Campbell, Laplace, Mecke, and Cox moment formulas

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.

Let $(K_n)$ be the compact exhaustion [F3], set $A_1=K_1$ and $A_i=K_i\setminus K_{i-1}$ for $i\ge 2$, and define ${\eta }_i(B)=\eta (B\cap A_i)$ and ${\eta }_{-i}(B)=\eta (B\setminus A_i)$. The sets $A_i$ are Borel by [F3]. Each ${\eta }_i$ is a finite-intensity Poisson process with intensity ${\nu }_i(B)=\nu (B\cap A_i)$, since $A_i\subseteq K_i$ and $\nu (K_i)<\mathrm{\infty }$. The sigma-algebras generated by evaluations of ${\eta }_i$ and ${\eta }_{-i}$ are independent: their generating cylinder events concern disjoint regions, so [F3]-[F4] extend the independent-count property to those sigma-algebras. Also $\eta ={\eta }_i+{\eta }_{-i}$.

For fixed $i$, integrate first over the independent laws of ${\eta }_{-i}$ and ${\eta }_i$. Applying the finite-intensity identity [F2] to $h_{\mu }(x,\zeta )=h(x,\mu +\zeta )$ for each fixed background $\mu$ gives $\mathbb{E}\int h(x,\eta ){\eta }_i(dx)=\int \mathbb{E}[h(x,\eta +{\delta }_x)]{\nu }_i(dx)$. Summing over $i$ and using Tonelli [F4] yields the add-one identity, since ${\sum }_i{\eta }_i=\eta$ and ${\sum }_i{\nu }_i=\nu$.

Now let $\eta$ be Cox directed by $\xi$ and use its supplied kernel $K$. For $P_{\xi }$-almost every $\mu$, $K(\mu ,\cdot )$ is a Poisson law of intensity $\mu$; [F6] and step 2.1 therefore give $\int \zeta (f)K(\mu ,d\zeta )=\mu (f)$, $\int e^{-\zeta (f)}K(\mu ,d\zeta )=e^{-\mu (1-e^{-f})}$, and $\int \zeta (f{)}^{2}K(\mu ,d\zeta )=\mu (f^2)+\mu (f{)}^2$. The conditional-law identity [F8] and nonnegative kernel integration average these formulas over $P_{\xi }$.

For nonnegative measurable $g$, define $H(x,\mu )=1_D(x,\mu )g(x,\mu -{\delta }_x)$ on $D$ and $H=0$ off $D$. This is measurable by [F5]. For $\eta$-almost every counted point $x$, $(x,\eta )\in D$, while $(x,\eta +{\delta }_x)\in D$ for every $x$; applying step 3.1 to $H$ gives $\mathbb{E}\int g(x,\eta -{\delta }_x)\eta (dx)=\int \mathbb{E}[g(x,\eta )]\nu (dx)$. A mass $m$ at $x$ contributes $m$ times the configuration with mass $m-1$, so the formula retains multiplicity.

By [F1], $\mathbb{E}\xi (f)=\nu (f)$ and $\mathbb{E}\xi (f^2)=\nu (f^2)$. Integrating step 3.2 gives $\mathbb{E}\eta (f)=\nu (f)$, $\mathbb{E}{e}^{-\eta (f)}=\mathbb{E}{e}^{-\xi (1-e^{-f})}$, and $\mathbb{E}\eta (f{)}^2=\nu (f^2)+\mathbb{E}\xi (f{)}^2$, all with extended nonnegative values.

Conversely, suppose the deletion identity holds for every nonnegative measurable $g$. For any such $h$, set $g(x,\mu )=h(x,\mu +{\delta }_x)$, which is measurable by [F5]. Then $g(x,\eta -{\delta }_x)=h(x,\eta )$ at every counted point and $g(x,\eta )=h(x,\eta +{\delta }_x)$ on the right, so the add-one identity follows.

Fix disjoint Borel sets $B_1,\dots ,B_r$ with finite $a_j=\nu (B_j)$ and write $N_j=\eta (B_j)$. Applying the add-one identity of step 5.1 to $h(x,\mu )=1_{B_i}(x)$ gives $\mathbb{E}{N}_i=a_i$, so each $N_i$ is finite almost surely. For a count vector $k\in {\mathbb{N}}_0^r$, put $p_k=\mathbb{P}(N_1=k_1,\dots ,N_r=k_r)$. Use $h_i(x,\mu )=1_{B_i}(x)1_{\{\mu (B_j)=k_j\mathrm{\forall }j\}}$ in the same identity. Disjointness makes insertion of $x\in B_i$ increase only the $i$th count, so $k_i{p}_k=a_i{p}_{k-e_i}$ for $k_i\ge 1$, and the right side is zero when $k_i=0$.

Iterating the recurrences of step 6.1 gives $p_k=p_0{\prod }_{j=1}^r{a}_j^{k_j}\mathrm{/}{k}_j!$, with the convention $0^0=1$. Since the probabilities sum to one and each scalar Poisson mass sequence is normalized [F9], finite product summation gives $p_0=e^{-{\sum }_j{a}_j}$ and $p_k={\prod }_j{e}^{-a_j}{a}_j^{k_j}\mathrm{/}{k}_j!$. Hence the counts on every finite family of disjoint finite-intensity sets are independent Poisson variables.

If $B$ has $\nu (B)=\mathrm{\infty }$, let $C_n=B\cap K_n$. Then $\nu (C_n)<\mathrm{\infty }$ and $\nu (C_n)\uparrow \mathrm{\infty }$. For each integer $M\ge 0$, $\mathbb{P}(\eta (B)\le M)\le e^{-\nu (C_n)}{\sum }_{k=0}^M\nu (C_n{)}^k\mathrm{/}k!\to 0$, because $e^{-t}{t}^k\to 0$ as $t\to \mathrm{\infty }$ (use $e^t\ge t^{k+1}\mathrm{/}(k+1)!$). Thus $\eta (B)=\mathrm{\infty }$ almost surely. For any finite disjoint family, infinite-mean counts are therefore constants $+\mathrm{\infty }$ and the finite-mean coordinates have the independent laws from step 7.1; this is the defining extended Poisson count convention.

A sum of two nonnegative extended numbers is finite exactly when both summands are finite, so step 4.2 gives $\mathbb{E}\eta (f{)}^2<\mathrm{\infty }$ exactly when $\nu (f^2)<\mathrm{\infty }$ and $\mathbb{E}\xi (f{)}^2<\mathrm{\infty }$. In that case $\eta (f)$ and $\xi (f)$ are square-integrable; [F9] and step 4.2 yield $\mathrm{Var}(\eta (f))=\nu (f^2)+\mathrm{Var}(\xi (f))$.