Unique coupling

Question

Let $X$ be a Polish metric space, and let $\mu,\nu$ be two Borel probability measures on $X$, when is the product measure the only coupling of $\mu$ and $\nu$. More formally, let $$\Gamma(\mu,\nu):=\{\pi\textrm{ Borel probability measure on }X\times X: \pi_1=\mu, \pi_2=\nu\}$$ Where $\pi_1,\pi_2$ are the marginal distributions of $\pi$ on the two factors. Are there known results providins sufficient conditions for $\Gamma$ to be a singleton.

PS pf course one trivial case where this happens is when $X$ is a singleton or when both measures are Diracs.

Answer 1

The only way this can happen is the situation that you described, where at least one of the measures gives full measure to a single point. In fact, the proof below does not require the two Polish spaces to be the same.

If not, let $0<\mu(A)<1$ and $0<\nu(B)<1$. We may assume, without loss of generality, that $\mu(A)\le \frac 12\le\nu(B)$: if this is not the case, replace $A$ and $B$ or both by their complements. Then define a coupling $\lambda$ by

$$\lambda(S)= \frac{\mu\times\nu(S\cap(A\times B))}{\nu(B)}+\frac{\mu\times\nu(S\cap(A^c\times B^c))}{\mu(A^c)} +\frac{\nu(B)-\mu(A)}{\mu(A^c)\nu(B)}\mu\times\nu(S\cap(A^c\times B)). $$

To verify this, we first observe that $\nu(B)-\mu(A)=\mu(A^c)-\nu(B^c)$. Then we see that for any set $C$, $$ \lambda(C\times X)=\frac{\mu(C\cap A)\nu(B)}{\nu(B)}+\frac{\mu(C\cap A^c)\nu(B^c)}{\mu(A^c)}+\frac{\nu(B)-\mu(A)}{\mu(A^c)\nu(B)}\mu(C\cap A^c)\nu(B) $$ $$ =\mu(C\cap A)+\mu(C\cap A^c)\left(\frac{\nu(B^c)}{\mu(A^c)} +\frac{\mu(A^c)-\nu(B^c)}{\mu(A^c)}\right). $$ This sums to $\mu(C)$. A similar calculation shows $\lambda(X\times D)=\nu(D)$.

Answer 2

For two measurable sets $A,B$, let $p=\mu(A)$ and $q=\nu(B)$. Consider any coupling of a Bernoulli$(p)$ and a Bernoulli$(q)$, say, $C:\{0,1\}^2 \to [0,1]$. Then we can find a coupling $(X,Y)$ of $\mu$ and $\nu$ with \begin{align} (X\in A, Y\in B) = q(1,1),\qquad (X\in A, Y\in B^c)= q(1,0), \\(X\in A^c, Y\in B^)= q(0,1),\qquad (X\in A^c, Y\in B^c)= q(0,0). \end{align} For convenience, denote $A^1=A$ and $A^0=A^c$, similarly $B^1=B$ and $B^0=B^c$.

Sample an iid sequence with the product measure $\mu\times \nu$, say $(X_t,Y_t)_{t\ge 1}$. Sample $(U,V)\sim q$, return $(X_T,Y_T)$ where $T=\min\{t\ge 1: X_t\in A^U, Y_t\in B^V\}$. This is a way to construct conditional distributions given that a an independent pair belong to one of $(A,B)$, $(A^c,B)$, $(A, B^c)$, $(A^c,B^c)$.

The conditional distribution is given by $P(X_T\in D|U=u, V=v)=\mu(D\cap A^u)/\mu(A^u)$. The law of $X_T$ is the desired marginal: $$ P(X_T\in D)=\sum_{u=0}^1\sum_{v=0}^1 P(X_T\in D|U=u, V=v) q(u,v)=\sum_{u=0}^1\sum_{v=0}^1 \frac{\mu(D\cap A^u)}{\mu(A^u)} q(u,v) $$ which is $\mu(D)$ as the denominator cancels with $\sum_v q(u,v)$. Furthermore the pushforward $(\mathbf 1\{X_T\in A\}, \mathbf 1\{Y_T\in B\})$ is distributed according to $q$ by construction.

As as long as we can find two sets $A,B$ with $\mu(A)\in (0,1)$ and $\nu(B)\in(0,1)$, we can construct infinitely many distinct couplings of $\mu$ and $\nu$ corresponding to distinct couplings of Bernoulli$(\mu(A))$ and Bernoulli$(\mu(B))$. One of these couplings is the same as that in the accepted answer, so the above construction is likely just different words for the same argument.