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I have a problem with following exercise (it comes from Geoffrey G. Grimmett, David R. Stirzaker, Probability and Random Processes, Oxford University Press 2001, page 161, ex. 3a):

Let $X$ have the Poisson distribution with parameter $Y$ where $Y$ has the Poisson distribution with parameter $\mu$. Show that $$G_{X+Y}(s) = e^{\mu(s e^{s-1} - 1)}$$


So $G_Y(s) = e^{\mu(s-1)}$ Now I want to compute $G_X$ (is this approach correct?) $$G_X(s) = \sum_{x=0}^{\infty} s^x P(X=x) = \sum_{x=0}\sum_{y=0} s^x \frac{e^{-y} y^x}{x!} \frac{e^{-\mu} \mu^y}{y!}$$ $$G_X(s) = e^{-\mu} \sum_{x=0} \frac{s^x}{x!} \sum_{y=0} \frac{y^x e^{-y}\mu^y}{y!}$$

And I don't know how to cope with this summation.

Secondly, are these variables independent? I.e. can I use then following formula? $$G_{X+Y}(s) = G_X(s) G_Y(s)$$

Thanks for your help.

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1 Answer

  1. $X$ and $Y$ are not independent, since $X$ has been defined using $Y$.

  2. The easiest way to compute $G_{X+Y}(s)={\Bbb E}s^{X+Y}$ is to first condition on $Y$. So, first, $$ {\Bbb E}[s^{X+Y}\mid Y]=s^Y{\Bbb E}[s^X\mid Y]. \qquad (*) $$ Since $X$ is Poisson with parameter $Y$, conditioned on $Y$, the pgf of $X$ is $e^{Y(s-1)}$. Plugging this into (*) gives $$ {\Bbb E}[s^{X+Y}\mid Y]=s^Y e^{Y(s-1)}. \qquad (**) $$ The only remaining step is to take the expectation of (**). Since the right-hand side of (**) is of the form $\alpha^Y$, this can be done by plugging $\alpha$ into the pgf for $Y$. This gives the desired result.

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