Showing that $s^2$ is an unbiased estimator of $\sigma^2$ [duplicate]

Question

$s^2 = \sum \frac{(x_i - \bar{x})^2}{n-1} $ which apparently equals $ \frac{\sum(x_i^2) +n\bar{x}^2 - 2n\bar{x}^2}{n-1}$. Does this just come from expanding the numerator and using the fact that $\bar{x}$ (the average) is not dependent on $i$? How might one continue this to show for certain $E(s^2) = \sigma^2$?

Answer 1

Yes, the second form simply comes from expanding the quadratic and from these (obvious) facts:

$$\sum_i \bar{x}^2 = n\bar{x}^2$$

and

$$\sum_i x_i \bar{x} = \bar{x}\sum_i x_i = \bar{x} \cdot n \bar{x}=\cdots $$

How might one continue ...

http://en.wikipedia.org/wiki/Variance#Sample_Variance (though you can do it a bit simpler than that, but take care to note that the expression there isn't quite identical to yours; you need to carry the ideas over, not the exact algebra line by line)

That is, use linearity of expectation, independence (when evaluating expectations of cross products with no common terms, for example) and known facts about $E(x_i^2)$.