Ordering Moments With Lyapunov's Inequality

Why Moment Orders Can Be Compared

A higher-order absolute moment gives more weight to large values of a random variable than a lower-order one does. Lyapunov's inequality makes this intuition precise: on a probability space, the pth-root absolute moment cannot decrease as the order $p$ increases. One important consequence is that a finite higher-order absolute moment guarantees finiteness of every lower positive-order absolute moment.

The probability-space condition matters. The comparison uses the fact that the total probability is one; it is not simply an algebraic comparison of powers. We will state the result for any real orders $0<r<s$, then use it to examine both finite and infinite-support examples.

Definition: For a random variable $X$ and a real order $p>0$, the absolute moment of order $p$ is $E[|X|^p]$, possibly infinite. Its pth-root absolute moment is $(E[|X|^p])^{1/p}$. When $p\geq 1$, this quantity is the $L^p$ norm of $X$; when $0<p<1$, it is not a norm in general.

The absolute value is essential in this definition. For an odd integer $p$, $E[X^p]$, when defined, is a signed moment and can be smaller than $E[|X|^p]$, or even zero when $X$ is not zero. Lyapunov's inequality concerns absolute moments.

Lyapunov's Inequality

We use Hölder's inequality in the following form: if $q>1$, $q'=q/(q-1)$, and $U,V$ have finite $q$th and $q'$th absolute moments, respectively, then $E[|UV|]\leq (E[|U|^q])^{1/q}(E[|V|^{q'}])^{1/q'}$. This standard form applies on the probability space supporting $X$.

Theorem (Lyapunov's inequality): Let $X$ be a random variable, and let $0<r<s$. If $E[|X|^s]<\infty$, then $$ (E[|X|^r])^{1/r}\leq (E[|X|^s])^{1/s}. $$ In particular, $E[|X|^r]<\infty$.

Proof. Put $q=s/r>1$, so that $q'=s/(s-r)$. Apply Hölder's inequality to $U=|X|^r$ and $V=1$. The required moments are finite: $E[|U|^q]=E[|X|^s]<\infty$, and $E[|V|^{q'}]=E[1]=1$. Therefore

E[|X|^r] =E[|X|^r\cdot 1] \leq (E[|X|^{rq}])^{1/q}(E[1])^{1/q'} =(E[|X|^s])^{r/s}.

Taking the $1/r$ power of both nonnegative sides gives the asserted inequality. The right side is finite by assumption, so the left side is finite as well. $\square$

If $E[|X|^s]=0$, the same argument gives $E[|X|^r]=0$; there is no need to divide by either moment. This covers the degenerate case $X=0$ almost surely. If $r=s$, the comparison is simply equality, so the strict condition $r<s$ identifies the nontrivial case.

Finiteness of Lower-Order Moments

The theorem immediately gives a useful integrability rule. Here, saying that an absolute moment is finite means its expectation is a finite real number, not merely that the integrand is finite at each outcome.

Corollary: If $E[|X|^s]<\infty$ for some $s>0$, then $E[|X|^r]<\infty$ for every $r$ with $0<r\leq s$.

Proof. For $r=s$, finiteness is the assumption. For $0<r<s$, Lyapunov's inequality gives

0\leq E[|X|^r]\leq (E[|X|^s])^{r/s}<\infty.

Thus the order-$r$ absolute moment is finite for every such $r$. $\square$

The conclusion includes noninteger orders. For example, a finite third absolute moment ensures that moments of orders $1/2$, $1$, $2$, and every other positive order below $3$ are finite. The converse does not follow: finiteness at a lower order does not guarantee finiteness at a higher one.

Worked Comparisons

Worked Example: Three Orders for a Discrete Random Variable

Let $X$ have the distribution $P(X=-2)=1/4$, $P(X=0)=1/2$, and $P(X=1)=1/4$. Compute its absolute moments of orders $1$, $2$, and $4$, and compare their fourth-, second-, and first-root values as appropriate.

For order $1$, only the nonzero outcomes contribute:

E[|X|]=\frac14(2)+\frac12(0)+\frac14(1)=\frac34.

For order $2$ and order $4$, the calculations are

E[|X|^2]=\frac14(2^2)+\frac12(0^2)+\frac14(1^2)=\frac54, \qquad E[|X|^4]=\frac14(2^4)+\frac12(0^4)+\frac14(1^4)=\frac{17}{4}.

The corresponding pth-root absolute moments are

(E[|X|])^1=0.750,\qquad (E[|X|^2])^{1/2}=\sqrt{1.25}\approx 1.118,\qquad (E[|X|^4])^{1/4}=(4.25)^{1/4}\approx 1.435.

They increase with the order, as Lyapunov's inequality requires. In particular, the fourth-root value is about $1.435$, since $(4.25)^{1/4}\approx 1.43527$.

Worked Example: A Mean Check and Equal Absolute Moments

Suppose $X=0$ with probability $1/2$ and $X=2$ with probability $1/2$. First calculate the mean, then compare the pth-root absolute moments at orders $1$, $2$, and $4$.

The mean is

E[X]=\frac12(0)+\frac12(2)=1.

Since $X$ is nonnegative, its absolute moments are $E[X^p]$. For each of the three orders,

E[|X|]=1,\qquad E[|X|^2]=\frac12(0^2)+\frac12(2^2)=2,\qquad E[|X|^4]=\frac12(0^4)+\frac12(2^4)=8.

Taking the corresponding roots gives

(E[|X|])^1=1,\qquad (E[|X|^2])^{1/2}=\sqrt{2}\approx 1.414,\qquad (E[|X|^4])^{1/4}=8^{1/4}=2^{3/4}\approx 1.682.

Again, the values increase with the order. The calculation of $E[X]$ also illustrates a simple check: the two outcomes have equal probability, so the mean is their average, $(0+2)/2=1$.

Worked Example: A Finite Second Moment but an Infinite Third Moment

Let $X$ have density $f(x)=3x^{-4}$ for $x\geq 1$, and $f(x)=0$ otherwise. First check that this is a probability density, then find which positive-order absolute moments are finite.

Its integral is

\int_1^\infty 3x^{-4}\,dx =3\left[-\frac{1}{3}x^{-3}\right]_1^\infty=1,

so it is normalized. Since $X\geq 1$, $|X|^p=X^p$. For $p>0$,

E[|X|^p]=\int_1^\infty x^p\,3x^{-4}\,dx =3\int_1^\infty x^{p-4}\,dx.

This improper integral is finite exactly when $p-4<-1$, or $p<3$. When $0<p<3$, direct integration gives

E[|X|^p]=3\left[\frac{x^{p-3}}{p-3}\right]_1^\infty =\frac{3}{3-p}.

For example, $E[X]=3/2$, while $E[X^2]=3$, so their root values are $3/2=1.5$ and $\sqrt{3}\approx 1.732$. At $p=3$, the integral becomes $3\int_1^\infty x^{-1}\,dx=\infty$; for $p>3$, it also diverges. Thus a finite second moment does not imply a finite third moment, while every positive-order moment below $3$ is finite, in agreement with the corollary.

What the Inequality Does and Does Not Say

Lyapunov's inequality compares moment values after taking the root corresponding to each order. It does not say that the unrooted moments $E[|X|^r]$ and $E[|X|^s]$ themselves must always be ordered in the same way. If typical values of $|X|$ lie below one, raising them to a higher power can reduce them. For instance, if $|X|=1/2$ with probability one, then $E[|X|]=1/2$ and $E[|X|^2]=1/4$, but the root values are both $1/2$, as the inequality predicts.

A second point is that the probability-space assumption is built into the comparison. In the Hölder step, the factor $1$ has moment $E[1]=1$. On a general measure space with total mass other than one, that factor need not be one, and the same statement cannot be applied unchanged.

Common Mistakes

Dropping the absolute value. The theorem concerns $E[|X|^p]$. A signed odd moment can involve cancellation and cannot replace the absolute moment in the comparison.
Reversing the implication. A finite higher-order moment guarantees lower-order finiteness, but a finite lower-order moment need not guarantee a finite higher one. The continuous example has a finite second moment and an infinite third.
Comparing unrooted moments as if they were the theorem's conclusion. Lyapunov compares $(E[|X|^p])^{1/p}$, not the raw expectations. The root is part of the order comparison.
Calling every pth-root absolute moment a norm. It is an $L^p$ norm for $p\geq1$, but for $0<p<1$ the triangle inequality can fail. In that range, call it a pth-root absolute moment, not a norm.

Key takeaway: On a probability space, if $0<r<s$ and $E[|X|^s]<\infty$, then $(E[|X|^r])^{1/r}\leq(E[|X|^s])^{1/s}$. Consequently, every positive-order absolute moment below a finite moment order is finite too.

Check Your Understanding

Use absolute moments and keep the order of each root explicit.

If $E[|X|^6]<\infty$, what does Lyapunov's inequality imply about $E[|X|^{2}]$ and $E[|X|^{5/2}]$?
For a random variable with $E[|X|^2]=9$, what upper bound does Lyapunov's inequality give for $E[|X|]$?
Why does a finite first absolute moment not, by itself, guarantee a finite second absolute moment?
For the density $f(x)=3x^{-4}$ on $x\geq1$, which positive orders have finite absolute moments, and what happens at order $3$?
Explain why the theorem compares pth-root absolute moments rather than simply asserting that the raw moments increase with their order.