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Randomized conditional p-values are uniform

Statement

Under the simple null and the supplied conditional law in Randomized conditional upper-tail p-values, for every $u \in [0, 1]$ , $P_{0} (p_{U} \leq u ∣ σ (S)) = u P_{0} -almost surely,$ while the nonrandomized upper-tail value satisfies $P_{0} (p_{0} \leq u ∣ σ (S)) \leq u P_{0} -almost surely .$ Thus the randomized value is conditionally uniform and the inclusive-tail value is conditionally superuniform, including when the conditional law has atoms.

Facts & Assumptions

Given: A simple null law $P_{0}$ , statistics $S, T$ , a supplied conditional law kernel $K (s, \cdot)$ of $T$ given $S$ , and an independent uniform randomizer $U$ .

[F1]

The randomized and nonrandomized values are $p_{U} = K (S, (T, \infty)) + U K (S, {T})$ and $p_{0} = K (S, [T, \infty))$ . Randomized conditional upper-tail p-values

[F2]

The kernel $K (s, \cdot)$ is a conditional law of $T$ given $S$ . Conditional law given a random element

[F3]

$U$ is independent of the pair $(S, T)$ . Independent random elements

[F4]

For sigma-finite measures, the product measure exists and has the rectangle formula. For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique

[F5]

Measurable rectangles generate the product sigma-algebra. The product sigma-algebra and its finite iterates

[F6]

Equality of finite measures on a generating pi-system extends to the generated sigma-algebra. Dynkin's pi-lambda theorem

[F7]

For any real law with CDF $F$ and independent uniform $W$ , $F (Y -) + W (F (Y) - F (Y -))$ is uniform. Randomizing across CDF jumps gives a uniform variable

[A1]

The randomizer has the uniform law on $[0, 1]$ , namely $P (U \leq u) = u$ for $u \in [0, 1]$ ; this is stipulated in the p-value definition.

Proof

1.1F2F3F4F5F6A1

For each $s$ , let $Λ_{s} : = K (s, \cdot) \otimes λ$ , where $λ$ is the uniform law of $U$ . By [F4] this is a probability measure with rectangle values $Λ_{s} (C \times D) = K (s, C) λ (D)$ . For $A \in σ (S)$ and Borel $C, D$ , the conditional-law identity [F2] gives $\int_{A} Λ_{S} (C \times D) d P_{0} = λ (D) P_{0} (A \cap {T \in C})$ . Independence in [F3] makes the right side $P_{0} (A \cap {T \in C, U \in D})$ . The class of product-Borel sets where the evaluation $s \mapsto Λ_{s} (E)$ is measurable and every such conditional identity holds is a Dynkin system: complements subtract from the probability-one whole-space evaluation, and disjoint unions follow by countable additivity and increasing limits. It contains the generating rectangles, so [F5] and [F6] extend measurability and the conditional identity to every product-Borel set. Thus $Λ_{S}$ is a conditional law of $(T, U)$ given $S$ .

2.1A1F1F7step 1.1algebra

For almost every conditioning value $s$ , the conditional law just constructed makes $T$ have law $K (s, \cdot)$ and keeps $U$ independent uniform. Write $F_{s} (t) : = K (s, (- \infty, t])$ and $Δ_{s} (t) : = F_{s} (t) - F_{s} (t -)$ . By [F7], $V = F_{s} (T -) + (1 - U) Δ_{s} (T)$ is uniform, since $1 - U$ is also uniform. The p-value formula in [F1] gives $p_{U} = 1 - V$ , so $p_{U}$ is conditionally uniform.

3.1F1step 2.1A1algebra∎

For the same $s$ , $p_{0} = 1 - F_{s} (T -)$ and $p_{U} = 1 - F_{s} (T -) - (1 - U) Δ_{s} (T) \leq p_{0}$ . Hence ${p_{0} \leq u} \subseteq {p_{U} \leq u}$ conditionally on $S = s$ . Step 2.1 gives $P_{0} (p_{0} \leq u ∣ S = s) \leq u$ . This holds for each $u \in [0, 1]$ and almost every conditioning value, proving conditional superuniformity.

Depends on

Used by

Fisher's exact test from two Bernoulli samples Example

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Sources

A. J. Brockwell, Universal Residuals: A Multivariate Transformation, Lemma 2 and Theorem 1 (standard reference, not scraped)
Jesse Hemerik and Jelle Goeman, Exact testing with random permutations, randomized p-value result (standard reference, not scraped)