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✓ 3 results · all verified · 0 also independently AI-judged

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Fairness criteria, impossibility boundaries, and limits — Examples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

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4 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

Equal base rates admit a constant calibrated score

Example

Let $Σ = {σ}$ and $B = {b}$ . In both groups take total mass $N_{t} = 1$ , feature proportion $a_{t σ} = 1$ , positive-outcome probability $p_{σ} = 1 / 2$ , and deterministic assignment $X_{σ b} = 1$ with score $v_{b} = 1 / 2$ . Each group then has base rate $1 / 2$ although the outcome is not determined by the feature, and the constant score is calibrated and balanced for both outcome classes.

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 one-feature, one-bin model in the example. Verify equal base rates, groupwise calibration, both class-balance criteria, and failure of perfect prediction.

[F1]

KMR's finite group-conditioned risk-score assignment and calibration: calibration is the binwise identity between positive mass and score times assigned mass; the base rate is the positive mass divided by group mass.

[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

1.1givenF1F2algebra

In either group, $ρ_{t} = a_{t σ} p_{σ} = 1 / 2$ . The sole bin has total mass $1$ and positive mass $1 / 2$ , so its calibration identity is $1 / 2 = (1 / 2) \cdot 1$ . Since $S$ is constantly $1 / 2$ , the positive and negative conditional mean scores are both $1 / 2$ in both groups, proving both balances.

2.1givenstep 1.1F1algebra∎

The outcome takes both values with probability $1 / 2$ , while $S = 1 / 2$ always; hence $S \neq Y$ almost surely. This is the equal-base-rate exception without perfect prediction.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

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.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

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.