# Probability generating function and binomial coefficients

I'm reading an article where the authors derive the mass function of a compound distribution by considering the generating function. The generating function of interest for a random variable $$N$$ is a composition of two generating functions $$G_Y(s)$$ and $$G_X(s)$$, where $$$$\tag{1} G_N(s)=G_Y(G_X(s))=e^{\lambda\left(\frac{(1-\rho)s}{1-\rho s}-1\right)}=e^{-\lambda} \sum_{m=0}^{\infty} \frac{1}{m !}\left(\lambda(1-\rho) s\right)^m\left(1-\rho s\right)^{-m}$$$$ They then state,

Now, expand the probability generating function of $$N$$ in (1) and then, collecting the coefficient of $$s^n$$, we find an explicit expression for the probability mass function of N as $$P(N=m)=e^{-\lambda} \sum_{i=0}^m \frac{1}{i !} {m-1\choose i-1}[\lambda(1-\rho)]^i \rho^{m-i}.$$

I don't see how this expression comes about from (1). I've tried using the negative binomial expansion identity of $$(1-\rho)^{-r}=\sum_{k=0}^{\infty} {{k+r-1} \choose k}\rho^k$$ to substitute into (1), but I'm not getting anywhere. Any thoughts on what "expand the probability generating function" means or how to recover this PMF?

• You are on the right track. The key is that the "$i$" in the result corresponds to "$m-k$" in the expansion you give.
– whuber
Aug 30, 2023 at 22:02
• The question has been posted on math.se before
– Ute
Aug 31, 2023 at 7:01
• I’m voting to close this question because this has already been asked and answered in Maths. Aug 31, 2023 at 14:14

Minor note: You have used the variable $$m$$ both as the index for your initial summation and also as the outcome for your random variable. This has the potential to cause confusion so I will use $$n$$ as the outcome of the random variable instead, and I will keep $$m$$ as the summation index. The result I get is what you want, but my $$n$$ is your $$m$$ and my $$m$$ is your $$i$$.

The probability generating function has the property that:

$$G_N(s) = \sum_n s^n \times \mathbb{P}(N=n).$$

Consequently, if you can rearrange the expression for $$G_N(s)$$ to state it in expanded polynomial form in $$s$$ then you get the probability values as the resulting coefficients of the polynomial. The negative binomial expansion here can be written as:

$$(1-\rho s)^{-m} = \sum_{k=0}^\infty {k+m-1 \choose k} \rho^k s^k,$$

and setting $$k=n-m$$ then gives the equivalent form:

$$(1-\rho s)^{-m} = \sum_{n=m}^\infty {n-1 \choose n-m} \rho^{n-m} s^{n-m} = \sum_{n=m}^\infty {n-1 \choose m-1} \rho^{n-m} s^{n-m}.$$

Using this identity you get:

\begin{align} G_N(s) &= e^{-\lambda} \sum_{m=0}^\infty \frac{1}{m!} (\lambda (1-\rho) s)^m (1-\rho s)^{-m} \\[6pt] &= e^{-\lambda} \sum_{m=0}^\infty \frac{1}{m!} (\lambda (1-\rho) s)^m \sum_{n=m}^\infty {m-1 \choose m-1} \rho^{n-m} s^{n-m} \\[6pt] &= e^{-\lambda} \sum_{m=0}^\infty \frac{1}{m!} (\lambda (1-\rho))^m \sum_{n=m}^\infty {m-1 \choose m-1} \rho^{n-m} s^n \\[6pt] &= \sum_{m=0}^\infty \sum_{n=m}^\infty e^{-\lambda} \frac{1}{m!} (\lambda (1-\rho))^m {n-1 \choose m-1} \rho^{n-m} s^{n} \\[6pt] &= \sum_{n=0}^\infty \sum_{m=0}^n e^{-\lambda} \frac{1}{m!} (\lambda (1-\rho))^m {n-1 \choose m-1} \rho^{n-m} s^{n} \\[6pt] &= \sum_{n=0}^\infty s^n \times \underbrace{e^{-\lambda} \sum_{m=0}^n \frac{1}{m!} {n-1 \choose m-1} (\lambda(1-\rho))^m \rho^{n-m}}_{\mathbb{P}(N=n)}. \\[6pt] \end{align}

Extracting the coefficients of this polynomial and setting these as the mass values gives the desired result.