# Cutting a Gaussian in two pieces that are maximally separated in the Wasserstein metric

Denote the standard Gaussian probability measure on $\mathbb R^n$ by $\gamma$. We partition $\mathbb R^n$ into two sets $A$ and $A^c$ such that $\gamma(A) = \gamma(A^c) = 1/2$.

Denote by $\gamma_{A}$ to Gaussian measure restricted to $A$, and normalized so that it is a probability measure. Similarly, define $\gamma_{A^c}$ to be the Gaussian measure restricted to $A^c$ and normalized.

My question is the following:

What is the optimal $A$ such that $\gamma_A$ and $\gamma_{A^c}$ are the farthest apart; i.e., solving

$$\arg\max_{A} W_2(\gamma_A, \gamma_{A^c}),$$

where $W_2$ is the 2-Wasserstein distance?

Possible generalization: Instead of constructing $\gamma_A$ and $\gamma_{A^c}$ as above, we could start with any two probability measures $\gamma_1$ and $\gamma_2$ such that $\gamma = \frac{\gamma_1 + \gamma_2}{2}$ and find $\arg \max_{\gamma_1, \gamma_2} W_2(\gamma_1, \gamma_2)$.

Finding upper bounds on the $W_2$ distance is also of interest. A natural conjecture, inspired by the Gaussian isoperimetric inequality, would be that $A$ should be a half-space. Counterexamples to this are also welcome!

• Have you compared a half-space to a spherically symmetric cap? Both Wasserstein distances should be computable. – Benoît Kloeckner Dec 15 '17 at 19:46
• I'm not sure how to compute those! (I assume by spherically symmetric cap you mean a ball around the origin that contains half the probability?) My guess is that there should be a constant upper bound on the largest possible Wasserstein distance, independent of $n$. The intuition is that most of the probability lies very close $\partial A$ on either side of it. So the optimal transport map would be to jump over the boundary to the other side, giving a small $W_2$ distance. But I'm not certain about any of this! – VSJ Dec 15 '17 at 20:42
• For the spherical cap (which you interpreted as I intended), the optimal map should be spherical so it reduces to a 1D problem. I realize now that the half space case seems non-trivial. – Benoît Kloeckner Dec 18 '17 at 20:03

Here are calculations for the two possibilities posed in the comments in the case $$n=2$$.
The element of probability for $$\gamma$$ is $$(1/\pi)\exp(-x^2-y^2)\,dx\, dy = (1/\pi)\exp(-r^2)\,r\,dr\, d\theta$$.
So the dividing lines can be either a line or a circle, $$|$$ or $$\circ$$, at $$x=0$$ or $$r=\sqrt{\ln(2)}$$.
In the circular case the probability between the circle and $$r$$ is the same as the probability between the circle and $$f(r)=\sqrt{-\ln(1-\exp(-r^2))}$$. These give
\begin{align} W_2^{^|} &= \int_{x=0}^{\infty}\int_{y=-\infty}^{\infty} 2x\frac{2}{\pi}\exp(-x^2-y^2)\,dx\, dy \\ &= \frac{2}{\sqrt{\pi}}\\ &\simeq 1.13\\ \\ W_2^{^{\Large\circ}} &= \int_{r=0}^{\sqrt{\ln(2)}}\int_{\theta=0}^{2\pi} \left(f(r)-r\right) \frac{2}{\pi}\exp(-r^2)\,r\,dr\, d\theta \\ &= \sqrt{\pi}(1-2\text{erf}(\sqrt{\ln 2}))+2\sqrt{\ln 2}\\ &\simeq 0.74 \end{align} So of these two options, the half-plane gives the larger Wasserstein distance.