ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pass

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.

Perfect prediction permits unequal base rates

Example

Let $Σ = {σ_{0}, σ_{1}}$ and $B = {b_{0}, b_{1}}$ . Set $p_{σ_{0}} = 0$ , $p_{σ_{1}} = 1$ , assign $σ_{i}$ deterministically to $b_{i}$ , and give $b_{0}, b_{1}$ scores $0, 1$ , respectively. Let group 1 have feature proportions $(a_{1 σ_{0}}, a_{1 σ_{1}}) = (3 / 4, 1 / 4)$ and group 2 have $(a_{2 σ_{0}}, a_{2 σ_{1}}) = (1 / 4, 3 / 4)$ , with $N_{1} = N_{2} = 1$ . The resulting base rates are $1 / 4$ and $3 / 4$ ; the score is perfectly predictive and satisfies all three criteria.

Example references

KMR's finite group-conditioned risk-score assignment and calibration: defines the finite group-conditioned law, calibration identity, base rate, and perfect-prediction convention.

Balance for positive and negative outcomes: defines the two outcome-conditioned mean-score balance criteria.

Facts & Assumptions

Given: The two-feature, two-bin model in the example. Verify the unequal base rates, calibration in each bin, perfect prediction, and both class-balance criteria.

[F1]

KMR's finite group-conditioned risk-score assignment and calibration: calibration is checked by the binwise positive-mass identity; perfect prediction means $S = Y$ almost surely in each group.

[F2]

Balance for positive and negative outcomes: positive and negative balance mean equality of the conditional mean score across groups, separately for the two outcomes.

Verification

1.1givenF1algebra

For each group, the positive mass is exactly its mass on $σ_{1}$ , so $ρ_{1} = 1 / 4$ and $ρ_{2} = 3 / 4$ . In bin $b_{0}$ , the positive mass is $0$ and the assigned mass is $a_{t σ_{0}}$ , giving $0 = 0 \cdot a_{t σ_{0}}$ ; in bin $b_{1}$ , the positive and assigned masses are both $a_{t σ_{1}}$ , giving $a_{t σ_{1}} = 1 \cdot a_{t σ_{1}}$ . Thus each group is calibrated.

2.1givenstep 1.1F2algebra∎

The score is $0$ on $σ_{0}$ and $1$ on $σ_{1}$ , exactly matching the deterministic outcome, so $S = Y$ in both groups. Therefore $α_{1} = α_{2} = 1$ and $β_{1} = β_{2} = 0$ , which proves both balances despite the unequal base rates.

Depends on

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