In section 3.6 of this paper: https://www.jstor.org/stable/25472639?read-now=1&seq=8#page_scan_tab_contents, the authors state that mixed Poisson distributions that are infinitely divisible can be represented as some Compound Poisson distribution.

An example of this is the Negative Binomial point process (which is a Poisson mixture with the rate parameter being gamma distributed). It can also be considered a Compound Poisson point process with a logarithmic distribution being the compounding distribution.

What I don't understand is how these two kinds of point processes can be the same. For example, consider the number of events from the point process in some interval, $\delta t$. As $\delta t \to 0$, the number of events with the Mixed Poisson process should be either $0$ or $1$. It should be very unlikely to encounter $2$ events in a very small interval. For a Compound Poisson, this shouldn't be so unlikely since we only need a Poisson arrival in the small interval and then the compounding distribution can easily be $2$ or more. What is wrong with my reasoning?

  • $\begingroup$ I think that the paper is not claiming they are the same. Infinitely-divisible mixed Poisson distribution make up only small fraction of all possible compound Poisson distributions. $\endgroup$ Commented Aug 2, 2020 at 1:59
  • $\begingroup$ What is claimed seems to be not that two random processes are the same, but that their probability distributions are the same. But the question of whether one process could be both of those things may be interesting. $\endgroup$ Commented Aug 2, 2020 at 2:40
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    $\begingroup$ ok, Maybe it is claimed that they don't just have the same distribution, but also that they are the same process. I'll have to look at this further. $\endgroup$ Commented Aug 2, 2020 at 4:30

1 Answer 1


I hope that I correctly grasped your question. And to answer the question, let us consider a concrete example.

Let $N=(N_t)_{t\geq0}$ and $\tilde{N}=(\tilde{N}_t)_{t\geq0}$ be independent Poisson processes and consider

$$X_t=N_{\tilde{N}_t}, \qquad \text{i.e.,}\qquad X_t(\omega)=N_{\tilde{N}_t(\omega)}(\omega).$$

Then each $X_t$ has a mixed Poisson distribution. In fact, we have a more general observation. Suppose that $T$ is any non-negative r.v. independent of $N$. Then $N_T$ always has a mixed Poisson distribution since $F_{N_{T}}(x) = \int_{0}^{\infty} F_{N_\lambda}(x) \, \mathrm{d}F_{T}(\lambda)$. This formula also shows that any mixed Poisson distribution can be realized in this way.

At the same time, $(X_t)_{t\geq0}$ is a compound Poisson process, since $X_t=\sum_{n=1}^{\tilde{N}_t} (N_n - N_{n-1})$ and $(N_n-N_{n-1})_{n\geq 1}$ are i.i.d. r.v.s.

Now we note that this realization contrasts with


in which each $Y_t$ has a mixed Poisson distribution that is also a compound Poisson distribution but $(Y_t)_{t\geq0}$ itself is not a compound Poisson process.

I think that your objection is based on the process of the form $Y_t$. However, this is not the only way to realize a process out of a given distribution.

  • $\begingroup$ sorry, what does $X_t=N_{\tilde{N}_t}$ mean? And what is $T$? $\endgroup$ Commented Aug 3, 2020 at 1:27
  • $\begingroup$ @RohitPandey, I updated my answer so that the notation is slightly more transparent. But the catch is that, since $N=(N_s)_{s\geq0}$ is a stochastic process parametrized by $s$, we can evaluate this at a random time as well, and in particular, at time $s=\tilde{N}_t$. Also, $T$ is just an arbitrary random variable satisfying the aforementioned condition. $\endgroup$ Commented Aug 3, 2020 at 3:45

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