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.

Rank deficiency gives singular information and nonunique weighted least squares

Example

Use the Bernoulli-logit model with $β = (0, 0)^{T}$ , design rows $x_{1}^{T} = (1, 1)$ and $x_{2}^{T} = (2, 2)$ , and responses $(y_{1}, y_{2}) = (0, 1)$ . The weights are positive, but the design columns are dependent.

Facts & Assumptions

Given: The two-row design and responses in the Example, the Bernoulli-logit family, current $β = 0$ , and $a (ϕ) = 1$ .

[F1]

General-link score and Fisher information: $I = X^{T} diag (w_{i}) X$ and $U = X^{T} ({\dot{μ}}_{i} (y_{i} - μ_{i}) / (a V_{i}))_{i}$ .

[F2]

One minimum-norm IRLS step: IRLS minimizes the weighted least-squares objective and applies the Moore-Penrose minimum-norm convention if minimizers are not unique.

[F3]

Each Bernoulli-logit response has the counting-measure mass $f (y; η) = \exp {y η - \log (1 + e^{η})}$ on ${0, 1}$ . It is normalized, has fixed support and finite moments, and its first two $η$ derivatives are locally bounded on this finite support. (Bernoulli random variables and binomial random variables as sums of independent Bernoulli trials, The exponential function is smooth and $(\exp)^{'} = \exp$ , The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t)

Verification

1.1F1F3givenalgebra

At $β = 0$ , both predictors are $0$ , both means are $1 / 2$ , and both logit derivatives are $1 / 4$ . Thus $w_{1} = w_{2} = 1 / 4$ by [F1]. Direct multiplication gives $I = X^{T} W X = \frac{5}{4} (\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array})$ , which has rank one; the score is $U = (1 / 2, 1 / 2)^{T}$ .

2.1F1F2step 1.1algebra∎

The working responses are $z_{1} = - 2$ and $z_{2} = 2$ . For $γ = (γ_{1}, γ_{2})^{T}$ , put $s = γ_{1} + γ_{2}$ ; the weighted objective is $J (s) = \frac{1}{4} {(- 2 - s)^{2} + (2 - 2 s)^{2}}$ . Differentiation gives $J^{'} (s) = (5 s - 2) / 2$ , so every $γ$ with $s = 2 / 5$ is a minimizer. These minimizers form a line, while $∥ γ ∥_{2}^{2} = s^{2} / 2 + (γ_{1} - γ_{2})^{2} / 2$ shows the Moore-Penrose convention selects $(1 / 5, 1 / 5)$ . The information inverse is undefined because $I$ is singular.

Depends on

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