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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 $D \sim \mathcal{Geometric}(\rho)$.

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

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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.

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