{"categories":["Stochastic Calculus"],"contentHtml":"

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

𝑑𝑊𝑖(𝑡)𝑑𝑊𝑖(𝑡)=𝑑𝑡

We still don’t know the cross-variation between two independent Brownian motions? 𝑑𝑊𝑖(𝑡)𝑑𝑊𝑗(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 𝐶Π the sample cross-variation:

𝐶Π=∑𝑘=0𝑛−1(𝑊𝑖(𝑡𝑘+1)−𝑊𝑖(𝑡𝑘))(𝑊𝑗(𝑡𝑘+1)−𝑊𝑗(𝑡𝑘))

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[𝐶Π]=𝔼[∑𝑘=0𝑛−1(𝑊𝑖(𝑡𝑘+1)−𝑊𝑖(𝑡𝑘))(𝑊𝑗(𝑡𝑘+1)−𝑊𝑗(𝑡𝑘))]

I have linearity, so I can move the expectation inside the summation.

=∑𝑘=0𝑛−1𝔼[(𝑊𝑖(𝑡𝑘+1)−𝑊𝑖(𝑡𝑘))(𝑊𝑗(𝑡𝑘+1)−𝑊𝑗(𝑡𝑘))]

Now we use independence. These two Brownian motions are independent from each other, so their increments are independent. Using this independence, we say

=∑𝑘=0𝑛−1𝔼[(𝑊𝑖(𝑡𝑘+1)−𝑊𝑖(𝑡𝑘))]𝔼[(𝑊𝑗(𝑡𝑘+1)−𝑊𝑗(𝑡𝑘))]

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[𝐶Π]

We could try the traditional expectation of squares method, but we would then have to work with 𝐶Π2. What is that?

𝐶Π2=∑𝑘=0𝑛−1(𝑊𝑖(𝑡𝑘+1)−𝑊𝑖(𝑡𝑘))(𝑊𝑗(𝑡𝑘+1)−𝑊𝑗(𝑡𝑘))⋅∑𝑘=0𝑛−1(𝑊𝑖(𝑡𝑘+1)−𝑊𝑖(𝑡𝑘))(𝑊𝑗(𝑡𝑘+1)−𝑊𝑗(𝑡𝑘))

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[𝛼𝑋+𝛽𝑌]=𝛼2𝕍[𝑋]+2𝛼𝛽𝑐𝑜𝑣(𝑋,𝑌)+𝛽2𝕍[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[𝐶Π]=∑𝑘=0𝑛−1𝕍[(𝑊𝑖(𝑡𝑘+1)−𝑊𝑖(𝑡𝑘))(𝑊𝑗(𝑡𝑘+1)−𝑊𝑗(𝑡𝑘))]

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[Inc𝑖2⋅Inc𝑗2]−𝔼[Inc𝑖⋅Inc𝑗]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.

=(𝑡𝑘+1−𝑡𝑘)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∥→0∑𝑘=0𝑛−1(𝑡𝑘+1−𝑡𝑘)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)+∫0𝑡Δ(𝑢)𝑑𝑊(𝑢)+∫0𝑡Θ(𝑢)𝑑𝑢

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 Θs? 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?

𝑑𝑋(𝑡)=𝜎1(𝑡)𝑑𝑊1(𝑡)+𝜎2(𝑡)𝑑𝑊2(𝑡)+Θ(𝑡)𝑑𝑡

Now what if we square it? Too long to write out, but it follows all the rules of multiplication. We proved earlier that (𝑑𝑊1(𝑡))2 is dt, and the same for 𝑑𝑊2. 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

=(𝜎12(𝑡)+𝜎22(𝑡))𝑑𝑡 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)=∫0𝑡(𝜎12(𝑢)+𝜎22(𝑢))𝑑𝑢

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

𝑑𝑋(𝑡)𝑑𝑌(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

=(𝜎𝑋,1+𝜎𝑋,2)(𝜎𝑌,1+𝜎𝑌,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(𝑡𝑘+1,𝑋𝑘+1,𝑌𝑘+1)=𝑓(𝑡𝑘,𝑋𝑘,𝑌𝑘)+𝑓𝑡(𝑡𝑘,𝑋𝑘,𝑌𝑘)(𝑡𝑘+1−𝑡𝑘)+𝑓𝑋(𝑡𝑘,𝑋𝑘,𝑌𝑘)(𝑋𝑘+1−𝑋𝑘)+𝑓𝑌(𝑡𝑘,𝑋𝑘,𝑌𝑘)(𝑌𝑘+1−𝑌𝑘) 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 𝑓𝑋𝑋, 𝑓𝑋𝑌, and 𝑓𝑌𝑌. 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.

𝑑𝑓(𝑡,𝑋,𝑌)=𝑓𝑡(𝑡,𝑋,𝑌)𝑑𝑡+𝑓𝑋(𝑡,𝑋,𝑌)𝑑𝑋+𝑓𝑦(𝑡,𝑋,𝑌)𝑑𝑌+12𝑓𝑋𝑋(𝑡,𝑋,𝑌)𝑑𝑋2+𝑓𝑋𝑌(𝑡,𝑋,𝑌)𝑑𝑋𝑑𝑌+12𝑓𝑌𝑌(𝑡,𝑋,𝑌)𝑑𝑌2

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(𝑋𝑌)=𝑋𝑑𝑌+𝑌𝑑𝑋+𝑑𝑋𝑑𝑌

Let’s do our partials to prove this. 𝑓(𝑡,𝑋,𝑌)=𝑋𝑌

𝑓𝑡=0,𝑓𝑋=𝑌,𝑓𝑌=𝑋,𝑓𝑋𝑋=0,𝑓𝑋𝑌=1,𝑓𝑌𝑌=0

Then using our formula, that’s

=0𝑑𝑡+𝑌𝑑𝑋+𝑋𝑑𝑌+12(0)𝑑𝑋2+1𝑑𝑋𝑑𝑌+12(0)𝑑𝑌2

=𝑋𝑑𝑌+𝑌𝑑𝑋+𝑑𝑋𝑑𝑌

Let’s find the differential of the discounted stock process

𝑒−𝑟𝑡𝑆(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−𝑟𝑡)=𝑒−𝑟𝑡𝑑𝑆+𝑆𝑑(𝑒−𝑟𝑡)+𝑑(𝑒−𝑟𝑡)𝑑𝑆

𝑑(𝑒−𝑟𝑡)=−𝑟𝑒−𝑟𝑡𝑑𝑡

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))+∫0𝑡𝑓𝑎(𝑢,𝑀(𝑢))𝑑𝑢+∫0𝑡𝑓𝑏(𝑢,𝑀(𝑢))𝑑𝑀(𝑢)+12∫0𝑡𝑓𝑏𝑏(𝑢,𝑀(𝑢))𝑑𝑀2(𝑢)

I know that M(0) is equal to 0 from rule 2, and we also know that 𝑑𝑀2(𝑢) is 𝑑𝑢 because of rule 4 (definition of quadratic variation).

𝑓(𝑡,𝑀(𝑡))=𝑓(0,0)+∫0𝑡(𝑓𝑎(𝑢,𝑀(𝑢))+12𝑓𝑏𝑏(𝑢,𝑀(𝑢)))𝑑𝑢+∫0𝑡𝑓𝑏(𝑢,𝑀(𝑢))𝑑𝑀(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)=𝑒𝑢𝑏−12𝑢2𝑎

with a nonrandom dummy variable 𝑢 that we don’t know the value of. Let’s get our partials.

𝑓𝑎=−12𝑢2𝑓(𝑎,𝑏)

𝑓𝑏=𝑢𝑓(𝑎,𝑏) 𝑓𝑏𝑏=𝑢2𝑓(𝑎,𝑏) If I plugged this in, I would get

𝑓(𝑡,𝑀)=𝑒𝑢𝑀−12𝑢2𝑡 I have now that (skipping some steps)

𝑒𝑢𝑀−12𝑢2𝑡=1+∫0𝑡𝑢𝑓(𝑢,𝑀)𝑑𝑀

If I take the expected value, I can factor out the nonrandom component

𝔼[𝑒𝑢𝑀]𝑒−12𝑢2𝑡=1+𝔼[∫0𝑡𝑢⋅𝑓(𝑢,𝑀)𝑑𝑀]

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[𝑒𝑢𝑀]=𝑒12𝑢2𝑡

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)=𝑒𝑢1𝑋+𝑢2𝑌−12(𝑢12+𝑢22)𝑡

which results in

𝔼[𝑒𝑢1𝑀1+𝑢2𝑀2]=𝑒12𝑢12𝑡𝑒12𝑢22𝑡

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:

𝑑𝑆1=𝛼1𝑆1𝑑𝑡+𝜎1𝑆1𝑑𝑊1

𝑑𝑆2=𝛼2𝑆2𝑑𝑡+𝜎2𝑆2(𝜌𝑑𝑊1+1−𝜌2𝑑𝑊2)

We can all agree that 𝑆1 follows Brownian motion, so we know its closed form solution.

𝑆2 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

𝐵1=𝑊1

And another

𝐵2=𝜌𝑊1+1−𝜌2𝑊2

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[𝜌𝑊1(𝑡)+1−𝜌2𝑊2(𝑡)|ℱ︀(𝑠)]

We can separate through linearity.

=𝜌𝔼[𝑊1(𝑡)|ℱ︀(𝑠)]+1−𝜌2𝔼[𝑊2(𝑡)|ℱ︀(𝑠)]

We already know that Brownian motion is a martingale, so we can use that information here.

=𝜌𝑊1(𝑠)+1−𝑝2𝑊2(𝑠)

Which by definition is

=𝐵𝑤(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 𝑑𝐵2

𝑑𝐵22=𝜌2𝑑𝑡+(1−𝑝2)𝑑𝑡

=𝑑𝑡

So it must be a Brownian motion. Now pulling us back to our original stock process.

𝑆1=𝑆1(0)𝑒(𝛼1−𝜎122)𝑡+𝜎1𝐵1

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[𝐵1|𝐵2]=∫0𝑡𝐵1𝑑𝐵2+∫0𝑡𝐵2𝑑𝐵1+∫0𝑡𝑑𝐵1𝑑𝐵2

We have Ito integral with respect to martingale, so that becomes 0

=𝔼[∫0𝑡𝜌𝑑𝑢]=𝜌𝑡

Therefore the covariance is

Cov(𝐵1,𝐵2)𝕍[𝐵1]𝕍[𝐵2]=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 𝑆1 follows GBM, and we know that structure

𝔼[𝑆1]=𝔼[𝑆1(0)𝑒(𝛼1−𝜎122)𝑡+𝜎1𝐵1]

Skipping some steps,

=𝑆1(0)𝑒𝛼1𝑡

How do we get the variance of 𝑆1?

𝕍[𝑆1]=𝔼[𝑆12]−𝔼[𝑆1]2

This would be

=𝑆12(0)𝑒2𝛼1𝑡+𝜎12𝑡−𝑆12(0)𝑒2𝛼𝑡

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 𝑆1 and 𝑆2, which is

𝔼[𝑆1𝑆2]−𝔼[𝑆1]𝔼[𝑆2]

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(𝑆1𝑆2)=𝑆1𝑑𝑆2+𝑆2𝑑𝑆1+𝑑𝑆1𝑑𝑆2

=𝑆1𝛼2𝑆2𝑑𝐵2

","contentSourceFormat":"typst","contentTypst":"== Week 7\n\n=== Lecture Notes\n\nWe'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!\n\nWe're mostly going to be doing applications and extensions today, which is not too bad.\n\nHow 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?\n\nLet'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.\n\nBecause 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\n\n$d W_i\\(t\\)d W_i\\(t\\)= d t$\n\nWe still don't know the cross-variation between two independent Brownian motions? $d W_i\\(t\\)d W_j\\(t\\)= ?$\n\nWe will prove that this is 0.\n\nSimilar 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 $C_Pi$ the sample cross-variation:\n\n$C_Pi = sum_(k = 0)^(n - 1) (W_i \\( t_(k + 1) \\) - W_i \\( t_k \\)) (W_j \\( t_(k + 1) \\) - W_j \\( t_k \\))$\n\nThis 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.\n\nIf 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.\n\nHow could I calculate the expected value? Let's just use the definition\n\n$bb(E)\\[C_Pi\\]= bb(E) [sum_(k = 0)^(n - 1) \\( W_i \\( t_(k + 1) \\) - W_i \\( t_k \\) \\) \\( W_j \\( t_(k + 1) \\) - W_j \\( t_k \\) \\)]$\n\nI have linearity, so I can move the expectation inside the summation.\n\n$= sum_(k = 0)^(n - 1) bb(E) [\\( W_i \\( t_(k + 1) \\) - W_i \\( t_k \\) \\) \\( W_j \\( t_(k + 1) \\) - W_j \\( t_k \\) \\)]$\n\nNow we use independence. These two Brownian motions are independent from each other, so their increments are independent. Using this independence, we say\n\n$= sum_(k = 0)^(n - 1) bb(E) [\\( W_i \\( t_(k + 1) \\) - W_i \\( t_k \\) \\)] bb(E) [\\( W_j \\( t_(k + 1) \\) - W_j \\( t_k \\) \\)]$\n\nWell what is that expected value? 0!\n\n$= 0$\n\nWhat is the limit as the norm of the partition goes to 0, of 0? 0!\n\nNow let's do variance $bb(V)\\[C_Pi\\]$\n\nWe could try the traditional expectation of squares method, but we would then have to work with $C_Pi^2$. What is that?\n\n$C_Pi^2 = sum_(k = 0)^(n - 1) (W_i \\( t_(k + 1) \\) - W_i \\( t_k \\)) (W_j \\( t_(k + 1) \\) - W_j \\( t_k \\)) dot.op sum_(k = 0)^(n - 1) (W_i \\( t_(k + 1) \\) - W_i \\( t_k \\)) (W_j \\( t_(k + 1) \\) - W_j \\( t_k \\))$\n\nWhich 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:\n\n$bb(V)\\[alpha X + beta Y\\]= alpha^2 bb(V)\\[X\\]+ 2 alpha beta c o v\\(X\\,Y\\)+ beta^2 bb(V)\\[Y\\]$\n\nIn 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.\n\nAside: The cross variation of a process with respect to itself is the same as quadratic variation.\n\nWe can then write\n\n$bb(V)\\[C_Pi\\]= sum_(k = 0)^(n - 1) bb(V) [\\( W_i \\( t_(k + 1) \\) - W_i \\( t_k \\) \\) \\( W_j \\( t_(k + 1) \\) - W_j \\( t_k \\) \\)]$\n\nbecause of this independence. Let's examine the variance of one of these elements.\n\nTo get this variance, we can take the expectation of squares here.\n\n$= bb(E) [upright(\"Inc\")_i^2 dot.op upright(\"Inc\")_j^2] - bb(E) [upright(\"Inc\")_i dot.op upright(\"Inc\")_j]^2$\n\nI know for any Brownian motion that the variance is length of time, because that's quadratic variation. Therefore it's just that value squared.\n\n$=\\(t_(k + 1) - t_k\\)^2$\n\nFor 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\n\n$lim_(parallel Pi parallel arrow.r 0) sum_(k = 0)^(n - 1)\\(t_(k + 1) - t_k\\)^2$\n\nBut that's just the formula for the quadratic variation of time, which must be 0 because it's a continuous differentiable function.\n\nIf it's converging to mean 0 and variance 0, then that's just 0.\n\nWe have proven that the cross variation is 0, and therefore the differential of the cross variation is also 0.\n\nIf you multiply a Brownian motion by itself, you get the variance not the mean. This result is only for independent Brownian motions.\n\n==== Two Dimensional Ito Process\n\nThis leads us to our two dimensional Ito process. Big distinction between just two Brownian motions and the two-dimensional, because dimensional means independent.\n\nWe could write a standard Ito process as\n\n$X\\(t\\)= X\\(0\\)+ integral_0^t Delta\\(u\\)d W\\(u\\)+ integral_0^t Theta\\(u\\)d u$\n\nBut now my $Delta$ and $W$ are both vectors, and we take the dot product of these, which is a scalar vector multiplication\n\nWhy don't we have a dot product with the $Theta$s? Can you have multi-dimensional time? No.~We don't want multidimensional time, so we don't have multidimensional $Theta$. $Theta$ could depend on the process, but you wouldn't decompose with reference to different aspects of movements in time.\n\nWhat if we take the differential?\n\n$d X\\(t\\)= sigma_1\\(t\\)d W_1\\(t\\)+ sigma_2\\(t\\)d W_2\\(t\\)+ Theta\\(t\\)d t$\n\nNow 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\n\n$= (sigma_1^2 \\( t \\) + sigma_2^2 \\( t \\)) d t$ 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.\n\n$\\[X\\,X\\]\\(t\\)= integral_0^t (sigma_1^2 \\( u \\) + sigma_2^2 \\( u \\)) thin d u$\n\nThere's a lot of different ways to do this. $Sigma$ 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.\n\nLet's say we have two well-defined multidimensional Ito processes in this way. Defined as such, what is\n\n$d X\\(t\\)d Y\\(t\\)= ?$\n\nLet'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\n\n$=\\(sigma_(X\\,1) + sigma_(X\\,2)\\)\\(sigma_(Y\\,1) + sigma_(Y\\,2)\\)$\n\nThis is all to build up to the concept of the two dimensional Ito formula. We are going to use the Taylor series expansion with #emph[three] variables now. That would say that\n\n$f\\(t_(k + 1)\\,X_(k + 1)\\,Y_(k + 1)\\)= f\\(t_k\\,X_k\\,Y_k\\)+ f_t\\(t_k\\,X_k\\,Y_k\\)\\(t_(k + 1) - t_k\\)+ f_X\\(t_k\\,X_k\\,Y_k\\)\\(X_(k + 1) - X_k\\)+ f_Y\\(t_k\\,X_k\\,Y_k\\)\\(Y_(k + 1) - Y_k\\)$ 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.\n\n#strong[For future reference: $X = X\\(t\\)$, $Y = Y\\(t\\)$, to save my fingers.]\n\n$d f\\(t\\,X\\,Y\\)= f_t\\(t\\,X\\,Y\\)d t + f_X\\(t\\,X\\,Y\\)d X + f_y\\(t\\,X\\,Y\\)d Y + 1 / 2 f_(X X)\\(t\\,X\\,Y\\)d X^2 + f_(X Y)\\(t\\,X\\,Y\\)d X d Y + 1 / 2 f_(Y Y)\\(t\\,X\\,Y\\)d Y^2$\n\nThese are all partials.\n\nWe have the product rule of Ito processes. And we still need to include the cross term. In normal calculus, this term is 0. $d\\(X Y\\)= X d Y + Y d X + d X d Y$\n\nLet's do our partials to prove this. $f\\(t\\,X\\,Y\\)= X Y$\n\n$f_t = 0\\,f_X = Y\\,f_Y = X\\,f_(X X) = 0\\,f_(X Y) = 1\\,f_(Y Y) = 0$\n\nThen using our formula, that's\n\n$= 0 d t + Y d X + X d Y + 1 / 2\\(0\\)d X^2 + 1 d X d Y + 1 / 2\\(0\\)d Y^2$\n\n$= X d Y + Y d X + d X d Y$\n\nLet's find the differential of the discounted stock process\n\n$e^(- r t) S\\(t\\)$ 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.\n\n$d (e^(- r t)) = e^(- r t) d S + S d\\(e^(- r t)\\)+ d\\(e^(- r t)\\)d S$\n\n$d\\(e^(- r t)\\)= - r e^(- r t) d t$\n\nWarning: 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.\n\nAny continuous process with t and W(t) is an Ito process. All of our results are only for continuous processes.\n\n==== Lévy\n\nThis is a very handy result. If we have a martingale M relative to filtration $cal(F)$, 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.\n\nQuadratic 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.\n\nWe 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.\n\nSo we can now use the Ito formula to look at an arbitrary function f on M $f\\(t\\,M\\(t\\)\\)= f\\(0\\,M\\(0\\)\\)+ integral_0^t f_a\\(u\\,M\\(u\\)\\)d u + integral_0^t f_b\\(u\\,M\\(u\\)\\)d M\\(u\\)+ 1 / 2 integral_0^t f_(b b)\\(u\\,M\\(u\\)\\)d M^2\\(u\\)$\n\nI 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).\n\n$f\\(t\\,M\\(t\\)\\)= f\\(0\\,0\\)+ integral_0^t (f_a \\( u \\, M \\( u \\) \\) + 1 / 2 f_(b b) \\( u \\, M \\( u \\) \\)) d u + integral_0^t f_b\\(u\\,M\\(u\\)\\)thin d M\\(u\\)$\n\nThis 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\n\n$f\\(a\\,b\\)= e^(u b - 1 / 2 u^2 a)$\n\nwith a nonrandom dummy variable $u$ that we don't know the value of. Let's get our partials.\n\n$f_a = - 1 / 2 u^2 f\\(a\\,b\\)$\n\n$f_b = u f\\(a\\,b\\)$ $f_(b b) = u^2 f\\(a\\,b\\)$ If I plugged this in, I would get\n\n$f\\(t\\,M\\)= e^(u M - 1 / 2 u^2 t)$ I have now that (skipping some steps)\n\n$e^(u M - 1 / 2 u^2 t) = 1 + integral_0^t u f\\(u\\,M\\)d M$\n\nIf I take the expected value, I can factor out the nonrandom component\n\n$bb(E)\\[e^(u M)\\]e^(- 1 / 2 u^2 t) = 1 + bb(E) [integral_0^t u dot.op f \\( u \\, M \\) d M]$\n\nExercise 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.\n\n$bb(E) [e^(u M)] = e^(1 / 2 u^2 t)$\n\nAnd that is the moment generating function of a normal variable with mean 0 and variance t.\n\nSo 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).\n\nNow let's do a two dimensional Lévy.\n\nThe same four properties are important to us, with the tweak that the cross variation has to be 0.\n\nWe 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\n\n$f\\(t\\,X\\,Y\\)= e^(u_1 X + u_2 Y - 1 / 2\\(u_1^2 + u_2^2\\)t)$\n\nwhich results in\n\n$bb(E) [e^(u_1 M_1 + u_2 M_2)] = e^(1 / 2 u_1^2 t) e^(1 / 2 u_2^2 t)$\n\nwhich is the joint moment generating function.\n\nWhat 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.\n\n==== Correlated Stock Prices\n\nLet'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:\n\n$d S_1 = alpha_1 S_1 d t + sigma_1 S_1 d W_1$\n\n$d S_2 = alpha_2 S_2 d t + sigma_2 S_2 (rho d W_1 + sqrt(1 - rho^2) d W_2)$\n\nWe can all agree that $S_1$ follows Brownian motion, so we know its closed form solution.\n\n$S_2$ is only a Brownian motion if that big term on the right is a Brownian motion.\n\nLet's think abstractly. We have a Brownian motion\n\n$B_1 = W_1$\n\nAnd another\n\n$B_2 = rho W_1 + sqrt(1 - rho^2) W_2$\n\nLévy 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.\n\n$bb(E)\\[rho W_1\\(t\\)+ sqrt(1 - rho^2) W_2\\(t\\)\\|cal(F)\\(s\\)\\]$\n\nWe can separate through linearity.\n\n$= rho bb(E)\\[W_1\\(t\\)\\|cal(F)\\(s\\)\\]+ sqrt(1 - rho^2) bb(E)\\[W_2\\(t\\)\\|cal(F)\\(s\\)\\]$\n\nWe already know that Brownian motion is a martingale, so we can use that information here.\n\n$= rho W_1\\(s\\)+ sqrt(1 - p^2) W_2\\(s\\)$\n\nWhich by definition is\n\n$= B_w\\(s\\)$\n\nIt also fulfills the zero property if you substitute. What about continuity? It's a linear combination of continuous processes, so it must be continuous.\n\nFinal: what is quadratic variation? We can solve that simply through $d B^2$\n\n$d B_2^2 = rho^2 d t +\\(1 - p^2\\)d t$\n\n$= d t$\n\nSo it must be a Brownian motion. Now pulling us back to our original stock process.\n\n$S_1 = S_1\\(0\\)e^(\\(alpha_1 - sigma_1^2 / 2\\)t + sigma_1 B_1)$\n\nWe have chosen our correlation coefficient to be between -1 and 1 so that we don't deal with imaginary numbers.\n\nIf we take the cross variation of these stock processes.\n\nThese are two Brownian motions, but they have correlation. Their cross variation is not 0! But what about their covariance?\n\nWhat is the correlation between our sources of noise? We need to know the covariance. We already know the variances of Brownian motion.\n\nI'm going to have that\n\n$bb(E)\\[B_1\\|B_2\\]= integral_0^t B_1 d B_2 + integral_0^t B_2 d B_1 + integral_0^t d B_1 d B_2$\n\nWe have Ito integral with respect to martingale, so that becomes 0\n\n$= bb(E)\\[integral_0^t rho d u\\]= rho t$\n\nTherefore the covariance is\n\n$frac(\"Cov\"\\(B_1\\,B_2\\), sqrt(bb(V)\\[B_1\\]) sqrt(bb(V)\\[B_2\\])) = rho$\n\nThis is the Kolesky decomposition of these two correlated Brownian motion into independent Brownian motions.\n\nWhat 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\n\n$bb(E)\\[S_1\\]= bb(E) [S_1 \\( 0 \\) e^(\\(alpha_1 - sigma_1^2 / 2\\)t + sigma_1 B_1)]$\n\nSkipping some steps,\n\n$= S_1\\(0\\)e^(alpha_1 t)$\n\nHow do we get the variance of $S_1$?\n\n$bb(V)\\[S_1\\]= bb(E)\\[S_1^2\\]- bb(E)\\[S_1\\]^2$\n\nThis would be\n\n$= S_1^2\\(0\\)e^(2 alpha_1 t + sigma_1^2 t) - S_1^2\\(0\\)e^(2 alpha t)$\n\nWe 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\n\n$bb(E)\\[S_1 S_2\\]- bb(E)\\[S_1\\]bb(E)\\[S_2\\]$\n\nI'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.\n\n$d\\(S_1 S_2\\)= S_1 d S_2 + S_2 d S_1 + d S_1 d S_2$\n\n$= S_1 alpha_2 S_2 d B_2$","dataUrl":"https://sharifhsn.dev/api/posts/multidimensional-ito-calculus.json","date":"2024-10-24","datePublished":"2024-10-24","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!…","site":"https://sharifhsn.dev","slug":"multidimensional-ito-calculus","source":"FE-610 | Stochastic Calculus","sourceUrl":null,"tags":["Stochastic Calculus","Multidimensional Models","Correlation"],"title":"Multidimensional Itô Calculus","url":"https://sharifhsn.dev/blog/multidimensional-ito-calculus/","version":"1","wordCount":3906}