Why a Triangle Inequality for Random Variables?
A sum can have a large value on some outcomes and a small value on others, so its typical size is not generally the sum of the typical sizes of its parts. The relevant comparison depends on how size is measured. For \(p\geq 1\), the \(L^p\) size of a random variable is based on the \(p\)th absolute moment. Minkowski’s inequality says that this size of a sum cannot exceed the sum of the separate sizes.
The result applies to dependent random variables as well as independent ones. Its proof for \(p>1\) illustrates a useful strategy: write the \(p\)th power of the size of the sum in a form that Hölder’s inequality can bound, then divide by a power of the quantity being bounded.
Definition and Statement
The hypothesis that both variables have finite \(p\)th absolute moments ensures that the quantities on the right side of the inequality below are defined and finite. In particular, it also ensures that their sum has a finite \(p\)th absolute moment.
For \(p=1\), the size is the expected absolute value. For \(p>1\), the proof uses Hölder’s inequality with conjugate exponents \(p\) and \(q=p/(p-1)\). In the form needed here, Hölder’s inequality states that if \(A\in L^q\) and \(B\in L^p\), then $$ E[|AB|]\leq \big(E[|A|^q]\big)^{1/q}\big(E[|B|^p]\big)^{1/p}. $$
Worked Examples
Worked Example: The \(L^1\) Inequality on a Finite Space
Suppose an outcome is chosen uniformly from three equally likely outcomes. Let \(X\) take the values \(2,-1,0\), and let \(Y\) take the values \(-1,2,0\), in that order. Compute their \(L^1\) sizes and the \(L^1\) size of their sum.
Each outcome has probability \(1/3\). Therefore, $$ \|X\|_1=E[|X|]=\frac13(|2|+|-1|+|0|)=1 $$ and $$ \|Y\|_1=E[|Y|]=\frac13(|-1|+|2|+|0|)=1. $$
The values of \(X+Y\) are \(1,1,0\). Thus $$ \|X+Y\|_1=\frac13(|1|+|1|+|0|)=\frac23\leq 2=\|X\|_1+\|Y\|_1. $$ The inequality is strict because \(X\) and \(Y\) have opposite signs on the first two outcomes, so adding them reduces the absolute value there.
Worked Example: A Strict \(L^2\) Inequality
Let the probability space have two equally likely outcomes. Suppose \(X\) takes values \(1,0\), while \(Y\) takes values \(0,1\), in that order. Verify Minkowski’s inequality for \(p=2\).
The second moments are $$ E[|X|^2]=\frac12(1^2+0^2)=\frac12, \qquad E[|Y|^2]=\frac12(0^2+1^2)=\frac12. $$ Consequently, \(\|X\|_2=\|Y\|_2=1/\sqrt{2}\).
On each outcome, \(X+Y=1\). Hence $$ \|X+Y\|_2=\left(\frac12(1^2+1^2)\right)^{1/2}=1. $$ The right side of Minkowski’s inequality is $$ \|X\|_2+\|Y\|_2=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}, $$ so \(1\leq\sqrt{2}\), as required. This is strict: the two variables are nonzero on disjoint outcomes, rather than being proportional in a way that would produce equality.
Worked Example: A Continuous Pair With a Constant Sum
Let \(U\) have the uniform distribution on \([0,1]\), and set \(X=U\) and \(Y=1-U\). For \(p=2\), compute each \(L^2\) size and compare the size of the sum with their total.
The density of \(U\) is \(1\) on \([0,1]\). Since \(X\) and \(Y\) are nonnegative, $$ E[|X|^2]=\int_0^1 u^2\,du=\frac13, \qquad E[|Y|^2]=\int_0^1(1-u)^2\,du =\left[-\frac{(1-u)^3}{3}\right]_0^1=\frac13. $$ Thus \(\|X\|_2=\|Y\|_2=1/\sqrt{3}\).
For every outcome, \(X+Y=U+(1-U)=1\), so $$ \|X+Y\|_2=\big(E[1^2]\big)^{1/2}=1. $$ Minkowski’s inequality holds because \(1\leq 2/\sqrt{3}\); squaring both positive sides gives the equivalent comparison \(1\leq 4/3\). The variables here are dependent, since \(Y=1-X\). The example makes clear that independence is not a condition for the inequality.
Proof From Hölder’s Inequality
First, \(X+Y\) has a finite \(p\)th absolute moment. For real numbers \(x,y\) and \(p\geq1\), the triangle inequality for absolute values and convexity of \(t\mapsto t^p\) on \([0,\infty)\) give $$ |x+y|^p\leq (|x|+|y|)^p \leq 2^{p-1}\big(|x|^p+|y|^p\big). $$ The second inequality follows by applying convexity at the midpoint of \(|x|\) and \(|y|\). Applying this pointwise with \(x=X\) and \(y=Y\), and taking expectations, shows $$ E[|X+Y|^p]\leq 2^{p-1}\big(E[|X|^p]+E[|Y|^p]\big)<\infty. $$
For every outcome, \(|X+Y|\leq |X|+|Y|\). Taking expectations, which are finite by the assumptions, gives \(E[|X+Y|]\leq E[|X|]+E[|Y|]\). This is exactly the claimed inequality for \(p=1\).
Let \(S=|X+Y|\), \(T=E[S^p]\), and \(q=p/(p-1)\), so \(1/p+1/q=1\). The preceding integrability argument gives \(T<\infty\). Pointwise, \(S^p=S^{p-1}|X+Y|\leq S^{p-1}(|X|+|Y|)\).
The random variable \(S^{p-1}\) belongs to \(L^q\), because \((p-1)q=p\) and \(E[(S^{p-1})^q]=T\). Applying Hölder separately to \(S^{p-1}|X|\) and \(S^{p-1}|Y|\) yields \(T\leq T^{1/q}(\|X\|_p+\|Y\|_p)\).
If \(T=0\), then \(\|X+Y\|_p=0\) and the inequality holds. If \(T>0\), divide by \(T^{1/q}\). Since \(1-1/q=1/p\), this gives \(T^{1/p}\leq\|X\|_p+\|Y\|_p\), or \(\|X+Y\|_p\leq\|X\|_p+\|Y\|_p\).
This proves the theorem for every \(1\leq p<\infty\). \(\square\)
What the Conditions Do—and Do Not—Say
The finite-moment assumptions matter: without them, one or more \(L^p\) sizes may be infinite, and the stated finite-valued comparison is no longer the theorem just proved. The range \(p<\infty\) is also part of this result; the proof uses the finite conjugate exponent \(q=p/(p-1)\) when \(p>1\).
No assumption about independence, identical distributions, or a particular distribution family appears in the argument. The inequality follows from the pointwise relation \(|X+Y|\leq |X|+|Y|\) and the moment bound supplied by Hölder. For any finite \(p\), it also has the expected homogeneity: for a constant \(c\), \(\|cX\|_p=|c|\|X\|_p\), directly from the definition. Together, homogeneity and Minkowski’s inequality are the key norm properties used when treating \(L^p\) sizes geometrically.
Common Mistake
- Assuming independence is needed. A tempting but incorrect argument is that the moments of a sum can only be controlled if the summands are independent. The proof does not factor any joint expectation into separate expectations: Hölder bounds each product directly. The uniform example above even uses \(Y=1-X\), so the variables are dependent.
- Skipping the finiteness check. Hölder’s inequality is applied to \(S^{p-1}\) and each of \(|X|\), \(|Y|\). This use requires \(S^{p-1}\in L^q\), which follows only after establishing \(E[S^p]<\infty\). The preliminary pointwise bound supplies that justification.
- Dividing by a quantity that might be zero. The proof divides by \(T^{1/q}\) only after treating \(T=0\) separately. Otherwise that division would not be justified.
Check Your Understanding
Use the definition and proof to check your understanding of the conditions and the argument.
- State the moment assumptions required to apply Minkowski’s inequality to \(X\) and \(Y\) for a specified finite \(p\).
- Why does \(|X+Y|^p\leq 2^{p-1}(|X|^p+|Y|^p)\) establish that \(X+Y\in L^p\)?
- For \(p>1\), verify that \((|X+Y|^{p-1})^q=|X+Y|^p\) when \(q=p/(p-1)\).
- Why must the proof treat \(E[|X+Y|^p]=0\) separately before dividing?
- Does Minkowski’s inequality require \(X\) and \(Y\) to be independent? Explain using the structure of the proof.