ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)

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.

Sensor placement along a high prior-variance direction

Example

Assume AC. Let $θ \sim N_{2} (0, C)$ with $C = diag (9, 1)$ . Compare two zero-cost one-sensor designs with independent unit-variance Gaussian noise: $d_{1}$ observes $A_{1} θ$ with $A_{1} = (1, 0)$ , and $d_{2}$ observes $A_{2} θ$ with $A_{2} = (0, 1)$ . Their expected information gains are $\frac{1}{2} \log 10$ and $\frac{1}{2} \log 2$ , respectively, so the sensor measuring the higher prior-variance direction is uniquely preferred.

Example references

Chaloner and Verdinelli, §2.2

MacKay, §3.1

Facts & Assumptions

Given: AC, the prior covariance $C = diag (9, 1)$ , the two row operators, and independent observation noise with covariance $1$ for both designs.

[A1]

AC is inherited from the EIG/KL convention. The Axiom of Choice

[F1]

For $C ⪰ 0$ and $Γ ≻ 0$ , $EIG = \frac{1}{2} \log \det (I + Γ^{- 1 / 2} A C A^{T} Γ^{- 1 / 2})$ . Linear Gaussian information gain has a log-determinant formula

[F2]

The natural logarithm is strictly increasing and $\log 1 = 0$ . Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm

Verification

Given: The hypotheses in the Example.

1.1A1F1givenalgebra

For $d_{1}$ , $A_{1} C A_{1}^{T} = 9$ and $Γ_{1} = 1$ , so [F1] gives $EIG (d_{1}) = \frac{1}{2} \log (1 + 9) = \frac{1}{2} \log 10$ . For $d_{2}$ , $A_{2} C A_{2}^{T} = 1$ and $Γ_{2} = 1$ , so [F1] gives $EIG (d_{2}) = \frac{1}{2} \log (1 + 1) = \frac{1}{2} \log 2$ .

2.1F2step 1.1algebra∎

The costs are equal and $10 > 2 > 0$ , so strict increase of the logarithm [F2] gives $EIG (d_{1}) > EIG (d_{2})$ . Thus $d_{1}$ is the unique maximizer among the two candidates.

Depends on

Used by

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Sources

Kathryn Chaloner and Isabella Verdinelli, Bayesian Experimental Design: A Review, Statistical Science 10(3) (1995), 273–304 (standard reference, not scraped)
David J. C. MacKay, Information-Based Objective Functions for Active Data Selection, Neural Computation 4(4) (1992), 590–604 (standard reference, not scraped)