Thinning: Cell Counts of a Poisson Number of Independent Points

Let $(\Omega,\mathcal{F},P)$ be a probability space, $\mathbb{N}$ the set of natural numbers with $\mathbb{N}_0=\mathbb{N}\cup\{0\}$, and $\mathbb{R}$ the set of real numbers. Let $\mu\ge0$ be real, let $K$ be a random variable with the Poisson distribution with parameter $\mu$, let $\nu$ be a probability measure on the Borel $\sigma$-algebra, and let $(V_i)_{i\in\mathbb{N}}$ be random variables each with distribution $\nu$, such that the family $(K,V_1,V_2,\dots)$ is independent.

Let $r\in\mathbb{N}$ and let $A_1,\dots,A_r$ be pairwise disjoint Borel sets, with $p_j=\nu(A_j)$. Define $\widetilde K(\omega)=K(\omega)$ if $K(\omega)\in\mathbb{N}_0$ and $\widetilde K(\omega)=0$ otherwise (so $\widetilde K=K$ with probability $1$, since a Poisson variable lies in $\mathbb{N}_0$ with probability $1$ by Poisson Distribution), and define the cell counts

where $\mathbf{1}_{E}$ is the function equal to $1$ on $E$ and $0$ off $E$, and an empty sum is $0$. Then:

1. each $C_j$ is a random variable with all values in $\mathbb{N}_0$, measurable with respect to the generated $\sigma$-algebra $\sigma\bigl(K,(V_i)_{i\in\mathbb{N}}\bigr)$;

2. for all $(n_1,\dots,n_r)\in\mathbb{N}_0^{r}$, with the factorial (convention $0!=1$), the convention $x^{0}=1$, the exponential function $\exp$, and the finite product notation,

3. the random variables $C_1,\dots,C_r$ are independent, and $C_j$ has the Poisson distribution with parameter $\mu p_j$.