TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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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.

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Design expectation and variance of a paired contrast

Statement

Consider one pair in the $t = 2, b = 1$ RCBD, with the two treatments labelled $0$ and $1$ . Let $Y_{ℓ} (z) \in R$ be unit $ℓ$ 's fixed potential outcome under its own treatment $z \in {0, 1}$ , and assume no interference: the outcome depends on the pair assignment only through that unit's own treatment. Define $δ_{ℓ} = Y_{ℓ} (1) - Y_{ℓ} (0)$ , $τ = (δ_{1} + δ_{2}) / 2$ , and the two possible observed contrasts $D_{1} = Y_{1} (1) - Y_{2} (0), D_{2} = Y_{2} (1) - Y_{1} (0) .$ If exactly one unit is assigned treatment $1$ uniformly at random, the observed treated-minus-control contrast $D$ obeys $E_{d} [D] = τ, {Var}_{d} (D) = \frac{(D_{1} - D_{2})^{2}}{4} = \frac{((δ_{1} - δ_{2}) + 2 (Y_{1} (0) - Y_{2} (0)))^{2}}{4} .$ Here $E_{d}$ and ${Var}_{d}$ average over treatment assignment only, holding all potential outcomes fixed. The variance depends on baseline outcomes as well as effect heterogeneity.

Facts & Assumptions

Given: Two fixed units, binary potential outcomes with no interference, and uniform randomization of exactly one unit to treatment $1$ .

[F1]

In the finite assignment model, units and potential outcomes are fixed while assignment is random; the model itself imposes no no-interference assumption (Finite randomized treatment-assignment models).

[F2]

The finite-population average treatment effect for two units is the average of their fixed unit-level effects (Finite-population average treatment effect and weak null).

[F3]

In a two-treatment, one-block RCBD, the two assignments are equally likely (Randomized complete-block design).

[F4]

On a finite probability space, expectation is the sum of values times their weights (Expectation of a real random variable on a finite probability space).

[F5]

Variance is $E [(X - E [X])^{2}]$ and covariance is $E [(X - E [X]) (Y - E [Y])]$ for square-integrable variables (Moments, variance, and covariance on a probability space).

[F6]

A statistic is a measurable map of the observed data, and a statistic with the estimand's target space is an estimator (Statistics and estimators).

[F7]

The finite-assignment experiment uses the data space $Z \times R^{I}$ with its full power-set sigma-algebra (Finite randomized treatment-assignment models).

[F8]

Covariance is bilinear and symmetric, including constant summands (Covariance is symmetric and bilinear in finite linear combinations).

Proof

1.1F1F3algebra

Embed the pair's design law in the finite binary assignment model on ${0, 1}^{2}$ by giving weight $1 / 2$ to $(1, 0)$ and $(0, 1)$ and weight zero to $(0, 0)$ and $(1, 1)$ . Zero-probability assignments are allowed by [F1]. The positive-weight assignments give the displayed $D_{1}, D_{2}$ , with average $(δ_{1} + δ_{2}) / 2$ and difference $(δ_{1} - δ_{2}) + 2 (Y_{1} (0) - Y_{2} (0))$ .

2.1F6F7step 1.1

Define $D$ by the treated-minus-control formula on the two positive-weight assignments and, for example, by zero on the two zero-weight assignments. This is a function on the full-power-set data space [F7], hence a measurable statistic and estimator of $τ$ by [F6]; its arbitrary zero-weight values do not affect design moments.

2.2F2F3F4step 1.1algebra

By [F4], $E_{d} [D] = (D_{1} + D_{2}) / 2 = (δ_{1} + δ_{2}) / 2 = τ$ , so the contrast estimates the pair's finite-population average effect without bias.

2.3F3F5F8step 1.1algebra

Let $X$ indicate that unit $1$ is assigned treatment $1$ , so $D = D_{2} + (D_{1} - D_{2}) X$ . Since the assignment space is finite, the variables are square-integrable. By [F5] and [F8], covariance with a constant is zero and ${Var}_{d} (D) = (D_{1} - D_{2})^{2} {Var}_{d} (X)$ .

3.1F3F4F5step 2.3algebra∎

The two equally weighted assignments give $E_{d} [X] = 1 / 2$ by [F4], hence ${Var}_{d} (X) = 1 / 4$ by [F5]. Therefore ${Var}_{d} (D) = (D_{1} - D_{2})^{2} / 4$ , which is the formula in the Statement.

Depends on

Used by

Observed paired-contrast variance is not determined by effect heterogeneity Counterexample

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Sources

Gary W. Oehlert, A First Course in Design and Analysis of Experiments, Chapter 13 §13.2 (standard reference, not scraped)
Susan Holmes and Wolfgang Huber, Modern Statistics for Modern Biology, Chapter 13 §13.4.4 (standard reference, not scraped)