CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-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.

Positive covariance does not imply PRDS

Statement refuted

Positive pairwise covariance of valid p-values is sufficient for their joint law to be PRDS on the true-null coordinates.

Facts & Assumptions

Given: A finite statistical experiment, two p-value statistics, and both null hypotheses true at its only parameter value.

[F1]

Statistical experiments: an experiment is a measurable sample space, a parameter-label set, and a probability law for each parameter.

[F2]

Valid p-values: a p-value statistic is a measurable map from the sample space into $[0, 1]$ .

[F3]

Null-valid p-values and familywise and false-discovery error rates: a p-value is valid for a null when $P (p_{i} \leq u) \leq u$ for every $u \in [0, 1]$ under each parameter in that null.

[F4]

Positive regression dependence on true-null coordinates (PRDS): PRDS requires conditional probabilities of every coordinatewise increasing Borel set to have a nondecreasing version on the support of each true-null coordinate.

[F5]

Conditional law given a random element: a conditional-law kernel satisfies $\int_{A} K (x, D) P_{P_{1}} (d x) = P (P_{1} \in A, P \in D)$ for Borel $A, D$ .

[A1]

Let $Ω = {ω_{1}, ω_{2}, ω_{3}, ω_{4}}$ with its power-set sigma-algebra, let $Θ = {θ_{0}}$ , and assign probabilities $1 / 10, 1 / 5, 2 / 5, 3 / 10$ to the four states. Set both null hypotheses equal to $Θ$ and define the p-values by the following state table:

State	Probability under $θ_{0}$	$p_{1}$	$p_{2}$
$ω_{1}$	$1 / 10$	$1 / 5$	$1 / 5$
$ω_{2}$	$1 / 5$	$3 / 5$	$4 / 5$
$ω_{3}$	$2 / 5$	$4 / 5$	$3 / 5$
$ω_{4}$	$3 / 10$	$1$	$1$

[A2]

The PRDS statement uses the standard-Borel conditional-law interface stated under the Axiom of Choice (The Axiom of Choice); the finite positive-mass conditional probabilities below are forced for every version.

Counterexample

1.1F1F2F3A1algebra

The four probabilities in [A1] sum to one, so they define the sampling law of a one-parameter experiment [F1]. Since the sample space is finite with its power-set sigma-algebra, each table coordinate is a measurable $[0, 1]$ -valued map and hence a p-value statistic by [F2]. The marginal masses of $p_{1}$ at $1 / 5, 3 / 5, 4 / 5, 1$ are $1 / 10, 1 / 5, 2 / 5, 3 / 10$ , giving cumulative probabilities $1 / 10, 3 / 10, 7 / 10, 1$ . Those are at most their corresponding support values. Between consecutive support values the cumulative probability stays fixed at the preceding value, and below $1 / 5$ it is zero; thus $P_{θ_{0}} (p_{1} \leq u) \leq u$ for every $u \in [0, 1]$ . For $p_{2}$ , the masses at $1 / 5, 3 / 5, 4 / 5, 1$ are $1 / 10, 2 / 5, 1 / 5, 3 / 10$ , giving cumulative probabilities $1 / 10, 1 / 2, 7 / 10, 1$ , again at most the support values and hence at most every intervening $u$ . Both p-values are valid by [F3].

1.2A1algebra

Direct summation in [A1] gives $E [p_{1}] = 19 / 25$ , $E [p_{2}] = 18 / 25$ , and $E [p_{1} p_{2}] = 74 / 125$ , hence $Cov (p_{1}, p_{2}) = 74 / 125 - (19 / 25) (18 / 25) = 28 / 625 > 0$ . Both marginal laws are nonconstant, so both variances are positive and the Pearson correlation is positive as well.

2.1F4F5A1A2step 1.1step 1.2algebra∎

The support of $p_{1}$ is ${1 / 5, 3 / 5, 4 / 5, 1}$ , with positive mass at every point. Let $D = {(x, y) \in [0, 1]^{2} : y \geq 4 / 5}$ , a coordinatewise increasing Borel set. At each support point $x$ , [F5] with conditioning event $A = {x}$ forces $K (x, D) = P (p_{1} = x, (p_{1}, p_{2}) \in D) / P (p_{1} = x)$ for every conditional-law version. The table makes these conditional probabilities at $x = 1 / 5, 3 / 5, 4 / 5, 1$ equal to $0, 1, 0, 1$ , respectively. Every version therefore decreases from $1$ at $3 / 5$ to $0$ at $4 / 5$ , violating the monotonicity in [F4] because $3 / 5 < 4 / 5$ . By steps 1.1 and 1.2, these are valid p-values with positive covariance; thus the law is not PRDS on the true-null set and refutes the statement. It makes no BH FDR claim for this law.

Remarks

The counterexample concerns the dependence premise only. Positive pairwise covariance does not verify the PRDS condition required by the ordinary BH theorem.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

25 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.