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Consider the product of two metric spaces $X\times Y$, and two probability distributions $\mu$ and $\nu$ on this product space. By the Kantorovich-Rubinstein duality, I can write the Wasserstein-1-distance $\mathcal{W}$ between $\mu$ and $\nu$ in terms of functions $f$ ranging over $\mathrm{Lip}_1(X\times Y)$:

$$ \mathcal{W} = {\sup}_f\big(\mathbb{E}_{\mu }[f(x,y)] - \mathbb{E}_{\nu }[f(x,y)]\big) $$

If the marginals of $\mu$ and $\nu$ on $X$ coincide, does the same result hold for $f$ ranging over a larger class of functions such as "functions continuous in $x$ and 1-Lipschitz in $y$"?

This seems plausible, because if $\mu(A\times Y) = \nu(A\times Y)=:m(A)$ for any $A$ in the $\sigma$-algebra of $X$, then I can rewrite

$$ \mathcal{W} = {\sup}_f\big(\mathbb{E}_{m}[\mathbb{E}_{\mu(x,\cdot)}[f(x,y)] - \mathbb{E}_{\nu(x,\cdot)}[f(x,\tilde{y})]]\big)$$

Having the same $x$ in both terms of the inner difference makes me feel that the Lipschitz continuity of $f$ in $x$ does not play a big role.

Does anyone have a result like this or a hint for the right direction?

Searching for keywords such as Wasserstein and marginal got me only hundreds of pages defining the Wasserstein metric.

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  • $\begingroup$ I think that the condition that the $X$-marginals are the same matters little, and that your conditioning on $x$ does not imply anything. It may still be optimal to transport a mass, say, from a point $(x,y)\in X\times Y$ to a point $(x_1,y_1)\in X\times Y$ with $x_1\ne x$. $\endgroup$
    – Iosif Pinelis
    Commented Mar 18, 2021 at 21:26
  • $\begingroup$ I also do not expect that your added assumption will lead to any simplifications. Consider the case where additionally the second marginal of $\mu$ and $\nu$ is the same. This still will not lead to simplifications in my opinion. The intuitive reason is that on your product space $X \times Y$, the marginal distributions account only for a tiny fraction of the information about $\mu$ and $\nu$, the essential information is encoded by the dependence structure of both $\mu$ and $\nu$ across the spaces $X$ and $Y$. $\endgroup$
    – Steve
    Commented Mar 19, 2021 at 10:10
  • $\begingroup$ Which metric are you using on the product space? $\endgroup$
    – Yuval Peres
    Commented Mar 20, 2021 at 14:35
  • $\begingroup$ @YuvalPeres, You are right, I should clarify this. I think the sum of the metrices $d_{X\times Y}((x,y), (x',y')) = d_{X}(x,x') + d_Y(y,y')$ is what I had in mind $\endgroup$
    – joemrt
    Commented Mar 22, 2021 at 7:18
  • $\begingroup$ As others said before: Equal marginals don't tell much. Consider $X\times Y = [0,1]^2$ and $\mu\equiv 1$ and $\nu$ being the line measure on the diagonal. Then $\nu$ is far apart from $\mu$ by very close to itself, although they all have the same marginals. $\endgroup$
    – Dirk
    Commented May 18, 2021 at 10:30

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