Variance of MLE $\sigma^2$ estimator

Question

Let $X_1,X_2,\dots,X_n$ be an i.i.d. random sample from $N(\mu,σ^2)$.

a. I found the estimator MLE of $\sigma^2$ $$\hat{\sigma^2}=\frac{1}{n}\sum_{i=1}^n (X_i-\overline{X})^2$$ But how to calculate: $$Var(\frac{1}{n}\sum_{i=1}^n (X_i-\overline{X})^2)$$ Please help me. I try write that $$\hat{\sigma^2}=\frac{1}{n}\sum_{i=1}^n (X_i-\overline{X})^2=\frac{1}{n}\sum_{i=1}^n X_i^2 - \overline{X}^2$$ and how compute it $$Var(\hat{\sigma^2})$$

Well you know that $(\sum_{i=1}^n (X_i-\overline{X})^2)/\sigma^2$ has a $\chi^2$ distribution, right? — , May 04 '17 at 22:47
You can solve this problem using the variance formula for a $\chi^2$ random variable. — , May 04 '17 at 22:54
I try to solve this problem by Fisher Information $$I(\sigma^2)=-E(\frac{\partial ^2}{\partial^2 \sigma^2} ln(f(x,\mu,\sigma^2)))=-E(-\frac{(x-\mu^2)}{2\sigma^6}+\frac{1}{2\sigma^4})$$ is it right? — alto de aitana, May 04 '17 at 23:02
ok I get that this variance equal $$ \frac{2 \sigma^4}{n}$$ but i still don't know how to compute this without fisher information, analytical method — alto de aitana, May 04 '17 at 23:16

score 1 · Answer 1 · answered May 05 '17 at 12:32

1

Since ${\sum_{i=1}^n(X_i-\bar X)^2\over\sigma^2}\sim \chi^2_{(n-1)},$ we get $V(\hat\sigma^2) = {\sigma^4\over n^2}\,V\left(\chi^2_{(n-1)}\right)=\displaystyle {2\sigma^4(n-1)\over n^2}.$

answered May 05 '17 at 12:32

Why my variance $$\frac{2\sigma^4}{n}$$ is other than yours? – alto de aitana May 05 '17 at 20:54
@altodeaitana You must have an error in your calculations. – May 05 '17 at 22:32
i'm sure my answer and calculations – alto de aitana May 05 '17 at 22:52

Variance of MLE $\sigma^2$ estimator

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