Expectation of sampling uniformly Consider the following situation, there are $m$ numbers $n_i$ in a set set $S$. At each time $t$, $k$ numbers are selected uniformly and without replacement and averaged. Then the points are put back into the set. What is the expected value of this average?
I believe it is:
$$
\frac{1}{m}\sum_{i=1}^m n_i
$$
But I cannot prove it. Is anyone able to confirm this? Clearly there are $m \choose k$ possible sets each having the same probability of being chosen but I cannot get much further than that.
 A: Denote the numbers that are selected by $X_{1},\ldots,X_{k}$. 
These are random variables with uniform distribution: $P\left\{ X_{j}=n_{i}\right\} =\frac{1}{m}$ for each $j$ and each $i$.
There is no independency but that is not relevant here.
Note that: 
$\mathbb{E}X_{j}=\frac{1}{m}\sum_{i=1}^{m}n_{i}$ for each $j$.
$\overline{X}:=\frac{1}{k}\left(X_{1}+\cdots+X_{k}\right)$
denotes the average and: 
$\mathbb{E}\overline{X}=\frac{1}{k}\left(\mathbb{E}X_{1}+\cdots+\mathbb{E}X_{k}\right)=\mathbb{E}X_{1}=\frac{1}{m}\sum_{i=1}^{m}n_{i}$
This because in general $\mathbb{E}\left(cY\right)=c\mathbb{E}Y$
and $\mathbb{E}\left(Y+Z\right)=\mathbb{E}Y+\mathbb{E}Z$.
A: Here's one way to do it:
Recall that
$$
S\equiv\left\{ n_{1},n_{2},\ldots,n_{m}\right\} .
$$
Consider the subsets of $S$ of size $k$ containing $n_{i}$ for
some $i$. There are exactly $\left(\begin{array}{c}
m-1\\
m-k
\end{array}\right)=\left(\begin{array}{c}
m-1\\
m-1-\left(k-1\right)
\end{array}\right)=\left(\begin{array}{c}
m-1\\
k-1
\end{array}\right)$ such subsets.
Now, consider a particular subset of $S$, $A$, with $\left|A\right|=k$.
The mean of this subset is
$$
\frac{1}{k}\sum_{a\in A}a.
$$
Hence the expectation of the procedure you described is
\begin{align*}
E & \equiv\frac{1}{\left(\begin{array}{c}
m\\
k
\end{array}\right)}\sum_{\substack{A\subset S\\
\left|A\right|=k
}
}\frac{1}{k}\sum_{a\in A}a\\
 & =\frac{1}{\left(\begin{array}{c}
m\\
k
\end{array}\right)k}\sum_{\substack{A\subset S\\
\left|A\right|=k
}
}\sum_{a\in A}a.
\end{align*}
We established that each $n_{i}$ appears in one of the subsets of
size $k$ exactly $\left(\begin{array}{c}
m-1\\
k-1
\end{array}\right)$ times. Therefore, the sum above can be simplified to
\begin{align*}
E & =\frac{1}{\left(\begin{array}{c}
m\\
k
\end{array}\right)k}\sum_{a\in S}\left(\begin{array}{c}
m-1\\
k-1
\end{array}\right)a\\
 & =\frac{\left(\begin{array}{c}
m-1\\
k-1
\end{array}\right)}{\left(\begin{array}{c}
m\\
k
\end{array}\right)k}\sum_{a\in S}a
\end{align*}
and you can verify that
$$\frac{\left(\begin{array}{c}
m-1\\
k-1
\end{array}\right)}{\left(\begin{array}{c}
m\\
k
\end{array}\right)k}=\frac{1}{m}.$$
