Chapter 4, Problem 199SE

a.

To determine

Prove that the expected values of all odd integer powers of Z are zero.

That is, $E\left(Z^{2i-1}\right)=0$.

Answer to Problem 199SE

It is proved that $E\left(Z^{2i-1}\right)=0$.

Explanation of Solution

The expected value of $Z^{2i-1}$ is as follows:

$\begin{array}{c}E\left(Z^{2i-1}\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{z}^{2i-1}}f\left(z\right)dz\\ ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{1}{\sqrt{2\pi }}{z}^{2i-1}{e}^{\left(-\frac{1}{2}\right)z^2}dz}\end{array}$

Define the function $g\left(z\right)=z^{2i-1}{e}^{\left(-\frac{1}{2}\right)z^2}$ then, $g\left(-z\right)={\left(-z\right)}^{2i-1}{e}^{\left(-\frac{1}{2}\right){\left(-z\right)}^2}=-g\left(z\right)$.

The function $g\left(z\right)$ is an odd function; thus, the integrand value becomes zero.

That is,

$\begin{array}{c}E\left(Z^{2i-1}\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{z}^{2i-1}}f\left(z\right)dz\\ ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{1}{\sqrt{2\pi }}{z}^{2i-1}{e}^{\left(-\frac{1}{2}\right)z^2}dz}\\ =0\end{array}$

Therefore, the expected values of all odd integer powers of Z are zero.

That is, $E\left(Z^{2i-1}\right)=0$.

b.

To determine

Show that $E\left(Z^{2i}\right)=\frac{2^i\Gamma \left(i+\frac{1}{2}\right)}{\sqrt{\pi }},\text{ for }i=1,2,3,\dots$.

Answer to Problem 199SE

It is proved that $E\left(Z^{2i}\right)=\frac{2^i\Gamma \left(i+\frac{1}{2}\right)}{\sqrt{\pi }},\text{ for }i=1,2,3,\dots$.

Explanation of Solution

The expected value of $Z^{2i}$ is as follows:

$\begin{array}{c}E\left(Z^{2i}\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{z}^{2i}}f\left(z\right)dz\\ ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{1}{\sqrt{2\pi }}{z}^{2i}{e}^{\left(-\frac{1}{2}\right)z^2}dz}\end{array}$

Define the function $g\left(z\right)=z^{2i}{e}^{\left(-\frac{1}{2}\right)z^2}$ then, $g\left(-z\right)={\left(-z\right)}^{2i}{e}^{\left(-\frac{1}{2}\right){\left(-z\right)}^2}=g\left(z\right)$.

Since the function $g\left(z\right)$ is an even function; thus, the integrand value can be written as follow:

$\begin{array}{c}E\left(Z^{2i}\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{z}^{2i}}f\left(z\right)dz\\ ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{1}{\sqrt{2\pi }}{z}^{2i}{e}^{\left(-\frac{1}{2}\right)z^2}dz}\\ =2{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{1}{\sqrt{2\pi }}{z}^{2i}{e}^{\left(-\frac{1}{2}\right)z^2}dz}\\ =\sqrt{\frac{2}{\pi }}{\displaystyle \underset{0}{\overset{\infty }{\int }}{\left(2t\right)}^i}{e}^{-t}\frac{1}{\sqrt{2t}}dt\\ =\frac{2^i}{\sqrt{\pi }}{\displaystyle \underset{0}{\overset{\infty }{\int }}{t}^{i+\frac{1}{2}-1}}{e}^{-t}dt\left\{\because \Gamma \left(i+\frac{1}{2}\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{t}^{i+\frac{1}{2}-1}}{e}^{-t}d\right\}\\ =\frac{2^i}{\sqrt{\pi }}\Gamma \left(i+\frac{1}{2}\right)\end{array}$

Hence, it is proved that $E\left(Z^{2i}\right)=\frac{2^i\Gamma \left(i+\frac{1}{2}\right)}{\sqrt{\pi }},\text{ for }i=1,2,3,\dots$.

c.

To determine

Find the values of $E\left(Z^2\right),E\left(Z^4\right),E\left(Z^6\right)$ and $E\left(Z^8\right)$ using the result in Part (b).

Answer to Problem 199SE

The values of $E\left(Z^2\right),E\left(Z^4\right),E\left(Z^6\right)$ and $E\left(Z^8\right)$ are 1, 3, 15, and 105, respectively.

Explanation of Solution

The result in Part (b) is as follows:

$E\left(Z^{2i}\right)=\frac{2^i\Gamma \left(i+\frac{1}{2}\right)}{\sqrt{\pi }},\text{ for }i=1,2,3,\dots$

The values of $E\left(Z^2\right),E\left(Z^4\right),E\left(Z^6\right)$ and $E\left(Z^8\right)$ can be obtained as follows:

Put $i=1$ then,

$\begin{array}{c}E\left(Z^2\right)=\frac{2\Gamma \left(1+\frac{1}{2}\right)}{\sqrt{\pi }}\\ =\frac{2\times \frac{1}{2}\Gamma \left(\frac{1}{2}\right)}{\sqrt{\pi }}\left\{\because \Gamma \left(n+1\right)=n\Gamma \left(n\right)\right\}\\ =1\end{array}$.

Put $i=2$ then.

$\begin{array}{c}E\left(Z^4\right)=\frac{4\Gamma \left(2+\frac{1}{2}\right)}{\sqrt{\pi }}\\ =\frac{4\Gamma \left(1+\frac{3}{2}\right)}{\sqrt{\pi }}\\ =\frac{4\times \frac{3}{2}\times \frac{1}{2}\Gamma \left(\frac{1}{2}\right)}{\sqrt{\pi }}\left\{\because \Gamma \left(n+1\right)=n\Gamma \left(n\right)\right\}\\ =3\end{array}$

Substitute $i=3$ then $E\left(Z^6\right)$ is as follows:

$\begin{array}{c}E\left(Z^6\right)=\frac{8\Gamma \left(3+\frac{1}{2}\right)}{\sqrt{\pi }}\\ =\frac{8\Gamma \left(1+\frac{5}{2}\right)}{\sqrt{\pi }}\\ =\frac{8\times \frac{5}{2}\times \frac{3}{2}\times \frac{1}{2}\Gamma \left(\frac{1}{2}\right)}{\sqrt{\pi }}\left\{\begin{array}{l}\because \Gamma \left(n+1\right)=n\Gamma \left(n\right)\\ \Gamma \left(\frac{1}{2}\right)=\sqrt{\pi }\end{array}\right\}\\ =15\end{array}$

Substitute $i=4$, then

$\begin{array}{c}E\left(Z^8\right)=\frac{16\Gamma \left(4+\frac{1}{2}\right)}{\sqrt{\pi }}\\ =\frac{16\Gamma \left(1+\frac{7}{2}\right)}{\sqrt{\pi }}\\ =\frac{16\times \frac{7}{2}\times \frac{5}{2}\times \frac{3}{2}\times \frac{1}{2}\Gamma \left(\frac{1}{2}\right)}{\sqrt{\pi }}\left\{\begin{array}{l}\because \Gamma \left(n+1\right)=n\Gamma \left(n\right)\\ \Gamma \left(\frac{1}{2}\right)=\sqrt{\pi }\end{array}\right\}\\ =105\end{array}$

d.

To determine

Show that $E\left(Z^{2i}\right)={\displaystyle \prod _{j=1}^i\left(2j-1\right)}$.

Explanation of Solution

From result in Part (b).

It is known that $E\left(Z^{2i}\right)=\frac{2^i\Gamma \left(i+\frac{1}{2}\right)}{\sqrt{\pi }},\text{ for }i=1,2,3,\dots$.

Hence,

$\begin{array}{c}\Gamma \left(i+\frac{1}{2}\right)=\Gamma \left(\frac{2i+1}{2}\right)\\ =\Gamma \left(\frac{2i-1}{2}+1\right)\\ =\left(\frac{2i-1}{2}\right)\Gamma \left(\frac{2i-1}{2}\right)\\ =\left(\frac{2i-1}{2}\right)\Gamma \left(\frac{2i-3}{2}+1\right)\\ =\left(\frac{2i-1}{2}\right)\left(\frac{2i-3}{2}\right)\Gamma \left(\frac{2i-3}{2}\right)\\ =\left(\frac{2i-1}{2}\right)\left(\frac{2i-3}{2}\right)\Gamma \left(\frac{2i-5}{2}+1\right)\\ =\left(\frac{2i-1}{2}\right)\left(\frac{2i-3}{2}\right)\cdots \left(\frac{1}{2}\right)\Gamma \left(\frac{1}{2}\right)\\ =\frac{\sqrt{\pi }}{2^i}{\displaystyle \prod _{j=1}^i\left(2j-1\right)}\end{array}$

Substitute $\Gamma \left(i+\frac{1}{2}\right)=\frac{\sqrt{\pi }}{2^i}{\displaystyle \prod _{j=1}^i\left(2j-1\right)}$ in $E\left(Z^{2i}\right)$ then it becomes,

$E\left(Z^{2i}\right)={\displaystyle \prod _{j=1}^i\left(2j-1\right)}$.