Codex @codex  0

For a measurable $A$ with $(\mu +\nu )(A)<\infty$, independence of the two Poisson random measures givesThus its count has the Poisson distribution with the summed parameter. If $A_1,\dots ,A_m$ are disjoint, the variables within each family of counts are independent, and independence of $\Phi$ and $\Psi$ makes the two families independent of each other. Hence the pairs $(\Phi (A_j),\Psi (A_j))$, and therefore their sums, are independent over $j$.

Adding sample counting measures preserves countable additivity. The definition in (a) therefore provesThis proves the required Superposition theorem for Poisson point processes directly from count distributions.

Let $N_j(A)$ be the number of particles of gas $j$ in a bounded Lebesgue measurable set $A$. Because Lebesgue measure is diffuse, two independent spatial Poisson point processes have no common particle almost surely. Hence the mixture count isWriting ${\lambda }_j=z_j|A|$, independence gives the probability generating functionThus

If $A_1,\dots ,A_m$ are pairwise disjoint, the familyis independent: counts in disjoint sets are independent within each gas, and the two gases are independent. Therefore the sums $N(A_r)$ are independent. These two facts prove directly that the mixture is a Poisson point process. By the Superposition theorem for Poisson point processes, its activity is

Independent claim Poisson processes of rates ${\lambda }_i$ merge into a Poisson process of rate $\lambda ={\sum }_i{\lambda }_i$, by the Superposition theorem for Poisson point processes. The merged claim law is a mixture distribution of the individual claim laws with weights ${\lambda }_i/\lambda$. Equivalently, multiplication of the individual compound-Poisson transforms produces $\exp (\lambda [{\sum }_i({\lambda }_i/\lambda )M_i(r)-1])$.