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!
Best Answer
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 $$\operatorname{E}[X^4] = \mu^4 + 6\mu^2 \sigma^2 + 3\sigma^4.$$