ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-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 one-observation GCV tie can select severe undersmoothing

Example

Take $A = [1]$ , observed datum $b = 1$ , true solution $x^{†} = 0$ (so the noise is $e = 1$ ), and grid $λ_{1} = 1 / 10 < λ_{2} = 10$ . The finite-grid GCV rule ties at these two parameters and chooses $λ_{1}$ , whose squared solution error is one hundred times the squared error at $λ_{2}$ .

Facts & Assumptions

Given: The scalar matrix $A = [1]$ , observation $b = 1$ , target $x^{†} = 0$ , and ordered grid $(1 / 10, 10)$ .

[F1]

For every $λ > 0$ , the Tikhonov objective is strictly convex and has the unique minimiser $(A^{*} A + λ I)^{- 1} A^{*} b$ : The Tikhonov solution is $(A^{T} A + λ I)^{- 1} A^{T} b$ .

[F2]

The ridge leave-one-out identity yields the generalized cross-validation score: The GCV score is $∥ (I - H_{λ}) b ∥^{2} / [m (1 - tr (H_{λ}) / m)^{2}]$ .

[F3]

Finite-grid generalized cross-validation for Tikhonov regularization: The rule selects the least-score grid point and breaks ties by the least index.

Verification

1.1givenF1F2algebra

Here $H_{λ} = 1 / (1 + λ)$ , the residual is $(1 - H_{λ}) b = λ / (1 + λ)$ , and $m = 1$ . Therefore the numerator and denominator of the score are both $(λ / (1 + λ))^{2}$ , so $G C V (λ) = 1$ for each of the two positive grid values.

2.1givenF1F3step 1.1algebra

By [F3], the tied scores select the smaller parameter $λ_{1} = 1 / 10$ . The corresponding estimate is $x_{1 / 10} = 10 / 11$ ; at the other grid value it is $x_{10} = 1 / 11$ .

3.1step 2.1algebra∎

Since the target is zero, the squared errors are $(10 / 11)^{2} = 100 / 121$ and $(1 / 11)^{2} = 1 / 121$ . The first is one hundred times the second, so this valid finite GCV selection need not minimize solution error on its grid.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Per Christian Hansen, Introduction to Inverse Problems, Chapter 5: Regularization Parameter Choice (standard reference, not scraped)