TITLE: Proving $E[X^4]=3σ^4$
QUESTION [17 upvotes]: Given a random variable $X\sim\mathcal N(0,\sigma^2)$, how can we prove that $E[X^4]=3\sigma^4$? I am having trouble even starting with the proof.

REPLY [6 votes]: @Yves Daoust already has a great answer, but we can omit a round of integration by parts, by just working it out in the most straight-forward way.
Applying integration by parts once,
$$
\mathbb{E} (x^4) = \int x^4 \varphi(x) dx = 0 + 3\sigma^2 \int x^2 \varphi(x) dx
$$
where $\varphi(x)$ is normal PDF and $\varphi(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{x^2}{2\sigma^2}}$
We recognise $\int x^2 \varphi(x) dx = \sigma^2$. So,
$$
\mathbb{E} (x^4) = 3\sigma^2 \sigma^2 = 3 \sigma^4
$$