ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge 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.

A scalar inverse with a conditioning-dependent shift modulus

Example

Fix $R, a_{0}, σ > 0$ and $0 \leq ρ_{A} < a_{0} / 2$ . The working inverse is $g (y) = y / a_{0}$ . Under a test operator $a = a_{0} + δ$ with $∣ δ ∣ \leq ρ_{A}$ , the observation is $Y = a θ + ε$ for $θ \in [- R, R]$ and independent $ε \sim N (0, σ^{2})$ . For any two truth laws $μ_{0}, μ$ on $[- R, R]$ , define $W_{1} (μ, μ_{0})$ as the infimum of $E ∣ Θ - Θ_{0} ∣$ over their couplings. The risk is $R_{μ, δ} (g) = E_{θ \sim μ, ε} ∣ g ((a_{0} + δ) θ + ε) - θ ∣^{2} = \frac{δ^{2}}{a_{0}^{2}} E_{μ} θ^{2} + \frac{σ^{2}}{a_{0}^{2}},$ and it satisfies $R_{μ, δ} (g) \leq (ρ_{A}^{2} R^{2} + σ^{2}) / a_{0}^{2}$ and $∣ R_{μ, δ} (g) - R_{μ_{0}, δ} (g) ∣ \leq \frac{2 δ^{2} R}{a_{0}^{2}} W_{1} (μ, μ_{0}) .$

Example references

The bounded operator and test-law shift frontier illustrated by a conditioned scalar inverse.

Facts & Assumptions

Given: $R, a_{0}, σ > 0$ , $0 \leq ρ_{A} < a_{0} / 2$ , $∣ δ ∣ \leq ρ_{A}$ , $θ \in [- R, R]$ , $ε \sim N (0, σ^{2})$ independent of $θ$ , and the working inverse $g (y) = y / a_{0}$ .

[A1]

The test law $μ$ and source law $μ_{0}$ are Borel probability measures on $[- R, R]$ ; $W_{1}$ is the infimum of expected absolute distance over their couplings.

Verification

1.1givenA1algebra

Substitution gives $g (Y) - θ = ((a_{0} + δ) θ + ε) / a_{0} - θ = δ θ / a_{0} + ε / a_{0}$ . Independence and $E ε = 0$ make the cross term vanish, so the expected squared error is $δ^{2} E_{μ} θ^{2} / a_{0}^{2} + σ^{2} / a_{0}^{2}$ . Since $∣ θ ∣ \leq R$ and $∣ δ ∣ \leq ρ_{A}$ , this is at most $(ρ_{A}^{2} R^{2} + σ^{2}) / a_{0}^{2}$ .

2.1A1step 1.1algebra∎

For any coupling $(Θ, Θ_{0})$ of $μ, μ_{0}$ , $∣ Θ^{2} - Θ_{0}^{2} ∣ = ∣ Θ - Θ_{0} ∣ ∣ Θ + Θ_{0} ∣ \leq 2 R ∣ Θ - Θ_{0} ∣$ . Taking expectations, then the infimum over couplings, gives $∣ E_{μ} θ^{2} - E_{μ_{0}} θ^{2} ∣ \leq 2 R W_{1} (μ, μ_{0})$ . Multiplication by $δ^{2} / a_{0}^{2}$ and step 1.1 yields the claimed risk-transfer bound.

Used by

Nothing in the library uses this result yet.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.