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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.

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    $\begingroup$ You're choosing $v_2$, but what is the assumed distribution for $v_1$? (It can't be "uniformly random" if there are arbitrarily large possible values). $\endgroup$ – Henning Makholm May 14 '15 at 18:33
  • $\begingroup$ @HenningMakholm Ah yes, $v_1$ is assumed fixed (that is, I ask the question for any $v_1$), I'll try to reformulate that, thanks. $\endgroup$ – Real May 14 '15 at 18:34
  • $\begingroup$ Also, it looks questionable to say on the one hand that the two "agents" are point-like, and on the other hand that at infinite velocity one of them "visits every point instantaneously". At any velocity, the set of all points ever visited will have measure zero -- in particular most points are never visited, even in the limit of velocity going towards infinity. $\endgroup$ – Henning Makholm May 14 '15 at 18:35
  • $\begingroup$ @HenningMakholm Would it make sense if I rephrased as "almost every point" instead? It was my understanding that ergodicity implies visiting every point. $\endgroup$ – Real May 14 '15 at 18:37
  • $\begingroup$ x @Real: It's not almost every point either. On the contrary almost all points remain unvisited forever. I'm not familiar with the precise definition of "ergodic", but if it implies that all (or even almost all) points are eventually visited, then it's incompatible with assuming that the objects being described are pointlike. $\endgroup$ – Henning Makholm May 14 '15 at 18:39
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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.

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  • $\begingroup$ Unfortunately I do not, I know mostly elemetary stochastic theory only. I would be quite satisfied with a proof for that case! I have a conviction that the answer should be simple for geometrical reasons though: for any closed and connected room the average minimum distance $\overline{D}_{min} = \iint D_{min}(p,p') \,dp\,dp'$ is of course a constant, so the expected minimum time $E[\tau_{min}(v_1,v_2)]$ must follow $E[\tau_{min}(v_1,\lambda_2v_2)] = \frac{1}{\lambda_2+c}E[\tau_{min}(v_1,v_2)] $ (by noting that $\tau_{min} = \frac{D_{min}-D}{v_1+v_2}$). Something analogous for the average case? $\endgroup$ – Real May 16 '15 at 1:58
  • $\begingroup$ You should specify you system more explicitly. What is $D_{min}$? Why your formula for $\tau_min$ is true and why does it make sense at all? In particular what does it mean $\lambda_2 v_2$? If you play standard billiard, your $v_2$ is a pair made of a point and a tangent vector, the vector you can multiply by a constant, the point -- no. Note also that if your billard flow is ergodic but not mixing (if you have a square billiard with no obstacles, for example), and you start two balls flowing in the same direction (and same speed), your $\tau_{min} = \infty$. $\endgroup$ – demitau May 16 '15 at 8:48
  • $\begingroup$ Ah ok I will add the requirement of being mixing, thanks. I was thinking in justifying the brownian motion already. $D_{min}$ is the minimum distance among two points. It's just an observation that if the agents were headed direcly toward eachother the anwer would be positive. The values $v_1$ and $v_2$ are the absolute values of the velocities (the direction would be chosen at random). $\endgroup$ – Real May 16 '15 at 20:25
  • $\begingroup$ If there is a line connecting $p$ and $p'$, then $D_{min}$ is simply $d(p,p')$ (I didn't define due to lack of space). The statement of the problem requires showing $E[\tau(v_1,v_2)]$ is a decreasing function of $v_2$. By multiplying $v_2$ by a constant $\lambda_2$ the value $E[\tau_{min}(v_1,v_2)]$ strictly decreases as we would want for the random direction case. $\endgroup$ – Real May 16 '15 at 20:25
  • $\begingroup$ If the billiard is mixing then the infimum (over time) of distances between two points starting with random directions is zero. Also I feel that I do not understand what do you mean by the symbol $E$. Is it an integral over time? Over all possible choices of the first speed (i.e. over $\mathbb{R}^2\setminus \{0\}$ )? Over direction of the first velocity (i.e. over $S^1$)? Over both coordinates? $\endgroup$ – demitau May 16 '15 at 21:25

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