I have to prove that the sample variance is an unbiased estimator. What is is asked exactly is to show that following estimator of the sample variance is unbiased:
$s^2=\frac{1}{n-1}\sum\limits_{i=1}^n(x_i-\bar x)^2$
I already tried to find the answer myself, however I did not manage to find a complete proof.
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.
The proof I used can be found under http://economictheoryblog.wordpress.com/2012/06/28/latexlatexs2/
The proof itself is not very complicated but rather long. That also the reason why I am not writing it down here and probably it is not fair towards the person who actually provided it in the first place.
For a shorter proof, here are a few things we need to know before we start:
$X_1, X_2 , ..., X_n$ are independent observations from a population with mean $\mu$ and variance $\sigma^{2}$
$\mathbb E(X_i) = \mu$ , $\mathbb{Var}(X_i)= \sigma^{2}$
$\mathbb E(X^2) = \sigma^{2} + \mu^{2}$
$\mathbb{Var}(X)=\mathbb E(X^2)-\mathbb [E(X)]^2$
$\mathbb E(\bar{X}^2) = \frac{\sigma^2}{n} + \mu^2$
Let's try to show that $\mathbb E(s^2)= \mathbb E\left(\frac{\sum_{i=1}^n (X_i - \bar X)^2}{n-1}\right) = \sigma^{2}$
To make my life easier, I will omit the limits of summation from now onwards, but let it be known that we are always summing from $1$ to $n$.
$\mathbb E\left(\sum (X_i - \bar X)^2 \right) = \mathbb E\left(\sum X_{i}^2 - 2 \bar X \sum X_i + n \bar X^2 \right) = \sum \mathbb E(X_{i}^2) - \mathbb E\left(n \bar X^2 \right)$
$\sum \mathbb E(X_{i}^2) - \mathbb E\left(n \bar X^2 \right) = \sum \mathbb E(X_{i}^2) - n \mathbb E\left(\bar X^2\right) = n \sigma^2 + n \mu^2 - \sigma^2 -n \mu^2$
This simplifies to $(n-1) \sigma^2$
So far, we have shown that $\mathbb E\left(\sum (X_i - \bar X)^2 \right) = (n-1)\sigma^2$
$\mathbb E(s^2)= \mathbb E\left(\frac{\sum (X_i - \bar X)^2}{n-1}\right) = \frac{1}{n-1} \mathbb E\left(\sum (X_i - \bar X)^2 \right)$
$\mathbb E(s^2) = \frac {(n-1)\sigma^2}{n-1} = \sigma^2$
We have now shown that the sample variance is an unbiased estimator of the population variance.
Let's improve the "answers per question" metric of the site, by providing a variant of @FiveSigma 's answer that uses visibly the i.i.d. assumption (showing also its necessity).
We want to prove the unbiasedness of the sample-variance estimator, $$s^2 \equiv \frac{1}{n-1}\sum\limits_{i=1}^n(x_i-\bar x)^2$$
using an i.i.d. sample of size $n$, from a distribution having variance $\sigma^2$,
$$E(s^2) =?\; \sigma^2$$
First, write
$$s^2 \equiv \frac{n}{n-1} \frac{1}{n}\sum_{i=1}^n(x_i-\bar x)^2$$
Then
$$\frac{1}{n}\sum_{i=1}^n(x_i-\bar x)^2 = \frac 1n \left(\sum_{n=1}^n(x_i^2- 2\bar x x_i + \bar x^2)\right) = \frac 1n \sum_{n=1}^nx_i^2- 2\bar x \frac 1n \sum_{n=1}^nx_i + \bar x^2$$
Since $\bar x = \frac 1n \sum_{n=1}^nx_i$ we get
$$\frac{1}{n}\sum_{i=1}^n(x_i-\bar x)^2 =\frac 1n \sum_{n=1}^nx_i^2- \bar x^2$$
We consider the expected value of the two components
$$E\left(\frac 1n \sum_{n=1}^nx_i^2\right) = \frac 1n \sum_{n=1}^nE(x_i^2)=E(X^2)$$
since the variables are identically distributed.
Also
$$\bar x ^2 = \left(\frac 1n \sum_{n=1}^nx_i\right)^2 = \frac 1{n^2}\left(\sum_{n=1}^nx_i^2 + \sum_{i\neq j}x_ix_j\right)$$
the second sum having $n^2-n$ elements. So $$E(\bar x^2) = \frac 1{n^2}(nE(X^2)) + \frac 1{n^2}\left[(n^2-n)E(x_i)E(x_j)\right]$$
We were able to write $E(x_ix_j) = E(x_i)E(x_j)$ because the sample is comprised of independent RVs. More over they are identical so $E(x_i)E(x_j) = [E(X)]^2$. Therefore
$$E(\bar x^2) = \frac 1nE(X^2) + \frac {n-1}{n}[E(X)]^2$$
Bringing it all together,
$$E(s^2) = \frac {n}{n-1}\cdot \left[E(X^2) - \frac 1nE(X^2) - \frac {n-1}{n}[E(X)]^2\right]$$
$$= \frac {n}{n-1}\cdot \left[\frac {n-1}{n}E(X^2) - \frac {n-1}{n}[E(X)]^2\right]$$
$$\implies E(s^2) = E(X^2) - [E(X)]^2 \equiv {\rm Var}(X)$$
