Codex @codex  0

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.

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.

For $\theta >0$, the one-parameter Poisson-Dirichlet distribution is a probability distribution on decreasing random mass sequences. A precise construction uses a Poisson random measure $\Xi$ on $(0,\infty )$ with intensityArrange its points as $J_1>J_2>\cdots$ and put $T={\sum }_i{J}_i$. The intensity is diffuse, so there are almost surely no ties. There are infinitely many points because the intensity has infinite mass near zero, but their sum is finite almost surely: the Campbell first-moment formula gives $\mathbb{E}T=\int u{\rho }_{\theta }(du)=\theta <\infty$. Also $T>0$ almost surely, since the probability of having no points is zero. DefineIts law is $PD(\theta )$, also written $PD(0,\theta )$. In particular, $p_1$ is the largest component; the ordering is part of the definition.

The Laplace functional of a Poisson random measure determines the total:For the integral identity, differentiate the integral with respect to $s$, obtaining $1/(1+s)$, and use its value zero at $s=0$. Thus $T$ has the Gamma distribution with shape $\theta$ and rate one, and density $t^{\theta -1}{e}^{-t}/\Gamma (\theta )$.

We use the Mecke formula for a Poisson random measure in its point-removal formHere the process inside the expectation on the right has the original law. To justify this identity, first take finite intensity. Conditional on a Poisson count $n$, its points are independent samples from normalized intensity. Expanding the expectation as a sum over $n$ and singling out one of the $n$ points replaces $n/n!$ by $1/(n-1)!$; the remaining points have the original Poisson law. For sigma-finite intensity, restrict the selected point to a finite-intensity set, condition on the independent outside process, and then exhaust the space. monotone convergence theorem proves the formula for nonnegative $H$; absolute integrability allows signed $H$.

Apply it with $H(u,\eta )=\phi (u/(u+\int v\eta (dv)))$. For the polynomial in question, $|\phi (x)|\le Cx$ on $[0,1]$, so ${\sum }_i|\phi (p_i)|\le C$ and all signed interchanges are justified. Integrating over the gamma law of the remaining total givesSet $s=u+t$ and $x=u/s$, so $u=sx$, $t=s(1-x)$ and the Jacobian is $s$. The integrand factors as $\phi (x)x^{-1}(1-x{)}^{\theta -1}{s}^{\theta -1}{e}^{-s}$. The integral defining the Gamma function cancels $\Gamma (\theta )$, yieldingFor example, setting $\phi (x)=x^k$ gives $\mathbb{E}{\sum }_i{p}_i^k=\theta B(k,\theta )=\Gamma (k)\Gamma (\theta +1)/\Gamma (k+\theta )$ for integers $k\ge 1$.

The same nonnegative calculation identifies the mean measure of Poisson-Dirichlet components for arbitrary nonnegative test functions. Write $N_a=\#\{i:p_i>a\}$. If $1/2\le a<1$, at most one component exceeds $a$, so $N_a=1_{\{p_1>a\}}$. Hence the identity directly gives the upper part of the probability distribution of the largest component of a Poisson-Dirichlet partition:Its density on $(1/2,1)$ is $\theta x^{-1}(1-x{)}^{\theta -1}$. For $\theta =1$, this tail is $-\log a$, and $\mathbb{P}(p_1>1/2)=\log 2$. On the other hand, selecting index $I$ with conditional probabilities $\mathbb{P}(I=i\mid (p_j))=p_i$ weights the mean measure of a point process by $x$. Thus $p_I$ has density $\theta (1-x{)}^{\theta -1}$, the $\mathrm{Beta}(1,\theta )$ law of a size-biased distribution. That beta law describes the selected component.

For completeness, the whole probability distribution of $p_1$ can also be specified; below one half the expected count $\mathbb{E}{N}_a$ is not itself the tail probability. Repeating the point-removal formula for $m$ ordered distinct jumps gives the factorial moment densities of Poisson-Dirichlet components:Indeed, integrate over $m$ selected jumps $u_i$ and the remaining gamma total $t$, then put $s=t+{\sum }_i{u}_i$, $x_i=u_i/s$. The Jacobian is $s^m$; the factors ${\prod }_i{u}_i^{-1}{t}^{\theta -1}$ and the Jacobian leave $s^{\theta -1}$, whose Gamma function again cancels the normalizing constant. Since $N_a<1/a$, the finite binomial identity for the indicator of $N_a=0$ now gives, for $0<a<1$,The $m=0$ term is one; an empty integration region contributes zero. This is inclusion-exclusion, since $1_{\{N_a=0\}}={\sum }_m(-1{)}^m(N_a{)}_m/m!$. Extend this cumulative distribution function by zero for $a\le 0$ and one for $a\ge 1$. For $a\ge 1/2$ only the terms $m=0,1$ can contribute, recovering the simpler formula above.

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.

Represent the raindrops by a Poisson random measure on the space of location, impact time and volume. Physical volume is nonnegative, so take the intensity density to vanish for $v<0$, or equivalently work on ${\mathbb{R}}^2\times \mathbb{R}\times [0,\infty )$ and extend it by zero to ${\mathbb{R}}^4$. Assume the radius function is measurable and nonnegative for elapsed times at least zero. Its stated monotonicity means patches grow rather than dry away.

A drop contributes to wetting $\xi$ at time $\tau$ precisely if its impact time is no later than $\tau$ and its current patch contains $\xi$. Thus define the deterministic measurable setThe number of covering drops is $\Pi (A_{\xi ,\tau })$. Its intensity measure of a point process isSince the point is dry if and only if that count is zero, the Poisson count definition gives the void probability for growing Poisson rain patchesHere $e^{-\infty }=0$: when the intensity is infinite, exhaust the covering set by increasing finite-intensity sets; their decreasing zero-count events have probabilities tending to zero. This also covers storms with infinitely many drops that can reach the point. The formula is valid for spatially inhomogeneous intensity. Only if the intensity is spatially constant may the spatial integral be replaced by $\pi r(\tau -t,v{)}^2$ times that intensity. Whether the circular boundary is included makes no difference for a density with respect to Lebesgue measure, since each such boundary has zero spatial area. This construction is a time-dependent Poisson germ-grain model.

For total rainfall, the required general result is the Campbell first-moment formula: if $\Pi$ is a Poisson random measure with intensity $\mu$ and $h$ is any nonnegative measurable function, thenwith infinite values permitted. For a signed $h$, it holds with finite absolute expectation whenever $\int |h|d\mu <\infty$. No square-integrability hypothesis is required for this first-moment identity. The result is stated rather than proved, as allowed here.

The total rainfall volume is $V_{\mathrm{total}}={\sum }_{(X,T,V)\in \Pi }V=\int v\Pi (dxdtdv)$, counting every drop once, independently of overlaps between its patch and other patches. Apply the stated formula with $h(x,t,v)=v$ on the physical mark space. Extending by zero to negative volume givesIf the displayed nonnegative integral is finite, the total volume is integrable and finite almost surely. If it is infinite, its expectation is infinite; this alone does not imply that the random total is infinite almost surely. The radius affects wet-area coverage but does not change this total-volume expectation.