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Poisson and Cox point processes — Examples

1 · Prerequisites

2 · Summary

On a finite window, a deterministic Poisson intensity and a two-point random directing intensity have the same mean measure and expected count but different count variances: 1 and 2. The example computes the conditional mixture explicitly and supplies its probability kernel.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-adaptedOpen item page →

Poisson and Cox counts with the same mean on a finite window

Statement

Let W be a compact Borel window in a locally compact, second-countable Hausdorff space S, and let μ be a finite Borel measure (Finite, sigma-finite, and semifinite measures) supported on W with μ(W)=1. A Poisson point process of intensity μ and the Cox process directed by ξ=Zμ, where P(Z=0)=P(Z=2)=1/2, both have mean count one in W. Their count variances differ: the Poisson variance is 1, whereas the Cox variance is 2.

Facts & Assumptions

Given: A compact Borel window W and a finite Borel probability measure μ supported on W.

[F1]

For each finite intensity measure ρ, a Poisson point process of intensity ρ exists and has independent counts on disjoint sets, each count being Pois⁡(ρ(B)). In particular its total count is Pois⁡(ρ(S)). (Finite-intensity Poisson-process existence and mixed-binomial law, Poisson and Cox point processes)

[F2]

If N∼Pois⁡(a), then EN=a and Var⁡(N)=a. (The scalar Poisson count law, Normalization, moments, and generating function of the Poisson count law)

[F3]

A supplied measurable probability kernel that is the regular conditional law of a point process given its directing random measure makes that process Cox when its conditional laws are Poisson. (Measure kernel and probability kernel, Conditional law given a random element, Poisson and Cox point processes)

Proof

technique · direct
1.1givenF1F3construct

Let P0 be the zero-process law and let P2μ be the finite-intensity Poisson law of intensity 2μ, both supplied by [F1]. On the disjoint union of a P0 branch and a P2μ branch, assign probability 1/2 to each branch, set Z=0 or 2 according to the branch, set ξ=Zμ, and use the branch process as ηC. For a locally finite measure ρ, define K(ρ,⋅)=P2μ when ρ(S)=2 and K(ρ,⋅)=P0 otherwise. Since ρ↦ρ(S) is an evaluation, this is a measurable probability kernel; on the support {0,2μ} of ξ, it is the conditional law of ηC and is Poisson with intensity ξ. Thus [F3] verifies the Cox property.

2.1givenstep 1.1F1algebra

Let ηP have the finite-intensity Poisson law of intensity μ from [F1]. Since EZ=1, for every Borel B, Eξ(B)=E[Z]μ(B)=μ(B), equal to the deterministic intensity measure of ηP.

2.2givenstep 1.1F1F2algebra

By [F1]-[F2], NP=ηP(W)∼Pois⁡(1), so ENP=1 and Var⁡(NP)=1. Conditional on Z=z, NC=ηC(W) is Pois⁡(z); hence ENC=EZ=1 and ENC2=E[Z+Z2]=12(0)+12(2+4)=3. Therefore Var⁡(NC)=3−12=2.

3.1step 2.1step 2.2algebra∎

The deterministic and directing mean measures agree by step 2.1, while steps 2.2 show equal expected window counts and different count variances. The difference comes from the random intensity Z: conditional Poisson variation contributes EZ=1, and variation in Z contributes Var⁡(Z)=1.

Sources