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Without null invariance a group-rank p-value can be anti-conservative

Statement refuted

For every finite group action and every null law, the inclusive group-average upper-tail rank is superuniform, even when the null law is not invariant under the group.

Facts & Assumptions

Given: A two-point finite sample space, a swapping group action, statistic $T (x) = x$ , the noninvariant null weights specified below, and an auxiliary $U \sim Unif [0, 1]$ independent of $X$ .

[F1]

A finite probability space assigns event probabilities by summing outcome weights. Finite probability spaces, outcome weights, events, and event probabilities

[F2]

A finite-group setup records the action, statistic, and invariance condition. Finite-group randomization setups

[F3]

The inclusive and tie-randomized orbit-rank formulas count transformed statistic values with multiplicity and divide by $∣ G ∣$ . Orbit upper-tail rank p-values

[A1]

We evaluate those numeric formulas on the stated action and statistic; this law deliberately falls outside the invariant-null setup of [F2], illustrating why its invariance premise is needed for validity.

Counterexample

Take $Ω = {0, 1}$ , $G = {e, s}$ with $s$ swapping $0$ and $1$ , and let $P (X = 1) = 9 / 10$ , $P (X = 0) = 1 / 10$ . This null law is not invariant because $P (s \cdot X = 1) = 1 / 10 \neq 9 / 10$ .

Given: The finite model and rank formula above.

1.1F1F2F3A1givenalgebra

For $x = 1$ , the transformed statistic values are $T (e \cdot 1) = 1$ and $T (s \cdot 1) = 0$ , so exactly one of the two group elements has value at least $T (1)$ and $p_{0} (1) = 1 / 2$ . For $x = 0$ , both transformed values $1, 0$ are at least $T (0) = 0$ , so $p_{0} (0) = 1$ .

2.1F1step 1.1givenalgebra

Therefore ${p_{0} \leq 1 / 2} = {X = 1}$ and $P (p_{0} \leq 1 / 2) = 9 / 10 > 1 / 2$ , contradicting superuniformity.

3.1F3A1step 1.1givenalgebra∎

The same example also defeats tie randomization: $p_{U} (1) = U / 2$ and $p_{U} (0) = (1 + U) / 2$ . Hence $P (p_{U} \leq 1 / 2) = 9 / 10$ , since at $X = 1$ the event always occurs and at $X = 0$ it requires $U = 0$ , an event of probability zero under the independent uniform law in the Given.

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Sources

Jesse Hemerik and Jelle Goeman, Exact testing with random permutations, Sections 2–3 (standard reference, not scraped)