Methods for Finding Raw Moments of the Normal Distribution I'm having some trouble with finding raw moments for the normal distribution. Right now I am trying to find the 4th raw moment on my own. So far, I know of two methods:  


*

*I can take the 4th derivative of the moment generating function for the normal distribution and evaluate it at 0.  

*I can use the fact that $E(x^4)$ is an expectation of a function of x to write  $$E({X}^{4})=\int_{Sx}^{} {x}^{4} f(x) dx=\int_{-\infty}^{\infty} {x}^{4}\frac{{e}^{\frac{{(x-\mu )}^{2}}{2{\sigma }^{2}}}}{\sqrt{2\pi }\sigma } dx$$


I'm wondering if there's a 3rd method. We haven't covered integrating the normal pdf in class, and taking the 4th derivative of ${e}^{\frac{{t}^{2}{\sigma }^{2}}{2}+t\mu }$ seems really messy/inelegant, so I'm wondering if there is some conceptual piece about moment generating functions I am missing. Thanks in advance!
 A: There is a nice recurrence for the raw moments of a normal distribution with mean $\mu$ and variance $\sigma^2$:
$$\mathbb E\left[X^{n+1}\right] = \mu \mathbb E\left[X^{n}\right] + n \sigma^2 \mathbb E\left[X^{n-1}\right] $$ starting at $\mathbb E\left[X^{0}\right]=1$ and $\mathbb E\left[X^{1}\right]=\mu$. So you get:
$$\begin{matrix}
\mathbb E\left[X^{0}\right]=&1 \\ 
\mathbb E\left[X^{1}\right]=&\mu \\
\mathbb E\left[X^{2}\right]=&\mu^2 +\sigma^2 \\ 
\mathbb E\left[X^{3}\right]=&\mu^3 +3\mu\sigma^2 \\ 
\mathbb E\left[X^{4}\right]=&\mu^4 +6\mu^2\sigma^2 +3\sigma^4\\
\mathbb E\left[X^{5}\right]=&\mu^5 +10\mu^3\sigma^2 +15\mu\sigma^4 \\
\mathbb E\left[X^{6}\right]=&\mu^6 +15\mu^4\sigma^2 +45\mu^2\sigma^4 +15\sigma^6
\end{matrix}$$
and so on
A: For a general normal random variable $X$ with mean $\mu$ and standard deviation $\sigma$, we can express the moments in terms of the moments of the standard normal, since $X = \mu + \sigma Z$; hence $$\operatorname{E}[X^k] = \operatorname{E}[(\mu + \sigma Z)^k] = \sum_{m = 0}^k \binom{k}{m} \mu^m \sigma^{k-m} \operatorname{E}[Z^{k-m}].$$  It can be shown in this answer that $$\operatorname{E}[Z^{2m}] = \frac{(2m)!}{2^m m!}$$ for positive integers $m$, and $0$ otherwise.  In particular, for $k = 4$, we find $$\operatorname{E}[X^4] = \mu^4 + 6\mu^2 \sigma^2 + 3\sigma^4.$$
A: It might help to remember the following identity when $X\sim N(\mu,\sigma^2)$:
$$\mathbb E\left[g(X)(X-\mu)\right]=\sigma^2 \mathbb E\left[g'(X)\right]$$
Here $g$ is any function for which both expectations above exist. The proof is based on integration by parts.
So for the third moment, choose $g(X)=X^2$:
$$\mathbb E\left[X^2(X-\mu)\right]=2\sigma^2 \mathbb E\left[X\right]$$
Combining with $\mathbb E\left[X^2\right]=\sigma^2+\mu^2$, rearrange to get
$$\mathbb E\left[X^3\right]=2\sigma^2\mu+\mu(\sigma^2+\mu^2)=\mu^3+3\mu\sigma^2$$
Similarly for the fourth moment, choose $g(X)=X^3$:
$$\mathbb E\left[X^3(X-\mu)\right]=3\sigma^2 \mathbb E\left[X^2\right]$$
Use the previous moments to get
$$\mathbb E\left[X^4\right]=3\sigma^2(\sigma^2+\mu^2)+\mu(\mu^3+3\mu\sigma^2)=\mu^4+6\mu^2\sigma^2+3\sigma^4$$
