Assign each of $N\sim \mathrm{Pois}(\lambda )$ claims with independent category labels to one of finitely many categories with probabilities $p_i$. Conditional counts have a multinomial distribution with probability generating function $({\sum }_i{p}_i{z}_i{)}^N$. Averaging over $N$ gives $\exp (\lambda ({\sum }_i{p}_i{z}_i-1))={\prod }_i\exp (\lambda p_i(z_i-1))$. This proves independent category counts and supports discrete-severity compound Poisson distribution representations.

Let $N$ be a nonnegative integer-valued random variable with finite expected value $\mu$, and independently label each of its items with either of two equally likely labels. If the two resulting counts are independent random variables, then $N$ has a Poisson distribution. Indeed, its probability generating function obeys $G(s)=G((1+s)/2{)}^2$. With $H(t)=G(1-t)$, iteration gives $H(t)=H(t/2^k{)}^{2^k}$. Finite expected value gives $H(h)=1-\mu h+o(h)$, hence $H(t)=e^{-\mu t}$ and $G(s)=e^{\mu (s-1)}$. Conversely, Poisson thinning produces independent Poisson distributions for the two counts.

Independent marking is the hypothesis that guarantees the Poisson thinning theorem. Equal marginal retention probabilities do not guarantee it. Consider one Bernoulli random variable $B$ with success probability $p\in (0,1)$, independent of a Poisson process, and retain every arrival when $B=1$ and none when $B=0$. Each arrival is retained with probability $p$, but the marks share the same coin. In an interval whose original count has Poisson distribution of mean $m>0$, the retained count is $R=BN$ and the rejected count is $D=(1-B)N$, soThe extra variance shows that $R$ is not Poisson distributed; the negative covariance proves dependence. Its no-arrival probability is $1-p+p{e}^{-m}$, rather than $e^{-pm}$. For the toy values $p=1/2,m=2$, both rules have retained mean $1$, but the common-coin rule has variance $2$ and no-arrival probability $(1+e^{-2})/2\approx 0.567668$, compared with variance $1$ and no-arrival probability $e^{-1}\approx 0.367879$ for independent marking. These formulas follow by conditioning on $B$ and using the law of total variance.

An arrival-dependent or adversarial retention rule can change the law just as strongly. Keeping every second arrival gives $R_t=\lfloor N_t/2\rfloor$. Its successive retained gaps are sums of two independent exponential random variables, hence have the Erlang distribution of shape two and rate $\lambda$, not an exponential distribution. The resulting renewal process and its count variance are analyzed in Every second Poisson arrival has underdispersed counts, and the two dependent streams in alternating split of a Poisson process. As an even more restrictive rule, keeping only the first arrival ever gives $R_t=1_{\{N_t\ge 1\}}$, a Bernoulli distribution at each $t>0$, so it cannot have a positive-mean Poisson distribution. A thinning rule must specify how decisions depend on arrivals and on one another; the independent-marking conclusion cannot be applied automatically to arbitrary deletion.

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 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 deterministic square-integrable $f$ and a Poisson random measure $\mu$ 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.

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.

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.

For unit volume intensity in the unit ball of ${\mathbb{R}}^d$, polar coordinates give intensity $r^{d-1}dr\sigma (d\omega )$. The separate radius and direction images are Poisson processes, but each counts the same random total number of original points. Their count covariance is the displayed intensity of the intersection of their two preimages. Conditional on the total count, radii and directions are independent samples, which does not imply unconditional independence of the two point processes.

A Poisson structure is an antisymmetric smooth bivector field whose induced function bracket satisfies the Jacobi identity. In coordinates its condition is the vanishing cyclic sum of ${\omega }^{dc}{\partial }_d{\omega }^{ab}$. Applying Jacobi to coordinate functions proves necessity; expanding arbitrary brackets cancels the second derivatives and proves sufficiency. A function bracket-commuting with every function is a Casimir function of a Poisson manifold.

A Poisson tensor is a smooth bivector field defining $\{f,g\}=\pi (df,dg)$ whose bracket obeys the Jacobi identity. In coordinates, ${\pi }^{i\ell }{\partial }_{\ell }{\pi }^{jk}+{\pi }^{j\ell }{\partial }_{\ell }{\pi }^{ki}+{\pi }^{k\ell }{\partial }_{\ell }{\pi }^{ij}=0$. Its Hamiltonian distribution integrates into symplectic leaves, even though its rank can vary across the manifold. The tensor specifies a bracket without requiring a globally nondegenerate symplectic form.

A smooth map $r:M\to N$ between Poisson manifolds is a Poisson map when $\{f\circ r,g\circ r{\}}_M=\{f,g{\}}_N\circ r$ for all smooth functions on $N$. When a Lie group acts by Poisson automorphisms and the quotient is smooth, brackets of invariant functions descend and make the quotient projection Poisson. The quotient can be Poisson without being symplectic.

The symplectic foliation of a Poisson manifold consists of maximal connected integral manifolds of its Hamiltonian distribution. Each leaf has a nondegenerate closed two-form inverse to the restricted Poisson structure, with compatible sign conventions. Ranks can differ between leaves, making this a singular foliation. Hamiltonian flows remain within a leaf, and Casimir functions of a Poisson manifold are constant along every leaf; a Casimir level set need not be a single leaf.

A uniform magnetic field defines a constant Poisson bivector on momentum space by the displayed components. For the Hamiltonian $E(\mathbf{p})$ and convention $\dot{f}=\{f,E\}$, it gives $\dot{\mathbf{p}}=e\nabla E\times \mathbf{B}$. Constancy makes every term in the coordinate Jacobi identity vanish. If $e\mathbf{B}\ne 0$, the symplectic leaves are the planes $\mathbf{B}\cdot \mathbf{p}=c$, and energy is conserved along the flow. The whole three-dimensional momentum space is not symplectic, and an energy surface need not itself be a Poisson submanifold.

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 $0<r<1$, the geometric series gives $P_r(t)=1+2{\sum }_{n\ge 1}{r}^n\cos nt$. Hence this positive even kernel has normalized integral one, and normalized convolution with it multiplies the $n$th Fourier coefficient by $r^{|n|}$. Its mass outside any fixed neighbourhood of zero tends to zero as $r\uparrow 1$. If an integrable function has finite one-sided limits $A,B$ at zero, evenness gives half the mass to each side and the convolution tends to $(A+B)/2$.

For $F\in L^1(\partial \mathbb{D})$ and a Lebesgue point ${\xi }_0$, its Poisson integral tends to $F({\xi }_0)$ through every fixed Stolz region. If $\delta =1-|z|$, the Poisson kernel there is at most $C_A\delta /({\delta }^2+d(\xi ,{\xi }_0{)}^2)$, where $d$ is shorter angular distance. Dyadic annuli about ${\xi }_0$ bound the near contribution by small local averages of $|F-F({\xi }_0)|$ times a summable sequence $C_A{2}^{-k}$. The contribution outside a fixed small arc is $O(\delta )\Vert F-F({\xi }_0){\Vert }_1$ and vanishes. This is the local kernel argument underlying the Fatou theorem for bounded holomorphic functions.

For drops with intensity $\lambda (x,t,v)$, impact time $t$ and nonnegative volume $v$, let their radius at time $\tau$ be $r(\tau -t,v)$. The dry probability at $\xi$ isThe exponent is the intensity of drops whose patch covers the point by that time. An infinite exponent means zero dry probability. No spatial-homogeneity assumption is needed.

Suppose the shell $2^{-n}<r<2^{1-n}$ has mass density $2^{n-1}{\rho }_0$. Its mass is $(14\pi {\rho }_0/3)4^{-n}$, by integrating $4\pi r^2\rho dr$. Summing the geometric series gives total mass $14\pi {\rho }_0/9$. At $r=2^{-N}$ the enclosed shells have indices $n\ge N+1$, so the enclosed mass is the total mass times $r^2$. The magnitude of the Newtonian gravitational field at every such boundary is therefore the same, $14\pi G{\rho }_0/9$. This does not assert that the field is constant throughout each shell.

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.