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Calibration and both outcome-balance criteria force equal base rates or perfect prediction

Statement

Let $Σ$ and $B$ be nonempty finite sets. In group $t \in {1, 2}$ , feature type $σ \in Σ$ has proportion $a_{t σ} \geq 0$ , where $\sum_{σ} a_{t σ} = 1$ , and total mass $N_{t} > 0$ . The positive binary outcome $Y = 1$ has probability $p_{σ} \in [0, 1]$ for feature type $σ$ , with $p_{σ}$ the same in both groups. A common randomized assignment chooses bin $b \in B$ from feature type $σ$ with probability $X_{σ b} \geq 0$ , where $\sum_{b} X_{σ b} = 1$ ; this choice is independent of group and outcome. Bin $b$ has score $v_{b} \in [0, 1]$ , and $S = v_{B}$ . Define $ρ_{t} = \sum_{σ} a_{t σ} p_{σ}$ and $μ_{t} = N_{t} ρ_{t}$ . Assume $0 < μ_{t} < N_{t}$ for both groups. Define the positive- and negative-outcome mean scores by

α_{t} = \frac{\sum_{σ, b} a_{t σ} p_{σ} X_{σ b} v_{b}}{ρ_{t}}, β_{t} = \frac{\sum_{σ, b} a_{t σ} (1 - p_{σ}) X_{σ b} v_{b}}{1 - ρ_{t}} .

Suppose $α_{1} = α_{2}$ and $β_{1} = β_{2}$ , and the assignment is calibrated within each group, meaning

\sum_{σ} a_{t σ} p_{σ} X_{σ b} = v_{b} \sum_{σ} a_{t σ} X_{σ b}

for every group $t$ and bin $b$ , and suppose the mean score among positive outcomes is equal across groups and the mean score among negative outcomes is equal across groups. Then either $ρ_{1} = ρ_{2}$ or $S = Y$ almost surely under each group law. In the latter case, every feature type with positive mass in either group has $p_{σ} \in {0, 1}$ , so the instance permits perfect prediction.

Statement references

Kleinberg, Mullainathan, and Raghavan, complete 23-page paper: Theorem 1.1 and the complete proof in §2 establish the equal-base-rate or perfect-prediction alternatives under these three named score criteria.

Applied Statistics: From Data to Decisions, full chapter: the section “The impossibility: you cannot satisfy them all” states the real-valued score theorem and distinguishes it from the separate binary-classifier theorem.

Facts & Assumptions

Given: The finite two-group risk-score model in the statement, groupwise calibration, and balance of the positive-class and negative-class mean scores. Prove equal base rates or perfect prediction.

[F1]

KMR's finite group-conditioned risk-score assignment and calibration: the group law is finite, and the same row-stochastic feature-to-bin assignment is used independently of group and outcome, with scores in $[0, 1]$ .

[F2]

KMR's finite group-conditioned risk-score assignment and calibration: for every group and bin, calibration gives $\Pr_{t} (Y = 1, B = b) = v_{b} \Pr_{t} (B = b)$ .

[F3]

Balance for positive and negative outcomes: the class-conditioned means are $α_{t} = E_{t} [S ∣ Y = 1]$ and $β_{t} = E_{t} [S ∣ Y = 0]$ ; balance means $α_{1} = α_{2}$ and $β_{1} = β_{2}$ .

[F4]

Class-conditional mean risk scores are areas under threshold rates: in each group, $α_{t} = \int_{0}^{1} {TPR}_{t} (u) d u$ and $β_{t} = \int_{0}^{1} {FPR}_{t} (u) d u$ .

Proof

1.1givenF2F3F4algebra

Set $α = α_{1} = α_{2}$ and $β = β_{1} = β_{2}$ by [F3]; since $S \in [0, 1]$ , both common means lie in $[0, 1]$ . By [F4], these means are the respective areas under the groups' true-positive-rate and false-positive-rate threshold functions. Summing the calibration identity [F2] over all bins gives $E_{t} [S] = \Pr_{t} (Y = 1) = ρ_{t}$ for each $t$ . Partitioning this finite expectation by $Y$ gives $ρ_{t} = ρ_{t} α + (1 - ρ_{t}) β$ , hence $ρ_{t} (1 - α) = (1 - ρ_{t}) β$ .

2.1givenstep 1.1algebra

If $α < 1$ , then $1 - α > 0$ and $1 - ρ_{t} > 0$ , so the last identity yields $ρ_{t} / (1 - ρ_{t}) = β / (1 - α)$ for both groups. Therefore $ρ_{1} / (1 - ρ_{1}) = ρ_{2} / (1 - ρ_{2})$ ; cross-multiplication gives $ρ_{1} = ρ_{2}$ . If instead $α = 1$ , the identity and $1 - ρ_{t} > 0$ give $β = 0$ .

3.1givenF1F3step 2.1algebra∎

In the branch $α = 1$ and $β = 0$ , each group has positive and negative mass by hypothesis. The score has finite support in $[0, 1]$ , so a conditional mean of one on $Y = 1$ is possible only when every positive-mass score value is $1$ ; a conditional mean of zero on $Y = 0$ is possible only when every positive-mass score value is $0$ . Thus $S = Y$ almost surely in both groups. If a feature type $σ$ with positive mass had $0 < p_{σ} < 1$ , both outcomes would have positive mass at $σ$ . The same assignment row $X_{σ b}$ is used for both outcomes [F1], so their conditional score laws would agree, contradicting concentration at scores $1$ and $0$ , respectively. Hence every occupied feature type has $p_{σ} \in {0, 1}$ , which permits perfect prediction.

Remarks

The proof is choice-free: it uses finite sums, finite conditional means, and algebra only. The theorem concerns the named calibration and class-conditioned score means; it does not assert a theorem about demographic parity, equalized odds, causal fairness, or all fairness criteria. The approximate result in KMR Theorem 1.2 is not asserted here; it requires its own exact approximation definitions and function $f (ε)$ .

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Sources

Jon Kleinberg, Sendhil Mullainathan, and Manish Raghavan, Inherent Trade-Offs in the Fair Determination of Risk Scores, arXiv:1609.05807v2 (2016), Theorem 1.1 and §2 proof (standard reference, not scraped)
Applied Statistics: From Data to Decisions, MS&E 125, Fairness in Algorithmic Decision-Making, section ‘The impossibility: you cannot satisfy them all’ (standard reference, not scraped)