CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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Observed paired-contrast variance is not determined by effect heterogeneity

Statement refuted

Define $δ_{i} = Y_{i} (1) - Y_{i} (0)$ , and let $D$ be the observed treated-minus-control contrast. The claimed formula is that in every two-unit paired randomization, ${Var}_{d} (D) = \frac{(δ_{1} - δ_{2})^{2}}{4} .$

Facts & Assumptions

Given: Two units with no-interference binary potential outcomes $Y_{1} (0) = 0$ , $Y_{1} (1) = 1$ , $Y_{2} (0) = 10$ , and $Y_{2} (1) = 10$ , and a uniform assignment of exactly one unit to treatment $1$ .

[F1]

The observed paired contrast has the design expectation and variance proved in the local corrected theorem (Design expectation and variance of a paired contrast).

[F2]

The effect of the unit selected for treatment $1$ has variance $(δ_{1} - δ_{2})^{2} / 4$ (Moments of the effect for the randomly selected treated unit).

Counterexample

1.1givenF1algebra

The fixed effects are $δ_{1} = 1$ and $δ_{2} = 0$ . If unit $1$ is treated, the observed contrast is $D_{1} = Y_{1} (1) - Y_{2} (0) = - 9$ ; if unit $2$ is treated, it is $D_{2} = Y_{2} (1) - Y_{1} (0) = 10$ . The two assignments each have weight $1 / 2$ .

2.1F1step 1.1algebra

The observed contrast has mean $(- 9 + 10) / 2 = 1 / 2$ and variance ${(- 9 - 1 / 2)^{2} + (10 - 1 / 2)^{2}} / 2 = 90.25$ , also given by [F1]. But the proposed expression is $(δ_{1} - δ_{2})^{2} / 4 = 1 / 4$ . Hence it is not the variance of the observed paired contrast.

3.1F2step 2.1algebra∎

The variance $1 / 4$ is instead the variance of the randomly selected unit effect $S \in {1, 0}$ from [F2]. This distinguishes that latent schedule quantity from the observed contrast, and completes the counterexample.

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