LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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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.

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Convergence in probability plus a higher-moment bound implies Lp convergence

Statement

Let $1 \leq q < s < \infty$ . Suppose random vectors $Z_{n}, Z$ in a fixed finite-dimensional Euclidean space satisfy $Z_{n} \to Z$ in probability, $E ∥ Z ∥^{s} < \infty$ , and $\sup_{n} E ∥ Z_{n} ∥^{s} < \infty$ . Then $E ∥ Z_{n} - Z ∥^{q} \to 0$ .

Facts & Assumptions

Given: The moment, dimension, and convergence hypotheses in the Statement.

[A1]

Convergence in probability means $P (∥ Z_{n} - Z ∥ > δ) \to 0$ for every $δ > 0$ .

[A2]

$1 \leq q < s$ and the stated $s$ th moments are uniformly bounded.

Proof

Given: The hypotheses in the Statement.

1.1A1A2givenalgebra

Put $D_{n} = ∥ Z_{n} - Z ∥$ . The triangle inequality and convexity of $t \mapsto t^{s}$ on $[0, \infty)$ give $D_{n}^{s} \leq 2^{s - 1} (∥ Z_{n} ∥^{s} + ∥ Z ∥^{s})$ , so [A2] implies $\sup_{n} E D_{n}^{s} = : C < \infty$ .

2.1A1A2step 1.1algebra∎

For any $0 < δ < M$ , split according to $D_{n} \leq δ$ , $δ < D_{n} \leq M$ , and $D_{n} > M$ . On these three events, respectively, $D_{n}^{q} \leq δ^{q}$ , $D_{n}^{q} \leq M^{q}$ , and $D_{n}^{q} \leq M^{q - s} D_{n}^{s}$ . Therefore $E D_{n}^{q} \leq δ^{q} + M^{q} P (D_{n} > δ) + C M^{q - s} .$ For fixed $δ, M$ , the probability term tends to zero by [A1]; then letting $M \to \infty$ and $δ ↓ 0$ proves $E D_{n}^{q} \to 0$ .

Used by

The perturbed-observation EnKF converges to the Kalman filter at fixed horizon Theorem

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Jan Mandel, Loren Cobb, and Jonathan D. Beezley, On the Convergence of the Ensemble Kalman Filter, §4 (standard reference, not scraped)