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.

SRSWOR inclusion probabilities, pairwise nonpositive indicator covariance, and the sample mean

Example

Let $N \geq 2$ , $1 \leq n \leq N$ , $U = {1, \dots, N}$ , and let $y_{i}$ be fixed real values. Let $S_{d}$ consist of all $n$ -element subsets of $U$ , with each such sample assigned probability $1 / M$ , where $M = ∣ S_{d} ∣$ . This is simple random sampling without replacement (SRSWOR). For distinct $i, j \in U$ , $π_{i} = \frac{n}{N}, π_{i j} = \frac{n (n - 1)}{N (N - 1)}, {Cov}_{d} (I_{i}, I_{j}) = - \frac{n (N - n)}{N^{2} (N - 1)} .$ The Horvitz–Thompson mean reduces to the ordinary sample mean $n^{- 1} \sum_{i \in s} y_{i}$ .

Facts & Assumptions

Given: $N \geq 2$ , $1 \leq n \leq N$ , a fixed finite population $(U, y)$ , and the uniform design on all $n$ -element subsets of $U$ .

[F1]

Finite populations and sampling designs: For a sampling design, $π_{i} = P_{d} (i \in S)$ , $π_{i j} = P_{d} (i, j \in S)$ , and the values $y_{i}$ are held fixed under the design law.

[F2]

The Horvitz–Thompson total and mean are design-unbiased under positive inclusion probabilities: If every $π_{i} > 0$ , the total and mean estimators with weights $1 / π_{i}$ are design-unbiased.

[F3]

Variance, standard deviation, and covariance on a finite probability space: On one finite probability space, $Cov (X, Y) = E [(X - E X) (Y - E Y)]$ .

[F4]

Expectation is linear for every finite family of random variables, without any independence hypothesis: Expectation is linear for finite families of random variables on one finite probability space.

[F5]

Indicators turn event probabilities, intersections, and finite counts into expectations and products: Indicator expectations equal event probabilities, and products of indicators equal indicators of event intersections.

[F6]

The indicator random variable of an event: An event indicator is $1$ on the event and $0$ outside it.

[F7]

Expectation of a real random variable on a finite probability space: Expectation is the finite weighted sum over sample outcomes.

Verification

1.1givenF1constructalgebra

Let $a$ be the number of $n$ -subsets containing a fixed unit. A permutation of $U$ taking one unit to another gives a bijection of the sample family, so every unit occurs in exactly $a$ samples. Counting pairs $(s, i)$ with $i \in s$ in two ways gives $N a = n M$ , hence $a / M = n / N$ and $π_{i} = n / N$ .

1.2givenF1constructalgebra

Let $b$ be the number of samples containing a fixed ordered pair of distinct units. Permutations of $U$ give the same $b$ for every such pair. Counting triples $(s, i, j)$ with $i \neq j$ and $i, j \in s$ gives $N (N - 1) b = n (n - 1) M$ , so $b / M = n (n - 1) / (N (N - 1))$ and $π_{i j} = n (n - 1) / (N (N - 1))$ . This also gives $π_{i j} = 0$ when $n = 1$ .

2.1F3F4F5F6F7step 1.1step 1.2algebra

Put $A_{i} = {s : i \in s}$ . By [F5] and [F6], $E_{d} [I_{i}] = P_{d} (A_{i}) = π_{i}$ and $E_{d} [I_{i} I_{j}] = P_{d} (A_{i} \cap A_{j}) = π_{i j}$ . Expanding the covariance in [F3] with linearity [F4] therefore gives ${Cov}_{d} (I_{i}, I_{j}) = π_{i j} - π_{i} π_{j} = \frac{n (n - 1)}{N (N - 1)} - \frac{n^{2}}{N^{2}} = - \frac{n (N - n)}{N^{2} (N - 1)}$ .

3.1givenF1F2step 1.1algebra∎

Since $π_{i} = n / N > 0$ , the estimator in [F2] has ${\hat{\overset{ˉ}{y}}}_{H T} (s, y ∣_{s}) = \frac{1}{N} \sum_{i \in s} \frac{y_{i}}{n / N} = \frac{1}{n} \sum_{i \in s} y_{i}$ . Its expectation is ${\overset{ˉ}{y}}_{U}$ by [F2].

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