ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-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 finite model with positive finite evidence

Example

Take $Θ = {a, b}$ with its full power-set sigma-algebra and prior masses $Π ({a}) = 1 / 3$ and $Π ({b}) = 2 / 3$ . Let the observation space be ${0, 1}$ with its full power-set sigma-algebra and counting measure. At parameter $a$ , assign probabilities $1 / 2, 1 / 2$ to $0, 1$ ; at parameter $b$ , assign probabilities $3 / 4, 1 / 4$ to $0, 1$ . For the observed value $y = 1$ , the evidence is positive and finite, and the posterior gives equal mass to $a$ and $b$ .

Facts & Assumptions

[F1]

Evidence normalizer and fiberwise posterior update: for a fixed observation, $Z_{Π} (y) = \int L_{y} d Π$ and the fiberwise update divides the unnormalized measure by $Z_{Π} (y)$ only when that value is finite and positive.

[F2]

Fiberwise posterior propriety is equivalent to finite positive evidence: a finite positive evidence normalizer gives a probability update after division by that normalizer.

Given: The finite parameter space, prior, observation probabilities, and observed value $y = 1$ in the Example.

Verification

1.1givenF1algebra

The prior has total mass $1 / 3 + 2 / 3 = 1$ , and both conditional probability vectors sum to one. The likelihood at $y = 1$ is $L_{1} (a) = 1 / 2$ and $L_{1} (b) = 1 / 4$ , so the evidence is $Z_{Π} (1) = (1 / 3) (1 / 2) + (2 / 3) (1 / 4) = 1 / 3 \in (0, + \infty)$ .

2.1step 1.1F1F2algebra∎

The unnormalized masses are $(1 / 3) (1 / 2) = 1 / 6$ at $a$ and $(2 / 3) (1 / 4) = 1 / 6$ at $b$ . Dividing each by $Z_{Π} (1) = 1 / 3$ gives posterior mass $1 / 2$ at each point, a probability update by [F2].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 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.

Sources

Gelman et al., Bayesian Data Analysis, 3rd ed., §1.3 (standard reference, not scraped)
Johnson, Ott & Dogucu, Bayes Rules!, §§2.1.3–2.1.4, 2.3.4–2.3.6 (standard reference, not scraped)