Variance of a MLE $\sigma^2$ estimator; how to calculate Let $X_1, X_2,...,X_n$ be an i.i.d. random sample from $N(0, \sigma^{2})$.
a. Find the variance of $\hat{\sigma}^{2}_{MLE}$  
So I found $\hat{\sigma}^{2}_{MLE}$ by taking the derivative of the log of the normal pdf function, but from there I am not sure how to proceed. $\hat{\sigma}^{2}_{MLE}$ comes out to $\frac{\sum_{i=1}^n X_i^{2}}{n}$. From there, would I do $\text{var}\left(\frac{\sum_{i=1}^n X_i^{2}}{n}\right)$ ? How do I compute this? Thanks!
 A: Hint: If $Y_1,\ldots,Y_n$ are independent random variables and $a_1,\ldots,a_n$ are real constants, then
$$
\mathrm{Var}\left(\sum_{i=1}^n a_iY_i\right)=\sum_{i=1}^n a_i^2\mathrm{Var}(Y_i).
$$
A: Let us use first principles and rederive from scratch while ignoring all prepackaged distributions (textbooks would tell you that a sum of squared standard Gaussian random variables $\sim$ a Chi-square distribution).
Let $X'= \frac{X}{\sqrt{n}}$. Hence  $X' \sim N(0,\frac{\sigma^2}{n})$. 
The pdf of a transformation $Y =X^{'2}$, becomes $f(y)= \frac{\sqrt{\frac{n}{y}} e^{-\frac{n y}{2 \sigma ^2}}}{\sqrt{2 \pi } \sigma }\,, y\in (0,\infty)$. The Characteristic Function of Y, $\mathcal{C}(t)=\frac{1}{\sqrt{1-\frac{2 i \sigma ^2 t}{n}}}$. The distribution of an $n$-summed variable has for characteristic function $\mathcal{C}(t)^n$. Now define the raw moment of the convolution $M(d)=-i^d \frac {\partial^d\mathcal {C(t)^n}} {\partial t^d}\bigg|_{t=0}$. So unless I made a mistake somewhere, $M(1)= \sigma ^2 $ and the variance $M(2)-M(1)^2=\frac{2 \sigma ^4}{n}$.
