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.

1.1givenalgebra

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.

2.1F1step 1.1algebra∎

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.

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)