ExampleConstruction: AI-adaptedVerification: 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 Rademacher sum

Example

Assume AC. Let $n \geq 1$ and let $X_{1}, \dots, X_{n}$ be independent Rademacher variables, meaning $P (X_{i} = 1) = P (X_{i} = - 1) = 1 / 2$ , and put $S_{n} = \sum_{i = 1}^{n} X_{i}$ . For every $t > 0$ , $P (∣ S_{n} ∣ \geq t) \leq 2 e^{- t^{2} / (2 n)} .$

Facts & Assumptions

Given: AC, an integer $n \geq 1$ , and independent real random variables $X_{1}, \dots, X_{n}$ with the two-point laws stated in the Example.

[A1]

Assume AC as required by the Hoeffding supplier used here (The Axiom of Choice).

[F1]

A real random variable is a measurable map to $(R, B (R))$ (Random elements and real random variables).

[F2]

Independence means independence of the sigma-algebras generated by the variables (Independent random elements).

[F3]

Expectation is integration against the probability law (Expectation of a nonnegative or integrable random variable).

[F4]

For independent variables in finite ranges $[a_{i}, b_{i}]$ , Hoeffding gives $P (∣ \sum_{i} (X_{i} - E X_{i}) ∣ \geq t) \leq 2 e^{- 2 t^{2} / \sum_{i} (b_{i} - a_{i})^{2}}$ when the denominator is positive (Hoeffding inequality for independent bounded sums).

Verification

1.1givenA1F1F2F3algebra

Each $X_{i}$ is a real random variable by [F1], and their given independence is understood by [F2]. Their two-point laws make them integrable with mean $(1 / 2) \cdot 1 + (1 / 2) \cdot (- 1) = 0$ by [F3], and they lie in $[- 1, 1]$ almost surely. Thus the Hoeffding width sum is $\sum_{i} (1 - (- 1))^{2} = 4 n$ .

2.1F4step 1.1algebra∎

Apply [F4] with those endpoints. The exponent is $- 2 t^{2} / (4 n) = - t^{2} / (2 n)$ , yielding the stated bound.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

34 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Philippe Rigollet and Jan-Christian Hütter, High-Dimensional Statistics: Lecture Notes, Chapter 1, Rademacher example and Theorem 1.9 (standard reference, not scraped)
John Duchi, Lecture Notes on Statistics and Information Theory, Chapter 4, Example 4.1.5 and Corollary 4.1.10 (standard reference, not scraped)