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I asked and bountied this question at math SE but didn't get any answers, so I suspect that only experts (if anyone) may know the answer.

The mixed Poisson distribution and compound Poisson distribution are two conceptually different ways to combine the discrete Poisson distribution with PMF $$f_{\lambda_0}(k) := \Pr(X_{\lambda_0}=k) = \frac{e^{-\lambda_0} \lambda_0^k}{k!}, \qquad \lambda_0 \in \mathbb{R}^+,\ k \in \mathbb{N} \tag{1}$$ with another probability distribution to get a new discrete PMF.

In a mixed Poisson distribution, the rate parameter $\lambda_0$ is generalized to a positive-real-valued continuous random variable $L$ with PDF $\pi(\lambda),\ \lambda\geq 0$, so that $f_{\lambda_0}(k)$ becomes the more general PMF $$f_L(k) = \frac{1}{k!} \int_0^\infty e^{-\lambda} \lambda^k \pi(\lambda)\, d\lambda. \tag{2}$$

For a discrete compound Poisson distribution, we take a Poisson-distributed discrete random variable $N_{\lambda_0}$ and an arbitrary discrete random variable $X$ (whose sample space is the natural numbers). We take a countably infinite number of variables $X_i$ that are all identically distributed with $X$ and independent of each other and with $N$, and we form the sum $$Y_{\lambda_0,X} = \sum_{i=1}^{N_{\lambda_0}} X_i. \tag{3}$$ Then the PMF $f_{\lambda_0,X}(k)$ of $Y_{\lambda_0,X}$ is a compound Poisson distribution.

A natural question is whether these families of PMF $f_L(k)$ and $f_{\lambda_0,X}(k)$ are distinct. Although I haven't been able to find any explicit statements about this question, I believe that I can prove that any mixed Poisson distribution whose random rate parameter $L$ is supported on only two positive numbers cannot be a compound Poisson distribution (see my math SE answer for the argument). Assuming my argument is correct, then not every mixed Poisson distribution is a compound Poisson distribution. What about the converse - is every discrete compound Poisson distribution a mixed Poisson distribution? (I propose a somewhat heuristic construction for re-expressing any compound Poisson distribution as a mixed Poisson distribution in my math SE question, but I'm not sure if it always works.)

We can naturally generalize this question beyond individual discrete random variables. The PMFs for mixed and compound Poisson distributions are PMFs for two different classes of Markovian jump processes (i.e. continuous-time and discrete-space stochastic processes) at a fixed point in time: a mixed Poisson process and a compound Poisson process, respectively. Is every compound Poisson process a mixed Poisson process?

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2 Answers 2

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Q: Is every compound Poisson distribution a mixed Poisson distribution?

A: No. Every mixed Poisson distribution has a variance greater than or equal to the mean. The compound Poisson distribution is not so constrained.

If the i.i.d. random variables that are compounded have first moment $\mu$ and second moment $\tau^2\geq\mu^2$, then the compound Poisson distribution has mean $\lambda t\mu$ and variance $\lambda t\tau^2$ (adding $N(t)$ variables with rate $\lambda$). So the variance will be smaller than the mean if $\tau^2<\mu$.

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  • $\begingroup$ But isn't a standard Poisson process just a compound Poisson process where the compounded variables are constant and equal to 1, so they have variance $\sigma^2 = 0$? Your argument would then imply that a standard Poisson process has zero variance at any time, which doesn't seem right. $\endgroup$
    – tparker
    Commented Jul 23, 2022 at 17:49
  • $\begingroup$ thanks; I mixed up second moment with variance; hopefully now it's correct. $\endgroup$ Commented Jul 23, 2022 at 18:09
  • $\begingroup$ Hm. I see your argument, but I was under the impression that any compound Poisson process (other than the simplest one where $X \equiv 1$) is over-dispersed. Any compound Poisson process can be canonically rewritten as another equivalent compound Poisson process (potentially with a different rate parameter) in which the compounded R.V. is only supported on the strictly positive integers. In this case $\tau^2 \geq \mu$, so your possibility can't occur. I'll need to think about how to reconcile this argument with yours (which contradicts mine, but I can't see any error on your part). $\endgroup$
    – tparker
    Commented Jul 23, 2022 at 19:21
  • $\begingroup$ So you're certainly correct that if we allow the compounded R.V. to take on negative integer values, that the resulting compound Poisson process isn't a mixed Poisson process. But I'm wondering about if we restrict the compounded R.V. to only take on nonnegative (or positive) values. $\endgroup$
    – tparker
    Commented Jul 23, 2022 at 19:52
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    $\begingroup$ @MichaelHardy I mean a discrete compound Poisson process, for which (as I said elsewhere in the question) the random variable $X_n$ is assumed to be natural-number-valued. As I said in the comments, I don't think it matters whether or not we include 0 in the set of natural numbers. I've edited the question to clarify that I'm only considering discrete compound Poisson distributions. $\endgroup$
    – tparker
    Commented Nov 30, 2022 at 15:24
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(This answer considers mixed/compound Poisson processes instead of mixed/compound Poisson distributions.)

I think that the only processes that are mixed Poisson and compound Poisson are ordinary Poisson processes. To see why, consider when each type of process has independent increments:

  • A mixed Poisson process should not have independent increments unless the rate $L$ is deterministic, meaning the process is an ordinary Poisson process. This is because if $L$ is random, two disjoint intervals of very long length $t$ will both have increment close to $Lt$, so their increments are correlated.
  • A compound Poisson process always has independent increments, inherited from the underlying ordinary Poisson process and the fact that $X_i$ are i.i.d.
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