A distribution which is Wasserstein-close to a compactly supported distribution is almost compactly supported

Question

I wonder whether this is true: If a distribution is very close (in the Wassertein sense) to another distribution with compact support, then the former must put only tiny amount of mass outside a neighborhood of the support of the latter.

More formally, let $\mu$ be a probability distributions on $\mathbb R^d$ whose support is a compact subset $A$ of $\mathbb R^d$. Recall that for $\epsilon > 0$, the $\epsilon$-blowup of $A$ is defined by $A_\epsilon := \{x \in \mathbb R^d | \operatorname{dist}(x, A) < \epsilon\}$, where $\operatorname{dist}(x, A) := \inf_{y \in A}\|x-y\|$.

Question

Does there exist constants $c_1,c_2 > 0$ such that for every $\epsilon \ge c_2$ and for every other distribution $\nu$ on $\mathbb R^d$ with $W(\mu,\nu) \le \epsilon$, it holds that $$\nu(A_\epsilon) \ge 1- \exp\left(-\frac{c_1^2\epsilon^2}{2}\right)? $$

Notes

Ultimately, I'm looking for anything of the form $\nu(A_\epsilon) \approx 1$. So even if the above Gaussian decreasing error turns out to be false, I'd be happy with something a bit slower (e.g exponential, etc.).

Some wild guesses

• $c_1 \asymp \operatorname{diam}(A)$

Answer 1

The claimed inequality is not true. The simplest possible counterexample works: let $x,y \in \mathbb{R}^n$ with $|x-y| = \epsilon$, and take $\mu = \delta_x$, $\nu = \delta_y$. Then $W_1(\mu,\nu) = \epsilon$. But $A = \{x\}$ and $y \notin A_\epsilon$, so $\nu(A_\epsilon) = 0$.

What is true is the following bound: for any $r > 0$, we have $$\nu(A_r) \ge 1 - r^{-1} W_1(\mu,\nu).$$ To see why, let $\eta$ be any coupling of $\mu$ and $\nu$. Then $$\begin{align*}\int_{\mathbb{R}^n \times \mathbb{R}^n} |x-y|\,\eta(dx,dy) &\ge \int_{A \times A_r^c} |x-y|\,\eta(dx,dy) \\ &\ge \int_{A \times A_r^c} r\,\eta(dx,dy) \\ &= r \eta(A \times A_r^c)\end{align*}$$ since $|x-y| \ge ra$ on $A \times A_r^c$ by definition of $A_r$.

Now since $A$ is the support of $\mu$, we have $\eta(A \times \mathbb{R}^n) = \mu(A) = 1$. Hence $$\begin{align*}\eta(A \times A_r^c) &= \eta((A \times \mathbb{R}^n) \cap (\mathbb{R}^n \times A_r^c)) \\ &= \eta(\mathbb{R}^n \times A_r^c) \\ &= \nu(A_r^c) \\ &= 1 - \nu(A_r). \end{align*}$$ We thus have $$\int_{\mathbb{R}^n \times \mathbb{R}^n} |x-y|\,\eta(dx,dy) \ge r(1-\nu(A_r)).$$ Taking the infimum over all couplings $\eta$ gives $W_1(\mu,\nu) \ge r(1-\nu(A_r))$ which is the desired inequality.

With the earth-moving interpretation, this just says that if the total haulage (in, say, tonne-meters) was $c$, then the amount that had to be moved at least $r$ meters must certainly be at most $c/r$ tonnes.

This bound is sharp, as can be seen by choosing $x,y$ with $|x-y| = r$, and taking $\mu = \delta_x$, $\nu = (1-\epsilon) \delta_x + \epsilon \delta_y$. Then $\nu(A_r)= 1-\epsilon$ while $W_1(\mu,\nu) = \epsilon r$.