{"categories":["Probability Theory"],"contentHtml":"<h3>Week 8</h3><h4>Lecture Notes</h4><h5>Chi-Squared</h5><p>Definition 5.36 says that X is a Chi squared distribution whose pdf is described by</p><p><math><mi>𝑓</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑥</mi><mo form=\"postfix\" stretchy=\"false\">)</mo><mo>=</mo><mfrac><mn>1</mn><mrow><msup><mn>2</mn><mrow><mi>𝑛</mi><mi>/</mi><mn>2</mn></mrow></msup><mi mathvariant=\"normal\">Γ</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mfrac><mi>𝑛</mi><mn>2</mn></mfrac><mo stretchy=\"false\">)</mo></mrow></mfrac><msup><mi>𝑥</mi><mrow><mfrac><mi>𝑛</mi><mn>2</mn></mfrac><mo lspace=\"0em\" rspace=\"0em\">−</mo><mn>1</mn></mrow></msup><msup><mi>𝑒</mi><mrow><mo>−</mo><mfrac><mi>𝑥</mi><mn>2</mn></mfrac></mrow></msup></math></p><p>where <math><mi>𝑋</mi><mo>∼</mo><msubsup><mi>𝑋</mi><mi>𝑛</mi><mn>2</mn></msubsup></math></p><p>This is actually linked to the gamma distribution</p><p><math><mi>𝑋</mi><mo>∼</mo><msubsup><mi>𝑋</mi><mi>𝑛</mi><mn>2</mn></msubsup></math>, then <math><mi>𝑋</mi><mo>∼</mo><mi mathvariant=\"normal\">Γ</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mfrac><mi>𝑛</mi><mn>2</mn></mfrac><mo>,</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo stretchy=\"false\">)</mo></math></p><p>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.</p><h5>Beta Distribution</h5><p>Described by 5.4.5</p><p>Gamma distribution is defined under the gamma function. We will need a two variable beta function. The intermediate function is defined as</p><p><math><mi mathvariant=\"normal\">B</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑎</mi><mo>,</mo><mi>𝑏</mi><mo form=\"postfix\" stretchy=\"false\">)</mo><mo>≔</mo><msubsup><mo form=\"prefix\" lspace=\"0em\">∫</mo><mn>0</mn><mn>1</mn></msubsup><msup><mi>𝑥</mi><mrow><mi>𝑎</mi><mo lspace=\"0em\" rspace=\"0em\">−</mo><mn>1</mn></mrow></msup><mo form=\"prefix\" stretchy=\"false\">(</mo><mn>1</mn><mo>−</mo><mi>𝑥</mi><msup><mo form=\"postfix\" stretchy=\"false\">)</mo><mrow><mi>𝑏</mi><mo lspace=\"0em\" rspace=\"0em\">−</mo><mn>1</mn></mrow></msup><mspace width=\"0.1667em\"></mspace><mi>𝑑</mi><mi>𝑥</mi></math></p><p>The integral is univariate. There’s an apparent symmetry. If we do a change of variable of <math><mi>𝑦</mi><mo>=</mo><mn>1</mn><mo>−</mo><mi>𝑥</mi></math> then</p><p><math><mi mathvariant=\"normal\">B</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑏</mi><mo>,</mo><mi>𝑎</mi><mo form=\"postfix\" stretchy=\"false\">)</mo><mo>=</mo><msubsup><mo form=\"prefix\" lspace=\"0em\">∫</mo><mn>0</mn><mn>1</mn></msubsup><msup><mi>𝑥</mi><mrow><mi>𝑏</mi><mo lspace=\"0em\" rspace=\"0em\">−</mo><mn>1</mn></mrow></msup><msup><mi>𝑦</mi><mrow><mi>𝑎</mi><mo lspace=\"0em\" rspace=\"0em\">−</mo><mn>1</mn></mrow></msup><mspace width=\"0.1667em\"></mspace><mi>𝑑</mi><mi>𝑥</mi></math></p><p>and</p><p><math><mo form=\"infix\" lspace=\"0em\">=</mo><msubsup><mo form=\"prefix\" lspace=\"0em\" rspace=\"0em\">∫</mo><mn>1</mn><mn>0</mn></msubsup><mo form=\"prefix\" stretchy=\"false\">(</mo><mn>1</mn><mo>−</mo><mi>𝑦</mi><msup><mo form=\"postfix\" stretchy=\"false\">)</mo><mrow><mi>𝑏</mi><mo lspace=\"0em\" rspace=\"0em\">−</mo><mn>1</mn></mrow></msup><msup><mi>𝑦</mi><mrow><mi>𝑎</mi><mo lspace=\"0em\" rspace=\"0em\">−</mo><mn>1</mn></mrow></msup><mo form=\"prefix\" stretchy=\"false\">(</mo><mo form=\"prefix\">−</mo><mi>𝑑</mi><mi>𝑦</mi><mo stretchy=\"false\">)</mo></math></p><p>Which when you cancel out the negative, they end up being equal. And in fact it can be shown that</p><p><math><mi mathvariant=\"normal\">B</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑎</mi><mo>,</mo><mi>𝑏</mi><mo form=\"postfix\" stretchy=\"false\">)</mo><mo>=</mo><mi mathvariant=\"normal\">B</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑏</mi><mo>,</mo><mi>𝑎</mi><mo stretchy=\"false\">)</mo></math></p><p>and a more useful result proposition 5.39, which is that</p><p><math><mi mathvariant=\"normal\">B</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑎</mi><mo>,</mo><mi>𝑏</mi><mo form=\"postfix\" stretchy=\"false\">)</mo><mo>=</mo><mfrac><mrow><mi mathvariant=\"normal\">Γ</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑎</mi><mo form=\"postfix\" stretchy=\"false\">)</mo><mi mathvariant=\"normal\">Γ</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑏</mi><mo stretchy=\"false\">)</mo></mrow><mrow><mi mathvariant=\"normal\">Γ</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑎</mi><mo lspace=\"0em\" rspace=\"0em\">+</mo><mi>𝑏</mi><mo stretchy=\"false\">)</mo></mrow></mfrac></math></p><p>this will be very useful later. Proving this is very tedious calculus.</p><p>We can now introduce the beta distribution through definition 5.40. If <math><mi>𝑥</mi></math> has the following pdf</p><p><math><msub><mi>𝑓</mi><mrow><mi>𝑎</mi><mo rspace=\"0em\">,</mo><mi>𝑏</mi></mrow></msub><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑥</mi><mo form=\"postfix\" stretchy=\"false\">)</mo><mo>=</mo><mfrac><mrow><msup><mi>𝑥</mi><mrow><mi>𝑎</mi><mo lspace=\"0em\" rspace=\"0em\">−</mo><mn>1</mn></mrow></msup><mo form=\"prefix\" stretchy=\"false\">(</mo><mn>1</mn><mo lspace=\"0em\" rspace=\"0em\">−</mo><mi>𝑥</mi><msup><mo stretchy=\"false\">)</mo><mrow><mi>𝑏</mi><mo lspace=\"0em\" rspace=\"0em\">−</mo><mi>𝑎</mi></mrow></msup></mrow><mrow><mi mathvariant=\"normal\">B</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑎</mi><mo rspace=\"0em\">,</mo><mi>𝑏</mi><mo stretchy=\"false\">)</mo></mrow></mfrac><msub><mi>𝐼</mi><mrow><mo stretchy=\"false\">[</mo><mn>0</mn><mo rspace=\"0em\">,</mo><mn>1</mn><mo stretchy=\"false\">]</mo></mrow></msub><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑥</mi><mo stretchy=\"false\">)</mo></math></p><p>then it’s in the beta distribution.</p><p>Proposition 5.41 says that the integral is 1.</p><p>Proposition 5.42 tells us the expected value and variance.</p><p><math><mi>𝔼</mi><mo form=\"prefix\" stretchy=\"false\">[</mo><mi>𝑋</mi><mo form=\"postfix\" stretchy=\"false\">]</mo><mo>=</mo><mfrac><mi>𝑎</mi><mrow><mi>𝑎</mi><mo lspace=\"0em\" rspace=\"0em\">+</mo><mi>𝑏</mi></mrow></mfrac></math></p><p><math><mi>𝕍</mi><mo form=\"prefix\" stretchy=\"false\">[</mo><mi>𝑋</mi><mo form=\"postfix\" stretchy=\"false\">]</mo><mo>=</mo><mfrac><mrow><mi>𝑎</mi><mi>𝑏</mi></mrow><mrow><mo stretchy=\"false\">(</mo><mi>𝑎</mi><mo lspace=\"0em\" rspace=\"0em\">+</mo><mi>𝑏</mi><msup><mo form=\"postfix\" stretchy=\"false\">)</mo><mn>2</mn></msup><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑎</mi><mo lspace=\"0em\" rspace=\"0em\">+</mo><mi>𝑏</mi><mo lspace=\"0em\" rspace=\"0em\">+</mo><mn>1</mn><mo stretchy=\"false\">)</mo></mrow></mfrac></math> To prove this, we take the definition of the expectation</p><p><math><mi>𝔼</mi><mo form=\"prefix\" stretchy=\"false\">[</mo><mi>𝑋</mi><mo form=\"postfix\" stretchy=\"false\">]</mo><mo>=</mo><msubsup><mo form=\"prefix\" lspace=\"0em\">∫</mo><mn>0</mn><mn>1</mn></msubsup><mi>𝑥</mi><msub><mi>𝑓</mi><mrow><mi>𝑎</mi><mo rspace=\"0em\">,</mo><mi>𝑏</mi></mrow></msub><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑥</mi><mo form=\"postfix\" stretchy=\"false\">)</mo><mi>𝑑</mi><mi>𝑥</mi></math></p><p><math><mo form=\"infix\" lspace=\"0em\">=</mo><mfrac><mn>1</mn><mrow><mi mathvariant=\"normal\">B</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑎</mi><mo rspace=\"0em\">,</mo><mi>𝑏</mi><mo stretchy=\"false\">)</mo></mrow></mfrac><msubsup><mo form=\"prefix\">∫</mo><mn>0</mn><mn>1</mn></msubsup><msup><mi>𝑥</mi><mi>𝑎</mi></msup><mo form=\"prefix\" stretchy=\"false\">(</mo><mn>1</mn><mo>−</mo><mi>𝑥</mi><msup><mo form=\"postfix\" stretchy=\"false\">)</mo><mrow><mi>𝑏</mi><mo lspace=\"0em\" rspace=\"0em\">−</mo><mn>1</mn></mrow></msup><mspace width=\"0.1667em\"></mspace><mi>𝑑</mi><mi>𝑥</mi></math></p><p>We can then adjust the <math><mi>𝑎</mi></math> to make it agree with the beta function</p><p><math><mo form=\"infix\" lspace=\"0em\">=</mo><mfrac><mrow><mi mathvariant=\"normal\">B</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑎</mi><mo lspace=\"0em\" rspace=\"0em\">+</mo><mn>1</mn><mo rspace=\"0em\">,</mo><mi>𝑏</mi><mo stretchy=\"false\">)</mo></mrow><mrow><mi mathvariant=\"normal\">B</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑎</mi><mo rspace=\"0em\">,</mo><mi>𝑏</mi><mo stretchy=\"false\">)</mo></mrow></mfrac></math></p><p>Now I can adjust it to be the gamma function instead.</p><p><math><mo form=\"infix\" lspace=\"0em\">=</mo><mfrac><mrow><mi mathvariant=\"normal\">Γ</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑎</mi><mo lspace=\"0em\" rspace=\"0em\">+</mo><mi>𝑏</mi><mo stretchy=\"false\">)</mo></mrow><mrow><mi mathvariant=\"normal\">Γ</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑎</mi><mo form=\"postfix\" stretchy=\"false\">)</mo><mi mathvariant=\"normal\">Γ</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑏</mi><mo stretchy=\"false\">)</mo></mrow></mfrac><mo lspace=\"0.2222222222222222em\" rspace=\"0.2222222222222222em\">⋅</mo><mfrac><mrow><mi mathvariant=\"normal\">Γ</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑎</mi><mo lspace=\"0em\" rspace=\"0em\">+</mo><mn>1</mn><mo form=\"postfix\" stretchy=\"false\">)</mo><mi mathvariant=\"normal\">Γ</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑏</mi><mo stretchy=\"false\">)</mo></mrow><mrow><mi mathvariant=\"normal\">Γ</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑎</mi><mo lspace=\"0em\" rspace=\"0em\">+</mo><mi>𝑏</mi><mo lspace=\"0em\" rspace=\"0em\">+</mo><mn>1</mn><mo stretchy=\"false\">)</mo></mrow></mfrac></math></p><p>We can cancel out some things, and then utilize the property of the gamma function that <math><mi mathvariant=\"normal\">Γ</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑛</mi><mo>+</mo><mn>1</mn><mo form=\"postfix\" stretchy=\"false\">)</mo><mo>=</mo><mi>𝑛</mi><mi mathvariant=\"normal\">Γ</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑛</mi><mo stretchy=\"false\">)</mo></math></p><p><math><mfrac><mi>𝑎</mi><mrow><mi>𝑎</mi><mo lspace=\"0em\" rspace=\"0em\">+</mo><mi>𝑏</mi></mrow></mfrac></math></p><p>Then for variance, skipping some steps, we can show that</p><p><math><mi>𝔼</mi><mo form=\"prefix\" stretchy=\"false\">[</mo><msup><mi>𝑋</mi><mn>2</mn></msup><mo form=\"postfix\" stretchy=\"false\">]</mo><mo>=</mo><mfrac><mn>1</mn><mrow><mi mathvariant=\"normal\">B</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑎</mi><mo rspace=\"0em\">,</mo><mi>𝑏</mi><mo stretchy=\"false\">)</mo></mrow></mfrac><msubsup><mo form=\"prefix\">∫</mo><mn>0</mn><mn>1</mn></msubsup><msup><mi>𝑥</mi><mrow><mi>𝑎</mi><mo lspace=\"0em\" rspace=\"0em\">+</mo><mn>1</mn></mrow></msup><mo form=\"prefix\" stretchy=\"false\">(</mo><mn>1</mn><mo>−</mo><mi>𝑥</mi><msup><mo form=\"postfix\" stretchy=\"false\">)</mo><mrow><mi>𝑏</mi><mo lspace=\"0em\" rspace=\"0em\">−</mo><mn>1</mn></mrow></msup><mspace width=\"0.1667em\"></mspace><mi>𝑑</mi><mi>𝑥</mi></math></p><p>Using a similar method, we can rewrite this as</p><p><math><mo form=\"infix\" lspace=\"0em\">=</mo><mfrac><mrow><mi mathvariant=\"normal\">B</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑎</mi><mo lspace=\"0em\" rspace=\"0em\">+</mo><mn>2</mn><mo rspace=\"0em\">,</mo><mi>𝑏</mi><mo stretchy=\"false\">)</mo></mrow><mrow><mi mathvariant=\"normal\">B</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑎</mi><mo rspace=\"0em\">,</mo><mi>𝑏</mi><mo stretchy=\"false\">)</mo></mrow></mfrac></math></p><p>Now we have from gamma</p><p><math><mo form=\"infix\" lspace=\"0em\">=</mo><mfrac><mrow><mi mathvariant=\"normal\">Γ</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑎</mi><mo lspace=\"0em\" rspace=\"0em\">+</mo><mn>2</mn><mo form=\"postfix\" stretchy=\"false\">)</mo><mi mathvariant=\"normal\">Γ</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑏</mi><mo stretchy=\"false\">)</mo></mrow><mrow><mi mathvariant=\"normal\">Γ</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑎</mi><mo lspace=\"0em\" rspace=\"0em\">+</mo><mi>𝑏</mi><mo lspace=\"0em\" rspace=\"0em\">+</mo><mn>2</mn><mo stretchy=\"false\">)</mo></mrow></mfrac><mo lspace=\"0.2222222222222222em\" rspace=\"0.2222222222222222em\">⋅</mo><mfrac><mrow><mi mathvariant=\"normal\">Γ</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑎</mi><mo lspace=\"0em\" rspace=\"0em\">+</mo><mi>𝑏</mi><mo stretchy=\"false\">)</mo></mrow><mrow><mi mathvariant=\"normal\">Γ</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑎</mi><mo form=\"postfix\" stretchy=\"false\">)</mo><mi mathvariant=\"normal\">Γ</mi><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑏</mi><mo stretchy=\"false\">)</mo></mrow></mfrac></math></p><p>canceling and substituting with our values, it’s</p><p><math><mo form=\"infix\" lspace=\"0em\">=</mo><mfrac><mrow><mo stretchy=\"false\">(</mo><mi>𝑎</mi><mo lspace=\"0em\" rspace=\"0em\">+</mo><mn>1</mn><mo form=\"postfix\" stretchy=\"false\">)</mo><mi>𝑎</mi></mrow><mrow><mo stretchy=\"false\">(</mo><mi>𝑎</mi><mo lspace=\"0em\" rspace=\"0em\">+</mo><mi>𝑏</mi><mo lspace=\"0em\" rspace=\"0em\">+</mo><mn>1</mn><mo form=\"postfix\" stretchy=\"false\">)</mo><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑎</mi><mo lspace=\"0em\" rspace=\"0em\">+</mo><mi>𝑏</mi><mo stretchy=\"false\">)</mo></mrow></mfrac></math></p><p>Then to calculate variance, we get</p><p><math><mo form=\"infix\" lspace=\"0em\">=</mo><mfrac><mrow><mo stretchy=\"false\">(</mo><mi>𝑎</mi><mo lspace=\"0em\" rspace=\"0em\">+</mo><mn>1</mn><mo form=\"postfix\" stretchy=\"false\">)</mo><mi>𝑎</mi></mrow><mrow><mo stretchy=\"false\">(</mo><mi>𝑎</mi><mo lspace=\"0em\" rspace=\"0em\">+</mo><mi>𝑏</mi><mo lspace=\"0em\" rspace=\"0em\">+</mo><mn>1</mn><mo form=\"postfix\" stretchy=\"false\">)</mo><mo form=\"prefix\" stretchy=\"false\">(</mo><mi>𝑎</mi><mo lspace=\"0em\" rspace=\"0em\">+</mo><mi>𝑏</mi><mo stretchy=\"false\">)</mo></mrow></mfrac><mo>−</mo><mfrac><msup><mi>𝑎</mi><mn>2</mn></msup><mrow><mo stretchy=\"false\">(</mo><mi>𝑎</mi><mo lspace=\"0em\" rspace=\"0em\">+</mo><mi>𝑏</mi><msup><mo stretchy=\"false\">)</mo><mn>2</mn></msup></mrow></mfrac></math></p><p>which will simplify into our answer</p>","contentSourceFormat":"typst","contentTypst":"== Week 8\n<week-8>\n=== Lecture Notes\n<lecture-notes>\n==== Chi-Squared\n<chi-squared>\nDefinition 5.36 says that X is a Chi squared distribution whose pdf is described by\n\n$f\\(x\\)= frac(1, 2^(n\\/2) Gamma\\(n / 2\\)) x^(n / 2 - 1) e^(- x / 2)$\n\nwhere $X tilde.op X_n^2$\n\nThis is actually linked to the gamma distribution\n\n$X tilde.op X_n^2$, then $X tilde.op Gamma\\(n / 2\\,1 / 2\\)$\n\nThere 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.\n\n==== Beta Distribution\n<beta-distribution>\nDescribed by 5.4.5\n\nGamma distribution is defined under the gamma function. We will need a two variable beta function. The intermediate function is defined as\n\n$upright(B)\\(a\\,b\\):= integral_0^1 x^(a - 1)\\(1 - x\\)^(b - 1)thin d x$\n\nThe integral is univariate. There's an apparent symmetry. If we do a change of variable of $y = 1 - x$ then\n\n$upright(B)\\(b\\,a\\)= integral_0^1 x^(b - 1) y^(a - 1) thin d x$\n\nand\n\n$= integral_1^0\\(1 - y\\)^(b - 1)y^(a - 1)\\(- d y\\)$\n\nWhich when you cancel out the negative, they end up being equal. And in fact it can be shown that\n\n$upright(B)\\(a\\,b\\)= upright(B)\\(b\\,a\\)$\n\nand a more useful result proposition 5.39, which is that\n\n$upright(B)\\(a\\,b\\)= frac(Gamma\\(a\\)Gamma\\(b\\), Gamma\\(a + b\\))$\n\nthis will be very useful later. Proving this is very tedious calculus.\n\nWe can now introduce the beta distribution through definition 5.40. If $x$ has the following pdf\n\n$f_(a\\,b)\\(x\\)= frac(x^(a - 1)\\(1 - x\\)^(b - a), upright(B)\\(a\\,b\\)) I_(\\[0\\,1\\])\\(x\\)$\n\nthen it's in the beta distribution.\n\nProposition 5.41 says that the integral is 1.\n\nProposition 5.42 tells us the expected value and variance.\n\n$bb(E)\\[X\\]= frac(a, a + b)$\n\n$bb(V)\\[X\\]= frac(a b, \\(a + b\\)^2\\(a + b + 1\\))$ To prove this, we take the definition of the expectation\n\n$bb(E)\\[X\\]= integral_0^1 x f_(a\\,b)\\(x\\)d x$\n\n$= frac(1, upright(B)\\(a\\,b\\)) integral_0^1 x^a\\(1 - x\\)^(b - 1)thin d x$\n\nWe can then adjust the $a$ to make it agree with the beta function\n\n$= frac(upright(B)\\(a + 1\\,b\\), upright(B)\\(a\\,b\\))$\n\nNow I can adjust it to be the gamma function instead.\n\n$= frac(Gamma\\(a + b\\), Gamma\\(a\\)Gamma\\(b\\)) dot.op frac(Gamma\\(a + 1\\)Gamma\\(b\\), Gamma\\(a + b + 1\\))$\n\nWe can cancel out some things, and then utilize the property of the gamma function that $Gamma\\(n + 1\\)= n Gamma\\(n\\)$\n\n$frac(a, a + b)$\n\nThen for variance, skipping some steps, we can show that\n\n$bb(E)\\[X^2\\]= frac(1, upright(B)\\(a\\,b\\)) integral_0^1 x^(a + 1)\\(1 - x\\)^(b - 1)thin d x$\n\nUsing a similar method, we can rewrite this as\n\n$= frac(upright(B)\\(a + 2\\,b\\), upright(B)\\(a\\,b\\))$\n\nNow we have from gamma\n\n$= frac(Gamma\\(a + 2\\)Gamma\\(b\\), Gamma\\(a + b + 2\\)) dot.op frac(Gamma\\(a + b\\), Gamma\\(a\\)Gamma\\(b\\))$\n\ncanceling and substituting with our values, it's\n\n$= frac(\\(a + 1\\)a, \\(a + b + 1\\)\\(a + b\\))$\n\nThen to calculate variance, we get\n\n$= frac(\\(a + 1\\)a, \\(a + b + 1\\)\\(a + b\\)) - frac(a^2, \\(a + b\\)^2)$\n\nwhich will simplify into our answer","dataUrl":"https://sharifhsn.dev/api/posts/gamma-beta-and-chi-squared-distributions.json","date":"2024-10-21","datePublished":"2024-10-21","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…","site":"https://sharifhsn.dev","slug":"gamma-beta-and-chi-squared-distributions","source":"FE-540 | Probability Theory","sourceUrl":null,"tags":["Probability Theory","Gamma Distribution","Beta Distribution"],"title":"Gamma, Beta, and Chi-Squared Distributions","url":"https://sharifhsn.dev/blog/gamma-beta-and-chi-squared-distributions/","version":"1","wordCount":723}