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.

  • $\begingroup$ Why use generating functions? One can compute directly $\mathbb P(X=n)$ for each $n$. $\endgroup$
    – Did
    Jan 17, 2013 at 21:56
  • $\begingroup$ And how to do this? $\endgroup$ Jan 18, 2013 at 1:05
  • $\begingroup$ See my answer... $\endgroup$
    – Did
    Jan 18, 2013 at 7:13

2 Answers 2


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 $p$.

  • $\begingroup$ After changing of variable I get different result, please correct me: $$x = \mu \lambda \\ dx = \mu d\lambda \\ \int_0^{\infty} e^{-\frac{x}{\mu}} \frac{x^n}{\mu^n n!} \mu e^{-x} \frac{1}{\mu} dx = \frac{1}{\mu^n} \int_0^{\infty} e^{-x} \frac{x^n}{n!} e^{-\frac{x}{\mu}} dx$$ $\endgroup$ Jan 18, 2013 at 17:26
  • $\begingroup$ And why does $\int_0^{\infty} e^{-x} \frac{x^n}{n!} dx$ equal $1$ ? $\endgroup$ Jan 18, 2013 at 17:45
  • $\begingroup$ Oh, I see - this is the Gamma Function, so I guess there is no simple explanation? :) Could you also explain me the first two equality: $$P(X=n|\Lambda) = e^{-\Lambda} \Lambda^n / n! \\ P(X=n) = \int_0^{\infty} e^{-\lambda} \frac{\lambda^n}{n!} f_{\Lambda} (\lambda) d\lambda$$ What formula or theorem it is? $\endgroup$ Jan 18, 2013 at 18:08
  • 1
    $\begingroup$ @LittleRookie General fact about conditional expectations/probabilities: $$E(P(A\mid Z))=P(A)$$ just like $$E(E(Y\mid Z))=E(Y)$$ $\endgroup$
    – Did
    Aug 29, 2017 at 23:19
  • 1
    $\begingroup$ @LittleRookie en.wikipedia.org/wiki/Law_of_total_expectation $\endgroup$
    – Did
    Aug 29, 2017 at 23:35

Be careful, $\Lambda$ is a random variable! So your computation only shows that $$ E[s^X \mid \Lambda] = \sum_{n=0}^\infty \frac{(s\Lambda)^n}{n!}e^{-\Lambda} = e^{\Lambda (s-1)}. $$

Now you should be able to compute $$G_X(s) = E[s^X] = E\left[E[s^X\mid \Lambda]\right].$$

  • $\begingroup$ I don't quite understand second equality, could you explain it? Why $s^X = E(s^X | \Lambda)$ ? Provided that, $G_X(s) = E(e^{\Lambda (s-1)}) = \int e^{x(s-1)} \mu e^{-\mu x} dx = \frac {\mu e^{x(s-1-\mu)}}{s-1-\mu}$. Is it okay? If yes, how to deduce that $X$ has a geometric distribution? $\endgroup$ Jan 17, 2013 at 22:26
  • 2
    $\begingroup$ I never wote such a thing as $s^X = E(s^X\mid\Lambda)$. I am only using the very basic property of conditional expectation $E[Z]=E\left[E[Z\mid \mathcal{F}]\right]$. Besides, the value of the integral cannot depend on the variable of integration $x$... $\endgroup$
    – Siméon
    Jan 18, 2013 at 11:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.