Moment-Estimate Workshop with Hölder and Minkowski

Why Exponent Bookkeeping Matters

A moment estimate for a sum of products usually has two distinct steps. First, Hölder’s inequality controls the moment of each product. Then Minkowski’s inequality combines the product terms into a bound for their sum. Each step has its own exponent requirements, and confusing them can lead to a plausible-looking but invalid estimate.

We will work through the bookkeeping explicitly. In particular, Minkowski’s inequality applies to the \(L^r\) norm when \(r\geq 1\). For \(0<r<1\), the corresponding moment expression is not a norm and does not satisfy the triangle inequality.

Moments, Norms, and the Two Inequalities

Let \((\Omega,\mathcal F,\mathbb P)\) be a probability space. For a measurable random variable \(X\) and \(p>0\), write \(X\in L^p\) when \(\mathbb E|X|^p<\infty\). For \(p\geq 1\), define its \(L^p\) norm by

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

When \(0<p<1\), the same expression is finite for \(X\in L^p\), but it is not a norm. This difference matters when estimating sums.

Hölder’s inequality: If \(a,b>1\), \(1/a+1/b=1\), \(U\in L^a\), and \(V\in L^b\), then \(\mathbb E|UV|\leq \|U\|_a\|V\|_b\). Minkowski’s inequality: If \(r\geq1\) and \(U,V\in L^r\), then \(\|U+V\|_r\leq\|U\|_r+\|V\|_r\).

For a product \(XY\), the useful move is to apply Hölder to \(|X|^r\) and \(|Y|^r\). If \(r\geq1\), \(a,b>1\) are conjugate exponents, \(X\in L^{ra}\), and \(Y\in L^{rb}\), then

\begin{aligned} \mathbb E|XY|^r &=\mathbb E\bigl(|X|^r|Y|^r\bigr)\\ &\leq \bigl(\mathbb E|X|^{ra}\bigr)^{1/a} \bigl(\mathbb E|Y|^{rb}\bigr)^{1/b}. \end{aligned}

Taking the \(r\)th root gives a product estimate. The integrability assumptions are exactly what make the two right-hand moments finite.

Product estimate: If \(r\geq1\), \(a,b>1\), \(1/a+1/b=1\), \(X\in L^{ra}\), and \(Y\in L^{rb}\), then \(XY\in L^r\) and

\|XY\|_r\leq \|X\|_{ra}\|Y\|_{rb}.

Worked Examples of Product Bounds

Worked Example: A Product in \(L^2\)

Suppose \(X\in L^6\) and \(Y\in L^3\). We want to bound \(\|XY\|_2\). Set \(r=2\), \(a=3\), and \(b=3/2\). These are conjugate because \(1/3+1/(3/2)=1/3+2/3=1\), and the required input spaces are \(L^{ra}=L^6\) and \(L^{rb}=L^3\). Therefore

\|XY\|_2\leq\|X\|_6\|Y\|_3.

For instance, if \(\|X\|_6\leq2\) and \(\|Y\|_3\leq3\), then \(\|XY\|_2\leq6\). The calculation confirms both the exponent match and the numerical bound.

A frequent special case takes \(a=b=2\). It says that two \(L^{2r}\) variables have a product in \(L^r\), with the bound \(\|XY\|_r\leq\|X\|_{2r}\|Y\|_{2r}\). Independence is not needed: Hölder’s inequality applies without any independence assumption.

Finite Sums of Products

After bounding each product separately, Minkowski’s inequality combines the terms, provided the target exponent is at least one. The following result records the assumptions term by term.

Theorem: Let \(1\leq r<\infty\), and let \(a,b>1\) satisfy \(1/a+1/b=1\). If \(X_i\in L^{ra}\) and \(Y_i\in L^{rb}\) for every \(i=1,\ldots,n\), then

\left\|\sum_{i=1}^n X_iY_i\right\|_r \leq \sum_{i=1}^n\|X_i\|_{ra}\|Y_i\|_{rb}.

Proof: By the product estimate, each \(X_iY_i\in L^r\) and \(\|X_iY_i\|_r\leq\|X_i\|_{ra}\|Y_i\|_{rb}\). Repeated application of Minkowski’s inequality gives \[ \left\|\sum_{i=1}^n X_iY_i\right\|_r \leq\sum_{i=1}^n\|X_iY_i\|_r. \] Substituting the product bounds into this inequality proves the claim. \(\square\)

The finite-sum estimate does not require the variables in different terms to be independent, nor does it require all the \(X_i\) to be equal or all the \(Y_i\) to be equal. It does require the stated moment assumptions for each term and \(r\geq1\) for the Minkowski step.

Worked Example: A Sum of Two Products in \(L^2\)

Suppose \(X_1,X_2,Y_1,Y_2\in L^4\), with \(\|X_1\|_4\leq2\), \(\|Y_1\|_4\leq3\), \(\|X_2\|_4\leq1\), and \(\|Y_2\|_4\leq5\). Use \(r=2\) and \(a=b=2\). Each product belongs to \(L^2\), since its two factors belong to \(L^{ra}=L^{rb}=L^4\). The theorem gives

\begin{aligned} \|X_1Y_1+X_2Y_2\|_2 &\leq\|X_1\|_4\|Y_1\|_4+\|X_2\|_4\|Y_2\|_4\\ &\leq (2)(3)+(1)(5)=11. \end{aligned}

Thus the sum has a finite second moment and its \(L^2\) norm is at most \(11\). No joint distribution information is needed for this bound.

Combining Different Kinds of Terms

The same approach handles expressions containing both single variables and products. To estimate \(A+BC+D^2\) in \(L^r\), it is enough to put each of the three terms in \(L^r\), then use Minkowski. The assumptions should be chosen to support the specific step used for each term.

Mixed-expression estimate: Let \(r\geq1\) and \(a,b>1\) be conjugate. If \(A\in L^r\), \(B\in L^{ra}\), \(C\in L^{rb}\), and \(D\in L^{2r}\), then

\|A+BC+D^2\|_r \leq \|A\|_r+\|B\|_{ra}\|C\|_{rb}+\|D\|_{2r}^{\,2}.

Indeed, the product estimate bounds \(\|BC\|_r\). Also, directly from the definition of the norm, \(\|D^2\|_r=(\mathbb E|D|^{2r})^{1/r}=\|D\|_{2r}^2\). Minkowski applied twice then gives the displayed bound.

Worked Example: A Cubic-Moment Bound for a Mixed Expression

Suppose \(A\in L^3\), \(B,C\in L^6\), and \(D\in L^6\). Take \(r=3\) and \(a=b=2\), which are conjugate exponents. The product estimate gives \(\|BC\|_3\leq\|B\|_6\|C\|_6\). Separately, \(\|D^2\|_3=\|D\|_6^2\). Hence

\|A+BC+D^2\|_3 \leq \|A\|_3+\|B\|_6\|C\|_6+\|D\|_6^2.

If the three right-hand quantities are at most \(2\), \(3\), and \(4\), respectively, then the bound is \(2+3+4=9\). In particular, all three terms are in \(L^3\), so the mixed expression is in \(L^3\).

What Changes Below the First Moment?

When \(0<r<1\), the expression \((\mathbb E|Z|^r)^{1/r}\) is finite for \(Z\in L^r\), but Minkowski’s inequality does not apply to it. Instead, use the elementary inequality \(|u+v|^r\leq |u|^r+|v|^r\). For completeness, if \(s,t\geq0\), then \((s+t)^r\leq s^r+t^r\): for \(t>0\), set \(x=s/t\) and note that \(g(x)=1+x^r-(1+x)^r\) satisfies \(g(0)=0\) and \(g'(x)=r(x^{r-1}-(1+x)^{r-1})\geq0\). The case \(t=0\) is equality. Combining this with \(|u+v|\leq|u|+|v|\) yields the stated inequality.

Sublinear-moment sum estimate: If \(0<r<1\) and \(Z_1,\ldots,Z_n\in L^r\), then

\mathbb E\left|\sum_{i=1}^n Z_i\right|^r \leq \sum_{i=1}^n\mathbb E|Z_i|^r.

Proof: The two-term inequality just established, applied repeatedly, gives \(\left|\sum_{i=1}^n Z_i\right|^r\leq\sum_{i=1}^n|Z_i|^r\) pointwise. Taking expectations proves the result. \(\square\)

Hölder can still estimate an individual product’s \(r\)th moment when \(0<r<1\): if \(a,b>1\) are conjugate, \(X\in L^{ra}\), and \(Y\in L^{rb}\), then \(\mathbb E|XY|^r\leq(\mathbb E|X|^{ra})^{1/a}(\mathbb E|Y|^{rb})^{1/b}\). What fails is applying Minkowski to combine those terms.

Worked Example: Why Minkowski Cannot Be Used for \(r=1/2\)

Let \(\Omega=\{\omega_1,\omega_2\}\), with probability \(1/2\) on each point. Set \(X_1=1_{\{\omega_1\}}\), \(X_2=1_{\{\omega_2\}}\), and \(Y_1=Y_2=1\). For each \(i\), \(\mathbb E|X_iY_i|^{1/2}=1/2\). But \(X_1Y_1+X_2Y_2=1\) everywhere, so

\mathbb E|X_1Y_1+X_2Y_2|^{1/2}=1 \quad\text{while}\quad \sum_{i=1}^2\mathbb E|X_iY_i|^{1/2}=\frac12+\frac12=1.

Thus the correct sublinear-moment inequality holds with equality in this example. After taking the power \(1/r=2\), however, the corresponding quasi-norm expression for the sum is \(1\), while the sum of the two quasi-norm expressions is \((1/2)^2+(1/2)^2=1/2\). This directly rules out the \(L^r\) triangle inequality here.

A Reliable Estimation Procedure

For a complicated expression, organize the argument in the order the inequalities will be applied. This keeps the target exponent visible and makes missing assumptions easier to detect.

Choose the target moment.
Write down the \(r\) for the desired \(L^r\) bound. If the plan uses Minkowski, check that \(r\geq1\).

Separate the summands.
List each single variable, product, or power that appears in the expression.

Match product exponents.
For a product in \(L^r\), choose conjugate \(a,b>1\) and require the factors to lie in \(L^{ra}\) and \(L^{rb}\).

Combine only after checking each term.
Once every summand belongs to \(L^r\), apply Minkowski and add the resulting bounds.

Common Pitfalls and Why These Bounds Are Useful

The most common mistake is to select convenient-looking exponents without checking their relationship to the target moment. For example, to bound a product in \(L^2\), assuming both factors are only in \(L^2\) does not meet the \(L^4\) and \(L^4\) requirements of the equal-exponent Hölder bound. Another mistake is to combine product terms by Minkowski when the target exponent is below one. In that range, use the powered moment estimate instead.

These estimates are useful whenever a random quantity is built from several components—for example, a finite sum of interactions or a polynomial expression in random variables. They turn assumptions about separate moments into a moment bound for the full expression. The bound is often not sharp, but it is explicit and does not require independence. If sharper control is needed, additional structure such as dependence information may matter; it should not be silently assumed in a Hölder–Minkowski estimate.

Check Your Understanding

For each question, identify the target exponent and the precise integrability assumptions needed before applying an inequality.

For \(X\in L^8\) and \(Y\in L^4\), what Hölder exponents give a bound for \(\|XY\|_{8/3}\)?
What assumptions on \(X_i,Y_i\) allow a Minkowski bound for \(\|X_1Y_1+X_2Y_2\|_3\) using equal Hölder exponents?
Why does the finite-sum theorem require \(r\geq1\)?
For \(0<r<1\), what inequality should replace Minkowski’s triangle inequality for a finite sum?
What moment assumption on \(D\) gives \(\|D^2\|_r=\|D\|_{2r}^2<\infty\)?