Expectation of $x^4$

Question

Can anyone help me prove that Expected Value of $X^4$ is $3\,($Var$(X))^4$, if the Expected Value of $X$ is zero and Var$(X)$ is the Variance of $X$ $(N(0,\sigma^2))$.

Answer 1

HINT :

We know that $$ \int_{-\infty}^\infty\frac{1}{\sigma\sqrt{2\pi}}e^{\Large-\frac{(x-\mu)^2}{2\sigma^2}}\ dx=1,\tag1 $$

$$ \text{E}[X]=\int_{-\infty}^\infty\frac{x}{\sigma\sqrt{2\pi}}e^{\Large-\frac{(x-\mu)^2}{2\sigma^2}}\ dx=\mu,\tag2 $$ and $$ \text{E}\left[X^2\right]=\int_{-\infty}^\infty\frac{x^2}{\sigma\sqrt{2\pi}}e^{\Large-\frac{(x-\mu)^2}{2\sigma^2}}\ dx=\text{Var}[X]+(\text{E}[X])^2=\sigma^2+\mu^2.\tag3 $$ Then differentiate $(3)$ with respect to $\mu$ twice. $$ \int_{-\infty}^\infty\frac{\partial^2}{\partial\mu^2}\left(\frac{x^2}{\sigma\sqrt{2\pi}}e^{\Large-\frac{(x-\mu)^2}{2\sigma^2}}\right)\ dx=\frac{\partial^2}{\partial\mu^2}\left(\sigma^2+\mu^2\right).\tag4 $$ Using $(1)$, $(2)$, $(3)$, and $\mu=0$, we can obtain the following result $$ \text{E}\left[X^4\right]=3\left(\text{Var}[X]\right)^2. =3*\sigma^4 $$