Generating function and binomial coefficients

Question

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?

Answer 1

We use the coefficient of operator $[s^k]$ to denote the coefficient of $s^k$ in a series. This way we can write for instance \begin{align*} [s^k](1+s)^n=\binom{n}{k}\tag{1} \end{align*}

We obtain \begin{align*} \color{blue}{[s^m]}&\color{blue}{e^{-\lambda} \sum_{i=0}^{\infty} \frac{1}{i !}\left(\lambda(1-\rho) s\right)^i\left(1-\rho s\right)^{-i}}\\ &=e^{-\lambda} \sum_{i=0}^{m} \frac{1}{i !}\left(\lambda(1-\rho) \right)^i[s^{m-i}]\left(1-\rho s\right)^{-i}\tag{2}\\ &=e^{-\lambda} \sum_{i=0}^{m} \frac{1}{i !}\left(\lambda(1-\rho) \right)^i[s^{m-i}]\sum_{k\geq 0}\binom{-i}{k} (-\rho s)^k\tag{3}\\ &=e^{-\lambda} \sum_{i=0}^{m} \frac{1}{i !}\left(\lambda(1-\rho) \right)^i[s^{m-i}]\sum_{k\geq 0}\binom{i+k-1}{k} (\rho s)^k\tag{4}\\ &\,\,\color{blue}{=e^{-\lambda} \sum_{i=0}^{m} \frac{1}{i !}\left(\lambda(1-\rho) \right)^i\binom{m-1}{i-1} \rho^{m-i}}\tag{5}\\ \end{align*} and the claim follows.

Comment:

• In (2) we use the linearity of the coefficient of operator and apply the rule $[s^{p-q}]A(s)=[s^p]s^qA(s)$. We also set the upper limit of the outer sum to $m$ since other indices do not contribute to $[s^m]$.

• In (3) we make a binomial series expansion.

• In (4) we apply the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

• in (5) we select the coefficient of $s^{m-i}$ according to (1) and use $\binom{p}{q}=\binom{p}{p-q}$.