Convex Order and Mean-Preserving Spreads

Why Compare Distributions by Convex Functions?

Two distributions can have the same mean but differ substantially in how much probability they place away from that mean. A comparison based only on averages cannot describe this difference. Convex functions provide a way to assess it: they penalize departures from the center increasingly, so a distribution that is more dispersed in the convex-order sense has at least as large an expectation for every convex test function.

This comparison is useful whenever the consequence of an outcome is convex in its value—for example, when a loss or penalty rises more sharply for extreme outcomes. The definition is deliberately demanding: a single convex function with the wrong expectation is enough to disprove a proposed ordering.

Definition and the Equal-Mean Requirement

Definition: Let $X$ and $Y$ be integrable real-valued random variables, not necessarily defined on the same probability space. Write $X\leq_{\mathrm{cx}}Y$ if $$ E[\phi(X)]\leq E[\phi(Y)] $$ for every finite, real-valued convex function $\phi:\mathbb{R}\to\mathbb{R}$. Expectations here may equal $+\infty$. They are well-defined: a finite convex function on $\mathbb{R}$ has an affine lower bound, and an integrable random variable makes that lower bound integrable.

The notation concerns the distributions of $X$ and $Y$, not a particular joint construction of them. Crucially, the definition tests all convex functions, including the affine functions $x\mapsto x$ and $x\mapsto -x$. Those two tests force the means to agree.

Theorem: If $X$ and $Y$ are integrable, then $X\leq_{\mathrm{cx}}Y$ implies $E[X]=E[Y]$. Equivalently, convex order can be expressed as equality of means together with the inequality $E[\phi(X)]\leq E[\phi(Y)]$ for every finite convex $\phi$.

Proof. The function $\phi(x)=x$ is convex, so the definition gives $E[X]\leq E[Y]$. The function $\phi(x)=-x$ is also convex, so it gives $-E[X]\leq -E[Y]$, or $E[Y]\leq E[X]$. Combining the inequalities yields $E[X]=E[Y]$. Conversely, equality of means together with the stated inequalities includes every convex-function test in the definition, and therefore gives $X\leq_{\mathrm{cx}}Y$. $\square$

Integrability is important here: both affine-function expectations must be finite for the two inequalities to imply equality of means. Convex order is therefore not simply a claim that one variable has a larger expected value or a larger variance. It compares distributions at a common mean.

Worked Example: A Constant Versus a Two-Point Distribution

Let $X=2$ with probability one, and let $Y$ equal $0$ or $4$, each with probability $1/2$. Determine the convex-order relation and check it using a particular convex function.

Both variables are integrable, and $$ E[X]=2,\qquad E[Y]=\tfrac12(0)+\tfrac12(4)=2. $$ For any finite convex $\phi$, convexity at the midpoint gives $$ \phi(2)=\phi\!\left(\tfrac12\cdot 0+\tfrac12\cdot 4\right) \leq \tfrac12\phi(0)+\tfrac12\phi(4)=E[\phi(Y)]. $$ Since $E[\phi(X)]=\phi(2)$, this proves $X\leq_{\mathrm{cx}}Y$ for every such $\phi$.

For the check $\phi(x)=x^2$, which is convex, $$ E[X^2]=4,\qquad E[Y^2]=\tfrac12(0^2)+\tfrac12(4^2)=8. $$ The constant distribution has no variation around its mean, whereas the two-point distribution places its mass on both sides of that same mean.

Mean-Preserving Spreads

The phrase mean-preserving spread describes an increase in dispersion that leaves the mean unchanged. At the level of distributions, the standard comparison is exactly convex order: $Y$ is a mean-preserving spread of $X$ when $X\leq_{\mathrm{cx}}Y$. One concrete way to produce such a spread is to replace some probability concentrated at a point by a distribution with the same mean.

Theorem: Let $\nu$ be an integrable probability distribution with finite mean, let $a\in\mathbb{R}$, and let $Q$ be an integrable probability distribution with mean $a$. For $0\leq p\leq1$, define $$ \mu'=(1-p)\nu+p\delta_a,\qquad \mu=(1-p)\nu+pQ, $$ where $\delta_a$ is unit mass at $a$. If $X$ has distribution $\mu'$ and $Y$ has distribution $\mu$, then $X\leq_{\mathrm{cx}}Y$.

Proof. Both mixture distributions are integrable. Their means agree because $Q$ has mean $a$: $$ \int x\,d\mu'(x)=(1-p)\int x\,d\nu(x)+pa =\int x\,d\mu(x). $$ If $p=0$, then $\mu=\mu'=\nu$, so the result holds with equality. If $p=1$, then $\mu'=\delta_a$ and $\mu=Q$. For every finite convex $\phi$, choose a supporting slope $s$ at $a$, so that $\phi(x)\geq\phi(a)+s(x-a)$ for all $x$. Integrating under $Q$ and using $\int x\,dQ(x)=a$ gives $\int\phi\,dQ\geq\phi(a)$. Hence $\int\phi\,d\mu\geq\int\phi\,d\mu'$.

It remains to consider $0<p<1$. The same supporting-line argument gives $\int\phi\,dQ\geq\phi(a)$. The integrals of a finite convex function under these integrable distributions are well-defined, possibly with value $+\infty$. Linearity for the mixtures therefore gives $$ \int\phi\,d\mu =(1-p)\int\phi\,d\nu+p\int\phi\,dQ \geq (1-p)\int\phi\,d\nu+p\phi(a) =\int\phi\,d\mu'. $$ Here both mixture weights are strictly positive, so no term of the form $0\cdot(+\infty)$ occurs. This holds for every finite convex $\phi$, proving $X\leq_{\mathrm{cx}}Y$. $\square$

The argument is Jensen's inequality in a form that also handles expectations equal to $+\infty$: a convex function lies above a supporting line, and averaging that line under a distribution with mean $a$ gives the value $\phi(a)$ as a lower bound. The replacement leaves the mean unchanged but cannot lower any convex-function expectation.

Worked Example: Replacing Some Mass at Zero

Use the theorem to compare a distribution that places some mass at $0$ with one that replaces that mass by equal probabilities at $-1$ and $1$. Let $\nu=\delta_2$, $a=0$, $p=1/3$, and $Q=\tfrac12\delta_{-1}+\tfrac12\delta_1$.

The original mixture is $$ \mu'=\tfrac23\delta_2+\tfrac13\delta_0, $$ and the spread mixture is $$ \mu=\tfrac23\delta_2+\tfrac16\delta_{-1}+\tfrac16\delta_1. $$ The replacement distribution $Q$ has mean $\tfrac12(-1)+\tfrac12(1)=0=a$, so the theorem applies. Directly, both mixture means are $$ \tfrac23(2)+\tfrac13(0)=\tfrac43, \qquad \tfrac23(2)+\tfrac16(-1)+\tfrac16(1)=\tfrac43. $$

For $\phi(x)=x^2$, the convex-function expectations are $$ \int x^2\,d\mu'(x)=\tfrac23(4)+\tfrac13(0)=\tfrac83, \qquad \int x^2\,d\mu(x)=\tfrac23(4)+\tfrac16(1)+\tfrac16(1)=3. $$ Thus this particular convex function increases from $8/3$ to $3$. The theorem establishes the inequality for every finite convex function, not just for $x^2$.

What Convex Order Does—and Does Not—Say

When second moments are finite, the test function $\phi(x)=x^2$ is convex. Consequently, $X\leq_{\mathrm{cx}}Y$ implies $E[X^2]\leq E[Y^2]$. Since convex order also requires equal means, it follows that $\operatorname{Var}(X)\leq\operatorname{Var}(Y)$. This is a consequence of convex order, not a replacement for its full collection of tests.

Worked Example: Equal Variances Do Not Settle Convex Order

Let $X$ take values $-1$ and $1$, each with probability $1/2$. Let $Y$ take value $0$ with probability $1/2$, and values $-\sqrt{2}$ and $\sqrt{2}$ with probability $1/4$ each. Check their means and variances, then test both possible convex-order directions.

Symmetry gives $E[X]=E[Y]=0$. Their second moments are $$ E[X^2]=\tfrac12(1)+\tfrac12(1)=1, \qquad E[Y^2]=\tfrac12(0)+\tfrac14(2)+\tfrac14(2)=1. $$ Thus both variances equal $1$.

For the convex function $\phi(x)=|x|$, $$ E[|X|]=1,\qquad E[|Y|]=\tfrac12(0)+\tfrac14\sqrt2+\tfrac14\sqrt2=\tfrac{\sqrt2}{2}<1. $$ This rules out $X\leq_{\mathrm{cx}}Y$. For the convex function $\phi(x)=x^4$, $$ E[X^4]=1,\qquad E[Y^4]=\tfrac12(0)+\tfrac14(4)+\tfrac14(4)=2. $$ This rules out $Y\leq_{\mathrm{cx}}X$. The distributions have equal means and equal variances, yet neither dominates the other in convex order.

Common Mistakes and a Practical Reading

Forgetting the mean condition. Convex order is not a comparison that allows different means. Testing $x$ and $-x$ shows that the means must be equal.
Using variance as the definition. A larger variance does not by itself establish convex order; the comparison must hold for every convex test function.
Checking only one convex function. A calculation with $x^2$ can confirm one necessary inequality, but cannot establish the full order. In contrast, one convex function with a reversed inequality disproves it.
Ignoring endpoint cases in a mixture argument. When a mixture weight is zero, a formal expression such as $0\cdot(+\infty)$ is undefined. Handle zero weights separately rather than applying a weighted-integral formula blindly.

To read $X\leq_{\mathrm{cx}}Y$, keep the comparison precise: the means agree, and no convex transformation has a larger expectation under $X$ than under $Y$. The statement concerns the distributions as a whole, not just their spread according to one summary.

Check Your Understanding

Use the definition and the worked arguments to answer these questions.

Which two convex functions establish that integrable variables in convex order must have equal means?
If $Q$ has mean $a$, why does a convex function satisfy $\int\phi\,dQ\geq\phi(a)$?
In the mass-replacement theorem, what changes when $p=1$, and why should that case be handled separately?
Give one convex-function test that can disprove a proposed convex-order comparison.
Why do equal means and equal variances not guarantee convex-order comparability?