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.

Split conformal coverage can be conservative when scores tie

Example

Take a constant score rule $s (z) = 0$ , two calibration examples, one exchangeable test example, and $α = 1 / 2$ . All three scores are tied. The ordinary split quantile set covers with probability $1$ , while the specified independent uniform tie-randomized rank set covers with probability exactly $1 / 2$ .

Facts & Assumptions

Given: $m = 2$ , $α = 1 / 2$ , all calibration and test scores equal $0$ , and an independent $U$ with Lebesgue probability law on $[0, 1]$ .

[F1]

The split algorithm sets $k = ⌈ (m + 1) (1 - α) ⌉$ , uses the $k$ th calibration score when $k \leq m$ , and otherwise uses $+ \infty$ . Split conformal prediction set

[F2]

The tie-randomized p-value counts strictly larger calibration scores and multiplies the full tie-group size, including the candidate, by $U$ . Conformal scores and rank p-values

[F3]

Under exchangeability, the randomized rank p-value is uniform. Conformal rank superuniformity under exchangeability

Verification

1.1givenF1

Here $k = ⌈ 3 / 2 ⌉ = 2 \leq m$ and the second calibration order statistic is $q = 0$ . The test score is also $0$ , so $0 \leq q$ and the nonrandomized prediction set always includes the true response.

1.2givenF2

There are no calibration scores strictly greater than the test score and there are two calibration ties; including the candidate gives tie count three. Thus $p_{U} = 3 U / 3 = U$ , and the tie-randomized set covers exactly when $U > 1 / 2$ , an event of probability $1 / 2$ .

2.1F3step 1.1step 1.2∎

The ordinary coverage $1$ is conservative relative to the nominal $1 - α = 1 / 2$ , whereas the randomized coverage is exact and agrees with [F3].

Depends on

Used by

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