LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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.

Tied upper ranks on a finite multiset

Statement

Let $a_{1}, \dots, a_{m} \in R$ with $m \geq 1$ . Let $J$ be uniform on ${1, \dots, m}$ and let $U$ be independent of $J$ and uniform on $[0, 1]$ . Set $R_{0} : = \frac{# {i : a_{i} \geq a_{J}}}{m}, R_{U} : = \frac{# {i : a_{i} > a_{J}} + U # {i : a_{i} = a_{J}}}{m} .$ Then $R_{U}$ is uniform on $[0, 1]$ , while $R_{0}$ is superuniform: $P (R_{0} \leq u) \leq u$ for every $u \in [0, 1]$ .

Facts & Assumptions

Given: A nonempty finite list of real numbers, a uniform random index $J$ , and an independent uniform variable $U$ .

[F1]

A uniform law on a nonempty finite set assigns equal weight $1 / m$ to each point. The uniform probability space on a nonempty finite set

[F2]

Every nonempty finite list of real values has a maximum; repeatedly removing all indices at that maximum sorts the distinct values. Every nonempty finite set of reals has a maximum and a minimum

[F3]

The tie breaker is independent of the selected index. Independent random elements

[A1]

$U$ has the uniform Borel law on $[0, 1]$ , so its probability on an interval is that interval's length.

Proof

1.1F1F2F3A1givenconstruct

Starting from the given list, repeatedly take the maximum of the remaining entries and remove every index carrying that value. By [F2] each nonempty remaining list has a maximum, and each iteration removes at least one of the original $m$ indices, so the process ends after at most $m$ iterations. It produces distinct values $b_{1} > b_{2} > \dots > b_{k}$ with block sizes $r_{j}$ . Put $c_{j} : = \sum_{ℓ < j} r_{ℓ}$ , so $\sum_{j} r_{j} = m$ . By [F1], the event that $a_{J} = b_{j}$ has probability $r_{j} / m$ . On that event, $R_{0} = (c_{j} + r_{j}) / m$ and, by [A1] and [F3], $R_{U} = (c_{j} + r_{j} U) / m$ , uniform on the interval $I_{j} = [c_{j} / m, (c_{j} + r_{j}) / m]$ of length $r_{j} / m$ .

2.1A1step 1.1algebra

By step 1.1, the intervals $I_{j}$ are consecutive, begin at $0$ , and end at $1$ . Conditional on $a_{J} = b_{j}$ , $R_{U} = (c_{j} + r_{j} U) / m$ , so for $u \in [0, 1]$ its conditional CDF is $0$ below the left endpoint, $(m u - c_{j}) / r_{j}$ between the endpoints, and $1$ above the right endpoint, by [A1]. Multiplying each conditional CDF contribution by $P (a_{J} = b_{j}) = r_{j} / m$ gives the length of $I_{j} \cap [0, u]$ . These lengths sum to $u$ because the intervals partition $[0, 1]$ up to shared endpoints. Hence $P (R_{U} \leq u) = u$ for every $u \in [0, 1]$ , so $R_{U}$ is uniform.

3.1step 1.1step 2.1algebra∎

For $u \in [0, 1]$ , $P (R_{0} \leq u)$ is the sum of $r_{j} / m$ over those blocks whose right endpoint $(c_{j} + r_{j}) / m$ is at most $u$ . These blocks form an initial segment; if the last is $j$ , the sum telescopes to $(c_{j} + r_{j}) / m \leq u$ , and if there is no such block the sum is zero. Hence $P (R_{0} \leq u) \leq u$ .

Depends on

Used by

Dependency tree · two levels

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Sources

Jesse Hemerik and Jelle Goeman, Exact testing with random permutations, randomized p-value result (standard reference, not scraped)