expected value of product of normal variable, unknown varience

Question

I thought I understood finding the expected value of functions of rvs, but it has been a long time since I went over it in class, and this problem had me second guessing a bit. I don't have an opportunity to check the answer since its not for a class so I was hoping someone could take a look.

$X\sim\mathcal N(\mu=0,var)$

$s_{X^2}$ is variance of $X^2$, $S_X$ is the space of $X$

1. $E[X^4]$. I tried the following
   a. Using $E[u(X)]= \int_{S_X} (u(x)f(x))\operatorname d x$, but that produces a result I can't integrate without knowing variance.
   b. Using $\mathsf {Cov}(X^2, X^2)= 1\cdot s_{X^2}\cdot s_{X^2}$, but again, the unknown variance stops me from getting a numerical answer.
   c. Solving for variance using a formula. All the formula I came across involved more than one unknown quantity.

Any pointers on what direction to take next?

Answer 1

Tip: $\mathsf E(X^4)$ is the Fourth Non-Central Moment of the Normal Distribution (well, the Central since this mean is zero).   This can be found by using the Moment Generating Function, or just looked up.

$$\begin{align}\mathsf E(X^4) ~=~& \mu^4 + 6\mu^2\sigma^2 + 3\sigma^4\\[1ex]~=~& 3\sigma^4 &\because~&\mu=0 = \mathsf E(X)\end{align}$$

It is a polynomial of $\mu$ and $\sigma^2$. (Well, of $\sigma^2$ since $\mu=0$, conveniently)

Now $\sigma^2 = \mathsf E(X^2)-\mathsf E(X)^2$ and $s_{X^2}=\mathsf E(X^4)-\mathsf E(X^2)^2$ so...

Put it together into an expression relating $\sigma^2$ to $s_{X^2}$