UMVUE for normal distribution $\sigma$

Question

Let $X_1,X_2,...,X_n$ be a random sample from a normal distribution with mean $\mu$ and variance $\sigma^2$.
I showed that $(\bar X,S^2)$ is jointly sufficient for estimating ($\mu$,$\sigma^2$) where $\bar X$ is the sample mean and $S^2$ is the sample variance.

Then assuming that$(\bar X,S^2)$ is also complete I have to show that $$\sqrt{ n-1\over 2}{\Gamma ({ n-1\over 2})\over\Gamma (\frac n2)} S$$ is a Uniformly Minimum Variance Unbiased Estimator for $\sigma$.

I think I have to use Lehman Scheffe theorem as $(\bar X,S^2)$ is jointly sufficient and complete for $\sigma$. But how can I find a function which is unbiased for $\sigma$ that contains both $(\bar X,S^2)$.

I don't understand how to work when there's a joint sufficiency and completeness.

Answer 1

Although the question was posted almost 4 years ago, I would like to answer this question. English is not my mother tongue and I am learning it so please don't mind my awkward sentences.

To solve this problem, we notice that $(n-1)S^2/ \sigma^2$ has a chisquare distribution with $n-1$ degree of freedom, while $S^2= \sum^n_{i=1}{(X-\bar{X})^2\over n-1}={{\sum^n_{i=1}X^2}-n \bar{X}^2\over n-1}$ and $X$ has a normal distribution with mean $\mu$ and variance $\sigma^2$.

Note that $S$ contains ${\sum^n_{i=1}X^2}$ and ${\sum^n_{i=1}X}$.

Let's evaluate $E[S]$. To simplify let $q=(n-1)S^2/ \sigma^2$, then $S=\sqrt{q \sigma^2 /(n-1)}$. $$E[S]=\int^{\infty}_0 \sqrt{ \sigma^2 \over (n-1)} q^{1 \over2}f_q dq \\=\int^{\infty}_0 \sqrt{ \sigma^2 \over (n-1)} q^{1 \over2} { q^{{n-1 \over 2} -1} e^{-q \over 2} \over \Gamma({n-1 \over 2}) 2^{n-1 \over 2}} dq \\ = \sqrt{ \sigma^2 \over (n-1)} \int^{\infty}_0 { q^{{n \over 2} -1} e^{-q \over 2} \over \Gamma({n-1 \over 2}) 2^{n-1 \over 2}} dq \\= \sqrt{ \sigma^2 \over (n-1)} { \Gamma({n \over 2}) 2^{1 \over 2} \over \Gamma({n-1 \over 2}) } $$

After some rearranging you can get the desired result.

It would be appreciated if someone corrects any grammatical or mathematical mistakes.