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Ripley K for a homogeneous Poisson process is Euclidean ball volume

Statement

Assume AC and Countable Choice as carried by the point-process supplier. Let η be a stationary homogeneous Poisson point process on Rd with intensity measure ρλd, where ρ>0. For every r≥0,

K(r)=∣B[0,r]∣.

The full Palm law is P0=L(η+δ0); adding and then deleting the distinguished atom means one occurrence, so no simplicity assumption is used.

Facts & Assumptions

Given: AC, Countable Choice, the stationary homogeneous Poisson process η, its full Palm law, and the closed ball convention.

[A1]

AC is carried by the transitive point-process supplier dependencies; this proof adds no choice. (The Axiom of Choice)

[A2]

Countable Choice is carried by the point-process and Lebesgue-measure supplier dependencies. (The Axiom of Countable Choice (ACω))

[F1]

The deletion-form Mecke identity holds for every nonnegative product-measurable integrand and removes one counted occurrence. (Mecke identity for locally finite Poisson point processes)

[F2]

The full Palm law satisfies the refined Campbell identity. (Existence and uniqueness of the full Palm law and the refined Campbell identity)

[F3]

Recentering is (θxμ)(B)=μ(B+x), is jointly measurable, and sends δx to δ0. (Recentering a point configuration at a spatial location, The spatial recentering action is jointly measurable)

[F4]
[F5]

K(r)=ρ−1E0[(μ−δ0)(B[0,r])] for the closed ball. (Pair-correlation density and Ripley K-function)

[F6]

For every compact C, the Poisson count η(C) has parameter ρ∣C∣<∞ and second moment (ρ∣C∣)2+ρ∣C∣. (Poisson and Cox point processes, Normalization, moments, and generating function of the Poisson count law)

Proof

Given: AC, Countable Choice, a stationary homogeneous Poisson process η of intensity ρλd, and its full Palm law P0.

1.1givenF5F6algebra

For each compact C⊆Rd, [F6] gives E[η(C)2]=(ρ∣C∣)2+ρ∣C∣<∞, so η is locally square integrable and satisfies the second-order premise of the K definition [F5].

1.2F1F3construct

Fix a configuration event A. In [F1] use hA(x,μ)=1Q(x)1A(θx(μ+δx)), which is product measurable by [F3] and the evaluation formula for adding one atom. At each occurrence x of η, deletion followed by addition restores η, so the left side is E∫1Q(x)1A(θxη)η(dx).

2.1step 1.2F1F2F3algebra

The right side of that Mecke identity is ρ∫QP(θx(η+δx)∈A) dx=ρ∣Q∣P(η+δ0∈A), because θxδx=δ0 and stationarity gives θxη=dη. The refined Palm identity [F2] applied to 1Q(x)1A(μ) says that the left side in step 1.2 is ρ∣Q∣P0(A). Since ∣Q∣=1 and ρ>0, P0(A)=P(η+δ0∈A) for every A, proving the full-Palm law identity.

3.1step 1.1step 2.1F4F5algebra

Under this Palm law, μ has the distribution of η+δ0. Removing the single distinguished occurrence gives (η+δ0)−δ0=η even if η has other occurrences at 0. Therefore E0[(μ−δ0)(B[0,r])]=E[η(B[0,r])]=ρ∣B[0,r]∣ by [F4]. Substituting in the K definition [F5] yields K(r)=∣B[0,r]∣.

4.1A1A2step 3.1F4F5∎

The endpoint r=0 is included: the closed ball is {0} and the same calculation gives K(0)=ρ−1Eη({0})=0=∣{0}∣. The proof uses the one-occurrence deletion convention and asserts neither diffuse-intensity simplicity nor any other simplicity property. AC and Countable Choice [A1]–[A2] are inherited supplier assumptions.

Sources

Last and Penrose, Chapters 4 and 9, Mecke equation and Theorem 9.4; Baddeley, Lecture 3, Example 3.3. The local proof derives the full Palm law directly from Mecke.

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Sources