Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is Po(yPo(x)) a Po(yx) distribution?

I got Po(yPo(x)) from moment generating functions but I'm not sure how or if to simplify from there...


share|cite|improve this question
Do you mean a Poisson distribution with a Poisson distributed parameter? – Raskolnikov Oct 7 '12 at 8:38
Exactly... Poisson with a parameter that is a constant y times a poisson distribution with parameter x. – Dirk Calloway Oct 7 '12 at 8:41
up vote 1 down vote accepted

No. As soon as $x$ and $y$ are both positive, this distribution is not Poisson (whether with parameter $xy$ or with any other one).

The Poisson distribution with parameter $yn$ has weight $\mathrm e^{-yn}(yn)^k/k!$ at $k$, hence the probability measure of interest has weight $p_k=\mathbb E(\mathrm e^{-yN}(yN)^k/k!)$ at $k$, where the distribution of $N$ is Poisson with parameter $x$. That is, $$ p_k=\mathrm e^{-x}\frac{y^k}{k!}\sum_{n\geqslant0}n^k\frac{(x\mathrm e^{-y})^n}{n!}. $$ In particular, $p_0=\mathrm e^{-z}$ with $z=x(1-\mathrm e^{-y})$, and $p_1=yx\mathrm e^{-y}\mathrm e^{-z}$. If the distribution $(p_k)_{k\geqslant0}$ is Poisson, then $p_1=z\mathrm e^{-z}$, that is, $z=yx\mathrm e^{-y}$. This condition is equivalent to $x=0$ or $\mathrm e^y=1+y$, that is, $x=0$ or $y=0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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