Bounding Moments of Sums With Minkowski

Why Bound a Sum One Summand at a Time?

A sum can be difficult to analyze directly, especially when its terms are dependent. Its distribution may not have a convenient form, and dependence can make independence-based shortcuts invalid. Minkowski’s inequality provides a general alternative: if each summand has a finite $p$th absolute moment, then the $p$th absolute moment of the sum is controlled by the sum of the individual $L^p$ sizes. No independence assumption is needed.

This bound is useful when the goal is an upper bound rather than an exact calculation. It can control the size of an aggregate, a centered sum, or the probability that a sum is unusually large. It is important to keep the form of the bound in view: Minkowski adds the $L^p$ norms first, and only then raises the result to the $p$th power.

Definition: For $p\geq 1$ and a random variable $X$ satisfying $E[|X|^p]<\infty$, its $L^p$ norm is $$ \|X\|_p=\bigl(E[|X|^p]\bigr)^{1/p}. $$ For random variables $X_1,\ldots,X_n$, the $p$th absolute moment of their sum is $E[|X_1+\cdots+X_n|^p]$, and its $L^p$ norm is $\|X_1+\cdots+X_n\|_p$.

The Finite-Sum Bound

The one-step result used here is Minkowski’s inequality for random variables: if $p\geq1$ and $E[|X|^p]$ and $E[|Y|^p]$ are finite, then $\|X+Y\|_p\leq\|X\|_p+\|Y\|_p$. Applying it repeatedly yields the finite-sum version.

Theorem: Let $p\geq1$, and let $X_1,\ldots,X_n$ be random variables on the same probability space such that $E[|X_i|^p]<\infty$ for each $i$. Then $$ \left(E\left[\left|\sum_{i=1}^n X_i\right|^p\right]\right)^{1/p} \leq \sum_{i=1}^n \left(E[|X_i|^p]\right)^{1/p}. $$ Equivalently, $$ E\left[\left|\sum_{i=1}^n X_i\right|^p\right] \leq \left(\sum_{i=1}^n \left(E[|X_i|^p]\right)^{1/p}\right)^p. $$ The variables need not be independent.

Proof. We use induction on $n$. For $n=1$, the inequality is equality. Suppose it holds for a sum of $n-1$ terms, and write $T_{n-1}=\sum_{i=1}^{n-1}X_i$. The induction hypothesis gives $\|T_{n-1}\|_p\leq\sum_{i=1}^{n-1}\|X_i\|_p<\infty$, so $T_{n-1}$ has a finite $p$th absolute moment. Minkowski’s inequality therefore applies to $T_{n-1}$ and $X_n$:

\left\|\sum_{i=1}^{n}X_i\right\|_p =\|T_{n-1}+X_n\|_p \leq \|T_{n-1}\|_p+\|X_n\|_p \leq\sum_{i=1}^{n}\|X_i\|_p.

This proves the norm inequality. Both sides are nonnegative, so raising them to the power $p$ gives the moment form. $\square$

The hypothesis $p\geq1$ matters: the displayed norm inequality is Minkowski’s inequality, and this form does not generally hold for $0<p<1$. The assumptions also ensure the right-hand side is finite, which in turn guarantees that the sum has a finite $p$th absolute moment.

Worked Example: Mutually Exclusive Summands

Let $B$ be Bernoulli with $P(B=1)=1/4$, and set $X_1=B$ and $X_2=1-B$. These variables are dependent: exactly one is 1. Find the $p=2$ Minkowski bound for $S=X_1+X_2$, and compare it with the actual second absolute moment.

First, $S=B+(1-B)=1$ for either possible value of $B$. Thus $E[|S|^2]=1$, so $\|S\|_2=1$.

For the first summand, $B^2=B$, giving $$ \|X_1\|_2=\left(E[B^2]\right)^{1/2}=(1/4)^{1/2}=1/2. $$ For the second, $1-B$ is 1 with probability $3/4$ and 0 otherwise, so $$ \|X_2\|_2=\left(E[(1-B)^2]\right)^{1/2}=(3/4)^{1/2}=\frac{\sqrt{3}}{2}. $$ Minkowski therefore gives $\|S\|_2\leq(1+\sqrt{3})/2$, or $$ E[|S|^2]\leq\left(\frac{1+\sqrt{3}}{2}\right)^2=1+\frac{\sqrt{3}}{2}\approx1.866. $$ The actual value is 1, so the bound is valid but not exact. The calculation uses the marginal moments of the summands and remains valid despite their dependence.

Coefficients, Centering, and a Tail Bound

Scalar coefficients can be included by applying the finite-sum result to the variables $a_iX_i$. Since $|a_iX_i|^p=|a_i|^p|X_i|^p$, their norms are $|a_i|\|X_i\|_p$. This gives the following useful version.

Corollary: Under the theorem’s conditions, for real constants $a_1,\ldots,a_n$, $$ \left(E\left[\left|\sum_{i=1}^n a_iX_i\right|^p\right]\right)^{1/p} \leq \sum_{i=1}^n |a_i|\left(E[|X_i|^p]\right)^{1/p}. $$

Proof. Apply the finite-sum theorem to $a_1X_1,\ldots,a_nX_n$. For each $i$, $\|a_iX_i\|_p=(E[|a_iX_i|^p])^{1/p}=|a_i|(E[|X_i|^p])^{1/p}$. Substitution gives the stated bound. $\square$

For sums of centered terms, the same inequality directly bounds the moment of their total deviation. If $E[|X_i|^p]<\infty$, then $E[X_i]$ exists for $p\geq1$, and $|X_i-E[X_i]|^p$ is integrable. Applying the theorem to these centered variables gives

\left(E\left[\left|\sum_{i=1}^n (X_i-E[X_i])\right|^p\right]\right)^{1/p} \leq \sum_{i=1}^n \left(E\left[|X_i-E[X_i]|^p\right]\right)^{1/p}.

To see why the individual centered moments are finite, use $|x-m|^p\leq 2^{p-1}(|x|^p+|m|^p)$ for $p\geq1$, with $m=E[X_i]$. Also, Jensen’s inequality gives $|E[X_i]|\leq(E[|X_i|^p])^{1/p}$, so each centered norm is at most $2\|X_i\|_p$. This latter bound is sometimes convenient when only uncentered moments are known, though it can be conservative.

A moment bound also gives a probability bound. For $t>0$, Markov’s inequality applied to the nonnegative random variable $\left|\sum_iX_i\right|^p$ gives

Result: If $p\geq1$, $E[|X_i|^p]<\infty$ for all $i$, and $t>0$, then $$ P\left(\left|\sum_{i=1}^nX_i\right|\geq t\right) \leq \frac{\left(\sum_{i=1}^n (E[|X_i|^p])^{1/p}\right)^p}{t^p}. $$

Derivation. For any nonnegative random variable $W$ and $c>0$, $P(W\geq c)\leq E[W]/c$. Set $W=|\sum_iX_i|^p$ and $c=t^p$; since $t>0$, the event $W\geq t^p$ is exactly $|\sum_iX_i|\geq t$. The finite-sum moment bound then yields $$ P\left(\left|\sum_iX_i\right|\geq t\right) \leq \frac{E[|\sum_iX_i|^p]}{t^p} \leq \frac{(\sum_i(E[|X_i|^p])^{1/p})^p}{t^p}. $$ No independence is required at either step. $\square$

Worked Example: Repeated Copies of One Random Variable

Let $Z$ be Bernoulli with $P(Z=1)=0.2$, and define four summands by $X_i=Z$ for $i=1,2,3,4$. They are perfectly dependent because all four equal the same random variable. Use $p=3$ to bound the third absolute moment of their sum.

Each $X_i$ has $E[|X_i|^3]=E[Z]=0.2$, so $\|X_i\|_3=0.2^{1/3}$. The finite-sum bound gives $$ \|X_1+X_2+X_3+X_4\|_3\leq4(0.2)^{1/3}. $$ Raising to the third power, $$ E[|X_1+X_2+X_3+X_4|^3]\leq 4^3(0.2)=12.8. $$ In fact the sum is $4Z$, so its third absolute moment is $E[|4Z|^3]=64(0.2)=12.8$. Equality holds here because the summands reinforce one another exactly. This example also shows why a bound for dependent terms cannot automatically be improved by treating them as independent.

Worked Example: Cancellation Between Dependent Summands

Let $Z$ take the values $-2$ and $2$, each with probability $1/2$, and set $X_1=Z$, $X_2=-Z$. For $p=4$, find the Minkowski bound and the actual fourth absolute moment of the sum.

Since $|Z|=2$ always, $E[|X_1|^4]=E[|X_2|^4]=16$, and each $L^4$ norm is $16^{1/4}=2$. Thus $$ \|X_1+X_2\|_4\leq 2+2=4, \qquad E[|X_1+X_2|^4]\leq4^4=256. $$ But $X_1+X_2=Z-Z=0$ for every outcome, so the actual fourth absolute moment is 0. The bound remains correct, but it does not use the cancellation between summands. Minkowski controls a sum using the sizes of its terms, not the detailed pattern of their dependence.

What the Bound Does—and Does Not—Say

Minkowski’s inequality is useful precisely because it makes no assumptions about the dependence structure. If individual moment bounds are available but the joint distribution is not, the inequality still supplies a valid upper bound. It applies to positive reinforcement, mutual exclusion, and exact cancellation alike.

That generality has a cost: the bound can be loose. In the cancellation example, the individual norms are large although the sum is identically zero. More generally, the inequality does not calculate the moment of the sum, and it does not claim that dependence is irrelevant to the actual value. Dependence affects the sum; it simply does not invalidate this particular upper bound.

Common mistake: adding moments instead of norms. The inequality is not $E[|\sum_iX_i|^p]\leq\sum_iE[|X_i|^p]$. The correct bound is $E[|\sum_iX_i|^p]\leq(\sum_i(E[|X_i|^p])^{1/p})^p$.
Common mistake: assuming independence is necessary. Minkowski requires the variables to be on the same probability space and their $p$th absolute moments to be finite; it does not require independence.
Common mistake: overlooking absolute values or coefficients. The summand norm uses $|X_i|$, and a scalar coefficient contributes its absolute value $|a_i|$, even if $a_i$ is negative.
Common mistake: using the result below $p=1$. This $L^p$ norm inequality is stated for $p\geq1$; the same proof and conclusion cannot simply be assumed for $0<p<1$.

When applying the bound, first check that every summand has a finite $p$th absolute moment, then take each $L^p$ norm, add those norms, and finally raise to $p$ if a moment rather than a norm is needed. That sequence keeps the inequality’s structure explicit and avoids confusing an upper bound with an exact calculation.

Check Your Understanding

Use the conditions and order of operations in the theorem to answer these questions.

For $p=2$, what finite-moment condition is needed for each of $X_1,\ldots,X_n$ to use the sum bound?
Write the Minkowski bound for $E[|aX+bY|^3]$ in terms of the third absolute moments of $X$ and $Y$.
Does the finite-sum inequality require independence? Explain what it does require.
If $X_1=Z$ and $X_2=-Z$, why can the bound be much larger than the actual moment of $X_1+X_2$?
How does Markov’s inequality turn a bound on $E[|\sum_iX_i|^p]$ into a bound on $P(|\sum_iX_i|\geq t)$?

Bounding Moments of Sums With Minkowski

What You'll Learn