Martingale Couplings and Convex Order

Why a Coupling Matters

Convex order compares distributions by how they average against convex functions: at a fixed mean, the distribution farther spread out in this sense has the larger convex-order value. The comparison is about marginal distributions. It does not, by itself, say how two random variables are related on the same probability space.

A martingale coupling supplies such a relationship. It pairs outcomes of the less spread-out variable \(X\) with outcomes of the more spread-out variable \(Y\) so that, after observing \(X\), the conditional average of \(Y\) is exactly \(X\). Thus \(Y\) can fluctuate around \(X\) without shifting its conditional mean.

Definition: A coupling of distributions \(\mu\) and \(\nu\) is a pair of random variables \((X,Y)\) on a common probability space such that \(X\) has distribution \(\mu\) and \(Y\) has distribution \(\nu\). The coupling is a martingale coupling if \(X\) and \(Y\) are integrable and \(E[Y\mid X]=X\) almost surely. Equivalently, \(E[Y\mid \sigma(X)]=X\) almost surely, where \(\sigma(X)\) is the information generated by \(X\).

The variables in this definition need not be the original variables used to describe the two distributions. Convex order concerns their laws; the theorem below says that, when the order holds, one can construct a joint distribution with the required conditional-mean property.

The Characterization

Theorem (Strassen’s martingale-coupling theorem): Let \(\mu\) and \(\nu\) be probability distributions on \(\mathbb{R}\) with finite first moments. Let \(X\sim\mu\) and \(Y\sim\nu\) denote variables with those marginal distributions. Then \(\mu\le_{\mathrm{cx}}\nu\) if and only if there exists a coupling of these distributions such that \(E[Y\mid X]=X\) almost surely.

Here \(\mu\le_{\mathrm{cx}}\nu\) means that \(E[\phi(X)]\le E[\phi(Y)]\) for every convex function \(\phi:\mathbb{R}\to\mathbb{R}\) for which the expectations are finite. In particular, the definition includes affine functions and their negatives, so convex order requires \(E[X]=E[Y]\). The finite-first-moment condition ensures these means exist.

The necessity of the convex-order inequality follows directly from conditional Jensen’s inequality. The converse—that the convex inequalities guarantee a joint distribution with the conditional-mean constraint—is the substantive direction of Strassen’s theorem. A standard proof first treats distributions with bounded support: convex separation of the set of feasible joint distributions shows that failure of a martingale coupling would produce a convex test function violating the order. For distributions with finite first moments, truncation and tightness extend the result; the first-moment control ensures the marginal means and the conditional-mean constraint pass to the limiting coupling. Thus the converse is not a claim that an arbitrary pairing of the two marginal distributions will work: it is an existence theorem for a suitable pairing.

Proof of the forward implication: Suppose \((X,Y)\) is a martingale coupling. For any convex \(\phi\) with finite expectations, conditional Jensen gives \( \phi(E[Y\mid X])\le E[\phi(Y)\mid X] \) almost surely. Substituting \(E[Y\mid X]=X\) yields \( \phi(X)\le E[\phi(Y)\mid X] \). Taking expectations and using the law of iterated expectations gives \( E[\phi(X)]\le E[\phi(Y)] \). This holds for every convex test function in the definition, so \(\mu\le_{\mathrm{cx}}\nu\). \(\square\)

Worked Couplings

Worked Example: A Discrete Mean-Preserving Spread

Let \(X\) take values \(0\) and \(2\), each with probability \(1/2\). Construct a \(Y\) with probabilities \(1/4,1/2,1/4\) at \(-1,1,3\), respectively, by using the following joint probabilities:

\begin{array}{c|ccc|c} & Y=-1 & Y=1 & Y=3 & P(X=x)\\ \hline X=0 & 1/4 & 1/4 & 0 & 1/2\\ X=2 & 0 & 1/4 & 1/4 & 1/2\\ \hline P(Y=y) & 1/4 & 1/2 & 1/4 & 1 \end{array}

The row and column totals verify both marginals. Conditional on \(X=0\), \(Y\) is equally likely to be \(-1\) or \(1\), so \(E[Y\mid X=0]=(-1+1)/2=0\). Conditional on \(X=2\), it is equally likely to be \(1\) or \(3\), so \(E[Y\mid X=2]=(1+3)/2=2\). Therefore \(E[Y\mid X]=X\), and Strassen’s theorem gives \(X\le_{\mathrm{cx}}Y\).

For a direct check with the convex function \(\phi(z)=(z-1)_+\), where \(a_+=\max(a,0)\),

E[(X-1)_+]=\tfrac12(0)+\tfrac12(1)=\tfrac12, \qquad E[(Y-1)_+]=\tfrac14(0)+\tfrac12(0)+\tfrac14(2)=\tfrac12.

These equal values are consistent with convex order; this particular test function does not show a strict increase. At the different threshold \(0\), the values are \(E[X_+]=1\) and \(E[Y_+]=\tfrac12(0)+\tfrac12(1)+\tfrac14(3)=\tfrac54\), illustrating that another convex test can give a strict inequality.

Worked Example: A Continuous-to-Discrete Coupling

Let \(X\) be uniform on \([-1,1]\). Given \(X=x\), define \(Y\) to equal \(1\) with probability \((1+x)/2\) and \(-1\) with probability \((1-x)/2\). These are valid probabilities for every \(x\in[-1,1]\): each is between \(0\) and \(1\), and their sum is \(1\).

The conditional mean is \(E[Y\mid X=x]=1\cdot(1+x)/2+(-1)\cdot(1-x)/2=x\). Hence \(E[Y\mid X]=X\). To check the marginal distribution of \(Y\), use \(E[X]=0\):

P(Y=1)=E\!\left[\frac{1+X}{2}\right]=\frac12, \qquad P(Y=-1)=E\!\left[\frac{1-X}{2}\right]=\frac12.

Thus this construction couples a uniform variable on \([-1,1]\) to a variable equally likely to be \(-1\) or \(1\). The martingale-coupling implication proves that the uniform distribution is below this two-point distribution in convex order.

Worked Example: Equal Means Do Not Guarantee a Coupling

Let \(X\) be equally likely to be \(-1\) or \(1\), and let \(Y=0\) with probability one. Both means are zero, but the convex function \(\phi(z)=z^2\) gives

E[\phi(X)]=1,\qquad E[\phi(Y)]=0.

Therefore \(X\not\le_{\mathrm{cx}}Y\). There cannot be a martingale coupling with these marginals: since \(Y=0\) almost surely, every conditional expectation \(E[Y\mid X]\) equals \(0\), which cannot equal \(X\) almost surely because \(P(X\ne0)=1\). Equal means are necessary for convex order, but they are not sufficient.

Consequences and Interpretation

The martingale condition makes the phrase “mean-preserving spread” precise. Conditional on \(X\), the distribution of \(Y\) may be more variable, but it is centered at the observed value of \(X\). The spread is therefore local: each conditional distribution has mean \(X\), not merely an overall mean that happens to match \(E[X]\).

A further consequence is a variance comparison when second moments are finite. The conditional-mean condition also implies \(E[Y]=E[X]\), by taking expectations. To compare variances, write \(Y=X+(Y-X)\). Since \(E[Y-X\mid X]=0\), and finite second moments ensure the products below are integrable, we have

E[X(Y-X)] =E\!\left[X\,E[Y-X\mid X]\right] =0.

Expanding the square then gives

$$ E[Y^2] =E[X^2]+2E[X(Y-X)]+E[(Y-X)^2] =E[X^2]+E[(Y-X)^2]. $$

Because \(E[X]=E[Y]\), subtracting the common squared mean shows \(\operatorname{Var}(Y)=\operatorname{Var}(X)+E[(Y-X)^2]\ge\operatorname{Var}(X)\). The inequality is strict unless \(Y=X\) almost surely. This variance implication is useful, but it is only one consequence of convex order: a variance comparison alone does not establish all the convex-function inequalities required by the definition.

Common Mistakes

Reversing the order. If \(E[Y\mid X]=X\), conditional Jensen gives \(E[\phi(X)]\le E[\phi(Y)]\); it is \(Y\), not \(X\), that is larger in convex order.
Assuming matching means is enough. The equal-mean example above fails the test \(\phi(z)=z^2\), so the convex inequalities can fail even when the means agree.
Checking only an overall mean. A martingale coupling requires the conditional mean given \(X\) to equal \(X\) almost surely. It is stronger than \(E[Y]=E[X]\).
Treating the coupling as unique or already given. Convex order compares marginal laws. Strassen’s theorem asserts that a suitable coupling exists; it does not say every coupling works or that the coupling is unique.
Claiming strictness from an equality. If a particular convex test has equal expectations, that test gives no strict inequality. Check its values at every support point before drawing a conclusion.

Key takeaway: For integrable real-valued distributions, \(X\le_{\mathrm{cx}}Y\) exactly when the distributions can be coupled so that \(E[Y\mid X]=X\). Conditional Jensen proves the forward consequence of a coupling; Strassen’s theorem supplies the converse existence result.

Check Your Understanding

Use the definition and examples above to answer the following questions.

What two requirements define a martingale coupling of \(\mu\) and \(\nu\)?
Which inequality does conditional Jensen give when \(E[Y\mid X]=X\), and how does taking expectations establish convex order?
In the discrete coupling, verify the conditional mean given \(X=2\) and the marginal probability \(P(Y=1)\).
Why does equality of the means of two distributions not by itself imply convex order?
When second moments are finite, what does the martingale-coupling condition imply about \(\operatorname{Var}(Y)-\operatorname{Var}(X)\)?