Timeline for Independence of variables predicted by the generator
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Apr 10, 2021 at 8:05 | comment | added | Mateusz Kwaśnicki | This is still nitpicking, but now one can consider a very rough Markov process, so that again virtually no function of $y$ is in the domain of the generator. If I remember correctly, if $\zeta$ is the Brownian motion on, say, the Sierpiński triangle, and $y$ is a non-constant function in the domain of the generator, then $y^2$ is not in the domain of the generator. So I suppose one can craft $y$ in the domain in such a way that the only functions of $y$ which are in the domain of the generator are linear functions of $y$. | |
| Apr 9, 2021 at 22:57 | comment | added | G. Panel | Thank you for this remark, I edited it ($y,z\in\mathcal{D}(\Omega)$ rather than $\mathcal{C}(X)$) hoping that it is now more relevant. (Sorry for the choice of notations, indeed form Liggett.) | |
| Apr 9, 2021 at 22:54 | history | edited | G. Panel | CC BY-SA 4.0 |
Edit: I assume that $y,z\in\mathcal{D}(\Omega)$ rather than $y,z\in\mathcal{C}(X)$
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| Apr 9, 2021 at 21:39 | comment | added | Mateusz Kwaśnicki | (By the way, perhaps Liggett uses this notation, but I find it rather confusing: $\Omega$ is typically the probability space, not an operator. Also, quite often $\zeta$ denotes the lifetime of a Markov process rather than the Markov process, which is commonly denoted by $X_t$ or another capital letter.) | |
| Apr 9, 2021 at 21:35 | comment | added | Mateusz Kwaśnicki | A trivial counter-example is when, say, $y$ is too rough, so that there are no non-constant functions of $y$ in the domain of the generator. (Say, if $\zeta$ is the Brownian motion and $y$ is nowhere differentiable.) I guess this is not the answer you are looking for, though. | |
| Apr 9, 2021 at 21:27 | history | asked | G. Panel | CC BY-SA 4.0 |