Expected value of position in subset. I have set of $n$ elements $[0, ... n-1]$.
I randomly pick a subset $S$ of $k$ elements (also ordered).
Assume I have $t \in \{1, .. k\}$.
What is expected value of $t$-th position in ordered subset?
I try to solve it like this:
$ E\big[S_{t}\big] = \sum_{i=1}^{n} \big((i - 1) * P(S_{t}=(i-1))\big) $
$P(S_{t} = i) = \cfrac{\binom{i - 1}{t-1} * \binom{n-i}{k-t}}{\binom{n}{k}} . $
Here:
$\binom{i - 1}{t-1}$ is for selecting prefix;
$\binom{n-i}{k-t}$ is for postfix;
But I can't simplify this equality much further (I guess that sum of product of combinations could be simplified).
 A: As in my comment above, I will consider the set to be $\{1,\ldots, n\}$, and let $t \in \{1,\ldots, k\}$ denote the position of interest, i.e. $S_t$ is the value of the $t$-th element of the random subset $S$, when in increasing order.
In this case, we can show that
$$P(S_t = i) = \frac{\binom{i-1}{t-1} \binom{n-i}{k-t} }{\binom{n}{k} }$$
Now substituting this into the formula for expectation:
\begin{align*} E[S_t] & = \sum_{i=1}^n i \frac{\binom{i-1}{t-1} \binom{n-i}{k-t} }{\binom{n}{k} } \\
& = \frac{1}{\binom{n}{k} } \sum_{i = 1}^n i \binom{i-1}{t-1} \binom{n-i}{k-t}
\end{align*}
Using the formula:
$$ i \binom{i-1}{t-1} = t \binom{i}{t}$$
then the summation can be written as:
\begin{align*}
\sum_{i=1}^n i \binom{i-1}{t-1}\binom{n-i}{k-t} &=
t \sum_{i=1}^n \binom{i}{t}\binom{n-i}{k-t} \\
& = t \left( \sum_{i=0}^n \binom{i}{t}\binom{n-i}{k-t} - \binom{0}{t}\binom{n}{k-t} \right) 
\end{align*}
where we have extended the sum over the range $i = 0,\ldots, n$ (instead of $i = 1, \ldots, n$). The first of the two terms is identified as the Vandermonde identity, whilst the second simplifies to $0$ so that
\begin{align*}
\sum_{i=1}^n i \binom{i-1}{t-1}\binom{n-i}{k-t} &=
t \binom{n+1}{k+1} \end{align*}
So over all, we have:
\begin{align*} E[S_t] &= t \frac{ \binom{n+1}{k+1} }{\binom{n}{k}} \\
& = t \frac{n+1}{k+1}
\end{align*}
