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 negative balanced-ANOVA moment estimate

Example

In the balanced one-way model take $k = r = 2$ , fix $μ = 0$ , and observe $Y_{11} = - 1$ , $Y_{12} = 1$ , $Y_{21} = - 1$ , and $Y_{22} = 1$ . The raw method-of-moments estimate of the between-group variance is negative. This observed estimate does not establish that the true between-group variance is zero.

Facts & Assumptions

Given: The balanced one-way model with $τ^{2} > 0$ , $σ^{2} > 0$ , $μ = 0$ , and the displayed observations.

[A1]

AC is assumed for the normal-law convention used here. The Axiom of Choice

[F1]

For $k, r \geq 2$ , the raw method-of-moments estimate is $(M S B - M S W) / r$ under the balanced one-way normal random-effects model. Expected mean squares in the balanced one-way normal random-effects model

[F2]

For $m \in R$ and $s > 0$ , $N (m, s^{2})$ has the strictly positive Lebesgue density $(2 π s^{2})^{- 1 / 2} \exp (- (x - m)^{2} / (2 s^{2}))$ . A positive-variance normal law has its location-scale density

[F3]

Independent random elements have independent generated sigma-algebras, so probabilities of finite coordinate rectangles factor. Independent random elements

Verification

Given: The model and observations in the Example.

1.1F1givenalgebra

Both group means and the grand mean are zero, so $M S B = 2 (0^{2} + 0^{2}) = 0$ . Each of the four within-group residuals is either $- 1$ or $1$ , hence $M S W = (1 + 1 + 1 + 1) / 2 = 2$ . By [F1], the raw estimate is $(0 - 2) / 2 = - 1$ .

2.1A1F2F3step 1.1construct∎

The map from the four observations to $(M S B - M S W) / 2$ is a polynomial divided by the fixed positive integer $2$ , hence continuous. At the displayed table its value is $- 1$ by step 1.1, so some open neighborhood of the table still has negative estimates. The latent vector $(a_{1}, a_{2}, ϵ_{11}, ϵ_{12}, ϵ_{21}, ϵ_{22}) = (0, 0, - 1, 1, - 1, 1)$ maps to that table. Since every coordinate has a strictly positive normal density by [F2], an open rectangle around this latent vector has positive probability by independence [F3]; it can be chosen inside the preimage of the negative-estimate neighborhood. Thus negative estimates occur with positive probability even when $τ^{2} > 0$ , so replacing them by zero changes the estimator and does not prove a structural zero.

Depends on

Used by

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Dependency tree · two levels

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Sources

Andrew Gelman et al., Bayesian Data Analysis, 3rd ed., §5.4 (standard reference, not scraped)
NIST/SEMATECH e-Handbook of Statistical Methods, §7.4.4 (standard reference, not scraped)