Timeline for Strong Markov property, independence, regular conditional probability
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 26, 2019 at 17:27 | vote | accept | sharpe | ||
| S Sep 26, 2019 at 17:27 | history | bounty ended | sharpe | ||
| S Sep 26, 2019 at 17:27 | history | notice removed | sharpe | ||
| Sep 26, 2019 at 9:24 | answer | added | Mateusz Kwaśnicki | timeline score: 1 | |
| Sep 24, 2019 at 12:16 | history | edited | sharpe | CC BY-SA 4.0 |
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| Sep 24, 2019 at 9:15 | history | edited | sharpe | CC BY-SA 4.0 |
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| Sep 24, 2019 at 9:13 | comment | added | Mateusz Kwaśnicki | Thanks for clarification! (By the way, I misread the formula with conditioning, please ignore the last point of my previous comment.) | |
| Sep 24, 2019 at 9:13 | history | edited | sharpe | CC BY-SA 4.0 |
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| Sep 24, 2019 at 8:52 | comment | added | sharpe | @MateuszKwaśnicki Thank you for your comment. $X$ is a absorbing BM on $S$ conditioned to hit $\{1\} \times (-1,1)$. So, $X$ does not hit $\{-1,1\} \times (1,\infty)$ before arriving at $\{1\} \times (-1,1)$. | |
| Sep 24, 2019 at 8:17 | comment | added | Mateusz Kwaśnicki | What happens to $X$ at the boundary, that is, at $\{-1,1\} \times (1,\infty)$? If it is reflected, everything works nicely; however, in this case one would rather speak about BM in $[-1,1] \times (1,\infty)$. If it is absorbed (or killed), the claimed result does not seem to be true. | |
| Sep 24, 2019 at 8:07 | answer | added | RaphaelB4 | timeline score: 1 | |
| S Sep 24, 2019 at 5:54 | history | bounty started | sharpe | ||
| S Sep 24, 2019 at 5:54 | history | notice added | sharpe | Authoritative reference needed | |
| Sep 24, 2019 at 5:54 | history | edited | sharpe | CC BY-SA 4.0 |
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| May 13, 2019 at 12:29 | history | edited | sharpe | CC BY-SA 4.0 |
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| May 13, 2019 at 3:10 | history | edited | sharpe | CC BY-SA 4.0 |
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| May 12, 2019 at 6:00 | history | asked | sharpe | CC BY-SA 4.0 |