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.

Identify an unoriented line not through the origin by $L(p,\theta )=\{z:z\cdot (\cos \theta ,\sin \theta )=p\}$, with $p>0$. A Poisson point process on this parameter space is a random collection of lines. The measure $cdpd\theta$ is invariant under rotations and translations; signed coordinates $t\in \mathbb{R}$, $0\le \theta <\pi$ give the same measure $cdtd\theta$. The number of lines meeting a disk of radius $r$ is Poisson with mean $2\pi cr$.

For an isotropic Poisson line process of intensity $cdpd\theta$, its pair intersections form a locally finite point process. Coincident intersections and parallel pairs have probability zero. Its intensity measure is $\pi c^2$ times area. Indeed the Poisson factorial moment measure and the change of variables from signed distances to intersection location giveThe factor $1/2$ counts unordered line pairs. Shared lines create dependence between intersections; this process is not a Poisson point process.

For $c,r>0$, an intersection inside a disk requires at least two lines to meet that disk. The latter count is Poisson with mean $2\pi cr$, giving the non-strict bound first. It is strict because two lines with nearby normal angles but separated perpendicular distances can both hit the disk while intersecting outside. Choose two positive-measure parameter boxes forcing this configuration; the event of exactly one line in each box and no other disk-hitting line has positive probability. This also yields a void probability too large to equal that of a Poisson point process with the intersection process's intensity measure.

For a Poisson point process with intensity measure $\mu$, ordered tuples of distinct atoms have factorial moment measure ${\mu }^{\otimes k}$. Thus for any nonnegative measurable function $g$,For $k=2$, separating diagonal terms from distinct pairs gives the second moment of a Poisson sum. Dividing by two instead counts unordered pairs, which is essential for geometric intersection intensities.

For a measurable map $f$ and a Poisson random measure of intensity $\nu$, image counts in disjoint sets equal original counts in their disjoint preimages. They are therefore independent Poisson variables with the pushforward intensity. The result is a Poisson random measure even if images collide; diffuse pushforward intensity gives a simple point process.

For unit volume intensity in the unit ball of ${\mathbb{R}}^d$, polar coordinates give intensity $r^{d-1}dr\sigma (d\omega )$. The separate radius and direction images are Poisson processes, but each counts the same random total number of original points. Their count covariance is the displayed intensity of the intersection of their two preimages. Conditional on the total count, radii and directions are independent samples, which does not imply unconditional independence of the two point processes.

Place equal masses at points of a homogeneous Poisson point process of intensity $\rho$ on the line, and truncate their signed inverse-power forces symmetrically at radius $R$. For force $m\mathrm{sign}(x)/|x{|}^p$, with $\alpha =1/p\in (0,2)$, the characteristic exponent tends to $-2\rho {\int }_0^{\infty }(1-\cos (mt/x^p))dx$. The cosine integral for a symmetric stable exponent evaluates it as displayed. Thus the weak limit is a symmetric stable distribution. Symmetric truncation is important when the far-field force series is not absolutely summable.

A Poisson germ-grain model assigns a measurable region to each point of a Poisson point process and takes the union of those regions. A location remains uncovered exactly when no Poisson point has a region containing it, so its coverage probability is determined by the intensity of the corresponding set of germs and marks.

For drops with intensity $\lambda (x,t,v)$, impact time $t$ and nonnegative volume $v$, let their radius at time $\tau$ be $r(\tau -t,v)$. The dry probability at $\xi$ isThe exponent is the intensity of drops whose patch covers the point by that time. An infinite exponent means zero dry probability. No spatial-homogeneity assumption is needed.

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 nonnegative measurable $H$ and a Poisson random measure of sigma-finite intensity $\rho$, removing the selected point from a sum leaves an independent copy of the original process under the integral. For finite intensity, condition on the Poisson count, represent the points as independent samples from normalized $\rho$, and shift the count index using $n/n!=1/(n-1)!$. For sigma-finite intensity, first select points in a finite-intensity set, condition on the independent outside process and then exhaust the space. Repetition gives the corresponding formula for ordered distinct tuples of points.

A sigma-finite measure on a standard Borel space is the intensity measure of a point process with Poisson counts. Partition the space into sets of finite intensity, draw an independent Poisson count on each piece, and conditional on that count draw independent points from its normalized intensity measure. The sum of their Dirac measures is the required Poisson random measure. Non-atomic intensity gives simplicity; local finiteness of the intensity gives local finiteness of the point configuration.

For deterministic square-integrable $f$ and a Poisson random measure of intensity $\nu$, first define the integral for simple functions on disjoint finite-intensity sets. Independence and the equality of Poisson distribution mean and variance prove the displayed isometry. Completion defines the integral for all $f\in L^2(\nu )$, even when $\int |f|d\nu =\infty$. A small-jump compensated martingale is thereby defined through convergence in Lp, not by subtracting two divergent ordinary integrals.

For nonnegative measurable $f$, independent Poisson counts first prove this formula for simple functions on disjoint measurable sets. Monotone approximation gives the general formula, allowing infinite integrals with $e^{-\infty }=0$. Conversely, the formula evaluated at $f={\sum }_j{u}_j{1}_{A_j}$ for disjoint finite-intensity sets gives the joint Laplace transform of independent Poisson counts, so characterizes the law of a Poisson random measure.

Let $M$ be a Poisson random measure of intensity $\mu$, let $\mu (A)<\infty$, and let $f\ge 0$ be measurable. Apply the Laplace functional of a Poisson random measure to $f+t{1}_A$ and differentiate from the right at $t=0$. On the random side, differentiation is dominated by the integrable random variable $M(A)$; on the intensity side, by $1_A$. This proves the displayed identity when $\int (1-e^{-f})d\mu$ is finite. When this integral is infinite, $e^{-M(f)}=0$ almost surely, and the same identity holds with both sides zero.

For a Poisson random measure on ${\mathbb{R}}^d$ with Lebesgue measure intensity and a nonnegative compactly supported continuous function $f$, put $m_r=\mu (B(0,r))$ and $a_r={\int }_{B(0,r)}{e}^{-f}d\mu$. Continuity gives $a_r/m_r\to e^{-f(0)}$. With $L=\mathbb{E}{e}^{-M(f)}$, the count-weighted Laplace functional of a Poisson random measure and independent inside/outside counts give respectivelyBoth converge to the displayed value. The conditioning is meaningful for every $r>0$; it is not conditioning on a positive-probability point at the origin.

A simple locally finite point measure on $[0,\infty )$ with void probability $e^{-\lambda (A)}$ on finite unions of bounded intervals is a Poisson random measure with Lebesgue measure intensity. Void-event intersections factor, hence the occupation indicators on disjoint cells are independent. The dyadic occupancy approximation of a simple point measure is binomial with rare-event probability $1-e^{-2^{-n}}$ and converges to a Poisson count of mean the interval length. Independence survives joint convergence for disjoint intervals. Extension of the mean measure and the Laplace functional of a Poisson random measure to Borel functions identifies all Borel counts, not just intervals.

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 $\int g^{2}d\mu$.

The expectation identity holds for a general point process with intensity measure $\mu$ whenever the sum is integrable. The stated variance identity uses the additional Poisson point process assumption.

For a Poisson random measure $\Pi$ with intensity $\mu$ and measurable nonnegative $h$, $\mathbb{E}\int hd\Pi =\int hd\mu$, allowing both sides to be infinite. For signed $h$, the same identity holds if $\int |h|d\mu <\infty$. This first-moment statement does not require square integrability. Applying it to a nonnegative volume mark gives the expected total volume in a marked Poisson model.

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.

If an intensity function of a point process has $\lambda (x)\sim c(b-x{)}^{-\alpha }$ at a finite endpoint, with $c>0$ and $\alpha >1$, the cumulative intensity satisfies $F(x)\sim c(b-x{)}^{1-\alpha }/(\alpha -1)$. The Poisson mapping theorem turns $F(X_n)$ into unit-rate arrivals. The strong law for Poisson arrival times gives $F(X_n)/n\to 1$ almost surely, hence the displayed accumulation rate. Reversing the coordinate gives the same argument at a left endpoint. Different singularity exponents produce different powers of the index.

For intensity $x^{-2}(1-x{)}^{-1}$ on $(0,1)$, the cumulative-intensity maps based at $1/2$ areThe left and right images are independent unit-rate Poisson processes. Labeling the original points by $X_{-k}<1/2<X_n$, with $k\ge 1,n\ge 0$, gives $L(X_{-k})/k\to 1$ and $F(X_n)/n\to 1$. Since $xL(x)\to 1$ at zero and $F(x)=1-\log (1-x)+o(1)$ at one, $k{X}_{-k}\to 1$ and $\log (1-X_n)/n\to -1$.

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

Attach conditionally independent nonnegative marks to a Poisson point process, with a measurable mark law depending only on location. If a mark at $z$ has mean $m(z)$ and variance $v(z)$, and the displayed integrals are finite, their total $W$ is finite almost surely and has the displayed moments. The Campbell first-moment formula gives the first moment and finiteness. The Poisson factorial moment measure formula gives the square of that first moment for distinct pairs; the diagonal contributes $\int (v+m^2)d\mu$. Their difference is the variance. Signed marks require absolute first-moment integrability.

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.

For a finite-intensity set of mean $m$, independent red marking with probability $r$ gives counts $R,G$ withDisjoint source sets give independent pairs of these counts. Refining overlapping test sets into disjoint cells then proves independence of the two whole Poisson point processes, not only their counts on one set.