Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate continuous distributions ▷ Beta distribution ▷ Relationship to chi-squared distribution

Theorem: Let $X$ and $Y$ be independent random variables following chi-squared distributions:

$\label{eq:chi2} X \sim \chi^2(m) \quad \text{and} \quad Y \sim \chi^2(n) \; .$

Then, the quantity $X/(X+Y)$ follows a beta distribution:

$\label{eq:beta-chi2} \frac{X}{X+Y} \sim \mathrm{Bet}\left( \frac{m}{2}, \frac{n}{2} \right) \; .$ $\label{eq:chi2-pdf} X \sim \chi^2(u) \quad \Rightarrow \quad f_X(x) = \frac{1}{\Gamma\left( \frac{u}{2} \right) \cdot 2^{u/2}} \cdot x^{\frac{u}{2}-1} \cdot e^{-\frac{x}{2}} \; .$

Define the random variables $Z$ and $W$ as functions of $X$ and $Y$

$\label{eq:ZW-XY} \begin{split} Z &= \frac{X}{X+Y} \\ W &= Y \; , \end{split}$

such that the inverse functions $X$ and $Y$ in terms of $Z$ and $W$ are

$\label{eq:XY-ZW} \begin{split} X &= \frac{ZW}{1-Z} \\ Y &= W \; . \end{split}$

This implies the following Jacobian matrix and determinant:

$\label{eq:XY-ZW-jac} \begin{split} J &= \left[ \begin{matrix} \frac{\mathrm{d}X}{\mathrm{d}Z} & \frac{\mathrm{d}X}{\mathrm{d}W} \\ \frac{\mathrm{d}Y}{\mathrm{d}Z} & \frac{\mathrm{d}Y}{\mathrm{d}W} \end{matrix} \right] = \left[ \begin{matrix} \frac{W}{(1-Z)^2} & \frac{Z}{1-Z} \\ 0 & 1 \end{matrix} \right] \\ \lvert J \rvert &= \frac{W}{(1-Z)^2} \; . \end{split}$

Because $X$ and $Y$ are independent, the joint density of $X$ and $Y$ is equal to the product of the marginal densities:

$\label{eq:f-XY} f_{X,Y}(x,y) = f_X(x) \cdot f_Y(y) \; .$

With the probability density function of an invertible function, the joint density of $Z$ and $W$ can be derived as:

$\label{eq:f-ZW-s1} f_{Z,W}(z,w) = f_{X,Y}(x,y) \cdot \lvert J \rvert \; .$

Substituting \eqref{eq:XY-ZW} into \eqref{eq:chi2-pdf}, and then with \eqref{eq:XY-ZW-jac} into \eqref{eq:f-ZW-s1}, we get:

$\label{eq:f-ZW-s2} \begin{split} f_{Z,W}(z,w) &= f_X\left( \frac{zw}{1-z} \right) \cdot f_Y(w) \cdot \lvert J \rvert \\ &= \frac{1}{\Gamma\left( \frac{m}{2} \right) \cdot 2^{m/2}} \cdot \left( \frac{zw}{1-z} \right)^{\frac{m}{2}-1} \cdot e^{-\frac{1}{2} \left( \frac{zw}{1-z} \right)} \cdot \frac{1}{\Gamma\left( \frac{n}{2} \right) \cdot 2^{n/2}} \cdot w^{\frac{n}{2}-1} \cdot e^{-\frac{w}{2}} \cdot \frac{w}{(1-z)^2} \\ &= \frac{1}{\Gamma\left( \frac{m}{2} \right) \Gamma\left( \frac{n}{2} \right) \cdot 2^{m/2} 2^{n/2}} \cdot \left( \frac{z}{1-z} \right)^{\frac{m}{2}-1} \left( \frac{1}{(1-z)} \right)^2 \cdot w^{\frac{m}{2}+\frac{n}{2}-1} e^{-\frac{1}{2} \left( \frac{zw}{1-z} + \frac{w(1-z)}{1-z} \right)} \\ &= \frac{1}{\Gamma\left( \frac{m}{2} \right) \Gamma\left( \frac{n}{2} \right) \cdot 2^{(m+n)/2}} \cdot z^{\frac{m}{2}-1} \cdot (1-z)^{-\frac{m}{2}-1} \cdot w^{\frac{m+n}{2}-1} \cdot e^{-\frac{1}{2} \left( \frac{w}{1-z} \right)} \; . \end{split}$

The marginal density of $Z$ can now be obtained by integrating out $W$:

$\label{eq:f-Z-s1} \begin{split} f_Z(z) &= \int_{0}^{\infty} f_{Z,W}(z,w) \, \mathrm{d}w \\ &= \frac{1}{\Gamma\left( \frac{m}{2} \right) \Gamma\left( \frac{n}{2} \right) \cdot 2^{(m+n)/2}} \cdot z^{\frac{m}{2}-1} \cdot (1-z)^{-\frac{m}{2}-1} \cdot \int_{0}^{\infty} w^{\frac{m+n}{2}-1} \cdot e^{-\frac{1}{2} \left( \frac{w}{1-z} \right)} \, \mathrm{d}w \\ &= \frac{1}{\Gamma\left( \frac{m}{2} \right) \Gamma\left( \frac{n}{2} \right) \cdot 2^{(m+n)/2}} \cdot z^{\frac{m}{2}-1} \cdot (1-z)^{-\frac{m}{2}-1} \cdot \frac{\Gamma\left( \frac{m+n}{2} \right)}{\left( \frac{1}{2(1-z)} \right)^{\frac{m+n}{2}}} \cdot \\ &\hphantom{=} \int_{0}^{\infty} \frac{\left( \frac{1}{2(1-z)} \right)^{\frac{m+n}{2}}}{\Gamma\left( \frac{m+n}{2} \right)} \cdot w^{\frac{m+n}{2}-1} \cdot e^{-\frac{1}{2(1-z)} \, w} \, \mathrm{d}w \; . \end{split}$

At this point, we can recognize that the integrand is equal to the probability density function of a gamma distribution with

$\label{eq:f-W-gam-ab} a = \frac{m+n}{2} \quad \text{and} \quad b = \frac{1}{2(1-z)} \; ,$ $\label{eq:f-Z-s2} \begin{split} f_Z(z) &= \frac{1}{\Gamma\left( \frac{m}{2} \right) \Gamma\left( \frac{n}{2} \right) \cdot 2^{(m+n)/2}} \cdot z^{\frac{m}{2}-1} \cdot (1-z)^{-\frac{m}{2}-1} \cdot \frac{\Gamma\left( \frac{m+n}{2} \right)}{\left( \frac{1}{2(1-z)} \right)^{\frac{m+n}{2}}} \\ &= \frac{\Gamma\left( \frac{m+n}{2} \right) \cdot 2^{(m+n)/2}}{\Gamma\left( \frac{m}{2} \right) \Gamma\left( \frac{n}{2} \right) \cdot 2^{(m+n)/2}} \cdot z^{\frac{m}{2}-1} \cdot (1-z)^{-\frac{m}{2}+\frac{m+n}{2}-1} \\ &= \frac{\Gamma\left( \frac{m+n}{2} \right)}{\Gamma\left( \frac{m}{2} \right) \Gamma\left( \frac{n}{2} \right)} \cdot z^{\frac{m}{2}-1} \cdot (1-z)^{\frac{n}{2}-1} \; . \end{split}$

With the definition of the beta function, this becomes

$\label{eq:f-Z-s3} f_Z(z) = \frac{1}{\mathrm{B}\left( \frac{m}{2}, \frac{n}{2} \right)} \cdot z^{\frac{m}{2}-1} \cdot (1-z)^{\frac{n}{2}-1}$

which is the probability density function of the beta distribution with parameters

$\label{eq:beta-chi2-para} \alpha = \frac{m}{2} \quad \mathrm{and} \quad \beta = \frac{n}{2} \; ,$

such that

$\label{eq:beta-chi2-qed} Z \sim \mathrm{Bet}\left( \frac{m}{2}, \frac{n}{2} \right) \; .$
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Metadata: ID: P356 | shortcut: beta-chi2 | author: JoramSoch | date: 2022-10-07, 13:20.