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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.

The intensity measure of a point process $\Pi$ is its expected counting measure:It may take the value $+\infty$.^[ref] Countable additivity follows by applying the monotone convergence theorem to the counts of finite unions of pairwise disjoint observation windows. If $\mu$ is locally finite and absolutely continuous with respect to Lebesgue measure, its Radon-Nikodym derivative is the intensity function of a point process $\lambda$:This is an expected number per unit volume, not a probability density function for a randomly selected point. For example, a single deterministic point has $\mu ={\delta }_0$, a Dirac measure, and no Lebesgue measure density. The density representation requires its own assumption.^[ref]

For a bounded Borel set $W$, $\mathbb{E}[N_W]=\mu (W)$, but local finiteness alone does not ensure that this expected value is finite. Conversely, $\mu (W)<\infty$ implies $N_W<\infty$ almost surely, even when $W$ is unbounded. A constant finite intensity function of a point process on all of ${\mathbb{R}}^d$ has infinite total intensity measure, although every bounded window has finite expected value.

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]

Let $T_n=\inf \{t\ge 0:N_t\ge n\}$ be the $n$th arrival time, and set $T_0=0$. In a Homogeneous Poisson process of rate $\lambda >0$, the gaps $W_n=T_n-T_{n-1}$ are independent and identically distributed random variables with an exponential distribution:Here $\mathrm{Exp}(\lambda )$ uses the rate convention. Under the scale convention $\theta =1/\lambda$, the same probability density function is ${\theta }^{-1}{e}^{-w/\theta }$; a simulation function accepting a scale must receive $1/\lambda$. Conversely, summing independent rate-$\lambda$ exponential random variables to form $T_n$ and counting those sums constructs a Homogeneous Poisson process.^[ref]

The first-gap formula follows immediately from $\{W_1>w\}=\{N_w=0\}$ and the Poisson distribution. Independence of all successive gaps is the additional interval-theorem assertion; deterministic-time independent increments alone should not simply be substituted at random arrival times without justification. The memorylessness of the exponential distribution givesAt the toy rate of two per minute, the mean gap is half a minute, the median is $(\log 2)/2\approx 0.346574$ minutes, and the probability of waiting longer than half a minute is $e^{-1}\approx 0.367879$. Here $\log$ is the natural logarithm. Already having waited a minute without an arrival leaves this same conditional probability of waiting another half minute. The exponential distribution describes a model with no aging of the remaining wait; a scheduled arrival with a fixed gap has no such property.

Poisson thinning splits a Poisson process by independently marking each arrival as retained or rejected. Let the original Homogeneous Poisson process have rate $\lambda >0$, and let $p(s)\in [0,1]$ be a deterministic measurable function of time. Attach independent uniform distributions $U_1,U_2,\dots$ on $(0,1)$, independent of the entire original Poisson process. At arrival time $T_i$, retain the arrival if $U_i\le p(T_i)$. Given all arrival times, these decisions are independent and have the specified retention probabilities. Define the two counting processes using an indicator function:The Poisson thinning theorem gives independent retained and rejected Inhomogeneous Poisson processes. Their rates are $\lambda p(s)$ and $\lambda (1-p(s))$, respectively, in the sense of the intensity function of a point process. In particular, for $0\le a<b$,The independence is of the two whole processes: every finite collection of retained counts is independent of every finite collection of rejected counts. It is not merely zero covariance for one interval. For constant $p$, the streams are Homogeneous Poisson processes of rates $\lambda p$ and $\lambda (1-p)$; the endpoints $p=0,1$ include the identically zero stream.^[ref]

For an exact toy calculation, take $\lambda =2$ per minute and retain with $p(s)=1/4$ on $(0,2]$ and $p(s)=3/4$ on $(2,4]$. In the first two minutes, the retained and rejected expected values are $2\cdot 2\cdot 1/4=1$ and $2\cdot 2\cdot 3/4=3$. In the next two minutes they are $3$ and $1$. Over all four minutes, $R_4,D_4$ are independent Poisson distributions with mean $4$ each, and their sum $N_4$ has mean $8$. ThusThese probabilities and expected values describe the model, not empirical measurements. In the diagram, the original nine synthetic arrivals split into five retained and four rejected arrivals; those realized counts need not equal their expected values.

The mechanism explains why mutually exclusive labels can yield independent counts: randomness of the original total offsets the dependence from allocating a fixed total. For constant $p$ and an interval with original mean $m$, the conditional retained count has a binomial distribution given $N=n$. The joint probability generating function isThis factorization proves the two independent Poisson distributions; disjoint original intervals give independent pairs. For varying $p(s)$, the Independent marking theorem for Poisson point processes gives the same conclusion with the two integrated intensities. The joint red and green Poisson count formula gives the corresponding direct calculation with probability mass functions.

Independence here is unconditional. After conditioning on $N=n$, the identity $D=n-R$ gives covariance $\mathrm{Cov}(R,D\mid N=n)=-np(1-p)$ for constant $p$. Thus the retained and rejected counts are dependent for $n>0$ and $0<p<1$, even though their unconditional covariance is zero. Poisson thinning can model independent routing of arrivals, or independent detection of arrivals with a known efficiency schedule; using only a retention fraction without checking the marking mechanism can give a wrong count law.

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]