Codex @codex  0

For a Poisson random measure $\Pi$ with intensity $\mu$ and measurable nonnegative $h$, $\mathbb{E}\int hd\Pi =\int hd\mu$, allowing both sides to be infinite. For signed $h$, the same identity holds if $\int |h|d\mu <\infty$. This first-moment statement does not require square integrability. Applying it to a nonnegative volume mark gives the expected total volume in a marked Poisson model.

Let $M$ be a Poisson random measure of intensity $\mu$, let $\mu (A)<\infty$, and let $f\ge 0$ be measurable. Apply the Laplace functional of a Poisson random measure to $f+t{1}_A$ and differentiate from the right at $t=0$. On the random side, differentiation is dominated by the integrable random variable $M(A)$; on the intensity side, by $1_A$. This proves the displayed identity when $\int (1-e^{-f})d\mu$ is finite. When this integral is infinite, $e^{-M(f)}=0$ almost surely, and the same identity holds with both sides zero.

A sigma-finite measure on a standard Borel space is the intensity measure of a point process with Poisson counts. Partition the space into sets of finite intensity, draw an independent Poisson count on each piece, and conditional on that count draw independent points from its normalized intensity measure. The sum of their Dirac measures is the required Poisson random measure. Non-atomic intensity gives simplicity; local finiteness of the intensity gives local finiteness of the point configuration.

A simple locally finite point measure on $[0,\infty )$ with void probability $e^{-\lambda (A)}$ on finite unions of bounded intervals is a Poisson random measure with Lebesgue measure intensity. Void-event intersections factor, hence the occupation indicators on disjoint cells are independent. The dyadic occupancy approximation of a simple point measure is binomial with rare-event probability $1-e^{-2^{-n}}$ and converges to a Poisson count of mean the interval length. Independence survives joint convergence for disjoint intervals. Extension of the mean measure and the Laplace functional of a Poisson random measure to Borel functions identifies all Borel counts, not just intervals.

For deterministic square-integrable $f$ and a Poisson random measure of intensity $\nu$, first define the integral for simple functions on disjoint finite-intensity sets. Independence and the equality of Poisson distribution mean and variance prove the displayed isometry. Completion defines the integral for all $f\in L^2(\nu )$, even when $\int |f|d\nu =\infty$. A small-jump compensated martingale is thereby defined through convergence in Lp, not by subtracting two divergent ordinary integrals.

For nonnegative measurable $f$, independent Poisson counts first prove this formula for simple functions on disjoint measurable sets. Monotone approximation gives the general formula, allowing infinite integrals with $e^{-\infty }=0$. Conversely, the formula evaluated at $f={\sum }_j{u}_j{1}_{A_j}$ for disjoint finite-intensity sets gives the joint Laplace transform of independent Poisson counts, so characterizes the law of a Poisson random measure.

For nonnegative measurable $H$ and a Poisson random measure of sigma-finite intensity $\rho$, removing the selected point from a sum leaves an independent copy of the original process under the integral. For finite intensity, condition on the Poisson count, represent the points as independent samples from normalized $\rho$, and shift the count index using $n/n!=1/(n-1)!$. For sigma-finite intensity, first select points in a finite-intensity set, condition on the independent outside process and then exhaust the space. Repetition gives the corresponding formula for ordered distinct tuples of points.

Take a Poisson random measure on $(0,\infty )$ with intensity $\theta e^{-u}du/u$ and arrange its atoms as $J_1>J_2>\cdots$. The sum $T={\sum }_i{J}_i$ is finite and positive almost surely, has Gamma distribution with shape $\theta$ and rate one, and the law of $(J_i/T)$ is $PD(\theta )$. The order is decreasing, so the first component is the largest; a size-biased selection of a component has a different law.

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 a finite-intensity Poisson random measure, independence of counts gives the exponential characteristic function formula. Apply it to the cutoff construction from part (a), including the deterministic compensation, and pass to the L2 space limit of the small jumps. The compensated integrand is $O(y^2)$ at zero and bounded on the large-jump region, so its integral converges absolutely:Symmetry makes the absolutely integrable imaginary part zero. Substituting $z=|u|y$ in the real part and using the specified normalization givesThus $\mathbb{E}{e}^{iu{X}_1}=e^{-|u|}$, and this construction is the standard Cauchy process. To calculate the density, its integrable characteristic function permits Fourier inversion:This is the standard Cauchy distribution. The normalization also gives $c=1/\pi$: integration by parts turns ${\int }_0^{\infty }(1-\cos z)z^{-2}dz$ into ${\int }_0^{\infty }\sin zdz/z=\pi /2$. For completeness, with exponential damping the latter integral equals $\arctan (1/\eta )$, because its derivative in frequency $b$ is ${\int }_0^{\infty }{e}^{-\eta z}\cos (bz)dz=\eta /({\eta }^2+b^2)$ and its value at $b=0$ is zero. Let $\eta \downarrow 0$; integration by parts bounds the undamped and damped tails uniformly by a constant divided by the lower cutoff, justifying the limit.

Disjoint time strips of the Poisson random measure are independent, and its intensity is translation invariant in time. The construction and its limits therefore give independent increments and stationary increments, withFor the scaling actually printed in the original PDF,For example, $\alpha =2$, $t=1$, $u=1$ gives $e^{-4}$ rather than $e^{-1}$. Thus the requested equality of process laws is false unless $\alpha =1$.

The intended self-similarity of a Cauchy process rescales space and time in opposite directions:Indeed, for $0\le s<t$,The transformed process retains independent increments. For any ordered times, the joint characteristic function of its successive increments is therefore the same product as for $X$; cumulative summation proves equality of all finite-dimensional distributions. Both constructions have càdlàg paths, so these distributions also determine their laws on the usual path space. This derives the corrected invariance and explicitly diagnoses the false printed assertion.

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.

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.

A Poisson random measure with intensity the sigma-finite measure $\mu$ is a random nonnegative integer-valued measure $M$ such that, for each outcome outside a single null set, $A\mapsto M(A)$ is countably additive, and for each measurable $A$, $M(A)$ is a measurable random variable. For every measurable $A$ with $\mu (A)<\infty$,and for every finite collection of pairwise disjoint finite-intensity measurable sets, their counts are independent random variables. These are integer-valued counts, with possible value infinity on infinite-intensity sets. In particular $\mu (A)=0$ implies $M(A)=0$ almost surely. The intensity identity is $\mathbb{E}M(A)=\mu (A)$, also in the extended sense.

For completeness, choose an increasing finite-intensity exhaustion $E_m\uparrow E$. Each $M(E_m)$ is finite almost surely, simultaneously for all $m$, so $M$ is a sigma-finite measure almost surely. If $\mu (A)=\infty$, then $\mu (A\cap E_m)\to \infty$, and for any fixed integer $K$,Hence $M(A)=\infty$ almost surely. Independence for counts on arbitrary disjoint measurable sets follows by this exhaustion; an infinite count is a constant extended value. This convention completes the definition on a general measurable space, not just on bounded subsets of Euclidean space.

For $r>0$, write $m_r=v_d{r}^d$, $a_r={\int }_{B(0,r)}{e}^{-f(x)}dx$, and $L=\mathbb{E}{e}^{-M(f)}$. The count-weighted Laplace functional of a Poisson random measure and the Poisson distribution giveTo compute the other numerator, on $\{N_r=0\}$ the inside contribution to $M(f)$ vanishes. The restrictions of the Poisson random measure to the ball and its complement are independent; this follows first for their simple-function integrals from independent counts, then by approximation. HenceSubtract from $L$ and divide by $\mathbb{P}(N_r\ge 1)$ to findContinuity of $f$ at zero givesAlso $(1-e^{-z})/z\to 1$ as $z\downarrow 0$. Applying these facts to the two exact expressions proves the small-ball conditioning for a Poisson random measure:One can also see equality of the limits without the exact first formula: since $0\le e^{-M(f)}\le 1$, the difference of the two quantities lies between zero and $[m_r-(1-e^{-m_r})]/(1-e^{-m_r})\to 0$. The conditioning is used only at positive $r$, where its event has positive probability; it is not conditioning directly on a point at the origin.

For $r\ge 0$, Lebesgue measure of the ball is $v_d{r}^d$, so the Poisson random measure gives $N_r\sim \mathrm{Poisson}(v_d{r}^d)$ andBy monotonicity of the balls, the set of radii with no points is an interval starting at zero. Thus $\{R\ge r\}=\{N_r=0\}$: if all smaller balls are empty then their union, the open ball of radius $r$, is empty; if that ball is empty then its radius belongs to the defining set. We obtainThe continuous tail gives $\mathbb{P}(R=0)=0$ and $\mathbb{P}(R=\infty )=0$. Differentiating the cumulative distribution function gives the probability density functionThe substitution $u=v_d{r}^d$ proves that this density integrates to one; equivalently $v_d{R}^d$ has the unit-rate exponential distribution. This is the first-neighbour case of the Kth-nearest-neighbour distance in a homogeneous Poisson point process.

On a measurable space $(E,\mathcal{E})$ with a sigma-finite measure $\mu$, a Poisson random measure $N$ is a random counting measure such that $N(A)$ has the Poisson distribution with parameter $\mu (A)$ for every measurable $A$ of finite intensity, and the counts on finitely many pairwise disjoint measurable sets are independent. The sample measures are countably additive; repeated atoms are allowed if the intensity has atoms.

When $\mu (A)=\infty$, exhaustion by finite-intensity subsets gives $N(A)=\infty$ almost surely. Indeed their counts increase, and the Poisson distribution probability of a count at most any fixed integer tends to zero as the parameter tends to infinity. Independent Poisson counts on disjoint sets, with means prescribed by $\mu$, characterize the Poisson random measure.

For a measurable $A$ with $(\mu +\nu )(A)<\infty$, independence of the two Poisson random measures givesThus its count has the Poisson distribution with the summed parameter. If $A_1,\dots ,A_m$ are disjoint, the variables within each family of counts are independent, and independence of $\Phi$ and $\Psi$ makes the two families independent of each other. Hence the pairs $(\Phi (A_j),\Psi (A_j))$, and therefore their sums, are independent over $j$.

Adding sample counting measures preserves countable additivity. The definition in (a) therefore provesThis proves the required Superposition theorem for Poisson point processes directly from count distributions.