CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge 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.

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Pair-count-normalized sample ACVF need not be positive semidefinite

Statement refuted

For every finite real data vector, the pair-count-normalized sample ACVF ${\tilde{γ}}_{n}$ has nonnegative Toeplitz quadratic forms.

Facts & Assumptions

Given: The pair-count-normalized sample ACVF definition for a finite data vector of length $n \geq 1$ .

[F1]

For $∣ h ∣ < n$ , the pair-count-normalized rescaling divides the centered lag-product sum by $n - ∣ h ∣$ and is zero outside that lag range (Autocovariance and sample autocovariance).

Counterexample

1.1F1givenalgebra

Take $n = 3$ and data $(x_{1}, x_{2}, x_{3}) = (1, 0, 0)$ . Its sample mean is $1 / 3$ , so its centered values are $(2 / 3, - 1 / 3, - 1 / 3)$ . By [F1], ${\tilde{γ}}_{3} (0) = 2 / 9$ , ${\tilde{γ}}_{3} (1) = - 1 / 18$ , and ${\tilde{γ}}_{3} (2) = - 2 / 9$ ; negative-lag values equal the corresponding positive-lag values.

2.1step 1.1algebra∎

In the Toeplitz matrix on indices $0, 1, 2$ , the diagonal entries are $2 / 9$ , the first off-diagonal entries are $- 1 / 18$ , and the outer off-diagonal entries are $- 2 / 9$ by step 1.1. For $v = (1, \frac{1}{2}, 1)^{T}$ , the quadratic form is $v^{T} T v = (2 / 9) (1 + 1 / 4 + 1) + 2 (- 1 / 18) (1 / 2 + 1 / 2) + 2 (- 2 / 9) = - 1 / 18 < 0$ . A positive-semidefinite Toeplitz form must be nonnegative on every real vector, so this data vector refutes the statement.

Depends on

Autocovariance and sample autocovariance

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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