Expected value of SRSWOR sample maximum If I draw a sample of size n without replacement from the set {1,2,3,...,N}- what is the expected value of the sample maximum? (n < N).
Possible to get a closed form solution?
 A: Just about all answers will have to be mathematically equivalent.  The point of this one is to develop a solution in the laziest possible way: that is, by pure reasoning unaccompanied by any calculation at all.

There are $\binom{N}{n}$ possible and equally likely samples, since each sample is a subset of $n$ of the $N$ elements.  (For those new to such notation, $\binom{N}{n}$ can be defined as the number of distinct samples of size $n$ without replacement from $N$ things: that is, the number of $n$-subsets of $N$ things.  In this answer you will not need to know any formulas for these quantities.)
A sample with maximum value $k \ge n$ consists of the number $k$ together with a subset of $n-1$ of the remaining $k-1$ elements smaller than $k$.  There are $\binom{k-1}{n-1}$ of these.
To obtain the expectation, by definition we must multiply the probability of each such sample, $1/\binom{N}{n},$ by the value of its maximum, $k,$ and add these up:
$$\mathbb{E}(\text{maximum}) = \frac{1}{\binom{N}{n}}\sum_{k=n}^N k\binom{k-1}{n-1}.$$

So much for the statistics.  The rest is combinatorics.  Our purpose is to obtain a succinct numerical formula for this rather abstract looking sum.
You can evaluate the sum by doing very little calculation indeed.  One way begins by interpreting the term $k\binom{k-1}{n-1}$ as a count: it is the number of ways you can pick one of $k$ things and independently select $n-1$ of the remaining $k-1$ things.  Equivalently, you could have selected all $n$ of those things (in $\binom{k}{n}$ ways) and then chosen one of those $n$ things as the "first" pick.  Since there are $n$ such choices,
$$k\binom{k-1}{n-1} = n\binom{k}{n}.\tag{*}$$
Take the constant factor of $n$ out of the sum:
$$\mathbb{E}(\text{maximum}) = \frac{1}{\binom{N}{n}}\, n\, \sum_{k=n}^N \binom{k}{n}.$$
What could this sum count?  Almost the same argument applies: associated with each $n+1$-subset of $N+1$ things are (1) its maximum, which I will call $k+1$, and (2) an $n$-subset of the remaining $k$ smaller things.  The sum counts these by partitioning the possibilities for such subsets by the values of their maxima.  Consequently, it counts all $n+1$-subsets of $N+1$ things and therefore equals $\binom{N+1}{n+1}$.  Plugging this into the expectation (and not forgetting the constant factor of $n$) gives
$$\mathbb{E}(\text{maximum}) = \frac{1}{\binom{N}{n}} \left(n\binom{N+1}{n+1}\right).\tag{**}$$
The stuff in parentheses looks a lot like the counting result we obtained in $(*)$.  We can make it so by multiplying by $n+1$ and dividing by the same value:
$$n\binom{N+1}{n+1} = \frac{n}{n+1} (n+1) \binom{N+1}{n+1} = \frac{n}{n+1}(N+1)\binom{N}{n}.$$
Plugging this into $(**)$ cancels the $\binom{N}{n}$ in the fraction (which is why we never needed a formula for it), leaving the simple result
$$\mathbb{E}(\text{maximum}) = \frac{n}{n+1}(N+1).$$
A: If you are sampling from the discrete uniform population (without replacement, as you have stipulated), then it is the German Tank Problem. 
Let the sample be $X_1, X_2, \ldots , X_n$ with $Y=\textrm{max} \left( X_1, X_2, \ldots , X_n \right).$ 
The joint mass function is $$f \left( x_1, x_2, \ldots , x_n \right)=\frac{1}{N \left( N-1 \right) \cdots \left( N-n+1 \right) }$$
The cdf of $Y$ is $$P[Y \le y]=P[X_1 \le y, X_2 \le y, \ldots , X_n \le y]=\frac{y \left( y-1 \right) \cdots \left( y-n+1 \right) }{N \left( N - 1 \right) \cdots \left( N-n+1 \right) } =\frac{{y \choose n}}{N \choose n}$$
Then the probability mass function will be $$g(y) = P[Y=y]=P[Y \le y] - P[Y \le y-1] = \frac{{y \choose n}-{y-1 \choose n}}{N \choose n}=\frac{{y-1} \choose {n-1}}{N \choose n} \ , \textrm{where} \  y \ge n $$
The expected value is then $$E[Y]=\sum_{y=n}^{N} y \ g(y)=\sum_{y=n}^N y \frac{{y-1} \choose {n-1}}{N \choose n}=\frac{\sum_{y=n}^N y {{y-1} \choose {n-1}}}{N \choose n}=\frac{n \sum_{y=n}^{N} {y \choose n}}{N \choose n}$$
The Hockey-stick identity is $$\sum_{y=n}^N {y \choose n} = {{N+1} \choose {n+1}}$$
So $$E[Y]= \frac{n {{N+1} \choose {n+1}}}{N \choose n}=\left( \frac{n}{n+1} \right) \left( N+1 \right)$$
This approach is given in Tenenbein (The Racing Car Problem, $\it{The 
\ American \ Statistician}$, February 1971). 
A: As with each and every sampling problem, the answer is "Yes, if you have access to the population and to an algorithm that efficiently enumerates all possible samples". So if you are talking about samples of size 3 out of population of size 7, then yes, you can probably derive that. If you are talking about realistic samples of size 1000 from a country's population, then you can probably forget it.
The central methodological issue is that finite population inference is non-parametric. Inference with respect to the sampling design works for the population $\{1,2,3,4,5,6,7\}$ as well as for the population $\{1,1,1,1,1,1,1,10^6\}$. By contrast, the existing methods of dealing with extreme values rely on assuming a smooth tail of a continuous distribution, for which one of the three functional forms established by Gnedenko would be applicable, and an i.i.d. sample (which isn't quite the same as SRS).
You can approximately fake an i.i.d. argument if you have a good approximation for the tail of your population, and if you have a simple enough design (SRSWR or SRSWOR with a negligible sampling fraction).
