# Wasserstein distance between rotated conditional distributions

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

• 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 $\theta$. – Tim Carson Aug 7 at 17:45
• "good factor of $\epsilon$" should read "good factor of $\theta$". – Tim Carson Aug 7 at 20:17
• 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 $\operatorname{supp}(\rho)$. – Terzo Aug 7 at 20:45
• Ah, I missed that assumption :) You're right. – Tim Carson Aug 7 at 21:13
• 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. – Tim Carson Aug 7 at 23:32