I don't know how to solve Exercise 8, Section 5.2 from Geoffrey G. Grimmett, David R. Stirzaker, Probability and Random Processes, Oxford University Press 2001. For those who don't have this book:
Let $X$ have a Poisson distribution with parameter $\Lambda$, where $\Lambda$ is exponential with parameter $\mu$. Show that $X$ has a geometric distribution.
$X \sim Poiss(\Lambda),\ \ \Lambda \sim Exp(\mu)$.
So we know that generating function of $X$ is $G_x(s) = \sum_{i=0} s^i \frac{\Lambda^i}{i!} e^{-\Lambda}= e^{\Lambda(s-1)}$.
Probability density function of $\Lambda$ is $f_{\Lambda} = \mu e^{-\mu x}$.
And I don't know what I should do next. How to decompose $\Lambda$ in $G_x$ (or maybe this is not a good idea?).
Thanks in advance for your help.
Best Answer
For every nonnegative integer $n$, $$\mathbb P(X=n\mid\Lambda)=\mathrm e^{-\Lambda}\frac{\Lambda^n}{n!}$$ hence $$ \mathbb P(X=n)=\mathbb E(\mathbb P(X=n\mid\Lambda))=\int_0^{+\infty}\left(\mathrm e^{-\lambda}\frac{\lambda^n}{n!}\right)\,f_\Lambda(\lambda)\,\mathrm d\lambda=\int_0^{+\infty}\left(\mathrm e^{-\lambda}\frac{\lambda^n}{n!}\right)\,\mu\mathrm e^{-\mu\lambda}\,\mathrm d\lambda $$ where the first equality comes from the Law of Total Expectation and Can we prove the law of total probability for continuous distributions?. The change of variable $x=(1+\mu)\lambda$ in the rightmost integral yields $$ \mathbb P(X=n)=\frac{\mu}{(1+\mu)^{n+1}}\int_0^{+\infty}\mathrm e^{-x}\frac{x^n}{n!}\mathrm dx=\frac{\mu}{(1+\mu)^{n+1}} $$ To sum up, $$ \mathbb P(X=n)=(1-p)p^n\qquad p=\frac1{1+\mu} $$ That is, the distribution of $X$ is geometric with parameter $p$.