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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.

Sampling with a homogeneous Poisson point process of mean comoving number density $\bar{n}$ adds a constant $1/\bar{n}$ to the dimensional galaxy power spectrum. Subtracting its expectation leaves sampling uncertainty. Correlated tracer populations and survey weights modify this simple formula; survey depth alone does not specify $\bar{n}$.

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

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 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.

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.

Write $z=(x,y)$ and $\mu (dz)=\lambda (z)dz$. Treat the nonnegative plant weights as independent marks conditional on the locations, with a measurable mark law $K_z(dw)$ depending only on $z$. Its first and second moments areThis is the independent-marking interpretation appropriate for location-dependent weights.

The general results needed are the following. The Independent marking theorem for Poisson point processes says that such marks turn a Poisson point process of intensity $\mu$ into a Poisson point process on location-mark space with intensity $\nu (dz,dw)=\mu (dz)K_z(dw)$. The Campbell first-moment formula says that for any nonnegative measurable $g$, $\mathbb{E}{\sum }_{u\in \Pi }g(u)=\int gd\nu$, allowing infinite values. The second Poisson factorial moment measure identity says that for nonnegative measurable $h$,These statements apply to sigma-finite intensities and therefore do not require finitely many plants in the entire field. We use these general results without proving them.

Define the total as the nonnegative sum $W={\sum }_{(z,w)\in \Pi }w$, or equivalently as the increasing limit of its finite-intensity truncations. the Campbell theorem givesSince $M<\infty$, the nonnegative random variable $W$ cannot be infinite on an event of positive probability. Thus $W$ is finite almost surely, even if the number of plants is infinite.

Separate the square into its diagonal and ordered distinct-pair terms. Nonnegativity allows the use of Tonelli's theorem and increasing limits, givingThe assumed second integral is finite, so $W$ is square-integrable. Subtracting $(\mathbb{E}W{)}^2$ yieldsThe term $v$ accounts for random weights at fixed locations; the term $m^2$ accounts for the Poisson fluctuations in the locations and count. Nonnegativity is natural for weights. For signed marks, absolute first-moment integrability, rather than just a conditionally convergent integral of the signed mean, would be needed for the ordinary sum.

The intensity function of a point process is positive and continuous inside the interval, so its measure is finite on every compact subinterval. Consequently the Poisson point process is locally finite there. Its nonatomic intensity makes it simple and gives no point at $0$ almost surely. At the endpoints,Both half-intervals therefore have infinite intensity. More explicitly, exhaust either half by increasing compact intervals with means $t_j\to \infty$. For any fixed integer $k$,Taking the countable union over $k$ proves infinitely many points on each side almost surely.

Local finiteness excludes accumulation in the interior, including at $0$. Hence there is a largest negative point and a smallest positive point. For example, given any negative point, the compact interval between it and $0$ contains finitely many points and therefore a largest one. Call it $X_0$; call the smallest positive point $X_1$, and enumerate consecutively in both directions. Infinite counts give every integer index, and the lack of interior accumulation implies $X_n\uparrow 1$ as $n\to \infty$ and $X_n\downarrow -1$ as $n\to -\infty$. Thus the specified two-sided labeling exists almost surely.

Use the cumulative-intensity transformationIt has $f(0)=0$, positive derivative, and limits $-\infty ,+\infty$ at the two endpoints, so it is an increasing bijection onto $\mathbb{R}$. One explicit expression isFor any bounded interval $(a,b)$ in the target, its preimage has intensityDisjoint target sets have disjoint preimages, so their counts are independent. This proves directly, or by the Poisson mapping theorem, that the transformed points form a unit-rate Poisson process on the real line.

Put $Y_n=f(X_n)$. On the positive half-line, the successive spacings are independent exponential random variables of mean $1$, so $Y_n=E_1+\cdots +E_n$. The strong law of large numbers gives $Y_n/n\to 1$ almost surely. On the negative half-line, independently, the reflected points satisfy $-Y_{-k}=E_1^{\prime }+\cdots +E_{k+1}^{\prime }$, because $Y_0$ is the first point to the left of zero. Thus $-Y_{-k}/k\to 1$, or $Y_n/n\to 1$ also as $n\to -\infty$. The shift by one has no effect on these limits.

From the explicit formula, or by integrating the endpoint intensity asymptotics,Substitute $x=X_n$ and use the strong law for Poisson arrival times. At the right endpoint,soAt the left endpoint the corresponding result isEquivalently $8|n|(1+X_n)\to 1$. The different powers of $|n|$ reflect the different endpoint singularity orders, an instance of power-law boundary accumulation of a Poisson point process.

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.

For $r>0$, let ${\kappa }_r$ be the probability law of a seed displacement from its tree. For the literal circular support, it is normalized arclength on $\{u:|u|=r\}$. The conditional mean measure of a point process isWe prove the conditional Poisson assertion directly by joint probability generating functions.

Fix disjoint bounded measurable sets $A_1,\dots ,A_k$. For a tree at $z$, put $q_j(z)={\kappa }_r(A_j-z)$ and $q_0(z)=1-{\sum }_j{q}_j(z)$. Conditional on its seed count, the allocation among these sets and their complement is multinomial distribution. Averaging over the Poisson seed count givesThus this tree contributes independent Poisson counts of means $\mu q_j(z)$ to the disjoint sets. Different trees contribute independently. Only trees within distance $r$ of the bounded union of the $A_j$ can contribute, and there are almost surely finitely many of them. Multiplying their probability generating functions givesThis is precisely the joint probability generating function of independent Poisson counts with the displayed means. Also ${\Lambda }_{\Pi }$ is locally finite, because each bounded set can receive seeds from only finitely many trees. This proves the conditional Poisson point process assertion and the Poisson offspring clusters directed by their parent process construction.

In arclength notation the directing measure is ${\Lambda }_{\Pi }=\mu {\sum }_{z\in \Pi }{\sigma }_{z,r}/(2\pi r)$, where ${\sigma }_{z,r}$ is arclength on the circle centered at $z$. If “circle” is used to mean the filled disc instead, take ${\kappa }_r(du)=1_{\{|u|\le r\}}du/(\pi r^2)$; thenThe generating-function proof and the following unconditional conclusion hold for either reading. The circular version has a singular conditional directing measure, giving the random directing measure of a Cox process formulation.

Unconditionally this is a Cox process, specifically a Poisson-offspring Neyman-Scott process. Its mean measure of a point process is nevertheless spatially uniform. With $q_A(z)={\kappa }_r(A-z)$, the Campbell first-moment formula and translation invariance givesince $\int q_A(z)dz=\int {\kappa }_r(du)\int 1_A(z+u)dz=|A|$. Uniform mean alone does not establish a Poisson law.

For a bounded set $A$ of positive area, the conditional count has mean and variance ${\Lambda }_{\Pi }(A)$. The law of total variance givesA Poisson integral of a deterministic function $q$ has variance $\lambda \int q^{2}dz$: for a simple function on disjoint sets this follows from independent Poisson counts, and square-integrable approximation gives the general formula. Here $0\le q_A\le 1$ and it has bounded support, soThe strict inequality holds when $\lambda ,\mu >0$, because $\int q_A=|A|>0$ implies $\int q_A^2>0$. A Poisson random variable has equal mean and variance. Hence the unconditional process is not a Poisson process in the nontrivial model. If $\lambda =0$ or $\mu =0$, there are no seeds and the process is the degenerate empty Poisson process. The overdispersion of nondegenerate mixed Poisson counts quantifies the additional clustering induced by the random parents.

Equip $S$ with a sigma-algebra $\mathcal{S}$. A Poisson random measure with intensity measure of a point process $\mu$ is a random counting measure $\Pi$ such that, for every measurable $A$ with $\mu (A)<\infty$,and the counts on every finite collection of pairwise disjoint measurable sets are independent. Thus $\mathbb{E}\Pi (A)=\mu (A)$. For a sigma-finite measure $\mu$, sets of infinite intensity contain infinitely many points almost surely: exhaust such a set by increasing finite-intensity sets and use the Poisson count tails as their means tend to infinity. The count is a measurable random variable for every measurable $A$, which is part of saying that $\Pi$ is a random measure.

Under the usual point representation write $\Pi ={\sum }_j{\delta }_{Z_j}$, counting multiplicities. If $\mu$ is a non-atomic measure, this is a simple Poisson point process; if $\mu$ has atoms, several point occurrences can have the same location, and each occurrence is counted separately. In a topological setting, local finiteness requires that the intensity be finite on compact sets.

Sufficient existence conditions are that $S$ is a standard Borel space and $\mu$ is a sigma-finite measure. This gives existence of a Poisson random measure on a standard Borel space. If a simple locally finite configuration is desired, additionally take a non-atomic, locally finite intensity on a locally compact second-countable Hausdorff space. These conditions are sufficient, not claimed necessary. In particular sigma-finiteness suffices for a countable family of occurrences that can each be independently coloured.

Let $R$ and $G$ be the red and green counting measures. The marking assumption means that, conditional on $\Pi$, the colours of all point occurrences are independent, with the same red probability $r$. For a measurable set $A$ of finite intensity $m=\mu (A)$, condition on its total count. To obtain $a$ red and $b$ green points the total must be $a+b$, so direct multiplication gives the joint red and green Poisson count formulaThus the two counts are independent Poisson variables with respective means $rm$ and $(1-r)m$. This computation includes zero intensity with the usual convention for the probability of zero counts.

For disjoint finite-intensity sets $A_1,\dots ,A_k$, the original counts are independent. Conditional marking in different sets uses disjoint collections of independent colours. Repeating the preceding calculation therefore givesThis proves the disjoint-count definition for each coloured Poisson point process and independence of their two count vectors.

To obtain independence of the whole processes, take any finite list of red test sets and any finite list of green test sets, initially all of finite intensity. Refine their union into the disjoint cells given by their membership patterns. Every test-set count is the sum of the counts of its cells. The entire red cell vector is independent of the entire green cell vector by the product formula, hence so are the corresponding test-set count vectors. For arbitrary measurable test sets, intersect first with a finite-intensity exhaustion of $S$ and pass to the increasing count limits. The count cylinder events generate the two evaluation sigma-algebras, and independence extended from generating pi-systems gives independence of those sigma-algebras. ConsequentlyThis is a direct count proof of the Poisson thinning theorem; no general marking or thinning result has been used to deduce it.

The mapping theorem for Poisson point processes applies provided $f$ is a measurable function and its pushforward measureis locally finite, meaningfor every bounded Borel set $B\subseteq {\mathbb{R}}^s$. If “Poisson process” means a simple point process, one also requires $f_{\ast }\Lambda$ to be atomless, equivalentlyUnder these conditions, $f(\Pi )$ is a Poisson point process with mean measure

Let $N_j(A)$ be the number of particles of gas $j$ in a bounded Lebesgue measurable set $A$. Because Lebesgue measure is diffuse, two independent spatial Poisson point processes have no common particle almost surely. Hence the mixture count isWriting ${\lambda }_j=z_j|A|$, independence gives the probability generating functionThus

If $A_1,\dots ,A_m$ are pairwise disjoint, the familyis independent: counts in disjoint sets are independent within each gas, and the two gases are independent. Therefore the sums $N(A_r)$ are independent. These two facts prove directly that the mixture is a Poisson point process. By the Superposition theorem for Poisson point processes, its activity is

A Poisson random measure $N$ of intensity $\mu$ assigns to disjoint measurable sets independent random variables, andwhenever $\mu (A)<\infty$, with the usual countable-additivity requirement.

Let $N$ be the homogeneous Poisson point process in ${\mathbb{R}}^3$ with intensity measure $\lambda dx$, and map each point $x$ to its radius $g(x)=|x|$. For an interval $(a,b]\subseteq {\mathbb{R}}_{+}$, the pushforward measure of the intensity isThe Mapping theorem for Poisson point processes therefore shows that the radii form a Non-homogeneous Poisson point process with intensity function

If $R$ is the distance to the closest star, the event $\{R>r\}$ says that the ball of radius $r$ contains no points. The zero-count probability of a Poisson distribution givesDifferentiating its cumulative distribution function yields the probability density function

Campbell's formula for a Poisson point process givesExpanding $S^2$ separates equal and distinct Poisson points. The factorial-moment formula givesSubtracting the square of the mean yields

A Poisson point process of intensity $\lambda (x)$ has independent counts on disjoint Borel sets, with $\Pi (A)\sim \mathrm{Poisson}({\int }_A\lambda (x)dx)$ whenever the integral is finite.

Consider a deliberately simplified model: fatal defects form a spatially uniform Poisson point process with intensity $D$ per unit area. A semiconductor die of area $A$ works exactly when it contains no fatal defect. Its defect count $N$ has a Poisson distribution with expected value $DA$, hence its probability of working, called its yield, isThe product $DA$ is dimensionless. The formula concerns fatal defects, not every visible imperfection. Spatial uniformity, independent defect counts in disjoint areas, and failure from even one fatal defect are model assumptions. Clustering, shared processing failures, repairable regions and defects that affect only some functions can invalidate them. This is a chosen probability model, not historical yield data.

Choose $D=0.005{\mathrm{mm}}^{-2}$. For semiconductor die areas $25$, $100$ and $200{\mathrm{mm}}^2$, the model gives $Y=e^{-0.125}=0.882497$, $e^{-0.5}=0.606531$ and $e^{-1}=0.367879$. Doubling the area squares the yield: $Y(2A)=Y(A{)}^2$. This follows either from the exponential function or by splitting the larger area into two disjoint parts and using independence. More functions on one semiconductor die can save assembly while making the no-defect requirement harder to satisfy.

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.