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Conditional and Finite-Group Exact Inference — Examples

1 · Prerequisites

2 · Summary

The examples derive Fisher's hypergeometric conditional law from two Bernoulli samples, compute all tail ranks in a small permutation orbit with ties and nontrivial stabilizers, and show that a group-rank p-value can be anti-conservative when the null law is not invariant.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

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.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

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].

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Fisher's exact test from two Bernoulli samples

Example

Take two independent Bernoulli samples of sizes $n_{1}, n_{2} \geq 1$ and test the one-sided alternative that the first success probability exceeds the second. Under the null, all $N = n_{1} + n_{2}$ trials have a common success probability $p \in [0, 1]$ . Let $K$ be the number of successes in the first sample and $S$ the total number in both samples. For every integer margin $s \in {0, \dots, N}$ , define the supplied conditional-law kernel $Q$ by $Q (s, {k}) = \frac{(\binom{n_{1}}{k}) (\binom{n_{2}}{s - k})}{(\binom{N}{s})}, \max (0, s - n_{2}) \leq k \leq \min (n_{1}, s),$ and set $Q (s, {k}) = 0$ outside the displayed range. For each $s$ , these weights sum to one because they count all $(\binom{N}{s})$ ways to choose the $s$ success positions according to how many lie in the first sample. At margins of positive probability this is the conditional law and it does not depend on $p$ ; at endpoint nulls $p = 0, 1$ , the formula also supplies a version at the other, probability-zero margins. Large values of $K$ support the one-sided alternative that the first sample has a larger success probability. For $n_{1} = n_{2} = 2$ and $s = 2$ , the conditional masses at $k = 0, 1, 2$ are $1 / 6, 4 / 6, 1 / 6$ ; the inclusive upper-tail p-values are $1, 5 / 6, 1 / 6$ .

Facts & Assumptions

Given: Positive integers $n_{1}, n_{2}$ , the null product Bernoulli model with common parameter $p$ , the statistic $K$ conditioned on $S$ , the supplied kernel $Q$ defined for every margin in the Example, and an auxiliary $U \sim Unif [0, 1]$ independent of all trials.

[F1]

A product of finite probability spaces assigns a sequence the product of its coordinate weights. The finite product of finite probability spaces

[F2]

Product weights normalize and the coordinate trials are independent. Product weights normalize, and coordinate events are mutually independent, Pairwise and mutual independence of finite-valued random variables

[F3]

For an event of positive probability, conditional probability is the ratio of the joint event probability to the conditioning-event probability. Conditional probability $P (A ∣ B)$ for $P (B) > 0$

[F4]

The nonrandomized conditional upper-tail value uses the inclusive tail; the randomized value uses the strict tail plus an independent uniform fraction of the tie mass. Randomized conditional upper-tail p-values

[F5]

For each fixed null value $p$ , under the supplied kernel the randomized upper-tail value is conditionally uniform and the inclusive value is conditionally superuniform. Randomized conditional p-values are uniform

[F6]

A test has unconditional size equal to the supremum of its null rejection expectations; conditional level at most $α$ implies unconditional level at most $α$ . Test size and power, Conditional level implies unconditional level

Verification

1.1F1F2givenalgebra

Write $Ω = {0, 1}^{N}$ and give each coordinate the two-point Bernoulli $(p)$ law. By [F1], a sequence with $s$ ones has weight $p^{s} (1 - p)^{N - s}$ ; the product weights sum to $(p + (1 - p))^{N} = 1$ , and [F2] gives the independent-trial model. For $0 < p < 1$ , every sequence with $S = s$ has the same positive weight.

2.1F1F3step 1.1givenconstructalgebra

There are $(\binom{N}{s})$ sequences with $s$ successes. Exactly $(\binom{n_{1}}{k}) (\binom{n_{2}}{s - k})$ of them have $K = k$ : choose $k$ of the first $n_{1}$ positions and $s - k$ of the other $n_{2}$ . These counts partition all size- $s$ subsets, so the displayed kernel weights sum to one for every $s$ , and when $P_{p} (S = s) > 0$ , dividing by the total count using [F3] gives the conditional mass; the feasible range is precisely where both binomial coefficients are nonzero. At $p = 0$ only $S = 0$ occurs and $K = 0$ ; at $p = 1$ only $S = N$ occurs and $K = n_{1}$ , so the supplied kernel has the required conditional law on the realized margin. At the other, probability-zero margins the same formula is simply the chosen kernel extension.

3.1F3F4step 2.1givenalgebra

For $n_{1} = n_{2} = 2$ and $s = 2$ , the feasible values are $k = 0, 1, 2$ , with counts $1, 4, 1$ among the six two-success sequences. Their probabilities are therefore $1 / 6, 4 / 6, 1 / 6$ . Summing the masses at or above each observed value gives inclusive p-values $1, 5 / 6, 1 / 6$ . The strict-tail-plus-tie formulas are respectively $5 / 6 + U / 6$ , $1 / 6 + 2 U / 3$ , and $U / 6$ ; their intervals $[5 / 6, 1]$ , $[1 / 6, 5 / 6]$ , and $[0, 1 / 6]$ have lengths matching the three conditional masses.

4.1F3F5F6step 2.1step 3.1given∎

For each margin $s$ with $P_{p} (S = s) > 0$ , [F5] makes the nonrandomized p-value conditionally superuniform and the randomized value conditionally uniform under the null. Rejecting when the nonrandomized value is at most $α$ is therefore a conditional level- $α$ test, so [F6] gives unconditional size at most $α$ . The test function of the randomized rule is $φ_{α} (x) = P_{U} (p_{U} (x, U) \leq α)$ ; it is measurable because the sample space is finite. By finite conditioning and independence of $U$ , $E_{p} [φ_{α} (X) ∣ S = s] = P_{p} (p_{U} \leq α ∣ S = s) = α$ , so the randomized rule has conditional and unconditional size exactly $α$ . At $p = 0$ and $p = 1$ the only observed margin has a degenerate conditional law; its inclusive value is $1$ and its randomized value is $U$ , so the same validity conclusions hold.

Conditional and Finite-Group Exact Inference — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

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

Statement refuted

Facts & Assumptions

Counterexample

A permutation orbit with tied ranks and nontrivial stabilizers

Example

Facts & Assumptions

Verification

Fisher's exact test from two Bernoulli samples

Example

Facts & Assumptions

Verification

Sources