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 $$Var(\hat{\sigma^2})$$

  • $\begingroup$ Well you know that $(\sum_{i=1}^n (X_i-\overline{X})^2)/\sigma^2$ has a $\chi^2$ distribution, right? $\endgroup$ – user940 May 4 '17 at 22:47
  • $\begingroup$ Yes i know that $\endgroup$ – alto de aitana May 4 '17 at 22:51
  • $\begingroup$ You can solve this problem using the variance formula for a $\chi^2$ random variable. $\endgroup$ – user940 May 4 '17 at 22:54
  • $\begingroup$ I totally don't understand how to solve this problem. $\endgroup$ – alto de aitana May 4 '17 at 22:57
  • $\begingroup$ 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? $\endgroup$ – alto de aitana May 4 '17 at 23:02

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

  • $\begingroup$ Why my variance $$\frac{2\sigma^4}{n}$$ is other than yours? $\endgroup$ – alto de aitana May 5 '17 at 20:54
  • $\begingroup$ @altodeaitana You must have an error in your calculations. $\endgroup$ – user940 May 5 '17 at 22:32
  • $\begingroup$ i'm sure my answer and calculations $\endgroup$ – alto de aitana May 5 '17 at 22:52

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