
Proving question
Hi,
given that:$\displaystyle S^2=\frac{1}{n1}\sum_{i=1}^{n}(X_i\overline X)^2$.
I want to show that $\displaystyle E[S^2]=\sigma^2$
However, when i reached: $\displaystyle \frac{n}{n1}(\frac{1}{n}\sum_{i=1}^{n}E[X_i^2]E[\overline X^2])$, i was puzzled on finding $\displaystyle E[\overline X^2]$?
Since $\displaystyle E[\overline X^2]=Var(\overline X) + [E(\overline X)]^2$ which variance should I use, s.r.s or i.i.d?
or maybe is there any better way to prove it?

$\displaystyle \frac{(n1)S^{2}}{\sigma^{2}} $ is $\displaystyle \chi^{2}_{n1} $ (chisquare distriubtion with (n1) degrees of freedom)
If you want to try and prove that statement, use the fact that $\displaystyle S^{2} $ and $\displaystyle \bar{X} $ are independent.
So $\displaystyle E(S^{2}) = E\Big(\frac{\sigma^{2}}{n1} \ \frac{(n1)S^{2}}{\sigma^{2}} \Big) = \frac{\sigma^{2}}{n1} E\Big(\frac{(n1)S^{2}}{\sigma^{2}} \Big) = \frac{\sigma^{2}}{n1} (n1) = \sigma^{2}$
since the expected value of a chisquare distribution with k degrees of freedom is k

Hi tks for the reply.
In addition to this, I've read book stating that the best estimator is $\displaystyle (1\frac{1}{N})s^2$ as compared to this answer showing that $\displaystyle E[s^2]=\sigma ^2$.
Any explaination for this?

Here's a much better proof that doesn't involve knowing anything about the chisquare distribution:
$\displaystyle \sum^{n}_{i=1} (X_{i} \overline{X})^{2} = \sum^{n}_{i=1} X_{i}^{2} n\overline{X}^{2}$
so $\displaystyle E\Big(\sum^{n}_{i=1} (X_{i} \overline{X})^{2}\Big) = E\Big(\sum^{n}_{i=1} X_{i}^{2}\Big)  E(n\overline{X}^{2}) = \sum^{n}_{i=1} E(X_{i}^{2})  nE(\overline{X}^{2})$
$\displaystyle = \sum^{n}_{i=1}(\sigma^{2}+ \mu^{2})n \Big(\frac{\sigma^{2}}{n}+ \mu^{2}\Big) = n(\sigma^{2}+ \mu^{2}) \sigma^{2} n\mu^{2} = (n1)\sigma^{2} $
then $\displaystyle E(S^{2}) = E\Big(\frac{1}{n1} \sum^{n}_{i=1} (X_{i} \overline{X})^{2}\Big) = \frac{1}{n1} E\Big(\sum^{n}_{i=1} (X_{i} \overline{X})^{2}\Big) = $ $\displaystyle \frac{1}{n1}(n1)\sigma^{2} = \sigma^{2} $