What is the expected value of the mean of a random subset? Lets say you have a set $A$ made up of $n$ integers. We then randomly choose $m$ distinct elements from $A$ and put them in a set $B$.
How would you prove that the expected value of the mean of $B$ is equal to the mean of $A$?
Namely $E\left[\dfrac{1}{m}\sum_ja_j\right] = \dfrac{1}{n}\sum_ia_i$ where $j$ represents each of the random integers in $B$ and $i$ represents the items in $A$.
 A: Lets write $b_i$ for the $i$th selected  element (also assume the subset is chosen uniformly at random); each of the elements $a_1,...,a_n$ are equally likely to be selected to be $b_i$ and therefore, $\mathbb Eb_i = \frac{1}{n}∑_{i=1}^n a_i$. By linearity of expectation,
$$ \mathbb E \frac{1}{m} \sum_{i=1}^m b_i = \frac{1}{m} \sum_{i=1}^m \mathbb E b_i = \frac{1}{m} \sum_{i=1}^m \left( \frac{1}{n}∑_{j=1}^n a_j\right) = \frac{1}{n}∑_{i=1}^n a_i $$
A: *

*Let $A = \{ a_1, a_2, \ldots, a_n \}$ and define
$$F(x_1, x_2, \ldots, x_n) = \frac{1}{\binom{n}{m}} \sum \left \{ \mathbb{M}( X_0 ) : X_0 \subseteq \{ x_1, \ldots, x_n \}, \ |X_0| = m \right\}$$
where $\displaystyle \mathbb{M}(X_0) = \frac{1}{m} \sum_{x \in X_0} x$ is the mean of $X_0$. Then $F(a_1, a_2, \ldots, a_n)$ is the expectation of the mean of a random subset $A_0 \subseteq A$ consisting of $m$ elements. 

*Expanding the above definition may be laborious, but it's evident that there exist $\alpha_1, \alpha_2, \ldots, \alpha_n \in \mathbb{R}$ such that 
$$(\forall x_1, \ldots, x_n \in \mathbb{Z}) \, F(x_1, x_2, \ldots, x_n) = \sum_{k=1}^n \alpha_k \cdot x_k,$$
because each $\mathbb{M}(X_0)$ is some linear combination of $x_i$'s within $X_0$, these are added up and divided by a constant $\binom{n}{m}$.
It's sufficient to show that $\alpha_i = \alpha_j$ for all $i, j \in \{ 1, \ldots, n \}$, for that leads to $\alpha_i = \frac{1}{n}$ and it follows that $F(a_1, \ldots, a_n) = \mathbb{M}(A)$.

*Let $\sigma : \{ 1, \ldots, n \} \to \{ 1, \ldots, n \}$ be a permutation. Then $\{ x_{\sigma(1)}, \ldots, x_{\sigma(n)} \} = \{ x_1, \ldots, x_n \}$, hence
$$\begin{align*}
F(x_{\sigma(1)}, \ldots, x_{\sigma(n)}) & = \frac{1}{\binom{n}{m}} \sum \left \{ \mathbb{M}( X_0 ) : X_0 \subseteq \{ x_{\sigma(1)}, \ldots, x_{\sigma(n)} \}, \ |X_0| = m \right\} \\[1ex] & = \frac{1}{\binom{n}{m}} \sum \left \{ \mathbb{M}( X_0 ) : X_0 \subseteq \{ x_1, \ldots, x_n \}, \ |X_0| = m \right\} = F(x_1, \ldots, x_n).
\end{align*}$$

*Let $i, j \in \{ 1, \ldots, n \}$ and let $\sigma$ be the transposition $(i \ j)$. Also let
$$z_k = \begin{cases} 1 & \text{for } k = j \\ 0 & \text{for } k \neq j \end{cases}$$
so 
$$z_{\sigma(k)} = \begin{cases} 1 & \text{if } k = i \\ 0 & \text{if } k \neq i \end{cases}$$
By 2. and 3. 
$$\alpha_j = F(z_1, \ldots, z_n) = F(z_{\sigma(1), \ldots, \sigma(n)}) = \alpha_i$$
which concludes the proof.
