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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 \begin{equation}\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} \end{equation} 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?

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

1 Answer 1

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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.

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