I am working through Stirzaker and Grimmett and found a problem and its solution that I was having difficulty understanding. It has been a while since I really played around with power series, so I might be missing some trick here.
The question is:
Find the generating function of the following mass function
$$ f(m) = \binom{n+m-1}{m}p^n(1-p)^m, \ \ \text{for} \ m \geq 0 $$
The solution, which is provided, is that the generating function $G(s)$ is:
$$ \Big\{\frac{p}{1 - s(1-p)}\Big\}^n $$
I was not sure how the authors arrived at this solution. It looks like they use the normal trick for a geometric series where:
$$ \sum^\infty_{k=1}ar^k = \frac{a}{1-r} $$
but I was not sure how they got the generating function above to fit into this form. Any help is appreciated.
For $n\geq 1$, note that $$ \frac{1}{(1-x)^n}=(1-x)^{-n} =\sum_{m=0}^{\infty}\binom{-n}{m}(-1)^{m}x^{m} =\sum_{m=0}^{\infty}\binom{n+m-1}{m}x^{m}\quad (|x|<1)\tag{1} $$ by the extended binomial theorem. In particular, it follows that $$ \frac{p^n}{[1-s(1-p)]^n}==\sum_{m=0}^{\infty}\binom{n+m-1}{m}p^n(1-p)^{m}s^{m} $$ by (1) which is the pgf of the negative binomial distribution as follows.