ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)

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.

No fixed sample budget uniformly protects arbitrarily rare modes

Example

For the two-mode posterior $π_{w} = w δ_{- b} + (1 - w) δ_{b}$ with $b > 0$ and $0 < w < 1 / 2$ , $M \geq 1$ independent exact draws miss the rare mode $- b$ with probability $(1 - w)^{M}$ . For any fixed $M$ , this probability tends to one as $w ↓ 0$ . If instead $w \in [w_{\min}, 1 - w_{\min}]$ with $0 < w_{\min} \leq 1 / 2$ , then the probability of missing either mode is at most $2 (1 - w_{\min})^{M}$ , so $M \geq \log (2 / α) / [- \log (1 - w_{\min})]$ suffices to make it at most $α \in (0, 1)$ .

Example references

A positive lower bound on posterior mode mass is one condition needed for uniform finite-budget guarantees.

Facts & Assumptions

Given: $b > 0$ , a mixture weight $0 < w < 1 / 2$ , $M \geq 1$ independent draws from $π_{w} = w δ_{- b} + (1 - w) δ_{b}$ , and, for the final bound, $w \in [w_{\min}, 1 - w_{\min}]$ with $0 < w_{\min} \leq 1 / 2$ .

[A1]

Each draw independently equals the rare atom $- b$ with probability $w$ and the other atom with probability $1 - w$ .

Verification

1.1givenA1algebra

The event that no draw equals $- b$ is the intersection of $M$ independent events, each of probability $1 - w$ , hence has probability $(1 - w)^{M}$ . Its limit as $w ↓ 0$ is one for every fixed finite $M$ .

2.1givenA1step 1.1algebra∎

The event that at least one of the two atoms is absent is contained in the union of the event “no $- b$ draw” and the event “no $b$ draw.” Their probabilities are $(1 - w)^{M}$ and $w^{M}$ , each at most $(1 - w_{\min})^{M}$ under the displayed weight interval. The union bound gives at most $2 (1 - w_{\min})^{M}$ . If $M \geq \log (2 / α) / [- \log (1 - w_{\min})]$ , then $(1 - w_{\min})^{M} \leq α / 2$ , proving the sufficient budget.

Used by

Nothing in the library uses this result yet.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.