Poisson point process

A Poisson point process has independent counts on disjoint sets and Poisson count in each set with mean given by its intensity measure.

A Poisson random measure $N$ with intensity measure $\mu$ assigns independent Poisson random variables to disjoint measurable sets, with $N(A)$ having mean $\mu (A)$. Its compensated version is $\tilde{N}=N-\mu$ on sets of finite intensity.

For unit-rate Poisson process arrivals $Z_i$, $X_p$ is finite almost surely when $p>1$. There are finitely many positive arrivals before time one; the expected contribution afterwards is ${\int }_1^{\infty }{z}^{-p}dz<\infty$, by the Poisson expectation identity proved on simple functions and extended by monotone convergence. Merging independent processes of rates $1$ and $\rho$ produces rate $1+\rho$; rescaling their arrivals by $1+\rho$ restores unit rate. Hence the sum of their inverse-power contributions has law $(1+\rho {)}^p{X}_p$. In particular two independent unit-rate inverse-square sums add to a variable with law $4X_2$.

For a Poisson random measure on $E$ with finite intensity $\nu (E)$, its integral of a finite-valued measurable mark function is the finite sum of the marks at its random atoms. Its characteristic function is $\mathbb{E}\exp (iu\int fdN)=\exp (\int (e^{iuf}-1)d\nu )$. On a time-mark space with intensity $ds\nu (dz)$ and finite mark intensity, restriction to each bounded time interval has finite intensity and defines a Compound Poisson process. Zero-valued marks do not affect the integral and need not appear in its Lévy measure.

The union of independent Poisson point processes with diffuse intensity measures ${\mu }_1,\dots ,{\mu }_m$ is a Poisson point process with intensity measureOn every finite collection of disjoint measurable sets, this follows by adding independent Poisson counts and checking that the resulting counts remain independent.

For a Poisson point process $\Pi$ with intensity measure $\mu$ and a square-integrable measurable function $g$,The second identity follows because distinct points contribute the square of the mean, while the diagonal contribution is $\int g^{2}d\mu$.

Let $\lambda$ be a nonnegative measurable intensity function on a space with reference measure $dx$. A Non-homogeneous Poisson point process has independent counts on disjoint measurable sets andwhenever the integral is finite.

The intensity function $\lambda$ is the density of the expected counting measure: $\mathbb{E}[N(A)]={\int }_A\lambda (x)dx$.

Let $\Pi$ be a Poisson point process with intensity measure $\mu$ and let $f$ be measurable. If the pushforwardis locally finite, then the image counting measure is Poisson with intensity $f_{\ast }\mu$. If one requires a simple point process, the pushforward must also be diffuse, so distinct source points do not produce coincident image points with positive probability.

For a homogeneous Poisson point process of intensity $\lambda$ in ${\mathbb{R}}^d$, mapping $x$ to $v_d|x{|}^d$, where $v_d$ is the unit-ball volume, gives a rate-$\lambda$ Poisson process on $[0,\infty )$. Mapping to $|x{|}^d$ instead gives rate $\lambda v_d$.

For intensity $\lambda$ in ${\mathbb{R}}^d$, let $R_k$ be the distance to the $k$th closest point and put $\rho =\lambda v_d$. Then $R_k^d$ is the $k$th arrival time of a rate-$\rho$ Poisson process, so

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 $\lambda (x)p_j(x)dx$.

Independently retaining a point at $x$ with probability $p(x)$ thins a Poisson point process of intensity $\lambda (x)$ to one of intensity $p(x)\lambda (x)$; retained and rejected processes are independent.