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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...

Thanks.

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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

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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 $y=0$.

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