In discrete versions of the problem, the rightmost bullet having a velocity of 1 means it stays there forever, of course. So it seems like the question is, how long does that take to happen? Consider Jim's discretization (rescaled) = the K=2 version of Dan's discretization. The RBV is 1 a little bit after the first time that the number of speed-1 bullets shot so far is greater than the number of speed-1/2 bullets. The "little bit after" is because there may still be a rightmost bullet of speed 1/2 that is doomed to be annihilated by a speed-1 bullet that hasn't quite caught up to it yet. In the worst case, this puts off the reign of the eventual rightmost bullet by a factor of 2. "More speed-1 bullets than speed-1/2 bullets" is just a redressing of a discrete random walk question: start at the origin, move left or right 1 unit at each time step, and stop when you hit -1. That happens with probability 1 but the expected time it takes is unbounded, right? And indeed if we simulate this system for time 100, we see that around 10% of the time, there's a long string of speed-1/2 bullets. Surely many people on this list know more than I do about the distribution of times at which the random walker first hits -1. How about K=3, though? Now the complicated when-do-bullets-reach-each-other dynamics of David's original version come back. If speed-1/3 and speed-2/3 bullets didn't annihilate one another often, then we have something like a random walk weighted 2-to-1 to move to the right, and the speed-1 bullets, only getting fired 1/3 of the time, would probably never make it through once a cloud of slower bullets had built up. It feels like there might be some inductive way to think about the behavior of the slower bullets (which are just a K=2 system, effectively). Can someone use this to argue that the rightmost bullet eventually has velocity 1 with probability 1 for all K? That would be a great result. Even so, though, what if the expected value of the first time when RBV=1 goes to infinity as a function of K? What, if anything, does that imply about the continuous version? --Michael On Sat, Jun 16, 2012 at 7:58 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Say with Michael K. that v[n] denotes the velocity of the farthest bullet right after time n.
The simplest discretization of the problem, a la Jim, may be this version:
Instead of the velocities being chosen from the uniform distribution on [0,1], let them be chosen from the set {1,2,3,...,K} with each having probability 1/K.
This may shed light on the original problem. -----
If the answer to the original question is Yes,
(*) v[n] -> 1 as n -> oo (with probability 1),
then it may be easier to first show that
(**) lim sup {v[1],v[2],v[3],...,v[n],...} = 1 (with probability 1).
Or, maybe (**) is true and (*) isn't.
((( If I had to guess, I'd say that (*) is probably not true, because the fastest bullets are not necessarily the ones that go farthest, but are often the ones that annihilate soonest. )))
--Dan
David wrote:
<< A gun sits on a line.
Every second, the gun shoots a bullet to the right at a random constant speed between 0 and 1.
If two bullets collide they annihilate.
It's probability 0, but if more than two bullets collide, the slowest bullets annihilate in pairs.
Can we expect the speed of the furthest bullet to approach 1 over time?
________________________________________________________________________________________ It goes without saying that .
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Forewarned is worth an octopus in the bush.