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Jun 26, 2018 at 4:49 history edited dohmatob CC BY-SA 4.0
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Jun 25, 2018 at 19:57 comment added dohmatob @NateEldredge Agreed. With such "tight" restrictions, I wonder if there is anything left beyond Karontovich-type generators ($\mathcal F = \{f = \operatorname{grad} (v)\}$, etc.). Yes I'm familiar with Kantorovich duality for Wasserstein distances $W_p$ induced by $\ell_p$ distances $c(x,y):= \|x-y\|_p$. I stated $W_1$ just for concreteness.
Jun 25, 2018 at 19:16 comment added Nate Eldredge It certainly makes sense as a definition. But the requirement to get $d_{\mathcal{F}}(\mu,\mu)=0$ is going to place tight restrictions on $\mathcal{F}$.
Jun 25, 2018 at 18:47 comment added dohmatob @NateEldredge I retract what I just said. The coupling I'm assuming is simply the tensor-product of $\mu$ and $\nu$ (the "independence coupling"). I've updated the question to make this more explicit. Lemme know if you're fine with this now.
Jun 25, 2018 at 18:39 history edited dohmatob CC BY-SA 4.0
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Jun 25, 2018 at 18:31 history edited dohmatob CC BY-SA 4.0
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Jun 25, 2018 at 18:10 comment added dohmatob @NateEldredge OK now I see, your first remark now makes sense. Indeed I need to further restrict the classes $\mathcal F$ to be made of only "separable functions", i.e functions for which that expectation is independent of the coupling.
Jun 25, 2018 at 18:06 comment added Nate Eldredge Are you familiar with the more general Kantorovich duality, which tells you the appropriate $\mathcal{F}$ to recover $\inf \mathbb{E} c(x,y)$ for appropriate cost functions $c$, the inf taken over all couplings? For instance you can recover the $W_p$ Wasserstein distances this way. See for instance Theorem 5.10 of Villani's Optimal Transport book.
Jun 25, 2018 at 18:02 comment added Nate Eldredge Well, the point is that for general functions $f$, the quantity $\mathbb{E}_{x \sim \mu, y \sim \nu} f(x,y)$ is not even well defined, because it depends on the particular coupling of $x,y$ being used. In the dual formulation of Wasserstein distance, $f$ is of the special form $f(x,y) = v(x) - v(y)$ and then this quantity is actually independent of the coupling chosen (linearity of expectation). So if you want to formulate your problem like this, you need restrictions on $\mathcal{F}$ just to get started.
Jun 25, 2018 at 17:36 comment added dohmatob @NateEldredge No you don't need to take such an inf over couplings (are you possibly confusing this with the primal formulation of Wasserstein distances ?).
Jun 25, 2018 at 16:19 comment added Nate Eldredge It seems to me that in your definition of $d_\mathcal{F}$, you need to take the inf over all possible couplings? When you write $\mu \otimes \nu$ it looks like you are specifically using product measure, but then you won't have $d_\mathcal{F}(\mu,\mu) =0$.
Jun 25, 2018 at 15:51 history edited dohmatob CC BY-SA 4.0
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Jun 25, 2018 at 15:43 history edited dohmatob CC BY-SA 4.0
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Jun 25, 2018 at 15:33 history edited dohmatob CC BY-SA 4.0
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Jun 25, 2018 at 15:26 history edited dohmatob CC BY-SA 4.0
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Jun 25, 2018 at 15:17 history edited dohmatob CC BY-SA 4.0
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Jun 25, 2018 at 15:08 history edited dohmatob CC BY-SA 4.0
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Jun 25, 2018 at 15:03 history edited dohmatob CC BY-SA 4.0
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Jun 25, 2018 at 14:58 history edited dohmatob CC BY-SA 4.0
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Jun 25, 2018 at 14:53 history asked dohmatob CC BY-SA 4.0