Timeline for Twisted random walks
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 7, 2013 at 11:45 | vote | accept | Joseph O'Rourke | ||
| Aug 6, 2013 at 19:55 | answer | added | Andreas Rüdinger | timeline score: 4 | |
| Aug 6, 2013 at 16:03 | comment | added | Joseph O'Rourke | @Yoav: Yes, analogous to runs of heads in a sequence of coin flippings: expectation zero, but possibly long runs. Cool re that surface suggestion---Thanks! | |
| Aug 6, 2013 at 15:43 | comment | added | Yoav Kallus | By symmetry the expectation would be zero winding, but the interesting question would be how the variance grows with time. Here is the surface on which I was suggesting to solve the diffusion equation: en.wikipedia.org/wiki/File:Riemann_surface_log.jpg | |
| Aug 6, 2013 at 15:39 | comment | added | Joseph O'Rourke | @Yoav: In some sense it is a single random walk on the unit circle that determines the winding. So I guess then the expectation should be zero winding, which seems a bit strange to me. Thanks for your suggestion. | |
| Aug 6, 2013 at 15:39 | comment | added | Joseph O'Rourke | @Liviu: Yes, two walks independent. | |
| Aug 6, 2013 at 15:31 | comment | added | Yoav Kallus | Can't you equivalently consider a single random walk's winding number about the origin? I think you must put some barrier at the origin, because (in the plane) I think it will hit the origin with non-zero probability. Once you figured out the right set-up, you can convert the discrete random walk into a diffusion equation on the Riemann surface for $\log z$, and if you can solve the diffusion equation you will have your answer. | |
| Aug 6, 2013 at 15:30 | comment | added | Liviu Nicolaescu | Are the two walks independent? | |
| Aug 6, 2013 at 13:24 | comment | added | Joseph O'Rourke | I mean to count just like the classical winding number: the total number of times the point travels $2 \pi$ counterclockwise around the circle, taking into account $\pm$ movements, i.e., $-$ cancels $+$. | |
| Aug 6, 2013 at 13:06 | comment | added | Stefan Kohl♦ | The normalization is clear -- but the issue is that the point won't wander around the unit circle in one direction. It will rather move randomly hence and forth, in both directions -- how do you count this? -- Or have I misunderstood something? | |
| Aug 6, 2013 at 12:51 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Tried to address Stefan's question...
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| Aug 6, 2013 at 12:49 | comment | added | Joseph O'Rourke | Sorry for not being clear. If one normalizes $b(t)-a(t)$, it can be viewed as a point on the unit circle centered on the origin. (Assume $a(t)=b(t)$ occurs with zero probability.) This point wanders around that unit circle. | |
| Aug 6, 2013 at 12:38 | comment | added | Stefan Kohl♦ | What do you mean by "turns around the origin"? -- For random walks $a(t)$ and $b(t)$, the difference $a(t)-b(t)$ will "go hence and forth" in some sense, I think -- I don't see how you get your winding number from this. | |
| Aug 6, 2013 at 12:08 | history | asked | Joseph O'Rourke | CC BY-SA 3.0 |
