Timeline for Infinitesimal generators of stochastic processes
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 28, 2011 at 22:23 | answer | added | Martin Gisser | timeline score: 1 | |
| Apr 20, 2011 at 15:22 | comment | added | Did | @John Baez: To be clear, some stochastic matrices $U$ cannot be written as $U=\exp(H)$, for any $H$ such that (1)+(2) holds, and some stochastic matrices $U$ can be written as $U=\exp(H)$ where $H$ is such that (1)+(2) does not hold. (For your interest, I discovered your comment by chance, so you might want to use this @ thing at the beginning of your comments to signal them to your interlocutor.) | |
| Apr 15, 2011 at 2:15 | comment | added | John Baez | The Hille-Yosida theorem is surely relevant, but that's a general theorem about continuous one-parameter semigroups on Banach spaces; I'm trying to characterize a particular class of such semigroups. It sounds like over at stackexchange they're saying that even in the finite-dimensional case not every stochastic operator $U$ is of the form $\exp(tH)$. That's unsurprising. I don't see them exhibiting square matrices $H$ obeying conditions 1) and 2) such that $U = \exp(tH)$ fails to be stochastic. That would shock and interest me. | |
| Apr 11, 2011 at 10:34 | answer | added | András Bátkai | timeline score: 2 | |
| Apr 11, 2011 at 10:31 | answer | added | Adam Skalski | timeline score: 6 | |
| Apr 11, 2011 at 10:17 | comment | added | Did | Surely I am missing something but isn't this called Hille-Yosida theorem? See math.brown.edu/~schhita/Semigroups.pdf. Note that even in the finite dimensional case, not every square matrix such that your conditions (1) and (2) hold is a generator. See math.stackexchange.com/questions/31174/… | |
| Apr 11, 2011 at 8:36 | comment | added | Yemon Choi | I think that if one applies the general machinery of C_0-semigroups en.wikipedia.org/wiki/C0-semigroup then you get $U(t)=\exp(tH)$ where $H$ is densely defined, and we should get $\int H\psi = 0$ whenever $\psi\in {\mathcal D}(H)$. But I also suspect that people who know about Markov processes will be able to give a better answer | |
| Apr 11, 2011 at 7:53 | history | asked | John Baez | CC BY-SA 3.0 |