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Independently assigning a mark to each point of a Poisson point process, with mark probabilities allowed to depend measurably on the point, produces independent Poisson point processes for the different marks. If the original intensity is $\lambda (x)dx$ and mark $j$ has probability $p_j(x)$, its intensity is $\lambda (x)p_j(x)dx$.

The mapping theorem for Poisson point processes applies provided $f$ is a measurable function and its pushforward measureis locally finite, meaningfor every bounded Borel set $B\subseteq {\mathbb{R}}^s$. If “Poisson process” means a simple point process, one also requires $f_{\ast }\Lambda$ to be atomless, equivalentlyUnder these conditions, $f(\Pi )$ is a Poisson point process with mean measure

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

A Poisson random measure $N$ of intensity $\mu$ assigns to disjoint measurable sets independent random variables, andwhenever $\mu (A)<\infty$, with the usual countable-additivity requirement.

Let $N$ be the homogeneous Poisson point process in ${\mathbb{R}}^3$ with intensity measure $\lambda dx$, and map each point $x$ to its radius $g(x)=|x|$. For an interval $(a,b]\subseteq {\mathbb{R}}_{+}$, the pushforward measure of the intensity isThe Mapping theorem for Poisson point processes therefore shows that the radii form a Non-homogeneous Poisson point process with intensity function

If $R$ is the distance to the closest star, the event $\{R>r\}$ says that the ball of radius $r$ contains no points. The zero-count probability of a Poisson distribution givesDifferentiating its cumulative distribution function yields the probability density function

Campbell's formula for a Poisson point process givesExpanding $S^2$ separates equal and distinct Poisson points. The factorial-moment formula givesSubtracting the square of the mean yields

A Poisson point process of intensity $\lambda (x)$ has independent counts on disjoint Borel sets, with $\Pi (A)\sim \mathrm{Poisson}({\int }_A\lambda (x)dx)$ whenever the integral is finite.