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Spatial summary estimators — Examples

1 · Prerequisites

2 · Summary

The boundary-area calculation uses Lebesgue measure under the inherited Countable Choice ${A C}_{ω}$ assumption from the A page. Two finite examples isolate common boundaries of the estimators on the A page. Rectangular windows can have diverging volume while their relative boundary does not vanish. A pair-count sample covariance can also violate a necessary positive-definiteness condition, even though it remains a defined finite-sample summary statistic.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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

Boundary-dominated windows fail the Van Hove condition

Example

The expanding rectangles $W_{n} = [0, n + 1] \times [0, 1] \subset R^{2}, n \geq 0,$ have volume tending to infinity, but their boundary remains a fixed proportion of the volume under a fixed-width enlargement.

Facts & Assumptions

Given: AC $_{ω}$ as required by the Van Hove definition, the definition of a Van Hove window sequence, and the compact set $K = [- 1, 1]^{2}$ .

[F1]

Van Hove sequences of spatial windows, The Axiom of Countable Choice ( ${A C}_{ω}$ ), Assuming countable choice, every Borel subset of $R^{n}$ is Lebesgue measurable, and Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume, A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included, and Measures on sigma-algebras: under AC $_{ω}$ the Borel rectangles are Lebesgue measurable, have the stated box volumes, and their restricted outer measure is a countably additive measure; Van Hove requires $∣ (W_{n} + K) ∖ W_{n} ∣ / ∣ W_{n} ∣ \to 0$ for every compact $K$ .

Verification

1.1F1givenalgebra

Each rectangle is bounded and Borel, and $∣ W_{n} ∣ = n + 1 > 0$ , so its volume tends to infinity as required in the first Van Hove condition.

2.1F1givenalgebra∎

Minkowski addition gives $W_{n} + K = [- 1, n + 2] \times [- 1, 2]$ . The two closed rectangles are nested, have finite volume by [F1], and Lebesgue measure is a measure by [F1], so countable additivity on $W_{n} ⊔ ((W_{n} + K) ∖ W_{n}) = W_{n} + K$ gives $\frac{∣ (W_{n} + K) ∖ W_{n} ∣}{∣ W_{n} ∣} = \frac{3 (n + 3) - (n + 1)}{n + 1} = 2 + \frac{6}{n + 1} ⟶ 2.$ This limit is not zero, contradicting [F1] for the compact $K$ ; hence the sequence is not Van Hove.

CounterexampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

A pair-count lattice sample covariance need not be positive definite

Statement refuted

For every finite scalar lattice sample, whenever the pair-count estimator from Lagged lattice covariance and variogram estimators is defined at all pairwise lags in a finite set of sites, the matrix of estimated lagged covariances on those sites is positive semidefinite.

Facts & Assumptions

Given: A finite scalar data vector on $V = {0, 1, 2} \subset Z$ and the estimator from Lagged lattice covariance and variogram estimators.

[F1]

Lagged lattice covariance and variogram estimators: for available lag $h$ , $\hat{C} (h)$ is the pair-count average of $Y_{j} Y_{j + h}$ minus the squared global sample mean.

[F2]

Positive definite functions on an abelian group: every finite quadratic form of a positive-definite covariance function must be nonnegative.

Counterexample

1.1F1givenalgebra

Take the data $(y_{0}, y_{1}, y_{2}) = (1, 2, 3)$ on $V = {0, 1, 2}$ . Its mean is $\overset{ˉ}{y} = 2$ . By [F1], the lag-zero estimator is $\hat{C} (0) = \frac{1^{2} + 2^{2} + 3^{2}}{3} - 2^{2} = \frac{2}{3},$ and the lag-two estimate, based on the single pair $(0, 2)$ , is $\hat{C} (2) = 1 \cdot 3 - 2^{2} = - 1.$ The negative-lag estimate equals $\hat{C} (- 2) = \hat{C} (2)$ because the reverse pair has the same product.

2.1F1F2step 1.1algebra∎

At the two sites $0$ and $2$ , all pairwise lag estimates are defined and the estimated covariance matrix is $(\begin{array}{cc} 2 / 3 & - 1 \\ - 1 & 2 / 3 \end{array}) .$ Its quadratic form at $v = (1, 1)^{T}$ equals $v^{T} (\begin{array}{cc} 2 / 3 & - 1 \\ - 1 & 2 / 3 \end{array}) v = - 2 / 3 < 0$ . This contradicts the necessary nonnegative quadratic-form condition [F2], so the stated universal claim is false.