Timeline for Solution to SDE conditional on high maxima of driving Brownian motion
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 27, 2022 at 13:00 | comment | added | Nate River | I believe I have proven the $L^1$ convergence desired. Will write it up sometime… | |
| Jun 27, 2022 at 4:28 | answer | added | Nate River | timeline score: 0 | |
| Jun 26, 2022 at 20:52 | answer | added | Christophe Leuridan | timeline score: 1 | |
| Jun 26, 2022 at 14:12 | comment | added | Martin Hairer | Got you, I thought $\epsilon$ was just a superscript (as it is for $\mathbb{P}^\epsilon$ in the same expression), not an actual exponent... | |
| Jun 26, 2022 at 13:18 | comment | added | Nate River | Well $X \approx \text{exp}(1/\varepsilon)$, so $X^\varepsilon \approx e$ right? | |
| Jun 26, 2022 at 12:37 | comment | added | Martin Hairer | Right, but if $\epsilon \log X \approx 1$, then $X \approx \exp(1/\epsilon)$, not $e$. (And the absolute error might be huge.) | |
| Jun 26, 2022 at 12:20 | history | edited | Nate River | CC BY-SA 4.0 |
added 4 characters in body
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| Jun 26, 2022 at 11:55 | comment | added | Nate River | Ah yes, I guess we’re both saying the same thing. I meant to consider $\varepsilon \log X$. | |
| Jun 26, 2022 at 10:22 | comment | added | Martin Hairer | You have to take $Y=\epsilon \log X$ for that, don’t you? | |
| Jun 26, 2022 at 9:39 | comment | added | Nate River | Hm what happens is that, taking $Y = \log X$, we get that $|\varepsilon Y - 1| \to 0$ in probability, and so $|X^\varepsilon - e| \to 0$ also in probability. Or did I misunderstand something? | |
| Jun 26, 2022 at 9:37 | comment | added | Martin Hairer | You missed the epsilon in the definition of $Y_\epsilon$ when trying to apply the earlier result, didn’t you? | |
| Jun 26, 2022 at 9:22 | history | asked | Nate River | CC BY-SA 4.0 |