$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 $E(s^2) = \sigma^2$?

  • $\begingroup$ It's the very same expansion as the one I gave on your other question. You need to pay more attention. $\endgroup$
    – Anubhav C
    Commented May 29, 2013 at 2:22
  • $\begingroup$ You are right Anubhav, I realized after but it is definitely something I should have seen beforehand. $\endgroup$
    – user26091
    Commented May 29, 2013 at 2:40
  • $\begingroup$ This question is also answered in full detail at stats.stackexchange.com/questions/47325/…: just set all the weights to be equal. $\endgroup$
    – whuber
    Commented May 29, 2013 at 14:13

1 Answer 1


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

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


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

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 $E(x_i^2)$.

  • $\begingroup$ Thanks Glen, I will try to work it out from here. I completely forgot about independence! $\endgroup$
    – user26091
    Commented May 29, 2013 at 1:00
  • $\begingroup$ @Glen_b Sorry for the meta question; I wasn't aware of the self-study tag and its rules. As your answer has been explained away and I only had two data points to start with, the issue is no longer worthy of a meta discussion. I have therefore deleted my post. $\endgroup$
    – Anubhav C
    Commented May 29, 2013 at 4:40
  • 1
    $\begingroup$ This one was a good example of what you were concerned about - it was originally just the first sentence - pretty much just a couple of links (but it's been edited since). $\endgroup$
    – Glen_b
    Commented May 29, 2013 at 9:18

Not the answer you're looking for? Browse other questions tagged or ask your own question.