Can anyone help me prove that Expected Value of $X^4$ is $3 \sigma^4$ if $X$ is normally distributed with zero expected value and variance equals $\sigma^2$?
I have tried to derive by myself using the hints in this post Expectation of $x^4$ but I haven't made any progress.
Differentiating w.r.t. $\mu$ any number of times preserves the exponential factor, so that the integral can be expressed as a linear combination of expectations:
$$\begin{align*} \frac{\partial^2}{\partial\mu^2} \frac{x^2}{\sigma\sqrt{2\pi}} e^{-\tfrac{(x-\mu)^2}{2\sigma^2}} &= \frac{x^4 - 2\mu x^3+\left(\mu^2-\sigma^2\right)x^2}{\sigma^5 \sqrt{2\pi}} e^{-\tfrac{(x-\mu)^2}{2\sigma^2}} \\ \implies \int_{-\infty}^\infty \frac{\partial^2}{\partial\mu^2} \frac{x^2}{\sigma\sqrt{2\pi}} e^{-\tfrac{(x-\mu)^2}{2\sigma^2}} \,dx &= \frac1{\sigma^4} \Bbb E\left[X^4\right] - \frac{2\mu}{\sigma^4} \Bbb E\left[X^3\right] + \frac{\mu^2-\sigma^2}{\sigma^4} \Bbb E\left[X^2\right] \end{align*}$$
Meanwhile, $\Bbb E\left[X^2\right]=\sigma^2+\mu^2$ so upon differentiating and plugging in $\mu=0$ we have the relation
$$2 = \frac1{\sigma^4} \Bbb E\left[X^4\right] - \frac1{\sigma^2} \Bbb E\left[X^2\right]$$
and the result follows.