Hölder's Inequality for Products of Moments

Why Bound an Expectation of a Product?

In many problems, the quantity of interest involves several random variables multiplied together, but their joint distribution is difficult to calculate. A useful alternative is to bound the product’s expectation using separate moments of its factors. Hölder’s inequality provides exactly this kind of control. It does not require independence, and its exponents specify which moments must be finite.

For two random variables, the key requirement is a pair of conjugate exponents. A larger exponent on one factor is balanced by a smaller exponent on the other, with the reciprocal exponents adding to one. The resulting bound is especially useful when the moments of the individual factors are known, even when their dependence is not.

Definition: Two real numbers $p$ and $q$ are conjugate exponents if $p>1$, $q>1$, and $\frac{1}{p}+\frac{1}{q}=1$. Equivalently, $q=\frac{p}{p-1}$. For a random variable $X$, its absolute $p$th moment is $E[|X|^p]$.

Young’s Inequality: The Pointwise Ingredient

The proof of Hölder’s inequality begins with a numerical inequality that holds for every pair of nonnegative real numbers. This pointwise bound will later be applied to normalized random variables.

Theorem: If $p$ and $q$ are conjugate exponents, then for all $a,b\geq 0$, $$ ab\leq \frac{a^p}{p}+\frac{b^q}{q}. $$ Equality holds precisely when $a^p=b^q$.

To prove this, fix $a\geq 0$ and consider $g(b)=ab-b^q/q$ for $b\geq 0$. If $a=0$, then $g(b)\leq 0=a^p/p$. If $a>0$, then $g(b)$ tends to $-\infty$ as $b$ tends to infinity, and

g'(b)=a-b^{q-1}.

Thus $g$ attains its maximum at $b=a^{1/(q-1)}=a^{p-1}$, where the relation $(p-1)q=p$ gives

g(b)=a^p-\frac{a^p}{q} =a^p\left(1-\frac{1}{q}\right) =\frac{a^p}{p}.

Therefore $ab-b^q/q\leq a^p/p$, which is the claimed inequality. When $a>0$, equality occurs exactly at the maximizing value $b=a^{p-1}$, or equivalently $a^p=b^q$. For $a=0$, equality in Young’s inequality requires $b=0$, which is the same condition. $\square$

Hölder’s Inequality for Two Random Variables

Theorem: Let $X$ and $Y$ be real-valued random variables, and let $p,q>1$ be conjugate exponents. If $E[|X|^p]<\infty$ and $E[|Y|^q]<\infty$, then $$ E[|XY|]\leq \bigl(E[|X|^p]\bigr)^{1/p} \bigl(E[|Y|^q]\bigr)^{1/q}. $$ In particular, $XY$ is integrable, and $|E[XY]|\leq E[|XY|]$ gives the same upper bound for $|E[XY]|$.

Here is a proof that also explains why finite moments are the relevant conditions. First suppose both $E[|X|^p]$ and $E[|Y|^q]$ are positive. Define the normalized variables

U=\frac{|X|}{(E[|X|^p])^{1/p}}, \qquad V=\frac{|Y|}{(E[|Y|^q])^{1/q}}.

They are measurable and nonnegative, and normalization gives $E[U^p]=E[V^q]=1$. Apply Young’s inequality pointwise to $U$ and $V$, then take expectations. Because the two terms on the right are integrable, this yields

E[UV]\leq \frac{E[U^p]}{p}+\frac{E[V^q]}{q} =\frac{1}{p}+\frac{1}{q}=1.

The pointwise inequality also ensures that $UV$ is integrable: it is bounded above by the integrable random variable $U^p/p+V^q/q$. Multiplying $E[UV]\leq 1$ by the two positive normalizing factors proves the stated bound. If either moment is zero, then the corresponding random variable is zero almost surely: for example, $E[|X|^p]=0$ for a nonnegative random variable $|X|^p$ implies $|X|^p=0$ almost surely. In that case $XY=0$ almost surely and the inequality holds as well. Finally, $|E[XY]|\leq E[|XY|]$ follows from the standard absolute-value bound for an integrable random variable. $\square$

The proof uses no assumption that $X$ and $Y$ are independent. Their dependence affects the actual value of $E[|XY|]$, but the inequality controls it using the two marginal moments alone.

Worked Example: A Bound from Third and Three-Halves Moments

Let $X$ and $Y$ be independent and nonnegative. Suppose $X=3$ with probability $1/3$ and $X=0$ otherwise; suppose $Y=1$ with probability $2/3$ and $Y=2$ with probability $1/3$. Use Hölder’s inequality with $p=3$ and $q=3/2$ to bound $E[|XY|]$, and compare the bound with the actual expectation.

The exponents are conjugate because $1/3+1/(3/2)=1/3+2/3=1$. For $X$,

E[|X|^3]=\frac{1}{3}(3^3)+\frac{2}{3}(0^3)=9,

so the stated distribution gives $E[|X|^3]=9$. For $Y$,

E[|Y|^{3/2}] =\frac{2}{3}(1^{3/2})+\frac{1}{3}(2^{3/2}) =\frac{2+2\sqrt{2}}{3}.

Consequently, the Hölder bound is

E[|XY|]\leq 9^{1/3}\left(\frac{2+2\sqrt{2}}{3}\right)^{2/3} \approx 2.86.

The approximation can be checked by cubing the expression: its cube is $9((2+2\sqrt{2})/3)^2$, approximately $23.3137$, whose cube root is approximately $2.86$. To calculate the actual expectation, independence gives $E[|XY|]=E[X]E[Y]$. Here $E[X]=1$ and $E[Y]=4/3$, so $E[|XY|]=4/3$, which is below the bound. The inequality is valid, but it need not be sharp.

Worked Example: Bounding the Probability of an Intersection

Let $A$ and $B$ be events with $P(A)=1/4$, $P(B)=1/2$, and $P(A\cap B)=1/8$. Use Hölder’s inequality with $p=4$, $q=4/3$ to bound $P(A\cap B)$.

Write the event indicators as $1_A$ and $1_B$, where $1_A=1$ on $A$ and $0$ otherwise. Then $1_A1_B=1_{A\cap B}$, and for any positive exponent $r$, $1_A^r=1_A$. The needed moments are therefore

E[1_A^4]=P(A)=\frac{1}{4}, \qquad E[1_B^{4/3}]=P(B)=\frac{1}{2}.

Since $1/4+3/4=1$, Hölder’s inequality applies and gives

P(A\cap B)=E[1_A1_B] \leq \left(\frac{1}{4}\right)^{1/4} \left(\frac{1}{2}\right)^{3/4} =2^{-5/4} \approx 0.4204.

The specified intersection probability, $1/8=0.125$, satisfies this bound. The bound uses only the separate probabilities of the events; it does not require their intersection probability as an input. It can therefore be useful when the dependence between events is unknown.

Products of More Than Two Factors

A three-factor version follows directly from the two-factor result. If $X,Y,Z$ have finite third absolute moments, apply Hölder first to $|X|^{3/2}$ and $|Y|^{3/2}$, using conjugate exponents $2,2$. This gives

E[|XY|^{3/2}] \leq \bigl(E[|X|^3]\bigr)^{1/2} \bigl(E[|Y|^3]\bigr)^{1/2}.

Now apply Hölder to $|XY|$ and $|Z|$, with conjugate exponents $3/2$ and $3$. Substituting the preceding bound gives the three-factor inequality

Corollary: If $X,Y,Z$ have finite third absolute moments, then $$ E[|XYZ|]\leq \bigl(E[|X|^3]\bigr)^{1/3} \bigl(E[|Y|^3]\bigr)^{1/3} \bigl(E[|Z|^3]\bigr)^{1/3}. $$

Indeed, the second application bounds $E[|XYZ|]$ by $(E[|XY|^{3/2}])^{2/3}(E[|Z|^3])^{1/3}$. Inserting the first bound and simplifying the exponents gives the displayed result. This argument also establishes integrability of the product.

Worked Example: A Three-Factor Bound for Event Indicators

On a sample space with four equally likely outcomes, let $A=\{1,2\}$, $B=\{1,3\}$, and $C=\{1,4\}$. Bound $P(A\cap B\cap C)$ using the three-factor inequality, and compare it with the actual probability.

The indicators $1_A,1_B,1_C$ each equal $1$ on two of the four outcomes. Thus each has third absolute moment $E[|1_A|^3]=E[|1_B|^3]=E[|1_C|^3]=1/2$. The corollary gives

P(A\cap B\cap C) =E[1_A1_B1_C] \leq \left(\frac{1}{2}\right)^{1/3} \left(\frac{1}{2}\right)^{1/3} \left(\frac{1}{2}\right)^{1/3} =\frac{1}{2}.

The intersection is $\{1\}$, so its actual probability is $1/4$. This verifies the bound directly and shows that it can be conservative when the three events overlap only partly.

Equality and a Common Pitfall

When both moments in the two-factor theorem are positive, the proof also identifies the equality condition. Young’s inequality is an equality precisely when $U^p=V^q$. Since the difference between the two sides of Young’s inequality is nonnegative, its expectation is zero only if that difference is zero almost surely. Therefore equality in the Hölder bound for $E[|XY|]$ holds precisely when

\frac{|X|^p}{E[|X|^p]} = \frac{|Y|^q}{E[|Y|^q]} \quad\text{almost surely}.

A frequent mistake is to treat Hölder’s inequality as an equality or formula for the product expectation. It provides an upper bound, and in general the bound can be strict, as the examples show. Another mistake is to apply it without checking the moments: the theorem requires $E[|X|^p]$ and $E[|Y|^q]$ to be finite. Finally, independence is not a requirement. It may help calculate an expectation in a particular example, but it is not part of the inequality’s hypotheses.

Key takeaway: For conjugate exponents $p,q>1$, finite absolute $p$th and $q$th moments guarantee that the product is integrable and that its absolute expectation is at most the product of the corresponding moment roots. The result controls dependence without requiring a joint-distribution calculation.

Check Your Understanding

Use the definitions and proof above to answer the following questions.

What exponent is conjugate to $p=5$, and what reciprocal identity verifies the answer?
State the moment assumptions needed to apply Hölder’s inequality to $E[|XY|]$ with exponents $p$ and $q$.
Why does the proof first divide each random variable by a moment root?
If $A$ and $B$ are events, express $P(A\cap B)$ as an expectation of indicators and give a Hölder bound using conjugate exponents.
Does Hölder’s inequality require independence? Explain what it does require instead.