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Exact-orbit and finite-B permutation p-value resolution

Example

Let $Ω = G = {e, a, a^{2}}$ with multiplication $a^{3} = e$ , so $a^{i} a^{j} = a^{i + j (m o d 3)}$ . The group acts on itself by left multiplication, the null law is uniform on $G$ , and $T (e) = 3$ , $T (a) = 2$ , $T (a^{2}) = 1$ . For the observed point $e$ , the full-orbit inclusive upper-tail rank is one out of three. Separately, define a finite- $B$ reported value for this example by adjoining the observed identity and using $p_{+} : = \frac{1 + \sum_{b = 1}^{B} 1 {T (g_{b} e) \geq T (e)}}{B + 1} .$ Take $B = 4$ and the realized sampled transformations $(g_{1}, g_{2}, g_{3}, g_{4}) = (a^{2}, a, e, a^{2})$ . This example calculates the two resolutions and their realized difference; it does not establish validity, unbiasedness, or a general error bound for the finite- $B$ formula.

Facts & Assumptions

Given: The group, action, statistic, invariant uniform null law, and finite list of sampled transformations in the Example. Prove the exact-orbit and realized finite- $B$ calculations.

[F1]

A finite-group randomization setup consists of a finite probability space, finite group action, statistic, and null laws invariant under that action. (Finite-group randomization setups)

[F2]

The inclusive orbit upper-tail rank counts group elements with $T (g x) \geq T (x)$ and divides by $∣ G ∣$ . (Orbit upper-tail rank p-values)

[F3]

Under an invariant null, the full-orbit inclusive rank p-value is superuniform. (Finite-group orbit-rank validity)

[F4]

For a finite scalar reference chain, the total realized error is the sum of its signed adjacent-stage errors. (Pathwise error decomposition along a reference chain)

[A1]

The operation $a^{i} a^{j} = a^{i + j (m o d 3)}$ is associative with identity $e = a^{0}$ and inverses $a^{3 - i}$ ; for each $g \in G$ , left multiplication $x \mapsto g x$ is a bijection of $G$ .

[A2]

The finite- $B$ formula is the explicitly defined calculation in the Example; no validity property is assumed for it.

Verification

1.1givenA1F1construct

Each map $x \mapsto g x$ permutes $G$ , so the uniform law is invariant and the data form the finite-group setup of [F1].

1.2givenA2algebra

The sampled statistic values are $1, 2, 3, 1$ , so the inclusive count among the four sampled transforms is one and $p_{+} = (1 + 1) / 5 = 2 / 5$ .

2.1givenF2F3step 1.1algebra

Exactly one of $e, a, a^{2}$ has statistic at least $T (e) = 3$ , so [F2] gives the exact-orbit rank $p_{0} (e) = 1 / 3$ ; [F3] supplies its superuniformity under this invariant null.

3.1A2F4step 1.2step 2.1algebra∎

The finite- $B$ output lies on the grid ${1 / 5, 2 / 5, 3 / 5, 4 / 5, 1}$ because its count is an integer from zero through four. Set $m = 1$ , $r_{0} = p_{0} (e)$ , and $r_{1} = \tilde{r} = p_{+}$ ; then the one-stage error is $2 / 5 - 1 / 3 = 1 / 15$ by [F4].

Depends on

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