Show that if $\eta \sim N(0, \sigma^2)$ then $\mathbb{E} [\eta^4] = 3 \sigma^4$? For the normally distributed random variable $\eta \sim N(0, \sigma^2)$ I am trying to show that
$$
\mathbb{E} [\eta^4] = 3 \sigma^4
$$
Can anyone help me to understand how this might be done?
 A: There are several different alternatives to show this depending on what we are allowed to use. One possibility is the following. If $Z\sim N(0,1)$, then $Z^2\sim\chi^2(1)$, where $\chi^2(1)$ is the chi-squared distribution with one degree of freedom. The expected value of $Z^2$ is $1$ and the variance is $2$ (see here). We have that
$$
\operatorname E[\eta^4]=\operatorname E[\sigma^4 Z^4]=\sigma^4\operatorname E[Z^4]=\sigma^4(\operatorname{Var}[Z^2]+(\operatorname E[Z^2])^2)=3\sigma^4
$$
using the fact that $\sigma Z\sim N(0,\sigma^2)$ and  $\operatorname{Var}[Z^2]=\operatorname E[Z^4]-(\operatorname E[Z^2])^2$.
A: Let's prove it for $\sigma=1$.
We have to compute the integral 
$$\int_{\mathbb{R}} x^4 \frac{e^{-x^2/2}}{\sqrt{2\pi}}dx=-x^3\frac{e^{-x^2/2}}{\sqrt{2\pi}}|_{-\infty}^{\infty}+\int3x^2\frac{e^{-x^2/2}}{\sqrt{2\pi}}dx=\\=0-3x\frac{e^{-x^2/2}}{\sqrt{2\pi}}|_{-\infty}^{\infty}+\int3\frac{e^{-x^2/2}}{\sqrt{2\pi}}dx=0+3=3.
$$
For general case consider that the thesis is $\mathbb{E}[(\eta/\sigma)^4]=3,$ and that if $\eta$ is $\mathcal{N}(0,\sigma^2),$ then $\eta/\sigma$ is $\mathcal{N}(0,1),$ and so you reconnect to the first case.
A: Although there are simpler ways to do this, you could use the moment-generating function (MGF) of $\eta \sim \operatorname{N}(0, \; \sigma^2)$:
$$\operatorname{M}_\eta(t)=\operatorname{exp} \Bigl(\mu t + \frac{\sigma^2 t^2}{2} \Bigr) = e^{\frac{\sigma^2 t^2}{2}}$$
Now, the $k^{\text{th}}$ moment of $\eta$, $\operatorname{E}(\eta^k)$ is the $k^{\text{th}}$ derivative of $\operatorname{M}_\eta(t)$ at $t = 0$, so $\operatorname{E}(\eta^4)$ is the $4^{\text{th}}$ moment of $\eta$ at $t = 0$:
$$\begin{align}
\operatorname{E}(\eta^4) & = \frac{d^4 \operatorname{M}_\eta(t)}{dt^4} \Biggr|_{t=0} \\
& = \frac{d^4}{dt^4} \bigl( e^{\frac{\sigma^2 t^2}{2}} \bigr) \Biggr|_{t=0} \\
& = \frac{d^3}{dt^3} \bigl( \sigma^2 t e^{\frac{\sigma^2 t^2}{2}} \bigr) \Biggr|_{t=0} \\
& = \frac{d^2}{dt^2} \bigl( \sigma^2 e^{\frac{\sigma^2 t^2}{2}} + \sigma^4 t^2 e^{\frac{\sigma^2 t^2}{2}}\bigr) \Biggr|_{t=0} \\
& = \frac{d}{dt} \bigl( \sigma^4 t e^{\frac{\sigma^2 t^2}{2}} + 2\sigma^4 t e^{\frac{\sigma^2 t^2}{2}} + \sigma^6 t^3 e^{\frac{\sigma^2 t^2}{2}}\bigr) = \frac{d}{dt} \bigl( 3\sigma^4 t e^{\frac{\sigma^2 t^2}{2}} + \sigma^6 t^3 e^{\frac{\sigma^2 t^2}{2}}\bigr) \Biggr|_{t=0} \\
& = 3\sigma^4 e^{\frac{\sigma^2 t^2}{2}} + 6\sigma^6 t^2 e^{\frac{\sigma^2 t^2}{2}} + \sigma^8 t^4 e^{\frac{\sigma^2 t^2}{2}} = \sigma^4 \bigl( 3e^{\frac{\sigma^2 t^2}{2}} + 6\sigma^2 t^2 e^{\frac{\sigma^2 t^2}{2}} + \sigma^4 t^4 e^{\frac{\sigma^2 t^2}{2}} \bigr) \Biggr|_{t=0}  \\
& = 3\sigma^4
\end{align}$$
And you're done.
