I am asked to show that the unbiased estimator $\hat{\sigma^2}=\frac{1}{n-1}\sum_{i=1}^n (X_i-\bar{X})^2$ is not efficient.

So far I was able to show that the Rao-Cramer Lower Bound is $\frac{2\sigma^4}{n}$ so I know that I need to find $Var[\hat{\sigma^2}]>RCLB$.

What I have tried here is to simply plug it in using the identity

$$\sum_{i=1}^n (X_i-\bar{X})^2 = \sum_{i=1}^n X_i^2 - n\bar{X}^2$$

What makes me uncomfortable is that unlike expectation, I need to know whether these two terms are independent or not, and I do not know how to find $Var[\bar{X}^2]$ even if they were.

I also tried to express the sum of square as

$$\sum_{i=1}^n (X_i-\mu +\mu-\bar{X})^2 = \sigma^2 \left[ \sum_{i=1}^n\left(\frac{X_i-\bar{X}}{\sigma}\right)^2 - 2\left(\frac{X_i-\bar{X}}{\sigma}\right)\sum_{i=1}^n\left( \frac{X_i-\mu}{\sigma} \right) + n\left( \frac{\bar{X}-\mu}{\sigma} \right)^2 \right]$$

hoping that $\chi_{n}^2$ and $N(0,1)$ could come into play but I face the same issue again.

I would appreciate your input.

  • $\begingroup$ If you want to find the variance of sample variance, this page could help (the accepted answer there has the variance in the case where the $X_i$ are normal, showing that it is $\frac{2\sigma^4}{n-1}$). Also this page. $\endgroup$ Jun 22, 2019 at 0:31
  • $\begingroup$ The author probably forgot to mention that the sample is i.i.d. and came from a Normal($\mu,\sigma^2$) distribution, otherwise nothing makes sense. $\endgroup$
    – bluemaster
    Aug 30, 2022 at 10:23

1 Answer 1


Because: $$\hat{\sigma^2} \sim \frac{\chi_{n-1}^2 \sigma^2}{n-1}$$ And since the variance of the Chi-squared distribution is $2k$ where $k$ is the number of the degrees of freedom we have: $$Var[\hat{\sigma^2}]=\frac{2(n-1)\sigma^4}{(n-1)^2}=\frac{2\sigma^4}{n-1}>\frac{2\sigma^4}{n}$$

  • $\begingroup$ Can you elaborate on why it is chi-squared? I thought $(\frac{X-\mu}{\sigma})^2$ was chi-squared, not $(\frac{X-\bar{X}}{\sigma})^2$ $\endgroup$
    – hyg17
    Jun 22, 2019 at 0:47
  • $\begingroup$ They both have Chi-squared distribution but the first one has $n$ degrees of freedom whereas the second one has just $n-1$. Please see stats.stackexchange.com/questions/121662/… $\endgroup$
    – Bartek
    Jun 22, 2019 at 0:50
  • $\begingroup$ thank you ! I will take a look at it $\endgroup$
    – hyg17
    Jun 22, 2019 at 1:09

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