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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!

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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 $$ \operatorname{var}\left( \,\widehat{\sigma^2} \, \right) = \frac{\sigma^4}{n^2} \operatorname{var}(\chi^2_{n-1}) = \frac{\sigma^4}{n^2}\cdot 2(n-1). $$

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