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

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A zero-variance stratum is outside the positive-variance allocation formula

Example

Let the first stratum have values $(5, 5)$ and the second have values $(0, 3, 6)$ . For total sample size three, the unqualified weights $N_{h} S_{h}$ assign no sample units to the first stratum, even though the stratified design requires at least one observation per stratum.

Facts & Assumptions

Given: Two fixed strata with those values, independent within-stratum SRSWOR, and total sample size $n = 3$ .

[F1]

Stratified sampling designs and the stratified total estimator: Each stratum sample size satisfies $1 \leq n_{h} \leq N_{h}$ ; $S_{h}^{2}$ is defined for noncertainty strata with $N_{h} > 1$ .

[F2]

Continuous fixed-size Neyman allocation minimizes stratified variance: The continuous Neyman formula requires positive variances in every variable noncertainty stratum.

[F3]

Unbiasedness and variance of the stratified total estimator: For a valid allocation, the total estimator is unbiased and its design variance is the sum of the noncertainty stratum contributions.

Verification

1.1givenchoosealgebra

The first stratum has $N_{1} = 2$ , ${\overset{ˉ}{y}}_{1} = 5$ , and $S_{1}^{2} = 0$ . The second has $N_{2} = 3$ , ${\overset{ˉ}{y}}_{2} = 3$ , and $S_{2}^{2} = ((- 3)^{2} + 0^{2} + 3^{2}) / 2 = 9$ , so $S_{2} = 3$ and the total is $19$ .

2.1F1F2step 1.1algebra

The unqualified proportionality weights would be $N_{1} S_{1} = 2 (0) = 0$ and $N_{2} S_{2} = 3 (3) = 9$ , giving formal continuous allocation $(x_{1}, x_{2}) = (0, 3)$ . But $S_{1} = 0$ violates [F2]'s positive-variance hypothesis, and $x_{1} = 0$ violates [F1]'s $n_{1} \geq 1$ condition.

3.1F1F3step 1.1step 2.1algebra∎

The valid allocation $(n_{1}, n_{2}) = (1, 2)$ has variance $2^{2} (1 - 1 / 2) 0 / 1 = 0$ in the first stratum and $3^{2} (1 - 2 / 3) 9 / 2 = 27 / 2$ in the second by [F3]. It is a valid design, whereas the formal zero allocation is not.

Depends on

Used by

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Dependency tree · two levels

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Sources

Shalabh, Sampling Theory, MTH 417 lecture notes, Chapter 4: Stratified Sampling, PDF pp. 8–9 and 16 (standard reference, not scraped)
Robin Arnold, STAT392 Sample Surveys, Chapter 8: Stratified Sampling, §§8.5 and 8.8 (standard reference, not scraped)