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.

Zero pair support can prevent universal unbiased variance estimation

Example

Let $U = {1, 2}$ have fixed values $y_{1} = a, y_{2} = b$ , and sample one unit by SRSWOR. Then $π_{1} = π_{2} = 1 / 2$ but $π_{12} = 0$ . The HT total is $2 a$ or $2 b$ , each with probability $1 / 2$ , so its variance is $(a - b)^{2}$ . The all-pairs inverse-inclusion formula cannot use the unsupported pair $(1, 2)$ . If one keeps only supported pairs, its expectation is $a^{2} + b^{2}$ , which is not the variance for all $a, b$ . In fact, no function of the observed singleton label and value is unbiased for this variance for every finite population.

Facts & Assumptions

Given: $U = {1, 2}$ , fixed real values $(a, b)$ , and a uniformly sampled singleton.

[F1]

Finite populations and sampling designs: The observed record contains the selected unit labels and their observed values.

[F2]

Simple random sampling without replacement: Each singleton has probability $1 / (\binom{2}{1}) = 1 / 2$ .

[F3]

Horvitz–Thompson variance and pair-inclusion estimate: The HT total uses weights $1 / π_{i}$ ; its variance has a pair term $Δ_{i j} y_{i} y_{j} / (π_{i} π_{j})$ , and the all-pairs inverse estimate is established under positive $π_{i j}$ for every ordered pair.

[F4]

Statistics and estimators: An estimator is a statistic, hence a function of the observed data record, with the target as its specified estimand.

Verification

1.1givenF2F3algebra

The two possible samples are ${1}$ and ${2}$ , each with probability $1 / 2$ by [F2]. Thus $π_{1} = π_{2} = 1 / 2$ and $π_{12} = 0$ ; the HT total is $2 a$ or $2 b$ , with mean $a + b$ and variance $\frac{1}{2} (2 a - a - b)^{2} + \frac{1}{2} (2 b - a - b)^{2} = (a - b)^{2}$ .

2.1givenF1F2F3step 1.1algebra

The coefficient $Δ_{12} / π_{12}$ is undefined. Restricting the sample-pair expression to supported ordered pairs leaves only diagonal terms: on ${1}$ the value is $\frac{Δ_{11}}{π_{11}} (a / π_{1})^{2} = (1 / 4) / (1 / 2) (2 a)^{2} = 2 a^{2}$ , and on ${2}$ it is $2 b^{2}$ . Its expectation is $a^{2} + b^{2}$ , which at $a = b = 1$ is $2$ while the variance in step 1.1 is $0$ ; omitting the unsupported cross pair is not universally unbiased.

3.1givenF1F2F4step 1.1algebracontradiction∎

Any estimator $G$ of the observed record has values $g_{1} (a)$ on ${1}$ and $g_{2} (b)$ on ${2}$ by [F1] and [F4], so universal unbiasedness would require $(g_{1} (a) + g_{2} (b)) / 2 = (a - b)^{2}$ for every $a, b$ . At $(0, 0), (0, 1), (1, 0), (1, 1)$ this gives respectively $g_{1} (0) + g_{2} (0) = 0$ , $g_{1} (0) + g_{2} (1) = 2$ , $g_{1} (1) + g_{2} (0) = 2$ , and $g_{1} (1) + g_{2} (1) = 0$ ; the first and fourth left sides sum to the same value as the second and third, but the right sides are $0$ and $4$ , a contradiction.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

20 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.