Constructing Couplings by Disintegration

Why Construct a Coupling?

A coupling places two random elements on the same probability space while preserving their individual laws. This lets us ask how they might behave together, not just how each behaves separately. Disintegration gives a particularly useful way to build such joint laws when both elements are related to a common conditioning variable.

Suppose one joint law describes an element \(X\) together with a variable \(S\), and another describes \(S\) together with an element \(Y\). If the two laws give the same marginal distribution to \(S\), their conditional laws can be combined over that shared distribution. The most direct combination makes \(X\) and \(Y\) conditionally independent given \(S\). Other combinations can preserve the same conditional marginals while imposing a different dependence structure.

More explicitly, for measurable \(A\subseteq E\) and \(B\subseteq F\), the marginal requirements are \(\pi(A\times F)=\alpha(A)\) and \(\pi(E\times B)=\beta(B)\). These requirements alone do not determine \(\pi\): many different couplings may have the same two marginals.

The Conditional-Product Construction

Let \(E,F,G\) be standard Borel spaces. Let \(P\) be a probability measure on \(E\times G\), and let \(Q\) be a probability measure on \(G\times F\). Assume their \(G\)-marginals agree, and call the common marginal \(\mu\). By disintegration, there are probability kernels \(K\) from \(G\) to \(E\) and \(L\) from \(G\) to \(F\) such that

Here a probability kernel \(K\) assigns a probability measure \(K(s,\cdot)\) to each \(s\), with \(s\mapsto K(s,A)\) measurable for every measurable \(A\subseteq E\). The analogous requirements apply to \(L\).

Proof. For each \(s\), the product measure \(K(s,\cdot)\otimes L(s,\cdot)\) is a probability measure on \(E\times F\). It is a measurable kernel: for a measurable rectangle \(A\times B\), its value is \(K(s,A)L(s,B)\), a measurable function of \(s\). The collection of sets in \(E\times F\) for which the value is measurable is closed under complements and countable disjoint unions. Since measurable rectangles generate the product sigma-algebra, the standard monotone-class argument gives measurability for every measurable subset of \(E\times F\). Integrating this kernel against \(\mu\) therefore defines a probability measure \(\Gamma\).

For measurable \(A\subseteq E\) and \(C\subseteq G\), its \((E,G)\)-marginal satisfies

because \(L(s,F)=1\). Equality on measurable rectangles establishes that this marginal is \(P\). Similarly, for measurable \(C\subseteq G\) and \(B\subseteq F\),

since \(K(s,E)=1\). Finally, for measurable \(A\subseteq E\), \(B\subseteq F\), and \(C\subseteq G\),

This is the conditional-independence factorization given \(S\). Thus the claimed marginals and conditional independence both hold. \(\square\)

The construction is called a conditional product because, at each fixed value \(s\), it combines the two conditional laws by their product measure. It does not generally make \(X\) and \(Y\) independent without conditioning: variation in \(S\) can induce dependence between them.

Dependence Can Be Chosen Within the Fibers

The conditional product is one choice, not the only one. A joint kernel can specify a different conditional relationship between \(X\) and \(Y\), as long as its two conditional marginals remain \(K\) and \(L\).

Proof. For measurable \(A\subseteq E\) and \(C\subseteq G\), the first marginal of \(H(s,\cdot)\) gives \(H(s,A\times F)=K(s,A)\) for \(\mu\)-almost every \(s\). Therefore

For measurable \(C\subseteq G\) and \(B\subseteq F\), the second marginal condition gives

Thus both required pairwise marginals are preserved. \(\square\)

The product kernel \(H(s,\cdot)=K(s,\cdot)\otimes L(s,\cdot)\) is one valid choice. When the spaces and kernels permit other joint kernels, those choices can encode dependence within each conditional distribution. The prescribed pairwise laws do not, by themselves, determine which joint law to use.

Worked Examples

What the Construction Does—and Does Not—Determine

Disintegration provides a way to preserve two existing relationships with the common variable \(S\). The conditional-product choice has a clear interpretation: after \(S\) is fixed, draw \(X\) and \(Y\) independently from their respective conditional laws. This is useful when conditional independence is part of the desired model or when a simple coupling is needed.

A common pitfall is to infer unconditional independence from conditional independence. In the Gaussian example, conditioning removes the dependence contributed by \(S\), but before conditioning both variables respond to the same random input. Another pitfall is to assume that knowing the two marginals identifies a unique coupling. The Bernoulli example shows explicitly that even the same conditional marginals over the same \(S\) can be joined in different ways.

In applications, the choice of joint conditional kernel is therefore a modeling decision. A product kernel is canonical when conditional independence is appropriate; another kernel is needed when the variables should be positively associated, negatively associated, or otherwise linked within each value of the conditioning variable. In every case, checking the kernel’s two conditional marginals is the essential step for ensuring the prescribed pairwise laws are retained.