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Unbiased estimator

by happyg1
Tags: estimator, unbiased
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happyg1
#1
Jan30-06, 12:21 PM
happyg1's Avatar
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{(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?
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happyg1
#2
Jan31-06, 09:03 AM
happyg1's Avatar
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:
[tex]E[S]=\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[S]=\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[S]=\sigma[/tex]
Any input will be appreciated.
CC
happyg1
#3
Jan31-06, 10:42 PM
happyg1's Avatar
P: 308
OK
Anyone who looked and ran away, here at last is the solution: (finally)
[tex]E[S]=\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
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