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Let $X,Y$ be Polish spaces, and $\mu$ and $\nu$ are probability measures on $X$ and $Y$ respectively. We say that $M$ is a coupling of $\mu$ and $\nu$ if it is a probability measure on $X\times Y$, with marginals $\mu$ and $\nu$.

The set $C(\mu, \nu)$ of all couplings of $\mu$ and $\nu$ is convex, and I wonder what are the extreme points of this set. My guess is that any coupling induced by a measurable map $f:X\to Y$ is an extreme point, but I am not sure whether all the extreme points can be represented in that way.

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  • $\begingroup$ Have you done the discrete case ? $\endgroup$
    – user83457
    Mar 3, 2016 at 12:23
  • $\begingroup$ @michael: afaik in the discrete case there is going to be a finite number of extreme points, however no, I don't know how to describe them nicely. $\endgroup$
    – SBF
    Mar 3, 2016 at 13:50
  • $\begingroup$ In ther discrete case the buzzword is "Birkhoff polytope". $\endgroup$
    – Dirk
    Mar 3, 2016 at 14:43
  • $\begingroup$ Relevant answer here: mathoverflow.net/a/152271/9652 $\endgroup$
    – Dirk
    Mar 3, 2016 at 14:47
  • $\begingroup$ (FYI) For the case when $\mu$ and $\nu$ are the uniform probabilities on $(0,1)$ it is known that the joinings supported by the graph of a map are dense in the set of joinings. See W. Gangbo, The Monge mass transfer problem and its applications. $\endgroup$ Apr 28, 2016 at 14:51

2 Answers 2

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The finite and infinite cases are actually quite different for this problem. In the simpler version of the finite case where $X=Y$ and $\mu=\nu$, you can show that the extreme points are introduced by maps $f: X\to X$ such that $\mu(f(x)) = \mu(x)$ almost everywhere, so for each distinct positive mass assumed by an atom under $\mu$, $f$ permutes the atoms with this mass.

In the infinite case, not all extreme points can be induced in this way. For an example, take $X = Y = [0,1/2]$ and let $M$ be the sum of one-dimensional Lebesgue measure on the lines $x = 1/4$ and $y=1/4$. Then $M$ has equal marginals, namely Lebesgue measure $\lambda$ plus a point mass $\delta$ of weight $1/2$ at $x = 1/4$.

To see that $M$ is extreme, write $M = pM_1 + (1-p)M_2$ for $M_1,M_2\in C(\lambda + \delta,\lambda + \delta)$ and $p\in [0,1]$. By the Radon-Nikodym theorem, the $M_i = u_iM$ for some density $u_i$. Since $M$ is supported on the two line segments, $M$-almost everywhere $u_i(x,y) = v_i(x) + w_i(y)$ for some $v_i$ and $w_i$ with $v_i(1/2) = w_i(1/2) = 0$. The marginals of $u_i M$ are $v_i\lambda + \left[2\int w_i(y)dy\right]\delta$ and $w_i\lambda + \left[2\int v_i(y)dy\right]\delta$. Since $M_1,M_2\in C$, these must be equal to $\lambda + \delta$, so $v_i = w_i = 1$ $\lambda$-almost everywhere. That is, $M_1 = M_2 = M$ and $M$ is extreme.

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  • $\begingroup$ Thanks, in your example - what if $M$ can be represented as a combination of some other elements w.r.t. general measure, not just finite linear combination? I highly doubt that, but unfortunately in that case we can't really easily use dominance, hence I am not sure how would you prove that. $\endgroup$
    – SBF
    Apr 28, 2016 at 15:17
  • $\begingroup$ @Ilya: Well, the definition of an extreme point x as I'm familiar with it is that if you write x as a convex combination of two points in the set, both those points are equal to x. $\endgroup$
    – Noah Stein
    Apr 28, 2016 at 15:47
  • $\begingroup$ You are right, I'm used to think in terms of integrals. $\endgroup$
    – SBF
    Apr 28, 2016 at 16:19
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A paper that gives examples of extremepoints not supported on the graph of a function and pointers to related iterature is:

Losert, Viktor. Counter-examples to some conjectures about doubly stochastic measures. Pacific Journal of Mathematics 99.2 (1982): 387-397.

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