Answer to Question #219685 in Statistics and Probability for joy

Question

Derive the variance of the F distribution.

Answer 1

Solution:

A random variable x has an F distribution if it can be written as a ratio

$X=\frac{Y_{1} / n_{1}}{Y_{2}\left/n_{2}\right.}$

between a Chi-square random variable Y₁ with n₁ degrees of freedom and a Chi-square random variable Y₂, independent of Y₁, with n₂ degrees of freedom (where each of the two random variables has been divided by its degrees of freedom).

Definition Let X be a continuous random variable. Let its support be the set of positive real numbers:

$R_{X}=[0, \infty)$

Let n₁, n₂ $\in \mathbb{N}$ . We say that X has an F distribution with n₁ and n₂ degrees of freedom if and only if its probability density function is

$f_{X}(x)= \begin{cases}c x^{n_{3} / 2-1}\left(1+\frac{n_{1}}{n_{2}} x\right)^{-\left(n_{1}+n_{2}\right) / 2} & \text { if } x \in R_{X} \\ 0 & \text { if } x \in R_{X}\end{cases}$

where c is a constant:

$c=\left(\frac{n_{1}}{n_{2}}\right)^{n_{1} / 2} \frac{1}{B\left(\frac{n_{1}}{2}, \frac{n_{2}}{2}\right)}$

and $B(\quad)$ is the Beta function.

Proof:

$\begin{aligned} & \mathrm{E}\left[X^{2}\right] \\=& \int_{-\infty}^{\infty} x^{2} f_{X}(x) d x \\=& \int_{0}^{\infty} x^{2} c x^{n_{1} / 2-1}\left(1+\frac{n_{1}}{n_{2}} x\right)^{-\left(n_{1}+n_{2}\right) / 2} d x \\=& c \int_{0}^{\infty} x^{n_{1} / 2+1}\left(1+\frac{n_{1}}{n_{2}} x\right)^{-\left(n_{1}+n_{2}\right) / 2} d x \\=& \left.c \int_{0}^{\infty}\left(\frac{n_{2}}{n_{1}} t\right)^{n_{1} / 2+1}(1+t)^{-\left(n_{1}+n_{2}\right) / 2} \frac{n_{2}}{n_{1}} d t \quad \text { (by a change of variable: } t=\frac{n_{1}}{n_{2}} x\right) \end{aligned}$

$=c\left(\frac{n_{2}}{n_{1}}\right)^{n_{1} / 2+2} \int_{0}^{\infty} t^{n_{1} / 2+1}(1+t)^{-n_{1} / 2-n \sqrt{2}} d t \\=c\left(\frac{n_{2}}{n_{1}}\right)^{n_{1} / 2+2} \int_{0}^{\infty} t^{\left(n_{1} / 2+2\right)-1}(1+t)^{-\left(n_{1} / 2+2\right)-\left(n_{2} / 2-2\right)} d t \\=c\left(\frac{n_{2}}{n_{1}}\right)^{n_{1} / 2+2} B\left(\frac{n_{1}}{2}+2, \frac{n_{2}}{2}-2\right) \quad \text{(integral representation of Beta function)} \\=\left(\frac{n_{1}}{n_{2}}\right)^{n / 2} \frac{1}{B\left(\frac{n_{1}}{2}, \frac{n_{2}}{2}\right)}\left(\frac{n_{2}}{n_{1}}\right)^{n / 2+2} B\left(\frac{n_{1}}{2}+2, \frac{n_{2}}{2}-2\right) \text{(substituting c )}$

$=\left(\frac{n_{2}}{n_{1}}\right)^{2} \frac{1}{B\left(\frac{n_{1}}{2}, \frac{n_{2}}{2}\right)} B\left(\frac{n_{1}}{2}+2, \frac{n_{2}}{2}-2\right) \\=\left(\frac{n_{2}}{n_{1}}\right)^{2} \frac{\Gamma\left(n_{1} / 2+n_{2} / 2\right)}{\Gamma\left(n_{1} / 2\right) \Gamma\left(n_{2} / 2\right)} \frac{\Gamma\left(n_{1} / 2+2\right) \Gamma\left(n_{2} / 2-2\right)}{\Gamma\left(n_{1} / 2+2+n_{2} / 2-2\right)} \text{(definition of Beta function)} \\=\left(\frac{n_{2}}{n_{1}}\right)^{2} \frac{\Gamma\left(n_{1} / 2+n_{2} / 2\right)}{\Gamma\left(n_{1} / 2\right) \Gamma\left(n_{2} / 2\right)} \frac{\Gamma\left(n_{1} / 2+2\right) \Gamma\left(n_{2} / 2-2\right)}{\Gamma\left(n_{1} / 2+n_{2} / 2\right)}$

$=\left(\frac{n_{2}}{n_{1}}\right)^{2} \frac{\Gamma\left(n_{1} / 2+2\right)}{\Gamma\left(n_{1} / 2\right)} \frac{\Gamma\left(n_{2} / 2-2\right)}{\Gamma\left(n_{2} / 2\right)} \\=\left(\frac{n_{2}}{n_{1}}\right)^{2}\left(n_{1} / 2+1\right)\left(n_{1} / 2\right) \frac{1}{\left(n_{2} / 2-1\right)\left(n_{2} / 2-2\right)} \quad (\because \left.\Gamma(z)=\Gamma(z-1)(z-1)\right) \\=\frac{n_{2}^{2}\left(n_{1}+2\right) n_{1}}{n_{1}^{2}\left(n_{2}-2\right)\left(n_{2}-4\right)} \\=\frac{n_{2}^{2}\left(n_{1}+2\right)}{n_{1}\left(n_{2}-2\right)\left(n_{2}-4\right)} \mathrm{E}[X]^{2}\\=\left(\frac{n_{2}}{n_{2}-2}\right)^{2} \\\operatorname{Var}[X]=\mathrm{E}\left[X^{2}\right]-\mathrm{E}[X]^{2}$

$=\frac{n_{2}^{2}\left(n_{1}+2\right)}{n_{1}\left(n_{2}-2\right)\left(n_{2}-4\right)}-\frac{n_{2}^{2}}{\left(n_{2}-2\right)^{2}} \\=\frac{n_{2}^{2}\left(\left(n_{1}+2\right)\left(n_{2}-2\right)-n_{1}\left(n_{2}-4\right)\right)}{n_{1}\left(n_{2}-2\right)^{2}\left(n_{2}-4\right)} \\=\frac{n_{2}^{2}\left(n_{1} n_{2}-2 n_{1}+2 n_{2}-4-n_{1} n_{2}+4 n_{1}\right)}{n_{1}\left(n_{2}-2\right)^{2}\left(n_{2}-4\right)} \\=\frac{n_{2}^{2}\left(2 n_{1}+2 n_{2}-4\right)}{n_{1}\left(n_{2}-2\right)^{2}\left(n_{2}-4\right)}=\frac{2 n_{2}^{2}\left(n_{1}+n_{2}-2\right)}{n_{1}\left(n_{2}-2\right)^{2}\left(n_{2}-4\right)}$