Convex Order Under Sums and Independent Noise

Why Convex Order Must Survive Aggregation

A comparison between two random quantities is often used to make a statement about a larger system. If one component is more variable in convex order than another, will that remain true after both are exposed to the same independent background variation? What if several independent components are added? These questions matter whenever risk, variability, or dispersion is assessed for an aggregate rather than for an isolated component.

The key is not that the added noise is small, or that the distributions have densities. The preservation results require finite means, convex order for the componentwise comparisons, and the appropriate independence. With those conditions in place, the argument follows from testing against convex functions.

Convex Order and the Required Conditions

Definition: Let \(X\) and \(Y\) be integrable real-valued random variables. We say that \(X\) is smaller than \(Y\) in convex order, written \(X\leq_{\mathrm{cx}}Y\), if \(E[X]=E[Y]\) and \[ E[\phi(X)]\leq E[\phi(Y)] \] for every finite-valued convex function \(\phi:\mathbb{R}\to\mathbb{R}\). The expectation of such a function may be \(+\infty\); it is well-defined because every finite convex function on \(\mathbb{R}\) has an affine lower bound and \(X,Y\) are integrable.

The equal-mean condition is part of the definition, not an optional consequence of the convex-function inequalities. It excludes a simple change in location from being mistaken for a change in dispersion. The convex functions used as tests include \(x\mapsto x^2\), when the relevant second moments exist, and \(x\mapsto (x-t)_+\), where \(u_+=\max(u,0)\).

For preservation under addition, “independent noise” means that the noise is independent of the variable it is added to. A statement comparing two sums requires that each sum have the intended independent summands. The distributional comparisons do not require a particular coupling between the two systems.

Adding the Same Independent Noise

Theorem: Suppose \(X\leq_{\mathrm{cx}}Y\), and let \(Z\) be integrable. If \(Z\) is independent of \(X\) and independent of \(Y\), then \[ X+Z\leq_{\mathrm{cx}}Y+Z. \] It is enough to take \(X,Y,Z\) mutually independent on one probability space; more generally, the conclusion depends only on the two marginal independence conditions and the distributions involved.

Independence ensures that adding \(Z\) applies the same averaging operation to the two distributions being compared. To prove the theorem, fix any finite convex test function \(\phi\). Choose an affine lower bound \(a x+b\leq \phi(x)\) for every real \(x\), and set \(\psi(x)=\phi(x)-a x-b\). Then \(\psi\) is convex and nonnegative. For each fixed \(z\), the function \(x\mapsto\psi(x+z)\) is convex, so convex order gives \(E[\psi(X+z)]\leq E[\psi(Y+z)]\). Independence and integration over the distribution of \(Z\) now give the result:

E[\psi(X+Z)] =\int E[\psi(X+z)]\,dF_Z(z) \leq \int E[\psi(Y+z)]\,dF_Z(z) =E[\psi(Y+Z)].

These integral identities are valid even if an expectation is infinite: the integrands are nonnegative, so Tonelli’s theorem applies. The affine part has finite expectation, and \(E[X+Z]=E[X]+E[Z]=E[Y]+E[Z]=E[Y+Z]\). Adding back the affine part therefore gives \(E[\phi(X+Z)]\leq E[\phi(Y+Z)]\). This holds for every convex \(\phi\), proving the theorem. \(\square\)

Worked Example: A Discrete Noise Distribution

Let \(X=0\) with probability one, let \(Y\) take the values \(-1\) and \(1\) with probability \(1/2\) each, and let \(Z\) take the values \(0\) and \(2\) with probability \(1/2\) each. Take \(Z\) independent of both \(X\) and \(Y\).

First, \(E[X]=0=E[Y]\). For any convex \(\phi\), Jensen’s inequality gives \(\phi(0)\leq \tfrac12\phi(-1)+\tfrac12\phi(1)\), so \(X\leq_{\mathrm{cx}}Y\). The theorem therefore gives \(X+Z\leq_{\mathrm{cx}}Y+Z\).

The distribution of \(X+Z\) assigns probability \(1/2\) to each of \(0\) and \(2\). The four equally likely combinations for \(Y+Z\) give values \(-1,1,1,3\). Both sums have mean \(1\). As a numerical check using the convex test \(\phi(x)=x^2\), \[ E[(X+Z)^2]=\tfrac12(0^2)+\tfrac12(2^2)=2, \qquad E[(Y+Z)^2]=\tfrac14\bigl((-1)^2+1^2+1^2+3^2\bigr)=3. \] This test confirms one required inequality; the theorem supplies the comparison for every convex test function.

Summing Independent Terms Ordered Term by Term

Theorem: Let \(X_1,\ldots,X_n\) be mutually independent integrable random variables, and let \(Y_1,\ldots,Y_n\) be mutually independent integrable random variables. If \(X_i\leq_{\mathrm{cx}}Y_i\) for every \(i\), then \[ \sum_{i=1}^{n}X_i\leq_{\mathrm{cx}}\sum_{i=1}^{n}Y_i. \] The two sequences may be constructed independently of one another; no particular cross-sequence dependence is needed.

For a proof, work on a product probability space on which all the variables in the two sequences are independent. This construction preserves their marginal distributions and the independence within each sequence. Replace the summands one at a time. At replacement \(i\), the sum of the other \(n-1\) terms is independent noise, and the theorem just proved applies to \(X_i\leq_{\mathrm{cx}}Y_i\). Chaining these \(n\) comparisons proves the claim.

X_1+\cdots+X_n \leq_{\mathrm{cx}}Y_1+X_2+\cdots+X_n \leq_{\mathrm{cx}}Y_1+Y_2+X_3+\cdots+X_n \leq_{\mathrm{cx}}\cdots \leq_{\mathrm{cx}}Y_1+\cdots+Y_n.

At each step, the noise is the sum of the terms already replaced and not yet replaced. These terms are independent of the pair being compared under the product construction. Finally, equal means at each component give equality of the total means: \(E[\sum_i X_i]=\sum_iE[X_i]=\sum_iE[Y_i]=E[\sum_iY_i]\). Transitivity of the convex-function inequalities completes the proof. \(\square\)

Worked Example: Summing Mean-Preserving Spreads

For \(i=1,2\), let \(X_i\) take \(0\) and \(2\) with equal probabilities, and let \(Y_i\) take \(-1\) and \(3\) with equal probabilities. Assume the \(X_i\)'s are independent and the \(Y_i\)'s are independent.

Each component has mean \(1\). To check the componentwise convex order, take any convex \(\phi\). The values \(0\) and \(2\) are convex combinations of \(-1\) and \(3\): \(0=\tfrac34(-1)+\tfrac14(3)\) and \(2=\tfrac14(-1)+\tfrac34(3)\). Convexity therefore gives \[ \phi(0)\leq\tfrac34\phi(-1)+\tfrac14\phi(3), \qquad \phi(2)\leq\tfrac14\phi(-1)+\tfrac34\phi(3). \] Averaging these inequalities yields \(E[\phi(X_i)]\leq E[\phi(Y_i)]\), so \(X_i\leq_{\mathrm{cx}}Y_i\).

The sum \(X_1+X_2\) takes \(0,2,4\) with probabilities \(1/4,1/2,1/4\). The sum \(Y_1+Y_2\) takes \(-2,2,6\) with the same respective probabilities. Both sums have mean \(2\). Their variances are \(2\) and \(8\), respectively, since the component variances are \(1\) and \(4\). For another convex-function check, use \(\phi(x)=(x-3)_+\): \[ E[(X_1+X_2-3)_+]=\tfrac14(4-3)=\tfrac14, \qquad E[(Y_1+Y_2-3)_+]=\tfrac14(6-3)=\tfrac34. \] The theorem guarantees the inequality for all convex tests, not just these two.

Worked Example: Constant Terms and a Random Aggregate

Let \(X_1=X_2=0\) with probability one, and let \(Y_1,Y_2\) be independent random variables taking \(-1\) and \(1\) with equal probabilities. By Jensen’s inequality, \(X_i\leq_{\mathrm{cx}}Y_i\) for each \(i\), and all four variables have mean zero.

The sum of the \(X_i\)'s is identically zero. The sum \(Y_1+Y_2\) takes \(-2,0,2\) with probabilities \(1/4,1/2,1/4\). Its mean is zero and its second moment is \[ E[(Y_1+Y_2)^2]=\tfrac14(4)+\tfrac12(0)+\tfrac14(4)=2, \] whereas \(E[(X_1+X_2)^2]=0\). For the convex test \(\phi(x)=(x-1)_+\), the expectations are also explicit: \[ E[(X_1+X_2-1)_+]=0, \qquad E[(Y_1+Y_2-1)_+]=\tfrac14(2-1)=\tfrac14. \] The result is not that every realization of the second sum is larger. One possible value is \(-2\). Convex order compares equal-mean distributions through all convex tests; it does not assert pointwise ordering.

Why Independence Matters

The independence condition cannot be dropped merely because the added noise has the same distribution in both comparisons. Let \(X=0\) and let \(Y\) take \(-1\) and \(1\) equally likely, so \(X\leq_{\mathrm{cx}}Y\). Let \(Z_X\) be an independent random sign, but set \(Z_Y=-Y\). Both noise variables have the same distribution as a random sign. Yet \(X+Z_X\) is a random sign while \(Y+Z_Y=0\) with probability one. The convex test \(\phi(x)=x^2\) gives \[ E[(X+Z_X)^2]=1>0=E[(Y+Z_Y)^2]. \] Thus the claimed order fails when the noise added to \(Y\) depends on \(Y\). It is independence, rather than equality of the noise marginals alone, that makes the convex averaging argument work.

Common Mistake

Pitfall: “Each \(X_i\) is smaller than \(Y_i\) in convex order, so their sums are ordered, whatever dependence the terms have.” This skips a necessary condition. The term-by-term summation theorem assumes independence within each sequence; without it, the distribution of a sum can change through dependence even when all component marginals stay fixed. Check the dependence structure of each sum before applying the theorem.

These preservation results are useful because they let a comparison scale from one component to an aggregate without restarting the convex-order analysis from scratch. Independent background variation can be included, and independent components can be compared one at a time. The assumptions also clarify what the conclusions do not say: convex order does not establish pointwise ordering, and these theorems do not justify comparisons for dependent sums.

Check Your Understanding

For each question, identify which assumptions are relevant before carrying out a calculation.

What two conditions are required in the definition of \(X\leq_{\mathrm{cx}}Y\)?
If \(X\leq_{\mathrm{cx}}Y\) and \(Z\) is integrable and independent of each, what is the convex-order relationship between \(X+Z\) and \(Y+Z\)?
Why can the independent-noise proof integrate the pointwise convex-function inequalities over the distribution of \(Z\), even when an expectation may be infinite?
In the term-by-term summation theorem, what independence is required, and why is a particular coupling between the \(X_i\)'s and \(Y_i\)'s unnecessary?
Give a convex test function that detects the failure in the dependent-noise example, and calculate the two expectations.