1. ## Bias of Estimator

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

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

[Hint: Recall the distribution of $\displaystyle \frac{(n-1)S^{2}}{\sigma^{2}}$ and the result $\displaystyle E(Y^{a})$$\displaystyle =\frac{\beta^{a}\Gamma(\alpha\\+a)}{\Gamma(\alpha) }]. A: Let \displaystyle U=\frac{(n-1)S^{2}}{\sigma^{2}}. Since \displaystyle U~\displaystyle \chi^{2}(n-1) we have that \displaystyle E(U)=n-1 and \displaystyle V(U)=2(n-1). Furthermore, using the result in the hint and the fact that \displaystyle U~\displaystyle Gamma(\frac{n-1}{2},2) we have that \displaystyle E(\sqrt{U})$$\displaystyle =E(U^{\frac{1}{2}})$

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

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

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

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

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:
$\displaystyle E(S)=\sigma\sqrt{\frac{2}{n-1}}\frac{\Gamma(n/2)}{\Gamma((n-1)/2)}\neq\sigma$

Which proves that $\displaystyle S$ 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:
$\displaystyle E(S)=\sigma\sqrt{\frac{2}{n-1}}\frac{\Gamma(n/2)}{\Gamma((n-1)/2)}\neq\sigma$

Which proves that $\displaystyle S$ 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 $\displaystyle \Gamma(a)=(a-1)\Gamma(a-1)$ which is just integration by parts.

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

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

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 $\displaystyle n$ is even, $\displaystyle \Gamma(n/2)$ will just be $\displaystyle (n/2-1)!$. If $\displaystyle n$ is odd, $\displaystyle \Gamma(n/2)$ will be $\displaystyle (n/2 - 1)(n/2 - 2)\ldots(0.5)\sqrt{\pi}$.