Question

Let X be a normal random variable with mean $\mu$and variance ${\sigma }^2$. Use the results of Theoretical Exercise 7.46 to show that

$E\left[X^n\right]=\sum _{j=0}^{[n/2]} \frac{\left(\begin{array}{c}n\\ 2j\end{array}\right){\mu }^{n-2j}{\sigma }^{2j}\left(2j\right)!}{2jj!}$

In the preceding equation, $[n/2]$ is the largest integer less than or equal to $n/2$. Check your answer by letting $n=1$ and $n=2$.

Short answer

It has been shown that$E\left(X^n\right)=\sum _{j=0}^{[n/2]} \left(\begin{array}{l}n\\ 2j\end{array}\right){\sigma }^{2j}\frac{\left(2j\right)!}{2^{j}j!}{\mu }^{n-2j}$

Step-by-step solution

Given information 

X be a normal random variable with mean $\mu$and variance ${\sigma }^2$.

In the preceding equation, $\left[n/2\right]$ is the largest integer less than or equal to$n/2$.

Solution 

We need to use the binomial theorem to obtain the result,

$E\left(X^n\right)=E\left((\mu +\sigma Z{)}^n\right)$

$=\sum _{j=0}^n \left(\begin{array}{c}n\\ j\end{array}\right){\sigma }^{j}E\left(Z^j\right){\mu }^{n-j}$

From theoretical exercise 46 , we need to obtain the following result:

$E\left[e^{tz}\right]=e^{\frac{t^2}{2}}$

$=\sum _{j=0}^{\mathrm{\infty }} \left(\frac{{\left(t^2/2\right)}^j}{j!}\right)$

Solution

Now we need to compare and substitute the given,

$E\left(X^n\right)=\sum _{j\le n,j\text{even}} \left(\begin{array}{c}n\\ j\end{array}\right){\sigma }^{j}E\left(Z^j\right){\mu }^{n-j}$

$=\sum _{j=0}^{[n/2} \left(\begin{array}{l}n\\ 2j\end{array}\right){\sigma }^{2j}E\left(Z^{2j}\right){\mu }^{n-2j}$

So as a result,

$E\left(X^n\right)=\sum _{j=0}^{[n/2} \left(\begin{array}{l}n\\ 2j\end{array}\right){\sigma }^{2j}\frac{\left(2j\right)!}{2^{j}j!}{\mu }^{n-2j}$

Hence,$E\left(X^n\right)=\sum _{j=0}^{[n/2]} \left(\begin{array}{l}n\\ 2j\end{array}\right){\sigma }^{2j}\frac{\left(2j\right)!}{2^{j}j!}{\mu }^{n-2j}$

Where,$[n/2]$is the greatest integer less than or equal to $n/2$.

Final answer 

From the above calculation it has been shown that$E\left(X^n\right)=\sum _{j=0}^{[n/2]} \left(\begin{array}{l}n\\ 2j\end{array}\right){\sigma }^{2j}\frac{\left(2j\right)!}{2^{j}j!}{\mu }^{n-2j}$.