shrink the interval to some acceptable level of error.

All this information comes from the landmark paper What Every Computer Scientist Should Know About Floating-Point Arithmetic

## Floating Point - URL: https://sharifhsn.dev/blog/floating-point/ - Structured data: https://sharifhsn.dev/api/posts/floating-point.json - Description: Floating point representation is a complex topic full of nuance. To understand it, we must focus on the abstract method used to represent non-integer numbers. Math is infinite.… - Date: 2022-05-28 - Exact published timestamp: 2022-05-28 - Topics: floating point, Computer Architecture - Categories: Computer Architecture - Source: Archive - Source URL: None - Content source format: typst

Floating point representation is a complex topic full of nuance. To understand it, we must focus on the abstract method used to represent non-integer numbers.

Numbers in Computers

Math is infinite. Or at least it can be. Numbers like $5$ can be represented finitely, but a number like $𝜋$ is not so simple to represent. $3.14159265358979 \dots$ but that still isn’t enough. It’ll never be enough.

Unfortunately, computers are not infinite. Although we would like to hold every digit of $𝜋$ in our computer’s memory, the fact is that there will always be limits to how numbers can be represented in computers.

Integers are numbers like $3$ , $15$ , $- 154$ , etc. They have no decimal point and can be negative. Representing these numbers in a computer is fairly simple, with the limit only being placed on the size of the number.

However, not all real numbers play so nicely. How do we represent the value $12.5$ in memory? We can’t have half bits. Can we go infinitely precise on decimals, like $7.2028301324$ ? What about the size? How do we set these bounds?

Floating Point

We have to have a different standard to define how these numbers will be represented in a computer. This standard is called floating point.

Generally, a floating point representation is composed of three parts: the base $𝛽$ , the significand $𝑝$ , and the largest/smallest allowed exponents $𝑒_{𝑚 𝑎 𝑥}$ and $𝑒_{𝑚 𝑖 𝑛}$ . This is the complete formula for floating point (ignoring specific details):

⌈ {log}_{2} (𝑒_{𝑚 𝑎 𝑥} - 𝑒_{𝑚 𝑖 𝑛} + 1) ⌉ + ⌈ {log}_{2} 𝛽^{𝑝} ⌉ + 1

For example, let’s say we had a floating point representation in base 2 which allowed for exponents of sizes up to $+ 128$ and down to $- 127$ , and had twenty-three bits of precision for the significand. This representation would need 32 bits.

A number written in floating point representation might look like this:

1.010001 \times 2^{7}

where the first part is the significand, the base of the exponent is the base, and it’s raised to its exponent. Floating point numbers are written like scientific notation to be normalized, where there is only one nonzero digit in front of the decimal point—this will become important for IEEE 754.

Imprecision

I mentioned bits of precision, which is finite. Some decimal numbers can’t be represented by a finite amount of precision. For example, the number $0.3$ is finitely representable in base 10, but becomes $1.001100110011 \dots \times 2^{- 2}$ in binary. This number will never be perfectly represented in binary, so we have to have approximations.

In order to understand how much error we encounter, we must be able to measure it. There are two techniques to measure error in floating point: ulps or “units in the last place” and relative error.

Let’s use an example here, with $𝛽 = 10$ and $𝑝 = 3$ . We want to approximate $2.781828$ to a floating point number with this precision. Since we can only encode 3 digits of precision, we end up with $2.78 \times 10^{0}$ . If we imagine a special decimal point after the precision stops for floating point, then the difference between these two is $0.1828$ . This is difference for ulps. The closest floating point number can still have a ulps error of up to $\frac{𝛽}{2} \cdot 𝛽^{- 𝑝}$ . This number is known as the machine epsilon $𝜖$ .

Relative error is a familiar concept in most sciences. It is simply the difference between measured and actual, divided by the actual value. In this case, our floating point approximation $2.78 \times 10^{0}$ is the “measured” value and our real number $2.781828$ is our actual value. In this case, it is $0.001828 / 2.781828 = 0.0006$ . Since this number can be small, it is often expressed in terms of $𝜖$ . In this example, $𝜖 = 5 \times 10^{- 3} = 0.005$ so our relative error is $0.13 𝜖$ .

Since we know the maximum ulp error for the closest floating point number, we should find the same for relative error. One difference about relative error is that it changes based on the size of the numbers, not just the absolute error in the significand. The largest possible error is $\frac{𝛽}{2} 𝛽^{- 𝑝} \times 𝛽^{𝑒}$ .

However, the relative error changes based on the size of the real number in question, and all real numbers within the same exponent range will have this error. So, the relative error for a number closer to $1.0 \times 𝛽^{𝑒}$ will be larger than for a number closer to $𝛽 \times 𝛽^{𝑒}$ . This variation is called wobble. The relative error is always bounded by $𝜖$ , as in the prior example. Depending on the significand, however, the wobble can be up to $𝛽$ for the same exponent. Crucially, this wobble can be expressed in either relative error or ulps, as long as the other is held fixed.

Typically, relative error is used, because it is more useful for compounding operations whereas ulps can vary wildly within $𝛽$ .

An important concept to understand here is contaminated digits. These are the least significant digits of a floating point number which may be error-prone and are therefore untrustworthy. The number of contaminated digits is ${log}_{𝛽} 𝑛$ where $𝑛$ is the factor of the relative error to $𝜖$ , like $0.13$ in the earlier example.

Error Mitigation

We have found how to measure error; now how do we reduce it? One method is to use guard digits. When performing calculations, a computer can “extend” the calculation by a few digits so that the extended digits are contaminated by the floating point operation. These digits will then be rounded out of the final result, reducing the contamination of the result.

Without a guard digit, the relative error can be large as $𝛽 - 1$ , which can put every digit in error! However, adding just one guard digit bounds the relative error to less than $2 𝜖$ !

Another method is to reduce the number of cancellations. A cancellation occurs when nearby quantities are subtracted. When this happens, the most significant and uncontaminated digits cancel out and the less significant, more contaminated digits are left. Cancellations can seriously magnify rounding errors in a series of cancellations; this is called catastrophic cancellation. However, cancellation can also be benign. If the two quantities being subtracted are exactly known, then the subtraction will have a tiny relative error if done with a guard digit.

Sometimes, we can rearrange formulas to have less or more benign cancellations. For example, the formula $𝑥^{2} - 𝑦^{2}$ has a catastrophic cancellation because $𝑥^{2}$ and $𝑦^{2}$ both suffer from rounding error. However, this formula can be rearranged to $(𝑥 \oplus 𝑦) \otimes (𝑥 ⊖ 𝑦)$ . The cancellation is now benign because $𝑥$ and $𝑦$ are presumably exact values without rounding error yet.

Notice that the operands in that formula are circled. This is a notation to indicate that these operations are performed by a computer and therefore may accrue rounding error, whereas the ordinary operands are used for exact calculations.

However, the impact of making a cancellation benign can be limited if the inputs to the equation are already inexact. This is common when converting between decimal and binary numbers. It is always worth eliminating a cancellation, though.

Exact Rounding

Although guard digits mitigate error, they will in many cases give a different value than the exactly rounded result. That is the result if the floating point number was computed exactly, then rounded to precision. Many algorithms require exact rounding to work properly.

The nature of rounding is controversial. The most commonly accepted rounding mode is round to even. This will round numbers that end in the half digit to whichever direction makes the number even. For example, $12.5$ will round to $12$ , not $13$ because $12$ is even. This achieves the result of rounding up and down being equal chance for the half digit, 5 in this case.

Exact rounding is useful for holding certain invariants in floating point calculation. For example, one way to increase precision in calculations is to split a multiple precision number into an array of single precision numbers. Adding these numbers back together with exact rounding will recover the multiple precision numbers.

We can represent a multiplication of two double precision floating point numbers $𝑥$ and $𝑦$ using this method. We can split $𝑥$ into $𝑥_{ℎ}$ and $𝑥_{𝑙}$ , each of which are single precision, and $𝑦$ into $𝑦_{ℎ}$ and $𝑦_{𝑙}$ likewise. Then, we can represent the multiplication as so:

𝑥 \cdot 𝑦 = (𝑥_{ℎ} + 𝑥_{𝑙}) (𝑦_{ℎ} + 𝑦_{𝑙}) = 𝑥_{ℎ} 𝑦_{ℎ} + 𝑥_{ℎ} 𝑦_{𝑙} + 𝑥_{𝑙} 𝑦_{ℎ} + 𝑥_{𝑙} 𝑦_{𝑙}

In this way, a multiplication of two double precision numbers can be represented as the sum of multiplications of single precision numbers. Because the numbers are exactly rounded, the number is completely recoverable. This means that double precision multiplication can be possible on a computer which only supports single precision multiplication.

All this information comes from the landmark paper What Every Computer Scientist Should Know About Floating-Point Arithmetic

## Network Security - URL: https://sharifhsn.dev/blog/network-security/ - Structured data: https://sharifhsn.dev/api/posts/network-security.json - Description: The internet has taken over the world, but with it have come malicious actors. People want to spy on you, scam you, steal from you, or harm you through the internet. Through… - Date: 2022-04-22 - Exact published timestamp: 2022-04-22 - Topics: Internet Technology - Categories: Internet Technology - Source: Archive - Source URL: None - Content source format: typst

The internet has taken over the world, but with it have come malicious actors. People want to spy on you, scam you, steal from you, or harm you through the internet. Through network security, we can shield potential victims from at least some of these harms.

Attacks

Attacks on a network can be performed passively, for example in eavesdropping. The network cannot tell the difference between this passive attack and ordinary traffic because no more traffic is being injected.

Active attacks, by contrast, are done by changing the kind or amount of traffic. This typically manifests as a DoS (Denial of Service) attack. This is usually a significant increase in traffic which shuts down the network due to congestion. If only one host is attacking the network, then the network can easily shut a particular malicious host out. DoS attacks are usually distributed, hence DDoS (Distributed Denial of Service).

Teardrop attacks are a modern form of DDoS. Instead of sending lots of useless traffic, all they do is open a new connection with a socket. Even though this is a small operation, when done on large scale, it can tax the memory of a server. We mentioned previously that in order to compensate for fragmentation, routers will keep a memory buffer to store fragments. This can be exploited by malicious packets that are all fragments which will run out the memory of the router.

The three main aspects of network security are authentication, confidentiality, and integrity.

Authentication: How do you prove that someone is who they say they are?
Confidentiality: How can you hide data from prying eyes?
Integrity: How can you prove that data has not been altered midstream?

In order to protect these, we need a security model.

An algorithm to ensure security that must be well-known.
Trusted parties know how to generate secret information.
Distribute that information through a secure method.
Specify a protocol for trusted parties to decrypt data.

Encryption

Encryption or cryptography is the encoding of a message in such a way that only the communicating parties can interpret it. That way, if a malicious party tries to read the message, they will be unable to understand it because they cannot decrypt it.

Encryption has two sides, the encryption algorithm and the decryption key. The algorithm should be publicly available in order to prove that it works. When you feed the decryption key into the algorithm, it will spit out a string associated with only that key.

The secret key itself should be long enough that the algorithm cannot easily be broken, but also short enough so that it can be transmitted easily. Fun fact: before minimum password lengths were instituted, the most common password was “god”.

The most basic kind of encryption algorithm is a substitution cipher. By simply replacing letters in a message with another letter with a known mapping, a message can be encoded and decoded. However, it is relatively trivial to break this cipher because the English language has patterns that can identify certain letters better than others.

You can increase the complexity of this algorithm through polyalphabetic encryption, where the cipher itself changes after every letter. Famously, this is how the Enigma cipher used by the Nazis during World War II worked.

Transposition ciphers will change the order of the words as well. This might seem like a really good cipher, unfortunately it can similarly be broken by looking at the structure of language.

In order to actually encrypt data, real-world encryption algorithms use a combination of these techniques.

But how do we generate the decryption key? There needs to be one for every user, but they need to be as random as possible and nigh impossible to crack, so we can’t use plaintext passwords.

We can use pseudorandom number generators in order to create a key. A pseudorandom number generator is an algorithm which, starting with some seed, will produce a string of numbers based on that seed which appear to be random. The seed will typically based on some environmental noise that can’t be easily predicted, such as the time of day at which the algorithm runs. The generator should be highly sensitive to changes in seed.

Symmetric Key

Using the same key, two parties should be able to both encrypt and decrypt the message. By applying the key to the message, the message should be encrypted, then decrypted.

There are two main kinds of symmetric ciphers: stream ciphers, and block ciphers. As the name implies, stream ciphers do bitwise operations on the message and the key in order to get the ciphertext. The bitwise operations can be symmetric, like ⊗. One example of a stream cipher is RC4 which can be used in SSL.

Block ciphers have a 1-to-1 mapping of certain blocks of $𝑘$ bits to other blocks, typically 64 bits. This is a good method, but the problem is that the number of mappings balloons massively: in general, it is $2^{𝑘}!$ , which is absolutely massive! It is both exponential and factorial. To simulate such a table, you can use a pseudorandom function.

DES

The Data Encryption Standard is the US encryption standard. It uses a 56-bit symmetric key for 64-bit plaintext input, and it is a block cipher. It is fairly secure for small attacks, but it can be brute force decrypted in less than a day. Its advantage is that there is no analytic exploit in the standard.

In order to make DES more secure, we can increase the number of keys to 3, making 3DES. The message is encrypted, decrypted, then encrypted again when run through all three keys. Effectively, there are $56 \times 3 = 168$ bits to crack, which is much stronger than 56.

Replay Attacks

This method of symmetric key is vulnerable to a certain kind of attack: the replay attack. Let’s say I encrypt a message and mail it to my boss. My boss decrypts the message, and sends me back a confidential message, so we can pass information to each other. But my boss doesn’t know my address, he only knows that he can trust me because the message is encrypted using our key.

What if a malicious hacker somehow obtains my encrypted message? If she sends that message to my boss, then my boss will email the hacker back with the confidential information because he thinks that only I have the ability to encrypt messages in that way.

The way to protect against this is by using a NONCE value, which is a Number used only ONCE. Let’s say I write a random number on my message every time I send a message to my boss. The hacker obtains my message and sends the same message later, with the same random number. My boss knows that the NONCE value cannot be the same and therefore does not trust the hacker with our confidential information.

The NONCE doesn’t have to be a random number. Many times, it can be a challenge query like asking for my mother’s maiden name. Having multiple challenges that are cycled through adds an extra layer of protection.

RSA

Nowadays, almost all cryptography is done by having two keys: a public key, and a private key. The public key is used to encrypt messages, and private key is used to decrypt messages. The public key is, as might be obvious, publicly available. Someone who wants to send a secure message to me can use my public key to encrypt it. However, only I can use my own private key to decrypt it. Importantly, you should not be able to determine the private key by analyzing the public key. They should be wholly different in nature.

RSA is the most commonly used implementation of this kind of cryptography. To encrypt a message, take the message and raise it to the power of the public key, mod some $𝑛$ . To decrypt said bit pattern, raise it to the power of the private key, again mod the same $𝑛$ .

Authentication can be done using the passing of challenge messages. A new secret key will be created which is encrypted and decrypted in order to encrypt the data.

One problem with RSA is that it’s really expensive and slow computation-wise, which is good for security but bad for convenience. In practice, most uses of RSA encryption only use the actual algorithm to encrypt some secret key, which is then used as a simpler cipher for the data between the two parties.

Hash

Hashing is the process of creating a unique-ish value from some data that can be used to verify that data as a signature. It should be fast, not easily irreversible, and not have collisions i.e. no two hashes should be the same.

The two most common hash functions are MD5 and SHA-1, which are 128 bits and 160 bits, respectively.

## A Crash Course in Python - URL: https://sharifhsn.dev/blog/python-crash-course/ - Structured data: https://sharifhsn.dev/api/posts/python-crash-course.json - Description: Python is a powerful jack-of-all-trades language that is much less limited than OCaml. Arithmetic works in a very simple and clear way in Python, basically like a calculator. 2 #… - Date: 2022-04-14 - Exact published timestamp: 2022-04-14 - Topics: Principles of Programming Languages - Categories: Principles of Programming Languages - Source: Archive - Source URL: None - Content source format: typst

Python is a powerful jack-of-all-trades language that is much less limited than OCaml.

Expressions

Arithmetic works in a very simple and clear way in Python, basically like a calculator.

2 # 2
2 + 4 # 6
(2 + 5) * (3 - 5) # -20

Unlike in OCaml, operations can be performed between floats and integers with no issues; Python will simply change the types behind the scenes. This is called type coercion.

2 + 3.5 # 5.5

This causes an interesting side effect: what is the return type of this function?

def add(a, b):
    return a + b

In OCaml, the equivalent function would have type int -> int -> int, since the operation + is only valid for int. However, this function is actually polymorphic in Python! In OCaml, you would call it type 'a. In Python, this type is called Any, and operation changes based on what the types are. The reason that this is possible is because everything in Python is an object. Essentially, the variables a and b are just boxes that could contain anything, and Python only checks whether the operation + is defined for the two variables at runtime.

Strings

String manipulation is a very common operation in Python, so there are some very useful ways to handle strings built into the language.

"hello " + "world" # "hello world"
"hello" * 3 # "hellohellohello"

You can also convert types to strings very easily.

str(5) # "5"
str(3.5) # "3.5"

And vice versa.

int("5") # 5
float("3.5") # 3.5

These are special built-in functions to make our lives earlier.

Variables

Like with OCaml, there is no need to specify types as in Java. Unlike OCaml, Python does not use type inference. Instead, every variable can contain any type in a box as mentioned earlier.

a = 3
b = "hello"

Variables in Python work differently under the hood than other languages. A variable is essentially just a name for an element.

c = b

c here is not just equal to b… it is actually b itself! And any change you make to b will reflect in c because of that. Actual copies must be explicit.

Slices

Slices are one of the most powerful tools in Python. In fact, it might be what Python is most known for and most useful for. Slicing allows for powerful manipulation of list-like data.

Ordinary list access uses bracket notation with a single number to access a single element. Slice notation works similarly, but it allows returning multiple elements as a sub-list of the original list.

x = "hello world"
x[1:7] # "ello w"

You might notice that I just performed this operation on a string; didn’t I just say that slices work on list-like types? In Python, strings are just fancy lists of chars!.

Tuples and Lists

A tuple is an immutable set of multiple elements, just like in OCaml. Slicing operations work the same way in that they return a subsequence tuple. There’s a weird side effect where if you slice in a way that returns a single element, you can get a single-element tuple.

Lists in Python are much more flexible in OCaml, as they can be heterogeneous with any type within. They are also mutable, so elements of lists can be reassigned.

Slicing can superpower this assignment by reassigning multiple values at once

Control

The traditional if statements are back, but with a bit of a twist. Python has significant whitespace, which means that the amount that you indent by affects the actual execution of the code. This is quite rare.

if x == 15:
    y = 0 # this tab is mandatory!
x = 3 # this line is outside the if block

There is a special keyword called pass which exists to allow empty blocks. Normally, this construct is forbidden:

if x == 15:
x = 3 # error!

But by using pass, this is possible:

if x == 15:
    pass # does nothing
x = 3

Like in C, boolean evaluation is 0 for false, true for everything else. However, you cannot assign a variable in a conditional! So this C construct would not be allowed:

while ((int x = some_func()) == 0) {
    // this is allowed in C
}

while (x = some_func()) == 0:
    pass # this is not allowed in Python!

Functions

Functions are defined using the def keyword.

def fac(n):
    if n < 0:
        return "negative!"
    elif n == 0:
        return 1
    else:
        return n * fac(n - 1)

Functional programming can be used similarly to OCaml where everything is a function.

def compose(f, g):
    def foo(x):
        return f(g(x))
    return foo

let compose f g =
    let foo x =
        f g x
    in foo x

## Disks - URL: https://sharifhsn.dev/blog/disks/ - Structured data: https://sharifhsn.dev/api/posts/disks.json - Description: Persistent storage is essential to every system that we can think of, and we can hardly think of a computer without it. However, managing disks is not as simple as it might… - Date: 2022-04-14 - Exact published timestamp: 2022-04-14 - Topics: Operating Systems Design - Categories: Operating Systems Design - Source: Archive - Source URL: None - Content source format: typst

Persistent storage is essential to every system that we can think of, and we can hardly think of a computer without it. However, managing disks is not as simple as it might appear on a surface level.

I/O

Before we dive deep into disks, we must first consider the general concept of input/output, or I/O. This is a general class of devices which externally connect to the computer and provide input and output. They include keyboards, mice, displays, speakers, etc. and are essential to any typical use of a computer.

All of these devices are very different, so how does the operating system communicate with each of them? In order to talk to external devices, the OS treats every device as a canonical device with three components: a status register, a command buffer, and a data buffer.

The status register tells the OS what the current status of the device is. There might be many types of status for a device, but the basic one we will consider is BUSY. If a device is busy, the OS will not command it to do anything and will wait until it is free. Once it’s free, the OS will put a command in the command buffer, and optionally some relevant data in the data buffer. The device will become busy again performing the command that the OS has given it.

Note that the device here is agnostic to the actual use of the device. The OS uses the same general procedure for every device.

One nice thing about I/O theoretically is that because its duties are separate from the computer, it can perform operations while the CPU is busy doing something else, increasing performance. However, this might not necessarily be the case. In order to initiate I/O, the CPU must, as previously stated, modify the command and data buffers. This can use up valuable CPU time.

In order to mitigate this, most computers have a separate DMA engine for direct memory access. This engine will communicate with I/O while the CPU is busy doing something else.

Although we do have this canonical device, the computer still needs to know what the different commands are and the type of data to put into the hardware. In order to communicate with the hardware effectively, all modern operating systems use device drivers. These are essentially low-level wrappers to the device which provide a convenient API for operating systems to use. In fact, almost all code in an operating system is devoted to these device drivers!

Hard Disks

For most of computing history, HDDs or hard disk drives have been the hardware for persistent storage, and their internal structure is important to making sure they are performant. Recently, SSDs or solid state drives have increased in popularity for faster storage access. However, they are still expensive for large amounts of data and HDDs are still used commonly, so let’s discuss them.

The basic hardware for a hard disk is a platter with several rotating wheels and an arm which moves between each wheel. Most hard disks are composed of thousands of such platters. Each wheel is sectioned into blocks of data. In order to access the data, the arm must seek the correct wheel and the wheel must rotate to bring the correct data block for the arm to read, which will transfer the data to the computer. These three steps form the basis of reading and writing from a hard disk. Seeking and rotating are slow and tend to be the bottleneck of access, while transferring is typically fast. The time taken for I/O is defined as

𝑇_{𝐼 / 𝑂} = 𝑇_{𝑠 𝑒 𝑒 𝑘} + 𝑇_{𝑟 𝑜 𝑡 𝑎 𝑡 𝑖 𝑜 𝑛} + 𝑇_{𝑡 𝑟 𝑎 𝑛 𝑠 𝑓 𝑒 𝑟}

and the rate of I/O is

𝑅_{𝐼 / 𝑂} = \frac{𝑆 𝑖 𝑧 𝑒_{𝑇 𝑟 𝑎 𝑛 𝑠 𝑓 𝑒 𝑟}}{𝑇_{𝐼 / 𝑂}}

These facts mean that when composing data, we want to reduce the amount of seeks and rotations as much as possible. This makes sequential data very important. When data is accessed sequentially, it tends to be on the same wheel and next to each other, so no seeks and very little rotation needs to be done.

Scheduling

Like with CPU cycles, disk requests can also be scheduled and reordered. This becomes important when trying to optimize for sequential data instead of random access.

The most obvious way to schedule a disk is to schedule requests that have the shortest seek time from the current wheel, so either on the same wheel or a close-by one. This increases performance, but it will starve requests that happen to be far away. We want to have some amount of fairness in our scheduling.

The currently accepted way to do this is SCAN or C-SCAN, also known as the elevator method. Essentially, the disk will be swept from beginning to end to check for any requests. If there is a request, then it will be performed. This preserves sequential data as the sweep goes in order, but makes sure that far away requests are also done. C-SCAN is an improvement over SCAN. Where SCAN simply goes back and forth like a real elevator, C-SCAN will treat the disk like a circle and sweep in the same direction. This prevents the sectors at the ends from being starved.

## Particles - URL: https://sharifhsn.dev/blog/particles/ - Structured data: https://sharifhsn.dev/api/posts/particles.json - Description: Tiny particles are affected in different ways than we are by waves. There are several different effects that must be discussed. Many people believe that Albert Einstein won the… - Date: 2022-04-12 - Exact published timestamp: 2022-04-12 - Topics: Physics - Categories: Physics - Source: Archive - Source URL: None - Content source format: typst

Tiny particles are affected in different ways than we are by waves. There are several different effects that must be discussed.

The Photoelectric Effect

Many people believe that Albert Einstein won the Nobel Prize for his discovery of special relativity, as that is what he is most well-known for. However, his actual claim to fame in that respect had to with the photoelectric effect. He found that when light of a certain wavelength is incident on certain metals, electrons are ejected from the surface of the metal.

The number of photoelectrons emitted, the photocurrent, increases with intensity, though the kinetic energy does not. The kinetic energy only increases with an increase in energy of the light i.e. a decrease in the wavelength. Above a certain wavelength, no electrons are ejected.

\frac{ℎ 𝑐}{𝜆_{0}} = 𝑊_{0} + 𝐾 𝐸_{𝑚 𝑎 𝑥}

where $𝑊_{0}$ is the work function for the metal in question. You can think of it as an intrinsic quality of the metal. $𝜆_{0}$ is the cutoff wavelength after which no more electrons will be ejected.

Pair Production/Annihilation

When a high energy photon collides with a nucleus, it may “split” into an electron and a positron. When this split happens, the charge, momentum, and the mass-energy of all parties must be conserved:

𝐸_{𝑝 ℎ 𝑜 𝑡 𝑜 𝑛} = 𝐾 𝐸_{𝑝 𝑜 𝑠 𝑖 𝑡 𝑟 𝑜 𝑛} + 𝐾 𝐸_{𝑒 𝑙 𝑒 𝑐 𝑡 𝑟 𝑜 𝑛} + 2 𝐸_{0}

where $𝐸_{0}$ is the rest mass-energy for each particle given by $𝐸 = 𝑚 𝑐^{2}$ , doubled since the electron and positron are identical.

The opposite operation can occur when an electron and a positron combining will annihilate the pair and create two identical photons with opposite momentum. The reason that there must be two photons and not just one is that the momentum must equal 0, so there has to be opposite momentum.

de Broglie Wavelength

All particles exhibit wave-like characteristics, which can be demonstrated by the double-slit experiment. The wavelength of a particle in this instance can be determined from the following equation:

𝜆 = \frac{ℎ}{𝑝} = \frac{ℎ}{𝑚 𝑣} = \frac{ℎ}{\sqrt{2 𝑚 𝐾 𝐸}}

The wavelength of a photon is a little different, since photons are completely massless. The momentum and energy of a photon are related by a simpler equation $𝐸 = 𝑝 𝑐$ , which applies to all massless particles, not just photons.

Uncertainty

One key problem for physicists studying particles is the Heisenberg Uncertainty Principle. In layman’s terms, the observation of a particle fundamentally changes its quality, so it can never be truly measured. In mathematical terms, the accuracy of a measurement of a position’s momentum and position at the same time is limited:

(Δ 𝑝_{𝑦}) (Δ 𝑦) \geq \frac{ℎ}{4 𝜋}

and the same function exists for simultaneous measurements of energy and time.

## Multimedia Networking - URL: https://sharifhsn.dev/blog/multimedia/ - Structured data: https://sharifhsn.dev/api/posts/multimedia.json - Description: Streaming multimedia is a relatively new concept in internet technology, but now has become central to the daily media consumption of people around the world. In contrast to the… - Date: 2022-04-07 - Exact published timestamp: 2022-04-07 - Topics: Internet Technology - Categories: Internet Technology - Source: Archive - Source URL: None - Content source format: typst

Streaming multimedia is a relatively new concept in internet technology, but now has become central to the daily media consumption of people around the world.

Multimedia

In contrast to the ordinary messages that encode text, multimedia streaming uses audio or video. Most of the traffic on the internet today is dominated by multimedia streaming, so it’s a big deal. Multimedia streaming is different from other kinds of streaming in that latency is of greatest importance, but it is also loss tolerant; if the streaming video is a bit distorted, it’s not a big deal. When the delay hits above 150 ms users begin to complain.

VoIP is a special case where there is media being streamed both ways, where two people speak on the same connection through data networks.

We must handle signals differently because we don’t have a simple Huffman coding like with text. There are three elements at play that determine the quality of the signal:

Sampling: how often do you sample the signal for data?
Quantization: how many levels/bits represent each sample?
Compression: how much are the quantized values compressed?

In order to minimize the amount of data being sent, video streaming will often only send the changes from one frame to the next. Otherwise, the level of fidelity would be infeasible on current internet technology.

Audio

Audio is generally ranges from 20 Hz to 22.05 kHz, and speech goes between 200 Hz and 8kHz. Sampling generally occurs up to 8kHz to capture speech, the most important element of audio in general, though it is possible to capture 16kHz for high-fidelity audio.

Quantization can either be 8 bits or 16 bits, which means there are either $2^{8}$ or $2^{16}$ levels. That’s a difference between 256 and 65,536!

The bit rate of samples typically depends on the type of audio being sent. For simple VoIP it will typically be 64Kbps, because it quantized at 8 bits and sampled 8k times per second. Audio will generally be 705.6Kbps from 16 bits of quantization and 44.1k samples per second. Stereo is the most fidelity because it is audio in both ears, so double the bit rate at 1.4112 Mbps. By removing frequencies that the human ear cannot distinguish, the amount of data can be reduced dramatically without altering the end sound.

Video

Images are 2D arrays of pixels at heart. The amount of pixels is resolution, which is typically expressed through a 16:9 aspect ratio as with 1080p or 4K/UHD.

Every color has 8 bits, and there are three colors RGB so each pixel will contain 24 bits. That means the raw data for a color image even of size 320×240 will be 7680 bytes! That’s a lot for one image. Compression through JPEG, Gif, etc. is typically used to help here.

Video is composed of images that are displayed at a certain frame rate. Movies run at 24 FPS, though some rare movies display at 48 FPS. Video games are commonly at either 30 FPS or 60 FPS, but can go even higher.

If we imagine a 4k television displaying color images at 60 FPS video, the amount of data transmitted would be $4096 \cdot 2160 \cdot 60 \cdot 24$ which is 12 Gbps!!! We obviously need some kind of compression.

Streaming

The client will typically downloading the first portion of the multimedia and begin consuming it. The rest is downloaded while the client consumes the data, so you don’t need to wait for the whole file to be downloaded. If you’ve ever watched a YouTube video, you can see this in real time as the gray bar overlaid on top of the timeline of the video. The client has a buffer to store that downloaded video which is constantly drained to show the video and filled at times corresponding to network delays.

However, this can cause buffering when the network delay is variable and the client consumes all of its data before the server can produce more. The video will be buffered until more data is received, and you see a still frame on the video while a loading symbol plays over it.

RTP

UDP streaming is the traditional type of streaming. As multimedia does not necessarily need to be reliable, UDP is better for reducing latency which is most important for the consumer. The packet has the timestamp of the multimedia so it can be ordered correctly in the client buffer. It also needs certain information like the encoding or whether the audio is for left or right microphone. The protocol used for multimedia that incorporates these is called RTP or Real-time Transport Protocol.

RTP is a flexible protocol that allows for many types of multimedia to be transmitted.

The header includes a timestamp field which increments per sample, which allows for correct timing. It also uses sequence numbers to distinguish between packet loss and silence.

HTTP

Most people stream video over their browsers nowadays, which uses HTTP. You might remember that HTTP uses TCP, which is slower than UDP but more reliable. The streaming is slower but more convenient.

DASH

Dynamic Adaptive Streaming allows for good bandwidth over HTTP. The client is the one that adjusts the request rate from multiple content formats/encoding. For example, the client can request data at a lower bit rate or quality like 480p if it feels that 720p is too slow.

The MPD or Media Presentation Descriptor is sent over HTTP and gives information about the different format segments. The client can then request each segment, and the segments can come from different ASes, which works well for CDNs. This is how ads are served over YouTube; the client is the one that decides whether to get an ad or not. This is also why blocking ads on YouTube can be done by an extension to your browser.

DNS

The ID for every YouTube video is a BASE64 11 character string, giving a space of $64^{11}$ possible videos. The ID is mapped to 192 DNS host names so that multiple servers can be accessed for videos. This ID is a special hash of the video ID.

Multiple ASes can broadcast the fact that they are all reachable to a BGP router, and a BGP router can then choose the shortest path. This is called anycast. This means that for accessing video servers, you can pick the country where the video is hosted quite easily. Let’s say you want to access a Japanese video in Japan. It’s obviously best to access the Japan server, so that’s the closest. However, all 192 servers that host the video will broadcast, so that if the Japan server goes down in an earthquake or something, the other servers are still available for the BGP router to pick from, so they can access, for example, the South Korean server.

VoIP

Traditional telephones work over PSTN (Public Switched Telephone Network), which is autonomous and only designed for telephones. However, as the Internet has gradually taken over the world, a demand was created to include telephony in the Internet. Thus, the VoIP (Voice IP) protocol was born.

The voice sound waves are converted into packets and transported as audio through lossy RTP, similar to other forms of multimedia.

In order to establish the connection to begin with, SIP or Session Initiation Protocol sets up the VoIP session between two hosts. SIP URLS are similar to email addresses that identify users on a network. Most of the work is lent to the caller in order to type in some identifying information as they typically have a display/keyboard. The URL is mapped to an IP address similar to DNS and email.

## Control in Prolog - URL: https://sharifhsn.dev/blog/prolog-control/ - Structured data: https://sharifhsn.dev/api/posts/prolog-control.json - Description: Conditional statements are very powerful and are in used in almost every language. How does Prolog implement the same idea? The basic idea behind a control statement is that… - Date: 2022-04-05 - Exact published timestamp: 2022-04-05 - Topics: Internet Technology, Principles of Programming Languages - Categories: Principles of Programming Languages - Source: Archive - Source URL: None - Content source format: typst

Conditional statements are very powerful and are in used in almost every language. How does Prolog implement the same idea?

Logic + Control

The basic idea behind a control statement is that there is logic, such as facts, rules, and queries, which are composed of clauses/goals, as well as control, which is how Prolog chooses the logic among several options. The control is implemented by the order of the facts/goals, which is extremely important. Prolog will always choose the first applicable rule in a program, and it will always choose the leftmost clause/goal in a query.

Brevity is key. If there are two rules that do the same thing, or one rule that is implied by another rule, you should probably take it out unless executing something more than once is your explicit intention.

Abstract Interpreter

A substitution $𝜎$ is a finite set of pairs of terms ${𝑋_{1} / 𝑡_{1}, . . ., 𝑋_{𝑛} / 𝑡_{𝑛}}$ where each $𝑡_{𝑖}$ is a term and each $𝑋_{𝑖}$ is a variable such that $𝑋_{𝑖} \neq 𝑡_{𝑖}$ and $𝑋_{𝑖} \neq 𝑋_{𝑗}$ if $𝑖 \neq 𝑗$ .

An empty substitution is denoted by the letter $𝜖$ .

Some important rules:

A variable cannot substitute itself e.g. $𝑍 / 𝑍$ is illegal.
Only a variable can be substituted e.g. $𝑚 / 𝑛$ is illegal because $𝑚$ is an atom.

The meaning of applying a term to a substitution is that every occurrence of $𝑋_{𝑖}$ in the compound term is replaced with the corresponding substituent in $𝜎$ simultaneously. This application is known as instantiation, and that new compound term $𝐸 𝜎$ is an instance.

Bringing back the earlier rules about unification, you can see how unification is derived from substitution. Variables can unify with anything, so they are the term that is substituted in Prolog. There is an exception to this, which is known as the occurs check.

We can use a substitution $𝜎$ as a unifier for two terms if the application of $𝜎$ to those terms makes them syntactically equal. This is distinct from semantic equality. When we talk about unifiers, we are only talking about the actual letters, not the meaning of the term. For $𝑆 = 𝑓 (𝑋, 𝑌)$ and $𝑇 = 𝑓 (𝑔 (𝑍), 𝑍)$ , if we have a $𝜎 = {𝑋 / 𝑔 (𝑍), 𝑌 / 𝑍}$ then $𝑆$ and $𝑇$ will be unified. Since the unification only cares about syntax, more than one unifier may exist for two terms. We could have just as well substituted the other way and it would still be a unifier.

Unifier Computation

unify(X, Y, θ) =
  X = Xθ
  Y = Yθ
  case X is a variable that does not occur in Y:
    return (θ{X/Y} ∪ {X/Y}) // this replaces X with Y in the unifier, and then adds a new substitution just in case X will appear later
  case Y is a variable that does not occur in X:
    return (θ{Y/X} ∪ {Y/X}) // this replaces Y with X in the unifier. this is a rare case when the variables substitute each other
  case X and Y are identical constants
    return Θ
  case X and Y are compound terms like f(X1, ..., Xn) and f(Y1, ..., Yn)
    return (fold_left (fun Θ (X,Y) -> unify(X, Y, Θ)) θ [(X1, Y1), ..., (Xn, Yn)]
    // this applies the unification process to the compound terms
    // the function is just folding left with θ as acc and the set of compound terms as the list

This is the basic pseudocode for the unify function. It’s recursive, and side-effect free. However, it’s not complete for writing an interpreter. For that, we also need backtracking.

Backtracking

The resolvent will maintain a list to satisfy our query. When we try to resolve a compound term, any goals within will be queued onto the resolvent to be resolved. This is non-deterministic, there is no order that will necessarily be followed other than that sub-goals within a single term will be inorder.

Missionaries and Cannibals

Let’s imagine a problem where we have three missionaries and three cannibals that need to cross a river in one boat. If the cannibals outnumber the missionaries, they will get eaten. How can we get every person across the river?

The concept of a safe state will be important here. A state is safe when no missionaries are eaten. By labelling certain states as unsafe, we can cut those paths out of our search algorithm. We should also define transitions between states, where a predicate moves a state from A to B.

Our states are essentially a pair representing position.

start(3-3-0-0-l).
finish(0-0-3-3-_).

The elements in order are missionaries on the original side, cannibals on the original side, missionaries on the target side, cannibals on the other side, and the location of the boat.

The safety of the state is based on whether there are more cannibals than missionaries.

safe(0-_-M2-C2-_) :- M2 >= C2.
safe(M1-C1-0-_-_) :- M1 >= C1.
safe(M1-C1-M2-C2-_) :- M1 >= C1, M2 >= C2.

Every state here represents a point at which there are at least as many missionaries as cannibals on each side of the river.

The change in state can be caused by a carry/2 predicate which details how many missionaries and cannibals are being moved, respectively. In our case, the boat can only move up to 2 people so we need to represent every possible case of that.

carry(2, 0).
carry(1, 1).
carry(0, 2).
carry(1, 0).
carry(0, 1).

However, we need a way to represent moving on both sides of the river, so we will have two transitions.

step(M1-C1-M2-C2-l, M3-C3-M4-C4-r) :-
    carry(X, Y),
    M1 >= X, M3 is M1 - X, M4 is M2 + X,
    C1 >= Y, C3 is C1 - Y, C4 is C2 + Y.

step(M1-C1-M2-C2-r, M3-C3-M4-C4-l) :-
    carry(X, Y),
    M2 >= X, M4 is M2 - X, M3 is M1 + X,
    C2 >= Y, C4 is C2 - Y, C3 is C1 + Y.

This step predicate checks every valid carry, then will execute that transition from the current state.

travel(A, A, _, []).
travel(A, C, Visited, [B | Steps]) :-

## Special Relativity - URL: https://sharifhsn.dev/blog/special-relativity/ - Structured data: https://sharifhsn.dev/api/posts/special-relativity.json - Description: Albert Einstein revolutionized the world of physics when he discovered the special theory of relativity. It is only possible due to his contributions that valuable technologies… - Date: 2022-04-05 - Exact published timestamp: 2022-04-05 - Topics: Physics - Categories: Physics - Source: Archive - Source URL: None - Content source format: typst

Albert Einstein revolutionized the world of physics when he discovered the special theory of relativity. It is only possible due to his contributions that valuable technologies like GPS function. But what is it exactly?

Dilation and Contraction

The theory behind special relativity is two-fold:

All inertial frames of reference of equivalent.
Observers must measure the same value for the speed of light in a vacuum.

This means that if even an observer has a huge difference in relative velocity, the speed of light is the same. This causes knock-on effects on the experience of time, and time dilation:

Δ 𝑡 = \frac{Δ 𝑡_{0}}{\sqrt{1 - \frac{𝑣^{2}}{𝑐^{2}}}}

The proper time $Δ 𝑡_{0}$ is the time interval with respect to an observer at rest. The dilated time is the time when the observer is at velocity $𝑣$ .

In the same way that time can be dilated, length can also be contracted:

𝐿 = 𝐿_{0} \sqrt{1 - \frac{𝑣^{2}}{𝑐^{2}}}

The relationship here is inverse! Time can only be dilated, and length can only be contracted. If the velocity was the speed of light (impossible) then the time interval would be infinite, and the length of an object would be 0.

Relativistic Addition

Two objects traveling at different relativistic velocities further complicate these calculations. If we want to calculate the proper velocity $𝑢$ of an object from its relativistic velocity $𝑢^{'}$ relative to an object at proper velocity $𝑣$ or vice versa, then we can use these equations:

𝑢 = \frac{𝑢^{'} + 𝑣}{1 + \frac{𝑣 𝑢^{'}}{𝑐^{2}}}

𝑢^{'} = \frac{𝑢 - 𝑣}{1 - \frac{𝑣 𝑢}{𝑐^{2}}}

As you can see, these are very similar equations, but their differences are important.

Mass and Energy

It’s the moment you’ve been waiting for, the famous equation!

𝐸 = \frac{𝑚_{0} 𝑐^{2}}{\sqrt{1 - \frac{𝑣^{2}}{𝑐^{2}}}}

Oh, it looks a little bit different. That’s because just like length and time are relativistic, mass is as well. In order to calculate total energy from rest mass, we must apply the same expression to it. There’s another way to calculate $𝐸$ :

𝐸 = 𝐾 𝐸 + 𝐸_{0}

where $𝐸_{0}$ comes from the classic $𝐸 = 𝑚 𝑐^{2}$ equation. We can also find the kinetic energy by rearranging these equations as

𝐾 𝐸 = (𝑚 - 𝑚_{0}) 𝑐^{2}

from finding the difference between relativistic mass and proper mass.

One more important attribute affected by relativity is momentum. As it depends on mass, it looks similar to the equation for $𝐸$ :

𝑝 = \frac{𝑚_{0} 𝑣}{\sqrt{1 - \frac{𝑣^{2}}{𝑐^{2}}}}

## Waves - URL: https://sharifhsn.dev/blog/waves/ - Structured data: https://sharifhsn.dev/api/posts/waves.json - Description: Light is curious because it is both a particle and a wave, which makes it behave strangely in certain contexts. When projected through a slit, some strange fringes will appear,… - Date: 2022-03-29 - Exact published timestamp: 2022-03-29 - Topics: Physics - Categories: Physics - Source: Archive - Source URL: None - Content source format: typst

Light is curious because it is both a particle and a wave, which makes it behave strangely in certain contexts. When projected through a slit, some strange fringes will appear, which is the subject of the famous double slit experiment/

Interference

When waves combine together, it is called interference. This interference can be either constructive or destructive. When light is projected through a slit, depending on the angle from the slit to the wall behind it, the light waves will either constructively interfere or destructively interfere, creating bright and dark fringes, respectively.

The position of the bright fringes is given by the following equation:

𝑑 sin 𝜃 = 𝑚 𝜆 \Rightarrow 𝑚 = 0, 1, 2, 3 . . .

where $𝑑$ is the distance between the slits, $𝜃$ is the angle of interference from the central max onward, and $𝑚$ is the order of the interference fringe. For bright fringes, the order of the interference fringes begins at 0 for the central bright fringe, then increments onward for each surrounding fringe. Note that there is no functional difference between a fringe on either side of the central fringe as long as they are the same order away from it.

Dark fringes have a similar equation:

𝑑 sin 𝜃 = (𝑚 + \frac{1}{2}) 𝜆

The ½ is important here because the location alternates with the bright fringes. $𝑚$ is also different because it starts at the first dark fringe surrounding the bright fringe. You can think like this:

… 3 2 2 1 1 0 0 0 1 1 2 2 3 …

where the italicized numbers are dark fringes and the bold are bright fringes.

Thin Film

Interference has a curious corollary with refraction if done through a thin film. Light that is transmitted between two media is partially reflected and partially passes through if passing from a medium of lower to higher index of refraction e.g. air to water. It is fully reflected if from higher to lower. The partially reflected light (but not the fully reflected light) undergoes a phase shift by a half wavelength.

Let’s say we want to find the distance that the light travels in the medium, the “height” of the medium layer $𝑡$ . If the light underwent a phase shift, the equation for interference is actually flipped:

2 𝑡 = (𝑚 + \frac{1}{2}) 𝜆_{𝑓 𝑖 𝑙 𝑚}

for constructive interference, and vice versa for destructive interference. However, if there is no net phase shift, then use the normal equations.

Note that $𝜆_{𝑓 𝑖 𝑙 𝑚}$ here is actually

𝜆_{𝑓 𝑖 𝑙 𝑚} = \frac{𝜆_{𝑣 𝑎 𝑐 𝑢 𝑢 𝑚}}{𝑛_{𝑓 𝑖 𝑙 𝑚}}

Diffraction

When a wave passes through a tiny slit, it spreads out. The slit has to be really small, as in on the order of the wavelength of the light, in order for this to happen, with more bending with a smaller slit.

Diffraction has interference as well, so the equation for the position of dark fringes is similar, although different:

𝑊 sin 𝜃 = 𝑚 𝜆 \Rightarrow 𝑚 = 1, 2, 3 . . .

where $𝑊$ is the width of the slit. Importantly, $𝜃$ is different here!!! It is the angle between the center of the slit and the center of the dark fringe in question. $𝑚$ is also different in that it cannot be 0! The first dark fringe in diffraction has order 1, because at order 0 there is constructive interference.

Diffraction grating has different properties than single slit diffraction, however, and is more similar to typical interference, following the constructive interference equation for bright fringes. Be careful about $𝑑$ in this context! It refers to the distance between each slit, which you might have to derive yourself from other values.

## The Link Layer - URL: https://sharifhsn.dev/blog/link-layer/ - Structured data: https://sharifhsn.dev/api/posts/link-layer.json - Description: Through all the layers of the internet, the most essential is the link layer : converting bits to and from signals, detecting errors, flow control, and addressing. We have spoken… - Date: 2022-03-28 - Exact published timestamp: 2022-03-28 - Topics: Internet Technology - Categories: Internet Technology - Source: Archive - Source URL: None - Content source format: typst

Through all the layers of the internet, the most essential is the link layer: converting bits to and from signals, detecting errors, flow control, and addressing.

Encoding

We have spoken mostly through the guise of bits per second being sent across the internet and have left it at that. But what does it mean to send a bit over air waves? This isn’t like a traditional electrical circuit or transistor in your computer; the internet uses signals.

The nuances of digital signal processing are beyond these notes, so we will simplify here. We can imagine signals like a pseudo-transistor where low signals are 0 and high signals are 1. However, this causes two issues. One is that time synchronization is impossible for waves that travel at the speed of light, and the other is that because signals are not emitting all the time, we need a way to differentiate between signals and simple noise.

The time synchronization is mandated by a clock standard such as the Manchester encoding, which is slower than the speed of light but allows each link to know exactly when a signal is coming.

Framing

A frame is a group of bits in sequence. Frames are useful for manipulating data at the link layer, but they can escalate small bit errors to big problems. If a frame is too big, then it will error out too often, so we will often have smaller frames.

We also need a way to delineate the beginning and ending of frames. There are two ways, character stuffing and bit stuffing.

With character stuffing, a special meta character will be used to delineate frames. ^ is used for the beginning of frame (BOF) and $ is used for EOF. If you need a $ in your text, then simply add an escape character, like another $ right beforehand.

With bit stuffing, a unique bit sequence 0x7E or 0b01111110 will delineate frames. If that sequence appears in the data, it will insert a 0 within like 0b011111010.

Error Control

Inevitably, bits will be corrupted on a physical link. This is a reality of circuit engineering. We have discussed several ways to handle errors on the application and protocol layer; what about on the link layer? We can either request retransmission as with TCP, or we can make corrections to the errors automatically, which may or may not be possible.

Parity bits can ensure a fixed sequence so that the extra bit can protect against errors. For example, with even parity, there will always be an even number of 1s, so if a bit flip changes that, you know that there was an error. But multiple bit errors can escape this detection.

Another method is when the sender will send a checksum computed from the message divided into chunks along with the message so that the receiver can use the checksum, just like with other layers.

Polynomial codes utilize the high degree of uncertainty when computing mod of a polynomial. The remainder must be added to the message and it can be checked by the receiver. This is a more complex form of error detection mathematically.

Addressing

We have heretofore defined a host as an IP address. But how do we identifies hosts at the link layer? For links we use a MAC address, which is a unique 48 bit address to a device which will communicate through the internet. MAC addresses are far less sophisticated than IP addresses. They cannot be grouped or categorized like with NAT. The MAC address of the destination is included in the frame that is sent. They are also unique to types of links, so a device will have different MAC addresses for Wi-Fi, Bluetooth, etc.

Hosts can then interact in multiple ways. One way is access point. Hosts must communicate with an access point which acts as an intermediary between hosts. Another is ad hoc. When you set up your Amazon Echo for the first time, you connect with it ad hoc via Bluetooth on your app and set it up from there to program it.

ARP

When connecting via LAN, hosts are part of the same subnet. In order to translate the IP address to a MAC address, the hosts use a protocol called ARP (Address Resolution Protocol. This is a simple protocol which defines its own identification and a source/destination address, as well as a broadcast FF packet. We know the sender IP, MAC, and the destination IP, but we don’t know the destination MAC address, so that field is blank. The ARP packet will traverse the entire link, and if a host recognizes its own IP address, then it will send an ARP reply back to the source with its MAC address.

LAN Extension

LANs seem pretty nice. Why don’t we just use LANs everywhere? Bandwidth is limited, so you need to limit the amount of hosts that can connect on one network.

A learning bridge is a type of access point that connects LAN segments. It maintains a table of hosts that are connected to the network. It intermediates between the hosts, receiving and sending packets as necessary if packets are sent between LANs. For each packet, the bridge stores the MAC and port of the packet and then floods all the LANs if it can’t find a match.

Multiple Access

If two packets are sent at the same time, that is called collision or interference. We can only have one pair communicating at a time. We must have a multiple access protocol to make sure that only one access method is used at a time. You can think of being in a class; if everyone spoke at once, it would be mayhem! There must be an extra overhead of each person raising their hand in order to ensure correct communication takes place.

Wireless

When a host sends a wireless signal that bounces off of a satellite to a receiver, how do we know that interference has occurred? We could use a TCP packet timeout, but that takes a while. A neat trick that you can do is leverage the fact that satellites reflect all signals to everyone, including your own. A host can receive its own signal and check if it was interfered with.

Originally, pure ALOHA was used, based on the needs of a university in Hawaii. This uses the broadcast method. If it receives garbage back, it will sleep for a random amount of time and retransmit. The reason it is random is because if both the hosts that sent garbage data send the packets again at the same time, it will cause another collision! For transmission time $𝑡$ , the vulnerable period where packets will be destroyed is $2 𝑡$ .

Slotted ALOHA has better checks to stop collisions from occurring in the first place by setting slots where hosts will send packets, which means that they wouldn’t collide in different slots.

But the best throughput is done if we check the actual channel before transmitting the packet, so we don’t have to guess or check after the fact if a collision happened. For this we need CSMA (Carrier Sense Multiple Access). If the channel is sensed and there are packets travelling, the host will wait to send the packet then immediately send a packet when free. This is a persistent system. The problem with this is that this will cause a collision when multiple hosts are waiting for a channel. We can solve this the way we solved ALOHA. If each host sleeps a random amount of time, even a small amount of time, it will give the hosts time to sense the channel.

Contention Access Methods

In order to sense the channel, there are many methods.

1-Persistent CSMA will constantly bug the channel to check if the channel is idle. If it’s busy, transmit and check again. This only really works when there is one host trying to connect.

Non-Persistent CSMA will transmit, then sleep for a random amount of time before trying again if the channel is busy. This means that that collisions won’t happen. This is better than the previous method, but can be slow.

$𝑃$ -Persistent CSMA has the same idea, but instead of being random it will transmit again with probability $𝑃$ . This fine-tunes the previous method for our specific needs and improves utilization.

Ethernet uses the Ethernet Backoff Algorithm, which utilizes binary exponential backoff. If a collision happens, then it will pick a time slot to transmit again out of $2^{𝑘}$ slots, where $𝑘$ is the number of collisions that have occurred. The length of each time slot is equivalent to. For example, let’s say two hosts collide on an Ethernet connection. Both hosts will both, on their own, divide the next few seconds into two slots and randomly pick between them. If they each end up in different slots, they’ll transmit fine. If they end up in the same slot again, repeat the process but with four slots. As you can see, the chance of collision will dramatically decrease with more slots. After a maximum of sixteen slots, if there are still collisions, then your host will give up and say that the link is down.

Ethernet

Ethernet is a wired multiple access protocol defined by IEEE 802.3. As previously stated, it uses 1-persistent CSMA with the backoff algorithm.

Ethernet frames have a preamble full of special information, as well as the source/destination MAC addresses already mentioned. The header will also say the type of protocol being used, whether it is IPv4, IPv6, ARP, RARP, etc. Finally, at the end, a checksum will be stored so that we can check if corruption has occurred.

Preamble	Source MAC	Dest. MAC	Type	Data	Checksum

Multiple Access Channel Partitioning

We have previously assumed that all hosts are fighting for the use of one channel and will have to retry connections if collisions occur. One way to solve this is to have predetermined allocation of channel access. This way, there is no chance of collision because each user will wait its turn on its own. However, this can introduce wastage if users are assigned to a channel partition but end up being idle. The division is governed by different techniques.

TDMA (Time Division Multiple Access) will divide the spectrum across time. Like how OS schedulers will give processes exclusive access to the CPU for a certain amount of time, so too will users be granted exclusive access to the channel for a certain amount of time.

FDMA (Frequency Division Multiple Access) will divide the spectrum across frequencies. The most common example of this is for radio. Radio channels are exclusive to certain frequencies.

CDMA (Code Division Multiple Access) will send signals in a coded format. Imagine a large group of people that are speaking all at the same time. If everyone is speaking in English, the message will quickly get garbled. However, if every conversation has its own language, then the message will be clear. Even if I hear a conversation next to me in Chinese, I will tune it out because I don’t know Chinese and be able to speak and hear in English. There must be some amount of power control so that one person does not speak too loud, but otherwise the communication will work.

LAN

Computers can be connected through a sort of wireless LAN. If connecting a computer that does not have a screen or an otherwise easily accessible user interface, it is often convenient to connect with it using ad-hoc mode to a device with a screen so it can be configured, like Amazon Echo.

Each local set of computers that can connect in these ways is known as a BSS (Basic Service Set). This is known commonly as a LAN. The sender will wait for the sense channel to be idle than transmit the entire frame, else do a random timeout multiplying by 2 as with Ethernet, and the access point will return an ACK.

Access control in this sense is mediated by the time waited to check if the channel is busy or not. However, this can still cause collisions if a router does not hear from another, which is the hidden terminal problem. An exposed terminal problem may occur when a terminal is not properly signaled to not send because it is busy.

In order to mediate these issues, senders send a small packet called RTS (request to send) and a receiver will send a small packet called CTS (clear to send). If senders hear an RTS from somewhere else, they’ll wait until the receiver is clear to send to. If there isn’t a CTS, they can transmit because the receiver is open in that time. In this context, the HTP and ETP are caused when neither RTS nor CTS are heard, or when RTS is heard but not CTS.

To avoid these issues, we can reserve channels in the same way that channels are reserved in CSMA. The small reservation packets are unlikely to collide compared to the large frame packets.

Bluetooth

Bluetooth is a short-range kind of Wi-Fi technology that operates in the ISM band of 2.4GHz to 2.8 GHz. It was initially created for use by wireless mice and keyboards. The data rate goes up to 721 Kbps, which is fine because it’s not intended for data transfer, just communicating with nearby wireless devices.

In order to code this message, frequency hopping will be used. A frequency sequence will be sent via CDMA about which frequencies will be used to transmit data in sequence. Then, the sender will send the sequence in parts at each frequency in order. The receiver which knows the frequency sequence will be able to tune into the right frequencies at the right time and receive the correct bitstream. Anyone else listening to the bitstream that doesn’t know the frequency sequence won’t understand it.

Personal area networks under Bluetooth are either Piconet or Scatternet. Piconet has master/slave nodes, where the master is the one that allocates the Bluetooth channels among the slaves. Scatternet allows devices to be either master or slave in different networks so that a device can act differently in Bluetooth in different settings.

Connection is established by beginning with an inquiry broadcast by the Bluetooth device. A potential slave will respond and the master will send back a page for the hopping sequence. The other device will establish itself as a slave through a slave response and the master device will do likewise, at which time the connection is established through ACK-DAC.

## Prolog - URL: https://sharifhsn.dev/blog/prolog/ - Structured data: https://sharifhsn.dev/api/posts/prolog.json - Description: Prolog is a special programming language that is not like many others. It exemplifies the paradigm of logic programming . Functional programming is based on the idea that all… - Date: 2022-03-24 - Exact published timestamp: 2022-03-24 - Topics: Principles of Programming Languages - Categories: Principles of Programming Languages - Source: Archive - Source URL: None - Content source format: typst

Prolog is a special programming language that is not like many others. It exemplifies the paradigm of logic programming.

Terms

Functional programming is based on the idea that all operations exist using functions that give the same output for a given input. Imperative programming is based on the idea that we can manipulate variables and order them around like soldiers. Logic programming takes propositional logic from discrete structures and turns it into a language.

Every program is made up of terms, which are constants, variables, and compound terms. A constant is any word that is lowercase, whether it be a number, a string, or just a word. A variable is a word that begins with a capital letter.

Compound terms are relations. Relations in the logical sense are a subset of the Cartesian product of two sets, where the result set consists of elements that are contained in both. The top level name is the functor, and the number of arguments is the arity. For example, a compound term might be male(robb) where male is the functor with an arity of 1. The order of arguments within the compound term matters!

Facts are a knowledge base for a particular program. Each fact is a compound term showing a relation, followed by a period. Rules are a generality on facts where they are facts given assumptions. A fact is just a rule that is true without assumptions. The order of facts/rules is very important to the execution of code! Prolog reads from top to bottom, so the most important facts/rules should be placed near the beginning.

father(rickard, ned).
father(rickard, brandon).
father(rickard, lyanna).
father(ned, robb).
father(ned, sansa).
father(ned, arya).

Queries

Once we have our knowledge base, we can query for information.

?- father(ned, sansa).

This query will return true because it is in our list of facts. However, what happens if we query something it doesn’t know?

?- father(ned, bran).

Prolog operates on the closed world assumption, which means that it only knows what it’s been told. If it doesn’t know something, it will assume falsity. So Prolog will say the Ned is not Bran’s father, despite it having no idea whether that is the case.

There are also existential queries. Instead of asking a boolean question, it asks if a fact exists.

?- father(ned, X).

This query asks for all the relations father where ned is the first argument. It will return

X = robb ;
X = sansa ;
X = arya .

in the order that the facts are laid out. The semicolon is the logical OR in Prolog, so it’s saying that either one of the results is true. It would return false if there are no facts that satisfies the query.

Rules

Rules in general can be any logical statements from facts that are inferred inductively. The general form is

H :- B1, B2, B3, ..., BN

means that the head H is true if B1 ∧ B2 ∧ B3 … etc.

parent(X, Y) :- father(X, Y).

means that the relation parent exists between X and Y as long as the relation father exists between X and Y. We can construct rules recursively as well:

ancestor(X, Y) :- parent(X, Y)
ancestor(X, Y) :- parent(X, Z), ancestor(Z, Y)

This defines ancestor in two ways. The first rule says that X is an ancestor of Y as long as X is a parent of Y. The second rule introduces an additional definition where X is an ancestor of Y as long as X is a parent of some Z which is an ancestor of Y.

We can almost think of this like a list, where parent means adjacent and ancestor means inorder. Of course two elements are inorder if they are adjacent, that’s the first rule. However, if we can imagine a long list, two elements are inorder if the adjacent element to X is inorder previous to Y! The logic applies in any general situation; this is the advantage of the logical paradigm.

Unification

The core of how Prolog will compute answers is through unification, which is similar to pattern matching. It has three unification conditions:

identical constants will be unified
variables will always unify
compound terms will unify if functor and arity match as well as their arguments, recursively (using the top two rules)

Lists

Data structures seem pretty incongruent with our idea of logical programming so far. Lists are defined similarly to OCaml as recursive data structures.

[1, 2, 3, 4] % list
[] % nil
[H | T] % head cons tail

To find the last element of a list, we can use this rule

last([H], H).
last([_ | T], V) :- last(T, V).

The first rule is the base case, where if there is only one element in a list, that element is the “last element”. Then, the second rule is recursively defined. If the first argument

Arithmetic

The arithmetic operators are built into Prolog, and are built as relations between terms. Arithmetic equality is not the same thing as unification. If you want to check for arithmetic equality, you must use is.

The is operator evaluates the right hand expression and unifies the expression with the left. There can be variables present, but the variables need to be defined as some ground value. You can’t create an algebraic equation using is.

Basic arithmetic operators are also built in, but only operate on numerals. If you try to apply, for instance, +, to lowercase word constants, there will be an error.

Backtracking

Prolog computes the length of a list very similar to OCaml. An example OCaml length function might be:

let rec len l =
    match l with
    | [] -> 0
    | h :: t -> len t + 1

The Prolog rules are similar:

len([], 0).
len([H | T], N) :- len(T, M), N is M + 1.

The relation len with arity 2 defined on the empty list maps to 0, obviously. The relation len with arity 2 and the first argument being a list with head and tail is the actual recursive part we are interested in.

You might raise an eyebrow at the use of is in the second rule; after all, didn’t I just say that the right hand side must have ground terms? We don’t know what M is!

Actually, we do know. You can think of the order of clauses as the imperative definition of variables. The variable M is declared and initialized with the first len clause, and it can then be used in the second clause. If we put the second clause first, Prolog would raise an error.

You might notice a pattern here. The function is almost identical to the OCaml function, except for one difference. In the functional paradigm, every function evaluates to an expression which is returned within the function, and those are the values that are used for every operation. In the logical paradigm, there is no implicit return. len(T, M) is the same thing as M = len(T)!

The length is computed using backtracking. It will try every possible rule for len and then evaluate every single one. You can think of a depth first search of the tree of rules.

Every point at which there are multiple branches is called a choice point. In the len example, it can choose to either check the fact for [] or the recursive rule.

Backtracking can cause unexpected results. Say we try to query

?- len(A, 2)

We’re querying for a list of size 2 which doesn’t exist. What would happen?

Well, Prolog will try to find a matching branch to an existing rule. It will fail the first rule most of the time, so it will match the second rule. Because M = N - 1, when trying to find M it will subtract from the second argument and backtrack. It will eventually match to return the result A = [h1, h2]; when it matches 0 as the length and backtracks to adding arbitrary variables to the list of T.

But then something unexpected happens. After printing that last result, Prolog would spin on and on, never returning again. That’s because although we have matched the first rule and returned a list, Prolog will try every rule to see if it matches. And in fact, the second rule does match [T2, 0] because a variable will match any value. Prolog will endlessly backtrack because a choice will always be available that can match the new recursion. This is much different from OCaml which will only execute the first match arm that matches. You need to be careful in Prolog in order to prevent this kind of result.

Tail Recursion

Our len function is not tail recursive, since it adds on to the recursive call. In order to optimize our code, including an accumulator will aid because of tail recursion which we have discussed before.

len2([], Acc, Acc).
len2([H | T], Acc, N) :- M is Acc + 1, len2(T, M, N).

What we’ve done here is shift the is clause before the recursive call, which means we don’t need to hold on to that choice when recursively calling len2, since it’s guaranteed to already be checked.

You might ask, why do we even need to have that is clause? We might as well directly call len2(T, Acc + 1, N). However, there’s an issue here. The len2 relation only unifies terms, which is a powerful operation, but it does not evaluate them. For a list [1, 5, 7, 3] it would show N = 0 + 1 + 1 + 1 + 1, not N = 4 which would be correct. In order to make this correct, we need to evaluate the addition before making the recursive call into the second clause because that clause does not perform any evaluation.

This means that when we call len2 we need to do len2([], 0, X). This is a little annoying; we always know that our accumulator will begin at 0, why do we need to include it? There is a way to elide this definition:

len2([)

Append

Another instructive example is the simple list append.

let rec append p q:
    match p with
    | [] -> q
    | [h :: t] -> h :: (append t q)

append([], Q, Q).
append([H | P], Q, [H | R]) :- append(P, Q, R).

Prefix/Suffix

For a given list, we can place a separator in the list. Every element prior to the separator is the prefix and every element after is the suffix. The separator can be anywhere, including before or after every element.

We can get all possible prefixes/suffixes for a list for all separators:

?- prefix(X, [1, 2, 3]).

X = [];
X = [1];
X = [1, 2];
X = [1, 2, 3].

Let’s think about how to build this out logically. We need to find all lists which are possible prefixes of this list. That means we don’t want to capture, for example, [2, 3], because that is not a possible prefix. We have to start from the left and include from there.

prefix([], []).

The empty list should map onto the empty list, naturally

Generate and Test

We need a general approach to solve problems from a Prolog standpoint. The paradigm is very different from imperative or functional, so we can’t just look at it the way that that OCaml or Java does.

Let’s make take, which will remove exactly one element x from a list.

take([H | T], H, T).
take([H | T], R, [H | S]) :- take(T, R, S).

The first rule is the most obvious. If the element is at the head of the list, we can just return the tail of the list. This is our base case.

The second rule maps onto the case in the middle of the list. To imagine this, let’s use an example

?- take([1, 2, 3], 2, T).

We try to unify this to the first rule, but it fails because 2 is not at the head of the first list. The second rule will unify because the first element is a list with multiple values, the second is a variable that will unify anything, and the third element is a variable that will unify anything. Upon match, it will query take([2, 3], 2, S). This query will unify with the first rule, which will replace S with the value [ 3 ]. Once that S is unified, the rule will backtrack to the initial query which asked for [H | S]. Well, H in that initial case was 1 so our final return T will be [1, 3].

By extending take, we can implement other functions like permutations.

perm([], []).
perm(L, [H | T]) :- take(L, H, R), perm(R, T).

Let’s permute [1, 2, 3] once again.

?- perm([1, 2, 3], X).

It will not unify with the first rule, but it will with the second, obviously. The second clause will permute the list with the head taken away. Now, you might notice that H is not an explicit value here as when we used it earlier. Prolog seeing this will try to unify it with every value in the list, so take will be queried three times, with each value in the list being used in H. The resulting R from each of those queries is permuted again. The H used in take will be the head that is cons the backtracked form.

[1, 2, 3] will become take([1, 2, 3], H, R). Prolog will run through every choice, starting with take([ 1,[1, 2, 3] R) which will unify R with [2, 3]. Now, the perm clause will be perm([2, 3], T). Let’s assume that this gives all valid permutations of [2, 3] i.e. [2, 3] and [3, 2]. Well, the resulting match on the original X would be H | [2, 3] and H | [3, 2] or for this particular unification, [1, 2, 3] and [1, 3, 2]. This conclusively shows the inductive step, and the base case is trivially the empty list.

As you might be able to tell from these examples, there is a general strategy to solving problems in Prolog.

Generate a solution.
Test if it is valid.
If not valid, backtrack and try another solution.

Quicksort

One way to sort a list in Prolog is to use the generate and test method. You can generate all permutations of a list, and test whether each list is sorted. This is obviously incredibly inefficient with a Big O of $𝑂 (𝑛!)$ .

partition([], Y, [], []).
partition([X|Xs], Y, {X|Ls], Rs) :- X =< Y, partition(Xs, Y, Ls, Rs).
partition([X|Xs], Y, Ls, [X|Rs]) :- X > Y, partition(Xs, Y, Ls, Rs).

quicksort([H | T], SL) :-
    partition(T, H, Ls, Rs), quicksort(Ls, SLs), quicksort (Rs, SRs), append(SLs, [H|SRs], SL).
quicksort([], []).

This quicksort method is a very declarative manifestation of how to do quicksort in plain English. Partition the list along the pivot, quicksort the left and right, then combine them.

8-Queens

One classic problem in math is to find all of the arrangements of queens on a chessboard where none of the queens threaten each other. We will find it for 8 queens on an 8×8 chessboard.

Let’s cut down on our sample space, because there are a lot of permutations. If a queen is next to another, it obviously threatens it. In fact, because queens can never share the same row or column, that massively cuts our sample space. We can describe our solution as an array of eight values from one through eight, where the index of the value is the row of the queen and the value itself being its column. This way of describing the sample space not only cuts our sample space but makes our computation simpler. However, we have not checked the diagonals. That is what we need Prolog for.

checkBoard([H | T]) :- L is H - 1, R is H + 1, checkRow(T, L, R), checkBoard(T).
checkRow([H | T], L, R) :- H =\= L, H =\= R, LN is L - 1, RN is R + 1, checkRow(T, LN, RN)
checkBoard([]).
checkRow([], _, _).

## Threads - URL: https://sharifhsn.dev/blog/threads/ - Structured data: https://sharifhsn.dev/api/posts/threads.json - Description: Concurrency is a powerful tool to increase performance in applications. This is usually accomplished using threads. Clock frequency in CPUs has massively increased over the past… - Date: 2022-03-23 - Exact published timestamp: 2022-03-23 - Topics: Operating Systems Design - Categories: Operating Systems Design - Source: Archive - Source URL: None - Content source format: typst

Concurrency is a powerful tool to increase performance in applications. This is usually accomplished using threads.

Motivation

Clock frequency in CPUs has massively increased over the past few decades, growing at an exponential rate from the tens of MHz to GHz. However, circa 2005, clock frequency began to stagnate, and for the past decade the clock frequency of consumer CPUs have not increased past around 4 GHz. Why is that?

Clock frequency is typically increased by increasing the number of transistors in a single CPU. This can be done either by increasing the size of the CPU, or decreasing the size of the transistors. Because we don’t want huge CPUs in our computers emitting tons of heat, we have opted to decrease the size of transistors. However, there is a limit to how small they can go. When a transistor starts getting smaller than 30 nm, power leakage will begin to affect the CPU, causing the same heat problems that a big CPU does.

In order to keep up with computing trends, new CPUs have multiple CPU cores with the same transistor density. However, utilizing multiple CPU cores requires parallelism in the CPU. An ideal CPU might run one process on each core, with each process getting full control over the CPU.

You can visualize this idea like a highway. Having multiple lanes allows cars to travel much faster than if there is only one lane. However, there can be issues when the lanes must combine, like in merging. This causes traffic like if there was only one lane, and can actually increase overhead as cars attempt to merge. Decreasing the amount of merging (or synchronization) is key to fast concurrency.

Threads

We have discussed processes before; threads are not much different. They contain information about an execution such as the register values, open file streams, etc., but are more lightweight than processes.

There are multiple frameworks for threading that are useful for different workflows. The producer/consumer model is typically used for data analysis. Multiple threads create some data which is handled by a synchronizing consumer thread. The pipeline model works like a real pipeline, where multiple threads are consumed in steps and pipeline flows. This is common in GPU programming. The background model will work similarly to the interactive/batch model in MLFQ. Background threads do behind-the-scenes dirty work when the CPU is idle, and foreground threads will run the actual important parts of a process.

Since threads exist within a single process, they share the same address space, including the stack and the heap. This can cause many, many bugs when it comes to accessing and modifying memory. We will discuss methods to solve these bugs later.

Threads do not share the same instruction pointer, however. Each thread can execute from different parts of the program. All register values are unique to each thread, as they can execute on any CPU.

Threads can technically share the same stack on the address space. However, this is pretty obviously a bad idea. Typically, threads will create their own stack so that each thread does not interfere with each other’s local variables and execution, with their own stack pointer %esp to point to their stack.

OS Support

It’s all well and good to have threads, but we need a way for the OS to know how we’re switching between threads. There are multiple ways to do this.

Userspace threads are a way to have concurrency regardless of what the OS thinks. A run-time will execute along with your program and swap between threads itself without the OS knowing. This is how pthread works, and also how Project 2 does. However, the power that concurrency grants for multicore CPUs is still dependent on the number of kernel threads generated, because those are the only ones that run on each CPU.

Kernel threads can be associated one-to-one with user-level threads. This way, threads can take advantage of multiprocessing because the kernel will parallelize operations across CPUs. However, this incurs an overhead for the syscalls needed to parallelize threads, and there will never be enough kernel threads for the user threads.

Scheduling

Small operations like incrementing a variable can be difficult when we have threads accessing the same data. The simple instruction balance++ where balance is stored at memory location 0x9cd4 would have this assembly output:

0x195    mov 0x9cd4, %eax
0x19a    add $0x1, %eax
0x19d    mov %eax, 0x9cd4

If we have a context switch when %eip is at 0x19a, the first thread will add but not store the new value back in memory. That means that the variable balance would be less than expected because the add by the first thread would not be stored into memory.

This can lead to non-determinism, where the same input will cause different outputs, as well as race conditions, where the result of a program will depend on the CPU timing of different operations.

We want these three instructions to be executed uninterrupted. This is known as atomicity. In this critical section, no process or thread can interrupt the currently executing thread. However, if we allow programs to do this, then they can take advantage of this property and essentially disable the timer for context switches by declaring the entirety of execution atomic. We can’t have full blocking on interrupts for threads, so we only lock other threads from executing at the critical section. When the timer interrupt hits, the scheduler will still take control, but no other thread will be able to execute at the critical section.

Synchronization Primitives

There are many high-level synchronization primitives in the OS that exist to ensure the correct order of instructions. Each kind of lock is designed to solve a specific problem; no single lock can solve all of them.

We want to have correctness in our concurrency. This means that we must have mutual exclusion where only one thread can access the critical section at a time. If multiple threads are waiting for something, they cannot be stuck forever, they must make progress. Similarly, threads cannot be forced to wait an unreasonably large amount of time. Concurrency must also be fair and not favor any thread over any thread. And obviously, we do not want to overly tax the CPU with the overhead of synchronization.

We need underlying hardware atomic operations in order to implement synchronization. When these hardware instructions execute, no instruction or interrupt can execute. Example instructions are Test&Set and Compare&Swap.

The reason we need this is because if we don’t make these instructions atomic, we can encounter race conditions. Imagine an unset lock with a while loop and two threads. The first thread successfully enters the loop because the lock is unacquired. However, before it can acquire the lock, the thread switches and the second thread executes and acquires the lock. Now the first thread has not acquired the lock but it is still in the loop, so when it runs again it will acquire the lock even though it has already been acquired, allowing two threads in the same critical section! In order to combat this, testing and setting are the same atomic operation so there isn’t a gap that can allow a switch in between those two operations.

Spinlocks

Making our implementation fair is not as easy. If we imagine the most basic spinlock which switches back and forth, a program that knows how long a context switch will take will acquire the lock right before every context switch so no other program will be obtain it.

In order to introduce fairness, the ticket system for schedulers can be used here. Every thread is assigned a ticket for every turn it gets, with the first thread getting ticket 1, second ticket 2, etc. Now when a thread releases a lock, the ticket increments to the next thread. Even if it wants to maliciously acquire the lock before a context switch occurs, it can’t because it has lost its turn.

We need a special atomic function to make this work like Test&Set with basic mutex. We should be able to get a ticket number and increment it atomically. The function used for this is Fetch&Add.

int FetchAndAdd(int *ptr) {
    int old = *ptr; // these two lines are
    *ptr = old + 1; // executed atomically!
    return old;
}

Spinlocks are fast when we have a short critical section because we’re not context switching very often, so when we switch between locks constantly we want to avoid that overhead as much as possible as that dominates execution time. However, in a situation where there is only one CPU, letting the threads spin on a lock is extremely wasteful. The CPU scheduler has no idea that a thread is waiting for a lock so it will ignorantly run it even though all it does is spin.

One alternative is to yield the thread instead of letting it spin, telling the CPU scheduler that the thread doesn’t need to run right now. In general, a shorter critical section is okay to spin, but a longer critical section is not okay to spin.

Condition Variables

Mutex locks are just a way of restricting access to a critical section. It’s synchronous, but there’s no order to the way that the threads have to run. That is up to the caprices of the scheduler. However, let’s say that ordering is significant to the execution of our threads. How can we ensure that threads are executed in a certain order?

We need some functions and concepts to implement this. There are two system call functions wait(cond_t, mutex_t) and signal(cond_t) that are used with condition variables cond_t.

Let’s assume that a thread has acquired the lock, but it has not satisfied the condition to run yet because another thread must run first to preserve ordering. In that case, the condition variable will indicate that the thread cannot run and it will call wait to sleep the thread until the condition variable changes, releasing the lock so that the other thread can run. The caller of signal wants to wake up that sleeping thread, unless there are no sleeping threads, in which case it does nothing.

The best time to use condition variables is in a producer/consumer system, like in a pipe. Producers write to a pipe, and consumers read from the pipe. The pipe buffer has a limited size, so the producer can only add data to it if the buffer is empty, and the consumer can only read from it if the buffer has contents. For simplicity, assume a buffer of single unit size for now.

void *producer(void *arg) {
    for (Int i = 0; i < loops; i++) {
        Mutex_lock(&m); // lock the mutex
        while (numfull == max) { // while the buffer is full,
            Cond_wait(&cond, &m;) // wait
        }
        do_fill(i); // put data in buffer
        Cond_signal(&cond); // signal conditional variable
        Mutex_unlock(&m); // unlock the mutex
    }
}

void *consumer(void *arg) {
    while(1) { // consume on and on forever
        Mutex_lock(&m); // lock the mutex
        while (numfull == 0) { // while buffer is empty,
            Cond_wait(&cond, &m); // wait
        }
        int tmp = do_get(); // get the data
        Cond_signal(&cond); // signal conditional variable
        Mutex_unlock(&m); // unlock the mutex
        printf("%d\n", tmp); // do something with data
    }
}

Let’s consider an example with a producer and two consumers, as well as a FIFO scheduler. The first consumer thread runs and waits immediately because the buffer is empty. The second consumer thread does the same. The producer thread runs through its entire loop, filling the buffer, then when it loops again it will wait (since our buffer is size 1). The first consumer thread will return to the while loop, and now that the buffer is no longer empty it will fetch the data, signal the thread, and finish.

However, there is a problem here. The Cond_signal function signal is extremely generic and does not signal a specific thread. Instead, it will signal a random thread, which may be the producer or other consumer thread. But the whole point of using conditional variables is so that we have ordering!

We need to have multiple conditional variables:

void *producer(void *arg) {
    for (Int i = 0; i < loops; i++) {
        Mutex_lock(&m); // lock the mutex
        while (numfull == max) { // while the buffer is full,
            Cond_wait(&empty, &m;) // wait until buffer is empty
        }
        do_fill(i); // put data in buffer
        Cond_signal(&fill); // signal to show buffer is somewhat filled
        Mutex_unlock(&m); // unlock the mutex
    }
}

void *consumer(void *arg) {
    while(1) { // consume on and on forever
        Mutex_lock(&m); // lock the mutex
        while (numfull == 0) { // while buffer is empty,
            Cond_wait(&fill, &m); // wait until buffer is somewhat filled
        }
        int tmp = do_get(); // get the data
        Cond_signal(&empty); // signal to show that buffer is empty
        Mutex_unlock(&m); // unlock the mutex
        printf("%d\n", tmp); // do something with data
    }
}

In this example, the consumer thread will only run when the producer has signaled that there is content in the buffer.

Semaphore

Conditional variables do not have any state. They only signal the condition to wake or sleep a thread. This can be useful if you have lots of threads that do the exact same thing, like in a producer-consumer model. However, if you want a more complex way to represent state, you need a semaphore.

Semaphores are essentially conditional variables that contain an integer value which can be changed by threads. This allows threads to have multiple options based on a single variable shared across multiple threads just by changing the value in the semaphore. The semaphore is incremented every time a thread is woken, and decremented every time a thread sleeps.

The atomic operations for semaphores are Allocate&Initialize, Wait, and Post. We must first create the semaphore by allocating/initializing it. The Wait operation decrements the semaphore if it is greater than 0, and Post will increment the semaphore and wake a sleeping thread.

In this context, joining and exiting threads have no complication with mutexes; all you need to do is wait and post the semaphore, respectively.

We can actually construct locks using semaphores. Acquiring and releasing locks work the same way as waiting and posting, so you can just use those same atomic operations instead of Test&Set. In reverse, you can build a semaphore using locks and conditional variables, though in practice that’s a little more complicated.

## Routing Algorithms - URL: https://sharifhsn.dev/blog/routing-algorithms/ - Structured data: https://sharifhsn.dev/api/posts/routing-algorithms.json - Description: When a host needs to find the best destination/prefix to another host, what routing algorithm does it use? We can use either intra or inter domain. As you may recall, the routing… - Date: 2022-03-21 - Exact published timestamp: 2022-03-21 - Topics: Internet Technology - Categories: Internet Technology - Source: Archive - Source URL: None - Content source format: typst

When a host needs to find the best destination/prefix to another host, what routing algorithm does it use? We can use either intra or inter domain.

Routing and Forwarding

As you may recall, the routing table is a large DRAM table containing all possible paths and the forwarding table is a small SRAM table containing only the best possible path. The router also caches previous best paths so that they don’t need to be retrieved for the same host over and over from DRAM.

You can visualize multiple routers in an undirected weighted graph. The weights matter so that a slow link will not be chosen even if the path is technically slower. By maintaining all paths, the router can pick a new best path if a link in the route crashes.

But how does the router actually update the forwarding table? There are several different algorithms we can use.

Link State

Link state is an intra-domain method that works fine for 50-100 routers, say on a university campus.

Let’s use the graph analogy again. It is often useful to abstract concepts in networking to graphs when thinking about algorithms, since we can apply well-known concepts from mathematics and algorithm design to the problem. There are multiple factors that play into the weight of each link, not just link speed, but routers assign their own holistic score to links for simplicity.

The obvious choice here is to use Dijkstra’s algorithm for shortest path, as it applies to undirected weighted graphs like the on we have just constructed. This algorithm generates the shortest path to every path up to and including the destination. It requires knowing all of the link speeds beforehand. This is accomplished through a link state broadcast i.e. pinging every router and checking the link speed of the ping. The algorithm is also iterative, which means the number of shortest destinations scales with iterations linearly. The time complexity of the algorithm is $Θ (| 𝐸 | + | 𝑉 | log | 𝑉 |)$ .

This algorithm works best in a centralized graph like at a university. Instead of sending a link state broadcast every time a connection is made, each node in a graph will send its own link state to a centralized location, and that location will be checked for recomputation of shortest path.

The centralized node is the one that does all the calculations for shortest path. This is known as a distance vector. Each router must calculate its own distance on top of the other links to the node that it must pass by. Think of a table that is passed from link to link which is updated according to Dijkstra’s algorithm. The initial distance vector table for a router only contains the link speeds for its adjacent routers, and all others are marked as infinity. Degenerate longer paths are thrown out when tables “merge” at a router.

If a link breaks, then the routers that have broken links get reset in the final table. When the broken router sends a new distance vector table, it updates the central table in the same process as the table generation.

However, this doesn’t work well when we have different domains in different autonomous systems controlled by different ISPs. We need some kind of inter-AS algorithm to manage this. However, as the intra-AS algorithms works well, we use federation in our algorithms. Inter-AS algorithms are used to route between ASes and intra-AS algorithms are then used within the AS.

BGP

There are some barriers to inter-AS routing. We might not know whether the host is reachable, so we need to ask nearby ASes if they know. The protocol that these ASes use is BGP, the Border Gateway Protocol. With BGP, ASes can

obtain subnet reachability information from neighboring ASes
propagate reachability information to all the routers within the AS
use this information to calculate the best route to another subnet

BGP works through persistent sessions between BGP routers over semi-permanent TCP connections. The sessions can be through any router, there doesn’t need to be a physical link. The two peers can communicate regardless of the physical router. This is the sequence of steps:

Establish TCP over port 179.
Exchange routing tables. At first, this is a lot of data, but after a connection is established only deltas (like Git) need to be exchanged.
Send four kinds of messages.

Open - establish session by exchanging AS numbers and the BGP identifier (arbitrary router IP address). There is a timer for how long to wait before just assuming the session is down i.e. in an outage.
Notification - report unusual conditions, usually an error. If there are header errors, or a timer has expired, etc. then the TCP session must be terminated.
Update - inform if there are new active or old inactive routers. The message includes the withdrawn routes, which are inactive routers, the length of that field as it can be variable, etc.
Keepalive - regular message that the session is alive. BGP needs this regular message otherwise the neighbor AS won’t know whether you’re alive or not.

When we say that a router is advertising a prefix, there are some necessary assumptions. The routing information must be valid, and does not need to be refreshed until necessary. The path that an AS advertises to a node is the same path that AS uses to communicate with that node.

The BGP protocol additionally has many attributes that indicate the characteristics of a prefix:

ORIGIN: advertises the origin of an announcement and prefix injection. This is manually configured by intra-routing protocols.
AS_PATH: a link of ASes which is the path of ASes that the prefix has traveled to. This is useful for detecting loops and selecting shorter routes of ASes.
NEXT_HOP: the hop field is when you cross the AS boundary. After you cross, the next hop value is the next AS you need to get to in order to get to the destination router. The reason this is important is because the IP packets within an AS can travel in any order irrespective of BGP, so the intra-routing algorithm can decide where to go based on this value. This becomes significant when a network experiences transit traffic and different networks are using that network as an intermediary. In order to get the packet to exit the network as quickly as possible, BGP can make the decision to send the packet through non-BGP routers if it’s a faster way to make the packet leave the network.
MED: stands for Multi-Exit Discriminator. If ASes are connected via multiple links, the AS that receives a prefix uses this value to discriminate between exits, with a lower value being better.
LOCAL_PREF: indicates a local preference for a specific prefix, is a number that is more for more preference, by default 100. For example, if outbound traffic is preferred to be a specific exit point, then that point will have high local preference. It can be used to break a tie between routers.

## Swapping - URL: https://sharifhsn.dev/blog/swapping/ - Structured data: https://sharifhsn.dev/api/posts/swapping.json - Description: Sometimes we don’t have a lot of memory to access. In those cases, we need a place to put our data. The natural solution is to put data in the disk, which has very high storage.… - Date: 2022-03-21 - Exact published timestamp: 2022-03-21 - Topics: Operating Systems Design - Categories: Operating Systems Design - Source: Archive - Source URL: None - Content source format: typst

Sometimes we don’t have a lot of memory to access. In those cases, we need a place to put our data. The natural solution is to put data in the disk, which has very high storage. This is known as swapping.

Mechanism

The actual process of copying memory to disks is not complicated, it is simply file I/O and is supported by hardware. We swap memory out to disk, and swap in data from disk to memory. The complicated part comes when we introduce paging to this idea.

In order to accommodate this, we add a present bit to our page table entry. If the bit is 0, then the page has been swapped out to disk and the physical frame is no longer valid. When there is an attempted translation of a page table entry with an unset present bit, a trap will occur and the hardware will swap in memory from the disk, setting the present bit with a new translation for the new physical frame number.

An important note is that a process will never directly access the disk! Instead, a trap will swap disk and memory and the process will access the memory as dictated by hardware and the OS.

AMAT

Swapping can be slow. We want to improve our Average Memory Access Time (AMAT). A hit in this context is an access that goes directly to RAM, and a miss is a trap that will get the memory from disk into RAM.

𝐴 𝑀 𝐴 𝑇 = (𝐻 𝑖 𝑡 % \cdot 𝑇_{𝑚}) + (𝑀 𝑖 𝑠 𝑠 % \cdot 𝑇_{𝑑})

Policies

When we swap in and out of memory, we need a way to find the most frequently accessed pages to reduce AMAT. There are several policies to accomplish this.

The optimal policy which has oracular knowledge of future page accesses can keep pages that will be accessed sooner and replace pages that will be accessed later. If we know that a page will be required soon, we’re definitely not going to swap it out. Alternatively, if we know that it’s going to be a while before the page is going to be accessed, we can swap it out, improving our AMAT. This is obviously infeasible when page access is random, which is almost all the time.

The simplest possible policy is, as with schedulers, FIFO. However, this does not work well. A better, also simple policy is LRU, or Least Recently Used. Most caches of this kind use LRU because of how effective it is despite its simplicity. The idea can be expressed in one line: swap out the page which has been least recently accessed. By keeping the pages of physical memory in a queue when they are created, it’s simple to dequeue and swap out the page for a new, freshly enqueued page. The counter is reset every time that a page is swapped out.

LRU adds more memory usage in order to maintain the queue

You would think that having more pages reduces the number of page misses. However, due to Belady’s Anomaly, this is not always the case for FIFO. LRU does not suffer from this phenomenon

However, LRU does not take into consideration how many times a page has been accessed. A large I/O scan that is only used once might flush memory because it is recent without regard to how often it is used. When you stream a movie, you are sequentially accessing that data. This is a common access pattern in computer science.

We can use pure LFU, or Least Frequently Used to combat this. However, this runs into the opposite problem: what if a page was very frequently accessed in the past? The policy will not forget that so it won’t be evicted despite it being very old.

A better approach is to combine both of the ideas of recency and frequency, through algorithms such as LRU-K or 2Q. However, these algorithms can be expensive due to their increased complexity compared to these other simpler algorithms.

Implementing LRU

There are two approaches: software and hardware, as always.

A software based LRU will be a sorted linked list. Accessing a linked list is slow, which means $𝑂 (𝑛)$ complexity for a memory reference, although replacing pages is very fast at $𝑂 (1)$ .

## Routers - URL: https://sharifhsn.dev/blog/routers/ - Structured data: https://sharifhsn.dev/api/posts/routers.json - Description: Everyone nowadays gets the Internet through the router in their home. But what is a router? It’s likely that you have an internet speed in the tens of Mbps. This unit is megabits… - Date: 2022-03-10 - Exact published timestamp: 2022-03-10 - Topics: Internet Technology - Categories: Internet Technology - Source: Archive - Source URL: None - Content source format: typst

Everyone nowadays gets the Internet through the router in their home. But what is a router?

Speed

It’s likely that you have an internet speed in the tens of Mbps. This unit is megabits per second, or one million bits per second. Some routers in important locations have speeds in the Gbps, which is 1 billion bits per second. This is incredibly quick. How do routers accomplish this?

Structure

The router is composed of a control plane and a data path. The control plane contains all the necessary external tooling to determine the paths. It contains routing protocols that need to be implemented, as well as a routing table. This table is used to look up all the possible paths to the destination and pick the shortest one.

This information is all stored on the data path, where each packet is actually processed. The routing table shortest path is put into the forwarding table which typically has a very small amount of SRAM in order to quickly mediate routing. You can think of it almost like a register that is close the process as opposed to the main memory where the calculations are made.

Remember this distinction! The routing table contains all possible routes from this router, kind of like Google Maps. It needs to be a big table, so it is usually stored in slow, inexpensive DRAM. The forwarding table only contains the best possible route, so it is a small table stored in fast, expensive SRAM.

Routers must also have input and output buffers. The input buffer is needed because the router may take in packets faster than it can process them, so it needs a place to put them before it can figure them out. The output buffer is needed because of the opposite problem; if the router processes the packets faster than the output link can output them, the packets need a place to be.

Forwarding Engine

Let’s examine the forwarding table in a little more detail. The forwarding table is a simple table that looks like this:

Dest-network	Port
65.0.0.0/8	3
128.9.0.0/16	1
149.12.0.0/19	7

It is a simple path from networks, remembering which port to send packets.

You might have noticed that our destination networks are in prefix form. The one on port 7 has the longest prefix because it is matching the first 24 bits. We must match the longest prefix form first when looking for the network. These prefixes are the most specific and therefore must be matched before the smaller, more general prefixes. Prefixes can overlap, so we must pick the more specific one. It is typically not explicitly linear, a hash table-like structure is used to make lookups faster.

In fact, hierarchical addressing is important for allowing us to aggregate routes. Different organizations within the same ISP network will have some bits after the network bits devoted to differentiating between organizations. Let’s imagine that Rutgers Newark and NJIT are on the same ISP network, but are different organizations, obviously. The first 19 bits of that prefix might be devoted to the ISP network, then the next 4 bits will be used for organizations within that network.

Switching Fabrics

Let’s consider each port as form of input and output. We will need a way to switch between devices. There are multiple ways to implement switching fabrics.

One way is through memory. This is the cheapest way of going about it. The I/O device will take the network and write it into memory. We can use DMA (Direct Memory Access) to do this more quickly. The header is processed, and the router checks the routing table for the best route. The packet must be copied from the port into memory, then processed, then loaded back into the output port. This is very slow. Most of the bottleneck is in the memory bus when copying memory. These routers are typically less than 10 Mbps.

Another way is through bus. The packet is transferred between input and output through a I/O bus, with a memory buffer to hold the packet on each end. There’s no CPU movement here, it’s all I/O instructions. But we still need some kind of command for the destination port number. The CPU will periodically calculate and cache the forwarding table on the input device (also known as line card). If there is a cache miss, then I/O will ask the CPU to update the cache from main memory, which is slow but doesn’t happen often. However, since there’s only one bus, there is only one communication possible between one port and another at a time. This speed is in the high Mbps.

The third way, which is the most high-end way, is the crossbar. It works like a bus, except that there exists a bus between every single port, which means that for $𝑛$ ports, we can have $\frac{𝑛}{2}$ simultanenous transfers. This is in the best-case scenario where each input goes to a different output. However, if two ports are trying to connect to the same port, then one of them will be bottlenecked because it’s still one bus. However, this complicated switching fabric is extremely expensive, which scales with the number of ports. The price can be worth it for a company like Google; their speeds can be in the Tbps!

## Mirrors - URL: https://sharifhsn.dev/blog/mirrors/ - Structured data: https://sharifhsn.dev/api/posts/mirrors.json - Description: Mirrors can interact with light in interesting ways. In fact, reflection and refraction are probably most people’s engagement with laws of physics surrounding light in their… - Date: 2022-03-08 - Exact published timestamp: 2022-03-08 - Topics: Physics - Categories: Physics - Source: Archive - Source URL: None - Content source format: typst

Mirrors can interact with light in interesting ways. In fact, reflection and refraction are probably most people’s engagement with laws of physics surrounding light in their daily life.

Reflection

On a plane mirror, like the kind that you’re used to in daily life, the angle of incidence equals the angle of reflection. To clarify some terms: incidence refers to the ray that is going towards the mirror, and reflection refers to the ray that is coming out of the mirror. The normal is a perpendicular line, in this case to the plane of the mirror.

Because of this characteristic of plane mirrors, the reflection of an object (or image) will be upright and the same size as the object. The image will be “inside” the mirror, which is called virtual. The distance “within” the mirror to the image is the same distance from the mirror to the object. This is useful for real life because we want to see most objects the same as they actually exist. However, in physics, sometimes we want to manipulate these variables, and for that we use spherical mirrors.

A spherical mirror is a mirror that is curved such that it could be cut out of a sphere. It can be either concave or convex (front and back of a spoon).

There are two points that are significant to a mirror, the focal point $𝐹$ and the center of curvature $𝐶$ . If the mirror was part of a sphere, then the center of curvature would be the center of that sphere. The focal point is where parallel rays intersect after reflection. Both points are on the principal axis of the mirror. The focal point is halfway between $𝐶$ and the center of the mirror.

In order to determine how the image of an object looks when reflected off of a mirror, it is useful to construct a ray diagram. The ray diagram is based on three rules (you only need two).

an incident ray parallel to the principal axis reflects through $𝐹$
an incident ray that passes through $𝐹$ reflects parallel to the principal axis
an incident ray which passes through $𝐶$ reflects on itself

Magnification

The calculation of the distance and size of an image are governed by the following equations:

\frac{1}{𝑓} = \frac{1}{𝑑_{𝑜}} + \frac{1}{𝑑_{𝑖}}

where $𝑑_{𝑜}$ is the distance from the mirror to the object and $𝑑_{𝑖}$ is the same for the image.

The magnification $𝑚$ of an image is the ratio of the image size to the object size:

𝑚 = \frac{ℎ_{𝑖}}{ℎ_{𝑜}} = - \frac{𝑑_{𝑖}}{𝑑_{𝑜}}

Now, you might be wondering about how the sign of each value affects these equations. The answer is, a LOT. To think about mirrors intuitively, think about what each sign must mean.

$𝑓$ is positive for concave mirrors because the focal point is front, but negative for convex mirrors because it’s behind the mirror.
$𝑑_{𝑜}$ is positive for a real object (in front of the mirror), but negative for a virtual object. A virtual object can’t generally exist, but you can simulate it using reflected light.
$𝑑_{𝑖}$ is the same, but a virtual image makes more sense.
$𝑚$ is obvious, positive for upright and negative for inversion

Refraction

If you put a straw in your water, the straw will strangely seem to bend at the water’s edge and go at an angle. The reason for this is refraction: light will change direction when it passes through a medium. The change in angle is described by Snell’s Law:

𝑛_{1} sin 𝜃_{1} = 𝑛_{2} sin 𝜃_{2}

where $𝜃_{1}$ is the angle of incidence, $𝜃_{2}$ is the angle of refraction, and $𝑛_{1}$ and $𝑛_{2}$ are the indices of refraction for the corresponding media. The index of refraction is expressed as the ratio between $𝑐$ and the speed of light in the material. If light is very fast in the material, $𝑛$ will be close to 1.

When light passes from a medium with a higher index of refraction to a lower one, there’s an angle at which the refracted light will be parallel to the media edge, known as the critical angle $𝜃_{𝑐}$ :

sin 𝜃_{𝑐} = \frac{𝑛_{2}}{𝑛_{1}}

At any angle greater than the critical angle, all the light will be reflected and none will be refracted.

The reason that this is the case can be derived from math. Let’s say that light passes from water (1.33) into air (1.00). The critical angle would be equivalent to ${sin}^{- 1} \frac{1.00}{1.33}$ or 48.75°.
What would happen if we tried to find the angle of refraction based on Snell’s Law if the incident angle is 50°, so greater than the critical angle?
$1.33 \cdot sin 50^{\circ} = 1.00 \cdot sin 𝜃_{2}$ $sin 𝜃_{2} = 1.33 \cdot sin 50^{\circ} = 1.019$
But there’s a problem here! The function ${sin}^{- 1}$ is not defined for $𝜃$ greater than 1, so the angle does not exist. This means that refraction does not occur, the light is only reflected.

Lens

The properties of magnification and refraction are used in lens that are located in glasses, telescopes, microscopes, and even your own eyeball! A lens has two mirrors on either side. If the mirrors are convex, the lens is converging, and if they are concave, the lens is diverging. Every lens has two focal points corresponding to each mirror.

The ray diagram for a converging lens follows similar rules to a concave mirror, but instead of thinking of reflected rays, the rays are refracted through the lens.

When a ray passes through the axis of a lens parallel, it will refract towards the other focal point.
When a ray passes through a focal point, it will refract parallel through the other side.
When a ray passes through the center of the lens, it will not bend.

The rules are the same for a diverging lens, though it may not appear that way. All of the rules for a diverging lens are based on a focal point on the other side of the mirror, but fundamentally have the same property.

All of the formulas are the same for lens like mirrors, however, the meaning of the signs changes in this context.

$𝑓$ is positive for converging lens because the focal point is front, but negative for diverging lens because it’s behind the lens
$𝑑_{𝑜}$ is positive for a real object (same side as light), but negative for a virtual object. A virtual object can’t generally exist, but you can simulate it using reflected light.
$𝑑_{𝑖}$ is positive for a real image on the opposite side of the light, and negative for a virtual image on the same side of the light.
$𝑚$ is obvious, positive for upright and negative for inversion

## Lambda Calculus - URL: https://sharifhsn.dev/blog/lambda-calculus/ - Structured data: https://sharifhsn.dev/api/posts/lambda-calculus.json - Description: When we discuss the principles of programming languages, lambda calculus is the bedrock of our discussion. It is the most abstract, formal way to describe functions and… - Date: 2022-03-03 - Exact published timestamp: 2022-03-03 - Topics: Principles of Programming Languages - Categories: Principles of Programming Languages - Source: Archive - Source URL: None - Content source format: typst

When we discuss the principles of programming languages, lambda calculus is the bedrock of our discussion. It is the most abstract, formal way to describe functions and application.

Turing Completeness

A programming language is said to be Turing complete if it can compute any function also computable on a Turing machine. It must either be able to emulate a Turing machine, or it must be able to emulate a Turing complete language.

Although this is a powerful definition, in truth it is a simple one to satisfy. Features like loops and currying are not related to Turing completeness.

Syntax

Lambda calculus is a very simple language which only uses functions and applications, but is still Turing complete.

An expression in lambda calculus is defined as

e ::= x    # variable
    | λx.e # abstraction (function definition)
    | e e  # application (function call)

Everything is a function in lambda calculus. We can make a simple interpreter in OCaml as with Imp:

type exp =
    | Var of string
    | Lam of string * exp
    | App of exp * exp

in the same order that was defined earlier. Lam here is similar to Fun for Imp. The lambda calculus expression

(λx.λy.x y) λx.x x

can be deconstructed to its AST as

App(
    Lam("x",
        Lam("y",
            App(Var "x", Var "y")
        )
    ),
    Lam("x",
        App(Var "x", Var "y")
    )
)

Let’s parse this out forwards. First we read that there is a parenthesis, which means that there is an expression enclosed within. There is a $𝜆$ , which means that the expression is a lambda. The next letter is x, which is the name of the Lam. The period following separates the function definition to the containing expression, which continues to another $𝜆$ . That means that the containing expression is also a Lam, which is called “y”. The containing expression is now an App which is composed of two Vars. This is the end of the nesting because we hit the end parenthesis. The same logic applies for the second expression but with less nesting. Since we have two expressions at the top-level, the full expression is an App.

The scope of $𝜆$ extends as far to the right as possible, excepting parentheses. This is why we needed the parentheses for the first term in the above statement, otherwise the first $𝜆$ would extend throughout the entire statement. The application, however, is left-associative, like OCaml.

Beta Reduction

A function call of type (λx.e1) e2 replaces all instances of x in e1 with e2. That means that we can substitute this statement with e1{e2/x}. This is called beta reduction. All we have done is apply the function and replace the formal parameters through substitutions. Beta reductions should always be idempotent for the statement. When no more beta reductions can be performed on a term, then it is said to be in beta normal form, for example λx.e.

Another example will be instructive here. Take the term (λx.λz.x z) y. This is a function application, since it has two terms. It follows the form that we stated earlier, so we can substitute all instances of y on the outside $𝜆$ . This would give us the final term λz.(y z) eliminating the outside λx and replacing the x in the inner term with y.

Alpha Conversion

Lambda calculus is statically scoped which means that variable definitions are only scoped locally. That means that within a function, you can rename bound variables with the same meaning. This is called alpha conversion.

An important distinction here is between a free and bound variable. Free variables are not contained within a lambda, while bound variables are. For example, in (λy.λz.y z x), y and z are bound to λz and λy, respectively, while x is free. λy contains the App of λz and z.

## More Internet Protocols - URL: https://sharifhsn.dev/blog/other-protocols/ - Structured data: https://sharifhsn.dev/api/posts/other-protocols.json - Description: Besides the primary Internet protocols, there are some others to learn about, like DHCP , NAT , IPv6 , etc. What happens if a device doesn’t have a permanent IP address? You take… - Date: 2022-03-03 - Exact published timestamp: 2022-03-03 - Topics: Internet Technology - Categories: Internet Technology - Source: Archive - Source URL: None - Content source format: typst

Besides the primary Internet protocols, there are some others to learn about, like DHCP, NAT, IPv6, etc.

DHCP

What happens if a device doesn’t have a permanent IP address? You take your phone around multiple mobile networks, there isn’t a consistent IP address between them. How do you access the internet without an IP address?

Dynamic Host Configuration Protocol is a client-server protocol that is used in these scenarios. As the name implies, it allows for dynamic IP address allocation that are leased for a certain amount of time. Configuring things like the subnet mask, gateway configuration, etc. to set up an IP is complicated and not feasible for the average user. Imagine if you had to do all that setup every time your phone moved to a new location!

DHCP has two main components: the protocol for delivering the bootstrapping information from the server to clients, and the algorithm for dynamically assigning addresses to new clients. How do you start a connection from nothing?

In order to allocate a new address, there are three modes. Automatic allocation gives permanent address, dynamic allocation leases addresses, and manual allocation is managed by a system administrator.

DHCP sends packets over a socket at port 67. Its IPv4 header indicates that it is protocol 17. The UDP packet is sent with no initial IP address because there is none. It is sent with the broadcast server IP address 255.255.255.255. The protocol must also interface with the link layer at the client’s MAC address; remember, we’re starting at zero. There is also space for the server hostname, boot filename, and several kinds of options.

The options indicate what the purpose of the UDP packet is. For example, option 1 is DHCPDISCOVER, which is the message that lets the server know that a client is looking for an IP address. The communication sequence is directly encoded into the options in the packet. Some others:

DHCP Offer: server response with parameter proposal
DHCP Request: like discover, but focused to a specific server
DHCP ACK: server gives IP address to client
DHCP NAK: server declines to give IP address to client
DHCP Decline: client declines the given IP address
DHCP Release: client gives up its IP address

DHCP typically applies within a subnet. Relay agents on routers, like with BOOTP, allow servers to handle requests from other subnets.

NAT

There are two kinds of IP addresses: public and private. Private addresses are reserved for 10.0.0.0 to 10.255.255.255. If you send a request to private address, your router will not send it out to the Internet. Private addresses are also free; you can hand out as many as you want without it costing anything.

However, requests sent from a private IP address cannot access the Internet. In order to access the Internet, we use a Network Address Translation box. This NAT box is assigned a single public IP address and it is the public-facing IP for all of the machines with private addresses on it.

This is likely how your home router works. Each device in your home only has a private address and every time it sends a request to the Internet, that request is sent to the NAT box in your router and translated to one single public IP for all of the machines in your house.

The NAT box will take in packets that are sent from private IP addresses and record the packet address in a table, then replace the IP address with its own and its port with some random port. When packets are sent back to that address, the NAT box consults its table and sends the packets back to the correct private IP address based on the port it was sent on.

Although there are technically $2^{24}$ addresses, in practice you can only have less than $2^{16}$ addresses because each of those needs to have an associated port, and that’s a 16-bit number (minus some reserved ports).

But why would you do this? Remember that IPv4 is a 32-bit number for addresses, which only allow for ~ 4 billion unique devices. Clearly, there will be, and perhaps already are, more devices that use the Internet than that. IPv6 was created in part to solve that problem, but as many devices only accept IPv4, this works in the meantime. It also provides security to users since ports aren’t accessible from the public, they are instead randomly assigned. Even if someone’s IP address is tracked, the malicious agent still doesn’t necessarily know what device the request is from.

IPv6

IPv6 is an updated protocol from IPv4 that is optimized for the needs of the modern day. It is significantly simplified compared to IPv4 as much of the machinery in an IPv4 header relates to problems that generally no longer exist. However, the header size is also 40 bytes, twice as large as the IPv4 header. This is because the new source and destination IP address that must be stored in the header are now 16 bytes, not 4, the vast majority of the header size is dominated by address size.

IPv6 does not protect against some problems that still exists, such as bit flips. If that happens, the packet is simply dropped because a different layer will do a checksum.

Version	Traffic Class	Traffic Class	Flow Label	Flow Label	Flow Label	Flow Label	Flow Label
Payload Length	Payload Length	Payload Length	Payload Length	Next Header	Next Header	Hop Limit	Hop Limit

followed by the source address and destination address, along 4-bit boundaries

## Electromagnetic Waves - URL: https://sharifhsn.dev/blog/electromagnetic-waves/ - Structured data: https://sharifhsn.dev/api/posts/electromagnetic-waves.json - Description: We have discussed electric fields and magnetic fields, and the way that they change and interact. The full interaction of these fields will create electromagnetic waves , the… - Date: 2022-03-01 - Exact published timestamp: 2022-03-01 - Topics: Physics - Categories: Physics - Source: Archive - Source URL: None - Content source format: typst

We have discussed electric fields and magnetic fields, and the way that they change and interact. The full interaction of these fields will create electromagnetic waves, the study of which will consume the rest of these notes.

The Electromagnetic Spectrum

It’s easiest to understand electromagnetic waves (EM waves) in the same way that we understand any other waves, through frequency $𝑓$ or wavelength $𝜆$ .

Throughout these notes, we will refer almost exclusively to the wavelength of an EM wave as identification for consistency’s sake. Note however that this is always mapped to a corresponding frequency, though inversely related.

Visible light is an EM wave, with wavelengths in the hundreds of nanometers. Stronger EM waves like X-rays or gamma rays go from $10^{- 8}$ to $10^{- 16}$ meters, which is extremely small! In contrast, weaker EM waves like radio waves can have wavelengths in the thousands of meters. As you can see, EM waves have a broad spectrum.

The wavelength of an EM wave in a vacuum is given by

𝑓 𝜆 = 𝑐

where $𝑐$ is the well-known speed of light at $3 \times 10^{8}$ meters per second. $𝑐$ can also be more precisely defined as

𝑐 = \frac{1}{\sqrt{𝜖_{0} 𝜇_{0}}}

where $𝜖_{0}$ is the permittivity of free space $8.85 \times 10^{- 12}$ and $𝜇_{0}$ is the permeability of free space $4 𝜋 \times 10^{- 7}$ .

Energy

EM waves transport energy, and that energy has a certain value depending on the electric and magnetic fields that created the EM wave. The electric energy density is:

𝑢_{𝐸} = \frac{1}{2} 𝜖_{0} 𝐸^{2}

In contrast, the magnetic energy density is given by:

𝜇_{𝐵} = \frac{1}{2 𝜇_{0}} 𝐵^{2}

Notice the different units here. The electric energy density is dependent on permittivity, while the magnetic energy density is dependent on the inverse permeability.

We can combine these two to get the total energy density:

𝑢 = \frac{1}{2} 𝜖_{0} 𝐸^{2} + \frac{1}{2 𝜇_{0}} 𝐵^{2} = 𝜖_{0} 𝐸^{2} = \frac{𝐵^{2}}{𝜇_{0}}

The direction of the wave must always be mutually perpendicular to the direction to the electric field and the magnetic field. This follows the right hand rule where the thumb is the magnetic field $𝐵$ , the pointer finger is the propagation of the wave, and the middle finger is the electric field $𝐸$ . You can remember this because the pointer finger points to where the wave is going, and $𝐵$ comes before $𝐸$ in the alphabet, so the first one is the thumb and the second is the middle finger.

An EM wave has a magnetic field with an rms value of $3.40 \times 10^{- 6} 𝑇$ . The wave passes perpendicularly through an opening that has an area of $0.35 𝑚^{2} .$
To get the electric field value $𝐸$ , we multiply the magnetic field $𝐵$ by $𝑐$ to get $3 \times 10^{8} 𝑚 / 𝑠 \cdot 3.4 \times 10^{- 6} 𝑇 = 1020 𝑁 / 𝐶$ .
To get the energy densities $𝜇_{𝐸}$ and $𝜇_{𝐵}$ , we simply apply the earlier formulas here.
$𝜇_{𝐸} = \frac{1}{2} 𝜖_{0} 𝐸^{2} = \frac{1}{2} (8.85 \times 10^{- 12}) (1020)^{2} = 4.604 \times 10^{- 6} 𝐽 / 𝑚^{3}$ $𝜇_{𝐵} = \frac{𝐵^{2}}{2 𝜇_{0}} = \frac{(3.4 \times 10^{- 6})^{2}}{2 \cdot 4 𝜋 \times 10^{- 7}} = 4.6 \times 10^{- 6} 𝐽 / 𝑚^{3}$
After getting the electric energy density and the magnetic energy density, getting the total energy density is as simple as adding them together to get $9.204 \times 10^{- 6} 𝐽 / 𝑚^{3}$ .
The intensity of a wave is a simple equation; you can think of it as the density multiplied by speed being how much is being transferred per unit time: $𝑆 = 𝑐 𝑢$ .
$𝑆 = 𝑐 𝑢 = (3 \times 10^{8} 𝑚 / 𝑠) (9.204 \times 10^{- 7} 𝐽 / 𝑚^{3}) = 2761.2 𝑊 / 𝑚^{2}$
Let’s say we want to find the energy that is carried through this opening over twenty seconds. Energy is in the unit joules $𝐽$ and $𝑊$ is $𝐽 / 𝑠$ . In order to convert a dimensions properly, we’re going to need multiply the intensity by $𝑠 \cdot 𝑚^{2}$ . With this in mind, it’s clear how we should form our equation. We’re multiplying the density by the time passed (20 seconds) and the area $0.35 𝑚^{2}$ .
$𝐸 = 𝑆 𝑡 𝐴 = (2761.2 𝑊 / 𝑚^{2}) (20 𝑠) (0.35 𝑚^{2}) = 19328.4 𝐽$

Polarization

Electromagnetic waves can be polarized in a a particular direction by passing throw a material which only allows for one vector to pass through. Unpolarized light passing through a polarizing material will have half the intensity that it originally had.

Malus’ Law states that an analyzer which alters the intensity and polarization direction of an EM wave will decrease the intensity as so:

\bar{𝑆} = \bar{𝑆_{0}} {cos}^{2} 𝜃

and changing the polarization direction to match the analyzer. $𝜃$ in this case is the difference between the polarized light and the analyzer. Say the light is polarized at angle of $30^{\circ}$ clockwise to the vertical and it passes through a filter that is at an angle of $15^{\circ}$ counterclockwise to the vertical. The $𝜃$ in this case will be $45^{\circ}$ .

A vertically polarized beam of intensity $𝑆_{0} = 60.0$ is incident through three polarizers $𝜃_{1} = {38.0}^{\circ}$ counter-clockwise, $𝜃_{2} = {17.0}^{\circ}$ clockwise, and $𝜃_{3} = {30.0}^{\circ}$ counter-clockwise.
$𝑆_{1} = 𝑆_{0} {cos}^{2} 𝜃_{1} = 60 {cos}^{2} 38^{\circ} = 37.258 𝑊 / 𝑚^{2}$ $𝑆_{2} = 𝑆_{1} {cos}^{2} 𝜃_{2} = 37.258 {cos}^{2} 55^{\circ} = 12.257 𝑊 / 𝑚^{2}$ $𝑆_{3} = 𝑆_{2} {cos}^{2} 𝜃_{3} = 12.257 {cos}^{2} 47^{\circ} = 5.701 𝑊 / 𝑚^{2}$

## ISP Addressing - URL: https://sharifhsn.dev/blog/isp-addressing/ - Structured data: https://sharifhsn.dev/api/posts/isp-addressing.json - Description: How does an ISP assign IP addresses to its customers? An ISP has a block of addresses that are partitioned to its customers. If an ISP network has an address like 200.8.4/24… - Date: 2022-02-28 - Exact published timestamp: 2022-02-28 - Topics: Internet Technology - Categories: Internet Technology - Source: Archive - Source URL: None - Content source format: typst

How does an ISP assign IP addresses to its customers?

An ISP has a block of addresses that are partitioned to its customers. If an ISP network has an address like 200.8.4/24 address, that is 256 addresses. /20 is for 4K hosts, and /16 is for 64K hosts. In fact, the calculation is $2^{32 - 𝑛}$ .

Subnetting

A network can be subdivided into subnets. This way you can have each router handling a smaller portion of the network, or have different kinds of routers i.e. wired/wireless handling different subnets.

In order to divide IP addresses, we use a subnet mask. For example, if our network is 128.64.32/24, a range we could have is 128.64.32.0-127 and 128-255. The easiest way to mask this is to look at the most significant bit. In the first range, the most significant bit is 0, and in the second range, the most significant bit is 1. The mask in this case would be 255.255.255.128 or in hex, F.F.F.8. If we bitwise AND this mask with the IP address, we will get the correct subnet.

We can follow the same procedure to get further divisions by the power of 2. We can mask over two bits to get four subnets of ranges 0-63, 64-127, 128-191, and 192-255 with a mask of 255.255.255.192.

Let’s think of an analogy. If we want to deliver some mail to our neighbor, the easiest way to do it is to go directly to his mailbox and give it to him instead of passing it off to the post office. In the same way, we can quickly send messages between hosts on the same subnet.

In order to facilitate this, the router will first check if the sender and destination (both first &ed with the subnet mask) are in the same subnet. If they are, it doesn’t bother sending the message over the Internet and it will actually just directly send the message to the destination itself.

In particular, this makes email between people on the same networks extremely quick. This is why it’s nice to have email between people with the scarletmail.rutgers.edu domain.

IPv4 Header

In order to get a packet to a destination host, we need a packet header with the identity of both the destination identity and the source identity. At this layer, we don’t worry about reliable exchange, dropping packets, sequence numbers, etc. That’s for TCP to handle, our only job is to send the packet to the destination.

However, we have to detect some kinds of problems. We don’t want to loop packets over and over, this will overload the network. There used to be a worry of fragmentation, if the size of the packet is greater than the maximum transmission unit (MTU) of the router. Nowadays, everyone has broadband so this is not an issue because everybody has high link speeds. When IPv4 was invented we needed checksums for verification if there are bit flips. IPv6 addresses have much less checksum machinery because it’s a waste nowadays, it only checks at the end. IPv4 has to calculate a checksum at every router, which catches errors early but is generally wasteful.

The time to live (TTL) is 1 byte long and is used to protect against loops. It starts at 255 and decrements every time it passes through a router. If the TTL is 0, the packet is dropped and a “time exceeded” error is thrown to sender. This prevents the packet from being looped around forever.

An entire 4 bytes is devoted to fragmentation machinery. This is not in IPv6, which simply throw an error if the packet is too large. The reason this is fine because it is a rare case nowadays and it cuts fat out of the header. Here are the steps of fragmentation:

router receives packet larger than MTU
if Don’t Fragment (DF) flag is set, throw Fragmentation Needed error
else, divide the packet into fragments maximum size outgoing MTU - header size (20 bytes)
in each new packet fragment, the length field is the size of the fragment, the More Fragments (MF) flag is set except for the last fragment, and the fragment offset field is set to the offset of the fragment in 8-byte blocks. The checksum is recomputed for the fragment. IPv4 - Wikipedia

ICMP

What do we do to propagate errors? We’re already using IP for sending messages, how do we send errors? This what the ICMP protocol is for. It is unreliable like UDP, and is tightly coupled with the implementation of IP. Its protocol ID is 1, which signifies its importance.

There are certain known error codes. The echo request/reply also known as ping is simply a request to check if the host is alive and responding. There is also the traceroute, which records the route that a packet takes by tracking TTL. As we know, when a router receives a packet, it decrements TTL. If we start TTL at 0 and slowly increase it, it will throw time exceeded back to us from each router in order. This way, we can see every router on the way to our destination.

## Making a Website with Zola, Github Pages, and Github Actions - URL: https://sharifhsn.dev/blog/making-a-website/ - Structured data: https://sharifhsn.dev/api/posts/making-a-website.json - Description: Making a website in the modern era is not easy to do for free. I did it using Zola, Github Pages, and Github Actions. I’ve always wanted to have a personal website where I can… - Date: 2022-02-26 - Exact published timestamp: 2022-02-26 - Topics: zola, github actions, website, Meta - Categories: Meta - Source: Archive - Source URL: None - Content source format: typst

Making a website in the modern era is not easy to do for free. I did it using Zola, Github Pages, and Github Actions. I’ve always wanted to have a personal website where I can upload what I do on my local computer to access remotely and have the world see. But it always seemed like too much of a hassle to set up and I didn’t have the capital to invest in a website that I didn’t need. However, when I started going back to university in-person this year, I found it much easier to take notes by typing them instead of using OneNote as I was accustomed, as the amount of code I had to write was drastically increased. Needing a way to access them remotely with a nice view, I thought a blog would be a good way to do that in addition to all the other things I had always wanted a website for. So, I embarked on a journey to create a website.

Github Pages

A website is no use unless we have somewhere to put it. Github Pages is a service offered by Github since 2008 that allows you to host your own website from a Github repository. You get one free website per Github account, which is called [username].github.io. All we have to do to enable it is create a repository named [username].github.io and enable Github Pages in the settings!

# should be above 2.28 to enable default branch name change
git --version

mkdir [username].github.io
cd [username].github.io

# personal git config
git config --global user.name "NAME"
git config --global user.email "EMAIL"
git config --global init.defaultBranch "main"
git init

gh repo create [username].github.io --public --source=. --remote-upstream

If you’re using Visual Studio Code as your editor, there’s a nicer way to do this than through the command line. After installing the Github extension, go to the Source Control button on the sidebar. There should be a button labeled “Publish to Github” which allows you to interactively initialize a Git repository in the current folder and publish it to Github.

Github Actions

We might have created our Github page, but we need a way to get all of the code from our repository to the website. This is called deployment. Luckily, we have a way to automatically deploy our website through Github Actions. Github Actions is another service offered by Github since 2019 that gives you free CI in public repositories. Although the free tier is fairly limited at 500 MB and 2000 minutes per month, it should be more than enough for a static blog that is not deployed very often.

There is an action automatically created for our Github page called pages-build-deployment which, as the name implies, builds and deploys the page you’ve created on push. The way that I organized my code, which is probably the simplest option, is that I hosted my code at the main branch and had a gh-pages branch that hosted the actual website which was built from the main branch. If you want to do the same, go to Settings -> Pages and make sure that the build target is the gh-pages branch at the root.

For now, this won’t do anything because we don’t have a gh-pages branch or anything in our main branch. So how do we actually make our website?

Zola

Zola is a static site generator written in Rust, and is one of the fastest out there. I decided to choose it for my website. If you’d prefer a different generator, this is where this guide diverges for you. There are plenty of tutorials for Hugo websites or others, but I have found a lack of Zola guides so I decided to create this.

To start, install Zola on your system. The documentation on the website is pretty stellar so I would recommend reading that to get a quick understanding on how to use Zola. I will explain the parts that I personally found difficult or unclear.

zola init
zola build # unnecessary as serve will also automatically build it
zola serve

Now you can see your new website at 127.0.0.1:1111!

Zola Deployment

Although we can build your website very simply on our local machine, it would be preferable to automatically build the website when we publish content so we don’t have to mess around with all of that. The Zola-approved way to do this is by using zola-deploy-action. All of you have to do is click the New Workflow button on the Actions page from your Github repository and follow the link to set up a workflow yourself, then copy-paste this into it:

# On every push this script is executed
on: push
name: Build and deploy GH Pages
jobs:
  build:
    runs-on: ubuntu-latest
    if: github.ref == 'refs/heads/main'
    steps:
      - name: checkout
        uses: actions/checkout@v2
      - name: build_and_deploy
        uses: shalzz/zola-deploy-action@v0.14.1
        env:
          # Target branch
          PAGES_BRANCH: gh-pages
          # Provide personal access token
          TOKEN: ${{ secrets.TOKEN }}

However, I wanted more configuration and control over my action. Github Actions operates using Jekyll by default, so unless you add a .nojekyll file to the build branch it will run unnecessary steps to build a Jekyll theme. In order to reduce complexity, I decided to make a similar action that adds that command. If you’re comfortable with the workflow as provided, then skip the next section.

Creating a Github Action

An action of the type we want here consists of three files: action.yaml, Dockerfile, and entrypoint.sh. Let’s break these down.

action.yaml: the configuration file that tells Github Actions what to do
Dockerfile: the configuration file for the Docker container that Github Actions will set up
entrypoint.sh: the shell script that will execute the commands we want

action.yaml is simple. It should follow this general format:

# action.yaml
name: 'ACTION_NAME'
description: 'DESC'
author: 'NAME'
runs:
  using: 'docker'
  image: 'Dockerfile'

The Dockerfile is more complex and has many more options. I kept mine simple to what is needed, you may have your own preference for Docker images.

# any Docker image is fine, I prefer debian
from debian:stable-slim
MAINTAINER NAME <EMAIL>

# for github actions
LABEL "com.github.actions.name"="ACTION_NAME"
LABEL "com.github.actions.description"="DESC"

# locale, I am in the U.S. so I use en_US
ENV LC_ALL C.UTF-8
ENV LANG en_US.UTF-8
ENV LANGUAGE en_US.UTF-8

# standard apt-get + wget and git for getting and building
RUN apt-get update && apt-get install -y wget git

# get zola on the docker image
RUN wget -q -O - \
"https://github.com/getzola/zola/releases/download/v0.15.3/zola-v0.15.3-x86_64-unknown-linux-gnu.tar.gz" \
| tar xzf - -C /usr/local/bin

COPY entrypoint.sh /entrypoint.sh

# give the entrypoint executable permissions
RUN chmod +x entrypoint.sh
ENTRYPOINT ["/entrypoint.sh"]

entrypoint.sh is where all the magic happens. Fundamentally, all that it does is call zola build on the main branch, which builds your website inside the public directory (you can configure this if you so wish). It then commits those website files to the gh-pages branch.

We need one more thing before we can create entrypoint.sh; a token. We need to authorize our action to be able to push to gh-pages. You can create a token by going to this page and creating a new token with at least repo rights. You can add the token to your repository by going to Settings -> Secrets -> Actions and creating a new repository secret called TOKEN (or any other name you like).

With that token, this is the basic necessities for entrypoint.sh:

#!/bin/bash
set -e
set -o pipefail

main() {
    git config --global url."https://".insteadOf git://
    git config --global url."$GITHUB_SERVER_URL/".insteadOf "git@github.com":

    # update git submodules (important if you have themes)
    git submodule update --init --recursive

    zola build

    cd public

    # if you want to add any commands do it here e.g. `touch .nojekyll`

    git init
    git config user.name "GitHub Actions"
    git config user.email "github-actions-bot@users.noreply.github.com"
    git add .

    git commit -m "Deploy ${GITHUB_REPOSITORY} to ${GITHUB_REPOSITORY}:gh-pages"
    git push --force "https://${GITHUB_ACTOR}:${TOKEN}@github.com/${GITHUB_REPOSITORY}.git" master:gh-pages
}

main "$@"

With all of these settings, this is how your workflow should look:

# .github/workflows/main.yml
on: push
name: Build and deploy GH Pages
jobs:
  build:
    runs-on: ubuntu-latest
    if: github.ref == 'refs/heads/main'
    steps:
      - name: checkout
        uses: actions/checkout@v2
      - name: build-and-deploy
        uses: ./ # wherever your action is in relation to the root of the repo
        env:
          TOKEN: ${{secrets.TOKEN}}

Themes

Zola requires a Tera template to render your site for the base site index.html as well as page.html for page-specific settings. You can also use Sass stylesheets if you enable it in your config.toml. Themes are a convenient way to have those built for you so you can get a website looking nice without excessive fiddling. I decided to use the after-dark theme which is based on the Hugo theme of the same name. Since I wanted to make my own modifications to it, I forked it and added the changes I wanted. The easiest way to add a theme for Github pages is to use submodules:

git submodule add https://github.com/getzola/after-dark.git themes/after-dark

The deploy action will take care of updating the submodule as necessary. Just add the theme to your config.toml file and voila!

Although I like the after-dark theme, I might change to a different theme or make my own in the future to accommodate my goals for this website. If that happens, I’ll detail that process in another post.

KaTeX

Most of the lecture notes I write incorporate KaTeX in some way. I find it an expressive way to write formulas and math expressions when reviewing for tests. after-dark does not provide KaTeX support by default, which is part of the reason I forked it.

My preferred option for KaTeX rendering would be server-side, as I don’t plan on pushing very often (perhaps once per day) and Zola compilation is extremely quick. However, after doing some research into previous attempts, I decided it wasn’t feasible for now. Perhaps in the future I’ll take a stab at implementing it myself, but for now I’ll settle for client-side.

The KaTeX docs give a pretty good description on how to incorporate it into your website. In Tera, all you have to do is enclose those stylesheets/scripts into CSS/JS blocks, respectively. My inspiration came from this pull request. I modified the standard auto-render.min.js script to add standard KaTeX $ $ tags.

## Currying Arguments in a Function - URL: https://sharifhsn.dev/blog/curry/ - Structured data: https://sharifhsn.dev/api/posts/curry.json - Description: Currying is the concept of having multiple arguments in a function. OCaml defaults to currying its functions. int -> int -> int is a function that takes two int s and returns an… - Date: 2022-02-21 - Exact published timestamp: 2022-02-21 - Topics: Principles of Programming Languages - Categories: Principles of Programming Languages - Source: Archive - Source URL: None - Content source format: typst

Currying is the concept of having multiple arguments in a function. OCaml defaults to currying its functions. int -> int -> int is a function that takes two ints and returns an int.

The -> is right-associated and the function application is left-associated. The last element of the function definition is always the return type, but calling a function always counts arguments from the left.

let f a b = a / b;;
let f = fun a -> (fun b -> a / b);;

These two lines are equivalent because the second line is just the uncurried function separated into two parts. The fun lambda only has one argument, and the -> keyword is right-associated so b is considered part of the arguments.

Currying allows you to pass only a portion of the expected arguments to the function, the same way that Python uses keyword arguments.

Another way to enable multiple arguments is by using a tuple that contains both arguments, which are destructured in the function definition. However, the advantage of currying is that you can separate the call of the function from the arguments.

let add a b = a + b;;
let addthree = add 3;;
addthree 4;; (* evaluates to 7 *)

This code allows addthree to exist as an implementation of add with a specific argument already given.

However, it’s not all roses with currying. Function need to retain state regardless of stack state, so the local variable 3 that is temporarily in the function addthree may not always be there. Anonymous functions may not have the same call stack.

In C-like languages, local variables are contained within their own stack frames. When a function calls another function, the new stack frame that is created contains its own local variables. If a variable is declared, then initialized through a function call, that variable contains junk until the function returns.

If we return a function that references a local variable in another function, reading it off the stack can get confusing. The first variable you look for is still uninitialized, so you need to evaluate where that variable is coming from to understand.

We solve this problem using static scoping. Nonlocal names refer to their nearest binding in the program text. This is also known as lexical scoping. If two variables have the same name in an inner scope and an outer scope, then we read the one in the inner scope first.

## HTTP Explained - URL: https://sharifhsn.dev/blog/http/ - Structured data: https://sharifhsn.dev/api/posts/http.json - Description: HTTP is probably the most visible Internet protocol to end users. It appears (as well as its cousin, HTTPS ) at the beginning of every URL we use to access the internet. But how… - Date: 2022-02-20 - Exact published timestamp: 2022-02-20 - Topics: Internet Technology - Categories: Internet Technology - Source: Archive - Source URL: None - Content source format: typst

HTTP is probably the most visible Internet protocol to end users. It appears (as well as its cousin, HTTPS) at the beginning of every URL we use to access the internet. But how does it work?

HTTP stands for HyperText Transfer Protocol. It defines the structure of messages that are passed between two programs: a client and a server. But before we talk about HTTP, let’s clarify some vocabulary

Web Terminology

object - a file that has an associated URL
Web page - a document that consists of multiple objects, typically a base HTML file and other objects. The HTML file will reference other objects through a path.
URL - address consisting of the following parts:

An example URL is split into these parts:

protocol	hostname	path name
http://	www.someSchool.edu	/someDepartment/picture.gif

Web browser - an application such as Firefox or Chrome that acts as the client in HTTP
Web server - an application such as Apache that acts as the server in HTTP and houses the Web objects in question

HTTP Low-Level

Under the hood, HTTP uses TCP to make a reliable connection between the server and client. The client and server both send requests and receive responses from their respective socket interfaces.

When a file is sent, there is no information that the server inherently remembers about the HTTP connection. Because of this, HTTP is considered a stateless protocol. This is not always desirable. Sometimes, websites want to remember clients in order to ease use of a website. We will discuss workarounds later.

Connections

The original HTTP 1.0 protocol was non-persistent. This means that every request made between a client and a server had its own connection that was opened and closed every time a request was made. Here’s the process:

A TCP connection is created between the client and server on port 80.
An HTTP request message is sent from the client to the server through the socket.
The server receives the message, retrieves the object that it is requesting, and sends the response message to the client through the socket.
The server “closes” the TCP connection (TCP will wait until it knows that the client has received the message)
The client receives the message and the connection terminates. It will extract the HTML file from the message and initiate a new request for each object referenced within, repeating these steps as needed.

Clearly, this method is inefficient. We are opening a TCP connection for every single request that is made between the same client and server. This inefficiency is mediated slightly by the ability to open serial TCP connections; typical web browsers open 5 to 10 parallel TCP connections. The response time for each connection is $2 \cdot 𝑅 𝑇 𝑇 + 𝑇_{𝑡 𝑟 𝑎 𝑛 𝑠}$ .

HTTP 1.1 introduced persistent connections. In step 4, the server doesn’t close the TCP connection after sending the response. The entire Web page is sent over the same TCP connection, and the requests can be pipelined for further performance gains. The connection is typically closed on timeout. HTTP/2 further builds on this by interleaving requests and responses in the same connection with a way to preserve ordering on each end.

Message Format

GET /http HTTP/1.1
Host: sharifhsn.github.io
Connection: close
User-agent: Mozilla/5.0
Accept-language: fr

The Host line is technically unnecessary, since the connection is already established by this point, but it’s useful for Web proxy caches. The Connection line means this connection should be non-persistent. The User-agent line specifies the user agent, which is the browser type that is making the request to the server. This is used so that the server can send different websites to different browsers. Accept-language expresses a preference for content in French if it exists, otherwise give the default. In general, the header lines exist to give information that is necessary for the request.

The HTTP response is different than the request. An example response to this message might be:

HTTP/1.1 200 OK
Connection: close
Date: Tue, 09 Aug 2011 15:44:04 GMT
Server: Apache/2.2.3 (CentOS)
Last-Modified: Tue, 09 Aug 2011 15:11:03 GMT
Content-Length: 6821
Content-Type: text/html

The status line at the top tells us some similar information, like the protocol we are using. The 200 OK is our status code for a good response. There are other status codes:

200 OK: request succeeded, information returned.
301 Moved Permanently: the address is moved, for example when a website links to its www address; there will be a location header telling the client where the new place is to go
400 Bad Request: request is written poorly and could not be understood
505 HTTP Version Not Supported: self-explanatory lol

There’s tons of valid header lines we can use. The HTTP specification allows for many header lines to be used from various software. Writing a correct HTTP implementation isn’t easy.

Cookies

I mentioned before that HTTP is stateless, which means that the server does not preserve any information about the client after the connection ends. However, websites really want ways to identify users for operations such as logins. In order to facilitate state, websites use cookies. You might have heard of them when you log in to a website and it asks you to accept cookies.

The typical creation method of a cookie might look like this:

A client makes a request to the server for the first time.
The server generates an identification number for the client.
The server sends a HTTP response to the client with a cookie in the header: Set-cookie: 1678
The client browser stores the cookie in a special file and sends a request to the server with the same cookie number.
Now, whenever the client requests the server, it will include a cookie in the header which the server will use.

Caching

You might have heard of a proxy server before. These servers are usually used by ISPs like your home ISP and your university ISP. These servers contain frequently-accessed data from several origin servers and are located much more closely to the client.

These servers are extremely helpful in improving response times for the client as well as reducing traffic for the server. The close location and LAN nature of proxy servers mean that the request/response time between clients and servers are quick. And since the origin server is not accessed as often, its burden of traffic is significantly reduced.

The delay in fetching a response from the Internet tends to be dominated by the time for the institutional server receiving a response from the Internet. However, if there is heavy traffic on the access link between the institutional server and the Internet router, that can slow down speeds to minutes per request, which is obviously unacceptable.

𝑟 𝑒 𝑞 𝑢 𝑒 𝑠 𝑡 𝑅 𝑎 𝑡 𝑒 \cdot 𝑟 𝑒 𝑞 𝑢 𝑒 𝑠 𝑡 𝑆 𝑖 𝑧 𝑒 / 𝑠 𝑝 𝑒 𝑒 𝑑 = 𝑡 𝑟 𝑎 𝑓 𝑓 𝑖 𝑐 𝐼 𝑛 𝑡 𝑒 𝑛 𝑠 𝑖 𝑡 𝑦

The intensity between client and LAN is usually negligible, but the access link between LAN and Internet dominates. With a Web cache, hit rate is generally between 0.2 and 0.7.

𝑐 𝑎 𝑐 ℎ 𝑒 𝐻 𝑖 𝑡 \cdot 𝐿 𝐴 𝑁 𝑠 𝑝 𝑒 𝑒 𝑑 + (1 - 𝑐 𝑎 𝑐 ℎ 𝑒 𝐻 𝑖 𝑡) \cdot 𝑖 𝑛 𝑡 𝑒 𝑟 𝑛 𝑒 𝑡 𝑆 𝑝 𝑒 𝑒 𝑑

The traffic intensity stops being an issue and Internet delay dominates again once a cache is introduced. It’s less expensive and faster than increasing access link speed.

## Paging - URL: https://sharifhsn.dev/blog/paging/ - Structured data: https://sharifhsn.dev/api/posts/paging.json - Description: The main problem that segments have introduced to managing memory space is that their variable size wastes memory through fragmentation. Fixed-size pieces that are easier to… - Date: 2022-02-20 - Exact published timestamp: 2022-02-20 - Topics: Operating Systems Design - Categories: Operating Systems Design - Source: Archive - Source URL: None - Content source format: typst

The main problem that segments have introduced to managing memory space is that their variable size wastes memory through fragmentation. Fixed-size pieces that are easier to handle are much more popular: these are known as pages.

Pages

Address spaces are split up into multiple pages, typically a power of 2. For example, let’s picture a tiny 6-bit address space that can only address 64 bytes. We could split it up into four 16-byte pages, that we’ll refer to in sequence.

The physical representation of these pages is through page frames, which are sequenced directly in memory and are ordered. The pages in the virtual address space map directly to page frames in physical memory, with no respect for order.

Memory management of the free space here can be done with a simple free list. All it needs to look for is four free page frames somewhere in memory and map each page to a page frame. This mapping is stored in a page table which is kept per process, since every process has its own address space.

Page Translation (IMPORTANT)

In our example from earlier, we worked with a 6-bit address space. The highest order bits are reserved for the virtual page number, and the lower bits are reserved for the offset within the page. For example, the virtual address 21 would be 010101 in binary. The top two bits 01 tell us that we are looking for virtual page 1. The OS checks the page table and sees that VP2 maps to physical frame 7, which is 111 in binary. The OS then translates the physical address by replacing the VP bits with the PF bits. In this example, the new address would be 1110101, or the 117th byte in memory.

Page Tables

We’ve talked about page tables a bit, but let’s go into details. They are a data structure like any other, but they can get very large. Each mapping in the table, which is called a page table entry (PTE), is typically $2^{2}$ bytes in size. It’s often simpler to think of these calculations in terms of the bits involved, since they will always be a power of two. With that in mind, this is the formula for page table size:

𝑝 𝑎 𝑔 𝑒 𝑇 𝑎 𝑏 𝑙 𝑒 𝑆 𝑖 𝑧 𝑒 = 𝑉 𝑃 𝑁 + 𝑃 𝑇 𝐸

That’s 22 bits for a typical 32-bit system, which is pretty massive at 4MB per page table, which is again per process. We obviously can’t keep this in the MMU, so we need to actually store it in memory.

How do we organize the page table as a data structure? The most obvious way is as a linear table, aka an array. The VPN is an index in the array, and the value at the index is the PTE which gets the PFN. The PTE itself contains the PFN plus some helpful bits for us. There is a valid bit checks if the mapping even exists, protection bits for privileged memory, and others.

This access is still extremely slow for load store operations. We need a better solution!

TLB

The hardware comes to the rescue here. MMUs provide a translation-lookaside buffer to cache commonly used translation. When a load-store instruction is executed, the CPU will first check the TLB if there exists a translation for the specified address if it exists. If it does, then it can quickly perform the operation without performance overhead of going back and forth on memory so often. If it doesn’t, the CPU hands the reins over to the OS to do its own page and offset translation and it will cache the translation after the OS gives it. Here’s the basic steps:

extract VPN from virtual address
check if VPN is in TLB
if hit, then perform the PFN concatenation and access the memory
if miss, hardware will check the page table to find the translation
update the TLB with the translation
retry TLB, now a guaranteed hit

All of this assumes that memory is valid, accessible, and unprivileged; these can cause exceptions that the OS will handle.

Step 4 is expensive and it is therefore the step that we want to avoid as much as possible. Luckily, many memory accesses reside on the same page. For example, arrays are almost guaranteed to have the same TLB hit because they are organized contiguously. Spatial locality helps us increase our hit rate.

Page size is also significant here. By having big pages, typically 4KB, we are unlikely to have TLB misses since the same page is accessed for this memory. Temporal locality will also help here, as the TLB will evict based on recency.

The TLB miss can be managed by either the CPU or the OS. Older CISC or complex-instruction set computers managed the TLB themselves, while modern RISC reduced computers raise an exception to be handled by the OS. In these computers, steps 4 and 5 are handled by the OS and step 6 only executes after the exception is handled.

The exception raised here is a little different from other exceptions. Typically, instructions that raise exceptions are skipped. Here, we want to retry the operation, so we need to get a different program counter.

The TLB cache is fully associative which means that any translation can be anywhere. This means that the VPN is encoded with the PFN and other bits.

One small note is that the TLB entries have valid bits just like the PTE, but they serve different purposes. In the TLB, it refers to a valid translation. If a context switch has occurred, for example, then all of the cache becomes invalidated. PTE valid bits refer to unallocated memory which results in a process kill.

Let’s examine that cache invalidation. Flushing the cache on every context switch seems to miss the point of a TLB since it happens so often. What are other ways we can manage this? Hardware will typically add an ASID (address space identifier), which is similar to a PID but contains less information. This way, a TLB can contain information for multiple processes without accidental contamination.

Smaller Tables

As discussed earlier, page tables can get real honking big, which is not good for memory consumption. TLBs can mitigate the performance problems of page table access, but we need better ways to mitigate memory usage of our page tables.

The obvious solution, also mentioned earlier, is to just increase the size of each page. The size of a page in $2^{𝑏 𝑖 𝑡 𝑠}$ is given by:

𝑎 𝑑 𝑑 𝑟 𝑒 𝑠 𝑠 𝑆 𝑝 𝑎 𝑐 𝑒 = 𝑛 𝑢 𝑚 𝑏 𝑒 𝑟 𝑂 𝑓 𝑃 𝑎 𝑔 𝑒 𝑠 + 𝑝 𝑎 𝑔 𝑒 𝑆 𝑖 𝑧 𝑒

so if we have a 32-bit address space, we can have 18 bits for our number of pages and 14 bits for our 4 KB page. 1 MB per page table is better because it’s smaller, but there’s a limit to this kind of strategy. If we make our pages too big, we will get internal fragmentation within each page, where an entire page is allocated but not that much memory within it is used, leading to waste. 4 KB is a good middle ground, which is why that’s what x86 uses.

Segmentation-Paging

We looked at segmentation earlier, but initially dismissed it due to its issues of variable size. What if we combined the two approaches into a hybrid? Instead of having one giant page table for the address space, what if we split it up into segments?

We can twist around the base/bounds registers to use them for a different purpose. The base register can hold the physical address of the actual page table for a particular segment, and the bounds will tell us where the physical end of the page table is. We can use these values to calculate the number of valid pages.

Again, let’s use the top two bits to refer to segment. We’ll figure out our base/bounds pair from the those bits, then get to our PTE by adding $𝑉 𝑃 𝑁 \cdot 𝑠 𝑖 𝑧 𝑒 𝑜 𝑓 (𝑃 𝑇 𝐸)$ to it. The bounds register can be used to track our number of valid pages so they don’t take up space in the page table if they are not used. However, this comes with the same issues of segmentation earlier: external fragmentation.

Multi-level Page Tables

As is common for many problems, the solution for page tables is to put them inside another page table. Multi-level page tables are the de facto solution for page tables that are used in x86. However, this introduces a significant amount of complexity in our page table search, so we will need to examine that.

We will chop up our page table into its own kind of pages, each of which only contain PTEs. If an entire page of PTEs is all invalid, which means that none of them have valid translations, don’t allocate that page. In order to manage this apparatus, we will introduce a new data structure: the page directory.

You can think of the page directory as a simple page table if the only memory being tracked was the sub-page table. Each entry has a valid bit which is true if any PTE in its mapped page is valid, as well as the PFN where the page of PTEs is stored. That PFN is only allocated if the valid bit is set.

If we organize this structure correctly, each chunk of the page table that we refer to as “pages” are actually page-sized and can fit into memory pages in kernel space. This greatly simplifies page table management.

However, there is a cost, as always. TLB misses require two memory loads in order to get the correct page because of the level of indirection that page directories introduce. Since TLB misses are rare, we take that tradeoff in return for significantly reduced memory consumption. This is a trade for space that sacrifices time.

Let’s get some numbers for a multi-level page table. Assume a 14-bit virtual address space split into 8 bits for VPN and 6 bits for offset. Remember that this translates to 256 entries per table and a page size of 64 bytes.

We have to do some special bit magic to manage these page levels as well. In this example, our page table size is 10 bits; remember, it’s VPN + PTE. We need to subtract our offset bits from that size to get 4 bits for number of pages, then subtract our PTE bits from that to get number of PTEs per page. In total, here is the formula:

𝑝 𝑎 𝑔 𝑒 𝑇 𝑎 𝑏 𝑙 𝑒 𝑆 𝑖 𝑧 𝑒 = 𝑛 𝑢 𝑚 𝑏 𝑒 𝑟 𝑂 𝑓 𝑃 𝑎 𝑔 𝑒 𝑠 + \underset{𝑃 𝑇 𝐸 𝑆 𝑖 𝑧 𝑒 + 𝑃 𝑇 𝐸 𝑃 𝑒 𝑟 𝑃 𝑎 𝑔 𝑒}{\underset{⏟}{𝑝 𝑎 𝑔 𝑒 𝑆 𝑖 𝑧 𝑒}}

$10 = 4 + (2 + 4)$ in this example.

The page directory size is the same as the number of pages, so it is also 4 bits. When doing our translation, the highest order bits of the VPN are reserved for the page directory index. After indexing using these higher bits, the rest of the bits of the VPN are the “offset” within the page that the higher bits point to. You can think of this is as a mini address with the VPN being the PDI and the offset being the page table index (PTI).

Infinity and Beyond

This example presented has only two levels, the page directory and the page table. But we can go deeper. Let’s reset our numbers to 30 bit address space and 9 bit page size. The VPN is therefore 21 bits.

If we apply the same formula as before here, we end up with 7 bits for PTEPerPage. The lowest bits of the VPN will be reserved here. Our page directory will now have 16 bits (14 + 2 for each entry), but this is WAY too much. We need every piece of our structure to fit into a page, so we need to get the PDI to 7 bits or less.

The solution here is to further split the PDI into another VPN/offset split. The offset here will be 7 bits, as this is what fits into the page.

The zeroeth index is the VPN part, which is 7 bits. It can address 128 pages of the second-level directory. The second-level directory will address 128 pages of PTEs. This way, both the first and second level directory are 9 bits (index + entry size) so they fit into a page!

This is really hard to understand! Try to think of it where every level is a split between index and offset. The first offset must always be the size of a page, and every inner offset must be the size of page minus the PTE size. In this case, the first offset was 9 bits, and every directory offset was 7 bits. To get the maximum level $𝑛$ , the formula is:

𝑎 𝑑 𝑑 𝑟 𝑒 𝑠 𝑠 𝑆 𝑝 𝑎 𝑐 𝑒 > (𝑝 𝑎 𝑔 𝑒 𝑆 𝑖 𝑧 𝑒 - 𝑃 𝑇 𝐸 𝑆 𝑖 𝑧 𝑒) \cdot (𝑛 - 1) + 𝑝 𝑎 𝑔 𝑒 𝑆 𝑖 𝑧 𝑒

Our maximum level in this example is 3, because a level of 4 would result in $(9 - 2) \cdot (4 - 1) + 9$ which is 30, the same as the address space, which must be greater.

Inverted Page Tables

A small coda to this discussion of paging is the inverted page table. Unlike other kinds of page tables, this is a page table that is shared between processes that maps every physical page. This massive table has information per entry about what process is using it, the virtual page number that the process using is referring to, and the physical page. PowerPC uses this model, with a hash table instead of an array to speed up lookups.

However, lookups are still slow, although they take up less memory. It is also difficult to implement sharing as specified earlier. You need to somehow chain multiple virtual addresses to one entry, which introduces serious complexity.

## TCP Explained - URL: https://sharifhsn.dev/blog/tcp/ - Structured data: https://sharifhsn.dev/api/posts/tcp.json - Description: TCP is the well-known protocol for reliable transfer of data. But how does it actually work? System of handshakes ensures reliable transmission of data. What happens if a packet… - Date: 2022-02-20 - Exact published timestamp: 2022-02-20 - Topics: Internet Technology - Categories: Internet Technology - Source: Archive - Source URL: None - Content source format: typst

TCP is the well-known protocol for reliable transfer of data. But how does it actually work?

System of handshakes ensures reliable transmission of data.

What happens if a packet is corrupted, and its bits are flipped?
- Use a checksum for each packet.
What happens if a packet is lost somewhere?
- Wait for acknowledgement from receiver, and if not received, resend.
What happens if a packet is duplicated? etc. etc.

There are lots of problems that can happen with UDP. Let’s create a protocol that doesn’t have this unreliability.

Our new protocol has a sequence number for each packet, either 0 or 1. This is known as an alternating bit protocol. When it sends a packet, it waits for ACK (acknowledgment) before it sends the next packet. This will take one RTT (round trip transmission) per packet. This is obviously pretty inefficient, but it does solve at least some of these issues. It solves the duplicate issue because if it sends the same packet, it will have the same bit sequence number and therefore will not be taken into account. You can think of this as an infinite loop which flips its break condition after being broken. This is possible because you’re only sending one packet at a time. Here is time and utilization.

𝑇_{𝑡 𝑟 𝑎 𝑛 𝑠 𝑚 𝑖 𝑡} = \frac{𝐿_{𝑝 𝑎 𝑐 𝑘 𝑒 𝑡 𝐿 𝑒 𝑛 𝑔 𝑡 ℎ}}{𝑅_{𝑡 𝑟 𝑎 𝑛 𝑠 𝑚 𝑖 𝑠 𝑠 𝑖 𝑜 𝑛 𝑅 𝑎 𝑡 𝑒}}

𝑈_{𝑠 𝑒 𝑛 𝑑 𝑒 𝑟} = \frac{\frac{𝐿}{𝑅}}{𝑅 𝑇 𝑇 - \frac{𝐿}{𝑅}}

If we can send W packets at aa time, then we can replace $\frac{𝐿}{𝑅}$ with $𝑊$ , which will improve our time by a factor of $𝑊$ ! However, if we send a stream of packets, then there are more issues. If you send a receiver too much data, then it will throw out the data it cannot receive. If packets get lost somewhere in the router, you will have no idea. Stopping the stream to check for issues wastes a lot of time.

We can solve this using a window time. We send packets up until the first ACK is received, then we check for any problems with ACK. But what is the window size?

𝐵 \cdot 𝑅 𝑇 𝑇 = 𝑊 \cdot 𝑝 𝑎 𝑐 𝑘 𝑒 𝑡 𝑆 𝑖 𝑧 𝑒

This way we can pipeline as many packets as are possible before anything can go wrong. This increases the complexity on the recv side, but keeps the connection simple and reliable. Let’s say the window size is 3. We can send 3 packets, then we must wait for ACK for the first packet. When the ACK is received for packet p, we send packet p + 3. If the timer expires, we have to resend the packet. We can make the assumption that everything to the left of the window has been ACKed, and everything to the right of the window has not yet been sent.

What is the efficiency of this new method?

𝑈_{𝑠 𝑒 𝑛 𝑑 𝑒 𝑟} = \frac{3 \cdot \frac{𝐿}{𝑅}}{𝑅 𝑇 𝑇 \cdot \frac{𝐿}{𝑅}}

This is an increase of a factor of 3!

But what if a packet is dropped and sent later? This is where the receiver comes in. There are two ways of dealing with this:

Go back N: simple, keep a buffer of 1 and throw out data with greater sequence numbers and ACK at that packet that was received. This will force the sender to resend all of the packets thata re now timed out. However, this also throws out correctly received packets.
Selective Repeat: complex, keep a buffer proportional to W and keep data that has been received, cancelling the timers for each. Once the data has been correctly received, you can move the window and keep the buffer open for anything that hasn’t been received. Let’s say packet 1 is timed out but 2 and 3 came in correctly. We can cancel those two timers and slide the window to 4 and 5 but still wait for 1. There are only ever W packets in transmission at any given point!

In order to demonstrate the benefits of selective repeat over go back N, we’ll work though an example. Say we are sending 5 packets of size 100 B each over a 100 MB/s link with a RTT of 100 ms, with packet timeout being send_time + RTT and the window buffer being of size 5. How long would it take to receive all of the packets if we drop the third packet?

The send_time here is 100 B / 100 MB/s or 0.001 ms. The time to receive a packet is half of RTT or 50 ms. The timeout length is 100.001 ms.

With go back N, packets 1 and 2 are received at times 50.001 ms and 50.002 ms. All good! But when the third packet is dropped, we must wait 100.01 ms to receive it again at 150.003 ms. Even though we received packets 4 and 5 already at 50.004 ms and 50.005 ms, we had to throw them out because we didn’t get packet 3. We resend them as soon as possible after packet 3 so they are received at 150.004 ms and 150.005 ms respectively. The whole transaction takes 150.005 ms.

With selective repeat, the four non-dropped packets are received in the same way. However, after packet 3 times out and is resent, it is the only packet that is sent at 150.003 ms, since every other packet was within our window. The whole transaction here will take 150.003 ms.

The protocol we have just created is known as TCP.

connection management
retransmission
flow control
congestion control
frame format

It has it all!

TCP is the default protocol for every web browser in existence and is the de facto protocol for reliable transmission. Chrome now also uses QUIC which is a Google-created protocol but the vast majority of Internet traffic is still TCP.

TCP Header

The TCP header is much more complex than the UDP header. It includes some of the same elements, such as the port numbers and the checksum, and others:

32-bit sequence number and acknowledgement number, which are used for reliability
16-bit receive window for flow control so that the bytes sent do not overwhelm the receiver
4-bit header length field to indicate the length of the header, needed because of variable options, but usually 20
variable options field needed when specific maximum segment size (MSS) is needed or other options for high-speed networks
6-bit flag field: ACK for a segment that is an acknowledgement; RST, SYN, and FIN for connections, PSH for immediate network sending, and URG to indicate some urgent data is located at the location where the 16-bit urgent data pointer points (these last three are typically unused)

The sequence and acknowledgment numbers are how TCP messages are ordered despite being sent in a full-duplex way. The sequence number orders the segments for reconstruction into a full message, and the acknowledgment number is a way of communicating the next byte that the sender needs. This way, if a segment is dropped, the sender of the dropped segment knows that the receiver didn’t get it because they’re still waiting on it in the acknowledgement number.

RTT

Round-trip time is an important concept when we consider making our protocol as fast as possible. This is how long it takes for a segment to be sent and an acknowledgment for that segment to be received. In order to get an accurate measure of the RTT, TCP takes the current RTT and refines it to an average as more segments are transmitted, since any one SampleRTT (SRTT) might be an outlier. In order to get the average EstimatedRTT we follow this formula:

𝐸 𝑅 𝑇 𝑇 = (1 - 𝛼) \cdot 𝐸 𝑅 𝑇 𝑇 + 𝛼 \cdot 𝑆 𝑅 𝑇 𝑇

$𝛼$ in this formula is typically 0.125. This means that average is weighted towards recent samples, since they better reflect the current weight.

We should also have a measure of variance in RTT, since averages can be deceiving. DevRTT is the measure of the variance, which is calculated like this:

𝐷 𝑒 𝑣 𝑅 𝑇 𝑇 = (1 - 𝛽) \cdot 𝐷 𝑒 𝑣 𝑅 𝑇 𝑇 + 𝛽 \cdot | 𝑆 𝑎 𝑚 𝑝 𝑙 𝑒 𝑅 𝑇 𝑇 - 𝐸 𝑠 𝑡 𝑖 𝑚 𝑎 𝑡 𝑒 𝑑 𝑅 𝑇 𝑇 |

Like ERTT, DRTT moves and is weighted recently, with a $𝛽$ value of usually 0.25.

The reason this is important is because the interval at which a TCP segment transmission times out is based on RTT. TCP calculates this way:

𝑇 𝑖 𝑚 𝑒 𝑜 𝑢 𝑡 𝐼 𝑛 𝑡 𝑒 𝑟 𝑣 𝑎 𝑙 = 𝐸 𝑅 𝑇 𝑇 + 4 \cdot 𝐷 𝑅 𝑇 𝑇

This starts out as one second, then changes over time. Timeouts cause this value to double because a new link may be chosen that has a longer RTT than what ERTT has estimated so far.

Congestion Control

A network is said to be congested when so much traffic is being sent through the network that many packets become lost. To combat this, TCP implements congestion control.

Every sender has a limited rate at which it can send traffic into its connection based on the way TCP thinks the network is congested. If it thinks it’s not very congested, then it increases the send rate, and vice versa. There are three elements to this implementation: how to limit the rate, how to perceive the congestion, and how mathematically the rate should be limited.

The mechanism at the sender keeps track of a congestion window, also known as cwnd. The amount of unacknowledged data that a sender sends must not be greater than either the congestion window or the receive window.

## Electromagnetic Induction - URL: https://sharifhsn.dev/blog/induction/ - Structured data: https://sharifhsn.dev/api/posts/induction.json - Description: The final piece to the puzzle of electromagnetism here is electromagnetic induction , where we use magnets to create current. Like with charges in a magnetic field, magnets do… - Date: 2022-02-18 - Exact published timestamp: 2022-02-18 - Topics: Physics - Categories: Physics - Source: Archive - Source URL: None - Content source format: typst

The final piece to the puzzle of electromagnetism here is electromagnetic induction, where we use magnets to create current.

Induced Emf/Current

Like with charges in a magnetic field, magnets do not generate current unless they are moving relative to a wire. The changing magnetic field $\vec{𝐵}$ is what creates the current. This current is called an induced current, and the “emf” that causes it which is the wire itself is an induced emf.

The emf can also be induced by changing the area of coil in a magnetic field, as that changes the relationship between the magnetic field and coil.

These examples are in a closed circuit, which has a current. An open circuit would not have a current, but it would still have the induced emf.

Motional Emf

An induced emf is created in a metal rod that moves through a magnetic field as long as the velocity is not parallel to the magnetic field. We can use RHR1 to determine where the positive and negative sides of the rod are when the rod has a velocity through a magnetic field. The movement of these charges to either end of the rod cause a charge separation to occur, which leads to a current running from the negative end to the positive end. This is called motional emf because it comes from the motion through a magnetic field.

We can find the motional emf when the length, velocity, and magnetic field are all perpendicular i.e. they all have their own axis:

𝐸 = 𝑣 𝐵 𝐿

There is another magnetic force that opposes the motional emf. The current creates its own magnetic field, which by RHR1 would actually oppose the velocity of a mutually perpendicular system. We have to have an external force moving the rod otherwise the magnetic field produced by its own current will cause it to stop.

Magnetic Flux

Remember electric flux? Magnetic flux is quite similar in that it is defined as amount of magnetic field passing through an area, so $Φ = 𝐵 𝐴 cos 𝜃$ in $𝑊 𝑏$ or $𝑇 \cdot 𝑚^{2}$ . We can define motional emf through magnetic flux like so:

𝐸 = - \frac{Δ Φ}{Δ 𝑡}

or, emf is the rate of change of magnetic flux. This is why we can induce an emf by changing the area of a coil. The reason that the equation is negative is because the induced current will create a magnetic force which will oppose its velocity direction.

Faraday’s Law

Faraday’s Law is precisely the equation we just laid out but with one additional component to account for loops:

𝐸 = - 𝑁 \frac{Δ Φ}{Δ 𝑡}

The emf is generated if the flux changes, which depends on $𝐵$ , $𝐴$ , or $𝜙$ , which are magnetic field, area, and angle of the magnetic field with respect to the normal of the surface.

Lenz’s Law

Lenz’s Law is less of an equation and more of a method to understand the polarity of an induced emf. Let’s follow the reasoning:

Check whether the flux is increasing or decreasing.
Find the direction of the induced magnetic field to oppose the change in flux.
Use RHR2 with the previous step to find the positive and negative end, with the polarity coming out of your palm.

This is best understood through an example. Let’s imagine we have a loop with a bar magnet approaching at north side head on.

The flux is increasing, because the magnet is getting closer.
The induced magnetic field must oppose the magnet field to decrease the flux.
RHR2 tells us that if the magnetic field is coming out of our palm at the magnet, the current must be going counter-clockwise. Therefore, the left point is positive and the right is negative, as the external circuit must have positive going to negative.

Transformers

A transformer is a device that increases or decreases ac voltage. It is responsible for reducing the voltage in your phone charge so you’re not sending 120 V to your phone that doesn’t need it. This change is called a step-up or step-down which increases or decreases the voltage, respectively, directly.

When we send power over power lines, there is a certain power loss that happens depending on the voltage and the resistance of the transmitting power wire:

𝑃_{𝑙 𝑜 𝑠 𝑡} = 𝐼^{2} 𝑅

which we subtract from our initial power to get the final power.

If we want to figure out how much voltage we need to step-up or step-down to get a specific power loss, we just resubstitute the desired power loss into that equation to get the current and proportion it to the voltage.

## Data Types in OCaml - URL: https://sharifhsn.dev/blog/data-types/ - Structured data: https://sharifhsn.dev/api/posts/data-types.json - Description: When we make our programs more and more complex, we need more complex data types as well. We have only used OCaml’s built-in data types, how can we construct our own data types?… - Date: 2022-02-17 - Exact published timestamp: 2022-02-17 - Topics: Principles of Programming Languages - Categories: Principles of Programming Languages - Source: Archive - Source URL: None - Content source format: typst

When we make our programs more and more complex, we need more complex data types as well. We have only used OCaml’s built-in data types, how can we construct our own data types?

`type`

The type keyword is similar to the typedef keyword in C, except more limited in scope. A type can only be multiple variants of arbitrary values.

(* coin is enum with variants Heads and Tails*)
type coin = Heads | Tails

Each variant can also contain data of other data types.

type shape =
 | Rect of float * float
 | Circle of float
let r = Rect (3.0, 4.0) (* r has type shape *)

shape here has two variants. It can either be a tuple of two floats when it is a Rect, or it can be a single float when it is a Circle.

These data types are also known as algebraic data types or tagged unions.

Option

ADTs can be useful when we want to ensure the complete handling of all cases. For example, if an object is nullable, it is useful to make sure that we must handle the null case instead of passing that off to the developer who might carelessly not handle it. This is where the option type comes from.

type 'a option =
 | Some of 'a
 | None

The 'a keyword means that that the type option is polymorphic, and the variant Some will contain whatever type that option is defined for. When handling an option, you must destructure it into its Some and None variants and handle both cases, otherwise OCaml will warn you for a non-exhaustive pattern match.

List

We can actually define our own list data type as a recursive data type, which is a data type which contains itself.

type 'a list =
 | Nil
 | Cons of 'a * 'a list

Here, list has two variants, Nil and a Cons tuple of an element and a list. If we think of the traditional list data type, this is actually just a more verbose version. [] is sugar for Nil and :: is sugar for Cons tuple.

let rec len l =
    match l with
    | Nil -> 0
    | Cons (_, t) -> 1 + (len t)
(* same as *)
let rec len l =
    match l with
    | [] -> 0
    | _ :: t -> 1 + (len t)

Exceptions

Exceptions are a special data type used for errors in OCaml. Exceptions are similar to type constructors in that they can take arguments or have none.

exception Sign of int
let f n =
    if n > 0 then
        raise (Sign n)
    else
        raise (Failure "foo")

We can raise an exception with arguments whenever we want, which will exit the function with the exception name and its arguments. Failure is a generic exception type that is used with strings.

There is also special try syntax used to catch exceptions.

let g n =
    try
        f n
    with Sign n ->
            Printf.printf "Caught %d\n" n
          | Failure s ->
            Printf.printf "Caught %s\n" s

The function g will try running f n, but if that raises an exception, it will be caught in with. It can be pattern matched for different exception types.

## Operational Semantics - URL: https://sharifhsn.dev/blog/operational-semantics/ - Structured data: https://sharifhsn.dev/api/posts/operational-semantics.json - Description: What are the formal semantics of how a programming language works, mathematically? That’s a broad topic. Operational semantics , which are how programs execute , are narrow… - Date: 2022-02-17 - Exact published timestamp: 2022-02-17 - Topics: Principles of Programming Languages - Categories: Principles of Programming Languages - Source: Archive - Source URL: None - Content source format: typst

What are the formal semantics of how a programming language works, mathematically? That’s a broad topic. Operational semantics, which are how programs execute, are narrow enough for one article.

Rules

The basis behind the mathematics of operation are based on using rules to define a judgment. The expression e will always evaluate to the value v. We can construct a micro-OCaml through the datatypes exp and value:

eval: exp -> value

This way of presenting semantics is called a definitional interpreter. eval means exp -> value, and we use interpretations to define the language’s meaning.

Grammar

It is useful to define meta-variables which represent categories of syntax. We will list a few here:

x: any identifier or variable name
n: a numeral value
e: any expression
::=: meta-syntax for definition
|: meta-syntax for variants

e ::= x | n | e + e | let x = e in e

This is an important line which defines what an expression exactly is. Up until now, we have used the term expression quite loosely, so it’s good to have an exact definition in operational semantics. An expression is either an identifier, a numeral or a combination of expressions. The last variant may seem a little strange to you, but remember, let expressions are simply replacing identifiers in an expression, so they are also expressions. This is a powerful definition because it fully encompasses the abstract syntax tree (AST).

We can define an evaluation as eval: exp -> value where evaluation is the process of turning an expression into a final value (in this case an int only).

Rules of Inference

In order to prove the veracity of a judgment, we check its composite rules. For example, let’s prove that 1 + 3 ⇒ 4 is true. 1 and 3 are expressions in the sum function which evaluate to their own numeral value. Two values being added together is their sum, which is 4.

The notation of rules of inference are used to present rules in formal mathematics:

\frac{𝐻_{1} \dots 𝐻_{𝑛}}{𝐶}

If all the hypotheses are true, then the conclusion is true. If there are no hypotheses, then the conclusion is automatically true (axiom).

Let’s express those same rules about numeral self evaluation and sum expressions in rules of inference:

\frac{𝑒_{1} \Rightarrow 𝑛_{1} 𝑒_{2} = > 𝑛_{2} 𝑛_{3} is 𝑛_{1} + 𝑛_{2}}{𝑒_{1} + 𝑒_{2} \Rightarrow 𝑛_{3}}

We can similarly describe the more complicated rules of let expressions:

\frac{𝑒_{1} \Rightarrow 𝑣 1 𝑒_{2} {𝑣_{1} / 𝑥} \Rightarrow 𝑣_{2}}{let 𝑥 = 𝑒_{1} in 𝑒_{2} \Rightarrow 𝑣_{2}}

Derivations

The derivation is a process in which we apply rules to an expression in succession. We take our conclusion, then break it up into its constituent rules. If any of the rules need more rules themselves, we break them up too. Think of it like a tree which expands out from the conclusion. Let’s use let x = 4 in x + 3 ⇒ 7 as an example.

\frac{4 \Rightarrow 4 \frac{4 \Rightarrow 4 3 \Rightarrow 3 7 is 4 + 3}{4 + 3 \Rightarrow 7}}{let 𝑥 = 4 in 𝑥 + 3 \Rightarrow 7}

Let’s look at this derivation step by step. First, we start at our conclusion. We say that x = 4 in the let expression. Because 4 is used an expression here, we need to generate the hypothesis that the expression 4 will evaluate to itself, which is true from our reflective axiom. Then, we need to show that the resulting expression 4 + 3 ⇒ 7 is valid, which requires the sum rule of inference. Knowing that 4 and 3 evaluate to themselves and that 7 is the sum of those two values, then we can confirm that hypothesis and therefore the conclusion!

The way that we’ve written this derivation is recursive in nature. This is how definitional interpreters will evaluate expressions; they will search the expression for any constituent and evaluate them in turn in order to evaluate the entire expression.

\frac{eval Num 4 \Rightarrow 4 Plus(Ident("x"), Num 3)}{eval Let("x", Num 4, Plus(Ident("x"), Num 3))}

All evaluation is mathematical proof. An expression e that provably evaluates to value v is v.

Environment

From a mathematical perspective, an environment is a partial function that maps identifiers to values. Because it’s partial, not all possible identifiers are mapped, so some identifiers are undefined.

The notation for an empty environment is •, which is undefined for every identifier. We can use notation here like in OCaml lists where we can define arguments as either the nil form • or a mapping cons the rest of the environment. Lookup of a value is similarly recursive like an OCaml list. In fact, in an OCaml definitional interpreter, the environment has type (id * value) list where we can search and match mappings like a regular association list.

We can add environment semantics to our previous understandings of judgments like so: A; e ⇒ v. Now, when we use identifiers in expressions, we can search the environment for the equivalent value and substitute it. The environment is the formal way of representing state in an operation.

Conditionals

So far all we’ve done in terms of operational semantics is define variables and perform simple sums on them. One of the greatest powers of programming is the ability to implement control flow: if this, then that. Since we’re programming functionally and not imperatively, these conditionals evaluate to expressions, not just operations, so we must have else as well to account for all cases.

To accommodate this, we must expand our earlier definition of expressions to include equality and the if expression. The if expression takes an equality expression as a conditional, and if it satisfies the boolean argument then the if body is returned as the evaluation, else the else body.

## Tail Recursion in OCaml - URL: https://sharifhsn.dev/blog/tail-recursion/ - Structured data: https://sharifhsn.dev/api/posts/tail-recursion.json - Description: Recursion in many languages can cause significant overhead. It might seem that the excess amount of recursion in OCaml would decrease its performance. But it actually doesn’t,… - Date: 2022-02-15 - Exact published timestamp: 2022-02-15 - Topics: Principles of Programming Languages - Categories: Principles of Programming Languages - Source: Archive - Source URL: None - Content source format: typst

Recursion in many languages can cause significant overhead. It might seem that the excess amount of recursion in OCaml would decrease its performance. But it actually doesn’t, and that’s thanks to tail recursion.

Implementing Reverse

In a functional language with a linked list, it’s not a trivial task to reverse a list. There are multiple ways to do it, and they have different performance implications.

The most basic implementation would just be to concatenate backwards.

let rec rev l =
    match l with
    | [] -> []
    | x :: xs -> (rev xs) @ [x]

Unfortunately, there’s a problem with this. Every time we call rev recursively, we must initialize a new stack frame for every recursive call, leaving the value x in [x] in the calling stack.

rev [1; 2; 3]
 → (rev [2; 3]) @ [1]
 → ((rev [3]) @ [2]) @ [1]
 → (((rev []) @ [3]) @ [2]) @ [1]
 → (([] @ [3]) @ [2]) @ [1]
 → ([3] @ [2]) @ [1]
 → [3; 2] @ [1]
 → [3; 2; 1]

As you can see, there are a total of three stack frames for a list with three elements. This is pretty bad. However, we can rewrite this function to use tail recursion.

let rec rev_helper l acc =
    match l with
    | [] -> acc
    | x :: xs -> rev_helper xs (x :: acc)
let rev l = rev_helper l []

There doesn’t need to be any stack frame here because there are no local variables. The only thing that’s being returned is the function call to rev_helper, so the stack can simply change to the rev_helper call without saving the previous stack frame. This tail recursion is incredibly powerful.

I’ll repeat it to be clear. Tail recursion works when the return value of a recursive function is only the recursive function call. The reason that the previous rev function didn’t work is because the return value was (rev xs) @ [x], so x must be saved in a stack frame in order to remember it.

However, you might have noticed that we had to include a new variable called acc. This accumulator variable is a common pattern for when we want to have tail recursion since the acc is located inside the function call as an argument.

The power of tail recursion can not be understated here. In a typical function, having excessive stack frames can easily cause a stack overflow for a large data set. With tail recursion, we can have both the lack of side effects associated with recursion and the performance associated with iteration.

General Tail Recursion Pattern

let f x =
    let rec aux arg acc =
        if (* base case *) then acc
        else
            let arg' = (* next argument *)
            let acc' = (* updated accumulator *)
            aux arg' acc' in
    aux x (* initial value of accumulator e.g. 0, []*)

## UDP Explained - URL: https://sharifhsn.dev/blog/udp/ - Structured data: https://sharifhsn.dev/api/posts/udp.json - Description: The simplest transport protocol that is used communicate between sockets is known as UDP , the User Datagram Protocol . Let’s say we want to design the simplest transport… - Date: 2022-02-14 - Exact published timestamp: 2022-02-14 - Topics: Internet Technology - Categories: Internet Technology - Source: Archive - Source URL: None - Content source format: typst

The simplest transport protocol that is used communicate between sockets is known as UDP, the User Datagram Protocol.

UDP

Let’s say we want to design the simplest transport protocol possible. All this protocol needs to do is get a message from an application at the socket and send it over the network to the other socket. However, there is one more function that it needs to perform: multiplexing/demultiplexing.

The segment (transport layer version of a datagram) must be directed to a specific socket. There’s multiple sockets that applications communicate with and they all need to be multiplexed together to create a segment, which must then be demultiplexed for the end host to understand what socket it is sent to.

You can think of receiving multiple envelopes in your mailbox, then demultiplexing by reading who each envelope is addressed to and giving it to the member of your household to whom it is addressed. Then, when your household members want to send all of their mail out, you multiplex it into the mailbox.

UDP attaches a header with source and destination port numbers as fields for multiplexing, a length field, and a checksum to protect against packet corruption. Each of these fields is two bytes in length, resulting in a tiny 8-byte header.

## Magnets - URL: https://sharifhsn.dev/blog/magnets/ - Structured data: https://sharifhsn.dev/api/posts/magnets.json - Description: The question of how magnets work has long puzzled many . We will endeavor to answer all of those questions today. Magnets work similarly to electric charges where like poles… - Date: 2022-02-11 - Exact published timestamp: 2022-02-11 - Topics: Physics - Categories: Physics - Source: Archive - Source URL: None - Content source format: typst

The question of how magnets work has long puzzled many. We will endeavor to answer all of those questions today.

Magnetic Fields

Magnets work similarly to electric charges where like poles repel, and unlike poles attract. However, magnets do not ever exist in isolation. Every magnet has a north and south pole, while you can have an isolated positive or negative charge.

Every magnet has a magnetic field around it, similar to the electric field around electric charges. The direction of the magnetic field is shown by the north pole. Every magnetic field points from south to north; this is why compasses point north. They are based off of the magnetic field of the polarity of the Earth’s own magnetic field.

You can apply many of the concepts of electric fields to magnetic fields, such as field lines and the way they curve and indicate field direction.

Magnetic Force

Charges experience electric force in a field, and they experience magnetic force in a magnetic field. However, a few conditions must be met:

The charge must be moving.
The velocity of the charge must have a component perpendicular to the direction of the magnetic field.

To visualize the second rule, imagine the magnetic force like a buffeting wind around the charge. The wind only has any force if it has something to push on. If you hold a piece of paper parallel to the direction to the wind, it’ll barely move because there’s nothing to push.

This is such an important point it’s in its own section. Right Hand Rule No. 1 (RHR1) will tell us the direction of a magnetic force based on the magnetic field and velocity of charge. Your right thumb points in the direction of the charge, and the fingers point in the direction of the magnetic field. The palm faces in the direction of the magnetic force.
This is only applies to positive charges. Negative charges will have the direction of the force as opposite.

The magnitude $𝐵$ of a magnetic field is defined in teslas $𝑇$ by

𝐵 = \frac{𝐹}{| 𝑞_{0} | (𝑣 sin 𝜃)}

where $𝐹$ is the magnitude of the magnetic force on test charge $𝑞_{0}$ with velocity $𝑣$ .

Motion

We have been comparing electric fields and magnetic fields a lot, but in regards to motion through them they are quite different. Electric fields attract towards its directions, while magnetic fields direct in the other perpendicular direction. If left to its devices, a moving charge would perpetually move in a circle remaining perpendicular to the magnetic field.

This means that the work done by a magnetic field is also different. In fact, the magnetic force cannot do work, it can only change the direction of the particle, not its speed nor its energy.

The force required to keep a charge moving in a circle is:

𝐹_{𝑐} = \frac{𝑚 𝑣^{2}}{𝑟} 𝑜 𝑟 𝑟 = \frac{𝑚 𝑣}{| 𝑞 | 𝐵}

Currents in a Magnetic Field

We’ve discussed charges moving through a magnetic field; what about a current? A current is just a collection of moving charges, after all. If we have a wire with a current running through it placed between two magnets, then we can consider the direction of the current as the charge direction in RHR1 to calculate the direction of the magnetic force.

We can use a bit of a trick to get the calculation of magnetic force for a current. We can rearrange the earlier equation for a magnetic field to get the equation for force. Current is the same thing as charge over time, so $\frac{Δ 𝑞}{Δ 𝑡}$ . Length of the current is the same thing as velocity multiplied by time $(\frac{𝑚}{𝑠} \cdot 𝑠)$ . If we multiply these, the $Δ 𝑡$ will cancel out and we will be left with the same expression as $| 𝑞_{0} | 𝑣$ ! Our new equation for force is:

𝐹 = 𝐼 𝐿 𝐵 sin 𝜃

The angle $𝜃$ here maximizes current when perpendicular and is zero when parallel, just as with a single charge.

Torque

As a refresher, torque describes the rate of change of the angular momentum of an object. Current-carrying wires have magnetic force applied to them which causes them to move. If the wire are in a loop, then the magnetic force is applied as torque which causes the loop to rotate. Its resting state is when the normal of the loop is aligned with the magnetic field. You can think of it like a compass needle, which will turn until it reaches its resting state of pointing towards the north pole.

To calculate the torque, we get the force of each side of the loop turning, which is half the width and so half the force. Summed together we get:

𝜏 = 𝑁 𝐼 𝐴 𝐵 sin 𝜃

$𝑁$ here is the number of loops in the wire and $𝐴$ is the area that the loops make. When the loop is parallel with the magnetic field it experiences the greatest torque, and when it is perpendicular it experiences none.

The collective expression $𝑁 𝐼 𝐴$ is known as the magnetic moment of the coil in $𝐴 \cdot 𝑚^{2}$ .

Motors operate using this principle of a coil turning in a magnetic field.

Magnetic Fields in a Current

Current-carrying wires create their own kinds of magnetic fields.

The second Right-Hand Rule (RHR2) is easier to remember, as it’s just a thumbs-up. The thumb points towards the direction of the current, and the fingers curl in the direction of the magnetic field generated.

The magnetic field magnitude is given by the following equation:

𝐵 = \frac{𝜇_{0} 𝐼}{2 𝜋 𝑟}

with $𝜇_{0}$ representing the permeability of free space with the value $𝜇_{0} = 4 𝜋 \times 10^{- 7} 𝑇 \cdot 𝑚 / 𝐴$ . This equation is for an infinitely long, straight wire, which is not necessarily the case.

Because of this property, currents can affect each other magnetically. Currents in the same direction are attrated to each other.

Currents in a loop have a slightly different magnetic field equation to the straight wire:

𝐵 = 𝑁 \frac{𝜇_{0} 𝐼}{2 𝑅}

in the center of the loop, where the field is strongest.

A useful visual for the magnetic field of a current loop is a bar magnet placed in the middle of the loop. If we use RHR2 here, the north pole is on the palm and the south pole is on the back of the hand.

Ampère’s Law

For any current geometry that produces a magnetic field that does not change in time,

Σ 𝐵_{| |} Δ 𝑙 = 𝜇_{0} 𝐼

where $Δ 𝑙$ is a small segment of length along a closed path of arbitrary shape around the current, $𝐵_{| |}$ is the component of the magnetic field parallel to $Δ 𝑙$ , $𝐼$ is the net current passing through the surface bounded by the path, and the $𝜇_{0}$ is the permeability of free space. The symbol $Σ$ indicates the sum of all $𝐵_{| |} Δ 𝑙$ terms must be taken around the closed path.

## SMTP Explained - URL: https://sharifhsn.dev/blog/smtp/ - Structured data: https://sharifhsn.dev/api/posts/smtp.json - Description: Email is the most popular way in the world to send large messages. This is a complicated process which requires its own protocol: SMTP . The Simple Mail Transfer Protocol is how… - Date: 2022-02-10 - Exact published timestamp: 2022-02-10 - Topics: Internet Technology - Categories: Internet Technology - Source: Archive - Source URL: None - Content source format: typst

Email is the most popular way in the world to send large messages. This is a complicated process which requires its own protocol: SMTP.

SMTP

The Simple Mail Transfer Protocol is how user agents communicate with mail servers. On a high level, email works similarly to physical mail. The mail server is like a post office with mailboxes for each user agent. When a person wants to send a message, they put the envelope in their own mailbox. The mailman user agent then takes the envelope to the post office and puts it in their server mailbox. When a person wants to retrieve a message, their mailman gets all the mail in their post office mailbox and delivers it to the person’s personal mailbox. The SMTP post office workers at the server are responsible for moving the mail between mailboxes.

To continue the post office analogy, we can imagine an email being sent between Alice and Bob. Alice lives in Hong Kong and Bob lives in San Juan, so they have different post offices. The workers at the Hong Kong post office see that there is an envelope in Alice’s message queue addressed to Hong Kong. They open a TCP connection as the client to the San Juan server and sends the envelope to the San Juan post office. The workers at San Juan now have an envelope, and they see that it is addressed to Bob. They then place the envelope in Bob’s mailbox. When Bob wants to check his email, he sends his mailman over to the post office, who returns with the email from Alice.

Let’s examine the TCP connection a little more closely. The connection is established on port 25, which is reserved for SMTP. It does some handshaking on the application layer in order to establish the email addresses of the sender and the recipient. The simple exchange goes like this:

server sends 220 with hostname, client responds with HELO and its own hostname
server confirms 250 that it’s ok
client tells server MAIL FROM and RCPT TO addresses and the server confirms each with 250
client sends DATA and server confirms with 354 that it’s ready to receive mail
the client sends the entirety of the message, ending with a lone period to finish the message, which the server confirms with 250
the client can repeat steps 3-5 for any additional messages, then sends QUIT for a server 221 response that closes the connection

vs. HTTP

Both HTTP and SMTP are protocols that transfer files between clients and servers, so there are natural parallels. However, there are also significant differences.

There is a difference in who initiates the connection which defines the two. In HTTP, the requesting client initiates the connection, making it a pull protocol; you can think of the client “pulling” the data from the server. In SMTP, it is the sending client that initiates the connection, making it a push protocol where the client “pushes” the data to the other server.

SMTP also has many restrictions. The message can only be in ASCII, so any data that is sent over that is not in ASCII, like multimedia, must be encoded and decoded. HTTP allows any data to be transferred. A result of this is that HTTP data of different types like multimedia must be in its own response message, while SMTP objects are all in the same ASCII message.

There is some peripheral data that SMTP messages can include in a header, just like HTTP, such as the sender, receiver, and the subject of the email. These are different from commands; this data is part of the actual email, not the protocol!

Mail Access

There’s a part of the post office analogy that has been ignored so far. How exactly does the mailman get to the post office? After all, this is another client-server connection, so there must also be a protocol that defines this interaction. We can use SMTP for the mailman delivering the envelope, as this is the same kind of “push” that the post office uses to send to the other post office. But this introduces an issue for the person on the other end; if the mailman can only push mail it already has to somewhere else, then how does Bob’s mailman get the mail on the server? There are many protocols that can define this interaction, such as POP3, IMAP, and HTTP.

In 2022, HTTP is the obvious choice for this exchange. After all, everyone accesses email through their web browser, which is making HTTP requests anyway. Why not just use an HTTP request to get the mail from the mail server? And as you might expect, almost every email provider today, from Gmail to Hotmail to Yahoo Mail uses Web-based email.

However, back when “user agent” and “web browser” weren’t synonymous, people had applications dedicated to accessing email that couldn’t access the web or use HTTP. In those times, Post Office Protocol—Version 3 and Internet Mail Access Protocol were used.

POP3 is a simple and therefore limited protocol. It operates over port 110 and has three steps: authorization, transaction, and update. The post office first authenticates the mailman by asking him for a username and password before he can check your mailbox. Then, the mailman retrieves the envelope from the post office. While the mailman is at the post office, they can also do some operations like marking emails for deletion and getting statistics for your account. When the mailman leaves the post office, the workers throw away all the mail that the mailman marked for deletion.

It’s common for users to want to organize emails through folders like Inbox, Junk, etc. This is of course possible through user agent, but POP3 does not allow for this organization in your post office mailbox. IMAP provides this as well as many more features. It is stateful so that the user can keep the folders that it creates at the mailbox. One more important feature is its ability to lazily fetch emails from the mailbox, so the user’s entire inbox is not just sent immediately. Imagine if you had to download every email in your inbox in order to access it! I know my connection wouldn’t be able to handle my 1k+ unread emails.

## Higher Order Functions in OCaml - URL: https://sharifhsn.dev/blog/higher-order-functions/ - Structured data: https://sharifhsn.dev/api/posts/higher-order-functions.json - Description: We can do a lot of cool things with functions besides calling them in OCaml. Values are a subset of expressions, as previously stated. All expressions can evaluate to values, but… - Date: 2022-02-08 - Exact published timestamp: 2022-02-08 - Topics: Principles of Programming Languages - Categories: Principles of Programming Languages - Source: Archive - Source URL: None - Content source format: typst

We can do a lot of cool things with functions besides calling them in OCaml.

Anonymous Functions

Values are a subset of expressions, as previously stated. All expressions can evaluate to values, but values are final.

Anonymous functions are also values. Sometimes, it’s more convenient not to create and name a whole new function for our purpose. Anonymous functions are ad hoc functions that exist as values in expressions. They are expressed using the keyword fun.

let y = fun x -> x + 3

This might not seem to have much benefit compared to a full function definition, but it is very useful within let expressions. Since anonymous functions are values, not just expressions, they can be manipulated far more powerfully than even general expressions.

let y = (fun x -> x + 1) 2 in
(fun z -> z - 2) y

This code might seem a little hard to parse, but it’s easier to think about if we rewrite it to use traditional function definitions.

let f x =
    x + 1
let g z =
    z - 2
let y = f 2 in
g y

Now we can tell that y is 3 in the function g, which then evaluates to 1. However, the former code snippet is a much terser way to write this expression if we don’t need the functions f and g anymore.

One good way to think about it is that anonymous functions are to regular functions as literals are to variables. If we only need to use the value "really_long_string" once, we don’t need to store it in a variable. On the other hand, it can be useful to store that literal in a variable s that is much shorter to write. Similarly, if we only need to use the function x -> x + 1 once, we don’t need to store it in a function variable.

In fact, this isn’t even an analogy. Functions are first-class in OCaml, so regular functions are just variables that store anonymous functions:

let f x = body
(* this is sugar for this *)
let f = fun x -> body

And in the same vein, we can name functions within let expressions in an anonymous ways.

let move l x =
    let left x = x - 1 in
    let right x = x + 1 in
    if l then left x
    else      right x
;;
(* same as *)
let move' l x =
    if l then (fun y -> y - 1) x
    else      (fun y -> y + 1) x

Note also that the local variable in the anonymous function doesn’t actually matter to the expression it’s used in; this is a consequence of the shadowing rules of OCaml.

There are several functions in the standard library of OCaml that use higher order functions.

Map

map is a function in the List module of OCaml. Like the name implies, this function maps a function onto every element of a list and returns that list. It has type ('a -> 'b) -> 'a list -> 'b list.

let rec map f l =
    match l with
    | [] -> []
    | h :: t -> (f h) :: (map f t)

This is a simple, yet powerful and useful function. That’s why it is included in the List module, although it’s trivial to write yourself.

Fold

fold is another function in the List module in OCaml, that iterates over a list. The essential idea is that you have an accumulator variable that you want to get based on the values in a list.

let rec fold f acc l =
    match l with
    | [] -> acc
    | h :: t -> fold f (f acc h) t

For example, this is a way to implement a sum function using fold.

let rec sum acc l =
    match l with
    | [] -> acc
    | h :: t -> sum (acc + h) t
sum 0 [2; 5; 100; 53];;
(* same as *)
fold (fun acc x -> acc + x) 0 [2; 5; 100; 53];;

Its type is ('a -> 'b -> 'a) -> 'a -> 'b list -> 'a. We can deconstruct that and understand each part of the function. The initial function f applies the type 'b to 'a and returns 'a. Then for the next two arguments, we keep the same 'a accumulator and iterate over the 'b list.

The use of an accumulator makes fold very versatile, since you can put anything in there.

We can combine map and fold to create the map/reduce framework which can be massively parallelized. We first map a function over our list, then we reduce the list into a single accumulator value.

There is also an alternative version of fold called fold_right that works in reverse, which can be better for certain problems. However, it comes with steep performance cost: every recursive call builds a new stack frame. The original fold is able to optimize this call away by using tail recursion.

## FTP Explained - URL: https://sharifhsn.dev/blog/ftp/ - Structured data: https://sharifhsn.dev/api/posts/ftp.json - Description: Before we had Google Drive, users needed a way to access files from other people quickly and easily. The method for that was FTP . The point of the File Transfer Protocol , was,… - Date: 2022-02-07 - Exact published timestamp: 2022-02-07 - Topics: Internet Technology - Categories: Internet Technology - Source: Archive - Source URL: None - Content source format: typst

Before we had Google Drive, users needed a way to access files from other people quickly and easily. The method for that was FTP.

FTP

The point of the File Transfer Protocol, was, as is obvious, to transfer files between two hosts. Like HTTP, FTP operates between a server and client, each of which have their own filesystem that they want to access. It also creates a TCP connection which has certain verification steps to establish a connection that allows the client to copy files stored in the server or vice versa.

Unlike HTTP however, FTP operates using two parallel connections, the control connection and the data connection. As the names imply, the control connection is used for verification and command information such as which file to get in which directory, and the data connection is the connection used to send the files. This is called an out-of-band connection, unlike HTTP and SMTP, which have the request information in a header in the same connection and are therefore in-band.

The session begins with a control connection initiated by the client to identify the user and commands. The server receives this information, and if it approves, it initiates a data connection with the client. Like HTTP 1.0, this is a non-persistent connection that is only open for the passing of a single file.

The nature of FTP authentication means that FTP must maintain state about the user account, unlike HTTP which is totally stateless by design. This introduces constraints on the number of simultaneous FTP connections.

Commands

FTP commands work like HTTP commands in request headers. They are ASCII and each lines ends with \r\n. Some example commands are:

USER username
PASS password
LIST: works like the ls command for the remote directory, sent over data connection
RETR filename: retrieves the file from the remote directory, over data
STOR filename: stores the file from current directory into remote, over data

Every command has a corresponding reply with a status code, similar to HTTP status codes.

## Memory Virtualization - URL: https://sharifhsn.dev/blog/memory-virtualization/ - Structured data: https://sharifhsn.dev/api/posts/memory-virtualization.json - Description: Accessing physical memory can cause big issues if we do it directly. How can we virtualize the memory like the CPU so we can use it efficiently and safely? Let’s think of a super… - Date: 2022-02-07 - Exact published timestamp: 2022-02-07 - Topics: Operating Systems Design - Categories: Operating Systems Design - Source: Archive - Source URL: None - Content source format: typst

Accessing physical memory can cause big issues if we do it directly. How can we virtualize the memory like the CPU so we can use it efficiently and safely?

Let’s think of a super simple one process system. Our OS is located in the lowest part of memory, and everything after that is reserved for the user. Within an address space, we have the stack-heap structure discussed before. There’s a big problem, that processes can access the memory that the OS is stored in and modify it. That’s not good!

If we want multiple processes as well, we want to give processes memory that is managed by the operating system so they do not interfere with each other. Every process should have a self-contained address space.

We need dynamic allocation from memory as well so that our programs can become very powerful. The stack is not always sufficient for the programs we want to write. The dynamic heap is where all this memory management magic (say that five times fast) happens.

Memory accesses are built into movl instructions in the ISA. There are special operations to dereference the location stored in a register and access memory. These are known as load/store operations.

Memory accesses also happen from the code area by the instruction pointer, though that’s typically not shown. If we want to count memory accesses, there is one fetch for every instruction, and another if the instruction is a load/store.

0x10: movl 0x8(%rbp), %edi
0x13: addl $0x3, %edi
0x19: movl %edi, 0x8(%rbp)

In this example, there are 5 total memory accesses. There is one for every instruction, and each movl is a load/store because one of the operands dereferences a memory pointer.

Typically, we want to reduce this as much as possible because memory access is really slow, and we can’t always rely on the cache.

So, what are some strategies to virtualize memory?

Time Sharing

When we virtualized the CPU, we gave each process the illusion that it had its own CPU that were all running at the same time. The way this was done was through context switches that preserved CPU state in memory. We could do something similar by saving memory to disk when the process isn’t running.

However, there is an immediate problem with this: it would be incredibly slow. Disk I/O even with SSDs is several orders of magnitude slower than DRAM, which is itself orders of magnitude slower than cache/register access. Considering how much processes play around with memory, this would be totally infeasible.

Static Relocation

This is an interesting solution. The instructions stored as static data refer to specific memory locations that are hard coded in the application. However, what we could do is change those pointers and memory to memory that is currently available every time we load the process into memory.

For example, imagine the previous example which has instructions at 0x10, 0x13, etc. We could imagine that those memory locations are no longer available, so the OS changes the static code portion to 0x1010, 0x1013, etc. This means that all jmp and load/store instructions would also have to be rewritten as well since they directly refer to memory locations.

However, this translation is pretty expensive, although it is simple to implement. There are also security concerns, as always. There is no reason that the new memory couldn’t be located somewhere that it can’t go.

Dynamic Relocation

We want the power of static relocation, but we need a way to manually protect each process from each other. This is such an important issue that it is usually provided as a hardware component: the Memory Management Unit (MMU). Whenever a process generates a virtual address in its address space, the MMU takes on the job for translating that address into a real address and giving it back to the process. This is a good example of modularization; we don’t want to have the process to worry about memory, so we offload that job onto a specialized unit designed for that purpose. The MMU is managed only by the OS, so the user can never access it.

There are two general operating modes, as discussed previously: kernel space, and user space. Privileged kernel operations have full power over all of memory and the MMU, and users have to ask the OS for memory through virtual translation.

Address Translation

To start out with, we will make some assumptions about how address spaces are laid out. They are all contiguous spaces in memory of the same size that are smaller than physical memory. As we will see later with segmentation, this assumption will not hold up, but it useful for now.

In order to translate a virtual address to a physical address in memory, we need some kind of reference space in memory. This is given by the base and bounds registers, which give the physical memory locations of where the virtual address space starts and ends, respectively. These registers are not part of the regular ISA and are instead part of the previously mentioned MMU, and operating on them requires privilege. The MMU can also throw an exception upon illegal memory access which is handled by our OS.

Our OS can manage memory in this simple way through the mechanism of a free list, which is a linked list which indicates free spaces, which is also used for malloc.

As discussed before, the address space has a structure where the stack grows downward and the heap grows upward. However, if you look at a diagram of this for more than five seconds, you might notice a giant chasm between the stack and heap. Relying on our assumption of a contiguous space of memory, this is a lot of wasted memory. Some address spaces fix this problem through segmentation.

Segmentation

As the name implies, a segmented address space is split into multiple segments with its own base/bounds pair to indicate its logical beginning and end. We can split every piece of the address space into its own segment, so stack goes in one segment, heap goes in another, etc. This way, although the virtual address space has this nice alignment, we are not wasting physical memory with our sparse address space.

In order to translate a virtual address, we treat the address as an offset with a segment. For example, if we were trying to access a virtual address in the heap, this would be the formula:

𝑝 ℎ 𝑦 𝑠 𝑖 𝑐 𝑎 𝑙 𝐴 𝑑 𝑑 𝑟 𝑒 𝑠 𝑠 = 𝑝 ℎ 𝑦 𝑠 𝑖 𝑐 𝑎 𝑙 𝐵 𝑎 𝑠 𝑒 + (𝑣 𝑖 𝑟 𝑡 𝑢 𝑎 𝑙 𝐴 𝑑 𝑑 𝑟 𝑒 𝑠 𝑠 - 𝑣 𝑖 𝑟 𝑡 𝑢 𝑎 𝑙 𝐵 𝑎 𝑠 𝑒)

Before we allow this translation to take place, however, we must check that the value does not exceed the physical bound. If this is the case, then we have caused the infamous segmentation fault.

A faster way to do with this masks is through bitwise operations. The top two bits of the address refer to the segment, either the code, stack, or heap. The rest of the address is the offset within the segment. This way, we can perform the bounds check before accessing the physical address by checking if the expression in the parentheses exceeds the virtual bounds.

However, one issue with this means that each segment gets the same maximum size, which is $2^{𝑜 𝑓 𝑓 𝑠 𝑒 𝑡}$ . If we want a bigger heap, we’re out of luck. We can solve this by tracking instructions instead of bits. For example, an instruction fetch for %rip will come from the code segment, so we don’t need to put that in the address that it is in the code segment.

Another issue is that the stack segment grows downwards instead of forwards, which means that its segment works differently. We need an extra bit for segments that indicates whether they grow forwards or backwards, which will be set to 0 for the stack and 1 for everything else.

If that bit is false, then we subtract the offset from the base instead of adding it as above.

Sharing :)

Sometimes, processes share the same code. It seems wasteful to copy the same code segment every time we have a new process, so modern operating systems implement sharing for code segments. Sharing can be implemented for any segment, but it is most common for code since it’s read-only. This is important for dynamically linked libraries since many processes will access the same library, like libc.

Hardware adds extra protection bits to each segment indicating its rwx value similar to a file. If a segment is indicated to be r--, then the OS can secretly share the segment between multiple processes, assured that they will only read it. This concept of a read guard allowing for multiple shared references as opposed to a write guard which only allows for one mutable reference will become very important when we discuss concurrency.

We have only been working on a few segments so far, but segmentation could theoretically be extended to as many segments as you want. In order to have fine-grained segments, a segment table is needed to quickly access thousands of segments.

There are a few more things that the OS needs to do in order to support segmentation. One is that it must preserve all base-bounds pairs upon a context switch, since the location of the address space is now more complicated. Another is that the growing and shrinking of segments must be managed through sbrk-like system calls that shift the bounds of the heap/stack. The final and most important issues that allocating variable size address spaces runs into the same problems of malloc where external fragmentation is difficult to avoid. Like with malloc, there is no perfect solution, ranging from algorithmic free lists to compact memory which rearranges segments every time a new one is created.

## Lets, Tuples, and Records in OCaml - URL: https://sharifhsn.dev/blog/lets-tuples-records/ - Structured data: https://sharifhsn.dev/api/posts/lets-tuples-records.json - Description: Lists are the most basic data structure in OCaml, but there are others that we need to be aware of. We have seen the keyword let used to define expressions and store values.… - Date: 2022-02-03 - Exact published timestamp: 2022-02-03 - Topics: Principles of Programming Languages - Categories: Principles of Programming Languages - Source: Archive - Source URL: None - Content source format: typst

Lists are the most basic data structure in OCaml, but there are others that we need to be aware of.

Let Expressions

We have seen the keyword let used to define expressions and store values. However, the same keyword can be used to create expressions which bind variables in other expressions. The let statements we used before do not evaluate to any value, while let expressions do evaluate.

let x = 5 in x * 3

The in keyword gives us a clue on what’s going on. You can think of it almost like a function where we replace variables in the inner expression with the values in the outer expression. The expression above evaluates to 15, because it is x * 3 in which x is 5.

We can type check this expression where x has the same type of the binding expression.

If you omit in, you can think of that as a let expression which is bound in the global scope instead of the scope of the body expression.

let x = 37;;

In this statement x is defined as 37 in the global scope and it can be used elsewhere.

I’ve used the word scope a lot here, so let’s define that a little more concretely. In the above let expression, the variable x is not visible in any other part of the program. Let’s imagine that we have both of these lines together, the let expression and the let statement. They’re both named x, so what would the value be?

We can imagine evaluating expressions right to left, and upon encountering a variable, act like it is a pointer to an expression in the outer scope. In the innermost scope, the expression is x * 3. This expression has no meaning because x is a variable, so we back up one scope and check if x is defined. And in fact, it is! So we will replace the x in that expression with 5. Note that even though x is defined as 37 in the global scope, the inner scope shadows the global scope.

Shadowing refers to when a variable name is rebound in an inner scope to have a different meaning. Some languages, such as Java, do not allow you to do this because of possible confusion. However, it is sometimes useful to use the same name for different things, so languages like C and OCaml permit shadowing.

You can also use let expressions inside a function, and this is often good style because it clarifies constants:

let area d =
    let pi = 3.14 in
    let r = d /. 2.0 in
    pi *. r *. r

Much better than C #define, right?

OCaml does not permit you to mutate variables. However, you can simulate this by shadowing a variable with a new value:

let x = 0;;
x = x + 1;; (* not allowed! *)
let x = x + 1; (* allowed, but discouraged *)

This is kind of an ugly hack so you should avoid it in real code, though it is technically possible under OCaml’s rules.

We can nest let expressions, but this is generally bad practice like shadowing. Realistically, it’s usually better to just write linear expressions.

let expressions don’t just have to use a plain variable. We can also use patterns to bind expressions, and if the binding expressions fails to match the pattern then we have an exception. This is useful when we want to extract a particular value from an expression.

let [x] = [1] in 1 :: x (*  evaluates to [1; 1*)

Tuples

Tuples represent collections, like lists, but they contain a fixed amount of values. The tradeoff is that they can be heterogeneous, which means they can have multiple types. The type of a tuple is the type of each of its component, separated by asterisks.

(1, 2) (* int * int *)
(1, "string", 3.5) (* int * string * float *)

Because each tuple has a distinct type, a list of tuples can only have one type of tuples in it.

Tuples lend themselves particularly well to pattern matching. Instead of having multiple function arguments like is typical in OCaml, we can have one argument that is a tuple and then pattern match it in order to destructure it. This is also a convenient way to return multiple variables from a function, which is otherwise not allowed.

Records

Each element of a tuple is referenced by its position. Sometimes, we want to reference elements by name, like in a dictionary. For this use, we use records. Records are a distinct type that must be pre-defined before being used.

type date = { month: string; day: int; year: int }

Now, we can construct records by using the same brace notation but giving each name a value.

let today = { day=3; year=2022; month="f"^"eb" };;

You might notice that this has a similar syntax to C-style structs. The fields can be accessed through . syntax as in today.day, and the order of the struct assignment doesn’t matter.

Records can also be conveniently destructured, like all other data structures.

## Circuits - URL: https://sharifhsn.dev/blog/circuits/ - Structured data: https://sharifhsn.dev/api/posts/circuits.json - Description: Circuits are the method by which all electricity travels throughout the world and power every single device in existence. It is therefore crucial to understand them while… - Date: 2022-02-01 - Exact published timestamp: 2022-02-01 - Topics: Physics - Categories: Physics - Source: Archive - Source URL: None - Content source format: typst

Circuits are the method by which all electricity travels throughout the world and power every single device in existence. It is therefore crucial to understand them while understanding electricity.

Electromotive Force (emf) and Current

The emf is similar to voltage, but not quite the same. It is the maximum voltage of a battery, signified by $𝐸$ (capital epsilon). It represents the maximum amount of joules of energy that can transfer from a battery based on the charge.

Current is formed when an electric field parallel to a wire moves free electrons from the positive terminal of a battery to the negative terminal. You can think of it mentally like the flow of charge over time, like water going through a hose, and in fact the equation is $𝐼 = \frac{Δ 𝑞}{Δ 𝑡}$ . For a non-constant flow, this is the average current. Current uses the unit $𝐴$ for ampere.

A direct current (dc) has electrons that go through the circuit in the same direction at all times, which is negative to positive. An alternating current (ac) alternates direction constantly from moment to moment.

Very important note: electron flow and conventional current $𝐼$ are not the same thing. Conventional current actually goes opposite to electron flow for historical reasons. When we draw current over circuits, we use conventional current notation, so current flows from positive to negative. This is because it was once thought that circuits worked by the passing of positive charges, not negative charges.

Ohm’s Law

The ratio $𝑉 / 𝐼$ is a constant, where $𝑉$ is the voltage applied across a piece of material (such as a wire) and $𝐼$ is the current through the material:

\frac{𝑉}{𝐼} = 𝑅 = 𝑐 𝑜 𝑛 𝑠 𝑡 𝑎 𝑛 𝑡 | 𝑉 = 𝐼 𝑅

$𝑅$ is the resistance of the piece of material in ohms ( $Ω$ ).

Again using our hose analogy, resistance is like how narrow the hose opening is which lets our water through. Resistance is typically discussed in the context of resistors, which are electrical devices that apply resistance to a circuit.

Resistance and Resistivity

The resistance of a piece of material is dependent on certain qualities. If you think in your head of a small plastic tic-tac as our resistor, this is the formula:

𝑅 = 𝜌 \frac{𝐿}{𝐴}

where $𝐿$ is length, $𝐴$ is cross-sectional area, and $𝜌$ is a constant called resistivity that is constant to a material. Conductors have very low resistivities, and insulators have very high resistivities.

Resistivity also typically depends on temperature, though in more complicated ways than can be expressed in the above formula. The most common way to express it as:

𝜌 = 𝜌_{0} [1 + 𝛼 (𝑇 - 𝑇_{0})]

You can also use resistance instead of resistivity here. The important constant here is $𝛼$ , which is the temperature coefficient of resistivity, which can be positive or negative depending on how the specific material relates resistivity to temperature.

Some materials, known as superconductors, have resistivity of zero at certain temperatures. This means you can circulate a current indefinitely within that circuit without needing a supply of emf from a battery. They can be used in MRI, maglevs, and computer chips.

POWER

Power is an important concept when we want to give energy to electrical appliances. The easiest way to think about it on a high level is the change in energy per unit time. Another way to think about it is the flow of voltage transferred.

𝑃 = \frac{𝐶 ℎ 𝑎 𝑛 𝑔 𝑒 𝑖 𝑛 𝑒 𝑛 𝑒 𝑟 𝑔 𝑦}{𝑇 𝑖 𝑚 𝑒 𝑖 𝑛 𝑡 𝑒 𝑟 𝑣 𝑎 𝑙} = \frac{(Δ 𝑞) 𝑉}{Δ 𝑡} = \frac{Δ 𝑞}{Δ 𝑡} 𝑉 = 𝐼 𝑉

The units of power are in watts $𝑊$ , which you might remember from lightbulbs. Current multiplied by voltage makes sense; it’s the flow multiplied by the quantity, which tells you change in quantity over time. You can also write it as:

𝑃 = 𝐼 𝑉 = 𝐼^{2} 𝑅 = \frac{𝑉^{2}}{𝑅}

which is often useful when we don’t have one of those elements.

Alternating Current

Most batteries in the world use ac instead of dc. Therefore, we must note the differences between it and dc. The voltage is not always the same, and fluctuates constantly according to this equation:

𝑉 = 𝑉_{0} sin 2 𝜋 𝑓 𝑡

where $𝑉_{0}$ is the maximum voltage and $𝑓$ is the frequency of isolation. This value is in radians when the sine function is applied.

Current oscillates at the same rate, and so does power. We can simply substitute the above equation for $𝑉$ to evaluate those.

We often want to understand average current and average voltage in relation to power, which are calculated by dividing the maximum of each by $\sqrt{2}$ . We can use these two values to get average power through the analogous power equations.

Series Wiring

All the circuits we’ve discussed have one device with a straight wire. Often we want to have multiple devices on the same circuit.

When we have multiple resistors in series, their resistances added up to get the total resistance of a circuit. A very useful method to solve these kinds of problems is to “combine” resistors by adding up their resistances. This becomes helpful when we start working with parallel wiring.

Another way we can think about voltage in this context is that we have voltage that goes through each resistor which is added up to get the final voltage. This will be relevant later when we talk about voltage drops.

Parallel Wiring

Parallel wiring is wiring where the voltage across each device is the same. Now, this might not necessarily make sense; wouldn’t the voltage be divided still? The element that is divided in parallel is actually the current, whereas in series it is the current that is preserved and the voltage that is divided. This is useful when we want multiple appliances connected to the same power source that do not interfere with each other with respect to voltage.

Parallel resistors, paradoxically, act as if they have less resistance. Instead of adding up the resistances straightforwardly, parallel resistances add up as so:

\frac{1}{𝑅_{𝑝}} = \frac{1}{𝑅_{1}} + \frac{1}{𝑅_{2}} + . . .

A more convenient way to calculate this is

𝑅_{𝑝} = \frac{𝑅_{1} \cdot 𝑅_{2}}{𝑅_{1} + 𝑅_{2}}

for only two circuits, though it gets more complex for more.

When we have multiple parts in series and parallel, the way to solve comes by combining different series and parallel parts together. Mind that the combinations must be only series or only parallel. The easiest way is to start furthest from the capacitor and work your way there.

Internal Resistance

We’ve separated devices into groups of resistors and conducting wire. However, this is a false dichotomy. All materials have some amount of internal resistance which resists current. Typically, batteries and wires have extremely low resistance to facilitate current.

Kirchhoff’s Rules

If resistors are in series or parallel, we can combine them fairly easily. There’s trouble when we can’t do that, however. We must use Kirchhoff’s Rules, specifically the junction rule and the loop rule to calculate resistance in these cases.

The junction rule states that the total current directed into a junction must equal the total current directed out of the junction. This is just an extension of conservation of electric charge that we discussed earlier. You can imagine an intersection of cars where you have four cars waiting at the light. They can either turn right or go straight. Either way they go, however, there must be four cars that are leaving the intersection unless something has gone horribly wrong.

The loop rule relates a similar idea but in relation to electric potential. The voltage rises at every capacitor, and it must have drops through every resistor that equal the rise through the resistor. For example, after a 12 V rise through a capacitor, a pass through a $5 Ω$ and a $1 Ω$ resistor must drop 10 V and 2 V respectively in order to have it be the same.

Some important notes:

the choice of direction is arbitrary, we will simply end up with a negative current if it is wrong
we must indicate the positive and negative directions of every capacitor/resistor in order to find our voltage drops/rises
our final equation will have the drops on one side and the rises on the other side, with every term either being voltage or a resistance multiplied by the variable $𝐼$
the answer is $𝐼$ which is solved for through normal algebra

In laboratory, there are various devices we use to measure current and voltage. You should remember that current is only the same in series and voltage is only the same in parallel. Therefore, ammeters must be connected in series and voltmeters in parallel.

Capacitors in Series/Parallel

Capacitors work opposite to resistors when they are in series and in parallel. They add up charge simply when placed in parallel, and have the inverse reciprocal relationship when placed in series. The reason this happens is because charge and capacitance are directly related while current and resistance are inversely related.

One note is that with regards to the charges on their plates, capacitors in series always have charges of the same magnitude because they will equalize out.

RC Circuits

We often have circuits that have both resistors and capacitors. In this situation, the capacitor’s time to charge depends on the resistance:

𝑞 = 𝑞_{0} [1 - 𝑒^{- 𝑡 / (𝑅 𝐶)}]

$𝑅 𝐶$ here is very simply, the resistance multiplied by the capacitance, which ends up being a unit in seconds known as $𝜏$ , the time constant. That is the amount of time for the capacitor to charge to 63.2%, which is $1 - 𝑒^{- 1}$ .

Discharging works with a similar equation:

𝑞 = 𝑞_{0} 𝑒^{- 𝑡 / (𝑅 𝐶)}

where $𝜏$ represents losing 63.2% charge.

## DNS Explained - URL: https://sharifhsn.dev/blog/dns/ - Structured data: https://sharifhsn.dev/api/posts/dns.json - Description: We need protocols to define the way that messages are interpreted. Messages are different based on what kind of information you are sending. For example, HTTP (HyperText… - Date: 2022-01-31 - Exact published timestamp: 2022-01-31 - Topics: Internet Technology - Categories: Internet Technology - Source: Archive - Source URL: None - Content source format: typst

Protocols

We need protocols to define the way that messages are interpreted.

Messages are different based on what kind of information you are sending. For example, HTTP (HyperText Transport Protocol) defines the way that webpages are served. Otherwise, the bytes that are sent between a client and a server are meaningless.

Identification

We can identify hosts using IP addresses and ports.

With cell phones, we use telephone numbers to identify each other and call each other. Sometimes, we can store names in contacts associated with the phone numbers that are easier to remember than unique numbers.

IP addresses work the same way. Each IP address uniquely identifies a host then can send and receive information through a network. IPv4 is a 32-bit number that was traditionally used, however due to its limited scope (only $2^{32} \approx 4$ billion possible IP addresses). IPv6 is the new type of address that is 128-bit number with $2^{128}$ possible addresses.

IPv4: 128.6.24.78

IPv6: 2001:4000:A000:C000:6000:B001:412A:8000

There can be more than one application running on a host, however. They can’t interfere with each other, so we need a way to distinguish them. This is done via port numbers, which are 16-bit numbers that can be bound to applications that communicate over the network. Some port numbers are reserved for special purposes, such as port 25 for email, port 80 for HTTP, port 443 for HTTPS. Port numbers are like different employees that work at a call center.

Both client and server must identify each other’s IP address and port numbers: a 4-tuple: $𝑆_{I P}, 𝑆_{P #}, 𝐷_{I P}, 𝐷_{P #}$ . This tuple is a uniquely defined bidirectional connection.

The OS manages a simple data structure of port numbers to track which are in use, no complexity involved.

Client-Server Architecture

Many clients communicate with one always-on server.

A server is a host that is always on and has a permanent IP address that is often public. For large servers, it is often necessary to use server farms that distribute servers geographically for faster communication across the globe.

A client is one of many hosts that makes a connection with the server to access data on the server. They might connect intermittently or have dynamic IP addresses. Clients do not communicate with one another directly, instead using the server as a medium.

For example, a website server might listen on a fixed, public IP address at port 80, waiting for a HTTP request. A client might request a webpage from this server and make a connection with its own IP address and port 80. Many clients will all have different connection tuples because of their own unique IP addresses. This way, many clients can make connections without confusing the server.

One domain name can map to different IP addresses. This is how google.com can be split across thousands of servers that serve billions of requests every day to the exact same domain. The domain name can be thought of as a contact with saved name that might have many phone numbers.

DNS

Domain names are converted to IP addresses through the DNS.

DNS is the Domain Name System which acts as a service to resolve IP addresses from a normal alphanumeric domain name, just like a telephone book. The DNS is itself an online service that must be accessed through a network.

The DNS listens on port 53 and the IP address is well-known. A simple DNS might be a large centralized database of names and IP addresses, around 4 billion for IPv4.

However, if this DNS crashes, that would destroy the whole Internet. Also, everybody on the Internet would be trying to access this server, which place tremendous load. It would be a big security issue and a large target for attack, both physical and virtual. Lookup in a server with billions of entries would be slow. Latency for hosts that are physically far away would be greatly increased depending on the server location. Every new host would need to be entered in the same location, which would be slow. This does not scale!

Real DNS is implemented as a distributed service across different countries. Top-level domains like .com, .org, .edu, are most popular and are the main domain names that people access. Each country also has it’s own top-level domain server named after it, like .de, .uk, .be. Many websites nowadays utilize country codes for special names that include the domain name e.g. youtu.be.

Second-level servers are the most commonly accessed domain names, like amazon.com, google.com, etc. These names are separated by periods, from right to left.

Larger domain names like universities can be further namespaced, like cs.rutgers.edu. Rutgers University has its own DNS server that manages these additional domain names. The final name that contains the actual data is called the authoritative domain.

## Packet Switching vs Message Switching - URL: https://sharifhsn.dev/blog/packet-switching/ - Structured data: https://sharifhsn.dev/api/posts/packet-switching.json - Description: How do we get the best performance in sending information over the internet? What is packet switching and how can it help? Performance is bottlenecked by propagation delay and… - Date: 2022-01-31 - Exact published timestamp: 2022-01-31 - Topics: Internet Technology - Categories: Internet Technology - Source: Archive - Source URL: None - Content source format: typst

How do we get the best performance in sending information over the internet? What is packet switching and how can it help? Performance is bottlenecked by propagation delay and transmission time.

Propagation delay is dictated by the physical distance between the client and server. If you were to send a message to server on Mars, there would be 200 million km to travel across, aka 1000s at the speed of light! New York to Los Angeles is about $20 m s$ at this same speed of $5 𝜇 s / k m$ .

Transmission time is dictated by bandwidth, which tells you how many bytes per second you can send. A $1 M B / s$ link speed will take $1 m s$ for $1000 B$ . The reason this happens is because as soon as the first bit is propagated, all the rest of the bits will follow which will be concentrated by bandwidth, $1 𝜇 s$ after each other with this link speed.

For small messages in the tens or hundreds of bytes like handshake acknowledgements, propagation delay is very significant. For larger messages, propagation delay is ignored and bandwidth is considered more significant.

$𝑅 𝑇 𝑇 = 2 𝑃 𝐷$ : Round-trip time is equal to twice the propagation delay.

Packet switching is faster because it is a continuous stream.

Assume 1000B/s link speed between sender and destination. If we do message switching, it is slow. A 1000B would take 1s (duh) to go from sender to router. Similarly, it would take another second to go from router to destination, so 2 seconds in total.

Let’s assume we split our 1000B message into 100B packets. The first packet will take 0.1s to arrive to the router and another 0.1s to arrive at the destination, so 0.2s. Each packet follows directly after the other, so the second packet takes only 0.1s total because it’s already at the router by the time the first packet is at the destination. In total, the message takes 1.1s to travel.

These benefits compound when messages are passing over multiple routers, because only the first packet has to deal with the overhead of the link speed for all the routers. All the rest of the packets will only take the time for the link speed between the last router and the destination.

However, bottlenecks can still occur when link speeds between routers are different. The weakest link will bottleneck the connection even in packet switching.

## Scheduling - URL: https://sharifhsn.dev/blog/schedulers/ - Structured data: https://sharifhsn.dev/api/posts/schedulers.json - Description: The way that the operating system decides which processes to run and when is a complicated process known as scheduling . This is how it works. We should understand a few… - Date: 2022-01-31 - Exact published timestamp: 2022-01-31 - Topics: Operating Systems Design - Categories: Operating Systems Design - Source: Archive - Source URL: None - Content source format: typst

The way that the operating system decides which processes to run and when is a complicated process known as scheduling. This is how it works.

We should understand a few vocabulary words before we discuss schedulers in detail.

job: an execution stream for a certain amount of time (not the same as process!)
workload: the jobs that must be executed
latency: time taken for one operation
throughput: number of total operations

Metrics

We can design schedulers that are designed to minimize certain metrics. No scheduler is perfect at everything, so we need to prioritize what metrics matter.

turnaround time: how long does the job take to complete?
- $𝑐 𝑜 𝑚 𝑝 𝑙 𝑒 𝑡 𝑖 𝑜 𝑛 𝑇 𝑖 𝑚 𝑒 - 𝑎 𝑟 𝑟 𝑖 𝑣 𝑎 𝑙 𝑇 𝑖 𝑚 𝑒$
response time: how long does the job take to start? e.g. game keystrokes
- $𝑠 𝑐 ℎ 𝑒 𝑑 𝑢 𝑙 𝑒 𝑇 𝑖 𝑚 𝑒 - 𝑎 𝑟 𝑟 𝑖 𝑣 𝑎 𝑙 𝑇 𝑖 𝑚 𝑒$
waiting time: how long does the job wait in the ready queue?
- same as response except when killing procs
throughput: jobs completed per unit time
resource utilization: manage small resources well e.g. battery
overhead: how strenuous is the scheduler itself?
fairness: how well is CPU time shared between jobs?

FIFO

FIFO stands for First In, First Out, which is a fairly self-explanatory title. A FIFO scheduler does jobs in the order that they arrive. This kind of scheduler is trivial to implement. However, we can run into issues if say, the first job takes a long time to complete. The other jobs that would complete much quicker are waiting even though it would be better if we could just get those out of the way first.

SJF

SJF stands for Shortest Job First, which again is self-explanatory. This is better than FIFO, but relies on the OS having oracle-like knowledge of all the jobs that will come, since a shorter job could come later that the scheduler can’t factor into its calculations.

These schedulers only work when you have access to all the jobs at once, which is almost never the case in real-world workloads. We need to use some kind of preemptive scheduling which will switch and split jobs as necessary.

STCF

STCF stands for Shortest To Completion First. It is similar to SJF, but it recalculates the job closest to completion every time a new job arrives. This way, if a super short job arrives while a long job is executing, STCF can switch to it and complete it quickly before finishing the long job, which improves completion time. This expands the machinery needed in the scheduler but has much better outcomes.

RR

RR stands for Round Robin. This approach is fairly different than the the previous two schedulers because it does not make any attempt to optimize for the shortest time. Instead, RR optimizes for fairness by executing every single job on the same exact time intervals. So it will run 10 ms of Job A, then 10 ms of Job B, then 10 ms of Job C, etc. in the fairest way possible regardless of the actual length of those jobs.

Preemptive scheduling is pretty cool, but we’re still treating all our jobs like they’re the same. If we introduce the notion of priority in jobs then we can use it for real-time jobs.

MLFQ

MLFQ stands for Multi-Level Feedback Queue. The implementation is similar to RR but it works on multiple priority levels. There are two job types: interactive and batch. Interactive processes are those like games and text editors that need immediate response and therefore low response time. Batch processes are those like daemons that are lower priority and care more about general turnaround time than response time.

For our implementation, we have multiple priority queues:

if a.priority > b.priority {
    a.run();
} else if a.priority == b.priority {
    rr(a, b);
}

MLFQ prioritizes nice processes. This is a technical term that means that a process is satisfied with getting a response and is willing to turn over the CPU as needed without the OS needing to force it to.

All jobs begin by having top priority, but if a job takes too long on the RR, then you demote it to a lower priority level. This way, smaller jobs that are contained within RR are executed quickly at top priority.

However, this system is incredibly easy to game. Processes control the jobs that they send to the CPU so you can split up jobs exactly aligned to RR to execute all the jobs at top priority even though you don’t actually need it.

Lottery

The lottery system is fairer, just like a real lottery. Every process gets a certain amount of lottery tickets associated with it, scaling with priority. Whichever process wins the lottery gets to run its job. This way, higher priority jobs have a higher probability of running than lower priority jobs, but it’s protected against gaming the system. The number of tickets that a process gets represents its CPU usage.

int counter = 0;
int winner = getrandom(0, total_tickets);
node_t *curr = head;
while curr {
    counter += curr->tickets;
    if (counter > winner) {
        break;
    }
    curr = curr->next;
}

Multiprocessing

Every scheduler we’ve discussed so far assumes only a single CPU can execute jobs, which was true for a long time. Now, however, multicore CPUs are becoming increasingly common in consumer electronics. How do we schedule jobs among different CPUs?

Cache affinity and coherence can become issues. CPUs have a cache located next to them different from main memory where commonly used memory is stored in order to speed up operations. Every core has its own caches, so what happens when multiple cores execute the same stream? They have to have cache coherence so that we don’t end up with bugs. The way to fix this is to copy caches between cores.

Obviously, this is pretty inefficient. We want to avoid this as much as possible by having each process run on its own CPU thread as much as possible; this is cache affinity. Basic FIFO scheduling is simple but is not good at preserving affinity. Having a scheduler that preserves affinity at the cost of some other inefficiencies can actually cause massive speedups because of the importance of cache affinity. However, this machinery can be complex.

CFS

The Completely Fair Scheduler is the scheduler that Linux actually uses. Instead of using time-slices to manage job usage, the scheduler assigns processes a proportion of the CPU. The fairness is absolute because each thread gets an absolute amount of CPU running. This also fixes multiprocessing issues because the scheduler is based around CPU cores instead of time. However, the switching rate might change a lot because we are no longer looking at time, and as we discussed context switching has its own costs that we might want to minimize. This is a tradeoff made for fairness.

I mentioned niceness earlier. CFS uses a more complex version of niceness which ranges from −20 to 19, with the default of 0, and higher values being worse. There is also a separate priority called real-time priority ranging from 0 - 99 which always execute before nice processes, regardless of how nice they are.

In order to calculate RR, we take the target latency e.g. 20 ms and split the time across the tasks equally among process of the same priority. However, because this is based around CPU and not constant time, this can be extremely fast. A tree data structure is used for priority in order to quickly get the highest priority and least time jobs.

Target Latency is the minimum amount of time required to get a task at least one turn on the processor. Within this window, every process gets some CPU. Minimum Granularity imposes small unfairness in CFS. There is a floor on the timeslice of 1 ms regardless of how much CPU we have. This means that for a large amount of jobs, smaller jobs get unfairly good treatment.

## Lists in OCaml - URL: https://sharifhsn.dev/blog/lists/ - Structured data: https://sharifhsn.dev/api/posts/lists.json - Description: Lists are the most basic data structure in OCaml. The most analogous structure is a vector in C++ or Rust. Lists are homogeneous and of arbitrary length . The most basic list is… - Date: 2022-01-27 - Exact published timestamp: 2022-01-27 - Topics: Principles of Programming Languages - Categories: Principles of Programming Languages - Source: Archive - Source URL: None - Content source format: typst

Lists are the most basic data structure in OCaml. The most analogous structure is a vector in C++ or Rust. Lists are homogeneous and of arbitrary length.

The most basic list is nil, the empty list: []. We can prepend elements to a list through the cons operator: ::. Every display of a list is actually just syntactic sugar for every element being cons with the empty list:

[1, 2, 3] = 1 :: 2 :: 3 :: []

Importantly, lists are immutable! When they are created, they cannot be changed. If you want to add an element to a list, you must construct a new list based on the old one.

This might seem inefficient, to create a new list every time we want to change it. But the semantics of a language do not necessarily correspond to its compiled execution. In particular, the invariant that lists are immutable can lead to optimizations where a compiler doesn’t need to keep track of mutated information.

It is convention to call the right hand side the tail and the left hand side the head. The cons operator has an element on the left side and an list on the right side.

Let’s think of lists in terms of expressions. [] is a value like 0 or 1. The cons operator evaluates the left expression to an element and the right expression to a list of type element. Like with function arguments, the order of this evaluation is irrelevant as long as we don’t introduce side effects.

Lists can also have the polymorphic type 'a. This is similar to a generic type in languages like Java, C++, and Rust. An 'a list can have any type. This is most useful when it comes to function types. If the function type accepts or returns a 'a type, then it is a generic function that can be used on a list of any type. Nil is a 'a list.

You can also have nested lists like this:

let m = [[1]; [2; 3]] ;; (* int list list *)

Unlike tuples, these lists do not have to be the same size. They do need to have the same type, however, so this is a list of int lists, or an int list list.

The cons operator does not work to concatenate two lists. It can only add one element to an existing list. This is a very common bug (at least for myself). OCaml has a special sugar operand @ to concatenate lists, although it is technically an ordinary function.

Pattern Matching

Pattern matching is an extremely important concept in OCaml for functional programming. It allows for powerful destructuring of enums, collections, and other complex objects.

let head l =
    match l with
    | (h :: _) -> h

This function head here matches the list l where there is an element h that is cons with an arbitrary expression. The _ indicates that some expression must be present, but we don’t care about what expression it is, and in fact we are throwing it away. h is also an arbitrary expression, but we are using it as the return type.

This match statement is not exhaustive. An exhaustive match means that there is an evaluation for every possible case of l. Here, there is no case for when l is nil. This does not work in OCaml; every match must be exhaustive. Here is an example of an exhaustive match:

let rec sum l =
    match l with
    | [] -> 0
    | h :: t -> h + sum t

This sum function works for every case of l. Either it is nil, or it is a head cons a tail. You will also notice here that the two cases resolve to the same type. This is another requirement for match statements: all patterns must evaluate to the same type, which is not necessarily the type of the initial expression. Here, the initial expression has the type int list and the return expression has the type int. This function only works for int list. This restriction is not present for our head function because h is polymorphic type 'a.

Pattern matching is generally much better than its alternatives. OCaml will warn you if your function is non-exhaustive, and it will throw an exception for an unhandled case. Also, duplicated cases are easy to avoid because OCaml will give a similar warning for unused cases.

Recursion with Lists

In order to manipulate lists in any significant way, we need to use recursion. Lists are immutable, so functions over them must be recursive.

let rec length l =
    match l with
    | [] -> 0
    | (_ :: t) -> 1 + length t

We can think of this as recursive because we have our base case of nil and our iterative step of applying the function to successive tails. This function does not use h, but we might, as in the sum function above. Recursion might seem like it incurs overhead here, as in C-like languages recursion typically costs stack frames. However, OCaml knows that even though you are writing your code recursively, all you’re doing is iterating over an immutable list, so it will elide those issues. Such recursive functions are called tail-recursive and are essential to the use of functional languages.

However, these functions work through the lists forward, like a linked list. How can we do something like reversing a list?

let rec rev_aux l acc =
    match l with
    | [] -> acc
    | x :: xs -> rev_aux xs (x :: acc)
let rev l = rev_aux l []

The key here is the variable acc, which is known as an accumulator. This is a variable that is included with each recursive function call that accumulates operations on it. This is a way of simulating side effects in a controlled way which is often useful. acc will accumulate heads while the tail grows smaller, and the actual rev function will get the accumulated nil that it passed to rev_aux.

## Electric Potential - URL: https://sharifhsn.dev/blog/electric-potential/ - Structured data: https://sharifhsn.dev/api/posts/electric-potential.json - Description: Besides electrical charge and field, there’s another force that is relevant to electricity: electric potential . Gravity is similar to the electrostatic force, apart from the… - Date: 2022-01-25 - Exact published timestamp: 2022-01-25 - Topics: Physics - Categories: Physics - Source: Archive - Source URL: None - Content source format: typst

Besides electrical charge and field, there’s another force that is relevant to electricity: electric potential.

Potential Energy

Gravity is similar to the electrostatic force, apart from the fact that it acts on mass instead of electric charge. The equations are practically identical. We’re familiar with the idea of gravitational potential energy; why not apply the concept to charge as well?

Recall that work is expressed as the difference between potential energy. The equation for work done by an electric force is:

𝑊_{𝐴 𝐵} = 𝐸 𝑃 𝐸_{𝐴} - 𝐸 𝑃 𝐸_{𝐵}

where $𝐴$ and $𝐵$ are the positions that the charges are in.

Electric Potential Difference

Remember that the electric force depends on the small test charge $𝑞_{0}$ . If we divide all parts of the previous equation by $𝑞_{0}$ , then we get the work per-unit-charge.

For the first time, we’re going to be talking about electric potential, expressed in volts $(𝑉)$ . This is different from electric potential energy, which is expressed in joules ( $𝐽)$ . The equation for electric potential derived from EPE is $𝑉 = \frac{𝐸 𝑃 𝐸}{𝑞_{0}}$ , showing that volts are the same as joule/coulomb.

$𝑉$ can only truly be measured as $Δ 𝑉$ , because it is relative measurement in terms of work, similar to energy. This potential difference $Δ 𝑉$ is known as voltage. When we talk about voltage in common parlance, that is the electric potential difference between two charges, typically the terminals of a capacitor.

We can think of positives and negatives here like gravity. Because electricity flows from positive to negative, that’s the direction of “gravity”. Positive charges accelerate to lower electric potential like massive objects accelerate to lower heights. Negative charges oppose this acceleration like the normal force to higher electric potential.

You might have also seen the unit $𝑒 𝑉$ , $𝑀 𝑒 𝑉$ , or $𝐺 𝑒 𝑉$ with reference to energy. This is the unit electron-volt, which is, as implied, the charge of an electron multiplied by one volt. This value is $1.60 \times 10^{- 19} 𝐽$ , and the mega and giga versions are six and nine order magnitudes greater than it, respectively.

Volts and Point Charges

If we apply this concept to the point charges we have discussed, we get a pretty simple formula:

Δ 𝑉 = \frac{- 𝑊_{𝐴 𝐵}}{𝑞_{0}} = \frac{𝑘 𝑞}{𝑟_{𝐵}} - \frac{𝑘 𝑞}{𝑟_{𝐴}}

And in order to determine $𝑉$ for the $𝑞$ at just point $𝐴$ , we consider point $𝐵$ to be infinitely far away and eliminate it, leaving us with the final potential equation:

𝑉 = \frac{𝑘 𝑞}{𝑟}

Note here that $𝑞$ is not absolute. $𝑉$ will have the same sign as $𝑞$ .

Like with electrostatic force, we can calculate $𝑉$ with respect to many point charges by adding them together, minding our signs .

We can calculate electric potential energy for multiple point charges in a similar way by adding together every pair EPE $\frac{𝑘 𝑞_{1} 𝑞_{2}}{𝑟}$ .

Equipotential Surfaces

An equipotential surface is one where the electric potential is the same everywhere. Think of a hollow sphere which surrounds a point charge. Every point on that surface is the same distance from the charge and so will have the same electric potential.

If you have a charge on an equipotential surface, then no work is done on it by the net electric force if it moves along it. This is because the potential is the same everywhere, looking at the equations from earlier.

On the same equipotential sphere, the electric field is perpendicular to the surface and points to decreasing potential, like gravity. The most clear example of this is a parallel plate capacitor. If we try to calculate the electric field from electric potential, it is simply

𝐸 = - \frac{Δ 𝑉}{Δ 𝑠}

where $Δ 𝑠$ is an expression of displacement, like distance. This is what we would use to find the electric field or distance for a parallel plate capacitor.

Capacitors and Dielectrics

We have already discussed one kind of capacitor, the parallel plate capacitor. A capacitor is simply two conductors placed close together, which are not touching. There will often be an insulator placed in between them called a dielectric.

The capacitor stores electric charge, with the positive plate having more electric potential. That potential difference is the voltage of the capacitor, although the plates have the same charge.

The magnitude $𝑞$ of the charge on each plate of a capacitor is directly proportional to the magnitude $𝑉$ of the potential difference between the plates:

𝑞 = 𝐶 𝑉

where $𝐶$ is the capacitance in farad $(𝐹)$ .

As you can tell from the equation, capitance is coulomb/volt and it represents how well the capacitor can store charge. Like coulombs, it is typically expressed in smaller forms like $𝜇 𝐹$ and $𝑝 𝐹$ which are six and twelve orders of magnitude less.

We can increase capacitance by adding a dielectric in the capacitor. The molecules in the dielectric will form dipole moments oriented towards the positive and negative plate as appropriate. These dipole moments will stop the electric field from passing through the dielectric, and instead some of it will begin and end at the surface. This reduces the strength of the electric field and allows it to store more charge.

The equation for capacitance of a parallel plate capacitor based on the dielectric constant $𝜅$ is:

𝐶 = \frac{𝜅 𝜖_{0} 𝐴}{𝑑}

There is also work done to fill a capacitor with charge, because it stores energy. The work done by a battery, for example, to fill up a capacitor’s charge is expressed in multiple ways as:

𝐸 𝑛 𝑒 𝑟 𝑔 𝑦 = \frac{1}{2} 𝑞 𝑉 = \frac{1}{2} 𝐶 𝑉 = \frac{𝑞^{2}}{2 𝐶} = \frac{1}{2} (\frac{𝜅 𝜖_{0} 𝐴}{𝑑}) (𝐸 𝑑)^{2}

## Functions in OCaml - URL: https://sharifhsn.dev/blog/functions/ - Structured data: https://sharifhsn.dev/api/posts/functions.json - Description: A function takes arguments, performs operations using them, and returns a value. In OCaml, we can write functions like so let rec fact n = (* the function is named fact and its… - Date: 2022-01-25 - Exact published timestamp: 2022-01-25 - Topics: Principles of Programming Languages - Categories: Principles of Programming Languages - Source: Archive - Source URL: None - Content source format: typst

A function takes arguments, performs operations using them, and returns a value. In OCaml, we can write functions like so

let rec fact n = (* the function is named fact and its arg is n *)
    if n = 0 then 1
    else n * fact (n - 1) ;;

This is a recursive function, so we must prepend the keyword rec to the name of the function. The reason for this is that the function name is not automatically in scope, so if we don’t use that we won’t necessarily know what fact refers to. rec adds the function name into scope.

As you can see, the declaration of a function is fairly similar to the declaration of a variable. In fact, we can use functions as variables in OCaml! This is because functions in OCaml are first-class functions.

You also might have noticed here that we haven’t mentioned the type of anything. This is because the type of the function can be inferred, also known as type inference. This process is a normal part of type checking for correctness. The compiler itself will look for a type that the code is correct for. In this example, n is compared with an integer, so it must only be an int. The return type is n multiplied by its own return type, so that must also be int. The compiler can see without us telling it that the function is of type int -> int.

But what is that int -> int that I just wrote? That is the constructor for a function type. The last type indicates the return type, and all other types are the arguments in order. The function type float -> int -> float, for example, takes a float and int as its arguments and returns a float. Try looking at different OCaml functions and guessing its type as a fun exercise! :smile:

Aside: you might question why we couldn’t just get a float as a result from the multiplication. The standard operands in OCaml only apply to int, and applying them to other types is an error. In order to add float, we must use +., not +.

Type checking works like a proof by contradiction. We assume that a function has some type, then check if there are any logical contradictions created by this assumption. If there are, then we reject this type.

Calling a Function

Function calls (also known as applications) look very similar to function declarations; they are simply the name of the function followed by its arguments, no parentheses necessary. The type checking involved is that we check if every expression supplied as an argument corresponds to the same type as it’s supposed to. We evaluate each expression from right to left to see if they’re correct or not (though this doesn’t matter unless you have side effects).

Once we find the value for each argument, we replace each argument in the function with its value, which creates an expression that can be evaluated to the final value.

Aside: Although we have type inference, we can use type annotations with OCaml. These increase the readability of code by making the type of the variable obvious: let (x : int) = 3

## Electrostatics - URL: https://sharifhsn.dev/blog/electrostatics/ - Structured data: https://sharifhsn.dev/api/posts/electrostatics.json - Description: What is electricity, and how does it work? Fundamentally, the negative charge comes from the electron , a subatomic particle. The positive charge comes from protons , also in the… - Date: 2022-01-18 - Exact published timestamp: 2022-01-18 - Topics: Physics - Categories: Physics - Source: Archive - Source URL: None - Content source format: typst

What is electricity, and how does it work?

Electrons

Fundamentally, the negative charge comes from the electron, a subatomic particle. The positive charge comes from protons, also in the atom. The unit of the charge is the Coulomb ( $𝐶$ ). An electron has a charge of $- 1.60 \times 10^{- 16} 𝐶$ , and a proton has the same charge but positive. However, the proton is much heavier than an electron, being about 1836 times as massive. This means that electrons are easy to displace, which is why their movement is the fundamental concept behind electricity.

The way that electric charge is generated is by rubbing two materials together in a way that transfers electrons between the two. For example, if you rub your clothes against a shag carpet, it will transfer electrons to your clothes.

However, charge is neither created nor destroyed, due to the law of conservation of electric charge!

As you might already know from playing around with magnets, like charges repel and unlike charges attract each other. This attraction is known as the electrostatic force.

Conductors/Insulators

Electric charge is not limited to its power to attract and repel. It can also move through objects; this is how electricity travels from power plants to your house, through wires. Not all materials are created equal in their ability to conduct electricity through them. Those that conduct poorly are called electrical insulators and those that can well are called electrical conductors. This material property is typically related to how easily valence electrons can move around in the molecules of the material.

Charging by Contact/Induction

I said before that electrical charge can be generated by rubbing, for example against a shag carpet. This is known as charging by contact. However, this is not the only way to charge an object. As we know, electrons can be attracted and repelled. If a negatively charged object is brought close to a neutrally charged object, the electrons in the neutrally charged object will be repelled to the other side, changing the distribution of charge in the object. This is known as charging by induction because the electrons are induced by the charge to move.

This effect will disappear when the inducing object leaves, unless we introduce grounding. A grounding wire is a way to disappear free electrons from a charged object. Typically, an object is wired to a large diffuse object such as the earth, so free electrons will leave the small object forever. In the example I used above, if there is a grounding wire attached to the other side where the electrons were repelled to, then those electrons would “disappear”. After the inducing object leaves, the formerly neutral object would become positively charged.

This can even occur in materials like plastic which are insulating. Although the free electrons cannot move as in materials like metals, positive charges can still be induced towards the surface, allowing for a surface positive charge that negatively charged objects to stick to.

Coulomb’s Law

The magnitude $𝐹$ of the electrostatic force exerted by one point charge $𝑞_{1}$ on another point charge $𝑞_{2}$ is directly proportional to the magnitudes $| 𝑞_{1} |$ and $| 𝑞_{2} |$ of the charges and inversely proportional to the square of the distance $𝑟$ between them:

𝐹 = 𝑘 \frac{| 𝑞_{1} | | 𝑞_{2} |}{𝑟^{2}}

where $𝑘$ is a proportionality constant: $𝑘 = 8.99 \times 10^{9} 𝑁 \cdot 𝑚^{2} / 𝐶^{2}$ in SI units.

So what does this law mean? Crucially, this law gives no mention of direction. The magnitude of the force is the same regardless of whether the charges attract or repel, so we don’t need to worry about that.

The constant $𝑘$ is sometimes written as $𝑘 = 1 / (4 𝜋 𝜖_{0})$ where $𝜖_{0}$ is the permittivity of free space with the value $8.85 \times 10^{- 12} 𝐶^{2} / 𝑁 \cdot 𝑚^{2}$ .

A Coulomb is an absolutely massive amount of charge that is typically only found in situations like lightning. It is far more common to encounter microcoulombs expressed as $𝜇 𝐶$ which are $10^{- 6} 𝐶$ .

This is a situation that is expressed between only two point charges. What happens when we have multiple point charges? We have to calculate the vector sum of all these forces.

This is easiest when all the charges are in a line, as this is simply a matter of adding and subtracting the forces depending on whether they are attractive or negative. We must calculate Coulomb’s law for each pair and then combine those as appropriate.

When we have points on a plane, things get a little more complicated. The easiest way to handle this is by separating each force into its $𝑥$ and $𝑦$ components. Pick one point to be set at the angle 0° degrees with respect to the point we are calculating for, then set the $𝜃$ for each other point with respect to that line. Then, get $𝐹_{𝑥}$ and $𝐹_{𝑦}$ through $cos 𝜃$ and $sin 𝜃$ , respectively. The point with 0° will only have an $𝑥$ component.

Then, once you have the $𝐹_{𝑥}$ and $𝐹_{𝑦}$ all parcelled out according to positives and negatives, get the total $𝐹$ by applying the Pythagorean theorem to them. The answer $𝜃$ is given by ${tan}^{- 1} \frac{𝐹_{𝑦}}{𝐹_{𝑥}}$ .

The Electric Field

We have so far been talking about how individual point charges affect other point charges. When we talk about an entire system of charges, we need a broader, high-level understanding of charge. For that purpose, the electric field is essential. Fundamentally, the electric field is defined as the electric force per coulomb $\vec{𝐸} = \vec{𝐹} / 𝑞_{0}$ . This calculation pretends that we have a small positive test charge $𝑞_{0}$ placed in the middle of this electric field, which is affected by the various charges to have a force pointing a direction. This test charge typically has a small magnitude so as not to disturb the surrounding field.

The electric field is independent of whatever point charge it’s affecting. We could imagine a field with a magnitude $2.0 𝑁 / 𝐶$ and direction upwards that causes a positive charge to go in the same direction as it, but a negative charge to go in the opposite direction. We can calculate the magnitude for each point charge with $𝐹 = | 𝑞_{0} | 𝐸$ .

Electric fields, just like forces, are vectors. When we add fields together, we must equally be mindful of $𝐸_{𝑥}$ and $𝐸_{𝑦}$ .

An electric field is composed of multiple point charges. The amount of electric field that an individual point charge $𝑞$ produces is expressed by the equation

𝐸 = \frac{𝑘 | 𝑞 |}{𝑟^{2}}

Notice how $𝑞_{0}$ the test charge is not included in this equation. Since this new equation only expresses magnitude, we must individually determine direction by the sign of the charge. If $𝑞$ is positive, $\vec{𝐸}$ is directed in the opposite direction of $𝑞$ , and vice versa for negative.

An additionally useful equation relates to parallel plate capacitors. These are two parallel plates of opposite charge that have an electric field directed from the positive plate to the negative plate. The equation for this electric field is

𝐸 = \frac{𝑞}{𝜖_{0} 𝐴} = \frac{𝜎}{𝜖_{0}}

As you can see in this equation, the symbol $𝜎$ denotes $𝑞 / 𝐴$ and is sometimes called charge density. Importantly, the distance between the plates is actually totally irrelevant to the electric field, unlike with point charges.

Electric Field Lines

It is often useful to draw fields as a set of lines directed out of or towards a point charge. In general, electric field lines are always directed away from positive charges and toward negative charges.

For convenience, we usually only draw them in two dimensions, though technically charges act in three dimensions. The number of lines is expressed as proportional the magnitude of the charge, so five times as many lines for five times the magnitude. In general, the electric field is stronger when lines are closer together, which is why parallel plate capacitors have the same electric field in the middle but bulge out at the ends.

The lines are not always straight, as in the case of an electric dipole. An electric dipole has two separated point charges of opposite signs and equal magnitude. The field lines in this case come from the positive charges and curve towards the negative charge, getting faster as they get closer. The electric field vector in this case is tangent to the electric field at any point. Between two charges of the same sign, there is practically no electric field because the lines curve away from each other.

Shielding

Electric fields can exist anywhere where electrons can move freely, which includes the inside of a conductor. It’s a logical conclusion: electrons hate being near each other, so they try to get away in the conductor. That leaves them all at the surface of the conductor. The same principle applies for positive charges; conductors generally have all their excess charge at their surface.

However, since all the excess charge is at the surface, that means that in the interior of the conductor, there is no movement of free electrons because they’re not there. This means that there is zero electric field inside a conductor.

If we imagine a conductor within a parallel plate capacitor, there is a lot of electric field going on. However, all of the charges are directed outside of it. Inside the conductor, any charge is completely shielded from the intense electric field outside the conductor.

Gauss’ Law

We have so far discussed electric fields created by point charges. In fact, an electric field is often composed by multiple charges that are spread out, also known as a charge distribution. Understanding an electric field created by a charge distribution is significantly more complex and requires understanding electric flux.

We can think of flux on a high level as similar to a vector, with a direction and magnitude, but with the additional dimension of area. This concept will be important when discussing magnets.

We can consider the formula for a point charge as a special form of Gauss’ Law which describes electric fields in general. Remember how we can substitute $1 / (4 𝜋 𝜖_{0})$ for $𝑘$ ? This means that our new equation is $𝐸 = 𝑞 / 4 (𝜋 𝜖_{0} 𝑟^{2})$ . If we multiply both sides by $𝐴$ which is $4 𝜋 𝑟^{2}$ for a sphere, then we get

𝐸 𝐴 = \frac{𝑞}{𝜖_{0}}

where $𝐸 𝐴$ is electric flux, also known as $Φ_{𝐸}$ .

However, we don’t always have a nice, neat distribution of charge like in a sphere. A Gaussian surface can be arbitrarily shaped. We have to think of this similarly to how we think about integrals in calculus. If we take an arbitrarily small area $Δ 𝐴$ , it has an electric field that is essentially constant in magnitude and direction. The surface has a normal which is a line perpendicular to the surface. The angle $𝜑$ is between the electric field vector and the normal. With all these variables, Gauss’ Law is expressed as so:

The electric flux $Φ_{𝐸}$ through a Gaussian surface is equal to the net charge $𝑄$ enclosed by the surface divided by $𝜖_{0}$ , the permittivity of free space:

Σ (𝐸 cos 𝜑) Δ 𝐴 = \frac{𝑄}{𝜖_{0}}

with the left side being $Φ_{𝐸}$ in $𝑁 \cdot 𝑚^{2} / 𝐶$ .

The flux has the same sign as $𝑄$ .

## CPU Virtualization - URL: https://sharifhsn.dev/blog/cpu-virtualization/ - Structured data: https://sharifhsn.dev/api/posts/cpu-virtualization.json - Description: We need a way to map on to the physical CPU through our operating system. We do this through virtualization. A process is an execution stream in the context of a process state.… - Date: 2022-01-04 - Exact published timestamp: 2022-01-04 - Topics: Operating Systems Design - Categories: Operating Systems Design - Source: Archive - Source URL: None - Content source format: typst

We need a way to map on to the physical CPU through our operating system. We do this through virtualization.

Processes

A process is an execution stream in the context of a process state. It is a self-contained stream of executing instructions on the CPU. Each process has a state which is composed by everything that the code can affect or be affected by, for example registers, address space, open files, etc. We need this state so that we can pause and resume processes without resetting the whole thing, just by picking up where the state left off.

You can think of address space as the region of memory in which the CPU operates, though this is a simplification due to memory virtualization.

A process is technically different than a program. When we talk about a program, it’s usually a collection of files on our computer. The process refers to the actually executing code which is dynamic with respect to its code. We can have multiple processes executed that are the same program.

Processes do not share information with other processes. This is distinct from the idea of threads. If a process examines the “same” memory address with respect to its address space as another process, they will see different values. Even though both process are looking at 0xFFE84264 the memory virtualization means that they are looking at different physical memory. Threads share data so they have the same address space; they are in a way a lightweight process.

Address Space

As mentioned, every process has its own address space. The OS assigns a chunk of memory towards it (again, this memory is virtual). At the high address, there is a stack of local variables that grows downwards. This stack contains the execution of the program, such as functions being called. The code segment which is the actual program is read from the absolute bottom. There are a few segments like .data and .bss which refer to global initialized/uninitialized variables, respectively. Then there is the heap, which is dynamic and grows upward. Usually, the heap is much larger than the stack because it contains allocated memory for the process.

The stack and the heap grow towards each other, so you might think they would collide at some point. In reality, there is a wide enough chasm between them that if they collide, you have bigger problems to worry about.

CPU Virtualization

Our problem is that a CPU can only execute one execution stream at one time. We want processes to be running at the same time, so how do we achieve the illusion that the process has full control of the CPU?

One solution is direct execution. We will run the process as simply as possible and execute all of its instructions in sequence. This is extremely simple to implement, but it means we can only run one process at once. If the process runs forever, then the CPU is stuck. Processes might write to other data that it’s not supposed to, or do some slow I/O operation, or execute privileged instructions it’s not allowed to access. In general, we don’t want to trust processes to behave. Operating systems use limited direct execution, which maintains some control over the execution of the process instead of giving it unfettered access to the CPU.

System Calls

We want to make sure user process can’t harm other processes. CPU hardware supports privilege levels for security reasons. These instructions should only be run in kernel space by the operating system, and user processes in user space can not access them. Privileges can have multiple levels. In order for processes to access privileges, we use system calls that can either pass the function off to the OS or give the process privilege, also known as a trap.

ID1 = syscall(SYS_getpid); // syscall

ID2 = getpid(); // call to libc

Let’s say a process wants to execute read(), which is a syscall. It will move the syscall ID for read 0x6 into the %eax register for execution, then run the instruction syscall. The operating system will then read the trap table to figure out what to do. The trap table is located in the hardware, which typically has an entry for system calls. Other traps could be illegal access for memory region that it doesn’t have access to. A trap just refers to a hardware operation that triggers on some software interaction. This is why Javascript can’t just randomly hack your computer from the internet; it is executing within a user process e.g. Chrome and is limited by the hardware traps in what it can do.

Multiprogramming

We want to make sure that multiple processes can run at once. This means we must switch between processes. There are two components to this; how to switch, and when to switch.

The dispatch loop has a simple structure, but the devil is in the details.

while (1) {
    // run process A for some time-slice
    // stop process A and save its context
    // CONTEXT SWITCH
    // load context of another process B
}

We have multiple ways to context switch. One way is cooperative multi-tasking. We trust the process to give up the CPU when it has judged some amount of work has been done or time has passed. We will provide yield() syscall for magnanimously donating CPU time to another process. However, this is annoying to program in, so most programs do not do this.

Modern programs use true multi-tasking, which gives OS control over which processes are running. The hardware will generate a timer interrupt every 10 ms, for example (Linux) that the OS will decide when to switch or not.

PCB

The process control block is a descriptor of a process that saves its context into a struct. There is a struct for every process that is running in the CPU. A PCB stores the following information:

PID - unique identification for a process
Process state (enum running, ready, or blocked)
Execution state (registers at time of pause)
Scheduling priority
Accounting information (parent/child procs)
Credentials (resource access/owner)
File pointers

When we save the context of the process in the dispatch loop, the PCB is where it is saved. It is typically less than a few kilobytes depending on the complexity of the process. The PCB is stored on the kernel stack which is a stack data structure located in kernel space. Switching a context means moving the stack pointer to the process that is going to resume and then retrieving that PCB from the stack. Then, we exit kernel space and resume executing B in user space by restoring the PCB information to the CPU.

When a process is doing I/O, it’s not using the CPU. This is an excellent point where the OS can block the process and run some other processes that actually need the CPU. Once the I/O is done, it is moved from the blocked state to the ready state. The convention is that only ready state programs can go to the running state.

Process Creation

One way to create a process is just creating one from scratch. We will load the code we’re running into memory at the bottom of the stack and create the stack. We also create a PCB with a kernel stack and everything, then put the process in the ready state. However, there are lots of complicated parts of processes that are difficult to initialize from scratch e.g. permissions, I/O, environment vars so this is generally not preferred.

A better way is to clone an existing process and change it to the appropriate information. fork() clones the calling process and exec(char *file) replaces the current process with the new process.

When we fork, we save the current state of the program as a new PCB that is added to the kernel stack, which is very optimized to use copy-on-write semantics.

The basic idea of copy-on-write is that it uses a pointer/reference for memory until we need to modify memory. This is an incremental process that trades off some overhead for a pay-as-you-go memory saving system.