1. Bias of Estimator

Suppose that $Y_{1},Y_{2},...,Y_{n}$ constitute a random sample from a normal distribution with parameters $\mu$ and $\sigma^{2}$.

Q: Show that $S=\sqrt{S^{2}}$ is a biased estimator of $\sigma$. Moreover, adjust $S$ to form an unbiased estimator for $\sigma$.

[Hint: Recall the distribution of $\frac{(n-1)S^{2}}{\sigma^{2}}$ and the result $E(Y^{a})$ $=\frac{\beta^{a}\Gamma(\alpha\\+a)}{\Gamma(\alpha) }$].

A: Let $U=\frac{(n-1)S^{2}}{\sigma^{2}}$. Since $U$~ $\chi^{2}(n-1)$ we have that $E(U)=n-1$ and $V(U)=2(n-1)$. Furthermore, using the result in the hint and the fact that $U$~ $Gamma(\frac{n-1}{2},2)$ we have that

$E(\sqrt{U})$ $=E(U^{\frac{1}{2}})$

$=\frac{(2)^{\frac{1}{2}}\Gamma(\frac{(n-1)}{2}+\frac{1}{2})}{\Gamma(\frac{(n-1)}{2})}$

$=\frac{\sqrt{2}\Gamma(\frac{n}{2})}{\Gamma(\frac{( n-1)}{2})}$.

So, $\sqrt{\frac{\sigma^{2}}{(n-1)}}\frac{\sqrt{2}\Gamma(\frac{n}{2})}{\Gamma(\fra c{(n-1)}{2})}$

$=\sigma\sqrt{\frac{2}{(n-1)}}\frac{\Gamma(\frac{n}{2})}{\Gamma(\frac{(n-1)}{2})}$.

I am not sure what do do fro the second part of the question.

Am I doing this correctly?

2. I don't know how much help this will be, but from my stats textbook I found in the section on estimators that:
$E(S)=\sigma\sqrt{\frac{2}{n-1}}\frac{\Gamma(n/2)}{\Gamma((n-1)/2)}\neq\sigma$

Which proves that $S$ is a biased estimator. However, note that it is asymptotically unbiased.

3. Originally Posted by cpbrunner
I don't know how much help this will be, but from my stats textbook I found in the section on estimators that:
$E(S)=\sigma\sqrt{\frac{2}{n-1}}\frac{\Gamma(n/2)}{\Gamma((n-1)/2)}\neq\sigma$

Which proves that $S$ is a biased estimator. However, note that it is asymptotically unbiased.
Thanks, I found the algebra error in my derivation. Even so, I am still stuck on the second part. My thinking is I need to multiply by the reciprocal of the first part of the result, but I don't know what to do about getting rid of the gamma stuff.

4. USE $\Gamma(a)=(a-1)\Gamma(a-1)$ which is just integration by parts.

Also use $\Gamma(.5)=\sqrt{\pi}$

So $\Gamma(2.5)=(1.5)(.5)\sqrt{\pi}$

5. Originally Posted by Danneedshelp
Thanks, I found the algebra error in my derivation. Even so, I am still stuck on the second part. My thinking is I need to multiply by the reciprocal of the first part of the result, but I don't know what to do about getting rid of the gamma stuff.

Yeah, you've got it. Multiply by the reciprocal for an unbiased estimator. The gamma stuff is just a constant as matheagle pointed out. If $n$ is even, $\Gamma(n/2)$ will just be $(n/2-1)!$. If $n$ is odd, $\Gamma(n/2)$ will be $(n/2 - 1)(n/2 - 2)\ldots(0.5)\sqrt{\pi}$.