How do you calculate the Expectation $E(X^4)$?

Question

Suppose $X$ is a normal random variable with, $\mu_x\ {\ne}\ 0\ \text{or}\ 1$ and $\sigma_x\ {\ne}\ 0\ \text{or}\ 1$. I would like to show that the expectation $E(X^4)$ has the following equality: $E(X^4)\ {=}\ \mu_x^4+6\sigma_x^2\mu_x^2+3\sigma_x^4$.

I tried to prove this by calculating $E[(x-\mu_x)^4]$ which leads me to

$E[(x-\mu_x)^4]\ {=}\ E(x^4-4x^3\mu_x+6x^2\mu_x^2-4x\mu_x^3+\mu_x^4)$

$E(x^4)\ {=}\ 4\mu_xE(x^3)-6\mu_x^2\sigma_x^2-3\mu_x^4+E[(x-\mu_x)^4]$

where I used $E(x^2)\ {=}\ \sigma_x^2+\mu_x^2$ and $E(x)\ {=}\ \mu_x$ to combine terms. This result didn't seem to help. I've looked into using the generating function approach but for $\mu_x\ {\ne}\ 0\ \text{or}\ 1$ and $\sigma_x\ {\ne}\ 0\ \text{or}\ 1$ these calculations become tedious very quickly. Is there an easy and quick way to show this?

Hint: write the integral for $E(X^3)$ and notice something... magical. — Sean Roberson, May 21 '23 at 19:36
A helpful lemma: https://en.wikipedia.org/wiki/Stein%27s_lemma — user51547, May 21 '23 at 20:04
Hint: write $X=\mu+\sigma Z$ with $Z\sim N(0,,1)$, then expand $E(X^4)$ in terms of $E(Z^n)$ with $n\le4$. — J.G., May 21 '23 at 21:55

user1172706 · Answer 1 · 2023-05-22T09:40:10.153

1

Let $X=\mu+\sigma Z$ where $Z$ is the standard normal random variable.

If you write the integral for $E(Z^n)$ and integrate by parts you obtain the simple relationship $$E(Z^{n+2})=(n+1)E(Z^n).$$

Therefore $E(Z^4)=3$. Then $$E(X^4)=E(\mu ^4+4\sigma \mu ^3Z+6\sigma ^2 \mu ^2Z^2+4\sigma ^3\mu Z^3+ \sigma ^4Z^4 )=\mu ^4+6\sigma ^2 \mu ^2+3 \sigma ^4.$$

edited May 22 '23 at 09:40

answered May 22 '23 at 09:34

user1172706

2,791

Perhaps worth adding the obvious starting point of $E[Z^0]=1$ and $E[Z^1]=0$ – Henry Apr 17 '25 at 14:20

How do you calculate the Expectation $E(X^4)$?

1 Answers1