# Unbiased estimator

by happyg1
Tags: estimator, unbiased
 P: 308 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
 P: 308 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
 P: 308 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