# How to find the joint distribution of sums of Poisson random variables

I am trying to determine the joint distribution of two sums of Poisson random variables.

Let's say $X \sim \text{Pois}(\lambda_{1})$, $Y \sim \text{Pois}(\lambda_{2})$, and $Z \sim \text{Pois}(\lambda_{3})$. Moreover assume that $X,Y$ and $Z$ are all mutually independent. I want to find $P(X=k|X+Y=n_{1}, X+Z=n_{2})$.

From this step, I know $P(X=k|X+Y=n_{1}, X+Z=n_{2}) = \frac{P(Y=n_{1}-k, Z= n_{2}-k) P(X)}{P(X+Y, X+Z)}$.

I know the numerator is a bivariate Poisson distribution and marginal Poisson. How would I go about finding the distribution for the denominator?

## 1 Answer

Sum over all possible values of $X$. You can sum because of countable additivity of $P(\cdot)$. The goal is to exploit the independence among your three random variables (note: I suggested this addition to your problem description in an edit just now...without this my answer is incorrect.).

\begin{align*} P(X+Y = a, X+Z=b) &= \sum_{k=0}^{\infty} P(X+Y=a, X+Z=b, X=k)\\ &= \sum_k P(X=k, Y=a-k, Z=b-k) \\ &= \sum_k P(X=k)P(Y=a-k)P(Z=b-k) \end{align*}

Also, you can further simplify the numerator by factoring the first factor into the two pmfs of $Y$ and $Z$.

• Thank you, Taylor! That was very helpful. I didn't realize it was that simple. – JayCEE Nov 11 '16 at 13:40