CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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An own-treatment-only exposure map can be misspecified

Statement refuted

With interference, positive probability for each own-treatment value alone suffices to justify an exposure-specific Horvitz–Thompson mean even when outcomes can also depend on peers' assignments.

Facts & Assumptions

Given: Two units in one group, the four binary assignments equally likely, exposure mappings $D_{1} = Z_{1}$ and $D_{2} = Z_{2}$ , and fixed outcomes $Y_{1} (z_{1}, z_{2}) = z_{2}$ and $Y_{2} (z_{1}, z_{2}) = 0$ .

[F1]

Exposure mappings and exposure consistency: an exposure mapping is sufficient only if assignments with the same mapped label give the same unit potential outcome.

[F2]

Horvitz–Thompson exposure-specific mean is design-unbiased: under exposure sufficiency and positive exposure probabilities, the finite-population exposure mean is estimated by inverse-probability weighting of the observed exposure indicators.

Counterexample

1.1givenF1algebra

Unit 1 has $D_{1} = 0$ under both assignments $(0, 0)$ and $(0, 1)$ , but its outcomes there are respectively $0$ and $1$ . Thus [F1] fails although $π_{1} (0) = π_{2} (0) = 1 / 2 > 0$ .

2.1givenF1F2step 1.1algebra∎

If one nevertheless sets the attempted exposure-zero values to $Y_{1}^{*} (0) = Y_{1} (0, 0) = 0$ and $Y_{2}^{*} (0) = 0$ , their mean is $0$ . The attempted Horvitz–Thompson estimator is $\frac{1}{2} {1 (Z_{1} = 0) Y_{1} / (1 / 2) + 1 (Z_{2} = 0) Y_{2} / (1 / 2)} = 1 (Z_{1} = 0) Z_{2}$ . It equals $1$ only at assignment $(0, 1)$ and zero at the other three assignments, so its expectation is $1 / 4 \neq 0$ . Hence positivity does not repair the misspecified exposure map, and [F2] cannot be applied without its sufficiency premise.

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Sources

Peter M. Aronow and Cyrus Samii, Estimating Average Causal Effects Under General Interference, Condition 1 and §4 (standard reference, not scraped)