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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.

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 $2\pi cr$.

Let $U$ have the uniform distribution on $[0,1)$ and place a point at every $k+U$, $k\in \mathbb{Z}$. This is a simple point process, locally finite, and a stationary point process: a translation only changes the phase modulo one, preserving its uniform distribution. For every Borel set $A$, the Tonelli theorem givesIts consecutive points are exactly one unit apart, despite having the same intensity measure as a unit-rate Poisson point process. This is the one-dimensional translated-grid construction.^[ref]

Every half-open interval of length one contains exactly one point. For $A=[0,1/2)$ and $B=[1/2,1)$, $\Pi (A)+\Pi (B)=1$, each count has the Bernoulli distribution with parameter $1/2$, andMore generally, an interval of length $L=m+r$, with $m$ a nonnegative integer and $0\le r<1$, contains $m$ or $m+1$ points with probabilities $1-r$ and $r$. Its count variance is $r(1-r)$, while a unit-rate Poisson point process has count variance $L$.

Draw one random variable $R$, taking values zero and two with equal probabilities. Conditional on $R$, generate a Poisson point process on the real line with constant rate $R$. This is a Cox process directed by the random measure $Rdx$: the same rate is shared by every window. It is a locally finite simple point process and a stationary point process. The law of total expectation gives $\mu (A)=\mathbb{E}[R]|A|=|A|$; the law of total variance and law of total covariance give, for bounded Borel sets $A,B$ with $A\cap B=\varnothing$,The conditional covariance vanishes by independence, but the shared conditional means fluctuate together. This distinguishes conditional from unconditional independence.^[ref]

For a window of length $L$, the exact probability mass function isIn particular, its void probability functional is $(1+e^{-2L})/2$, compared with $e^{-L}$ for the unit-rate Poisson point process. Whole realizations can be empty or comparatively crowded, while their mean density remains one. This example demonstrates shared-rate statistical dependence, rather than a model of short-range offspring clusters.

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.

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.

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.