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I have read the topic on relaxing constraint on relaxing marginal constraints Optimal transport with relaxed constraint on marginals, where the constraint is expressed as the difference of initial and final distribution.

Question: I wonder if there exists a problem: instead of using \begin{equation} \begin{array}{c} \pi^1_*\Pi&=\mu\\ \pi^2_*\Pi&=\nu \end{array}\qquad\qquad \qquad (\#) \end{equation}, the marginals are changed to \begin{equation} \begin{array}{c} \pi^1_*\Pi&=\mu\\ \pi^2_*\Pi&\simeq\nu \end{array}\qquad\qquad \qquad (\ast) \end{equation}, where '$\simeq$' means that two distributions are similar, say if $\nu \sim \mathcal{N}(0,1)$ while $\pi^2_*\Pi \sim \mathcal{N}(2,1)$ (mean has changed) or $\pi^2_*\Pi \sim \mathcal{N}(2,4)$ (both mean and variance have changed). Any help would be appreciated.

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You can define the constraints by taking the the first marginal to be some given $\mu$ and the second marginal to be in some subset of distributions (in your case this subset should be all distribution "similar" to a given distribution $\nu$, however you wish to define this "similarity").

You can view this problem as a minimum over many optimal transport problems, where the minimum is taken over all distributions "similar" $\nu$.

More formally:

Let $Nu$ to be a set of distributions, then

$$\min\{\int c\ \mathrm{d}\Pi;\ \pi_*^1 \Pi=\mu,\ \pi_*^2 \Pi\in Nu\}=\min_{\nu'\in Nu}\min\{\int c\ \mathrm{d}\Pi;\ \pi_*^1 \Pi=\mu,\ \pi_*^2 \Pi=\nu'\}$$

If the subset $Nu$ is weak* compact and $c$ is lower semi-continuous then you know the minimum is attained for some $\nu'\in Nu$.

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  • $\begingroup$ Thanks, MOMO. I have noticed that the original OPT is an LP problem and is convex in initial and final distributions. I think the problem you formulated is quite general and may be difficult to handle. How to solve this or is there any reference I can check? Thanks~ $\endgroup$ – Kun Cao Dec 30 '18 at 8:39
  • $\begingroup$ Sorry I am not familiar with a relevant reference. A way to solve such a problem is, as I suggested above, by solving the original problem for each similar distribution. $\endgroup$ – MOMO Dec 30 '18 at 14:52

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