We have three independently Poisson distributed random variables, $U, V, W$, with intensities $\lambda, \mu, \nu$. Now, define $X = U + W$, and $Y = V + W$. The task is now to find the joint pmf of $(X, Y)$. What I started doing was the following: \begin{equation} P(X \leq x, Y \leq y) = \sum_i \sum_j P(X = i, Y = j) = \sum_i \sum_j P(U + W = i, V + W = j), \end{equation} but as they are not independent, I'm not really sure on how to continue. Any pointers?
Edit As someone seems to have downvoted I'm guessing it's for lack of trying. As such, my continued try resulted in \begin{equation} \sum_i \sum_j P(U + W = i, V + W = j) = \\ \sum_i \sum_j \sum_k P(U + W = i, V + W = j \mid W = k)P(W = k) = \\ \sum_i \sum_j \sum_k P(U = i - k, V = j - k)P(W = k). \end{equation} But still not sure at all, that this is even remotely correct.
The joint PMF is \begin{align} P(U+W = i, V+W = j) &= \sum_k P(U+W=i,V+W=j∣W=k)P(W=k) \\ &= \sum_k P(U=i−k,V=j−k)P(W=k) \\ &= \sum_k P(U=i−k)P(V=j−k)P(W=k) \\ &= \sum_k \frac{e^\lambda \lambda^{i-k}}{(i-k)!}\frac{e^\mu\mu^{j-k}}{(j-k)!}\frac{e^\nu \nu^{k}}{k!} \end{align} Use the fact $k \leq i, k\leq j \implies k \leq \min\{i, j\}$ to rewrite the sum as \begin{align} P(U+W = i, V+W = j)&=e^{\lambda+\mu+\nu}\lambda^i\mu^j\sum_{k=1}^{\min\{i, j\}}\frac{\lambda^{-k}\mu^{-k}\nu^{k}}{(i-k)!(j-k)!k!} \end{align}