A statistics problem (normal distribution)

Question

If $X\sim N(0, \sigma^2)$, how can I compute $\operatorname{Var}(X^2)$? Here is my idea... but I cannot get there. $$\operatorname{Var}(X^2) = E(X^4) - (E(X^2))^2$$

Answer 1

There are many ways to do this, but here's one. The moment-generating function of $X$ is $\exp\frac{\sigma^2t^2}{2}$. Its power series begins $1+\frac{\sigma^2t^2}{2}+\frac{\sigma^4t^4}{8}$. Multiplying the $t^4$ coefficient by $4!$ gives $E(X^4)=3\sigma^4$. I assume you know $E(X^2)=\sigma^2$, so the result is $2\sigma^4$.

Answer 2

Hint:

$$\int_{-\infty}^\infty x^{2n}e^{-x^2/2}dx=2\int_{-\infty}^\infty 2^{n-1/2}t^{n-1/2}e^{-t}dx=2^{n+1/2}\Gamma\left(n+\frac12\right).$$