LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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.

Levy maximal bound from uniform tail bounds

Statement

Let $X_{1}, \dots, X_{n}$ be independent real random variables, $n \geq 1$ , with $S_{k} = \sum_{j = 1}^{k} X_{j}$ . Let $l > 0$ and $0 \leq δ < 1$ . If $P (∣ \sum_{j = i}^{n} X_{j} ∣ \geq l / 2) \leq δ (1 \leq i \leq n),$ then $P (\max_{k \leq n} ∣ S_{k} ∣ \geq l) \leq \frac{δ}{1 - δ} .$ No centering or moment assumption is required.

Facts & Assumptions

[F1]

Disjoint groups of an independent sigma-algebra family remain independent: Let $(F_{i})_{i \in I}$ be an independent family of sigma-algebras on a probability space, and let $J_{0}, \dots, J_{m - 1} \subseteq I$ be pairwise disjoint index sets. For each $r < m$ , define $G_{r} : = σ (⋃_{i \in J_{r}} F_{i}) .$ Then the sigma-algebras $G_{0}, \dots, G_{m - 1}$ are independent.

[F2]

Arithmetic and lattice operations preserve measurability whenever they are defined: Let $(X, A)$ be a measurable space and let $f, g : X \to \overline{R}$ be measurable. Then: 1. $c f$ is measurable for every real scalar $c$ ; 2. $\max (f, g)$ , $\min (f, g)$ , $∣ f ∣$ , $f^{+}$ , and $f^{-}$ are measurable; 3. if $f + g$ is pointwise defined, then $f + g$ is measurable; 4. with the convention of rem-zero-times-infinity-convention-for-pointwise-products, the pointwise product $f g$ is measurable.

Proof

Given: The objects and hypotheses of the statement.

1.1

Let $A_{k} = {∣ S_{j} ∣ < l (j < k), ∣ S_{k} ∣ \geq l}$ and $A = ⋃_{k} A_{k}$ . These are measurable disjoint first-crossing events. For $k < n$ , $A_{k}$ is independent of the remaining tail $R_{k} = S_{n} - S_{k}$ by grouping. For $k = n$ , $R_{n} = 0$ , so its probability of magnitude at least $l / 2$ is zero.

F2F1given

2.1

On $A_{k} \cap {∣ S_{n} ∣ \leq l / 2}$ , the triangle inequality gives $∣ R_{k} ∣ \geq l / 2$ . Thus $P (A \cap {∣ S_{n} ∣ \leq l / 2}) \leq \sum_{k} P (A_{k}) P (∣ R_{k} ∣ \geq l / 2) \leq δ P (A)$ . The complementary part has probability at most $P (∣ S_{n} ∣ > l / 2) \leq δ$ , using the hypothesis at $i = 1$ . Hence $(1 - δ) P (A) \leq δ$ , and division by the positive $1 - δ$ proves the assertion. This includes $δ = 0$ and $n = 1$ .

step 1.1givenalgebra∎

Depends on

Used by

Convergence in probability and almost surely agree for independent series Theorem

Dependency tree · two levels

12 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

Lemma 3.8 and proof, pp. 62–63 (standard reference, not scraped)