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A permutation orbit with tied ranks and nontrivial stabilizers

Example

Let $G = S_{3}$ permute the coordinates of $Ω = {e_{1}, e_{2}, e_{3}} \subset {0, 1}^{3}$ , give the three points equal probability, take $T (x) = x_{1}$ , and use an independent $U \sim Unif (0, 1)$ . Each point has a stabilizer of size two. The group-average inclusive ranks are $p_{0} (e_{1}) = 1 / 3$ and $p_{0} (e_{2}) = p_{0} (e_{3}) = 1$ . The randomized values are $p_{U} (e_{1}) = U / 3$ and $p_{U} (e_{2}) = p_{U} (e_{3}) = 1 / 3 + 2 U / 3$ , illustrating how tied ranks fill adjacent intervals even when each orbit point has multiple group preimages.

Facts & Assumptions

Given: The uniform finite sample space, coordinate-permutation action, statistic, and independent $U \sim Unif (0, 1)$ .

[F1]

A finite-group randomization setup requires an invariant null law and defines the orbit. Finite-group randomization setups

[F2]

Orbit ranks count transformed group elements with multiplicity. Orbit upper-tail rank p-values

[F3]

Under an invariant null, the inclusive orbit rank is superuniform and the randomized rank is uniform on each positive-probability orbit. Finite-group orbit-rank validity

[F4]

Conditional probability on a positive-probability finite event is the ratio of event probabilities. Conditional probability $P (A ∣ B)$ for $P (B) > 0$

Verification

1.1F1F4givenalgebra

The group $S_{3}$ permutes the three one-hot vectors transitively. Each vector has weight $1 / 3$ , so every permutation preserves the law and the null is invariant. The permutations fixing a given coordinate position are the identity and the transposition of the other two positions; hence each orbit point is produced by exactly two of the six group elements.

2.1F2step 1.1givenalgebra

At $e_{1}$ , $T (e_{1}) = 1$ ; two transformed values equal $1$ and four equal $0$ . Thus $p_{0} (e_{1}) = 2 / 6 = 1 / 3$ and $p_{U} (e_{1}) = 2 U / 6 = U / 3$ . At $e_{2}$ or $e_{3}$ , the observed statistic is $0$ ; two transformed values are strictly greater and four tie, so $p_{0} = 6 / 6 = 1$ and $p_{U} = (2 + 4 U) / 6 = 1 / 3 + 2 U / 3$ .

3.1F3step 2.1givenalgebra∎

Under the uniform null, $e_{1}$ occurs with probability $1 / 3$ , and $e_{2}, e_{3}$ together with probability $2 / 3$ . Conditional on the first case, $p_{U}$ is uniform on $[0, 1 / 3]$ ; conditional on the second, it is uniform on $[1 / 3, 1]$ . For $0 \leq u \leq 1 / 3$ , $P (p_{U} \leq u) = (1 / 3) (3 u) = u$ . For $1 / 3 \leq u \leq 1$ , $P (p_{U} \leq u) = 1 / 3 + (2 / 3) ((u - 1 / 3) / (2 / 3)) = u$ . Hence $p_{U}$ is uniform on $[0, 1]$ . The inclusive rank is $1 / 3$ with probability $1 / 3$ and $1$ with probability $2 / 3$ , so $P (p_{0} \leq u)$ is $0$ for $u < 1 / 3$ , $1 / 3$ for $1 / 3 \leq u < 1$ , and $1$ for $u = 1$ ; thus it is superuniform. These are the single-orbit conclusions of [F3].

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Sources

Peter B. Stark, Statistics 240 Lecture Notes, Part 8: Resampling Methods, §1.3, PDF pp. 10–11 (standard reference, not scraped)
Jesse Hemerik and Jelle Goeman, Exact testing with random permutations, Sections 2–3 (standard reference, not scraped)