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Integrability and Uniform Integrability · Tutorial 140 of 946

Lyapunov’s Inequality and Moment Comparison

Learn how probability-space moment comparisons follow from Jensen’s inequality, including their integrability consequences and limitations.

Intermediate 10 min read

What You'll Learn

  • State Lyapunov’s inequality for absolute moments and \(L^p\) norms.
  • Derive lower-moment bounds from a finite higher moment.
  • Extend the comparison to \(L^\infty\).
  • Recognize when equality holds in the moment comparison.
  • Distinguish moment bounds on probability spaces from those on general measure spaces.
  • Identify why a finite lower moment need not imply a finite higher moment.

Why Compare Moments?

A random variable may have several finite moments, and those moments quantify different aspects of its size. A basic question is whether knowing a higher moment is finite or bounded tells us anything about lower moments. On a probability space, it does: the total probability is one, and that normalization gives a useful comparison unavailable on arbitrary measure spaces.

This relationship is called Lyapunov’s inequality. It orders \(L^p\) norms by their exponent and gives a direct test for integrability: a finite moment of order \(s\) guarantees finite absolute moments of every smaller positive order. The direction matters. A finite lower moment alone does not, in general, control a higher one.

Moments and Norms

Let \((\Omega,\mathcal F,\mathbb P)\) be a probability space and let \(X\) be a real-valued random variable. For \(r>0\), the \(r\)-th absolute moment is \(\mathbb E|X|^r\), when finite. For \(p\geq 1\), the \(L^p\) norm is \(\|X\|_p=(\mathbb E|X|^p)^{1/p}\), provided \(X\in L^p\). The moment comparison itself applies to all positive exponents, including those below one; the term “\(L^p\) norm” is reserved here for \(p\geq 1\).

Definition: For positive exponents \(0<r<s\), Lyapunov’s inequality is the comparison $$ \bigl(\mathbb E|X|^r\bigr)^{1/r} \leq \bigl(\mathbb E|X|^s\bigr)^{1/s}, $$ whenever \(\mathbb E|X|^s<\infty\). In particular, the inequality also asserts that the lower moment \(\mathbb E|X|^r\) is finite.

The probability-space assumption is essential to the standard comparison. It gives \(\mathbb E1=1\), so expectations can be treated as averages with respect to a probability measure. We will use Jensen’s inequality for the concave function \(t\mapsto t^\alpha\), where \(0<\alpha<1\).

Lyapunov’s Inequality

Theorem: Let \(0<r<s<\infty\), and suppose \(X\) is a random variable on a probability space with \(\mathbb E|X|^s<\infty\). Then $$ \mathbb E|X|^r \leq \bigl(\mathbb E|X|^s\bigr)^{r/s}, \qquad \bigl(\mathbb E|X|^r\bigr)^{1/r} \leq \bigl(\mathbb E|X|^s\bigr)^{1/s}. $$
Proof: Set \(Y=|X|^s\) and \(\alpha=r/s\). Then \(Y\geq0\), \(\mathbb EY<\infty\), and \(0<\alpha<1\). The function \(t\mapsto t^\alpha\) is concave on \([0,\infty)\), so Jensen’s inequality gives $$ \mathbb E|X|^r = \mathbb E(Y^\alpha) \leq (\mathbb EY)^\alpha = \bigl(\mathbb E|X|^s\bigr)^{r/s}. $$ The right-hand side is finite, so the lower moment is finite as well. Taking the \(r\)-th root of both nonnegative sides yields $$ \bigl(\mathbb E|X|^r\bigr)^{1/r} \leq \bigl(\mathbb E|X|^s\bigr)^{1/s}. $$ If \(\mathbb E|X|^s=0\), then \(|X|^s=0\) almost surely, and both moments are zero; the same inequalities hold. \(\square\)

For exponents \(1\leq p<s<\infty\), this says that membership in \(L^s\) implies membership in \(L^p\), with the norm bound \(\|X\|_p\leq\|X\|_s\). The result is stronger than a comparison for integer moments: neither exponent needs to be an integer.

Worked Example: Moments of a Uniform Random Variable

Worked Example: Moments of a Uniform Random Variable

Let \(X\) be uniformly distributed on \([0,1]\). Its density is one on this interval, so for every \(u>0\), $$ \mathbb E|X|^u = \int_0^1 x^u\,dx = \frac{1}{u+1}. $$ Consequently, $$ \|X\|_u = \left(\frac{1}{u+1}\right)^{1/u} \quad (u\geq1). $$ For example, \(\mathbb E|X|=1/2\), while \(\mathbb E|X|^2=1/3\). The corresponding norms are $$ \|X\|_1=\frac12, \qquad \|X\|_2=\frac{1}{\sqrt{3}}. $$ Since \(1/2\leq1/\sqrt{3}\), this concrete comparison agrees with Lyapunov’s inequality. More generally, the formula gives finite moments for every positive exponent, and the theorem guarantees that \(\|X\|_r\leq\|X\|_s\) whenever \(1\leq r<s\).

The Norm Comparison, Including \(L^\infty\)

The finite-exponent result gives monotonicity of \(L^p\) norms. The comparison also extends to the essential supremum. Recall that \(X\in L^\infty\) means that there is a finite constant \(M\) such that \(|X|\leq M\) almost surely; the \(L^\infty\) norm is the least such bound, denoted \(\|X\|_\infty\).

Corollary: If \(1\leq p<q<\infty\) and \(X\in L^q\), then \(X\in L^p\) and \(\|X\|_p\leq\|X\|_q\). If \(X\in L^\infty\) and \(1\leq p<\infty\), then \(X\in L^p\) and \(\|X\|_p\leq\|X\|_\infty\).
Proof: The assertion for finite \(q\) is Lyapunov’s inequality with \(r=p\) and \(s=q\). For the other assertion, let \(M=\|X\|_\infty\). By the definition of essential supremum, \(|X|\leq M\) almost surely. Since \(\mathbb P(\Omega)=1\), $$ \mathbb E|X|^p \leq \mathbb E(M^p) = M^p\mathbb P(\Omega) = M^p. $$ Taking \(p\)-th roots gives \(\|X\|_p\leq M=\|X\|_\infty\). This also covers \(M=0\), in which case \(X=0\) almost surely. \(\square\)

Thus, on a probability space, finite \(L^q\) norms increase as the exponent increases, up to and including the \(L^\infty\) endpoint. This conclusion should not be transferred unchanged to a measure space whose total measure is not one: the proof uses \(\mathbb E1=1\).

Worked Example: A Variable Supported on a Rare Event

Worked Example: A Variable Supported on a Rare Event

Let \(A\in\mathcal F\) have probability \(a=\mathbb P(A)\), where \(0\leq a\leq1\), and let \(X=c\,1_A\) for a constant \(c\geq0\). Since \(|X|=c\) on \(A\) and is zero off \(A\), for every \(u>0\), $$ \mathbb E|X|^u = c^u a. $$ If \(c>0\) and \(a>0\), this gives $$ \bigl(\mathbb E|X|^u\bigr)^{1/u} = c\,a^{1/u}. $$ For \(0<a<1\) and \(r<s\), we have \(1/r>1/s\), so \(a^{1/r}\leq a^{1/s}\). Therefore \(c\,a^{1/r}\leq c\,a^{1/s}\), exactly as Lyapunov’s inequality requires. If \(a=0\) or \(c=0\), both norms are zero; if \(a=1\), both equal \(c\). The example also shows how a small event probability can reduce all these norms, while preserving their ordering.

Moment Bounds Imply Lower-Moment Bounds

Lyapunov’s inequality is often used when a moment is bounded uniformly over a family of variables, or when a particular higher moment has an explicit bound. The inequality transfers that information to every smaller positive exponent.

Corollary: Suppose \(0<r<s<\infty\) and \(\mathbb E|X|^s\leq C\) for a finite constant \(C\geq0\). Then $$ \mathbb E|X|^r\leq C^{r/s}. $$ In particular, \(X\) has a finite \(r\)-th absolute moment.
Proof: Apply Lyapunov’s inequality to obtain $$ \mathbb E|X|^r \leq \bigl(\mathbb E|X|^s\bigr)^{r/s}. $$ Because \(r/s>0\), the function \(t\mapsto t^{r/s}\) is increasing for \(t\geq0\). The bound \(\mathbb E|X|^s\leq C\) therefore gives $$ \mathbb E|X|^r \leq \bigl(\mathbb E|X|^s\bigr)^{r/s} \leq C^{r/s}. $$ For \(C=0\), the same inequalities imply \(\mathbb E|X|^r=0\). \(\square\)

Worked Example: A Fourth-Moment Bound

Worked Example: A Fourth-Moment Bound

Suppose \(X\) is a random variable on a probability space and \(\mathbb E|X|^4\leq16\). Taking \(s=4\) and \(r=2\) gives $$ \mathbb E|X|^2 \leq 16^{2/4} = 4. $$ Taking \(r=1\) instead gives $$ \mathbb E|X| \leq 16^{1/4} = 2. $$ Thus the fourth-moment bound guarantees both square integrability and integrability, with explicit bounds. It also bounds the corresponding norms: \(\|X\|_2\leq2\) and \(\|X\|_1\leq2\). These conclusions rely on the probability-space assumption.

Equality and a Common Pitfall

For \(0<r<s\), the concave function \(t\mapsto t^{r/s}\) is strictly concave. The equality case in Jensen’s inequality therefore shows that equality in Lyapunov’s moment comparison holds precisely when \(|X|^s\) is almost surely constant, equivalently when \(|X|\) is almost surely constant. This includes the case \(X=0\) almost surely. A random variable can have either sign and still attain equality, provided its absolute value does not vary.

A frequent mistake is to reverse the implication. A finite lower moment does not guarantee a finite higher moment. For instance, let \(c=(\sum_{n=1}^{\infty}n^{-3})^{-1}\), and define a positive integer-valued random variable \(N\) by \(\mathbb P(N=n)=c n^{-3}\) for \(n\in\mathbb N\). These probabilities sum to one by the definition of \(c\). Then $$ \mathbb E N = c\sum_{n=1}^{\infty}\frac{1}{n^2} < \infty, \qquad \mathbb E N^2 = c\sum_{n=1}^{\infty}\frac{1}{n} = \infty. $$ So finite first moment does not imply finite second moment. Lyapunov’s inequality provides a one-way implication: a finite higher moment controls lower moments.

The distinction between probability spaces and general measure spaces is another important caution. The inequality for norms uses \(\mathbb P(\Omega)=1\). On a measure space with total measure greater than one, even a constant function can have an \(L^p\) norm that depends on the total measure; the probability-space comparison cannot simply be assumed without checking the measure.

Summary

For \(0<r<s\), a finite \(s\)-th absolute moment on a probability space yields a finite \(r\)-th absolute moment and the bound $$ \bigl(\mathbb E|X|^r\bigr)^{1/r} \leq \bigl(\mathbb E|X|^s\bigr)^{1/s}. $$ For exponents at least one, this orders the \(L^p\) norms, with the same comparison extending to \(L^\infty\). Higher-moment bounds therefore give lower-moment bounds, but the reverse implication can fail.

Check Your Understanding

Use the definitions and comparisons in this tutorial to answer the following questions.

  1. If \(\mathbb E|X|^6\leq64\), what upper bounds follow for \(\mathbb E|X|^2\) and \(\mathbb E|X|\)?
  2. Which concavity property of \(t\mapsto t^{r/s}\) is used to prove Lyapunov’s inequality?
  3. Why does the proof use the assumption that the underlying measure is a probability measure?
  4. For \(X=c\,1_A\), what is \(\mathbb E|X|^u\) in terms of \(c\), \(u\), and \(\mathbb P(A)\)?
  5. Does a finite first absolute moment imply a finite second absolute moment? Give a reason for your answer.
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