# Sum of Dependent Poisson Random Variables

I am working with the distribution of the sum of two dependent random variables. In my problem, there are two unobserved events, X and Y, where X precedes Y and Y is a function of the outcome of X, but I only observe the sum of the two outcomes, i.e:

$$Z=X+Y|X$$ $$Pr(Z=x)=\sum_{z=0}^xPr(X=z)*Pr(Y=x-z|X=z)$$

It looks like a standard convolution distribution, except that Y and X are not independent. In my particular case, X and Y have a Poisson distribution, where the mean of Y is an increasing function of the value of the outcome of X. I am unaware of a closed form for the representation of the PMF of this distribution, and so need to derive it directly through the above sum. For my particular problem, I need to compute the PMF thousands of times, and am looking for ways to reduce the computational burden of doing so. As such, I was hoping someone would either

1) Know if there is a closed form representation for the PMF

2) Know of any approximations to the PMF

Any help would be greatly appreciated.

Thanks!

• Your questions cannot be answered with the information given: you need to supply a formula expressing how $E[Y]$ depends on $X$. – whuber Jun 27 '14 at 21:53
• There's an infinity of ways of having dependent Poissons. – Glen_b Jun 28 '14 at 9:48

You are on the right track, the conditional distribution $X|Y$ is a so-called mixed poisson distribution (the Poisson mean is just equal to its parameter $\lambda$).