CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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 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.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 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.