TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07

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$L^{p}$ convergence implies convergence in probability

Statement

Let $1 \leq p < \infty$ . If $X_{n} \to X$ in $L^{p}$ , then $X_{n} \to X$ in probability.

Facts & Assumptions

Given: $1 \leq p < \infty$ and $X_{n} \to X$ in $L^{p}$ .

[L1]

Markov's inequality bounds $P (Z \geq a)$ by $E Z / a$ for nonnegative $Z$ and $a > 0$ (Markov's inequality for random variables).

[L2]

$L^{p}$ convergence means $E ∣ X_{n} - X ∣^{p} \to 0$ ( $L^{p}$ convergence for random variables).

Proof

technique · direct

1.1L1

For $ε > 0$ , apply [L1] to $Z = ∣ X_{n} - X ∣^{p}$ with $a = ε^{p}$ to obtain [L1] $P (∣ X_{n} - X ∣ > ε) \leq ε^{- p} E ∣ X_{n} - X ∣^{p} .$

2.1step 1.1L2∎

The bound in step 1.1 tends to $0$ by [L2], so the definition of convergence in probability applies.

Depends on

Used by

Dependency tree · two levels

9 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

S. Roch, Lecture 3: Modes of convergence, Theorem 3.12 (standard reference, not scraped)

LpLp convergence implies convergence in probability