Timeline for Is the truncated Brownian motion of the class DL?
Current License: CC BY-SA 2.5
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 4, 2010 at 15:57 | vote | accept | kenneth | ||
| Jun 4, 2010 at 15:57 | comment | added | kenneth | Thank you, Nate. I agree that $W^1$ is martingale, and optional sampling does not apply here. Hence, although $\mathbb{E} [W^1(T^1)] = 1 > W^1(0)$ is correct, it shall not become a reason for the strict local martingale. | |
| Jun 4, 2010 at 14:46 | comment | added | Nate Eldredge | Perhaps the OP is thinking that if $W^a$ were a martingale, we should have $E[W^a(T^a)]=E[W^a(0)]=0$ by the optional sampling theorem (e.g. Karatzas and Shreve 1.3.22). But the optional sampling theorem is not applicable here because $W^a$ does not have a "last element". | |
| Jun 4, 2010 at 13:40 | history | answered | Steve Huntsman | CC BY-SA 2.5 |