Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $$ \mathsf P(\xi^{-1}(A)\cap \eta^{-1}(B)) = \mu(A)\nu(B). $$

share|cite|improve this question
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.