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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?

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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.

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  • $\begingroup$ Thanks! now I see what I have done wrong. It should be sqrt((n-2)/(n-3)) $\endgroup$
    – mathstat
    Jan 30 '19 at 16:43

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