Proof of Thinning: Cell Counts of a Poisson Number of Independent Points

Write $\mathcal{G}=\sigma(K,(V_i)_{i\in\mathbb{N}})$ for the generated $\sigma$-algebra, $A_0=\mathbb{R}\setminus(A_1\cup\dots\cup A_r)$ (a Borel set), $p_0=\nu(A_0)$, and for $n\in\mathbb{N}_0$, $S_j^{(n)}=\sum_{i=1}^{n}\mathbf{1}_{\{V_i\in A_j\}}$ ($S_j^{(0)}=0$). Since $\nu$ is a probability measure and $A_0,\dots,A_r$ are pairwise disjoint with union $\mathbb{R}$, finite additivity gives $p_0+p_1+\dots+p_r=1$.

Step 1 (the truncated count $\widetilde K$). The set $\mathbb{N}_0$ is Borel: each singleton $\{n\}=\bigcap_{q\in\mathbb{N}}(n-1/q,\,n+1/q)$ is Borel, and $\mathbb{N}_0$ is a countable union of singletons. For $n\in\mathbb{N}$ we have $\{\widetilde K=n\}=K^{-1}(\{n\})$, and $\{\widetilde K=0\}=K^{-1}(\{0\}\cup(\mathbb{R}\setminus\mathbb{N}_0))$; in every case the level set is a preimage under $K$ of a Borel set, hence an event lying in $\sigma(K)\subseteq\mathcal{G}$. Since $\widetilde K$ takes all its values in $\mathbb{N}_0$, the preimage of any Borel set is the countable union of its level sets, so $\widetilde K$ is a random variable, $\mathcal{G}$-measurable. By Poisson Distribution, $P(K\in\mathbb{R}\setminus\mathbb{N}_0)=0$ and $P(K=n)=\exp(-\mu)\mu^{n}/n!$ for $n\in\mathbb{N}_0$; hence

$$P(\widetilde K=n)=\exp(-\mu)\frac{\mu^{n}}{n!}\qquad(n\in\mathbb{N}_0),$$

where for $n=0$ the extra null set changes nothing by finite additivity.

Step 2 (Part 1). On the event $\{\widetilde K=n\}$ we have $C_j=S_j^{(n)}$. Hence for $c\in\mathbb{N}_0$,

$$\{C_j=c\}=\bigcup_{n\in\mathbb{N}_0}\bigl(\{\widetilde K=n\}\cap\{S_j^{(n)}=c\}\bigr),$$

a countable union. As shown in the proof of Multinomial Distribution of Cell Counts for Independent Identically Distributed Points, each $\{S_j^{(n)}=c\}$ is an event lying in $\sigma(V_i:1\le i\le n)\subseteq\mathcal{G}$ (for $n=0$ it is $\Omega$ if $c=0$ and $\emptyset$ otherwise), so $\{C_j=c\}\in\mathcal{G}$. All values of $C_j$ lie in $\mathbb{N}_0$ (each $C_j(\omega)$ is a finite sum of zeros and ones), so as in Step 1, $C_j$ is a random variable, $\mathcal{G}$-measurable. This proves Part 1.

Step 3 (Part 2). Fix $(n_1,\dots,n_r)\in\mathbb{N}_0^r$ and put $s=n_1+\dots+n_r$. Since the cells $A_0,\dots,A_r$ partition $\mathbb{R}$, we have $\sum_{j=0}^{r}S_j^{(n)}=n$ identically; thus on $\{\widetilde K=n\}$, the event $\bigcap_{j=1}^{r}\{C_j=n_j\}$ forces $S_0^{(n)}=n-s$, and it is empty for $n<s$. Therefore

$$\bigcap_{j=1}^{r}\{C_j=n_j\}=\bigcup_{n\ge s}\Bigl(\{\widetilde K=n\}\cap Q_n\Bigr),\qquad Q_n=\{S_0^{(n)}=n-s\}\cap\bigcap_{j=1}^{r}\{S_j^{(n)}=n_j\},$$

a countable disjoint union (the $\{\widetilde K=n\}$ are disjoint). For $n\ge\max(s,1)$: the subfamily $V_1,\dots,V_n$ of the independent family is independent (Independence of Events and of Random Variables), each with distribution $\nu$, and the $r+1$ cells $A_0,\dots,A_r$ are pairwise disjoint with union $\mathbb{R}$, so Multinomial Distribution of Cell Counts for Independent Identically Distributed Points gives

$$P(Q_n)=\frac{n!}{(n-s)!\,n_1!\cdots n_r!}\,p_0^{\,n-s}\prod_{j=1}^{r}p_j^{\,n_j}.$$

If $n=0$ (possible only when $s=0$), then $Q_0=\Omega$ and the displayed formula also holds, reading $0!/0!=1$ and empty exponents as $1$. Moreover $Q_n\in\sigma(V_i:1\le i\le n)$ and $\{\widetilde K=n\}\in\sigma(K)$; by Grouping Lemma for Independent Random Variables applied to the independent family $(K,V_1,V_2,\dots)$ with the two disjoint blocks (the index of $K$) and (the indices of $V_1,\dots,V_n$), these two $\sigma$-algebras are independent, so $P(\{\widetilde K=n\}\cap Q_n)=P(\widetilde K=n)P(Q_n)$ (for $n=0$, $Q_0=\Omega$ makes this trivial). By countable additivity and Step 1,

$$P\Bigl(\bigcap_{j=1}^{r}\{C_j=n_j\}\Bigr)=\sum_{n\ge s}\exp(-\mu)\frac{\mu^{n}}{n!}\cdot\frac{n!}{(n-s)!\,n_1!\cdots n_r!}\,p_0^{\,n-s}\prod_{j=1}^{r}p_j^{\,n_j}.$$

Substituting $n=s+t$, $t\in\mathbb{N}_0$, and $\mu^{n}=\mu^{s}\mu^{t}$, the series becomes

$$\exp(-\mu)\,\Bigl(\prod_{j=1}^{r}\frac{(\mu p_j)^{n_j}}{n_j!}\Bigr)\sum_{t=0}^{\infty}\frac{(\mu p_0)^{t}}{t!}=\exp(-\mu)\exp(\mu p_0)\prod_{j=1}^{r}\frac{(\mu p_j)^{n_j}}{n_j!},$$

since $\mu^{s}\prod_j p_j^{n_j}=\prod_j(\mu p_j)^{n_j}$ (as $s=\sum_j n_j$) and $\sum_{t\ge0}(\mu p_0)^{t}/t!$ is the defining series of $\exp(\mu p_0)$ from The Real Exponential Function. By Basic Properties of the Exponential Function and $1-p_0=p_1+\dots+p_r$,

$$\exp(-\mu)\exp(\mu p_0)=\exp(-\mu(1-p_0))=\exp\Bigl(-\sum_{j=1}^{r}\mu p_j\Bigr)=\prod_{j=1}^{r}\exp(-\mu p_j).$$

Combining the last two displays yields Part 2.

Step 4 (Part 3). For each $j$ let $g_j(c)=\exp(-\mu p_j)(\mu p_j)^{c}/c!$, $c\in\mathbb{N}_0$. Each $g_j$ maps into $[0,1]$ and $\sum_{c=0}^{\infty}g_j(c)=1$: this is the total-mass computation of Poisson Distribution for the parameter $\mu p_j\ge0$. By Part 1 the $C_j$ take all values in $\mathbb{N}_0$, and by Part 2 their joint probability mass function factorizes as $\prod_j g_j$. Hence Factorized Joint Probability Mass Function Implies Independence applies: $C_1,\dots,C_r$ are independent, and for every Borel set $B$,

$$P(C_j\in B)=\sum_{c\in B\cap\mathbb{N}_0}\exp(-\mu p_j)\frac{(\mu p_j)^{c}}{c!},$$

which is exactly the Poisson distribution with parameter $\mu p_j$ evaluated at $B$, by Poisson Distribution. So $C_j$ has the Poisson distribution with parameter $\mu p_j$. $\blacksquare$