Timeline for random walk and Brownian motion on Riemannian manifold
Current License: CC BY-SA 3.0
15 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Jul 23, 2013 at 20:56 | comment | added | shu | @GeorgeLowther, yes, this is exact what I would like to do! But not so evident... | |
Jul 23, 2013 at 20:37 | answer | added | Did | timeline score: 7 | |
Jul 23, 2013 at 20:35 | comment | added | George Lowther | This comes down to discrete approximation of a continuous diffusion. You will be guaranteed weak convergence as long as the generator of the discrete process approximates the generator of BM on the manifold (i.e., the Laplacian) closely enough. | |
Jul 23, 2013 at 16:33 | comment | added | shu | In fact, in the case if we take account of length of each edge, the triangulation will converge as metric space to the manifold $X$. So if we can define a random walk(which take account of length of the edge), the limit should be universal. | |
Jul 23, 2013 at 16:24 | comment | added | shu | What I mean by random walk is not to say at each point, we take 1/2 chance to one of the other next points. In the case you metioned, we shoud take account of length of the edge to modifier the proba to the next point. | |
Jul 23, 2013 at 16:21 | comment | added | Noah Stein | The limit process is not universal. Consider triangulating a circle by slicing the top half into $M$ segments and the bottom into just one. Then the scale of the motion in the top half will be $1/M$ times the scale of the motion in the bottom half at every iteration, so also in any reasonable limit. | |
Jul 23, 2013 at 15:53 | comment | added | shu | @Noah Stein, but the limit process might be a universal objet... | |
Jul 23, 2013 at 15:44 | comment | added | Noah Stein | I'm not quite sure what you want out of the construction you propose. Certainly it depends on the triangulation, whereas Brownian motion doesn't, right? | |
Jul 23, 2013 at 15:40 | history | edited | shu | CC BY-SA 3.0 |
added 362 characters in body
|
Jul 23, 2013 at 3:15 | comment | added | Theo Johnson-Freyd | Hi shu, Since Nate has marked this as a possible duplicate, perhaps you might expand your question to clarify why Donsker's theorem doesn't do what you want? | |
Jul 22, 2013 at 22:01 | comment | added | shu | hi, Nate Eldredge. I am afraid Donsker's theorem is nothing to do with the triangulation. | |
Jul 22, 2013 at 21:36 | review | Close votes | |||
Jul 23, 2013 at 3:15 | |||||
Jul 22, 2013 at 21:30 | history | edited | Nate Eldredge |
probability tag
|
|
Jul 22, 2013 at 21:20 | comment | added | Nate Eldredge | See Reference needed: Donsker's Invariance Principle for Riemannian Manifolds; I think the references given there should answer your question. I'm tentatively marking this as duplicate. | |
Jul 22, 2013 at 19:24 | history | asked | shu | CC BY-SA 3.0 |