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

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    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.2012-08-24
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    @Ahriman: may be, provided that for the coupling of two measures it's sufficient on consider a product space.2012-08-24
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    If you consider the law of the random variable $(\xi, \eta)$, I don't see any restrictions to consider a product space.2012-08-24
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    This happens if and only if $\mu$ or $\nu$ is a Dirac measure.2012-08-24
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    @did: thank's, would you hint on how to prove it?2012-08-24
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    Hint: Bernoulli case.2012-08-24
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    @did: Thank you - I am following your hint - but just to clarify it's meaning. Do you advise to assume that $\mu(A)\in (0,1)$ for some $A\in \mathscr A$ and to construct $\mathsf P$ for which $\xi,\eta$ are not independent by first considering $X = \{a,b\}$?2012-08-24

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