Find probability generating function We have $Y \sim \mathrm{Poisson}(X)$ and $X \sim \mathrm{Exp}(a-1)$, with $a >0$. Now I want to show that the probability generating function of $Y$ is equal to $G_Y (z) = \frac{a-1}{a-z}$.
I guess I have to do something with conditional expectations/probabilities, but I am struggling. Any hints?
 A: $\newcommand{\E}{\operatorname{E}}$You must have meant $a-1>0$ since otherwise the exponential distribution makes no sense.
\begin{align}
\Pr(Y=y) & = \E(\Pr(Y=y\mid X)) = \E\left( \frac{X^y e^{-X}}{y!} \right) \\[10pt]
& = \frac 1 {y!} \int_0^\infty x^y e^{-x}\Big( e^{-(a-1)x} ((a-1)\  dx) \Big) = \frac {a-1} {y!} \int_0^\infty x^y e^{-ax} \,dx \\[10pt]
& = \frac {a-1} {y!} \cdot \frac 1 {a^{y+1}} \int_0^\infty (ax)^y e^{-ax} (a\,dx) \\[10pt]
& = \frac{a-1}{y!}\cdot \frac 1 {a^{y+1}} \int_0^\infty u^y e^{-u}\,du = \frac{a-1}{y!} \cdot \frac 1 {a^{y+1}} \cdot y! = \underbrace{ \frac{a-1}{a^{y+1}}}_{ \begin{smallmatrix} \text{geometric} \\ \text{distribution} \\  \text{on }\{0,1,2,\,\ldots\,\} \end{smallmatrix}}
\end{align}
$$
G_Y(z) = \sum_{y=0}^\infty \Pr(Y=y) z^y = \underbrace{\sum_{y=0}^\infty \frac{a-1}{a^{y+1}}\cdot z^y = \frac{a-1}{a(1-\frac z a)}}_\text{sum of a geometric series} = \frac{a-1}{a-z}.
$$
By the way, it is hazardous to write $X\sim\mathrm{Exp}(b)$ since sometimes that means $X$ has the exponential distribution
$$
e^{-bx} (b\ dx) \text{ for } x>0,
$$
with expected value $1/b$, and sometimes it means $X$ has the exponential distribution
$$
e^{-x/b}\, \frac{dx} b \text{ for }x>0,
$$
with expected value $b$.
A: Another related (but perhaps easier to remember) route to the solution: We can treat $G_Y(z)$ as the probability generating function for the number of Poisson arrivals at rate $\lambda = 1$ into an interval of time that is exponentially distributed with mean $\frac{1}{a-1}$ (with $a > 1$).
The general form of this expression when the interval has a PDF with Laplace transform $A^*(s)$ and the Poisson arrivals are at rate $\lambda$ is $A^*(\lambda-\lambda z)$.  In this case, we have $A^*(s) = \frac{a-1}{s+a-1}$, and $\lambda = 1$, so
$$
G_Y(z) = A^*(1-z) = \frac{a-1}{1-z+a-1} = \frac{a-1}{a-z}
$$
The general rule can be obtained as
\begin{align}
P(z) & = \sum_{k=0}^\infty \int_{t=0}^\infty f(t) p_k z^k \, dt \\
     & = \sum_{k=0}^\infty \int_{t=0}^\infty f(t) \frac{(\lambda t)^k}{k!} e^{-\lambda t} z^k \, dt \\
     & = \int_{t=0}^\infty f(t) e^{-\lambda t} \sum_{k=0}^\infty \frac{(\lambda tz)^k}{k!} \, dt \\
     & = \int_{t=0}^\infty f(t) e^{-\lambda t} e^{\lambda tz} \, dt \\
     & = \int_{t=0}^\infty f(t) e^{-(\lambda-\lambda z)t} \, dt \\
     & = A^*(\lambda-\lambda z)
\end{align}
where the last line follows from the definition of the (one-sided) Laplace transform, and $f(t) \leftrightarrow A^*(s)$.
