Variance of $X^2$ if $X \sim \mathcal{N}(0,1)$? I'm trying to find the variance of $X^2$, considering that $X$ is distributed as a standard normal $(X \sim \mathcal{N}(0,1))$.
I can find that $E(X^2)=1$, by using that $\sigma^2(X)=E(X^2)-(E(X))^2=1$, but I'm stuck trying to find the variance.
 A: If $X\sim N(0,1)$, then $Z=X^2 \sim \chi^2(1)$.
So, we can use the Gamma($\frac{1}{2},\frac{1}{2}$) distribution to get:
$$\mathbb{E}[Z] = \frac{\left(\frac 12\right)}{\frac 12} = 1; \quad 
\mathbb{E}[Z^2] = \frac{\left(\frac 12+1\right)\left(\frac 12\right)}{\left(\frac 12\right)^2} = 3$$ 
$\therefore Var(X^2)  = 3 - 1 = 2$ 
A: The comment posted by angryavian is the way to go. I'll give you a few more details. We know that $\sigma^2(X)=E(X^2)-E(X)^2$, so by extension, $\sigma^2(X^2)=E(X^4)-E(X^2)^2$. Now for a standard normal distribution, we know that $E(X^2)=\sigma^2(X)=1$, so we must just find the value of $E(X^4)$. By definition,
$$E(X^4)=\left(\int_{-\infty}^{\infty} x^4e^{-x^2/2} dx \right)/\sqrt{2 \pi}$$ 
(The $\sqrt{2 \pi}$ in the denominator comes from the function for the standard normal distribution.) Using $u=x^3$ and $dv=xe^{\frac{-x^2}{2}} dx$, we just have to integrate a couple times and use the well-known fact:
$$\sqrt{\pi}=\int_{-\infty}^{\infty} e^{-x^2} dx$$
Carrying this out, we will see that $E(X^4)=3$ and therefore $\sigma^2(X^2)=E(X^4)-E(X^2)^2=2$. You can further show that $E(X^{2n})=(2n-1)!!$, where the double factorial function is the product of the integers less than or equal to a number of the same parity. For example, $5!!=(5)(3)(1)=15$ and $6!!=(6)(4)(2)=48$.
