Expectation of the variance of the sampling set without replacement Select $n$ numbers without replacement from the set $\{1,2,...,m\}$, and generate the set $S=\{a_1,a_2,...,a_n\}$. I want to calculate the expectation of the variance for the sampling set $\mathbb{E}[Var(S)]$ and the maximum variance among all samples : $\max{Var(S)}$.
Besides, what's the distribution of the sample variance?
 A: I will give the hints for the first question. I assume that you are sampling without a replacement from the set $A=\{a_1,...,a_m\}$. We have that 
$$P(S=\{a_1,...,a_n\})=\frac{1}{m\choose n}$$
So
$EVar(S)=\sum_{\{i_1,...,i_n\}\subset \{1,...,m\}}\left(\frac{1}{n}\sum_{j=1}^n a_{i_j}^2-\left(\frac{1}{n}\sum_{j=1}^na_{i_j}\right)^2\right)\frac{1}{m\choose n}$
Now 
$$\sum_{\{i_1,...,i_n\}\subset \{1,...,m\}}a_{i_1}^2={{m-1}\choose{n-1}}\sum_{i=1}^ma_i^2 $$
and
$$\sum_{\{i_1,...,i_n\}\subset \{1,...,m\}}a_{i_1}a_{i_2}={{m-2}\choose {n-2}}\sum_{i\neq j}^ma_ia_j$$
So the first term in expectation will be
$$\sum_{\{i_1,...,i_n\}\subset \{1,...,m\}}\frac{1}{n}\sum_{j=1}^n a_{i_j}^2\frac{1}{m\choose n}=\sum_{i=1}^na_i^2\frac{1}{m\choose n}{{m-1}\choose {n-1}}=\frac{1}{m}\sum_{i=1}^ma_i^2$$
I'll leave the second term as an exercise. The end result should be that 
$$EVar(S)=Var(a_1,...,a_m)$$
maybe with some constants missing. As this is a standard result in survey sampling theory, you can look it up in appropriate book.
As for the second question, I do not think there is a closed formula. The case with $m=3$ and $n=2$ illustrates this. Then there are 3 possible samples and $Var S$ can get three values $(a_1-a_2)^2/4$, $(a_2-a_3)^2/4$ and $(a_1-a_3)^2/4$. The maximum depends on the set $A=\{a_1,a_2,a_3\}$.
A: We know that
$$\widehat{Var}(\mathbf{a}) = \frac{1}{n-1}\left(\sum_{i=1}^n a_i^2 - \frac{1}{n}\left(\sum_{i=1}^n a_i \right)^2 \right)$$
is an unbiased estimator of the population variance, which is easily computed as $(m+1)m/12$.  This, therefore, answers the first question concerning the expected variance.
I will only sketch how to maximize the variance.  I claim it is maximized when the $a_i$ are in two contiguous blocks: that is, $\mathbf{a}$ is in the form
$$\mathbf{a} = (1, 2, \ldots, k, m-l+1, m-l+2, \ldots, m).$$
(Evidently $k+l = n$.)  To prove this claim, suppose $\mathbf{a}$ is not in this form: then you can find a gap in one of the end sequences and increase the variance by changing one of the components of $\mathbf{a}$ to that gap.  It remains only to maximize the variance among these special forms of $\mathbf{a}$; this is done by making the end sequence lengths as balanced as possible; that is, by setting $k=l$ when $n$ is even and otherwise by setting either $k=l+1$ or $l=k+1$.  When $n=2k$ is even, the maximum variance equals
$$ n \frac{\left(3 m^2-3 m n+n^2 -1\right)}{12 (n-1)}.$$
When $n=2k+1$ is odd, the maximum variance is
$$(n+1) \frac{\left(3 m^2-3 m n+n^2\right)}{12 n}.$$
