Poisson and Cox counts with the same mean on a finite window

Let $P_0$ be the zero-process law and let $P_{2\mu }$ be the finite-intensity Poisson law of intensity $2\mu$, both supplied by [F1]. On the disjoint union of a $P_0$ branch and a $P_{2\mu }$ branch, assign probability $1\mathrm{/}2$ to each branch, set $Z=0$ or $2$ according to the branch, set $\xi =Z\mu$, and use the branch process as ${\eta }_C$. For a locally finite measure $\rho$, define $K(\rho ,\cdot )=P_{2\mu }$ when $\rho (S)=2$ and $K(\rho ,\cdot )=P_0$ otherwise. Since $\rho \mapsto \rho (S)$ is an evaluation, this is a measurable probability kernel; on the support $\{0,2\mu \}$ of $\xi$, it is the conditional law of ${\eta }_C$ and is Poisson with intensity $\xi$. Thus [F3] verifies the Cox property.

Let ${\eta }_P$ have the finite-intensity Poisson law of intensity $\mu$ from [F1]. Since $\mathbb{E}Z=1$, for every Borel $B$, $\mathbb{E}\xi (B)=\mathbb{E}[Z]\mu (B)=\mu (B)$, equal to the deterministic intensity measure of ${\eta }_P$.

By [F1]-[F2], $N_P={\eta }_P(W)\sim \mathrm{Pois}(1)$, so $\mathbb{E}{N}_P=1$ and $\mathrm{Var}(N_P)=1$. Conditional on $Z=z$, $N_C={\eta }_C(W)$ is $\mathrm{Pois}(z)$; hence $\mathbb{E}{N}_C=\mathbb{E}Z=1$ and $\mathbb{E}{N}_C^2=\mathbb{E}[Z+Z^2]=\frac{1}{2}(0)+\frac{1}{2}(2+4)=3$. Therefore $\mathrm{Var}(N_C)=3-1^2=2$.

The deterministic and directing mean measures agree by step 2.1, while steps 2.2 show equal expected window counts and different count variances. The difference comes from the random intensity $Z$: conditional Poisson variation contributes $\mathbb{E}Z=1$, and variation in $Z$ contributes $\mathrm{Var}(Z)=1$.