# Coupling of $\mu$ and $\nu$ is always independent

Let $\mu,\nu$ be two probability measures on a measurable space $(X,\mathscr A)$. The coupling of $\mu$ and $\nu$ consists of constructing a new probability space $(\Omega,\mathscr F,\mathsf P)$ together with two random variables \begin{align} \xi:(\Omega,\mathscr F)&\to(X,\mathscr A)\quad \\ \eta:(\Omega,\mathscr F)&\to(X,\mathscr A) \end{align} such that $\xi_*(\mathsf P) = \mu_i$ and $\eta_*(\mathsf P) = \nu$. I.e. for example $\mathsf P(\xi^{-1}(A)) = \mu(A)$ for any $A\in \mathscr A$.

I wonder if there are any sufficient/necessary conditions on $\mu,\nu$ which assure that no matter which coupling is chosen, $\xi\perp \eta$ in the sense that $$\mathsf P(\xi^{-1}(A)\cap \eta^{-1}(B)) = \mu(A)\nu(B).$$

-
It seems to be equivalent to the following condition : $\mu \otimes \nu$ is the only probability measure on $(X \times X, \mathscr{A} \otimes \mathscr{A})$ such that first and second marginals are respectively given by $\mu$ and $\nu$. Correct me if I'm wrong. – Ahriman Aug 24 '12 at 13:48
@Ahriman: may be, provided that for the coupling of two measures it's sufficient on consider a product space. – Ilya Aug 24 '12 at 13:50
If you consider the law of the random variable $(\xi, \eta)$, I don't see any restrictions to consider a product space. – Ahriman Aug 24 '12 at 13:51
This happens if and only if $\mu$ or $\nu$ is a Dirac measure. – Did Aug 24 '12 at 13:57
@did: thank's, would you hint on how to prove it? – Ilya Aug 24 '12 at 14:00

The necessary and sufficient conditions are that either $\mu$ gives all measurable sets probability $0$ or $1$ or $\nu$ gives all measurable sets probability $0$ or $1$.

To show that these conditions are sufficient, assume for example that $\mu$ gives all measurable sets probability $0$ or $1$. Then for all sets $A$ and $B$ in $\mathscr A$, either $\mu(A)=0$ or $\mu(A)=1$. If $\mu(A)=0$, then $$\mathsf P(\xi^{-1}(A)\cap \eta^{-1}(B))\le\mathsf P(\xi^{-1}(A))=\mu(A)=0$$ so $$\mathsf P(\xi^{-1}(A)\cap \eta^{-1}(B))=0=\mu(A) \nu(B).$$ If $\mu(A)=1$, then $\mu(A^C)=0$, so \begin{eqnarray*} \mathsf P(\xi^{-1}(A)\cap \eta^{-1}(B))&=&\mathsf P(\eta^{-1}(B))-\mathsf P(\xi^{-1}(A^C)\cap \eta^{-1}(B))\\ &=&\mathsf P(\eta^{-1}(B)), \ \ \text{by the above}\\ &=&\nu(B)\\ &=&\mu(A)\nu(B). \end{eqnarray*}

To show that the conditions are necessary, I will assume that they don't hold, and follow Did's comment above to construct a counterexample coupling. If the conditions don't hold, there is $A\in\mathscr A$ with $\mu(A)=p\in(0,1)$ and $B\in\mathscr A$ with $\nu(B)=q\in(0,1)$. Assume without loss of generality that $p\le q$.

To couple the event $A$ with $B$, you can construct a $2$ by $2$ contingency table, $$\begin{array}{lll} & {\Bbb P}(A)=p & {\Bbb P}(A^C)=1-p \\ {\Bbb P}(B)=q & t & u \\ {\Bbb P}(B^C)=1-q & v & w \\ \end{array}$$ and fill in its entries, $t$, $u$, $v$, and $w$, with any nonnegative numbers so that the marginals are correct. The marginals force the values $u=q-t$, $v=p-t$, and $w=1-p-q+t$, so you must pick a value of $t$ in the closed interval $[\max(0, p+q-1),p]$. Once you have done this, you can extend the coupling by constructing product measures, as follows:

Define product subprobability measures $\omega_1$, $\omega_2$, $\omega_3$, and $\omega_4$ on $(X\times X, {\mathscr A} \times {\mathscr A})$ by

$$\omega_1(D\times E)=t \frac{\mu(D\cap A) \nu(E\cap B)}{pq},$$

$$\omega_2(D\times E)=(q-t)\frac{ \mu(D\cap A^C) \nu(E\cap B)}{(1-p)q},$$

$$\omega_3(D\times E)=(p-t)\frac{ \mu(D\cap A) \nu(E\cap B^C)}{p(1-q)},$$

$$\omega_4(D\times E)=(1-p-q+t)\frac{ \mu(D\cap A^C) \nu(E\cap B^C)}{(1-p)(1-q)}.$$

By construction, these have total measure $t$, $q-t$, $p-t$, and $1-p-q+t$, respectively. Therefore, if $\mathsf P:=\omega_1+\omega_2+\omega_3+\omega_4$, $\mathsf P$ is a probability measure. Now, let $\Omega:=X\times X$, ${\mathscr F}:={\mathscr A}\times {\mathscr A}$, and let $\xi$ and $\eta$ be projections on the first and second coordinates. To show that, for example, $\xi_*(\mathsf P) = \mu$, observe that if $D\in\mathscr A$, \begin{eqnarray*} \mathsf P(\xi^{-1}(D))&=&\mathsf P(D\times X)\\ &=& \omega_1(D\times X)+\omega_2(D\times X)+\omega_3(D\times X)+\omega_4(D\times X)\\ &=& t\frac{\mu(D\cap A)q}{pq}+(q-t)\frac{\mu(D\cap A^C)q}{(1-p)q} +(p-t) \frac{\mu(D\cap A)(1-q)}{p(1-q)}+(1-p-q+t)\frac{\mu(D\cap A^C)(1-q)}{(1-p)(1-q)}\\ &=& \frac{t+(p-t)}{p}\mu(D\cap A) + \frac{(q-t)+(1-p-q+t)}{1-p}\mu(D\cap A^C)\\ &=& \mu(D). \end{eqnarray*} The proof that $\eta_*(\mathsf P)=\nu$ is similar. However, now \begin{eqnarray*} \mathsf P(\xi^{-1}(A)\cap \eta^{-1}(B))&=&\mathsf P(A\times B)\\ &=&\omega_1(A\times B), \qquad \text{since the other $\omega_i$s vanish}\\ &=& t\frac{pq}{pq}\\ &=& t. \end{eqnarray*} Since we can choose $t$ to be any number in $[\max(0,p+q-1),p]$, it is plainly not necessary that $t$ equal $pq=\mu(A)\nu(B)$. This completes the proof.

-
Thanks for the answer, let me check it. Do you know of any book references to this fact/ – Ilya Feb 24 '13 at 17:19