I'm trying to prove that the sample variance is an unbiased estimator.

I know that I need to find the expected value of the sample variance estimator $$\sum_i\frac{(M_i - \bar{M})^2}{n-1}$$ but I get stuck finding the expected value of the $M_i\bar{M}$ term. Any clues?

I would also like to calculate the variance of the sample variance. In short I would like to calculate $\mathrm{Var}(M_i - \bar{M})^2$ but again that term rears its ugly head.

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    $\begingroup$ Assuming the $M_i$ are iid, hint: let $\mu = E(M_1)$. Write $\sum (M_i - \bar M)^2 = \sum (M_i - \mu)^2 - n(\bar M - \mu)^2$ by throwing $\pm \mu$ into the difference and expanding. $\endgroup$ – guy Apr 3 '12 at 3:43
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    $\begingroup$ See the first part of my answer here: math.stackexchange.com/questions/72975/… $\endgroup$ – user940 Apr 3 '12 at 4:01

I know that during my university time I had similar problems to find a complete proof, which shows exactly step by step why the estimator of the sample variance is unbiased. I took it from http://economictheoryblog.wordpress.com/2012/06/28/latexlatexs2/ but they use a different notation, however I think you can take it from there


The only full and complete proof I could find on the internet can be found under http://economictheoryblog.wordpress.com/2012/06/28/latexlatexs2/

As goes over 70 (!) steps and shows every little formula manipulations I prefer rather not to post it here.


A well explained proof, based on maximum likelihood estimation is explained here: http://www.visiondummy.com/2014/03/divide-variance-n-1/

It first explains how to derive the sample variance formula, and then proofs it is unbiased.


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