Say I'm interested in estimating the $\sigma^2$ of a normal population, which has a Cramer-Rao Lower Bound of $\frac{2\sigma^4}{n}$ if my calculations are correct. I found that the uniformly minimum variance unbiased estimator of $\sigma^2$ should be $\frac{1}{n} \sum_{i = 1}^n (x_i - \mu)^2$, using the equality condition for the Cauchy-Schwarz Inequality. Does this mean that if the population mean is unknown, then I have no UMVUE?

I have verified that $\frac{1}{n-1} \sum_{i = 1}^n (x_i - \bar{X})^2$ does not reach the CRLB stated above (although it achieves the bound asymptotically).

  • $\begingroup$ Sorry, I mean $\sigma^2$ is also supposed to be unknown... $\endgroup$ Commented Oct 20, 2016 at 6:32
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    $\begingroup$ When $\mu$ and $\sigma$ are unknown, $\frac{1}{n-1} \sum_{i = 1}^n (x_i - \bar{X})^2$ is the UMVUE, even though it does not reach the Cramer-Rao lower bound. Which is a lower bound, not a minimal value that some estimator should reach. $\endgroup$
    – Xi'an
    Commented Oct 20, 2016 at 6:36
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    $\begingroup$ Addendum: Using the Cramer-Rao lower bound is a way to prove an estimator is UMVUE, not a necessary property of the UMVUE. $\endgroup$
    – Xi'an
    Commented Oct 20, 2016 at 6:41
  • $\begingroup$ I see, thanks. In that case, how can I show that $\frac{1}{n-1} \sum_{i = 1}^n (x_i - \bar{X})^2$ is the UMVUE? (Sorry but I don't seem to have the right to vote up comments in this SE.) $\endgroup$ Commented Oct 20, 2016 at 6:45
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    $\begingroup$ You can use the Lehmann-Scheffé theorem. $\endgroup$
    – Xi'an
    Commented Oct 20, 2016 at 10:14


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