# MSE for MLE of normal distribution's ${\sigma}^2$

So I've known $$MLE$$ for $${\sigma}^2$$ is $$\hat{{\sigma}^2}=\frac{1}{n}\sum_{i=1}^{n} (X_{i} -\bar{X})^2$$, and I'm looking for $$MSE$$ of $$\hat{{\sigma}^2}$$. But I'm having trouble to get the result.

What I tried goes like below:

By definition, $$MSE$$ = $$E[(\hat{{\sigma}^2}$$ - $${\sigma}^2$$)$$^2$$], which is = $$Var(\hat{{\sigma}^2} - {\sigma}^2)+(E(\hat{{\sigma}^2} - {\sigma}^2))^2$$ = $$Var(\hat{{\sigma}^2})-Var({{\sigma}^2})+(E(\hat{{\sigma}^2} - {\sigma}^2))^2$$.

From here, I tried to find $$Var(\hat{{\sigma}^2})$$, which is = $$Var(\frac{1}{n}\sum_{i=1}^{n} (X_{i} -\bar{X})^2$$) = $$\frac{1}{n^2}Var(\sum_{i=1}^{n} X_{i}^2 -n\bar{X}^2)$$ = $$\frac{1}{n^2}(\sum_{i=1}^{n} Var (X_{i}^2) -n^2Var(\bar{X}^2))$$

But I'm not sure how to get $$Var (X_{i}^2)$$ and $$Var(\bar{X}^2)$$. I tried $$Var (X_{i}^2)$$ = $$E(X_i^4) - (E(X_i^2))^2$$, But I'm not quite sure what $$E(X_i^4)$$ would be.

Could anyone help me with this? Am I on the correct path to solve this? Thanks in advance!

With $$\displaystyle \widehat{\sigma^2} = \frac 1 n \sum_{i=1}^n \left( X_i - \overline X \right)^2$$ you have $$\displaystyle\frac{n\widehat{\sigma^2}}{\sigma^2} \sim \chi^2_{n-1},$$ so $$\operatorname{var}\left( \,\widehat{\sigma^2} \, \right) = \frac{\sigma^4}{n^2} \operatorname{var}(\chi^2_{n-1}) = \frac{\sigma^4}{n^2}\cdot 2(n-1).$$