Chapter 7.2, Problem 34E

Suppose that W₁ and W₂ are independent χ²-distributed random variables with ν₁ and ν₂ df, respectively. According to Definition 7.3,

$F=\frac{W_1/v_1}{W_2/v_2}$

has an F distribution with ν₁ and ν₂ numerator and denominator degrees of freedom, respectively. Use the preceding structure of F, the independence of W₁ and W₂, and the result summarized in Exercise 7.30(b) to show

1. a E(F) = ν₂/(ν₂ − 2), if ν₂ > 2.
2. b $V(F)=[2v_2^2(v_1+v_2-2)]/[v_1{(v_2-2)}^2(v_2-4)]$, if ν₂ > 4.

^*7.30    Suppose that Z has a standard normal distribution and that Y is an independent χ²-distributed random variable with ν df. Then, according to Definition 7.2,

$T=\frac{Z}{\sqrt{Y/v}}$

has a t distribution with ν df.

1. a If Z has a standard normal distribution, give E(Z) and E(Z²). [Hint: For any random variable, E(Z²) = V(Z) + (E(Z))².]
2. b According to the result derived in Exercise 4.112(a), if Y has a χ² distribution with ν df, then

   $E(Y^a)=\frac{\Gamma ([v/2]+a)}{\Gamma (v/2)}{2}^a,\text{if }v>-2a.$

   Use this result, the result from part (a), and the structure of T to show the following. [Hint: Recall the independence of Z and Y.]

   1.                                 i.            E(T) = 0, if ν > 1.
   2.                               ii.            V(T) = ν/(ν − 2), if ν > 2.

To determine

Prove that $E\left(F\right)=\frac{{\nu }_2}{\left({\nu }_2-2\right)},\text{if}{\nu }_2>2$.

Explanation of Solution

Let $F=\frac{W_1/{\nu }_1}{W_2/{\nu }_2}$

Here, $W_1$ has a ${\chi }^2$-distributed random variables with ${\nu }_1$ df and $W_2$ has a ${\chi }^2$-distributed random variables with ${\nu }_2$ df. $W_1$ and $W_2$ are independent.

$\begin{array}{c}E\left(F\right)=E\left(\frac{W_1/{\nu }_1}{W_2/{\nu }_2}\right)\\ =\frac{{\nu }_2}{{\nu }_1}E\left(\frac{W_1}{W_2}\right)\\ =\frac{{\nu }_2}{{\nu }_1}E\left(W_1\right)E\left(\frac{1}{W_2}\right)\end{array}$

From the exercise 7.30(b), if Y has a ${\chi }^2$ distribution with $\nu$ df, then $E\left(Y^a\right)=\frac{\Gamma \left(\left[\nu /2\right]+a\right)}{\Gamma \left(\nu /2\right)}{2}^a,\text{if}\nu >-2a$.

Here,

$\begin{array}{c}E\left(\frac{1}{W_2}\right)=E\left(W_2^{-1}\right)\\ =\frac{\Gamma \left(\left[{\nu }_2/2\right]-1\right)}{\Gamma \left({\nu }_2/2\right)}{2}^{-1}\\ =\frac{\Gamma \left(\left[{\nu }_2/2\right]-1\right)}{\left(\frac{{\nu }_2}{2}-1\right)\Gamma \left(\frac{{\nu }_2}{2}-1\right)}{2}^{-1}\\ =\frac{2\times 2^{-1}}{{\nu }_2-2}\\ =\frac{1}{{\nu }_2-2}\end{array}$

Then,

$\begin{array}{c}E\left(F\right)=\frac{{\nu }_2}{{\nu }_1}\times {\nu }_1\times \frac{1}{{\nu }_2-2}\\ =\frac{{\nu }_2}{{\nu }_2-2},{\nu }_2>2\end{array}$

Hence proved.

To determine

Prove that $V\left(F\right)=\frac{2{\nu }_2^2\left({\nu }_1+{\nu }_2-2\right)}{{\nu }_2{\left({\nu }_2-2\right)}^2\left({\nu }_2-4\right)},\text{if}{\nu }_2>4$.

Explanation of Solution

The variance of F is calculated as follows:

$V\left(F\right)=E\left(F^2\right)-{\left\{E\left(F\right)\right\}}^2$

$\begin{array}{c}E\left(\frac{1}{W_2^2}\right)=E\left(W_2^{-2}\right)\\ =\frac{\Gamma \left(\left[{\nu }_2/2\right]-2\right)}{\Gamma \left({\nu }_2/2\right)}{2}^{-2}\\ =\frac{\Gamma \left(\left[{\nu }_2/2\right]-2\right)}{\left(\frac{{\nu }_2}{2}-1\right)\left(\frac{{\nu }_2}{2}-2\right)\Gamma \left(\frac{{\nu }_2}{2}-2\right)}{2}^{-2}\\ =\frac{2^{-2}}{\left(\frac{{\nu }_2}{2}-1\right)\left(\frac{{\nu }_2}{2}-2\right)}\\ =\frac{2^{-2}\times 2^2}{\left({\nu }_2-2\right)\left({\nu }_2-4\right)}\\ =\frac{1}{\left({\nu }_2-2\right)\left({\nu }_2-4\right)}\end{array}$

$\begin{array}{c}E\left(W_1^2\right)=V\left(W_1\right)+{\left\{E\left(W_1\right)\right\}}^2\\ =2{\nu }_1+{\left({\nu }_1\right)}^2\\ ={\nu }_1\left({\nu }_1+2\right)\end{array}$

$\begin{array}{c}E\left(F^2\right)=E{\left(\frac{W_1/{\nu }_1}{W_2/{\nu }_2}\right)}^2\\ =\frac{{\nu }_2^2}{{\nu }_1^2}E\left(\frac{W_1^2}{W_{{}_2}^2}\right)\\ =\frac{{\nu }_2^2}{{\nu }_1^2}E\left(W_1^2\right)E\left(\frac{1}{W_{{}_2}^2}\right)\\ =\frac{{\nu }_2^2}{{\nu }_1^2}\times {\nu }_1\left({\nu }_1+2\right)\times \frac{1}{\left({\nu }_2-2\right)\left({\nu }_2-4\right)}\\ =\frac{{\nu }_2^2}{{\nu }_1^2}\frac{{\nu }_1\left({\nu }_1+2\right)}{\left({\nu }_2-2\right)\left({\nu }_2-4\right)}\end{array}$

Then,

$\begin{array}{c}V\left(F\right)=\frac{{\nu }_2^2}{{\nu }_1^2}\frac{{\nu }_1\left({\nu }_1+2\right)}{\left({\nu }_2-2\right)\left({\nu }_2-4\right)}-{\left(\frac{{\nu }_2}{{\nu }_2-2}\right)}^2\\ =\frac{{\nu }_2^2}{{\nu }_1^2}\frac{{\nu }_1\left({\nu }_1+2\right)}{\left({\nu }_2-2\right)\left({\nu }_2-4\right)}-\frac{{\nu }_2^2}{{\left({\nu }_2-2\right)}^2}\\ =\frac{{\nu }_2^2}{\left({\nu }_2-2\right)}\left[\frac{{\nu }_1\left({\nu }_1+2\right)}{{\nu }_1^2\left({\nu }_2-4\right)}-\frac{1}{\left({\nu }_2-2\right)}\right]\\ =\frac{{\nu }_2^2}{\left({\nu }_2-2\right)}\left[\frac{{\nu }_1\left({\nu }_1+2\right)\left({\nu }_2-2\right)-{\nu }_1^2\left({\nu }_2-4\right)}{{\nu }_1^2\left({\nu }_2-4\right)\left({\nu }_2-2\right)}\right]\\ =\frac{{\nu }_2^2}{\left({\nu }_2-2\right)}\left[\frac{{\nu }_1\left({\nu }_1+2\right)\left({\nu }_2-2\right)-{\nu }_1^2\left({\nu }_2-4\right)}{{\nu }_1^2\left({\nu }_2-4\right)\left({\nu }_2-2\right)}\right]\\ =\frac{{\nu }_2^2}{\left({\nu }_2-2\right)}\left[\frac{{\nu }_1\left\{\left({\nu }_1+2\right)\left({\nu }_2-2\right)-{\nu }_1\left({\nu }_2-4\right)\right\}}{{\nu }_1^2\left({\nu }_2-4\right)\left({\nu }_2-2\right)}\right]\\ =\frac{{\nu }_2^2}{\left({\nu }_2-2\right)}\left[\frac{\left({\nu }_1{\nu }_2-2{\nu }_1+2{\nu }_2-4\right)-\left({\nu }_1{\nu }_2-4{\nu }_1\right)}{{\nu }_1\left({\nu }_2-4\right)\left({\nu }_2-2\right)}\right]\\ =\frac{{\nu }_2^2}{\left({\nu }_2-2\right)}\left[\frac{\left({\nu }_1{\nu }_2-2{\nu }_1+2{\nu }_2-4\right)-{\nu }_1{\nu }_2+4{\nu }_1}{{\nu }_1\left({\nu }_2-4\right)\left({\nu }_2-2\right)}\right]\\ =\frac{{\nu }_2^2}{\left({\nu }_2-2\right)}\left[\frac{2{\nu }_1+2{\nu }_2-4}{{\nu }_1\left({\nu }_2-4\right)\left({\nu }_2-2\right)}\right]\\ =\frac{2{\nu }_2^2\left({\nu }_1+{\nu }_2-2\right)}{{\nu }_1{\left({\nu }_2-2\right)}^2\left({\nu }_2-4\right)},{\nu }_2>4\end{array}$

Hence proved.