ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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Positive pair covariances and a negative SYG estimate

Example

Let $U = {1, 2, 3, 4}$ and use the fixed-size design $p_{d} ({1, 2}) = p_{d} ({3, 4}) = \frac{2}{5}, p_{d} ({1, 3}) = p_{d} ({1, 4}) = p_{d} ({2, 3}) = p_{d} ({2, 4}) = \frac{1}{20} .$ All units have $π_{i} = 1 / 2$ . The pair differences are $Δ_{12} = Δ_{34} = 3 / 20 > 0$ and $Δ_{i j} = - 1 / 5$ for the four cross pairs. Thus a positive pair difference does not force a negative Sen–Yates–Grundy (SYG) estimate for every fixed population: for $y = (0, 0, 1, 1)$ all six sample estimates are nonnegative. But for $y = (0, 1, 0, 0)$ the supported sample ${1, 2}$ produces the negative estimate $- 3 / 2$ , even though the true design variance is $1$ . In both schedules the SYG estimates average to the corresponding true variance.

Facts & Assumptions

Given: The six two-unit samples and their probabilities are as stated, with fixed population values chosen separately for each schedule.

[F1]

Finite populations and sampling designs: A sampling design is a finite support of subsets with positive weights summing to one; $π_{i}$ and $π_{i j}$ are the first- and second-order inclusion probabilities.

[F2]

Inclusion-covariance difference: $Δ_{i j} = π_{i j} - π_{i} π_{j}$ .

[F3]

Sen–Yates–Grundy variance identity: For fixed sample size and positive first-order inclusion probabilities, the SYG sample estimate is unbiased when all $π_{i j} > 0$ ; pairwise nonpositive distinct-unit $Δ_{i j}$ is a sufficient guarantee of nonnegative sample estimates.

Verification

1.1givenF1F2algebra

The six weights sum to $2 (2 / 5) + 4 (1 / 20) = 1$ and every sample has size two; each unit's incident weights sum to $1 / 2$ , so $π_{i} = 1 / 2$ . Pair probabilities are $π_{12} = π_{34} = 2 / 5$ and $1 / 20$ for cross pairs; by [F2], $Δ_{12} = Δ_{34} = 2 / 5 - 1 / 4 = 3 / 20$ and $Δ_{i j} = 1 / 20 - 1 / 4 = - 1 / 5$ for cross pairs. All $π_{i j}$ are positive.

2.1F1F3step 1.1algebra

For $y^{A} = (0, 0, 1, 1)$ , $x = (0, 0, 2, 2)$ . Samples ${1, 2}$ and ${3, 4}$ give SYG estimate zero; each cross sample gives $- Δ_{i j} (x_{i} - x_{j})^{2} / π_{i j} = 16$ . The six estimates are $0, 0, 16, 16, 16, 16$ and have expectation $4 (1 / 20) (16) = 16 / 5$ . HT totals are $0, 4, 2$ with probabilities $2 / 5, 2 / 5, 1 / 5$ , so their mean is $2$ and variance is $(2 / 5) 4 + (2 / 5) 4 = 16 / 5$ .

3.1F1F2F3step 1.1step 2.1algebra∎

For $y^{B} = (0, 1, 0, 0)$ , $x = (0, 2, 0, 0)$ . The HT total is $2$ or $0$ with probability $1 / 2$ each, so its variance is $1$ . Sample ${1, 2}$ has probability $2 / 5$ and SYG estimate $- Δ_{12} (x_{1} - x_{2})^{2} / π_{12} = - (3 / 20) (4) / (2 / 5) = - 3 / 2$ ; samples ${2, 3}$ and ${2, 4}$ have estimate $16$ and the other three have zero. The expectation is $(2 / 5) (- 3 / 2) + 2 (1 / 20) (16) = 1$ , equal to the variance despite the negative sample value; thus the nonpositive-Delta condition is sufficient but not necessary for all supported estimates to be nonnegative for a particular fixed population, as schedule $y^{A}$ in step 2.1 shows.

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