PMF of compound Poisson process?

Can I obtain an analytic expression for PMF of compound Poisson process?

$$Y_t = \sum \limits_{i=1}^{X_t} D_i$$,

where $$X_t \sim \mathcal{Poisson}(\lambda)$$ and $$D \sim \mathcal{Geometric}(\rho)$$.

• You need to assume $\{D_i\}$ and $\{X_t\}$ are independent, right? Commented Jul 15, 2015 at 17:28
• @Zhanxiong: Yes, that is assumed indeed. Commented Jul 17, 2015 at 8:47
• @user3817794 One more concern, I guess $X_t \sim \text{Poisson}(\lambda t)$ instead of $\lambda$, otherwise the notation $t$ looks superfluous. Commented Jul 17, 2015 at 14:49

1 Answer

First, I will simplify your notation and let $$Y = \sum_{i=1}^N D_i$$ (with the understanding that the sum is zero if $$N=0$$.) The $$D_i$$ are iid geometric random variables and $$N \sim \mathcal{Pois}(\lambda)$$, independently of the $$D_i$$'s. This can be called a compound Poisson-geometric distribution, but is also known as the Polya-Aeppli distribution.

You did not specify which version of the geometric distribution to use, I will assume the shifted geometric, with support $$\{1,2,3,\dotsc\}$$ and probability mass function $$\DeclareMathOperator{\P}{\mathbb{P}} \P(D_i=k)= (1-p)^{k-1} p, \quad k=1,2,3,\dotsc$$ as this choice seems to be the usual one in this context.

There is an R package (on CRAN) for this distribution polyaAeppli and the theory behind that package is in this paper, where it is shown that the pmf (probability mass function) is given by $$\P(Y=y) = \sum_{k=1}^y e^{-\lambda} \frac{\lambda^k}{k!}(1-p)^{y-k} p^k \binom{y-1}{k-1}$$ for positive $$y$$, and $$\P(Y=0)=e^{-\lambda}$$. This distribution also has an Wikipedia article.