Variance of the Square Suppose $X_1, \cdots, X_n$ are a sample of independent variables taken from a normally distributed population with mean $\mu$ and variance $\sigma^2$.
I would like to determine the variance of the squares $X_1^2, \cdots, X_n^2$.
From a Monte Carlo simulation it seems that when $\mu$ is large enough, it is close to $\operatorname{Var}(X^2)= (2\mu\sigma)^2$, but I have no proof, and I'm struggling to come up with one. I understand it has to do with the Chi-Square distribution, but I have not connected the dots.
 A: $Var(X^2)=E(\{X^2\}^2)-[E(X^2)]^2=E(X^4)-[E(X^2)]^2$
Wikipedia has the non-central moments of a normal distribution.
$$E(X^4)=\mu^4+6\mu^2\sigma^2+3\sigma^4, E(X^2)=\mu^2+\sigma^2$$
Thus$$\begin{align}Var(X^2)&=\mu^4+6\mu^2\sigma^2+3\sigma^4-(\mu^2+\sigma^2)^2\\&=\mu^4+6\mu^2\sigma^2+3\sigma^4-\mu^4-2\mu^2\sigma^2-\sigma^4\\&=4\mu^2\sigma^2+2\sigma^4\end{align}$$
A circuitous answer involving the chi-squared distribution with one degree of freedom:
$$\left(\frac{X-\mu}{\sigma}\right)^2\sim \text{Chi Squared}(1)$$
The variance of a chi-squared random variable with 1 df is $Var(\chi^2_{1})=2$ so $$\begin{split}2&=\frac 1 {\sigma^4}Var(X^2-2X\mu+\mu^2)\\
2\sigma^4&=Var(X^2)+4\mu^2Var(X)-4\mu Cov(X^2, X)\\
&=Var(X^2)+4\mu^2\sigma^2-4\mu\left(E(X^3)-E(X^2)E(X)\right)\\
&=Var(X^2)+4\mu^2\sigma^2-4\mu(2\mu\sigma^2)\end{split}$$
and so $Var(X^2)=2\sigma^4+4\mu^2\sigma^2$.
A: $$var(x) = E(x^2) - E(x)^2 = \sigma^2$$
$$ E(x^2) = \sigma^2 + \mu^2$$
$$var(x^2) = E(x^4) - E(x^2)^2$$
$$= [\mu^4 + 6 \mu^2 \sigma^2+ 3 \sigma^4 ] - [\sigma^2 + \mu^2]^2 $$
$$ \mu^4 + 6 \mu^2 \sigma^2 + 3 \sigma^4 - \sigma^4 - 2\sigma^2 \mu^2 - \mu^4 = 4\sigma^2 \mu^2 +2 \sigma^4   $$
