expected value for max of a set of random variables we randomly pick $M$ elements from a set of $N$ real numbers, $A=\{ a_1,a_2,\ldots,a_N \}$. Then we sort these $M$ numbers, what is the expected value for the largest element?  (let say $N=1000000$, $M=1000$), I am interested in the solution for general $A$, and/or the case that elements of $A$ are taken from a normal distribution.
 A: Note:  Since you're concerned with subsets, I'm going to assume that you're choosing your elements without repetition.  As per your question, I'll also assume that you're picking your elements with uniform probability (ie completely at random).
Adding a bit of notation, suppose that our chosen subset is $B=\{b_1,b_2,\cdots,b_M\}$ with $b_1<b_2<\cdots <b_M$.  If we assume also that $a_1<a_2<\cdots<a_N$, then clearly $P(b_M<a_M)=0$, since the smallest $M$ elements that we can choose from $A$ are $\{a_1,a_2,\cdots,a_M\}$.
Now, $P(b_M=a_M)=\frac{1}{{N \choose M}}$, since the only possible such subset is $\{a_1,a_2,\cdots, a_M\}$.
To compute $P(b_M=a_{M+1})$, the only way this can happen is if $\{b_1,\cdots, b_{M-1}\}\subset \{a_1,a_2,\cdots, a_M\}$, hence $P(b_M=a_{M+1})=\frac{{M \choose M-1}}{{N \choose M}}$.
And by a similar argument, for $1\leq i \leq N-M$, $P(b_M=a_{M+i})=\frac{{M +(i-1) \choose {M-1}}}{{N \choose M}}$.
So, from there, the expected value is $$\frac{1}{{N \choose M}}\sum\limits_{i=0}^{N-M}a_i {{M+(i-1)} \choose {M-1}}$$
A: If you want to pick $M$ random numbers, you can pick the highest one $m$ in $N$ ways and then the rest in $\binom{m-1}{M-1}$:
$$\binom{N}{M} = \sum_m\binom{m-1}{M-1}\,,$$ 
so the probability that $m$ is the maximum is $$\binom{m-1}{M-1}\Big/\binom{N}{M}\,.$$
A: Hint 1: For every nonnegative random variable $X$, $\mathrm E(X)=\int\limits_0^{+\infty}\mathrm P(X\geqslant x)\mathrm dx$.
Hint 2: If $M=\max\{\xi_1,\xi_2,\ldots,\xi_n\}$, then, for every $x$, $[M\leqslant x]=\bigcap\limits_{k=1}^n[\xi_k\leqslant x]$.
