Chapter 7.2, Problem 20E

1. a If U has a ${\chi }^2$ distribution with v df, find $E\left(U\right)$ and $V\left(U\right)$.
2. b Using the results of Theorem 7.3, find $E\left(S^2\right)$ and $V\left(S^2\right)$ when $Y_1,Y_2,\dots ,Y_n$ is a random sample from a normal distribution with mean μ and variance ${\sigma }^2$.

To determine

Compute $E\left(U\right)$.

Compute $V\left(U\right)$.

Answer to Problem 20E

$E\left(U\right)$ is $\nu$.

$V\left(U\right)$ is $2\nu$.

Explanation of Solution

From the given information, U follows ${\chi }^2$ distribution with $\nu$ df.

The probability density function of U is

$f\left(u\right)=\frac{{\left(\frac{1}{2}\right)}^{\frac{\nu }{2}}}{\Gamma \left(\frac{\nu }{2}\right)}{e}^{-\frac{u}{2}}{u}^{\frac{\nu }{2}-1};u>0$

Then,

$\begin{array}{c}E\left(u\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}uf\left(u\right)}du\\ ={\displaystyle \underset{0}{\overset{\infty }{\int }}u\frac{{\left(\frac{1}{2}\right)}^{\frac{\nu }{2}}}{\Gamma \left(\frac{\nu }{2}\right)}{e}^{-\frac{u}{2}}{u}^{\frac{\nu }{2}-1}}du\\ =\frac{{\left(\frac{1}{2}\right)}^{\frac{\nu }{2}}}{\Gamma \left(\frac{\nu }{2}\right)}{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-\frac{u}{2}}}{u}^{\left(\frac{\nu }{2}+1\right)-1}du\\ =\frac{{\left(\frac{1}{2}\right)}^{\frac{\nu }{2}}}{\Gamma \left(\frac{\nu }{2}\right)}\times \frac{\Gamma \left(\frac{\nu }{2}+1\right)}{{\left(\frac{1}{2}\right)}^{\frac{\nu }{2}+1}}\\ =\frac{{\left(\frac{1}{2}\right)}^{\frac{\nu }{2}}}{\Gamma \left(\frac{\nu }{2}\right)}\times \frac{\frac{\nu }{2}\Gamma \left(\frac{\nu }{2}\right)}{{\left(\frac{1}{2}\right)}^{\frac{\nu }{2}}\left(\frac{1}{2}\right)}\\ =2\left(\frac{\nu }{2}\right)\\ =\nu \end{array}$

$\begin{array}{c}E\left(u\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}uf\left(u\right)}du\\ ={\displaystyle \underset{0}{\overset{\infty }{\int }}u\frac{{\left(\frac{1}{2}\right)}^{\frac{\nu }{2}}}{\Gamma \left(\frac{\nu }{2}\right)}{e}^{-\frac{u}{2}}{u}^{\frac{\nu }{2}-1}}du\\ =\frac{{\left(\frac{1}{2}\right)}^{\frac{\nu }{2}}}{\Gamma \left(\frac{\nu }{2}\right)}{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-\frac{u}{2}}}{u}^{\left(\frac{\nu }{2}+1\right)-1}du\\ =\frac{{\left(\frac{1}{2}\right)}^{\frac{\nu }{2}}}{\Gamma \left(\frac{\nu }{2}\right)}\times \frac{\Gamma \left(\frac{\nu }{2}+1\right)}{{\left(\frac{1}{2}\right)}^{\frac{\nu }{2}+1}}\\ =\frac{{\left(\frac{1}{2}\right)}^{\frac{\nu }{2}}}{\Gamma \left(\frac{\nu }{2}\right)}\times \frac{\frac{\nu }{2}\Gamma \left(\frac{\nu }{2}\right)}{{\left(\frac{1}{2}\right)}^{\frac{\nu }{2}}\left(\frac{1}{2}\right)}\\ =2\left(\frac{\nu }{2}\right)\\ =\nu \end{array}$

$\begin{array}{c}E\left(u^2\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{u}^{2}f\left(u\right)}du\\ ={\displaystyle \underset{0}{\overset{\infty }{\int }}{u}^2\frac{{\left(\frac{1}{2}\right)}^{\frac{\nu }{2}}}{\Gamma \left(\frac{\nu }{2}\right)}{e}^{-\frac{u}{2}}{u}^{\frac{\nu }{2}-1}}du\\ =\frac{{\left(\frac{1}{2}\right)}^{\frac{\nu }{2}}}{\Gamma \left(\frac{\nu }{2}\right)}{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-\frac{u}{2}}}{u}^{\left(\frac{\nu }{2}+2\right)-1}du\\ =\frac{{\left(\frac{1}{2}\right)}^{\frac{\nu }{2}}}{\Gamma \left(\frac{\nu }{2}\right)}\times \frac{\Gamma \left(\frac{\nu }{2}+2\right)}{{\left(\frac{1}{2}\right)}^{\frac{\nu }{2}+2}}\\ =\frac{{\left(\frac{1}{2}\right)}^{\frac{\nu }{2}}}{\Gamma \left(\frac{\nu }{2}\right)}\times \frac{\left(\frac{\nu }{2}+1\right)\frac{\nu }{2}\Gamma \left(\frac{\nu }{2}\right)}{{\left(\frac{1}{2}\right)}^{\frac{\nu }{2}}{\left(\frac{1}{2}\right)}^2}\\ =\left(\frac{\nu }{2}+1\right)\frac{\nu }{2}\left(4\right)\\ =2\nu \left(\frac{\nu +2}{2}\right)\\ =\nu \left(\nu +2\right)\end{array}$

Then, the $V\left(U\right)$ is

$\begin{array}{c}V\left(U\right)=E\left(u^2\right)-{\left\{E\left(u\right)\right\}}^2\\ =\nu \left(\nu +2\right)-{\nu }^2\\ ={\nu }^2+2\nu -{\nu }^2\\ =2\nu \end{array}$

Thus, $E\left(U\right)$ is $\nu$ and $V\left(U\right)$ is $2\nu$.

To determine

Compute $E\left(S^2\right)$ and $V\left(S^2\right)$ when $Y_1,Y_2,\mathrm{.........},Y_n$ is a random sample from a normal distribution with mean $\mu$ and variance ${\sigma }^2$ by using the results of theorem 7.3.

Answer to Problem 20E

$E\left(S^2\right)$ is ${\sigma }^2$.

$V\left(S^2\right)$ is $\frac{2{\sigma }^4}{\left(n-1\right)}$.

Explanation of Solution

Let

$\begin{array}{l}{S}^2=\frac{{\sigma }^2}{n-1}\times \frac{n-1}{{\sigma }^2}{S}^2\\ E\left(S^2\right)=\frac{{\sigma }^2}{n-1}E\left(\frac{n-1}{{\sigma }^2}{S}^2\right)\end{array}$

From the theorem 7.3, if $Y_1,Y_2,\mathrm{........},Y_n$ follow normal distribution with mean $\mu$ variance ${\sigma }^2$, then $\frac{\left(n-1\right)S^2}{{\sigma }^2}=\frac{1}{{\sigma }^2}{\displaystyle \sum _{i=1}^n{\left(Y_i-\overline{Y}\right)}^2}$ follows ${\chi }^2$ distribution with $\left(n-1\right)$ df. So $E\left(\frac{n-1}{{\sigma }^2}{S}^2\right)$ is $\left(n-1\right)$ and $V\left(\frac{n-1}{{\sigma }^2}{S}^2\right)$ is $2\left(n-1\right)$.

Then,

$\begin{array}{c}E\left(S^2\right)=\frac{{\sigma }^2}{n-1}E\left(\frac{n-1}{{\sigma }^2}{S}^2\right)\\ =\frac{{\sigma }^2}{n-1}\left(n-1\right)\\ ={\sigma }^2\end{array}$

Thus, $E\left(S^2\right)$ is ${\sigma }^2$.

$\begin{array}{c}V\left(S^2\right)={\left(\frac{{\sigma }^2}{n-1}\right)}^{2}V\left(\frac{n-1}{{\sigma }^2}{S}^2\right)\\ =\frac{{\sigma }^4}{{\left(n-1\right)}^2}\left[2\left(n-1\right)\right]\\ =\frac{2{\sigma }^4}{\left(n-1\right)}\end{array}$

Thus, $V\left(S^2\right)$ is $\frac{2{\sigma }^4}{\left(n-1\right)}$.