Alphabeta Math
TheoremStatement: 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.

Benjamini–Hochberg FDR under independence

Statement

Let E=((X,A),Θ,(Pθ)θ∈Θ) be a statistical experiment, let m≥1, and fix a prespecified family of nonempty null hypotheses H01,…,H0m⊆Θ with valid p-value statistics pi:X→[0,1]. For each θ, put

I0(θ)={i:θ∈H0i},m0(θ)=∣I0(θ)∣.

Suppose that under every Pθ the m p-values are mutually independent and that

Pθ({x∈X:pi(x)≤u})=u(i∈I0(θ), 0≤u≤1).

False-null p-value laws may otherwise be arbitrary. For every q∈(0,1), the BH rule in Benjamini–Hochberg step-up procedure satisfies

FDR⁡θ=m0(θ)mq≤q.

This is an FDR guarantee; no FWER bound is asserted.

Facts & Assumptions

Given: The fixed finite family and statistical experiment in the Statement, q∈(0,1), and the stated independence and exact-uniformity conditions under each Pθ.

[F1]

False discovery proportion, FDR, and FWER: for true-null index set I0, V counts rejected true nulls, R is the number rejected, and FDR⁡=E[V/(R∨1)].

[F2]

Benjamini–Hochberg step-up procedure: for sorted values p(k), BH takes the largest R with p(R)≤Rq/m and rejects the first R ranks.

[F3]

Valid p-values: a p-value statistic is a measurable statistic into [0,1]; validity requires Pθ(pi≤u)≤u under every null law.

[F4]

Independent random elements: random elements on one probability space are independent when their generated sigma-algebras are independent.

[F5]

Disjoint groups of an independent sigma-algebra family remain independent: disjoint groups of an independent family of sigma-algebras remain independent after taking the sigma-algebras they generate.

[F6]

Independent random elements have product joint law: a finite independent family of random elements has joint law equal to the finite product of its marginal laws.

[F7]

Leave-one-out identity for the BH step-up rule: for each i, the leave-one-out variable Si is measurable with respect to the other p-values, and Ii/(R∨1)=1{pi≤(Si+1)q/m}/(Si+1).

[F8]

The expectation of an indicator is the probability of the event: for every event A, E[1A]=P(A).

[F9]

Linearity, monotonicity, and the modulus bound for expectation: finite sums of integrable random variables have expectation equal to the sum of their expectations.

[F10]

Expectation of a nonnegative or integrable random variable: expectation is the integral against the specified probability law.

[A1]

For every fixed θ, the p-values are mutually independent under Pθ, and each i∈I0(θ) has the exact uniform law, so Pθ({x∈X:pi(x)≤u})=u on [0,1].

Proof

1.1givenA1F2F3F4F5F6F7choose

Fix θ and work on the probability space (X,A,Pθ), regarding each Pi:=pi as a random element; [F3] makes these functions measurable. For k∈{1,…,m} and a∈R, {P(k)≤a}=⋃J⊆[m], ∣J∣=k⋂j∈J{Pj≤a}, so each rank-threshold event is measurable. The events {R=r} are finite Boolean combinations of the rank-threshold events, so R is measurable. If 0<R<m, maximality in [F2] gives P(R)≤Rq/m and P(R+1)>(R+1)q/m>Rq/m, so the rejected hypotheses are exactly those with Pi≤Rq/m. If R=m, all values are at most q=Rq/m; if R=0, no value is zero because then rank 1 would qualify. Hence the same characterization holds in every case. The rejection events are measurable because {Pi≤Rq/m}=⋃r=0m({R=r}∩{Pi≤rq/m}), so the BH rule is a measurable map into the finite set 2[m]. Now fix i∈I0(θ). [A1] and [F4] give independence; [F5] groups Pi and the remaining Pj, while [F7] makes Si measurable with respect to the latter group. Thus Pi and Si are independent random elements, and [F6] gives their product joint law.

2.1A1F6step 1.1algebra

For r∈{0,…,m−1} put tr=(r+1)q/m≤q<1. By [A1], Pθ(Pi≤tr)=tr; independence from step 1.1 and the product law [F6] give Pθ(Pi≤tr, Si=r)=trPθ(Si=r).

3.1F7F8F9F10step 2.1algebra

By [F7], partitioning on the finite-valued Si expresses Ii/(R∨1) as ∑r=0m−11{Pi≤tr,Si=r}/(r+1). Each event is measurable, and [F8] and [F10] give the absolute expectation of its summand as Pθ(Pi≤tr,Si=r)/(r+1)≤1, so each summand is integrable. Then [F8], [F9], and step 2.1 give Eθ[Ii/(R∨1)]=∑r=0m−1trPθ(Si=r)/(r+1)=(q/m)∑r=0m−1Pθ(Si=r)=q/m.

4.1F1F9step 1.1step 3.1algebra∎

The measurable BH rejection rule from step 1.1 lets us apply [F1] with P=Pθ and I0=I0(θ). Thus V=∑i∈I0(θ)Ii and FDR⁡θ=Eθ[V/(R∨1)]; repeated finite linearity [F9] and step 3.1 give FDR⁡θ=∑i∈I0(θ)q/m=(m0(θ)/m)q≤q, since m0(θ)≤m. This is the FDR defined in [F1], not an FWER conclusion.

Remarks

No choice axiom is used: the family is finite, each leave-one-out index is fixed in the proof, and the product-law supplier is applied only to finite families.

Depends on

Used by

Dependency tree · two levels

35 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