Poisson and Cox point processes

Fix an at most countable basis $\mathcal{B}$ for $S$ and let $\mathcal{U}=\{U\in \mathcal{B}:\bar{U}\text{ is compact}\}$. For any $x\in S$, [F2] with $O=S$ gives an open $V$ containing $x$ whose closure is compact. The basis property gives $U\in \mathcal{B}$ with $x\in U\subseteq V$; then $\bar{U}\subseteq \bar{V}$, and $\bar{U}$ is closed in the compact space $\bar{V}$, hence compact by [F3]. Thus $\mathcal{U}$ is an at most countable open cover of $S$.

If $S=\mathrm{\varnothing }$, take $K_n=\mathrm{\varnothing }$. Otherwise enumerate the subfamily $\mathcal{U}$ by filtering the fixed countable basis in its given enumeration; if the subfamily is finite, repeat its last member. Put $K_n={\bigcup }_{j=1}^n{\bar{U}}_j$. An open cover of $K_n$ restricts to an open cover of each of the finitely many compact sets ${\bar{U}}_j$; taking the union of their finite subcovers gives a finite subcover of $K_n$. Thus $K_n$ is compact. The sequence is increasing, and its union is $S$ because $\mathcal{U}$ covers $S$.

Let $C\subseteq S$ be compact. The open cover $\mathcal{U}$ of $C$ has a finite subcover by the ambient-cover criterion in [F4]. Let $n$ be at least every index in that finite subcover. Then $C\subseteq {\bigcup }_{j=1}^n{U}_j\subseteq K_n$. Thus every compact $C$ is contained in some $K_n$.

Steps 2.1 and 3.1 give the required increasing compact exhaustion and its cofinal containment property. The construction uses only a fixed countable basis and finite subcovers, so it invokes no choice principle.

The function $\nu (B)=\mathbb{E}[\zeta (B)]$ is defined for every Borel set $B$ by [F1] and [F2], and $\nu (\mathrm{\varnothing })=0$ because each measure ${\zeta }_{\omega }$ vanishes at $\mathrm{\varnothing }$.

Let $f\ge 0$ be Borel and choose $s_n\uparrow f$ as in [F4]. If $s_n$ is simple, its integral against $\zeta$ is a finite nonnegative linear combination of the measurable evaluations in [F1], hence is measurable. For each $\omega$, monotone convergence for the measure ${\zeta }_{\omega }$ gives ${\zeta }_{\omega }(s_n)\uparrow {\zeta }_{\omega }(f)$. The pointwise-limit result [F5] therefore makes $\omega \mapsto {\zeta }_{\omega }(f)$ measurable.

Let $(B_j{)}_{j\ge 1}$ be pairwise disjoint Borel sets. For every $\omega$, ${\zeta }_{\omega }({\bigcup }_j{B}_j)={\sum }_j{\zeta }_{\omega }(B_j)$ by [F1] and [F6]. The finite partial sums increase, so [F3] gives $\nu ({\bigcup }_j{B}_j)={\lim }_n{\sum }_{j=1}^n\mathbb{E}[\zeta (B_j)]={\sum }_j\nu (B_j)$. Together with step 1.1 this proves that $\nu$ is a measure.

For a nonnegative simple $s={\sum }_{j=1}^m{a}_j{1}_{B_j}$, the definition $\nu (B)=\mathbb{E}[\zeta (B)]$ and finite linearity give $\mathbb{E}[\zeta (s)]=\int sd\nu$. For the increasing sequence from step 1.2, both $\zeta (s_n)\uparrow \zeta (f)$ pathwise and $\int s_{n}d\nu \uparrow \int fd\nu$; applying [F3] to both nonnegative integrals proves the expectation identity.

For compact $K$, local finiteness of $\nu$ at $K$ means exactly $\nu (K)=\mathbb{E}[\zeta (K)]<\mathrm{\infty }$ by its definition. Requiring this for all compact $K$ proves the final equivalence.

Let ${\bar{\mathbb{N}}}_0:={\mathbb{N}}_0\cup \{+\mathrm{\infty }\}$, the range of every count evaluation, and let $\mathcal{C}$ consist of the cylinder sets $\{\mu :\mu (B_j)\in E_j,1\le j\le m\}$ for finite Borel families $B_1,\dots ,B_m$ and subsets $E_j\subseteq {\bar{\mathbb{N}}}_0$. Each such constraint is measurable because this countable range consists of Borel singletons in $[0,+\mathrm{\infty }]$. The family is a pi-system: intersecting two cylinders concatenates their finite coordinate lists, and if a Borel set occurs twice its two constraints are replaced by their intersection. Every evaluation preimage is a one-coordinate cylinder of this form, so by definition of the evaluation sigma-algebra $\sigma (\mathcal{C})$ is the full configuration sigma-algebra.

The hypothesis says the laws $P_{\eta }$ and $P_{{\eta }^{\mathrm{\prime }}}$ assign the same probability to every member of $\mathcal{C}$, since membership in a cylinder is a finite-vector count event. Both are probability measures by [F1]. Applying [F2] to this generating pi-system gives $P_{\eta }=P_{{\eta }^{\mathrm{\prime }}}$.

If $\lambda =0$, [F1] puts all mass at $0$, so the masses sum to one and the stated moments and generating function have their asserted values.

Suppose $\lambda >0$. By [F1] and the defining exponential series [F2], ${\sum }_{k=0}^{\mathrm{\infty }}{e}^{-\lambda }\frac{{\lambda }^k}{k!}=e^{-\lambda }{e}^{\lambda }=1.$ Thus the displayed masses normalize to a probability law.

Using [F1] and reindexing the nonnegative series, $\mathbb{E}[N]=e^{-\lambda }{\sum }_{k=1}^{\mathrm{\infty }}k\frac{{\lambda }^k}{k!}=\lambda e^{-\lambda }{\sum }_{j=0}^{\mathrm{\infty }}\frac{{\lambda }^j}{j!}=\lambda \mathrm{.}$

Reindexing with $j=k-2$ gives $\mathbb{E}[N(N-1)]=e^{-\lambda }{\sum }_{k=2}^{\mathrm{\infty }}k(k-1){\lambda }^k\mathrm{/}k!={\lambda }^2{e}^{-\lambda }{\sum }_{j=0}^{\mathrm{\infty }}{\lambda }^j\mathrm{/}j!={\lambda }^2$ by [F2]. Since $N^2=N(N-1)+N$, this and step 2.1 yield $\mathbb{E}[N^2]={\lambda }^2+\lambda$ and hence $\mathrm{Var}(N)=\mathbb{E}[N^2]-(\mathbb{E}N{)}^2=\lambda$.

For $0\le s\le 1$, [F1] and [F2] give $\mathbb{E}[s^N]=e^{-\lambda }{\sum }_{k=0}^{\mathrm{\infty }}\frac{(\lambda s{)}^k}{k!}=e^{-\lambda }{e}^{\lambda s}=e^{\lambda (s-1)}\mathrm{.}$ Together with step 1.1 this proves the generating-function formula also at $\lambda =0$.

On the parameter space $S\times {\mathcal{N}}_{\mathrm{l}\mathrm{f}}(S)$ define $Q((x,\mu ),B):=\mu (B)$. For each parameter this is a measure, its evaluations are measurable by [F2], and $(K_n)$ is a common finite-mass exhaustion because $\mu (K_n)<\mathrm{\infty }$ for every locally finite $\mu$. Thus $Q$ is a uniformly sigma-finite kernel.

The diagonal $\mathrm{\Delta }=\{(x,y):x=y\}$ is closed in $S\times S$ because $S$ is Hausdorff. The product topology has the countable basis consisting of rectangles from countable bases of the two copies of $S$; every open set is a countable union of these rectangles. Hence the product Borel sigma-algebra equals $\mathcal{B}(S)\otimes \mathcal{B}(S)$, so $1_{\mathrm{\Delta }}$ is product measurable by [F4].

For $(x,\mu )$, $\mu (\{x\})=\int 1_{\mathrm{\Delta }}(x,y)Q((x,\mu ),dy)$. The measurability theorem in [F3] makes $(x,\mu )\mapsto \mu (\{x\})$ measurable; therefore its inverse image $D$ of $[1,\mathrm{\infty }]$ is measurable.

On $D$, $\mu -{\delta }_x$ is again a locally finite counting measure: in a representation of $\mu$ as a countable sum of point masses, remove one occurrence at $x$; no compact-set count increases. For every Borel $B$, $R(x,\mu )(B)=\mu (B)-1_B(x),$ where $\mathrm{\infty }-1$ is interpreted as $\mathrm{\infty }$. This evaluation is measurable on $D$, since both terms are measurable and on $D$ the subtraction is nonnegative. The evaluations generate the target sigma-algebra [F2], so $R$ is measurable.

Steps 2.1-3.1 prove measurability of the domain and deletion map, including the multiplicity convention.

If $\lambda =0$, take a one-point probability space and let $\eta$ be the zero measure. For every Borel $B$, $\nu (B)=0$, so $\eta (B)=0$ has the required Poisson law and every finite vector of counts is independent.

Now suppose $\lambda >0$ and put $Q=\nu \mathrm{/}\lambda$. Let $\mathrm{\Omega }={\coprod }_{n\ge 0}(\{n\}\times S^n)$ with the sigma-algebra whose sections in every $S^n$ are product-measurable. For each $n$, the section map defines a probability measure $Q_n(A)=Q^{\otimes n}(A_n)$ on $\mathrm{\Omega }$, by [F3]. The countable sum $P={\sum }_{n\ge 0}{e}^{-\lambda }{\lambda }^n{Q}_n\mathrm{/}n!$ is a probability measure by [F2] and [F4].

On the component $(n,x_1,\dots ,x_n)$ define $\eta ={\sum }_{i=1}^n{\delta }_{x_i}$. For each Borel $B$, $\eta (B)={\sum }_{i=1}^n{1}_B(x_i)$ is measurable on each component, so $\eta$ is a random locally finite counting measure.

Let $B_1,\dots ,B_m$ be pairwise-disjoint Borel sets and let $B_0=S\setminus {\bigcup }_{j=1}^m{B}_j$. Put $q_j=Q(B_j)$. Conditional on the component $S^n$, independence of the $n$ coordinates gives, for $k_0+\cdots +k_m=n$, $P\{\eta (B_j)=k_j,0\le j\le m\mid n\}=\frac{n!}{{\prod }_{j=0}^m{k}_j!}{\prod }_{j=0}^m{q}_j^{k_j}$, where $0^0=1$. This is the multinomial count of coordinate assignments to the partition $B_0,\dots ,B_m$.

Put $K={\sum }_{j=1}^m{k}_j$. Summing the conditional multinomial masses in step 1.4 over $k_0\ge 0$ gives $P\{\eta (B_j)=k_j,1\le j\le m\}=e^{-\lambda }{\prod }_{j=1}^m\frac{(\lambda q_j{)}^{k_j}}{k_j!}{\sum }_{k_0=0}^{\mathrm{\infty }}\frac{(\lambda q_0{)}^{k_0}}{k_0!}={\prod }_{j=1}^m{e}^{-\lambda q_j}\frac{(\lambda q_j{)}^{k_j}}{k_j!}={\prod }_{j=1}^m{e}^{-\nu (B_j)}\frac{\nu (B_j{)}^{k_j}}{k_j!}$, since the exponential series gives the sum as $e^{\lambda q_0}$ and $1-q_0={\sum }_{j=1}^m{q}_j$. Thus these counts are independent Poisson variables with respective means $\nu (B_j)$, proving existence with the required count laws.

For uniqueness in distribution, take any finite family of Borel sets and partition $S$ into the finitely many atoms generated by that family. Each original count is a sum of counts on these disjoint atoms. For either process, [F1] gives independent Poisson counts on those atoms with the same means $\nu (C)$, so the atom-count vectors have the same product law and every original finite evaluation vector has the same law. The point-process law characterization [F5] then gives equality of the laws.

Steps 1.1-3.1 prove the zero-mass case, construct the mixed-binomial law for positive finite mass, and show that every finite-intensity Poisson process has that law. No choice principle is used; only finite products and a specified countable sum occur.

First let $s={\sum }_{j=1}^m{c}_j{1}_{B_j}$ be a nonnegative simple function with disjoint Borel sets $B_j$ and coefficients $c_j>0$, after discarding zero coefficients. Write $a_j=\nu (B_j)$. If some $a_j=\mathrm{\infty }$, [F1] gives $\eta (s)=\mathrm{\infty }$ almost surely and both displayed right-hand quantities are infinite. If all $a_j<\mathrm{\infty }$, [F1]-[F2] give $\mathbb{E}\eta (s)={\sum }_j{c}_j{a}_j=\nu (s)$ and $\mathbb{E}\eta (s{)}^2={\sum }_j{c}_j^2(a_j+a_j^2)+2{\sum }_{i<j}{c}_i{c}_j{a}_i{a}_j=\nu (s^2)+\nu (s{)}^2$. The zero simple function also satisfies both identities.

For arbitrary $f\ge 0$, choose simple $s_n\uparrow f$ by [F3]. Then $\eta (s_n)\uparrow \eta (f)$, $\eta (s_n{)}^2\uparrow \eta (f{)}^2$, $\nu (s_n)\uparrow \nu (f)$, and $\nu (s_n^2)\uparrow \nu (f^2)$. Applying monotone convergence [F3] to the two expectation integrals and to the two measure integrals, and using step 1.1 for each $s_n$, yields both asserted identities, including the extended-infinite cases.

The addition map is measurable because for every Borel $B$ its output evaluation is $(\mu +{\delta }_x)(B)=\mu (B)+1_B(x)$, a measurable function of $(x,\mu )$; therefore $(x,\mu )\mapsto h(x,\mu +{\delta }_x)$ is measurable. The left integral is measurable by the evaluation-kernel integration assertion [F4].

If $\lambda =0$, then $\nu =0$ and [F1] makes $\eta$ the zero measure almost surely, so both sides vanish. Suppose $\lambda >0$ and write $Q=\nu \mathrm{/}\lambda$, $p_n=e^{-\lambda }{\lambda }^n\mathrm{/}n!$. By [F1], the left side equals ${\sum }_{n\ge 0}{p}_n{\int }_{S^n}{\sum }_{i=1}^{n}h(x_i,{\sum }_{j=1}^n{\delta }_{x_j})Q^{\otimes n}(d(x_1,\dots ,x_n))$.

For each $n\ge 1$, every summand indexed by $i$ has the same integral: coordinate permutation sends the $i$th coordinate to the first, preserves $Q^{\otimes n}$ by [F2], and leaves the sum of Dirac masses unchanged. Tonelli [F3] and $n{p}_n=\lambda p_{n-1}$ therefore turn step 1.2 into $\lambda {\sum }_{m\ge 0}{p}_m{\int }_{S^{m+1}}h(x_1,{\delta }_{x_1}+{\sum }_{j=2}^{m+1}{\delta }_{x_j})Q^{\otimes (m+1)}(d(x_1,\dots ,x_{m+1}))$.

Integrating first in $x_1$ in step 2.1 and using $\nu =\lambda Q$, the result is ${\int }_S{\sum }_{m\ge 0}{p}_m{\int }_{S^m}h(x,{\delta }_x+{\sum }_{j=1}^m{\delta }_{x_j})Q^{\otimes m}(d(x_1,\dots ,x_m))\nu (dx)$. By the mixed-binomial representation [F1], the sum and inner integral are $\mathbb{E}[h(x,\eta +{\delta }_x)]$ for each fixed $x$, so this is the right side of the claimed identity.

Put $K_0=\mathrm{\varnothing }$, $A_1=K_1$, and $A_i=K_i\setminus K_{i-1}$ for $i\ge 2$. By [F6], each $K_i$ is closed, so the disjoint sets $A_i$ are Borel and cover $S$. Since $A_i\subseteq K_i$, countable additivity of the measure gives $\nu (A_i)\le \nu (K_i)<\mathrm{\infty }$, hence ${\nu }_i(B):=\nu (B\cap A_i)$ is finite. If $i>m$, then $A_i\cap K_m=\mathrm{\varnothing }$.

For each $i$, let ${\mathrm{\Omega }}_i$ be the explicit finite-intensity mixed-binomial space from [F2] for ${\nu }_i$ (a one-point space when ${\nu }_i=0$), with law $P_i$ and coordinate counting measure ${\eta }_i$. The countable product theorem [F3], under [A1]-[A2], supplies a probability space on which all these coordinates are independent. Each ${\eta }_i$ is supported on $A_i$ almost surely: its count on $S\setminus A_i$ is Poisson with mean zero by [F2], hence zero almost surely.

Let $E={\bigcap }_{i\ge 1}\{{\eta }_i(S\setminus A_i)=0\}$. Each event in this intersection is measurable by the coordinate evaluation maps and has probability one by step 2.1; countable subadditivity applied to their null complements gives $\mathbb{P}(E)=1$. For Borel $B$, define $\eta (B)={\sum }_{i\ge 1}{\eta }_i(B)$ on $E$ and $\eta (B)=0$ off $E$. On $E$ the countable sum is a measure by [F4], while off $E$ it is the zero measure; its evaluations are measurable as the restriction to the measurable event $E$ of increasing limits of finite sums. On $E$ it is a counting measure because the coordinate construction lists finitely many point locations per cell and the countable union of those lists is countable; off $E$ the empty configuration is a counting measure. On $E$, if $C$ is compact, [F1] gives $C\subseteq K_m$ for some $m$; then $\eta (C)={\sum }_{i=1}^m{\eta }_i(C)<\mathrm{\infty }$, because later cells avoid $K_m$ and each of the first $m$ coordinates contains finitely many points. Off $E$ all compact counts are zero. Thus $\eta$ is a measurable point process with locally finite realizations everywhere, and agrees almost surely with the unmodified superposition, preserving all its count laws.

Fix pairwise-disjoint Borel sets $B_1,\dots ,B_r$ and write $N_{ij}={\eta }_i(B_j)$ and $a_{ij}={\nu }_i(B_j)$. By [F2], within each row $i$ the $N_{ij}$ are independent Poisson variables of means $a_{ij}$; rows are independent by [F3] and [A1]-[A2], so all finitely many variables in the first $n$ rows are independent. For independent $X\sim \mathrm{Pois}(a)$ and $Y\sim \mathrm{Pois}(b)$, direct convolution gives $\mathbb{P}(X+Y=k)=e^{-(a+b)}{\sum }_{l=0}^k{a}^l{b}^{k-l}\mathrm{/}(l!(k-l)!)=e^{-(a+b)}(a+b{)}^k\mathrm{/}k!$ by the binomial theorem. Iterating this identity shows that $N_j^{(n)}={\sum }_{i=1}^n{N}_{ij}$ are independent across $j$ with laws $\mathrm{Pois}(a_j^{(n)})$, where $a_j^{(n)}={\sum }_{i=1}^n{a}_{ij}$.

On the probability-one event $E$ from step 3.1, the counts $N_j^{(n)}$ increase to $N_j=\eta (B_j)$ and $a_j^{(n)}\uparrow a_j:=\nu (B_j)$. For finite $a_j$, the events $\{N_j^{(n)}\le k_j\text{ for every }j\}$ decrease to $\{N_j\le k_j\text{ for every }j\}$; continuity from above follows by applying [F4] to their complementary increasing indicators. Their probabilities, computed from step 3.2, converge to ${\prod }_j{\sum }_{k=0}^{k_j}{e}^{-a_j}{a}_j^k\mathrm{/}k!$. Inclusion-exclusion over the finitely many lower rectangles then gives the product of Poisson point masses for every count vector, proving independence and the desired laws when all $a_j<\mathrm{\infty }$.

If $a_j=\mathrm{\infty }$, then for each integer $M\ge 0$, continuity from above gives $\mathbb{P}(N_j\le M)={\lim }_n{e}^{-a_j^{(n)}}{\sum }_{k=0}^M(a_j^{(n)}{)}^k\mathrm{/}k!=0$. The limit is zero because $e^{-t}{t}^k\to 0$ for each fixed $k$, as follows from $e^t\ge t^{k+1}\mathrm{/}(k+1)!$. Hence $N_j=\mathrm{\infty }$ almost surely. Applying the finite-coordinate argument of step 4.1 to the coordinates with finite $a_j$, while the infinite-mean coordinates are almost surely constant at $+\mathrm{\infty }$, gives the required extended independent-count convention for the whole family.

Steps 3.1, 3.2, 4.1, and 5.1 prove local finiteness and the independent Poisson count laws. The only choice principles used are [A1]-[A2] for the countable product in step 2.1.

For every Borel $B$ with $\nu (B)<\mathrm{\infty }$, the defining Poisson law and the scalar mean identity [F9] give $\mathbb{E}\eta (B)=\nu (B)$; if $\nu (B)=\mathrm{\infty }$, the extended-count convention gives the same equality. Apply [F1] with $\zeta =\eta$ to obtain Campbell's formula for every nonnegative Borel $f$.

First take a nonnegative simple $s={\sum }_{j=1}^m{c}_j{1}_{B_j}$ on disjoint Borel sets, discarding zero coefficients. Independence of the counts and the scalar generating function [F9] give $\mathbb{E}{e}^{-\eta (s)}={\prod }_j\exp (\nu (B_j)(e^{-c_j}-1))=\exp (-\nu (1-e^{-s}))$; when $c_j>0$ and $\nu (B_j)=\mathrm{\infty }$, both sides are zero by the infinite-count convention.

The point-addition map is measurable because each output evaluation is $(\mu +{\delta }_x)(B)=\mu (B)+1_B(x)$, and configuration addition is measurable because $(\mu +\zeta )(B)=\mu (B)+\zeta (B)$; both follow from the evaluation sigma-algebra. Thus $h_{\mu }(x,\zeta ):=h(x,\mu +\zeta )$ is measurable for fixed $\mu$. For a random locally finite counting measure $\zeta$, the map $\omega \mapsto \int h(x,\zeta (\omega ))\zeta (dx,\omega )$ is measurable by applying the kernel-integration theorem to the evaluation kernel with parameter $\zeta$. This makes the random integrals below well-defined.

Choose simple $s_n\uparrow f$ using [F7]. Then $1-e^{-\eta (s_n)}\uparrow 1-e^{-\eta (f)}$ and $1-e^{-s_n}\uparrow 1-e^{-f}$. Monotone convergence [F7] on the probability space and on $(S,\nu )$ passes the simple identity of step 1.2 to the asserted Laplace formula.