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For $r>0$, let ${\kappa }_r$ be the probability law of a seed displacement from its tree. For the literal circular support, it is normalized arclength on $\{u:|u|=r\}$. The conditional mean measure of a point process isWe prove the conditional Poisson assertion directly by joint probability generating functions.

Fix disjoint bounded measurable sets $A_1,\dots ,A_k$. For a tree at $z$, put $q_j(z)={\kappa }_r(A_j-z)$ and $q_0(z)=1-{\sum }_j{q}_j(z)$. Conditional on its seed count, the allocation among these sets and their complement is multinomial distribution. Averaging over the Poisson seed count givesThus this tree contributes independent Poisson counts of means $\mu q_j(z)$ to the disjoint sets. Different trees contribute independently. Only trees within distance $r$ of the bounded union of the $A_j$ can contribute, and there are almost surely finitely many of them. Multiplying their probability generating functions givesThis is precisely the joint probability generating function of independent Poisson counts with the displayed means. Also ${\Lambda }_{\Pi }$ is locally finite, because each bounded set can receive seeds from only finitely many trees. This proves the conditional Poisson point process assertion and the Poisson offspring clusters directed by their parent process construction.

In arclength notation the directing measure is ${\Lambda }_{\Pi }=\mu {\sum }_{z\in \Pi }{\sigma }_{z,r}/(2\pi r)$, where ${\sigma }_{z,r}$ is arclength on the circle centered at $z$. If “circle” is used to mean the filled disc instead, take ${\kappa }_r(du)=1_{\{|u|\le r\}}du/(\pi r^2)$; thenThe generating-function proof and the following unconditional conclusion hold for either reading. The circular version has a singular conditional directing measure, giving the random directing measure of a Cox process formulation.

Unconditionally this is a Cox process, specifically a Poisson-offspring Neyman-Scott process. Its mean measure of a point process is nevertheless spatially uniform. With $q_A(z)={\kappa }_r(A-z)$, the Campbell first-moment formula and translation invariance givesince $\int q_A(z)dz=\int {\kappa }_r(du)\int 1_A(z+u)dz=|A|$. Uniform mean alone does not establish a Poisson law.

For a bounded set $A$ of positive area, the conditional count has mean and variance ${\Lambda }_{\Pi }(A)$. The law of total variance givesA Poisson integral of a deterministic function $q$ has variance $\lambda \int q^{2}dz$: for a simple function on disjoint sets this follows from independent Poisson counts, and square-integrable approximation gives the general formula. Here $0\le q_A\le 1$ and it has bounded support, soThe strict inequality holds when $\lambda ,\mu >0$, because $\int q_A=|A|>0$ implies $\int q_A^2>0$. A Poisson random variable has equal mean and variance. Hence the unconditional process is not a Poisson process in the nontrivial model. If $\lambda =0$ or $\mu =0$, there are no seeds and the process is the degenerate empty Poisson process. The overdispersion of nondegenerate mixed Poisson counts quantifies the additional clustering induced by the random parents.