CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-04

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Lyapunov's moment inequality on a probability space

Statement

Let $(Ω, F, P)$ be a probability space and let $1 \leq p \leq q \leq \infty$ .

If $q < \infty$ and $X \in L^{q} (P)$ , then $X \in L^{p} (P)$ and $∥ X ∥_{p} \leq ∥ X ∥_{q} .$
If $q = \infty$ and $X \in L^{\infty} (P)$ , then $X \in L^{p} (P)$ and $∥ X ∥_{p} \leq ∥ X ∥_{\infty} .$

Facts & Assumptions

Given: A probability space and exponents $1 \leq p \leq q \leq \infty$ .

[L1]

A probability measure has total mass $1$ (Probability measures and probability spaces).

[L2]

On a finite measure space, $L^{q}$ includes into $L^{p}$ with factor $μ (Ω)^{1 / p - 1 / q}$ for finite $q$ , and $L^{\infty}$ includes into $L^{p}$ with factor $μ (Ω)^{1 / p}$ (Finite-measure $L^{r}$ includes into $L^{p}$ for $p < r$ ).

Proof

technique · direct

1.1

If $p = q$ , the inequality is equality. If $p < q < \infty$ , apply [L2] with $μ = P$ and use [L1] to collapse the factor $P (Ω)^{1 / p - 1 / q}$ to $1$ .

L1L2

1.2

If $q = \infty$ , the same specialization of [L2] and [L1] gives $∥ X ∥_{p} \leq P (Ω)^{1 / p} ∥ X ∥_{\infty} = ∥ X ∥_{\infty} .$

L1L2

2.1

Steps 1.1 and 1.2 are exactly the finite- $q$ and $q = \infty$ cases of Lyapunov's moment inequality on a probability space.

step 1.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

J. R. Norris, Probability and Measure, Theorem 4.4.2 discussion (standard reference, not scraped)