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

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Unbiasedness and variance of the SRSWOR cluster-total estimator

Statement

Let a fixed finite population have $M \geq 2$ PSUs with fixed cluster totals $T_{1}, \dots, T_{M}$ . Select $m$ PSUs by uniform SRSWOR, where $1 \leq m \leq M$ . For $S_{1}$ the selected set, define ${\hat{T}}_{C} = \frac{M}{m} \sum_{i \in S_{1}} T_{i}, S_{T}^{2} = \frac{1}{M - 1} \sum_{i = 1}^{M} (T_{i} - \overset{ˉ}{T})^{2}, \overset{ˉ}{T} = \frac{1}{M} \sum_{i = 1}^{M} T_{i} .$ Then $E_{d} [{\hat{T}}_{C}] = T : = \sum_{i = 1}^{M} T_{i}$ and ${Var}_{d} ({\hat{T}}_{C}) = \frac{M^{2} (1 - m / M)}{m} S_{T}^{2} .$

Facts & Assumptions

Given: $M \geq 2$ fixed PSU totals, a uniform SRSWOR sample $S_{1}$ of size $1 \leq m \leq M$ , and design expectation $E_{d}$ over that sample only.

[F1]

Cluster sampling units, cluster totals, and one-stage SRSWOR: For a uniform SRSWOR of $m$ PSUs, the first-stage sample space is the set of all $m$ -element subsets of $[M]$ , each with probability ${(\binom{M}{m})}^{- 1}$ ; $T_{i}$ is fixed.

[F2]

Moments, variance, and covariance on a probability space: For square-integrable real random variables, variance and covariance are the expectations of the centered square and centered product.

[F3]

Covariance is symmetric and bilinear in finite linear combinations: Covariance is symmetric and bilinear in finite linear combinations of square-integrable variables.

[F4]

Variance of a finite sum as the sum of all variances and covariances: The variance of a finite sum of square-integrable real random variables is the sum of all pairwise covariances.

Proof

1.1givenF1choosealgebra

For each PSU define $I_{i} = 1 {i \in S_{1}}$ . Counting $m$ -subsets gives $π_{i} = (\binom{M - 1}{m - 1}) / (\binom{M}{m}) = m / M$ ; for distinct $i, r$ , $π_{i r} = 0$ when $m = 1$ and $π_{i r} = (\binom{M - 2}{m - 2}) / (\binom{M}{m}) = m (m - 1) / (M (M - 1))$ when $m \geq 2$ . Hence ${Var}_{d} (I_{i}) = π_{i} (1 - π_{i}) = m (M - m) / M^{2}$ and ${Cov}_{d} (I_{i}, I_{r}) = π_{i r} - π_{i} π_{r} = - m (M - m) / (M^{2} (M - 1))$ .

2.1givenstep 1.1algebra

Since ${\hat{T}}_{C} = (M / m) \sum_{i} I_{i} T_{i}$ , linearity of finite expectation and $E_{d} [I_{i}] = m / M$ from step 1.1 give $E_{d} [{\hat{T}}_{C}] = (M / m) \sum_{i} (m / M) T_{i} = \sum_{i} T_{i} = T$ .

3.1givenF2F3F4step 1.1algebra∎

Put $x_{i} = T_{i} - \overset{ˉ}{T}$ , so $\sum_{i} x_{i} = 0$ and fixed sample size gives ${\hat{T}}_{C} - T = (M / m) \sum_{i} I_{i} x_{i}$ . By [F4] and [F3], its variance is $(M^{2} / m^{2}) (\sum_{i} Var (I_{i}) x_{i}^{2} + \sum_{i \neq r} Cov (I_{i}, I_{r}) x_{i} x_{r})$ . Since $\sum_{i \neq r} x_{i} x_{r} = - \sum_{i} x_{i}^{2}$ , substitution from step 1.1 gives ${Var}_{d} ({\hat{T}}_{C}) = (M^{2} / m^{2}) (m (M - m) / M^{2}) (1 + 1 / (M - 1)) \sum_{i} x_{i}^{2} = M^{2} (1 - m / M) S_{T}^{2} / m$ .

Remarks

The proof is design-based and uses no distributional model for the fixed $T_{i}$ , no normal approximation, and no axiom of choice. At $m = M$ the sample is a census and the displayed variance is zero; $m = 1$ is covered by the zero pair-inclusion probability in step 1.1. The statement excludes $M = 1$ because $S_{T}^{2}$ with denominator $M - 1$ is then undefined; that sole-PSU census has zero variance directly.

Depends on

Used by

Average cluster size does not determine the unequal-size design effect Counterexample

Dependency tree · two levels

8 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.

Sources

Shalabh, Sampling Theory, Chapter 9: Cluster Sampling, PDF pp. 4–5, IIT Kanpur (standard reference, not scraped)
Sharon L. Lohr, Sampling: Design and Analysis, 2nd ed., §5.2.1, printed pp. 136–138 (standard reference, not scraped)