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Unequal assignment weights under the Fisher sharp null

Example

Two fixed units have two allowed assignments, a=(1,0) and b=(0,1), with known assignment probabilities P(Z=a)=0.9 and P(Z=b)=0.1. Suppose the potential-outcome schedule satisfies the Fisher sharp null and the predeclared score is T(a,y)=1, T(b,y)=0. The design-based inclusive randomization p-value is 0.9 when a is observed and 1 when b is observed.

Facts & Assumptions

Given: The finite assignment law, fixed sharp-null potential outcomes, and the predeclared score above. Compute the weighted upper-tail p-value and identify which validity premise is being used.

[F1]

Finite randomized treatment-assignment models: the units and potential-outcome schedule are fixed, the known finite assignment law supplies the randomness, and the observed response vector is determined by the realized assignment.

[F2]

Fisher sharp null and predeclared test orderings: under the Fisher sharp null, one response vector is imputable under every allowed assignment; the ordering score is fixed before observing assignment and outcomes.

[F3]

Fisher sharp-null randomization p-values are valid: the inclusive p-value sums the actual assignment probabilities over assignments whose imputed score is at least the observed score, and is valid under every Fisher-sharp schedule.

[F4]

Exact finite-group permutation p-values and tests: the full-orbit group-rank validity conclusion assumes each null law is invariant under every group transformation.

[F5]

Finite-population average treatment effect and weak null: the weak zero-average-effect null constrains only the average unit effect and is weaker than the Fisher sharp null.

Verification

Given: Assignments a,b with weights 0.9,0.1, a Fisher-sharp schedule, and the score ordering T(a,y)=1>T(b,y)=0.

1.1F1F2F3given

The assignment weights are nonnegative and sum to one. Under the sharp null the same fixed response vector y is available at both assignments, so the two score values in the Example are imputable before observing Z. The published weighted randomization theorem [F3] therefore applies to this schedule and this predeclared ordering.

1.2F1F2F3givenalgebra

If Z=a, the only assignment with score at least 1 is a, so prand(a,y)=0.9. If Z=b, both assignments have score at least 0, so prand(b,y)=0.9+0.1=1.

2.1F1F3F4F5step 1.2algebra∎

The p-value is 0.9 with probability 0.9 and 1 with probability 0.1. Its CDF is 0 for u<0.9, 0.9≤u for 0.9≤u<1, and 1 at u=1, hence it is superuniform. The assignment law is not invariant under the swap a↔b, since the two assignment masses differ, so the group-invariance theorem [F4] is not the justification here. This example uses the Fisher sharp null and makes no claim under the weaker average-effect null [F5].

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