Timeline for Integral of $M^\text{*} - M$ with respect to $M^\text{*}$ is zero for $M^\text{*}$ the running maximum of $M$ a continuous local martingale
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jun 15, 2023 at 10:46 | vote | accept | George | ||
| Jun 15, 2023 at 10:46 | history | edited | George |
Added measure/analysis tags, as the question is inherently analytic as the accepted answer describes
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| Jun 15, 2023 at 5:14 | comment | added | mike | I thought , without saying, that if you use the reflected process $W^*_t - W_t$, for which the running max, $W^*_t$ is the local time, then it was exactly the same problem for brownian motion. | |
| Jun 14, 2023 at 19:22 | answer | added | Christophe Leuridan | timeline score: 4 | |
| Jun 14, 2023 at 16:11 | comment | added | George | @mike Thank you for your comment! I'm not seeing how to apply it in this case - I'm not very familiar with local times. It feels like I would want to apply this with (in their notation) $a=0$ and $X_t = M^\text{*}_t - M_t$ but I'm not sure how this corresponds to $dL_t^a$... Do you have any quick words about intuition in interpreting this local time? | |
| Jun 14, 2023 at 15:45 | comment | added | mike | This problem for Brownian motion is I think prop vi.1.3 in revuz and yor, although it is stated for local time | |
| Jun 14, 2023 at 13:39 | history | edited | George | CC BY-SA 4.0 |
title wording & added almost surely quantification
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| S Jun 14, 2023 at 13:31 | review | First questions | |||
| Jun 14, 2023 at 14:39 | |||||
| S Jun 14, 2023 at 13:31 | history | asked | George | CC BY-SA 4.0 |