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## unbiased estimator

Hi, I'm working on the following problem and I need some clarification:
Suppose that a sample is drawn from a $$N(\mu,\sigma^2)$$ distribution. Recall that $$\frac{(n-1)S^2}{\sigma^2}$$ has a $$\chi^2$$ distribution. Use theorem 3.3.1 to determine an unbiased estimator of $$\sigma$$
Thoerem 3.3.1 states:
Let X have a $$\chi^2(r)$$ distribution. If $$k>-\frac{r}{2}$$ then $$E(X^k)$$ exists and is given by:
$$E(X^k)=\frac{2^k(\Gamma(\frac{r}{2}+k))}{\Gamma(\frac{r}{2})}$$
My understanding is this:
The unbiased estimator equals exactly what it's estimating, so $$E(\frac{(n-1)S^2}{\sigma^2})$$is supposed to be$$\sigma^2$$ which is 2(n-1).
Am I going the right way here?
CC

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Ok, So after hours of staring at this thing, here's what I did: I let k=1/2 and r=n-1, so the thing looks like this: $$E[S]=\sigma(\sqrt{\frac{2}{n-1}}\frac{\Gamma\frac{n}{2}}{\Gamma\frac{n-1}{2}}$$ so I use the property of the gamma function that says: $$\Gamma(\alpha)=(\alpha-1)!$$ which leads to: $$E[S]=\sigma\sqrt\frac{2}{n-1}(n-1)$$ So now do i just flip over everything on the RHS,leaving $$\sigma$$ by itself and that's the unbiased estimator, i.e. $$\sqrt{2(n-1)}E[S]=\sigma$$ Any input will be appreciated. CC

OK Anyone who looked and ran away, here at last is the solution: (finally) $$E[S]=\sigma\sqrt{\frac{2}{n-1}} \frac{\Gamma\frac({n}{2})}{\Gamma\frac({n-1}{2})}$$ is indeed correct, however my attempt to reduce the RHS with the properties of the Gamma function is wrong. The unbiased estimator is obtained by isolating the $$\sigma$$ on the RHS and then using properties of the Expectation to get: $$E\left(\sqrt\frac{n-1}{2}\frac{\Gamma(\frac{n-1}{2})}{\frac\Gamma(\frac{n}{2})}S\right)=\sigma$$ So at last it has been resolved. WWWWEEEEEEEEEEEeeeeeeee CC

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