ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-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 standardized Laplace ICA model has signed-permutation alternatives

Example

Let $S_{1}, S_{2}$ be independent with common density $f (s) = 2^{- 1 / 2} e^{- \sqrt{2} ∣ s ∣}$ , which has mean zero, variance one, and is non-Gaussian. Put $S = (S_{1}, S_{2})^{T}$ , $A = I_{2}$ , and $Q = (\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array})$ . Then $X = A S$ and $X = (A Q) (Q^{T} S)$ are two square independent-component representations of the same law, with standardized non-Gaussian sources.

Facts & Assumptions

Given: The independent Laplace sources and signed-permutation matrix in the example statement.

[F1]

Square independent-component model with standardized sources: a square ICA representation has an invertible mixing matrix and mutually independent centered unit-variance sources.

Verification

1.1givenalgebra

Put $a = \sqrt{2}$ . Direct integration gives $\int_{0}^{\infty} e^{- a s} d s = 1 / a$ and $\int_{0}^{\infty} s^{2} e^{- a s} d s = [- e^{- a s} (s^{2} / a + 2 s / a^{2} + 2 / a^{3})]_{0}^{\infty} = 2 / a^{3}$ . Hence $\int_{R} f (s) d s = 1$ , symmetry gives mean zero, and $E S_{j}^{2} = 2 (1 / \sqrt{2}) (2 / a^{3}) = 1$ . The one-sided derivatives of $f$ at zero are $1$ and $- 1$ , unlike the smooth standard Gaussian density; since both densities are continuous, this law is non-Gaussian.

2.1F1givenstep 1.1algebra

The matrix $Q$ is a signed permutation, so $Q^{T} S = (S_{2}, - S_{1})^{T}$ . The two coordinates remain independent, centered, and unit variance by step 1.1; the even Laplace density is unchanged by sign reversal, so both remain non-Gaussian.

3.1F1step 2.1algebra∎

Since $Q^{T} Q = I_{2}$ , $(A Q) (Q^{T} S) = A S$ exactly. Thus the observed law is unchanged while the source order/sign representation changes.

Depends on

Square independent-component model with standardized sources

Used by

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