Alphabeta Math
ExampleConstruction: 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.

Fallback when the calibration correction exceeds the target

Example

Let n,M≥1, B>0, and 0≤α<B/(n+1). Consider any loss-nested finite-grid family with calibration losses in [0,B] and endpoint losses Li(λM)≤α almost surely. The B-corrected calibration threshold is too large to pass at any grid point, so the rule falls back to λM. The endpoint condition then bounds the realized test loss, and the expected test risk is at most α.

Example references

Finite nested decision families and bounded risk: nonnegative bounded losses and the safe maximum-grid endpoint.

Finite-grid conformal risk control: the pass threshold includes the B correction and otherwise returns the maximum grid point.

Angelopoulos et al., Conformal Risk Control, §1.1

Facts & Assumptions

Given: A finite grid, n≥1, B>0, 0≤α<B/(n+1), calibration losses in [0,B], and endpoint test loss at most α almost surely.

[F1]

Losses are nonnegative and endpoint-safe, with Ln+1(λM)≤α almost surely. Finite nested decision families and bounded risk

[F2]

At grid point λj, the algorithm tests (∑i=1nLi(λj)+B)/(n+1) and uses λM when no point passes. Finite-grid conformal risk control

[F3]

Expectation preserves almost-sure inequalities for integrable real random variables. Linearity, monotonicity, and the modulus bound for expectation

Verification

1.1givenF2algebra

Since every calibration loss is nonnegative, for each j the algorithm's threshold satisfies (∑i=1nLi(λj)+B)/(n+1)≥B/(n+1)>α. No point passes, so the rule returns λM.

2.1givenF1F3step 1.1∎

The endpoint condition gives 0≤Ln+1(λM)≤α almost surely. The loss is bounded and integrable, so [F3] implies E[Ln+1(λM)]≤α.

Depends on

Used by

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Dependency tree · two levels

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Sources