Let $m\in \mathbb N$ and suppose that $Z_n, n\in \mathbb N$, is a sequence of independent random vectors in $\mathbb R^m$ with common distribution $\mathbb P(Z_1 = e_i) = p_i, i\in \{ 1,…,m\}$, where $e_i$ is the $i-$th unit vector in $\mathbb R^m$ and $p_1+ \cdots+ p_m = 1$. Let $Z \sim$ Po$(\gamma)$, independent of $(Z_1, Z_2,…).$ Show that the components of the random vector $X := \sum_{j=1}^Z Z_j$ are independent and Poisson distributed with parameters $p_1\gamma, …, p_m \gamma$.
This can be proved if we show that $$\mathbb P(X_1 = k_1, … , X_n = k_n) = \prod_{j=1}^n \mathbb P(X_j = k_j) = \prod_{j=1}^n \text{Po}(k_j; p_i \gamma).$$
How to prove the above equation? Is there any other easy way to prove the statement?
Can we argue that the components of $X$ are indepedent by that $Z_n, n\in \mathbb N$, is a sequence of independent random vectors?
Best Answer
Given that the random vector $\textbf X$ has certain number of entries $n$, it follows the multinomial distribution parameters $n, \textbf p$. This is because $X$ is created by incrementing each coordinate with the vector of probabilities $\textbf p$, $n$ times. Then the joint distribution is found by:
$$\begin{split}f(x_1,...,x_m,z)&=f(x_1,...,x_m|z)\Pr(z)\\ &=\left(\frac{z!}{x_1!...x_m!}p_1^{x_1}...p_m^{x_m}\right)\left(\frac{e^{-\gamma}\gamma^z}{z!}\right)\mathbb 1_{\sum_i x_i=z}\\ &=e^{-\gamma(p_1+...+p_m)}\frac{(p_1\gamma)^{x_1}...(p_m\gamma)^{x_m}}{x_1!...x_m!}\mathbb 1_{\sum_i x_i=z}\\ &=\frac{e^{-\gamma p_1}(p_1\gamma)^{x_1}}{x_1!}...\frac{e^{-\gamma p_m}(p_m\gamma)^{x_m}}{x_m!}\mathbb 1_{\sum_i x_i=z}\end{split}$$
Marginalize $z$ out:
$$\begin{split}f(x_1,...,x_m)&=\sum_{z=0}^\infty\frac{e^{-\gamma p_1}(p_1\gamma)^{x_1}}{x_1!}...\frac{e^{-\gamma p_m}(p_m\gamma)^{x_m}}{x_m!}\mathbb 1_{\sum_i x_i=z}\\ &=\frac{e^{-\gamma p_1}(p_1\gamma)^{x_1}}{x_1!}...\frac{e^{-\gamma p_m}(p_m\gamma)^{x_m}}{x_m!}\end{split}$$
Therefore $X_i\sim \text{Poisson}(\gamma p_i)$ for $i=1,...,m$.