TITLE: Why does this expected value simplify as shown?
QUESTION [9 upvotes]: I was reading about the german tank problem and they say that in a sample of size $k$, from a population of integers from $1,\ldots,N$ the probability that the sample maximum equals $m$ is:
$$\frac{\binom{m-1}{k-1}}{\binom{N}{k}}$$
This make sense. But then they take expected value of the sample maximum and claim:
$$\mu = \sum_{m=k}^N m \frac{\binom{m-1}{k-1}}{\binom{N}{k}} = \frac{k(N+1)}{k+1}$$
And I don't quite see how to simplify that summation. I can pull out the denominator and a $(k-1)!$ term out and get:
$$\mu = \frac{(k-1)!}{\binom{N}{k}} \sum_{m=k}^N m(m-1) \ldots (m-k+1)$$
But I get stuck there...

REPLY [9 votes]: Here is another way of looking at the problem that uses the natural symmetry. 
The samples of size $k$ are in one-to-one correspondence
with the sequence of "gaps" between the observations 
(via the sequence of order statistics):
$$\{x_1,x_2,\dots, x_k\}\leftrightarrow x_{(1)}<x_{(2)}<\cdots <x_{(k)}
\leftrightarrow (G_1,G_2,\dots, G_{k+1})$$
Here we define $G_1=x_{(1)}$, $G_j=x_{(j)}-x_{(j-1)}$ for $2\leq j\leq k$,
  and $G_{k+1}=(N+1)-x_{(k)}$. This 
correspondence shows that $G_1,G_2,\dots, G_{k+1}$ are 
exchangeable random variables, in particular, they are identically distributed.
Since 
$\sum_{j=1}^{k+1} G_j=N+1$ 
identically,  the  expectations of these $k+1$ identically distributed
random variables must be $\mathbb{E}(G_j)={N+1\over k+1}$.
In particular, $\mathbb{E}(G_{k+1})=\mathbb{E}((N+1)-x_{(k)})={N+1\over k+1}$
which implies 
$$\mathbb{E}(\mbox{Maximum})=\mathbb{E}(x_{(k)})=(N+1)- {N+1\over k+1}={k\over k+1} (N+1).$$
Similarly, we see that the expected values of the order statistics $\mathbb{E}(x_{(1)}),\mathbb{E}(x_{(2)}),\dots, 
\mathbb{E}(x_{(k)})$ are evenly spread out from ${1\over k+1} (N+1)$
 to ${k\over k+1} (N+1).$ 
See this post for more questions about these gaps.