A question about sampling distribution Assume there are $n$ independent random variables $X_1,X_2,\ldots,X_n$ and i wonder why the sample variance is $S^2=\frac{\sum\limits_{i=1}^n \ (X_i-X)^2}{n-1}$ where $X$=$\frac{X_1+X_2+\cdots+X_n}{n}$ instead of $S^2=\frac{\sum\limits_{i=1}^{n}\ (X_i-X)^2}{n}$.
 A: There is a Wikipedia article about this: Bessel's correction.  (I wrote some of it myself, but in this case others did most of it.)
Bessels correction eliminates bias.  Eliminating bias is sometimes a very bad idea, as I explained in this paper: "An Illuminating Counterexample, American Mathematical Monthly, volume 110, number 3 (March, 2003), pages 234-238.
A: You have said that $X_1, X_2, \ldots, X_n$ are independent, but not added, as most people do, that they are also identically distributed (or at the very least, have the same mean and variance).  With this added condition, we have
$$E\left[\sum_{i=1}^n (X_i-X)^2\right] = (n-1)\sigma^2$$
where $\sigma^2$ is the common variance of the $n$ random variables. Thus,
defining the sample variance as
$$S^2 = \frac{1}{n-1}\sum_{i=1}^n (X_i-X)^2$$
has the nice property that $E[S^2] = \sigma^2$.
As Didier Piau has already suggested to you in a comment, you should
work out the expectation calculation for yourself. 
You might also want
to try and work out what happens when the $X_i$ have the same mean $\mu$
but different variances $\sigma_i^2$, and also when they have
different means $\mu_i$ and variances $\sigma_i^2$.
