3
$\begingroup$

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).

$\endgroup$
6
  • $\begingroup$ Sorry, I mean $\sigma^2$ is also supposed to be unknown... $\endgroup$ Commented Oct 20, 2016 at 6:32
  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ You can use the Lehmann-Scheffé theorem. $\endgroup$
    – Xi'an
    Commented Oct 20, 2016 at 10:14

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.