I've recently seen the following question on the internet:
If I wanted to randomly find someone in an amusement park, would my odds of finding them be greater if I stood still or roamed around?
Which I formulate more precisely as follows:
Take a closed euclidean space $S$ with billiard-ball mixing and ergodic dynamics of two point-like agents with uniformly random starting positions and directions (but with given velocities). Is the expected meeting time $E[\tau(v_1,v_2)]$, a function of the velocities of agent $1$ and agent $2$ respectively, a decreasing function of $v_2$ for any $v_1$?
Where the meeting time $\tau$ is defined as the minimum time $t>0$ for which the distance among agents $d(p_1,p_2)$ satisfies $d(p_1,p_2)<D$.
It seems like a fairly difficult thing to show or disprove, at least with my limited knowledge. However, the limiting behavior seems very simple:
$$E[\tau(v_1,\infty)] = 0$$
Which follows directly from the fact that the dynamics is ergodic and mixing (at infinite velocity the agent visits every point instantaneously). Also,
$$E[\tau(0,0)] = \infty$$
For sufficiently small $D$. There's also the fact that
$$E[\tau(v_1,v_2)] = E[\tau(v_2,v_1)]$$
Is there are way to conclude from those observations? Any solution is also welcome.
If you believe that the system of two balls is ergodic (which is true for strictly convex scatterers) and mixing fast (which nobody is able to prove by the moment but is also probably true), (very) roughly speaking you can think of it as of two independent (dicrete) random walks.
Do you know the answer for this simplified case of random walks? What about two independent Brownian motions?
If it works for random walks and Brownian motions then there is a chance that it works for billiard as well. But I am pretty sure it is way beyond the current technology of proofs for billiards.