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.

Threshold rates can vary despite balanced mean scores

Example

Let group 1 have two feature types $σ_{1}, σ_{2}$ , each of proportion $1 / 2$ , with outcome probabilities $p_{σ_{1}} = 1 / 5$ and $p_{σ_{2}} = 4 / 5$ . Let group 2 have three feature types $τ_{1}, τ_{2}, τ_{3}$ with proportions $9 / 32, 14 / 32, 9 / 32$ and outcome probabilities $p_{τ_{1}} = 1 / 10$ , $p_{τ_{2}} = 1 / 2$ , and $p_{τ_{3}} = 9 / 10$ . Each feature type is assigned deterministically to its own bin, whose score equals that type's outcome probability; this is one common feature-based assignment for both groups, with $N_{1} = N_{2} = 1$ . Both groups have base rate $1 / 2$ , and the assignment is calibrated in each group with positive- and negative-outcome mean scores $17 / 25$ and $8 / 25$ , respectively. At threshold $u = 1 / 2$ , the group 1 true-positive and false-positive rates are $4 / 5$ and $1 / 5$ , while the group 2 rates are $151 / 160$ and $79 / 160$ . At threshold $u = 9 / 10$ , group 1's two rates are $0, 0$ and group 2's are $81 / 160, 9 / 160$ . Thus the threshold rates change with the threshold even though both class-conditioned mean-score criteria hold.

Example references

KMR's finite group-conditioned risk-score assignment and calibration: defines the finite group law, shared feature-to-bin assignment, calibration, and score convention.

Balance for positive and negative outcomes: defines the positive- and negative-outcome conditional mean scores.

Class-conditional mean risk scores are areas under threshold rates: identifies those class-conditioned means with areas under the threshold-rate functions.

Facts & Assumptions

Given: The finite distributions and deterministic feature-to-bin assignment in the example. Verify calibration, class-mean balance, and the threshold-rate values at $u = 1 / 2$ and $u = 9 / 10$ .

[F1]

KMR's finite group-conditioned risk-score assignment and calibration: groupwise calibration is the identity between positive mass in a bin and its score times the bin's assigned mass; scores lie in $[0, 1]$ .

[F2]

Balance for positive and negative outcomes: $α_{t}$ and $β_{t}$ are the class-conditioned mean scores, and equality across groups separately defines the two balance criteria.

[F3]

Class-conditional mean risk scores are areas under threshold rates: the mean scores $α_{t}$ and $β_{t}$ are the integrals of ${TPR}_{t}$ and ${FPR}_{t}$ over thresholds.

Verification

1.1givenF1algebra

In both groups the weighted score mean is $E_{t} [S] = 1 / 2$ . For group 1, $E_{1} [S^{2}] = \frac{1}{2} ((1 / 5)^{2} + (4 / 5)^{2}) = 17 / 50$ . For group 2, $E_{2} [S^{2}] = \frac{9}{32} (1 / 10)^{2} + \frac{14}{32} (1 / 2)^{2} + \frac{9}{32} (9 / 10)^{2} = 17 / 50$ . Each bin contains one feature type and its score equals that type's $p_{σ}$ , so its positive mass is its score times its assigned mass; hence calibration holds. The common positive mass/base rate is $1 / 2$ .

2.1givenstep 1.1F2F3algebra

By calibration, $E_{t} [S 1_{{Y = 1}}] = E_{t} [S^{2}] = 17 / 50$ and $E_{t} [S 1_{{Y = 0}}] = E_{t} [S] - E_{t} [S^{2}] = 8 / 50$ . Dividing by the positive and negative masses $1 / 2$ gives $α_{t} = 17 / 25$ and $β_{t} = 8 / 25$ for both groups, so both balances hold. By [F3], the two groups' TPR areas agree and their FPR areas agree.

3.1givenF1step 2.1algebra

At $u = 1 / 2$ , group 1 has positive mass $(1 / 2) (4 / 5) = 2 / 5$ and negative mass $(1 / 2) (1 / 5) = 1 / 10$ at scores at least $1 / 2$ , so its rates are $(2 / 5) / (1 / 2) = 4 / 5$ and $(1 / 10) / (1 / 2) = 1 / 5$ . Group 2 has positive mass $(14 / 32) (1 / 2) + (9 / 32) (9 / 10) = 151 / 320$ and negative mass $(14 / 32) (1 / 2) + (9 / 32) (1 / 10) = 79 / 320$ at scores at least $1 / 2$ , so its rates are $151 / 160$ and $79 / 160$ .

4.1givenF1step 3.1algebra∎

At $u = 9 / 10$ , group 1 has no score at least $9 / 10$ , so both rates are zero. Group 2's only qualifying score is $9 / 10$ , contributing positive mass $(9 / 32) (9 / 10) = 81 / 320$ and negative mass $(9 / 32) (1 / 10) = 9 / 320$ ; dividing each by $1 / 2$ gives rates $81 / 160$ and $9 / 160$ . These computations show threshold dependence despite the mean balances.

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