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BY arbitrary dependence and BH PRDS — Examples

1 · Prerequisites

2 · Summary

The counterexample comes first: its valid p-values have positive covariance but fail PRDS, showing why pairwise dependence summaries alone do not establish the BH premise. It makes no BH FDR claim for that joint law. The second example uses a finite independent discrete null model, verifies validity and PRDS, and compares the BH and BY rejection counts on a positive-probability event.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

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.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

BY and BH thresholds for independent valid p-values

Example

Let $S = {1 / 100, 1 / 25, 13 / 100, 29 / 100, 1}$ and put masses $1 / 100, 3 / 100, 9 / 100, 16 / 100, 71 / 100$ on these five values. On the finite sample space $Ω = S^{4}$ equipped with its power-set sigma-algebra $F = P (S^{4})$ (equivalently, the finite product of the discrete sigma-algebras), use the product probability law. Let the four coordinate projections be p-values and make all four null hypotheses true in the one-point parameter experiment. Each p-value is measurable and valid, and the coordinates are independent. For $q = 1 / 5$ , the observed ordered vector

$(\frac{1}{100}, \frac{1}{25}, \frac{13}{100}, \frac{29}{100})$

has positive probability. BH rejects three hypotheses on this outcome, whereas BY rejects two. The independent model supplies the PRDS premise for BH, and BY's theorem needs only validity.

Facts & Assumptions

Given: The finite sample space and product law described in the Example, four true null hypotheses, and $q = 1 / 5$ .

[F1]

Statistical experiments: a statistical experiment is a measurable sample space, a parameter-label set, and a family of probability laws.

[F2]

Null-valid p-values and familywise and false-discovery error rates: p-value validity is the superuniform bound $P (p_{i} \leq u) \leq u$ at every $u \in [0, 1]$ under a true null.

[F3]

Positive regression dependence on true-null coordinates (PRDS): PRDS requires each true-null coordinate's conditional probabilities of increasing Borel sets to be nondecreasing on its support.

[F4]

Conditional law given a random element: a conditional-law kernel satisfies the joint-probability identity for every conditioning event and target event.

[F5]

Benjamini–Hochberg step-up procedure: BH uses critical values $k q / m$ and rejects the first $R$ ordered hypotheses, where $R$ is the largest passing rank.

[F6]

Benjamini–Yekutieli step-up procedure: BY uses critical values $k q / (m H_{m})$ and rejects the first $R$ ordered hypotheses.

[F7]

Harmonic numbers for set-cover analysis: $H_{4} = 1 + 1 / 2 + 1 / 3 + 1 / 4 = 25 / 12$ .

[F8]

BY controls FDR under arbitrary dependence: valid p-values under arbitrary dependence give BY FDR at most $(m_{0} / m) q$ .

[F9]

BH controls FDR under PRDS: valid p-values with PRDS on the true-null set give BH FDR at most $(m_{0} / m) q$ .

[A1]

The sample space is $Ω = S^{4}$ with $F = P (S^{4})$ ; equivalently, $F$ is the finite product of the discrete sigma-algebras on $S$ . The four coordinates have the specified independent common marginal law, and all four null hypotheses are true.

[A2]

Assume the Axiom of Choice (The Axiom of Choice) when invoking the general PRDS theorem's conditional-law convention; the finite kernel constructed below is explicit.

Verification

technique · direct finite calculation

Given: The product experiment, the marginal distribution, all four true nulls, and $q = 1 / 5$ .

1.1A1F1F2algebra

The five masses sum to one, and the sum of the product point masses on $S^{4}$ is $(\sum_{s \in S} μ ({s}))^{4} = 1$ , so these masses define a probability law on the finite sample space. For any subsets $B_{1}, \dots, B_{4} \subseteq S$ , summing the product masses over $B_{1} \times \dots \times B_{4}$ factors as $\prod_{i} μ (B_{i})$ ; thus the coordinate p-values are mutually independent. Each coordinate has support $1 / 100, 1 / 25, 13 / 100, 29 / 100, 1$ and cumulative probabilities $1 / 100, 4 / 100, 13 / 100, 29 / 100, 1$ , respectively. At each support value the cumulative probability equals that value; between successive support values it stays at the preceding value, and below $1 / 100$ it is zero. Thus $P (p_{i} \leq u) \leq u$ for every $u \in [0, 1]$ , so all four p-values are valid by [F2].

1.2A1F3F4givenconstruct

Fix a coordinate $i$ . For $x \in S$ , define $K_{i} (x, \cdot)$ as the law of the four-vector obtained by fixing coordinate $i$ at $x$ and drawing the other three coordinates independently from the specified marginal; for $x \notin S$ , use a fixed point mass. Since $S$ is finite, each evaluation is Borel measurable and each section is a probability measure. For Borel $A \subseteq [0, 1]$ and $D \subseteq [0, 1]^{4}$ , product-mass factorization gives $\int_{A} K_{i} (x, D) P_{p_{i}} (d x) = P (p_{i} \in A, p \in D)$ , so this is a conditional law by [F4]. If $D$ is coordinatewise increasing and $x \leq x^{'}$ in $S$ , then fixing any same remaining triple makes the vector at $x$ coordinatewise at most the vector at $x^{'}$ , so its indicator of $D$ is no larger. Averaging over the same three-coordinate product law yields $K_{i} (x, D) \leq K_{i} (x^{'}, D)$ . This holds for every $i$ , and [F3] gives PRDS on the all-true-null set.

1.3A1F5F6F7givenalgebra

The event that the four coordinates take the four distinct values $1 / 100, 1 / 25, 13 / 100, 29 / 100$ in any order has probability $4! (1 / 100) (3 / 100) (9 / 100) (16 / 100) > 0$ . On this event $H_{4} = 25 / 12$ . BH's critical values are $1 / 20, 1 / 10, 3 / 20, 1 / 5$ , so the first three ordered p-values pass and the fourth does not, giving $R_{B H} = 3$ . BY's critical values are $3 / 125, 6 / 125, 9 / 125, 12 / 125$ , so ranks one and two pass and rank three does not, giving $R_{B Y} = 2$ .

2.1A1A2F8F9step 1.3algebra∎

Here $m_{0} = m = 4$ . By [F8] and [F9], each rule has FDR at most $q = 1 / 5$ under this model; step 1.3 gives a positive-probability event on which their rejection counts differ.