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.

Repeated method differences can share a random intercept

Example

Let $Ω = {- 1, 1}^{3}$ have uniform probability, and let $U, ϵ_{1}, ϵ_{2}$ be its coordinate variables. For two repeated paired measurements on one target, suppose the differences follow the random-intercept model $D_{j} = U + ϵ_{j}$ , $j = 1, 2$ . Then each difference has mean $0$ and variance $2$ , but $Cov (D_{1}, D_{2}) = 1$ . In fact, $P (D_{1} = 2, D_{2} = 2) = \frac{1}{8} \neq \frac{1}{16} = P (D_{1} = 2) P (D_{2} = 2) .$ The repeated differences are dependent and hence are not an iid sample.

Facts & Assumptions

Given: The eight equally likely triples in $Ω$ , coordinate functions $U, ϵ_{1}, ϵ_{2}$ , and differences $D_{j} = U + ϵ_{j}$ .

[F1]

Independent random elements are those whose generated sigma-algebras are independent. Independent random elements

[F2]

An iid family is independent and has a common law. Identical distribution and IID families

[F3]

Finite-space expectations are weighted sums over atoms. Expectation agrees with the published finite weighted sum

[F4]

Variance and covariance are the expectations of centered squares and centered products. Moments, variance, and covariance on a probability space

[F5]

Covariance is bilinear on finite linear combinations. Covariance is symmetric and bilinear in finite linear combinations

[F6]

Independent square-integrable real random variables have zero covariance. Independence forces covariance to vanish

Verification

Given: $Ω = {- 1, 1}^{3}$ with uniform weights and $D_{j} = U + ϵ_{j}$ . Verify the marginal moments and the within-target dependence.

1.1F1F3F4givenalgebra

Each triple has mass $1 / 8 = (1 / 2)^{3}$ , so for arbitrary subsets $S_{1}, S_{2}, S_{3} \subseteq {- 1, 1}$ , $P (U \in S_{1}, ϵ_{1} \in S_{2}, ϵ_{2} \in S_{3}) = \prod_{j = 1}^{3} ∣ S_{j} ∣ / 2$ . Thus [F1] makes the coordinates independent. Each coordinate has mean $0$ and variance $1$ by the finite weighted sum [F3] and centered-square definition [F4].

2.1F3F4F5F6step 1.1algebra

By [F5] and [F6], $E [D_{j}] = 0$ , $Var (D_{j}) = Var (U) + Var (ϵ_{j}) + 2 Cov (U, ϵ_{j}) = 2$ , and $Cov (D_{1}, D_{2}) = Var (U) = 1$ ; all cross terms vanish because they are covariances of distinct independent coordinates.

3.1F1F2F3givenalgebra∎

The event $D_{1} = D_{2} = 2$ requires the single triple $(1, 1, 1)$ , so its probability is $1 / 8$ . Each event $D_{j} = 2$ requires $U = 1$ and $ϵ_{j} = 1$ and has probability $1 / 4$ , making the product of marginal probabilities $1 / 16$ . The factorization required for independence therefore fails; by [F2] the two differences are not iid.

Depends on

Used by

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