9
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ Have you compared a half-space to a spherically symmetric cap? Both Wasserstein distances should be computable. $\endgroup$ – Benoît Kloeckner Dec 15 '17 at 19:46
  • $\begingroup$ 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! $\endgroup$ – VSJ Dec 15 '17 at 20:42
  • $\begingroup$ 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. $\endgroup$ – Benoît Kloeckner Dec 18 '17 at 20:03
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.