# moments of Normal distribution squared

let $$X \sim N(0,\sigma^2)$$

now let $$Y=X^2$$

what is $$E(Y)$$ and $$V(Y)$$

so $$E(Y)=E(X^2)=\sigma^2$$

but I can't find $$V(Y)$$

• Surely you can find $V(Y)$ from the definition, because the moments of $Y$ bear a simple relationship to those of $X:$ $$E[Y^k] = E[(X^2)^k] = E[X^{2k}]$$ shows the raw moments of $Y$ are the corresponding raw even moments of $X.$ – whuber Mar 13 at 13:47

Using the definition of variance, you have: $$\operatorname{var}(Y)=E[X^4]-E[X^2]^2$$ You need the 4-th non-central moment of normal RV. Since, the mean is $$0$$, you can also use central moment, which is $$3\sigma^4$$. Subtracting $$E[X^2]^2=\sigma^4$$ makes variance of $$Y$$ $$2\sigma^4$$.
Another way is to use the relation between normal, chi-squared and gamma distributions. The squared standard normal is chi-squared distribution with $$k=1$$, which is at the same time Gamma RV. So, let $$X=\sigma Z$$, then $$Z^2\sim\chi^2(1)\rightarrow \sigma^2 Z^2=X^2 \sim \Gamma(k=1/2,\theta=2\sigma^2)$$. So, $$Y$$ is actually Gamma distributed. Gamma variance is $$\operatorname{var}(Y)=k\theta^2=2\sigma^4$$.
Another way can be simply evaluating the integral (let $$c=\frac{1}{\sqrt{2\pi}\sigma}$$):
\begin{align}E[X^4]&=c\int_{-\infty}^\infty x^4 \exp(-x^2/2\sigma^2)dx\\&=c\underbrace{\int_{-\infty}^\infty x^4 \exp(-x^2/2\sigma^2)dx}_{u=x^3,dv=x\exp(-x^22\sigma^2)}\\&=3\sigma^2\underbrace{\int_{-\infty}^{\infty}cx^2\exp(-x^2/2\sigma^2)dx}_{E[X^2]}\\&=3\sigma^4\end{align}