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.

Binary laws with a rare event are not contiguous

Statement

For $n \geq 2$ , define probability laws on ${0, 1}$ by $P_{n} {1} = 1 / n, Q_{n} {1} = 1 / 2.$ Then $Q_{n}$ is not contiguous with respect to $P_{n}$ . The likelihood ratios $L_{n} = d Q_{n} / d P_{n}$ exist, but their $P_{n}$ -distributional limit has mean less than one; therefore the matching-mean hypothesis in A mean-one likelihood-ratio limit implies contiguity is essential.

Facts & Assumptions

Given: The two-point measurable space, the probability laws above, and the event-supremum definition of contiguity.

[F1]

Contiguous sequences of probability measures: $Q_{n} ◃ P_{n}$ would require $Q_{n} (A_{n}) \to 0$ whenever $P_{n} (A_{n}) \to 0$ .

[F2]

A mean-one likelihood-ratio limit implies contiguity: A nonnegative likelihood ratio with mean one and a nonnegative distributional limit of mean one gives contiguity.

Proof

technique · exhibit a rare event and compute the likelihood ratio

1.1F1givenalgebra

Let $A_{n} = {1}$ . Then $P_{n} (A_{n}) = 1 / n \to 0$ , while $Q_{n} (A_{n}) = 1 / 2$ for every $n$ . By [F1], this disproves $Q_{n} ◃ P_{n}$ .

2.1F2givenalgebra∎

On the two atoms, the density is $L_{n} (1) = \frac{Q_{n} {1}}{P_{n} {1}} = \frac{n}{2}, L_{n} (0) = \frac{Q_{n} {0}}{P_{n} {0}} = \frac{n}{2 (n - 1)} .$ Under $P_{n}$ , the value $n / 2$ occurs with probability $1 / n$ and the other value tends to $1 / 2$ , so $L_{n} \Rightarrow 1 / 2$ . Each $E_{P_{n}} L_{n} = 1$ , but $E (1 / 2) = 1 / 2 \neq 1$ ; hence the matching-mean condition of [F2] fails.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 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

John Duchi, A few notes on contiguity, asymptotics, and local asymptotic normality, §1.2 (standard reference, not scraped)