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wu :: forums    riddles    putnam exam (pure math) (Moderators: towr, Icarus, william wu, SMQ, Eigenray, Grimbal)    2nd order random walks « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    Author  Topic: 2nd order random walks  (Read 1483 times) JocK Uberpuzzler     Gender: Posts: 877 2nd order random walks   « on: Jul 9th, 2005, 3:39am » Quote Modify A standard (1st order) random walk is characterised by random positional changes:   xn+1 = xn +- 1   with both signs occurring randomly with equal probability.   Starting with x0 = 0, the asymptotic behaviour of such a 1st order random walk is characterised  by the fact that, averaged over many random walks, the squares of the displacements  (xn)2  grow proportional to n, the number of steps.     Here we consider 2nd order random walks for which the velocities (rather than the positions)undergo random changes:   xn+1 - xn = xn - xn-1 +- 1   Can you characterise the asymptotic behaviour of the squares of the displacements for this 2nd order random walk?   (You may assume as starting condition the random walker to be at rest at the origin: x-1 = x0 = 0.)         -- Bonus questions --   What if you make the size of the random velocity changes dependent on the position? For instance:   xn+1 - xn = xn - xn-1 +- xn   or even:   xn+1 - xn = xn - xn-1 +- (xn)2     Furthermore, we have restricted ourselves to 1D random walks. Extending the 1st order random walk to higher dimensions, leaves the proportionality of the squares of the displacements with the number of steps unchanged. But what about the various 2nd order random walks listed above? Is the asymptotic behaviour different in different dimensions?           « Last Edit: Jul 9th, 2005, 3:46am by JocK » IP Logged solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it. xy - y = x5 - y4 - y3 = 20; x>0, y>0. ThudnBlunder Uberpuzzler The dewdrop slides into the shining Sea     Gender: Posts: 4489 Re: 2nd order random walks   « Reply #1 on: Jul 9th, 2005, 4:56pm » Quote Modify Quote: Can you characterise the asymptotic behaviour of the squares of the displacements for this 2nd order random walk?   (You may assume as starting condition the random walker to be at rest at the origin: x-1 = x0 = 0.) :How about 4n3/9?   « Last Edit: Jul 9th, 2005, 4:57pm by ThudnBlunder » IP Logged THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you. Barukh Uberpuzzler     Gender: Posts: 2276 Re: 2nd order random walks   « Reply #2 on: Jul 9th, 2005, 11:12pm » Quote Modify on Jul 9th, 2005, 4:56pm, THUDandBLUNDER wrote: :How about 4n3/9? I get almost the same: n3/3. IP Logged Barukh Uberpuzzler     Gender: Posts: 2276 Re: 2nd order random walks   « Reply #3 on: Jul 12th, 2005, 6:22am » Quote Modify So, two close answers were given. What will the judges say?   IP Logged ThudnBlunder Uberpuzzler The dewdrop slides into the shining Sea     Gender: Posts: 4489 Re: 2nd order random walks   « Reply #4 on: Jul 12th, 2005, 7:21am » Quote Modify For vn - vn-1 = +/- xn I think that E(|xn|) => oo as n => oo IP Logged THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you. Barukh Uberpuzzler     Gender: Posts: 2276 Re: 2nd order random walks   « Reply #5 on: Jul 12th, 2005, 9:04am » Quote Modify on Jul 12th, 2005, 7:21am, THUDandBLUNDER wrote: For vn - vn-1 = +/- xn I think that E(|xn|) => oo as n => oo Given the starting conditions in the original problem, I think E(xn2) = 0 for every n.   IP Logged Pages: 1  Reply Notify of replies Send Topic Print Forum Jump: ----------------------------- riddles -----------------------------  - easy   - medium   - hard   - what am i   - what happened   - microsoft   - cs => putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth   - complex analysis   - wanted   - psychology   - chinese « Previous topic | Next topic »

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