Are shy couplings necessarily rigid?

Definitions and mathematical statement

Shared notation.

$\mathbb N=\{1,2,\ldots\}$, and $\mathbb Z,\mathbb Q,\mathbb R,\mathbb C$ denote the integers, rationals, reals and complex numbers. Any other conventions are specified locally.

Let $D\subset\mathbb R^d$, $d\ge2$, be a bounded connected open set. A coupling of reflected Brownian motions in $D$ means two such processes with their prescribed individual laws defined on the same probability space. Call a coupling shy if some $\varepsilon>0$ has \[\mathbb P\left(\inf_{t\ge0}\|X_t-Y_t\|_2\ge\varepsilon\right)>0.\] A rigid coupling here means that a deterministic function $f:\overline D\to\overline D$ satisfies $Y_t=f(X_t)$ simultaneously for all $t\ge0$ almost surely.

Statement recorded in the supplied source.

If a shy coupling exists in $D$, must there exist another shy coupling $(X'_t,Y'_t)$ that is rigid? The separation constant for the new coupling may differ from that for the original one. [S2]

Short English statement

Does the existence of a pair of reflected Brownian motions that stay apart with positive probability imply the existence of such a pair related by a deterministic map?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, UnsolvedMath: A Curated Collection of Open Mathematics Problems, record 9500007 (AMR-094-0007). CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Krzysztof Burdzy, My favorite open problems, Problem 7, Are shy couplings necessarily rigid? (with Wilfrid Kendall). https://sites.math.washington.edu/~burdzy/open_mathjax.php.