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.

  • $\begingroup$ Hint: Differentiate $$\sum^\infty_{k=0}ar^k = \frac{a}{1-r}$$ on both sides with respect to $r$, $n-1$ times and see what you get. Notice the lower limit on that sum is $k=0$ btw. $\endgroup$ – N. Shales Aug 25 '17 at 22:59
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    $\begingroup$ Thanks @N.Shales I will give that a shot. I appreciate it. $\endgroup$ – krishnab Aug 25 '17 at 23:03
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    $\begingroup$ If you manage to solve your problem then I suggest you post it as an answer here as I believe that is looked upon favorably by the MSE community. $\endgroup$ – N. Shales Aug 25 '17 at 23:08
  • $\begingroup$ The final formula in this table may help: en.wikipedia.org/wiki/… $\endgroup$ – user940 Aug 25 '17 at 23:23
  • $\begingroup$ Have you learn about the negative binomial distribution and probability generating function? en.wikipedia.org/wiki/Negative_binomial_distribution Essentially for a given pmf, it can be generated by the corresponding pgf. The pgf is given by $\displaystyle G(s) = E[s^X] = \sum s^x f_X(x)$ $\endgroup$ – BGM Aug 26 '17 at 7:19

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.


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