Proving question

**noob mathematician** · August 22nd 2009, 06:08 AM

Hi,

given that: $S^2=\frac{1}{n-1}\sum_{i=1}^{n}(X_i-\overline X)^2$ .

I want to show that $E[S^2]=\sigma^2$

However, when i reached: $\frac{n}{n-1}(\frac{1}{n}\sum_{i=1}^{n}E[X_i^2]-E[\overline X^2])$ , i was puzzled on finding $E[\overline X^2]$ ?

Since $E[\overline X^2]=Var(\overline X) + [E(\overline X)]^2$ which variance should I use, s.r.s or i.i.d?

or maybe is there any better way to prove it?

**Random Variable** · August 22nd 2009, 07:18 AM

$\frac{(n-1)S^{2}}{\sigma^{2}}$ is $\chi^{2}_{n-1}$ (chi-square distriubtion with (n-1) degrees of freedom)

If you want to try and prove that statement, use the fact that $S^{2}$ and $\bar{X}$ are independent.

So $E(S^{2}) = E\Big(\frac{\sigma^{2}}{n-1} \ \frac{(n-1)S^{2}}{\sigma^{2}} \Big) = \frac{\sigma^{2}}{n-1} E\Big(\frac{(n-1)S^{2}}{\sigma^{2}} \Big) = \frac{\sigma^{2}}{n-1} (n-1) = \sigma^{2}$

since the expected value of a chi-square distribution with k degrees of freedom is k

**noob mathematician** · August 23rd 2009, 04:58 AM

Hi tks for the reply.

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

Any explaination for this?

**Random Variable** · August 23rd 2009, 08:23 AM

Here's a much better proof that doesn't involve knowing anything about the chi-square distribution:

$\sum^{n}_{i=1} (X_{i} -\overline{X})^{2} = \sum^{n}_{i=1} X_{i}^{2} -n\overline{X}^{2}$

so $E\Big(\sum^{n}_{i=1} (X_{i} -\overline{X})^{2}\Big) = E\Big(\sum^{n}_{i=1} X_{i}^{2}\Big) - E(n\overline{X}^{2}) = \sum^{n}_{i=1} E(X_{i}^{2}) - nE(\overline{X}^{2})$

$= \sum^{n}_{i=1}(\sigma^{2}+ \mu^{2})-n \Big(\frac{\sigma^{2}}{n}+ \mu^{2}\Big) = n(\sigma^{2}+ \mu^{2})- \sigma^{2}- n\mu^{2} = (n-1)\sigma^{2}$

then $E(S^{2}) = E\Big(\frac{1}{n-1} \sum^{n}_{i=1} (X_{i} -\overline{X})^{2}\Big) = \frac{1}{n-1} E\Big(\sum^{n}_{i=1} (X_{i} -\overline{X})^{2}\Big) =$ $\frac{1}{n-1}(n-1)\sigma^{2} = \sigma^{2}$

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