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I have two independent variables $X\sim \mathcal B(n,p)$, Binomial and $Y\sim \mathcal P(\lambda)$, Poisson. How would I go about finding the distribution of $Z=XY$ and the couple $(Z,S)$, where $S=X+Y$?

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  • $\begingroup$ Is that binomial and Poisson? $\endgroup$ – Jean-Sébastien Nov 7 '12 at 18:12
  • $\begingroup$ @Jean-Sébastien yeah! I thought that was kind of obvious. I'm gonna add it. $\endgroup$ – user31280 Nov 7 '12 at 18:15
  • $\begingroup$ I think it is obvious, but there isn't really a standard notation so who knows $\endgroup$ – Jean-Sébastien Nov 7 '12 at 18:16
  • $\begingroup$ Do you have any reason to believe that anything fancy can come out of it? $\endgroup$ – Did Nov 7 '12 at 22:43
  • $\begingroup$ @did Yes because, I'm studying for an exam and I encountered it in a past question paper which made me believe that the Prof was looking for something "fancy". $\endgroup$ – user31280 Nov 7 '12 at 22:45
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Note that $(Z,S)$ is almost surely in a subset $D_0\cup D_1$ of $\mathbb Z\times\mathbb Z$, defined by $$ D_0=\{(n_0(s,t),2s)\mid 0\leqslant t\leqslant s\},\qquad n_0(s,t)=s^2-t^2, $$ and $$ D_1=\{(n_1(s,t),2s+1)\mid 0\leqslant t\leqslant s\},\qquad n_1(s,t)=s^2+s-t^2-t. $$ Then, for every $s\geqslant0$ and $0\leqslant t\leqslant s$, $$ [Z=n_0(s,t),S=2s]=[X=s-t,Y=s+t]\cup[X=s+t,Y=s-t], $$ and $$ [Z=n_1(s,t),S=2s+1]=[X=s-t,Y=s+t+1]\cup[X=s+t+1,Y=s-t], $$ from which explicit formulas can be deduced.

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