CounterexampleConstruction: 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.

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.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 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.