ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A finite window makes the border correction weight visible

Example

Take $W = [0, 1] \subset R$ , radius $r = 1 / 4$ , and observed occurrences at $0.1, 0.4, 0.5, 0.9$ . The border window is $W_{- r} = [1 / 4, 3 / 4]$ : the two boundary anchors are excluded because their full closed $r$ -neighborhoods are not observed. The two eligible anchors each have one observed partner within distance $r$ .

Verification

Given: The border-estimator procedure of Border-corrected estimator for Ripley K, $W = [0, 1]$ , $r = 1 / 4$ , and $ξ = δ_{0.1} + δ_{0.4} + δ_{0.5} + δ_{0.9}$ .

[F1] $W_{- r} = {x \in W : [x - r, x + r] \subseteq W}$ ; the anchor weight is $1$ on $W_{- r}$ and $0$ outside, and the output is the eligible ordered-pair sum divided by $\hat{ρ} N (W_{- r})$ . (Border-corrected estimator for Ripley K)

[F2] For an eligible anchor, the entire closed radius- $r$ interval is inside the observation window, so all radius- $r$ partner occurrences are observed. (Border-corrected estimator for Ripley K)

[A1] AC and Countable Choice are inherited through the estimator's K and point-process dependencies; this finite calculation uses no choices. (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ ))

1.1F1constructalgebra

For $x \in [0, 1]$ , the condition $[x - 1 / 4, x + 1 / 4] \subseteq [0, 1]$ is equivalent to $x \geq 1 / 4$ and $x \leq 3 / 4$ . Therefore $W_{- r} = [1 / 4, 3 / 4]$ and the anchor weights at $0.1, 0.4, 0.5, 0.9$ are respectively $0, 1, 1, 0$ .

2.1step 1.1F1algebra

The eligible anchor at $0.4$ has only $0.5$ as a distinct occurrence within distance $1 / 4$ ; the eligible anchor at $0.5$ has only $0.4$ . Thus the ordered-pair numerator is $2$ , $N (W_{- r}) = 2$ , $N (W) = 4$ , and $\hat{ρ} = 4 / ∣ W ∣ = 4$ . The border procedure returns $2 / (4 \cdot 2) = 1 / 4$ .

3.1A1step 1.1step 2.1F2∎

The excluded anchor $0.1$ has a radius- $1 / 4$ interval extending to $- 0.15$ outside $W$ , so its complete neighborhood is not observed; the anchor $0.4$ has $[0.15, 0.65] \subseteq W$ , as guaranteed by [F2]. This calculation verifies the procedure's edge-exposure rule only and proves no finite-sample unbiasedness claim. AC and Countable Choice [A1] are inherited supplier assumptions.

Sources

Baddeley, Lecture 2, §2.6, formula (18); Last and Penrose, Chapter 8, spatial K-function discussion. This is a geometric example, not an unbiasedness result.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

17 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Adrian Baddeley, Spatial Point Processes and their Applications (standard reference, not scraped)
Günter Last and Mathew Penrose, Lectures on the Poisson Process (2017 author-hosted version) (standard reference, not scraped)