Q:

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:

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 $\sigma_i^2$.

