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

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Strong law for independent nonidentical variables

Statement

For independent square-integrable real random variables $(X_{n})_{n \geq 1}$ , the condition $\sum_{n \geq 1} Var (X_{n}) / n^{2} < \infty$ implies $\frac{S_{n} - E S_{n}}{n} ⟶ 0 almost surely, S_{n} = \sum_{k = 1}^{n} X_{k} .$ The variables need not have a common law or a common mean.

Facts & Assumptions

[F1]

Strong law under summable normalized variances: Let $(X_{n})_{n \geq 1}$ be independent square-integrable real random variables. Let $0 < b_{n}$ be deterministic and nondecreasing with $b_{n} \to \infty$ . If $\sum_{n \geq 1} \frac{Var (X_{n})}{b_{n}^{2}} < \infty,$ then $\frac{1}{b_{n}} \sum_{k = 1}^{n} (X_{k} - E X_{k}) ⟶ 0 almost surely .$ In particular, for IID centered square-integrable variables and any $ε > 0$ , $S_{n} / [\sqrt{n} (\log n)^{1 / 2 + ε}] \to 0$ almost surely (the displayed normalization is used for $n \geq 2$ ).

Proof

Given: The objects and hypotheses of the statement.

1.1

Take $b_{n} = n$ for $n \geq 1$ . This sequence is positive, increasing, and tends to infinity, and the required normalized variance series is exactly the one in the hypothesis.

givenalgebra

2.1

The normalized-variance strong law gives $n^{- 1} \sum_{k = 1}^{n} (X_{k} - E X_{k}) \to 0$ almost surely. Finite linearity identifies the numerator with $S_{n} - E S_{n}$ . This includes zero variance and deterministic variables and requires no relation between different means.

F1step 1.1algebra∎

Depends on

Strong law under summable normalized variances

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Theorems 2.5.6 and 2.5.9, pp. 84–86 (standard reference, not scraped)