Methods for Finding Raw Moments of the Normal Distribution

Question

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:

1. I can take the 4th derivative of the moment generating function for the normal distribution and evaluate it at 0.
2. 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!

Answer 1

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

Answer 2

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

Answer 3

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

$$\mathbb E\left[X^4\right]=3\sigma^2(\sigma^2+\mu^2)+\mu(\mu^3+3\mu\sigma^2)=\mu^4+6\mu^2\sigma^2+3\sigma^4$$

Answer 4

In general, if $X\sim N(\mu, \sigma^2),$ then for any integer $n\geq 1$, $$E(X^n) = \sum_{j=0}^{\lfloor \frac{n}{2}\rfloor}\binom{n}{2j}\dfrac{\mu^{n-2j}\sigma^{2j}(2j)!}{j!2^j}.$$