Timeline for Brownian bridge interpreted as Brownian motion on the circle
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
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| Jun 28 at 14:22 | comment | added | Michael Hardy | @Wicher : You can use MathJax in comments and write $\operatorname{cov}(X_s,X_t) = 1/6 - |s-t|(1-|s-t|).$ Or $$ \operatorname{cov}(X_s,X_t) = \frac16 - |s-t|(1-|s-t|). $$ | |
| Jun 28 at 11:25 | comment | added | Wicher | @WillSawin Yes this gives a covariance function cov(X_s,X_t)=1/6-|s-t|(1-|s-t|). | |
| May 2, 2012 at 1:17 | comment | added | Will Sawin | One could normalize so that the average value of the BB over the entire circle was always $0$. | |
| Nov 23, 2010 at 0:28 | comment | added | Simon Lyons | Thanks for your reply. After some more research, I stumbled across this article: emis.de/journals/EJP-ECP/_ejpecp/ECP/include/getdocf87c.pdf Apparently Levy developed the theory of Brownian motion indexed by spheres. The variance of the process in Levy's definition doesn't quite match what I had in mind, so I guess I need to rethink things. | |
| Nov 22, 2010 at 22:37 | comment | added | Michael Hardy | It's not clear that we could still speak of the value of the BB at a point, but we could still speak of the difference between the values of the BB at two different points. If the circumference of the circle is 1, then there are two distances between the two points---the lengths of each of the two arcs connecing them---and the sum of the two distances is 1. It can then be deduced that the variance of the difference between the values of the BB at the two points is just the product of those two distances. | |
| Nov 22, 2010 at 20:26 | history | answered | Michael Hardy | CC BY-SA 2.5 |