ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge 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 three-node Gaussian chain has conditionally independent endpoints

Example

Assume AC. Let $Θ = (\begin{array}{ccc} 2 & - 1 & 0 \\ - 1 & 2 & - 1 \\ 0 & - 1 & 2 \end{array})$ , $Σ = Θ^{- 1}$ , and $X \sim N_{3} (0, Σ)$ . Put an edge between $i$ and $j$ exactly when $Θ_{i j} \neq 0$ . Then the graph is the path $1 - 2 - 3$ , and $X_{1}$ and $X_{3}$ are conditionally independent given $X_{2}$ .

Example references

The Elements of Statistical Learning, §17.3.1

Evans, Graphical Models, §5.1

Graph diagram

The vertices are the variables $X_{1}, X_{2}, X_{3}$ ; an edge is drawn exactly when the corresponding off-diagonal precision entry is nonzero. The missing $1 - 3$ edge records $Θ_{13} = 0$ . The verification uses the quadratic form and the proved Gaussian theorem, not the diagram.

Facts & Assumptions

Given: The Gaussian vector, precision matrix and graph convention in the Example.

[F1]

The Gaussian precision-zero theorem says that $Θ_{i j} = 0$ exactly when $X_{i}$ and $X_{j}$ are conditionally independent given all other coordinates, for positive-definite covariance. Gaussian precision zeros characterize pairwise conditional independence.

[F2]

In a finite undirected graph, an edge is the unordered pair of its endpoints. Finite undirected graphs, Markov properties, and clique factorization.

[A1]

AC is the explicit hypothesis of the Gaussian theorem; here it means every family of nonempty sets has a choice function. Gaussian precision zeros characterize pairwise conditional independence The Axiom of Choice.

Verification

technique · direct

1.1givenalgebra

For $x = (x_{1}, x_{2}, x_{3})^{T}$ , $x^{T} Θ x = x_{1}^{2} + x_{3}^{2} + (x_{1} - x_{2})^{2} + (x_{2} - x_{3})^{2}$ . This is positive for every nonzero $x$ : if the sum vanished, each square would vanish and successively $x_{1} = x_{3} = 0$ , $x_{2} = x_{1} = x_{3} = 0$ . Hence $Θ$ is positive definite and $Σ = Θ^{- 1}$ is a valid positive-definite covariance.

1.2F2givenalgebra

The only nonzero off-diagonal entries are $Θ_{12} = Θ_{21} = - 1$ and $Θ_{23} = Θ_{32} = - 1$ , while $Θ_{13} = Θ_{31} = 0$ . By [F2], the graph is exactly the path $1 - 2 - 3$ .

2.1A1F1step 1.1step 1.2∎

Theorem [F1] applies with $p = 3$ , covariance $Σ ≻ 0$ from step 1.1 and precision $Θ = Σ^{- 1}$ . Since $Θ_{13} = 0$ , it gives $X_{1} ⊥ ⁣ ⁣ ⁣ ⊥ X_{3} ∣ σ (X_{2})$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

24 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

The Elements of Statistical Learning, second edition (standard reference, not scraped)
Robin Evans, Graphical Models course notes (standard reference, not scraped)