Chapter 4.6, Problem 112E

a.

To determine

Find an expression for $E\left(y^a\right)$ if $\nu >-2a$.

Answer to Problem 112E

The expression for $E\left(y^a\right)=\frac{2^a\Gamma \left(\frac{\nu }{2}+a\right)}{\Gamma \left(\frac{\nu }{2}\right)}\left\{\because \nu >-2a\right\}$

Explanation of Solution

Note that a random variable is said to be a chi-square distribution with $\nu$ degrees of freedom if and only if Y is a gamma-distributed random variable with parameters $\alpha =\frac{\nu }{2}$ and $\beta =2$.

Consider,

$\begin{array}{c}E\left(y^a\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{y}^{a}f\left(y\right)}dy\\ ={\displaystyle \underset{0}{\overset{\infty }{\int }}{y}^a\frac{y^{\alpha -1}{e}^{-\frac{y}{\beta }}}{{\beta }^{\alpha }\Gamma \left(\alpha \right)}dy}\\ =\frac{1}{{\beta }^{\alpha }\Gamma \left(\alpha \right)}{\displaystyle \underset{0}{\overset{\infty }{\int }}{y}^{a+\alpha -1}}{e}^{-\frac{y}{\beta }}dy\\ =\frac{1}{{\beta }^{\alpha }\Gamma \left(\alpha \right)}\times {\beta }^{a+\alpha }\Gamma \left(a+\alpha \right)\\ =\frac{{\beta }^a\Gamma \left(a+\alpha \right)}{\Gamma \left(\alpha \right)}\end{array}$

Therefore, it is proved that $E\left(y^a\right)=\frac{{\beta }^a\Gamma \left(\alpha +a\right)}{\Gamma \left(\alpha \right)}$.

Now, substitute $\alpha =\frac{\nu }{2}$ and $\beta =2$ in the above expression.

$E\left(y^a\right)=\frac{2^a\Gamma \left(\frac{\nu }{2}+a\right)}{\Gamma \left(\frac{\nu }{2}\right)}\left\{\because \nu >-2a\right\}$

b.

To determine

Explain the reason why Part a requires $\nu >-2a$.

Explanation of Solution

The answer in Part a has to be satisfied that $\nu >-2a$ because the gamma function is defined only if

$\begin{array}{l}\frac{\nu }{2}+a>0\\ \frac{\nu }{2}>-a\\ \nu >-2a\end{array}$

Thus, the function $\Gamma \left(\frac{\nu }{2}+a\right)$ is valid only if $\nu >-2a$.

c.

To determine

Find an expression for $E\left(\sqrt{Y}\right)$ using the results in Part a.

Find one assumption about $\nu$ used for the calculation.

Answer to Problem 112E

The expression for $E\left(\sqrt{y}\right)=\frac{\sqrt{2}\Gamma \left(\frac{\nu +1}{2}\right)}{\Gamma \left(\frac{\nu }{2}\right)}$

Explanation of Solution

The answer in Part a is as follows:

$E\left(y^a\right)=\frac{2^a\Gamma \left(\frac{\nu }{2}+a\right)}{\Gamma \left(\frac{\nu }{2}\right)}\left\{\because \nu >-2a\right\}$

Substitute $a=\frac{1}{2}$ in the above expression; then, one can get the expression for $E\left(\sqrt{Y}\right)$.

$\begin{array}{c}E\left(\sqrt{y}\right)=E\left(y^{\frac{1}{2}}\right)\\ =\frac{2^{\frac{1}{2}}\Gamma \left(\frac{\nu }{2}+\frac{1}{2}\right)}{\Gamma \left(\frac{\nu }{2}\right)}\\ E\left(\sqrt{y}\right)=\frac{\sqrt{2}\Gamma \left(\frac{\nu +1}{2}\right)}{\Gamma \left(\frac{\nu }{2}\right)}\text{, }\nu >\text{0 }\end{array}$

Therefore, the value of $\nu >0$ should be assumed.

d.

To determine

Use the results in Part a to find an expression for $E\left(\frac{1}{Y}\right)$, $E\left(\frac{1}{\sqrt{Y}}\right)$, and $E\left(\frac{1}{Y^2}\right)$.

Identify the value of $\nu$ needed to be assumed in each case.

Explanation of Solution

The result in Part a is as follows:

$E\left(y^a\right)=\frac{2^a\Gamma \left(\frac{\nu }{2}+a\right)}{\Gamma \left(\frac{\nu }{2}\right)}\left\{\because \nu >-2a\right\}$

Case1:

Substitute $a=-1$ in the above expression to get the expression for $E\left(\frac{1}{Y}\right)$.

$\begin{array}{c}E\left(y^{-1}\right)=\frac{2^{-1}\Gamma \left(\frac{\nu }{2}-1\right)}{\Gamma \left(\frac{\nu }{2}\right)}\left\{\because \Gamma \left(\frac{\nu }{2}-1\right)\text{value is defined only if}\nu >2\right\}\\ =\frac{\Gamma \left(\frac{\nu }{2}-1\right)}{2\Gamma \left(\frac{\nu }{2}\right)}\end{array}$

Thus, in this case, one needs to assume the value of $\nu >2$.

Case 2:

Substitute $a=-\frac{1}{2}$ in the above expression to get the expression for $E\left(\frac{1}{\sqrt{Y}}\right)$.

$\begin{array}{c}E\left(\frac{1}{\sqrt{Y}}\right)=E\left(y^{-\frac{1}{2}}\right)\\ =\frac{2^{-\frac{1}{2}}\Gamma \left(\frac{\nu }{2}-\frac{1}{2}\right)}{\Gamma \left(\frac{\nu }{2}\right)}\\ \text{= }\frac{\Gamma \left(\frac{\nu -1}{2}\right)}{\sqrt{2}\Gamma \left(\frac{\nu }{2}\right)}\left\{\because \Gamma \left(\frac{\nu -1}{2}\right) \text{value is defined only if }\nu >1\right\}\end{array}$

Therefore, the value of $\nu >1$ should be assumed.

Case3:

Substitute $a=-2$ in the above expression to get the expression for $E\left(\frac{1}{Y^2}\right)$.

$\begin{array}{c}E\left(\frac{1}{y^2}\right)=E\left(y^{-2}\right)\\ =\frac{2^{-2}\Gamma \left(\frac{\nu }{2}-2\right)}{\Gamma \left(\frac{\nu }{2}\right)}\\ \text{= }\frac{\Gamma \left(\frac{\nu }{2}-2\right)}{4\Gamma \left(\frac{\nu }{2}\right)}\\ =\frac{\Gamma \left(\frac{\nu }{2}-2\right)}{4\left(\frac{\nu }{2}-1\right)\left(\frac{\nu }{2}-2\right)\Gamma \left(\frac{\nu }{2}-2\right)}\left\{\because \Gamma \left(\nu \right)= \left(\nu -1\right)\Gamma \left(\nu -1\right)\right\}\\ =\frac{1}{\left(\nu -2\right)\left(\nu -4\right)}\text{, }\left\{\nu >4\right\}\end{array}$

Thus, value of $\nu >4$ should be assumed in this case.