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Unbiased estimator 
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#1
Jan3006, 12:21 PM

P: 308

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{(n1)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{(n1)S^2}{\sigma^2})[/tex]is supposed to be[tex]\sigma^2[/tex] which is 2(n1). Am I going the right way here? CC 


#2
Jan3106, 09:03 AM

P: 308

Ok, So after hours of staring at this thing, here's what I did:
I let k=1/2 and r=n1, so the thing looks like this: [tex]E[S]=\sigma(\sqrt{\frac{2}{n1}}\frac{\Gamma\frac{n}{2}}{\Gamma\frac{n1}{2}}[/tex] so I use the property of the gamma function that says: [tex]\Gamma(\alpha)=(\alpha1)![/tex] which leads to: [tex]E[S]=\sigma\sqrt\frac{2}{n1}(n1)[/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(n1)}E[S]=\sigma[/tex] Any input will be appreciated. CC 


#3
Jan3106, 10:42 PM

P: 308

OK
Anyone who looked and ran away, here at last is the solution: (finally) [tex]E[S]=\sigma\sqrt{\frac{2}{n1}} \frac{\Gamma\frac({n}{2})}{\Gamma\frac({n1}{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{n1}{2}\frac{\Gamma(\frac{n1}{2})}{\frac\Gamma(\frac{n}{2})}S\right)=\sigma[/tex] So at last it has been resolved. WWWWEEEEEEEEEEEeeeeeeee CC 


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