Timeline for Sweep-segment bot: Will this random walk sweep the plane?
Current License: CC BY-SA 2.5
6 events
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| Dec 10, 2010 at 23:24 | comment | added | George Lowther | @Didier: By Theorem 9.1 and 9.2 of Kallenberg (Foundations of Modern Probability), any random walk whose increments have finite variance is recurrent. That is, its set of cluster points forms a closed additive subgroup of $\mathbb{R}^2$. In this case, the support of the possible steps of $A_n$ spans all of $\mathbb{R}^2$, so its set of cluster points is $\mathbb{R}^2$. | |
| Dec 10, 2010 at 22:52 | comment | added | George Lowther | Didn't notice when leaving the comment to Hugh J's answer: that one and this are essentially the same. The points $A_n$ just follow a random walk on the plane. | |
| Dec 10, 2010 at 21:05 | comment | added | Did | OK, I am stupid: $U_{n+1}$ is uniform on the circle and independent of $(A_n,V_n)$, hence $U_{n+1}V_n$ is also uniform on the circle and also independent of $(A_n,V_n)$. The dynamics of $(A_n)_n$ is simply $A_{n+1}=A_n+W_n$ with $(W_n)_n$ i.i.d. and uniform on the circle. This is a random walk in the plane with uniformly bounded increments hence it is recurrent, at least in the sense that $|A_n|$ does not converge to infinity and probably also in the sense that $(A_n)_n$ visits infinitely often every ball of positive radius--which should be enough to show that the whole plane is sweeped. | |
| Dec 10, 2010 at 20:46 | comment | added | Joseph O'Rourke | @Didier: Nice idea to interpret the rotations as complex-number multiplications! Not sure I yet see the consequences of your analysis... | |
| Dec 10, 2010 at 20:37 | history | edited | Did | CC BY-SA 2.5 |
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| Dec 10, 2010 at 20:32 | history | answered | Did | CC BY-SA 2.5 |