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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.

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    $\begingroup$ you're almost there. remove sums on $i$ and $j$ else you'll just be proving $1 = 1$. Split the last sum into two as $U$ and $V$ are independent. Let me know if you're still not able to finish $\endgroup$
    – learner
    Sep 5, 2020 at 14:56
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    $\begingroup$ also what you're calculating above is the CDF which is clearly not needed. The statement $P(X \leq x, Y \leq y) = \sum_i \sum_j P(X = i, Y = j)$ is not true. RHS is $1$, LHS is not $\endgroup$
    – learner
    Sep 5, 2020 at 15:03
  • $\begingroup$ You're comment really helped, I think I'm onto something. And sorry, the sums are from $i = 0$ to $x$ and vice versa. But as you say, not really needed to consider the CDF $\endgroup$
    – Tingiskhan
    Sep 5, 2020 at 15:07
  • $\begingroup$ okay great, let me know if you can find the PMF $\endgroup$
    – learner
    Sep 5, 2020 at 15:08

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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}

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