I'm trying to show that variance of $\hat{\sigma}^{UMVUE}$(which is estimator of $\sigma$ in $N(\mu, \sigma^2)$) is larger than cramer-rao lower bound, which I have found to be $\frac{\sigma^2}{2n}$. Below is my calculation of variance.

\begin{align*} \Rightarrow~ Var(\hat{\sigma_n}^{UMVUE}) &= \sigma^2\left(\frac{n-1}{2}\frac{\Gamma((n-1)/2)^2}{\Gamma(n/2)^2}-1\right)\\ &= \sigma^2\left\{\frac{n-1}{2}\left(\frac{(\frac{n-3}{2})^{\frac{n-2}{2}} e^{-\frac{n-3}{2}}\sqrt{2\pi}(1+o(1/\sqrt{n}))}{(\frac{n-2}{2})^{\frac{n-1}{2}} e^{-\frac{n-2}{2}}\sqrt{2\pi}(1+o(1/\sqrt{n}))}\right)^2-1\right\}\\ &= \sigma^2\left\{ \frac{n-1}{n-2}e\left(\frac{1+o(1/\sqrt{n})}{1+o(1/\sqrt{n})}\right)^2 \left(\frac{n-3}{n-2}\right)^{n-2}-1 \right\}\\ &= \sigma^2\left\{ \frac{n-1}{n-2} e \left(\frac{1+o(1/\sqrt{n})}{1+o(1/\sqrt{n})}\right)^2 \frac{1}{e}(1+o(1/\sqrt{n})) -1 \right\}\\ &= \frac{\sigma^2}{n-2} (1+o(1/\sqrt{n})) \end{align*}

The second equality is Lanczos approximation, and o() is small-o notation.

The problem is, according to above result, $\lim_{n\to\infty}n\cdot Var(\hat{\sigma_n}^{UMVUE})=\sigma^2$.

However, this result is not true. I used R(program) to calculate the limit and the result was approximately(slightly larger than) $\sigma^2/2$. Can you tell me the mistake I have made in the calculation?


1 Answer 1


Are you certain that your $\frac{1+o(1/\sqrt{n})}{1+o(1/\sqrt{n})} =1$? Both parts of the fraction are derived from approximations to the gamma function in two distinct points. In general, I think these terms should not be equal, unless I am perhaps missing something about this particular approximation, please correct me in that case.

  • $\begingroup$ Thanks! now I see what I have done wrong. It should be sqrt((n-2)/(n-3)) $\endgroup$
    – mathstat
    Jan 30, 2019 at 16:43

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.