Show that Y − X|X ∼ Poisson[λ(1 − π)]

Question

Let X|Y ∼ Binomial(Y,π), and let Y ∼ Poisson(λ). So, I need to show that Y − X|X ∼ Poisson[λ(1 − π)]. I already proved that X~Poisson(λπ). But the X and Y are not independents, so how can I find Y-X|X?

Answer 1

For $Y-X$ unconditionally, define:

$$G(s,t):= \mathbb{E}[s^Xt^Y]=\mathbb{E}_Y\mathbb{E}[s^Xt^Y|Y]=\mathbb{E}_Y[t^Y\mathbb{E}[s^{X|Y}]]$$

Now, $X|Y\sim Bin(Y,\pi)\implies \mathbb{E}[s^{X|Y}]=(1-\pi+\pi s)^Y$, so:

$$G(s,t)=\mathbb{E}[t^Y(1-\pi+\pi s)^Y]$$

Here, we know $Y\sim Poi(\lambda)\implies \mathbb{E}[r^Y]=e^{\lambda(r-1)}$. This gives:

$$G(s,t)=e^{\lambda(t(1-\pi+\pi s)-1)}$$

Now, $G(1/u,u)=\mathbb{E}[u^{Y-X}]$ by definition, whence

$$\mathbb{E}[u^{Y-X}]=e^{\lambda(u(1-\pi+\pi/u)-1)}=e^{\lambda(1-\pi)(u-1)}$$

Thus, $Y-X$ has the same probability generating function as a $Poi(\lambda(1-\pi))$ random variable, and thus is equal in distribution.

For $Y-X|X$, note that by taking $s=1$ and $t=1$ separately, once can see $X\sim Poi(\lambda\pi),Y\sim Poi(\lambda)$ (in fact, we knew $Y$ already - this is more of a sanity check step). Using Bayes' theorem, we have:

$$\mathbb{P}(Y=y|X=x)=\frac{\mathbb{P}(X=x|Y=y)\mathbb{P}(Y=y)}{\mathbb{P}(X=x)} = \frac {{y \choose x}\pi^x (1-\pi)^{y-x} \frac{\lambda^ye^{-\lambda}}{y!}} {\frac{(\lambda\pi)^xe^{-\lambda\pi}}{x!}}$$

$$=\frac{(\lambda(1-\pi))^{y-x}e^{-\lambda(1-\pi)}}{(y-x)!}$$

It thus follows naturally that $Y-X|X\sim Poi(\lambda(1-\pi))$ also.