Find marginal distribution if conditional is multinomially distributed

Question

I would like to find the marginal distribution of $X=\left(X_1,\dots,X_k\right)$ if $X|N=n$ is multinomially distributed with parameters $n$ and $p_1, \dots_, p_k$ and $N$ has Poisson distribution with parameter $\lambda$. I was thinking of using the law of total probability but I didn't see how to proceed with it. Could somebody show me how to do this?

I don't know if this is correct, but here's my attempt:

Since $y_1 + \dots y_m = n$ we get $y_m =n - \sum_{j=1}^{m-1}y_j$ and so

$$ P\left( Y_1 = y_1, \dots , Y_m = y_m \right) = \sum_{n=0}^{\infty} P\left(Y_1 = y_1, \dots, Y_m = y_m | N=n \right)P\left(N=n \right) = \sum_{n=0}^{\infty} \frac{n!}{y_1!\dots y_{m-1}!(n - \sum_{j=1}^{m-1}y_j)!} \prod_{j=1}^{m-1}p_j^{y_j} \cdot p_m^{n - \sum_{j=1}^{m-1}y_j} \frac{e^{-\lambda}\lambda^n}{n!} $$

$n$ has to start from $n= y_1+\dots y_{m-1}$ because of the factorial so this becomes (after making the change of variable $j = n -y_1 - \dots -y_{m-1}$ $$ e^{-\lambda} \cdot \frac{\prod_{j=1}^{m-1}(p_j\lambda)^{y_j}}{y_1!\dots y_{m-1}!} \sum_{j=0}^{\infty}\frac{(\lambda p_m)^j}{j!} = e^{-\lambda(1-p_m)} \cdot \frac{\prod_{j=1}^{m-1}(p_j\lambda)^{y_j}}{y_1!\dots y_{m-1}!} = \prod_{j=1}^{m-1}\frac{(p_j\lambda)^{y_j}e^{-\lambda p_j}}{y_j!}. $$

What confuses me is that (if my calculations/reasoning is correct) I don't get the marginal distribution of $Y_m$.

Answer 1

Each $X_i$ has Poisson distribution with parameter $\lambda p_i$ and the $X_i$ are independent. This is the Poisson splitting property, that extends Poisson thinning. See e.g. Splitting Poisson process formal proof for a detailed discussion of the case $k=2$. The general case follows by iteration, or see https://probabilityandstats.wordpress.com/2011/07/17/splitting-a-poisson-distribution/ and https://kristianeschenburg.github.io/2018/11/poisson-multinomial

Directly: Given non-negative integers $x_1,\ldots, x_k$ that sum to $n$, we have

$$P(X_1=x_1,\ldots, X_k=x_k)=\frac{e^{-\lambda} \lambda^n}{n!} \cdot \frac{n!}{\prod_{i=1}^k x_i!} \prod_{i=1}^k p_i^{x_i} =\prod_{i=1}^k \frac{e^{-\lambda p_i} (\lambda p_i)^{x_i}}{x_i!} \,,$$ and the RHS is the pdf of independent Poisson variables with parameters $\lambda p_i$ evaluated at $x_1,\ldots x_k$.