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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.

The same intensity measure can belong to very different point processes. A Poisson point process with a locally finite intensity measure $\mu$ adds two requirements: each finite-intensity window count has the Poisson distribution of mean $\mu (W)$, and counts in pairwise disjoint observation windows are mutually independent. These requirements determine its probability law; they do not follow from the formula for $\mu$.^[ref]

A randomly shifted lattice point process, a unit-rate Poisson point process, and the shared random rate gives dependent point counts construction all satisfy $\mu (A)=|A|$ on the real line. Nevertheless, their counts in $[0,1)$ have respectively variance zero, one, and two. Counts in the disjoint observation windows $[0,1/2)$ and $[1/2,1)$ have respectively covariance $-1/4$, zero, and $1/4$. These exact calculations show why measuring mean density cannot identify a spacing or statistical dependence model.

Observation windows are measurable sets used to count points of a point process. They are pairwise disjoint when $W_i\cap W_j=\varnothing$ for every $i\ne j$. Half-open intervals such as $[0,1)$ and $[1,2)$ meet at a boundary without sharing any point, whereas the closed intervals $[0,1]$ and $[1,2]$ both contain one. A disjoint union makes window counts additive; it does not make their random variables independent.

An elementary outcome is one member of the sample space. For the ordered two-toss model, $HT$ is one outcome; it contains both toss results. By contrast, “at least one head” is an event containing three outcomes. The outcome $HT$ and the singleton set $\{HT\}$ are different objects: when that singleton is measurable, the notation $\mathbb{P}(\{HT\})$ gives its probability. In the finite model below, every subset is measurable.

The word elementary refers to the chosen level of detail. An outcome can be an ordered pair, a sequence, or an entire realization of a point process. A random variable then extracts a numerical feature of that outcome; the window count of a point process is one example.

For a homogeneous isotropic point process of galaxies with mean number density $\bar{n}$, the galaxy two-point correlation function gives the excess pair probability at distinct positions. Dividing the factorial pair density by the probability of a central galaxy gives conditional neighbouring-galaxy density $\bar{n}[1+\xi (r)]$. Choosing a central location uniformly in volume does not perform this conditioning.

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.

Here is an explicit example even within simple point processes. Let $K$ be a positive integer with probability mass function $\mathbb{P}(K=k)=1/[k(k+1)]$. The telescoping series sums to one, $K$ is finite almost surely, but $\mathbb{E}[K]={\sum }_{k\ge 1}1/(k+1)=\infty$. Conditional on $K$, sample $K$ independent points with the uniform distribution on $(0,1)$. The resulting point process has finitely many distinct points almost surely, while $\mu ((0,1))=\infty$. Thus finite counts in each realization and finite mean counts are separate requirements; local finiteness of the intensity measure is an integrability assumption.

Identify a line with its unique parameter pair in $E=(0,\infty )\times [0,2\pi )$, usingA Poisson line process means a Poisson point process on the Borel parameter space $E$: for each measurable $B$ of finite mean measure, the number of its parameter points is Poisson with mean $\mu (B)$, and counts in finitely many disjoint sets are independent. Infinite-measure sets are handled by a countable finite-measure exhaustion. Here $\mu (dp,d\theta )=dpd\theta$ is sigma-finite and nonatomic, so there are countably many distinct lines almost surely. It is not an assertion that line counts are Poisson with respect to planar area.

Fix $r>0$ and let $D_r=\{z:|z|<r\}$. Exactly the lines with $0<p<r$ hit this disk, apart from the zero-probability tangent case. Thus the count of disk-hitting lines isEvery intersection in $D_r$ uses two of these lines. Writing $N_r=\#(\Phi \cap D_r)$, we haveThis also proves local finiteness of the intersection point process. Initially it gives a non-strict bound $\mathbb{P}(N_r\ge 1)\le 1-(1+2\pi r)e^{-2\pi r}$.

To establish strictness, choose the two parameter boxesEach has positive measure, and both lie among the disk-hitting lines. If one line from each box intersected at $z\in D_r$, thenBut their distance parameters differ by more than $r/4$, a contradiction. The event of exactly one line in each box and no other disk-hitting lines has probabilityOn this event $M_r=2$ but $N_r=0$. ThereforeThe following original diagram illustrates why two disk-hitting lines need not produce an intersection inside the disk.

To decide whether $\Phi$ is Poisson, compute its intensity measure and compare its void probability. Use the equivalent signed line coordinates $L(t,\theta )$ with $t\in \mathbb{R}$ and $0\le \theta <\pi$. If $t>0$ this is the original pair $(t,\theta )$; if $t<0$ it is $(-t,\theta +\pi )$. The measure is $dtd\theta$, with no extra factor of two. Translation by $a$ sends $t$ to $t+a\cdot n_{\theta }$, preserving this measure. Thus both the line and intersection processes are stationary.

Almost surely no two lines are parallel and no three are concurrent. In each finite-intensity parameter region, conditional line parameters have continuous joint densities. Parallel pairs and concurrent triples satisfy lower-dimensional equations and have probability zero; a countable exhaustion proves the assertion for all lines. Hence different unordered pairs have distinct intersections almost surely. For a bounded Borel set $A$, the second Poisson factorial moment measure identity givesFor nonparallel normals, their intersection $z$ satisfies $t=z\cdot n_{\theta }$, $s=z\cdot n_{\varphi }$. The absolute Jacobian determinant of this transformation is $|\sin (\varphi -\theta )|$. Consequentlysince the inner integral is $2$ for every $\theta$. In particular $\mathbb{E}{N}_r={\pi }^2{r}^2$.

If $\Phi$ were a Poisson point process, the count $N_r$ would therefore be Poisson with that mean, giving $\mathbb{P}(N_r=0)=e^{-{\pi }^2{r}^2}$. But the earlier necessary-two-lines argument givesAt $r=2/\pi$ this lower bound is $5e^{-4}$, whereas the supposed Poisson void probability is $e^{-4}$. This contradiction proves $\Phi$ is not a Poisson point process. Its points share parent lines, and the explicit void/mean comparison demonstrates the resulting dependence rather than merely asserting it.

A point process records random locations by a random measure that counts them. Examples include arrival times of a queue, positions of stars, and the finite eigenvalue configuration of a random matrix. The count in an observation window is a random variable; its expected value is supplied by the intensity measure of a point process, while its fluctuations and relationships with other windows require more information.

For a precise spatial convention, take $E={\mathbb{R}}^d$ with its Borel sigma-algebra and a probability space $(\Omega ,\mathcal{F},\mathbb{P})$. A locally finite point process $\Pi$ is a measurable function from $\Omega$ to the space ${\mathcal{N}}_{\mathrm{lf}}(E)$ of locally finite point measures, equipped with the sigma-algebra generated by evaluations $\nu \mapsto \nu (A)$ for Borel sets $A$. Each realization has the formwhere $J(\omega )$ is finite or countable and repeated locations are allowed; ${\delta }_x$ is a Dirac measure. Here a point measure is a counting measure with possible multiplicities, rather than the counting measure of a set of distinct locations.^[ref]

Local finiteness means ${\Pi }_{\omega }(B)<\infty$ for every bounded Borel set $B$, on a single almost sure event. In ${\mathbb{R}}^d$ it suffices to check the countable collection of balls of integer radius centered at the origin, and this is equivalent to finiteness on every compact set. On a general measurable space there is no notion of boundedness without additional structure: state a metric or an explicit family of observation windows. Another convention on locally compact spaces requires finiteness on compact sets; this need not coincide with finiteness on all bounded sets outside ${\mathbb{R}}^d$. Local finiteness does not require finitely many points in all of $E$.^[ref]

A point process is simple when, on a single almost sure event, every location has multiplicity at most one. Its realizations are simple locally finite point measures. For instance, ${\delta }_{0.4}+2{\delta }_{1.1}+{\delta }_{1.8}$ is locally finite but not simple: it has four points counted with multiplicity at three distinct locations. Simplicity is a statement about each entire realization. Checking $\mathbb{P}(\Pi (\{x\})>1)=0$ separately for every fixed $x$ is insufficient: $2{\delta }_U$ with $U$ having the uniform distribution on $(0,1)$ passes every such fixed-location check, yet always has a double point.^[ref]

A point process on ${\mathbb{R}}^d$ is stationary if every translation preserves its probability law. If its mean count in a unit cube is finite, its intensity measure is $\mu (A)=\rho |A|$ for some $\rho \ge 0$, where $|A|$ denotes Lebesgue measure. This follows because stationarity makes the mean measure invariant under translations. A constant intensity function of a point process describes only these mean counts, and by itself asserts neither stationarity of the entire law nor independence.^[ref]

Thinning removes points from a point process according to a retention rule. Independent retention with probability $p(x)$ changes the intensity measure to $p(x)\mu (dx)$. For a Poisson point process, the retained and removed point processes are independent Poisson processes.

For a bounded Borel set $W\subset {\mathbb{R}}^d$, the window count $N_W$ of a locally finite point process is a finite nonnegative integer random variable. It counts multiplicities, and the convention for window boundaries matters. On the time half-line, $N(t)=\Pi ((0,t])$ is a counting process and $N(t)-N(s)=\Pi ((s,t])$. For pairwise disjoint observation windows $W_1,\dots ,W_m$, the measure identityis valid for every realization. Additivity of counts supplies no independence assumption.