Product Integrability from Hölder Bounds

Why moments control products

A product can be large when several of its factors are large at the same time. Independence can sometimes help analyze that behavior, but it is not necessary for a useful general bound. Hölder’s inequality controls a product using moment information about its individual factors, regardless of how those factors depend on one another.

The key is to match the moment exponents to the number of factors. If the available moments are strong enough, the product has a finite expected absolute value. The same argument also gives an explicit bound, not just an integrability conclusion.

Moment spaces and the product question

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$, and write $\mathbb{E}$ for expectation. For $1\leq p<\infty$, a random variable $X$ belongs to $L^p$ if $\mathbb{E}|X|^p<\infty$, and its $L^p$ norm is

\|X\|_p=\bigl(\mathbb{E}|X|^p\bigr)^{1/p}.

A random variable belongs to $L^\infty$ if it is essentially bounded; its norm $\|X\|_\infty$ is the essential supremum of $|X|$. The question here is: given moment information about $X$ and $Y$, when can we conclude that $XY$ belongs to $L^1$?

Definition: A product $XY$ is integrable when $\mathbb{E}|XY|<\infty$, or equivalently when $XY\in L^1$. Integrability concerns the absolute value: cancellation between positive and negative values does not by itself establish this condition.

For finite exponents $p,q\geq 1$, the basic sufficient condition is that their reciprocals add to at most one. On a probability space, any spare moment strength can be used to pass from an $L^s$ bound to an $L^1$ bound. This last step uses that the total probability is one.

The product bound

Theorem: Let $1\leq p,q\leq\infty$, with the convention $1/\infty=0$. Suppose $X\in L^p$, $Y\in L^q$, and $$ \frac{1}{p}+\frac{1}{q}\leq 1. $$ Then $XY\in L^1$, and $$ \mathbb{E}|XY|\leq \|X\|_p\|Y\|_q. $$

Proof. First suppose $p,q<\infty$. Set $a=1/p+1/q$, which is positive and at most one, and define $s=1/a$. Thus $1\leq s<\infty$. The exponents $p/s$ and $q/s$ are at least one, and their reciprocals add to one. Apply Hölder’s inequality to $|X|^s$ and $|Y|^s$ with these exponents:

\mathbb{E}|XY|^s \leq \bigl(\mathbb{E}|X|^p\bigr)^{s/p} \bigl(\mathbb{E}|Y|^q\bigr)^{s/q}.

Taking the $s$-th root gives $\|XY\|_s\leq \|X\|_p\|Y\|_q$. Since $\mathbb{P}(\Omega)=1$, Lyapunov’s inequality gives $\|XY\|_1\leq\|XY\|_s$ when $s\geq1$. Therefore $XY\in L^1$ and the claimed expectation bound follows.

If $p=\infty$ and $q<\infty$, then $|XY|\leq\|X\|_\infty|Y|$ almost surely. Taking expectations and using $\mathbb{E}|Y|\leq\|Y\|_q$ gives the result. The case $q=\infty$, $p<\infty$ follows by interchanging the factors. If both exponents are infinite, $|XY|\leq\|X\|_\infty\|Y\|_\infty$ almost surely, which also gives the result. $\square$

The theorem needs no independence assumption. Hölder’s inequality is applied directly to the joint values of the factors, so it remains valid when they are strongly dependent. It also gives a quantitative conclusion: the expected absolute product is bounded by the product of the available moment norms.

Worked Example: Dependent factors with second moments

Let $Z$ be a standard normal random variable, and define $X=Z$ and $Y=Z$. These factors are completely dependent: knowing $X$ determines $Y$. Since $\mathbb{E}Z^2=1$, both variables belong to $L^2$, with $\|X\|_2=\|Y\|_2=1$. The theorem with $p=q=2$ gives

\mathbb{E}|XY| = \mathbb{E}Z^2 \leq \|X\|_2\|Y\|_2 = 1.

In fact, $\mathbb{E}|XY|=\mathbb{E}Z^2=1$, so equality holds. This calculation verifies integrability despite complete dependence. No factorization of an expectation into a product of expectations has been used.

When the reciprocal sum is strictly less than one

If $p,q$ are finite and $1/p+1/q<1$, the proof above shows something stronger than integrability: the product belongs to $L^s$, where $1/s=1/p+1/q$. Because $s>1$, this is stronger moment information than $L^1$. The expectation bound follows by passing from $L^s$ to $L^1$ on a probability space.

Worked Example: A strict moment margin

Let $U$ be uniformly distributed on $(0,1)$, and set $X=U^{-1/8}$ and $Y=U^{-1/8}$. Again the factors are dependent, in fact identical. For $p=q=4$,

\mathbb{E}|X|^4 = \int_0^1 u^{-1/2}\,du = 2, \qquad \|X\|_4=\|Y\|_4=2^{1/4}.

The reciprocal sum is $1/4+1/4=1/2$, so the theorem applies. Direct calculation gives

\mathbb{E}|XY| = \int_0^1 u^{-1/4}\,du = \frac{4}{3} \leq 2^{1/4}2^{1/4} = \sqrt{2}.

The integral formula used here is $\int_0^1u^{-c}\,du=1/(1-c)$ for $c<1$; substituting $c=1/4$ gives $4/3$, and substituting $c=1/2$ gives $2$. Thus the factors have more moment control than is needed merely to make their product integrable.

More than two factors

The same reasoning extends to a finite collection of factors. If the reciprocal moment exponents add to at most one, Hölder’s inequality bounds the expected absolute product. This statement can be obtained by repeatedly applying the two-factor theorem, as follows.

Theorem: Let $n\in\mathbb{N}$, let $1\leq p_i\leq\infty$ for $i=1,\ldots,n$, and suppose $X_i\in L^{p_i}$ for each $i$. If $$ \sum_{i=1}^n\frac{1}{p_i}\leq1, $$ then $\prod_{i=1}^n X_i\in L^1$, and $$ \mathbb{E}\left|\prod_{i=1}^n X_i\right| \leq \prod_{i=1}^n\|X_i\|_{p_i}. $$

Proof. We use induction on $n$. The case $n=1$ follows from $\mathbb{E}|X_1|\leq\|X_1\|_{p_1}$, since the underlying measure is a probability measure. Suppose the claim holds for $n-1$ factors. Put $a=\sum_{i=1}^{n-1}1/p_i$ and $b=1/p_n$. If $a=0$, all the first $n-1$ exponents are infinite, so their product is essentially bounded, with norm at most $\prod_{i=1}^{n-1}\|X_i\|_{p_i}$. Apply the two-factor theorem to that product and $X_n$.

If $a>0$, define $r=1/a$. The induction claim, applied to the first $n-1$ factors, yields $\prod_{i=1}^{n-1}X_i\in L^r$ and an $L^r$ norm at most $\prod_{i=1}^{n-1}\|X_i\|_{p_i}$; this follows by the same Hölder argument, or by applying that argument to the powers of these factors. Since $a+b\leq1$, we have $1/r+1/p_n\leq1$. The two-factor theorem applied to $\prod_{i=1}^{n-1}X_i$ and $X_n$ proves the asserted bound and integrability. This completes the induction. $\square$

Worked Example: Three dependent factors

Let $U$ again be uniform on $(0,1)$, and set $X_1=X_2=X_3=U^{-1/6}$. For each factor,

\mathbb{E}|X_i|^3 = \int_0^1 u^{-1/2}\,du = 2, \qquad \|X_i\|_3=2^{1/3}.

The reciprocal exponents sum to $1/3+1/3+1/3=1$. The three-factor bound therefore gives

\mathbb{E}|X_1X_2X_3| \leq 2^{1/3}2^{1/3}2^{1/3} = 2.

Indeed, their product is $U^{-1/2}$, whose expectation is $\int_0^1u^{-1/2}\,du=2$. All three factors are identical, so this is also a direct check that dependence does not invalidate the bound.

Why the moment condition matters

The condition on reciprocal exponents is sufficient, and in general it cannot simply be discarded. If the sum is greater than one, individual moment assumptions need not control the product. Here is a dependent example with finite first moments but a nonintegrable product.

Worked Example: Finite factor moments, infinite product expectation

Let $U$ be uniform on $(0,1)$, and set $X=Y=U^{-3/5}$. Each factor has a finite first moment:

\mathbb{E}|X| = \int_0^1u^{-3/5}\,du = \frac{5}{2}, \qquad \mathbb{E}|Y|=\frac{5}{2}.

But the product is $XY=U^{-6/5}$, and

\mathbb{E}|XY| = \int_0^1u^{-6/5}\,du = \infty.

The integral diverges because its exponent $6/5$ is greater than one. Here the available exponents are $p=q=1$, whose reciprocal sum is $2$, so the theorem’s condition fails. This example shows why separate integrability of the factors does not in itself imply integrability of their product.

Practical interpretation and common pitfalls

To use the product bound, identify the moment available for each factor, then check the reciprocal sum. For two factors with finite moments, the familiar boundary case is $1/p+1/q=1$; when the sum is smaller, the probability-space assumption supplies the additional step from a higher product moment to an expectation bound. Infinite exponents represent essentially bounded factors and are handled by the same reciprocal convention.

Two distinctions are especially important. First, the conclusion concerns $\mathbb{E}|XY|$, not just a formal or conditionally convergent value of $\mathbb{E}[XY]$. Second, the result does not assert that $\mathbb{E}[XY]=\mathbb{E}[X]\mathbb{E}[Y]$. That identity requires additional conditions, such as independence and suitable integrability. Hölder’s bound instead works directly with the joint product and needs no independence.

These estimates are useful whenever a later argument involves products of random variables—for example, bounding an expectation by controlling moments of its factors. They convert moment assumptions into a concrete integrability guarantee and a numerical upper bound, while making clear exactly how much moment control the argument uses.

Check Your Understanding

Answer the following questions using the product bounds and examples above.

If $X\in L^3$ and $Y\in L^{3/2}$, what bound does the theorem give for $\mathbb{E}|XY|$?
If $X,Y\in L^4$ on a probability space, which $L^s$ space does the proof place $XY$ in?
Does the product bound require $X$ and $Y$ to be independent? Explain briefly.
Why does $X,Y\in L^1$ alone fail to guarantee that $XY\in L^1$?
For three factors with exponents $p_1=2$, $p_2=4$, and $p_3=4$, does the reciprocal-sum condition hold?

Product Integrability from Hölder Bounds

What You'll Learn

Why moments control products

Moment spaces and the product question

The product bound

Worked Example: Dependent factors with second moments

Worked Example: Dependent factors with second moments

When the reciprocal sum is strictly less than one

Worked Example: A strict moment margin

More than two factors

Worked Example: Three dependent factors

Why the moment condition matters

Worked Example: Finite factor moments, infinite product expectation

Practical interpretation and common pitfalls

Check Your Understanding

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