Geometric sum of geometric random variables I am looking to find the probability mass function of $Y=\sum_{i=1}^NX_i$ where $X_i\sim\textrm{Geometric}(a)$ and $N\sim\textrm{Geometric}(b)$. I attempted to do this by finding the probability generating function of $Y$ and comparing it to known probability generating functions to take advantage of the uniqueness property. (In my searches online, it sounds like I should find that $Y\sim\textrm{Geometric}(ab)$.)
(Note that I am using the geometric distribution with support $0,1,\dots$)
I calculated the probability generating function of $X$, $g_X(s)=\frac{a}{1-(1-a)s}$ and correspondingly, $g_N(s)=\frac{b}{1-(1-b)s}$, so the law of total expectation implies that $g_Y(s)=g_N(g_X(s))=\frac{b}{1-(1-b)\frac{a}{1-(1-a)s}}$, however I cannot manipulate this to look like any probability generating function that I recognize, and certainly not $g_Z(s)=\frac{ab}{1-(1-ab)s}$ where $Z\sim\textrm{Geometric}(ab)$. Is there something wrong with my thinking here?
Thanks for the help.
 A: I'd go with a more low-tech approach. 
Assume that
$$ \mathbb{P}[X_i=k]=p(1-p)^{k-1}, \qquad k=1,2,\ldots \tag{1}$$
and $\mathbb{P}[N=n]=r(1-r)^{n-1}$ for $r=1,2,\ldots$. Then:
$$\begin{eqnarray*} \mathbb{P}[Y=m] &=& \sum_{n\geq 1}\mathbb{P}[N=n]\cdot\mathbb{P}[X_1+\ldots+X_n=m]\\&=&\frac{r}{1-r}\sum_{n\geq 1}(1-r)^n\left(\frac{p}{1-p}\right)^n(1-p)^m\cdot r(m,n)\tag{2}\end{eqnarray*} $$
where:
$$\begin{eqnarray*} r(m,n) &=& \#\{(a_1,\ldots,a_n): a_i\in\mathbb{N}_{>0},a_1+\ldots+a_n=m\}\\&=&[x^m]\left(x+x^2+x^3+\ldots\right)^n\\&=&[x^m] \frac{x^n}{(1-x)^n}=\binom{m-1}{n-1}\tag{3}\end{eqnarray*} $$
hence:
$$\begin{eqnarray*} \mathbb{P}[Y=m]&=&rp(1-p)^{m-1}\sum_{n\geq 1}\binom{m-1}{n-1}\left(\frac{p-pr}{1-p}\right)^{n-1}\\&=&rp(1-p)^{m-1}\left(1+\frac{p-pr}{1-p}\right)^{m-1}\\&=&pr(1-pr)^{m-1}\tag{4}\end{eqnarray*}$$
proving your claim for geometric distributions supported on $1,2,\ldots$. There is little to change in order to deal with geometric distributions supported on $0,1,\ldots$, too.
