CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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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.
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Marginal split coverage need not give rare-subgroup coverage

Statement refuted

Marginal split-conformal coverage at level $1 - α$ forces coverage at least $1 - α$ conditional on every feature subgroup.

Facts & Assumptions

Given: The feature group space ${A, B}$ and response space ${0, 1}$ have their power-set sigma-algebras. The group variable satisfies $P (G = A) = 0.01$ and $P (G = B) = 0.99$ , and the response is always $Y = 0$ . There are nine calibration examples and one test example, all mutually independent and identically distributed according to this law. The score is $s (G, y) = 1 {G = A}$ , and $α = 0.1$ .

[F1]

Independence of random elements means their generated sigma-algebras are independent, so events determined by distinct examples factor. Independent random elements

[F2]

The split algorithm uses $k = ⌈ (m + 1) (1 - α) ⌉$ and predicts $y$ when its score is at most the calibration threshold. Split conformal prediction set

[F3]

The split conformal theorem guarantees marginal coverage at least $1 - α$ under exchangeability. Split conformal marginal coverage

Counterexample

1.1givenF1F2

For $m = 9$ and $α = 1 / 10$ , $k = ⌈ 10 \cdot 9 / 10 ⌉ = 9$ , so $q$ is the maximum of the nine calibration scores. Each score is $1$ with probability $1 / 100$ and $0$ with probability $99 / 100$ . The events ${G_{i} = B}$ are measurable from their respective examples, so [F1] makes them mutually independent and independent of ${G_{10} = A}$ . Conditional on $G_{10} = A$ , no calibration example is in group $A$ with probability $\prod_{i = 1}^{9} P (G_{i} = B) = (99 / 100)^{9}$ . The test score is $1$ and is covered exactly when at least one calibration example has group $A$ ; thus the conditional coverage is $1 - (99 / 100)^{9} \approx 864827 / 10^{7} < 9 / 10$ .

2.1givenF2F3step 1.1∎

A group- $B$ test has score $0 \leq q$ and is always covered. Hence marginal coverage is $99 / 100 + (1 / 100) (1 - (99 / 100)^{9}) \approx 9908648 / 10^{7} \geq 9 / 10$ , consistent with [F3], while the rare-group conditional coverage violates the refuted claim.

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