# Joint distribution of dependent Poisson Random Variables

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?

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}