Hi, I'm working on the following problem and I need some clarification: Suppose that a sample is drawn from a [tex]N(\mu,\sigma^2)[/tex] distribution. Recall that [tex]\frac{(n-1)S^2}{\sigma^2}[/tex] has a [tex]\chi^2[/tex] distribution. Use theorem 3.3.1 to determine an unbiased estimator of [tex]\sigma[/tex] Thoerem 3.3.1 states: Let X have a [tex]\chi^2(r)[/tex] distribution. If [tex] k>-\frac{r}{2}[/tex] then [tex]E(X^k)[/tex] exists and is given by: [tex] E(X^k)=\frac{2^k(\Gamma(\frac{r}{2}+k))}{\Gamma(\frac{r}{2})}[/tex] My understanding is this: The unbiased estimator equals exactly what it's estimating, so [tex]E(\frac{(n-1)S^2}{\sigma^2})[/tex]is supposed to be[tex]\sigma^2[/tex] which is 2(n-1). Am I going the right way here? CC
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: [tex]E=\sigma(\sqrt{\frac{2}{n-1}}\frac{\Gamma\frac{n}{2}}{\Gamma\frac{n-1}{2}}[/tex] so I use the property of the gamma function that says: [tex]\Gamma(\alpha)=(\alpha-1)![/tex] which leads to: [tex]E=\sigma\sqrt\frac{2}{n-1}(n-1)[/tex] So now do i just flip over everything on the RHS,leaving [tex]\sigma[/tex] by itself and that's the unbiased estimator, i.e. [tex]\sqrt{2(n-1)}E=\sigma[/tex] Any input will be appreciated. CC
OK Anyone who looked and ran away, here at last is the solution: (finally) [tex]E=\sigma\sqrt{\frac{2}{n-1}} \frac{\Gamma\frac({n}{2})}{\Gamma\frac({n-1}{2})}[/tex] 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 [tex]\sigma[/tex] on the RHS and then using properties of the Expectation to get: [tex]E\left(\sqrt\frac{n-1}{2}\frac{\Gamma(\frac{n-1}{2})}{\frac\Gamma(\frac{n}{2})}S\right)=\sigma[/tex] So at last it has been resolved. WWWWEEEEEEEEEEEeeeeeeee CC