ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pass

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A tied two-sample Bernoulli permutation test

Example

Let $Ω = {0, 1}^{2}$ , and for $q \in [0, 1]$ assign the four vectors masses $P_{q} (0, 0) = (1 - q)^{2}, P_{q} (0, 1) = q (1 - q), P_{q} (1, 0) = q (1 - q), P_{q} (1, 1) = q^{2} .$ These are two independent Bernoulli $(q)$ observations, one in each sample. Let $G = S_{2}$ act by swapping coordinates and take the signed two-sample score $T (x) = x_{1} - x_{2}$ . Under the equal-distribution null, the law is invariant under this swap. For the discordant sample $x = (1, 0)$ the full-orbit p-value is $1 / 2$ ; for $x = (0, 1)$ it is $1$ , and either tied sample has inclusive p-value $1$ .

For a uniform-subset Monte Carlo example without replacement, let $G = Z / 6 Z$ act on itself by addition with its full power-set sigma-algebra, let $X$ be uniform on $G$ , and take $T (x) = 1 {x = 0}$ . Set $B = 2$ , choose $S$ uniformly from the ten two-element subsets of $G ∖ {0}$ independently of $X$ , and use $W = {0} \cup S$ . This is the identity-adjoined design of Identity-corrected random-permutation Monte Carlo p-value, with $p_{2, w o r} (x) = \frac{# {g \in W : T (g + x) \geq T (x)}}{3} .$

The equal-distribution null above is stronger than equality of means. For a weak mean-difference null, use the data space $D^{2}$ with full power-set sigma-algebra, where $D = {- 2, - 1, 0, 1, 2}$ , and let $S_{2}$ act by swapping coordinates. On $D^{2}$ assign $(x, y)$ mass $P {x} Q {y}$ , where $P$ puts mass $1 / 2$ on each of $- 1, 1$ and $Q$ puts masses $1 / 4, 1 / 2, 1 / 4$ on $- 2, 0, 2$ , respectively. Use the score $T (x, y) = x - y$ . Both coordinate means are zero, so the weak mean-difference null $E [X] = E [Y]$ holds. The pooled law is not invariant under the swap.

Facts & Assumptions

Given: The Bernoulli product law, its coordinate-swap action and score, and the full-orbit rank definitions; the uniform-subset design on $Z / 6 Z$ ; and the weak-null product law on $D^{2}$ with the coordinate-swap action. Verify invariance and compute the ranks and design p-value.

[F1]

Measurable finite-group permutation setups: a finite group acts measurably on the experiment data space, and the theorem's null premise is invariance under every group element.

[F2]

Full-orbit and tie-randomized permutation p-values: the inclusive rank counts all group transformations with $T (g \cdot x) \geq T (x)$ ; the randomized rank interpolates over the tied block using an independent uniform variable.

[F3]

Exact finite-group permutation p-values and tests: under an invariant null, inclusive ranks are superuniform and tie-randomized ranks are exactly uniform.

[F4]

Finite probability spaces, outcome weights, events, and event probabilities: nonnegative weights on a finite set that sum to one define a finite probability law, and event probabilities are sums of the weights.

[F5]

Identity-corrected random-permutation Monte Carlo p-value: without replacement, adjoin the identity to a uniformly selected $B$ -subset of the nonidentity group elements, selected independently of the data $X$ , and divide the inclusive count by $B + 1$ .

Verification

Given: The Bernoulli product law and $S_{2}$ swap score, the $Z / 6 Z$ design, and the weak mean-difference model and swap action stated in the Example.

1.1F1F4givenalgebra

All four masses are nonnegative and sum to $(q + (1 - q))^{2} = 1$ , so [F4] makes them a finite probability law. The swap fixes $(0, 0)$ and $(1, 1)$ and exchanges $(0, 1)$ with $(1, 0)$ , which have equal mass. Thus $P_{q} (s^{- 1} A) = P_{q} (A)$ for every $A \subseteq Ω$ , and the group action is invariant for every $q \in [0, 1]$ .

1.2F1F4F5givenalgebra

The subset $S$ is uniform and independent of $X$ by the Example, so it satisfies the without-replacement design premise in [F5]. In the $Z / 6 Z$ example, addition by any fixed group element permutes the six equally likely outcomes, so the null law is invariant. If $X = 0$ , only the identity in $W$ has score at least $T (X) = 1$ , and the without-replacement p-value is $1 / 3$ . If $X \neq 0$ , then $T (X) = 0$ and every element of $W$ has score at least zero, so the p-value is $1$ . Its distribution is therefore $1 / 3$ with probability $1 / 6$ and $1$ with probability $5 / 6$ ; its CDF is superuniform.

2.1F2step 1.1algebra

At $(0, 0)$ and $(1, 1)$ , both transformed scores are $0$ , giving $p_{o r b} = 1$ and $p_{t i e} = U$ . At $(1, 0)$ the two scores are $1, - 1$ , giving $p_{o r b} = 1 / 2$ and $p_{t i e} = U / 2$ . At $(0, 1)$ they are $- 1, 1$ , giving $p_{o r b} = 1$ and $p_{t i e} = (1 + U) / 2$ .

3.1F2F3step 2.1algebra

Put $a = q (1 - q)$ . Then $p_{o r b} = 1 / 2$ with probability $a$ and equals $1$ otherwise. Its CDF is $0$ for $u < 1 / 2$ , $a \leq 1 / 4 \leq u$ for $1 / 2 \leq u < 1$ , and $1$ at $u = 1$ , so it is superuniform. For $0 \leq u \leq 1 / 2$ , the tie-randomized CDF is $2 a u + (1 - 2 a) u = u$ : the $(1, 0)$ branch contributes $2 a u$ , the $(0, 1)$ branch contributes zero except at a probability-zero endpoint, and the tied branches contribute $(1 - 2 a) u$ . For $1 / 2 \leq u \leq 1$ , it is $a + a (2 u - 1) + (1 - 2 a) u = u$ . Thus $p_{t i e}$ is uniform, in agreement with [F3].

4.1F1F4givenalgebra∎

In the weak mean-difference example, $E [X] = E [Y] = 0$ . But the coordinate swap sends $(1, 0)$ to $(0, 1)$ , and the product law assigns these points probabilities $P (X = 1) Q (Y = 0) = 1 / 4$ and $P (X = 0) Q (Y = 1) = 0$ , respectively. Hence this weak null does not make the pooled law invariant under the swap, so the invariant-null premise of [F3] cannot be inferred from equality of means.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

25 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.