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

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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.

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Finite-group rank validity on a measurable sample space

Statement

Let a finite group $G$ act on a measurable space $(S, S)$ by measurable maps, and let $P$ be an invariant probability law: $P (g^{- 1} A) = P (A)$ for every $g \in G$ and $A \in S$ . For a measurable statistic $T : S \to R$ , define

$p_{0} (x) = \frac{# {g \in G : T (g \cdot x) \geq T (x)}}{∣ G ∣}, p_{U} (x, u) = \frac{# {g : T (g \cdot x) > T (x)} + u # {g : T (g \cdot x) = T (x)}}{∣ G ∣} .$

If $X \sim P$ and $U$ is independent uniform on $[0, 1]$ , then $p_{0} (X)$ is superuniform and $p_{U} (X, U)$ is uniform on $[0, 1]$ . The sample space need not be finite; only the acting group is finite.

Facts & Assumptions

Given: A finite group action by measurable maps, an invariant probability law, a measurable real statistic, and an independent uniform tie randomizer.

[F1]

Group and abelian group and Left group actions, transitive actions, and faithful actions: $G$ is a group acting with $e \cdot x = x$ and $(g h) \cdot x = g \cdot (h \cdot x)$ ; hence right multiplication by a fixed element is a bijection of $G$ .

[F2]

A measurable function between measurable spaces: the action maps and $T$ are measurable. The product sigma-algebra and its finite iterates makes the coordinate projections on $S \times [0, 1]$ measurable, and Arithmetic and lattice operations preserve measurability whenever they are defined preserves measurability under the finite sums and products in the rank formulas.

[F3]

Tied upper ranks on a finite multiset: for a finite list of real numbers, an independent uniform index gives a superuniform inclusive upper rank and a uniform tie-randomized upper rank.

[F4]

Independent random elements: the uniform group element and the tie randomizer are independent of the data and of each other.

Proof

Given: The hypotheses in the Statement. Prove both rank conclusions.

1.1F1F3construct

Fix $x \in S$ and form the finite list $(T (g \cdot x))_{g \in G}$ . For any $h \in G$ , the list $(T (g \cdot (h \cdot x)))_{g \in G}$ has exactly the same multiset of values, because right multiplication $g \mapsto g h$ is a bijection of $G$ by [F1]. Therefore, when $H$ is uniform on $G$ , $p_{0} (H \cdot x)$ and $p_{U} (H \cdot x, U)$ are exactly the inclusive and tie-randomized ranks of a uniform index in this list.

1.2F2givenalgebra

For each $g$ , $x \mapsto T (g \cdot x)$ is measurable by [F2]. Its difference from $T (x)$ is measurable, so the indicators of the strict and equal comparisons are measurable; finite sums and products with the second coordinate $u$ remain measurable by [F2]. Thus $p_{0}$ and the joint map $(x, u) \mapsto p_{U} (x, u)$ are measurable.

1.3F4givenalgebra

Let $H$ be uniform on $G$ , independent of $(X, U)$ . For every measurable $A$ , $P (H \cdot X \in A) = ∣ G ∣^{- 1} \sum_{h \in G} P (X \in h^{- 1} A) = P (A)$ by invariance. Since $H$ is independent of $U$ , the pair $(H \cdot X, U)$ has the same law as $(X, U)$ .

2.1F3step 1.1step 1.2step 1.3algebra∎

For each fixed $x$ , steps 1.1–1.2 and [F3] give $P_{H} (p_{0} (H \cdot x) \leq t) \leq t$ for $0 \leq t \leq 1$ , and $p_{U} (H \cdot x, U)$ is uniform. Integrating these statements over $X$ , then using equality in law from step 1.3, proves the two conclusions for $p_{0} (X)$ and $p_{U} (X, U)$ . No choice principle is used: all group averages and ranks are finite.

Depends on

Used by

Studentized Wilcoxon validity under heterogeneous laws Theorem

Dependency tree · two levels

30 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Jesse Hemerik and Jelle Goeman, Exact Testing with Random Permutations, Sections 2–3 (standard reference, not scraped)
Peter B. Stark, Statistics 240 Lecture Notes, Part 8: Resampling Methods, §1.3 (standard reference, not scraped)