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Assignment randomization and sharp-null tests — Examples

1 · Prerequisites

2 · Summary

The examples calculate a two-sided test under unequal assignment probabilities with tied scores, showing the conservative weighted p-value and the exact uniform value obtained by randomizing within the atom. A separate two-unit schedule has zero Finite-population average treatment effect and weak null while each unit's outcome changes, demonstrating why the Fisher sharp null is a stronger hypothesis.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

Zero average effect need not be a Fisher sharp null

Statement refuted

In a binary no-interference treatment model, a zero finite-population average treatment effect implies the Fisher sharp null.

Facts & Assumptions

Given: Two units with binary treatment vector $z = (z_{1}, z_{2})$ in $Z = {(0, 0), (1, 0), (0, 1), (1, 1)}$ , assignment probability $1 / 4$ at each vector, and fixed responses $Y_{1} (z) = z_{1}$ and $Y_{2} (z) = - z_{2}$ .

[F1]

Finite randomized treatment-assignment models: $Y_{i} (z)$ is the fixed response of unit $i$ under assignment $z$ ; assignment probability is separate from this schedule.

[F2]

Finite-population average treatment effect and weak null: the weak null is zero average of the unit-level effects in the no-interference binary-treatment submodel.

[F3]

Fisher sharp null and predeclared test orderings: the Fisher sharp null requires each unit's response to be invariant over every allowed assignment.

Counterexample

1.1givenF1F2algebra

Under the stated no-interference schedule, unit 1's treatment effect is $Y_{1} (1) - Y_{1} (0) = 1$ , while unit 2's is $Y_{2} (1) - Y_{2} (0) = - 1$ . Their finite-population average from [F2] is $(1 + (- 1)) / 2 = 0$ . The four assignment weights are nonnegative and sum to one, so they define a valid finite assignment law.

2.1F2F3step 1.1algebra∎

But $Y_{1} (0, 0) = 0 \neq 1 = Y_{1} (1, 0)$ and $Y_{2} (0, 0) = 0 \neq - 1 = Y_{2} (0, 1)$ . Thus both units' responses vary across allowed assignments, contrary to [F3]. The average-effect null [F2] does not imply the Fisher sharp null.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

A weighted two-sided randomization test with tied scores

Example

Let three units have fixed outcomes $y = (0, 1, 2)$ under every assignment, and randomize exactly one unit to treatment. The assignments select unit 1, 2, or 3 with probabilities $1 / 2$ , $1 / 3$ , and $1 / 6$ , respectively. Use the treatment-minus-control mean as signed score and its absolute value as the predeclared two-sided ordering about zero. The upper-tail p-value is $2 / 3$ for either outer assignment and $1$ for the middle assignment. It is superuniform despite the unequal assignment weights and the tied outer scores; an independent uniform tie randomizer gives an exactly uniform p-value.

Facts & Assumptions

Given: The three-unit complete randomization described above, its fixed potential-outcome schedule, and an independent $U$ uniform on $[0, 1]$ .

[F1]

Finite randomized treatment-assignment models: the outcomes are fixed under the design and the only assignment randomness has the specified finite law.

[F2]

Fisher sharp null and predeclared test orderings: a two-sided ordering about null center zero uses $T (z, y) = ∣ S (z, y) ∣$ and includes ties in the upper tail.

[F3]

Fisher sharp-null randomization p-values are valid: the weighted upper-tail value is superuniform under every Fisher sharp-null schedule, and independent uniform randomization within tied mass is exactly uniform.

[A1]

The auxiliary $U$ is independent of the treatment assignment and satisfies $P (U \leq u) = u$ for $u \in [0, 1]$ .

Verification

Given: The outcomes, assignment weights, treatment-minus-control mean, and auxiliary randomizer stated above.

1.1givenF1algebra

The assignment weights are nonnegative and sum to one, and each unit's outcome is the same under all three assignments, so this is a finite assignment law satisfying the Fisher sharp null.

2.1givenF2step 1.1algebra

If unit 1, 2, or 3 is treated, the signed differences are respectively $0 - (1 + 2) / 2 = - 3 / 2$ , $1 - (0 + 2) / 2 = 0$ , and $2 - (0 + 1) / 2 = 3 / 2$ . The two-sided scores are therefore $3 / 2, 0, 3 / 2$ .

3.1F1step 2.1algebra

Each outer assignment has inclusive upper-tail mass $1 / 2 + 1 / 6 = 2 / 3$ ; the middle score is the minimum, so its inclusive upper-tail mass is $1$ . The nonrandomized p-value therefore equals $2 / 3$ with probability $2 / 3$ and $1$ with probability $1 / 3$ . Its distribution function is $0$ below $2 / 3$ , $2 / 3$ on $[2 / 3, 1)$ , and $1$ at $1$ , which is at most its argument throughout $[0, 1]$ .

3.2givenA1step 2.1algebra

At either outer assignment the strict-tail mass is zero and the tied mass is $2 / 3$ , so $p^{+} = (2 / 3) U$ . At the middle assignment the strict-tail mass is $2 / 3$ and the tied mass is $1 / 3$ , so $p^{+} = 2 / 3 + U / 3$ .

4.1A1F3step 3.1step 3.2algebra∎

For $0 \leq u \leq 2 / 3$ , the outer assignments contribute $(2 / 3) P (U \leq 3 u / 2) = u$ . For $2 / 3 \leq u \leq 1$ , they contribute $2 / 3$ and the middle assignment contributes $(1 / 3) P (U \leq 3 u - 2) = u - 2 / 3$ ; hence $P (p^{+} \leq u) = u$ on the entire unit interval. Together with step 3.1, this verifies the theorem's superuniform and uniform conclusions for this assignment law.