Timeline for Brownian bridge interpreted as Brownian motion on the circle
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Nov 26, 2010 at 1:33 | comment | added | Omer | if $PQRS$ is a rectangle circumscribed in $S^1$ with sides parallel to the axes, then this construction gives $B_P+B_R=B_Q+B_S$. | |
| Nov 24, 2010 at 7:13 | comment | added | Ori Gurel-Gurevich | The new construction is not stationary. However, if you change the definition to $B_t=C_{a^2}+D_{b^2}$ it will be. But then it is symmetric, so maybe it's better to just define it on the $\ell_1$ circle instead. I still don't think it will be a Brownian bridge, but I'm not sure why right now. | |
| Nov 23, 2010 at 10:43 | history | edited | Gideon Schechtman | CC BY-SA 2.5 |
added 112 characters in body
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| Nov 23, 2010 at 8:20 | comment | added | Gideon Schechtman | Following Ori's two dimensionality objection, I editted my answer. | |
| Nov 23, 2010 at 8:18 | history | edited | Gideon Schechtman | CC BY-SA 2.5 |
added 256 characters in body; deleted 3 characters in body
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| Nov 23, 2010 at 7:55 | comment | added | Gideon Schechtman | As somebody above already remarked, it can't be an actual Brownian bridge. | |
| Nov 23, 2010 at 6:32 | comment | added | Ori Gurel-Gurevich | Actually, now that I re-read the question, I'm no sure what it is, so I guess you probably didn't mean this to define the Brownian bridge. | |
| Nov 23, 2010 at 6:27 | comment | added | Ori Gurel-Gurevich | But this is not a Brownian bridge - the distribution of the values on any (say, finite) subset is two dimensional. More specifically, $B_{(-1,0)}=-B_{(1,0)}$. | |
| Nov 23, 2010 at 6:17 | history | answered | Gideon Schechtman | CC BY-SA 2.5 |