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Suppose we have a probability distribution $\rho$ on $\mathbb{R}^d$. Let $ E \subset \operatorname{supp}(\rho) $, and $R_\theta$ a rotation of angle $\theta$ such that $ R_\theta E \subset \operatorname{supp}(\rho) $. Let $ \rho(x \mid x \in E) $ be the conditional distribution on $E$.

Is it true that $$ W_1(\rho(x \mid x \in E),\rho(x \mid x \in R_\theta E)) \lesssim \operatorname{diam}(\operatorname{supp}(\rho)) \cdot \theta \qquad ? $$

I think this is true at least if $\rho$ is rotational invariant, but I was wondering if it can be generalized.

Thanks

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  • $\begingroup$ I don't think you can expect the good factor of $\epsilon$ in general. Consider $E = [-1, 1] \times [-\epsilon, \epsilon]$ and $\rho$ the uniform distribution on $E$. Note the diameter is independent of $\epsilon$. For any $\theta$, we can send $\epsilon$ to zero and the distribution of the rotated measure gets all in a little region around the origin (think of the intersection of a line and a rotated line). This should show we can make the LHS at least [something] independent of $\theta$. $\endgroup$
    – Tim Carson
    Commented Aug 7, 2019 at 17:45
  • $\begingroup$ "good factor of $\epsilon$" should read "good factor of $\theta$". $\endgroup$
    – Tim Carson
    Commented Aug 7, 2019 at 20:17
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    $\begingroup$ Thanks for your comment. I am not sure to follow, though. In your example, $\operatorname{supp}(\rho)=E$, right? If so, $R_\theta E$ is not contained in $\operatorname{supp}(\rho)$. $\endgroup$
    – Terzo
    Commented Aug 7, 2019 at 20:45
  • $\begingroup$ Ah, I missed that assumption :) You're right. $\endgroup$
    – Tim Carson
    Commented Aug 7, 2019 at 21:13
  • $\begingroup$ What if we take the same $E$, but we take $\rho$ to be $(1-\delta)$ times the uniform distribution on $E$, plus $\delta$ times the uniform distribution on the circle of radius 2? Now for fixed $\epsilon$ and $\theta$ if you send $\delta$ to zero the mass of the rotated-conditional-distribution should concentrate in the intersection of $E$ and $R_\theta E$. So you may be able to get back to my previous comment. $\endgroup$
    – Tim Carson
    Commented Aug 7, 2019 at 23:32

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