Timeline for A variant of Dubins–Schwarz's theorem
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 20, 2023 at 19:46 | comment | added | Thomas Kojar | then i suggest making a new question and we will try to help. | |
| Feb 20, 2023 at 10:57 | comment | added | Fawen90 | @ThomasKojar Thanks. The answer seems to be highly relevant to my question, while the reasoning is not 100% clear to me... | |
| Feb 20, 2023 at 6:15 | comment | added | Thomas Kojar | If your setting is more tricky or different, I would suggest making a new question about this. | |
| Feb 20, 2023 at 6:10 | comment | added | Thomas Kojar | that is a more doable question and you don't need a common Brownian motion for that. See second answer there: mathoverflow.net/questions/405610/a-comparison-of-diffusions/… | |
| Feb 19, 2023 at 7:44 | comment | added | Fawen90 | @mike Fortunately I don't care about the joint law of $(X,Y)$. I only need that, after changing a suitable probability space, $Y$ is "later" than $X$. The motivation is that I wish to show the exit time of $(-1,1)$ for $X$ is "smaller" than that of $Y$. So I do want this answer, but I don't quit follow your construction... | |
| Feb 19, 2023 at 6:50 | comment | added | mike | I really think you don't want this answer. Thie biggest thing wrong with it is that it ignores the joint distribution of the 2 processes, and while your question does not ask for the same joint distribution, I think you must want it. Consider the case of $\alpha, \beta$ 2 different constants, so $X,Y$ are just $\alpha W_t, \beta W_t$. What do you want done with them ? | |
| Feb 18, 2023 at 12:14 | comment | added | Fawen90 | @mike Thank you mike for the answer, while it is not completely clear to me. Do you mind detailing this embedding procedure by providing an answer? I do appreciate | |
| Feb 18, 2023 at 7:32 | comment | added | mike | You can do this up to any finite time T, but it is a cheat. Just imbed the first process, the wait for the brown Ian motion to return to zero, and imbed the second, even using a different Brownian motion if you like. Of course, you could them in either order. | |
| Feb 18, 2023 at 2:15 | history | edited | LSpice | CC BY-SA 4.0 |
`\DeclareMathOperator`
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| Feb 17, 2023 at 23:08 | comment | added | Thomas Kojar | If there was a common Brownian motion, it would be need to be adaptable to two different filtrations and thus, for example, contradict the Markov property. | |
| Feb 17, 2023 at 23:07 | comment | added | Thomas Kojar | In the DS theorem, the two Brownian motions constructed are adapted to different filtrations. | |
| Feb 17, 2023 at 16:24 | comment | added | Fawen90 | @IosifPinelis Yes. I think "taking values in $\mathbb R_+$" means that $\alpha_t\ge 0$ (if my English is not that bad :)) | |
| Feb 17, 2023 at 15:56 | history | asked | Fawen90 | CC BY-SA 4.0 |