I'm lost and frustrated. I don't know how the author (Karl Sigmund; The Calculus of Selfishness) transforms 6.37 in the book pages imaged below:
$$ P_y = \sigma w^{N-1} + rc\frac{x}{1-w}\sum_{h=0}^{N-1} \binom{N-1}{h} (1-w)^h w^{N-h-1} \left(\frac{h}{h+1}\right) \;\; (6.37) $$
to 6.39 in the picture:
$$ P_y = \sigma w^{N-1} + rc\frac{x}{1-w}\left( 1 - \frac{1-w^N}{N(1-w)}\right) \;\; (6.39) $$
Can anybody help me? I am a lawyer by training, so I am learning the mathematics required for evolutionary game theory bit by bit. So far I have always managed to put things together eventually; this however seems to elude me. My own thoughts so far are probably not helpful, but I'll try:
I am assuming I am supposed to use the expected value of binomial distributions. However, if I use 6.38:
$$ \binom{N-1}{h} = \binom{N}{h+1} \frac{h+1}{N} \;\; (6.38) $$
my binomial coefficient would not fit the sum. My next guess would be that I should try to fit the sum to the binomial coefficient: As h=0 can be ignored, maybe one could add to the sum so it goes from $h=1$ to $N$, and then subtract it in another term), but puzzling around with it has not given me any answer that looks like 6.39. My guess is that somehow the fact that the sum from $i$ to $(N-1)$ of $w^i$ is $(1-w^N)/(1-w)$ has a role to play, but so far I am just stuck.
In going from 6.37 to 6.39 we need to prove that:
$$\sum_{h=0}^{N-1} \binom{N-1}{h} (1-w)^h w^{N-h-1} \left(\frac{h}{h+1}\right) = 1 - \frac{1-w^N}{N(1-w)} \qquad\qquad\text{(*)}$$
\begin{eqnarray*} && \\ \text{LHS} &=& \dfrac{1}{N} \sum_{h=0}^{N-1} h\binom{N}{h+1} (1-w)^h w^{N-h-1} \qquad\qquad\qquad\qquad\qquad\qquad\text{(using 6.38)} \\ &=& \dfrac{1}{N(1-w)} \sum_{h=0}^{N-1} h\binom{N}{h+1} (1-w)^{h+1} w^{N-h-1} \\ &=& \dfrac{1}{N(1-w)} \left[ \sum_{h=0}^{N-1} (h+1)\binom{N}{h+1} (1-w)^{h+1} w^{N-h-1} - \sum_{h=0}^{N-1} \binom{N}{h+1} (1-w)^{h+1} w^{N-h-1} \right] \\ &=& \dfrac{1}{N(1-w)} \left[ N(1-w) - \left[ ((1-w)+w)^N-w^N \right] \right] \qquad\qquad\qquad\text{(see (a), (b) below)} \\ &=& 1- \dfrac{1-w^N}{N(1-w)} \\ &=& RHS. \end{eqnarray*}
Notes:
(a) $\sum\limits_{h=0}^{N-1} (h+1)\binom{N}{h+1} (1-w)^{h+1} w^{N-h-1}$ is the expansion of the expectation of a binomial random variable with parameters $N$ and $1-w$. Hence it has value $N(1-w)$. That is, if $X\sim Bin(n,p)$ then $np=E(X)=\sum\limits_{x=1}^{n}x\binom{n}{x}p^x(1-p)^{n-x}=\sum\limits_{x=0}^{n-1}(x+1)\binom{n}{x+1}p^{x+1}(1-p)^{n-x-1}$.
(b) $\sum\limits_{h=0}^{N-1} \binom{N}{h+1} (1-w)^{h+1} w^{N-h-1}$ is the binomial expansion of $((1-w)+w)^N$ but the sum is missing the term for $h=-1$ which has value $w^N$. So $\sum\limits_{h=0}^{N-1} \binom{N}{h+1} (1-w)^{h+1} w^{N-h-1} = ((1-w)+w)^N-w^N$.