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Suppose you have:

$X|N \sim \text{MN}(N,p_1,p_2...p_J)$

$N \sim \text{Poisson}(\lambda)$

Where is the marginal distribution of $X$?

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    $\begingroup$ The univariate version of what you wrote is not a Poisson-Binomial distribution but rather a Binomial distribution with $N \sim \text{Poisson}(\lambda)$, which in turn is just two Poisson distributions, one with mean $p\lambda$ and one with mean $(1-p)\lambda$, for successes and failures respectively. $\endgroup$
    – jbowman
    Commented May 7, 2018 at 23:26
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    $\begingroup$ It's just a bunch of Poisson distributions with means $p_i\lambda$. $\endgroup$
    – jbowman
    Commented May 7, 2018 at 23:29
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    $\begingroup$ Now, now, integrating out $N$ is hardly "probability theory". More precisely, then, it's a collection of $J$ Poisson distributions with means $p_j\lambda, j = 1, \dots, J$. $\endgroup$
    – jbowman
    Commented May 7, 2018 at 23:32
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    $\begingroup$ Note that there is a different level of information and detail that one puts into the rather limited space of a comment than what one puts into a complete answer! $\endgroup$
    – jbowman
    Commented May 8, 2018 at 0:06

1 Answer 1

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Just to formalise what jbowman is telling you in the comments, for any valid argument vector $\boldsymbol{x}$ with $\sum x_j = n$, application of the law-of-total-probability gives:

$$\begin{equation} \begin{aligned} \mathbb{P}(\boldsymbol{X}=\boldsymbol{x}) &= \mathbb{P}(\boldsymbol{X}=\boldsymbol{x} | N=n) \mathbb{P}(N=n) \\[6pt] &= \text{Mu}(\boldsymbol{x} | n,\boldsymbol{p}) \text{Pois}(n|\lambda) \\[6pt] &= \frac{n!}{\prod x_j!} \Big( \prod_{j=1}^J p_j^{x_j} \Big) \frac{\lambda^n}{n!} \exp(-\lambda) \\[6pt] &= \prod_{j=1}^J \frac{(\lambda p_j)^{x_j}}{x_j !} \exp(-\lambda p_j) \\[6pt] &= \prod_{j=1}^J \text{Pois}(x_j | \lambda p_j). \\[6pt] \end{aligned} \end{equation}$$

(Note that the first line of this working is the law-of-total-probability, but with the sum being taken over only one value of $n$ that is consistent with the vector $\boldsymbol{x}$.) Hence, we can see that the elements of $\boldsymbol{X}$ are independent, with marginal distributions $x_j \sim \text{Pois}(\lambda p_j)$.

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