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Let,$\mu $ and $\nu$ be two probability measures on $\Omega$ such that $|\Omega| < \infty$. let $(X,Y)$ be an optimal coupling. How to prove that the optimal coupling is not unique , by finding a counter example?

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Are you looking to show that there exist $\mu$ and $\nu$ such that there are two optimal couplings, or that for every $\mu$ and $\nu$, there are two optimal couplings? – Aaron Nov 2 '11 at 2:00
@Aaron I interpreted it the first way. For instance, if $\mu=\nu$ there is only one optimal coupling, the one with support on the diagonal. – Byron Schmuland Nov 2 '11 at 18:25

Here's a counterexample under my interpretation of the problem. If this isn't what you want, you should add more detail to your question so that we know precisely what you are asking.

Let $\Omega=\{0,1,2,3\}$ and let $\mu,\nu$ be the uniform distribution over $\{0,1\}$ and $\{0,1,2,3\}$, respectively. A coupling of these measures is a measure $m$ on the product space $\Omega\times\Omega$ whose marginals are $\mu$ and $\nu$. An optimal coupling is such a measure $m$ that minimizes $$\inf\{\varepsilon>0: m((x,y): |x-y|\geq \varepsilon) \leq \varepsilon\}.$$ This infimum is achieved and the minimum value is $d(\mu,\nu)$, the distance between the two measures under the Lévy-Prohorov metric (see Theorem 1.2 in Chapter 3 (page 96) of Markov Processes: Characterization and Convergence by Ethier and Kurtz).

In our case, we want a probability measure $m$ on $\Omega\times\Omega$ that maximizes the probability of the diagonal, subject to $m(0,0)+m(1,0)+m(2,0)+m(3,0)=1/2$, $m(0,0)+m(0,1)=1/4$, $m(1,0)+m(1,1)=1/4$, and $m(2,0)+m(2,1)=1/4$.

Here are two distinct solutions: $$ m_1(0,0)=m_1(1,1)=m_1(2,0)=m_1(3,1)=1/4, \text{ otherwise zero} $$ $$ m_2(0,0)=m_2(1,1)=m_2(2,1)=m_2(3,0)=1/4, \text{ otherwise zero} .$$

In other words, outside the diagonal we can move the mass around in any way, as long as we respect the marginal distributions.

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A more usual sense for a coupling to be optimal is that P(X=Y) is maximal. Your example is optimal in this sense as well. – Did Nov 2 '11 at 18:32
@Didier Yes, I'm not sure what the OP wants. It's my good luck that my post accidentally answers the question in the usual sense too! – Byron Schmuland Nov 2 '11 at 18:35

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