Joint distribution of dependent Poisson Random Variables

Question

Let $A \sim \text{Poisson}(\beta \gamma)$, $B \sim \text{Poisson}(\beta)$ and $C \sim \text{Poisson}(\gamma)$. They are mutually independent. Define $X=A+B$ and $Y=A+C$. I am interested in finding the joint distribution of $(X,Y)$.

Of course, $X \sim \text{Poisson}(\beta (\gamma+1))$ and $Y \sim \text{Poisson}(\gamma(\beta+1))$ and they are not independent.

So, the joint distribution should be $P(X=x,Y=Y)=P(X=x|Y=y)P(Y=y)$. Now, if we look at $P(X=x|Y=y)$, we see that it is equivalent to writing $P(A+B=x|A+C=y)$. Now, the event $\{A+B=x|A+C=y\} = \cup_{a} \{B=x-a|C=y-a\}$. But since $B$ and $C$ are independent, we have $P(A+B=x|A+C=y) = \sum_{a=0}^{x}P(B=x-a)$. Multiplying with $P(Y=y)$ we get the joint distribution. Is this approach correct? Or am I missing something?

Answer 1

It isn't quite clear to me what your last passages about independence are supposed to accomplish and I suspect you've done some $\sum a_nb_n=(\sum a_n)(\sum b_n)$ kind of mistake. Anyways, you can compare your result with this computation and see if it's the same.

$\def\P{\operatorname{\mathsf P}}\qquad\begin{align}\P(X=x, Y=y)&=\P(A+B=x, A+C=y)\\[1ex]&=\mathrm e^{-\gamma-\beta-\beta\gamma}\sum_{a,b,c\ge 0\\ a+b=x\\ a+c=y}\frac{(\beta\gamma)^{a}\beta^b\gamma^c}{a!b!c!}\\[1ex]&=\mathrm e^{-\gamma-\beta-\beta\gamma}\sum_{a,b,c\ge 0\\ b=x-a\\ c=y-a}\frac{(\beta\gamma)^{a}\beta^b\gamma^c}{a!b!c!}\\[1ex]&=e^{-\beta-\gamma-\beta\gamma}\sum_{a=0}^{\min\{x,y\}}\frac{(\beta\gamma)^a\beta^{x-a}\gamma^{y-a}}{a!(x-a)!(y-a)!}\\[1ex]&=\frac{\beta^{x}\gamma^{y}}{e^{\beta+\gamma+\beta\gamma}}\sum_{a=0}^{\min\{x,y\}}\frac{1}{a!(x-a)!(y-a)!}\end{align}$