Why Convex and Stochastic Orders Are Incomparable

Introduction

Usual stochastic order and convex order both compare probability distributions, but they ask different questions. Usual stochastic order compares every threshold: does one variable have no more probability above any given level? Convex order compares averages of convex functions: does one distribution produce at least as much spread in the sense relevant to convex loss or payoff? Because these criteria test different features, neither order implies the other.

The mean is central to the distinction. A convex-order comparison between integrable variables forces their means to agree. By contrast, usual stochastic order allows unequal means; indeed, if two integrable variables are ordered stochastically and have the same mean, their distributions must be identical.

Two Different Definitions

Definition: For real-valued random variables $X$ and $Y$, the usual stochastic order $X\le_{\mathrm{st}}Y$ means $$ F_X(t)\ge F_Y(t)\quad\text{for every }t\in\mathbb{R}. $$ Equivalently, $P(X>t)\le P(Y>t)$ for every $t$. Thus $X$ has no more probability above each threshold, or equivalently no less probability at or below each threshold.

The CDF direction can seem counterintuitive: the variable further to the right has the smaller CDF at a fixed threshold. The tail form follows directly from $P(X>t)=1-F_X(t)$.

Definition: Suppose $X$ and $Y$ are integrable real-valued random variables. The convex order $X\le_{\mathrm{cx}}Y$ means $$ E[\phi(X)]\le E[\phi(Y)] $$ for every real-valued convex function $\phi$ for which the expectations are finite. In particular, affine functions and their negatives are among the test functions.

The two definitions test different collections of functions. Usual stochastic order is characterized by threshold indicators, which are increasing but generally not convex. Convex order tests every convex function, including functions that are not increasing. Neither test class simply replaces the other.

Worked Counterexamples

Worked Example: Convex Order Without Usual Stochastic Order

Let $X=0$ with probability one, and let $Y$ take values $-1$ and $1$, each with probability $1/2$. Show that $X\le_{\mathrm{cx}}Y$, but that neither direction of usual stochastic order holds.

For any convex $\phi$, convexity at the midpoint gives $$ \phi(0)\le \frac{\phi(-1)+\phi(1)}{2}. $$ The left side is $E[\phi(X)]$, and the right side is $E[\phi(Y)]$. Therefore $X\le_{\mathrm{cx}}Y$.

For $X\le_{\mathrm{st}}Y$, the CDF criterion would require $F_X(t)\ge F_Y(t)$ at every threshold. At $t=-1/2$, $F_X(-1/2)=0$, while $F_Y(-1/2)=1/2$, so the required inequality fails. For $Y\le_{\mathrm{st}}X$, the criterion at $t=0$ would require $F_Y(0)\ge F_X(0)$; instead, $1/2<1$. Thus convex order holds, but usual stochastic order holds in neither direction.

Worked Example: Usual Stochastic Order Without Convex Order

Let $B\sim\operatorname{Bernoulli}(0.4)$, and set $X=B$ and $Y=B+1$. Verify that $X\le_{\mathrm{st}}Y$, but $X\not\le_{\mathrm{cx}}Y$.

On this coupling, $X\le Y$ for every outcome. In particular, $P(X>t)\le P(Y>t)$ for every $t$, so $X\le_{\mathrm{st}}Y$. Their means are $$ E[X]=0.4,\qquad E[Y]=1.4. $$ Convex order would require its expectation inequality for every convex function, including $\phi(x)=-x$, which is affine and hence convex. But $$ E[\phi(X)]=-0.4,\qquad E[\phi(Y)]=-1.4, $$ and $-0.4\le -1.4$ is false. Thus $X\not\le_{\mathrm{cx}}Y$. The unequal means already rule out convex order; the test function $-x$ exhibits the failure directly.

Why Convex Order Forces Equal Means

Theorem: If $X$ and $Y$ are integrable and $X\le_{\mathrm{cx}}Y$, then $E[X]=E[Y]$.

Proof. The function $\phi(x)=x$ is convex, so the definition gives $E[X]\le E[Y]$. The function $\psi(x)=-x$ is also convex, so it gives $-E[X]\le -E[Y]$, or $E[X]\ge E[Y]$. Combining the two inequalities yields $E[X]=E[Y]$. Integrability ensures that both expectations are finite. $\square$

This is not an optional feature of convex order: it is built into the definition when all convex functions are tested. A common alternative convention states equal means explicitly and then compares convex functions, or uses equivalent stop-loss criteria together with equal means. Under either convention, distributions with different means cannot be in convex order.

Worked Example: Equal Means Do Not Give Stochastic Comparability

Let $U=1$ with probability one, and let $V$ take values $0$ and $2$, each with probability $1/2$. Both means are $1$. Show that $U\le_{\mathrm{cx}}V$, but that neither direction of usual stochastic order holds.

For every convex $\phi$, midpoint convexity gives $$ E[\phi(U)]=\phi(1)\le \frac{\phi(0)+\phi(2)}{2}=E[\phi(V)]. $$ Hence $U\le_{\mathrm{cx}}V$. To test $U\le_{\mathrm{st}}V$, use $t=0$: $F_U(0)=0$ and $F_V(0)=1/2$, violating $F_U(0)\ge F_V(0)$. For $V\le_{\mathrm{st}}U$, use $t=1$: $F_V(1)=1/2$, whereas $F_U(1)=1$, violating $F_V(1)\ge F_U(1)$.

This example has equal means and a genuine convex-order comparison, yet no stochastic ordering. The spread in $V$ is visible to convex functions, but its probability is not uniformly shifted to one side of every threshold.

Equal-Mean Stochastic Order Is Only Equality

Theorem: Let $X$ and $Y$ be integrable real-valued random variables. If $X\le_{\mathrm{st}}Y$ and $E[X]=E[Y]$, then $X$ and $Y$ have the same distribution.

Proof. For a distribution function $F$, define its quantile function for $0<u<1$ by $$ Q_F(u)=\inf\{x\in\mathbb{R}:F(x)\ge u\}. $$ This is a nondecreasing, and therefore measurable, function of $u$. If $U$ is uniform on $(0,1)$, then $Q_F(U)$ has distribution function $F$: for each real $t$, the set of $u$ for which $Q_F(u)\le t$ has length $F(t)$. Thus the quantile functions provide a coupling with the specified marginal distributions.

The assumption $X\le_{\mathrm{st}}Y$ means $F_X(t)\ge F_Y(t)$ for all $t$. Consequently, for every $u\in(0,1)$, the first point at which $F_X$ reaches level $u$ is no larger than the first point at which $F_Y$ reaches that level. Hence $Q_{F_X}(u)\le Q_{F_Y}(u)$.

Use the same uniform $U$ to define $X'=Q_{F_X}(U)$ and $Y'=Q_{F_Y}(U)$. They have the distributions of $X$ and $Y$, respectively, and $Y'-X'\ge0$ almost surely. Integrability gives $$ E[Y'-X']=E[Y]-E[X]=0. $$ A nonnegative random variable with expectation zero is zero almost surely, so $Y'=X'$ almost surely. Their distributions are therefore identical, proving the claim. $\square$

In particular, if two integrable laws are distinct but have the same mean, neither can dominate the other in usual stochastic order. One direction of stochastic dominance would, by the theorem, force the laws to be identical.

Common Mistake and Practical Interpretation

A tempting but incorrect argument is: “Convex order tests more functions, so it must imply usual stochastic order.” The phrase “more functions” does not settle the issue because the two sets of test functions are different. Convex functions need not be increasing, and increasing threshold indicators need not be convex. The examples above verify the failures directly rather than relying on an informal comparison of test classes.

The practical distinction is also important. Usual stochastic order concerns the direction of every threshold comparison, making it appropriate when one wants a uniform statement about exceeding levels. Convex order concerns expectations under convex criteria and, because means agree, isolates differences in spread rather than a shift in location. A distribution can be more spread out in convex order without being uniformly larger at thresholds; a variable can also be stochastically larger simply because it is shifted upward, without satisfying convex order.

Key takeaway: Neither usual stochastic order nor convex order implies the other. Convex order requires equal means; usual stochastic dominance between integrable variables with equal means is possible only when their distributions are identical.

Check Your Understanding

Use the definitions and results above to answer these questions.

State the CDF inequality that defines $X\le_{\mathrm{st}}Y$, and explain how it translates into a comparison of upper-tail probabilities.
Why does testing the convex functions $x$ and $-x$ force equal means under convex order?
For the pair $X=0$ and $Y\in\{-1,1\}$ with equal probabilities, identify one threshold that disproves $X\le_{\mathrm{st}}Y$ and one that disproves $Y\le_{\mathrm{st}}X$.
Explain why the shifted Bernoulli variables in the second worked example are stochastically ordered but not convex ordered.
Can two distinct integrable distributions with equal means be ordered by usual stochastic order? State the result that justifies your answer.