ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pass

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 zero-gradient quadratic limit

Example

Let $Ω = {- \sqrt{2}, 0, \sqrt{2}}$ with probabilities $1 / 4, 1 / 2, 1 / 4$ , let $Z (ω) = ω$ , and set $T_{n} = Z / \sqrt{n}$ for $n \geq 1$ . For $g (t) = t^{2}$ at $θ = 0$ , the first derivative vanishes and the second derivative is $2$ . The second-order delta method gives $n T_{n}^{2} \Rightarrow Z^{2}$ , which equals $0$ or $2$ with probability $1 / 2$ each.

Facts & Assumptions

Given: The finite three-point law and sequence in the example.

[A1]

$Ω = {- \sqrt{2}, 0, \sqrt{2}}$ has full sigma-field and assigns probabilities $1 / 4, 1 / 2, 1 / 4$ respectively; $Z (ω) = ω$ .

[F1]

First- and second-order delta method for random vectors: if $r_{n} > 0$ is deterministic with $r_{n} \to \infty$ , $r_{n} (T_{n} - θ) \Rightarrow Z$ , the Borel map $g$ has zero first derivative at $θ$ , and $g (θ + h) - g (θ) = \frac{1}{2} H [h, h] + o (∥ h ∥^{2})$ for a continuous symmetric bilinear map $H$ , then $r_{n}^{2} (g (T_{n}) - g (θ)) \Rightarrow \frac{1}{2} H [Z, Z]$ .

Verification

Given: The three-point law, $r_{n} = \sqrt{n}$ , $θ = 0$ , and $g (t) = t^{2}$ .

1.1A1givenalgebra

By [A1], $Z$ has the stated three-point law. We have $\sqrt{n} (T_{n} - 0) = Z$ exactly and $g (h) - g (0) = h^{2} = \frac{1}{2} (2) h^{2}$ for every $h$ , so the first derivative is zero, the second derivative is $2$ , and the second-order remainder vanishes.

2.1A1F1givenstep 1.1algebra∎

Apply [F1] to obtain $n T_{n}^{2} \Rightarrow Z^{2}$ . By [A1], $Z^{2} = 0$ at $ω = 0$ and $Z^{2} = 2$ at $ω = \pm \sqrt{2}$ , so each value has probability $1 / 2$ .

Depends on

First- and second-order delta method for random vectors

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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