# A question about Poisson and Binomial distributions

I'm struggling with understanding why the following statement is true:

Let $X$ be a Random Variable with Poisson distribution. Let $Z$ be a Random Variable independent from $X$, whose distribution is $P(Z=0.9)=0.2=1-P(Z=0.6)$. Let $Y$ be a Random Variable such that $Y |( X=x , Z=z)\sim \text{Binom}(x,z)$.

Then given $Z=0.9, Y\sim \text{Poisson}$.

I'm told that it is a consequence of some general fact about split Poisson variables but I couldn't make much of that fact, or wasn't able to see why it's true by myself. I don't really know how to go about this so any help would be greatly appreciated.

Thanks!

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To prove that the distribution of $Y$ conditional on $Z$ is Poisson, you could do the following \begin{eqnarray*} \Pr \left( Y = y|Z = z \right) & = & \sum_{x = y}^{\infty} \Pr \left( Y = y|Z = z, X = x \right) \Pr \left[ X = x \right]\\ & = & \sum_{x = y}^{\infty} \left( \begin{array}{c} x\\ y \end{array} \right) z^y \left( 1 - z \right)^{x - y} \frac{\lambda^x}{x!} e^{- \lambda}\\ & = & e^{- \lambda} z^y \lambda^y \frac{1}{y!} \sum_{x = y}^{\infty} \frac{\left\{ \left( 1 - z \right) \lambda \right\}^{x - y}}{(x - y) !}\\ & = & e^{- \lambda} z^y \lambda^y \frac{1}{y!} \underbrace{\sum_{w = 0}^{\infty} \frac{\left\{ \left( 1 - z \right) \lambda \right\}^w}{w!}}_{= e^{\left( 1 - z \right) \lambda}}\\ & = & e^{- z \lambda} \frac{\left( z \lambda \right)^y}{y!} \end{eqnarray*}
The first line comes from $\Pr \left( Y = y | Z = z, X = x \right) = 0$ for $y >x$ (One thousand thanks for Stefan Hansen for pointing that out). The second line comes from inputting the formulas for the binomial and Poisson probabilities and the fourth line comes from applying the change of variable the change of variable $w = x - y$. By the last line we conclude that $Y$ conditional on $Z = z$ is Poisson with parameter $\lambda z$.
Shouldn't your sum be from $x=y$, since $P(Y=y\mid Z=z,X=x)=0$ when $y>x$? –  Stefan Hansen Jan 30 '13 at 13:53