Timeline for Martingale representation of a stopped Brownian motion
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
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| Jul 31, 2021 at 18:27 | comment | added | GJC20 | @MateuszKwaśnicki So, for this case, the existence of $\sigma$ is ensured for this particular case. As the general case, see my previous post mathoverflow.net/questions/398025/…, do you believe it is still true? | |
| Jul 31, 2021 at 18:25 | comment | added | GJC20 | @MateuszKwaśnicki Your observation is amazing! It's absolutely right. What I thought is to compare the quadratic variation, which yields $t\wedge \tau=\int_0^t |\sigma(s,B_{s\wedge \tau})|^2ds$. But then I don't know how to get $\sigma$... | |
| Jul 31, 2021 at 16:59 | comment | added | Mateusz Kwaśnicki | I guess one can simply set $B_t = W_t$, $\sigma(s,x)=1$ if $|x|<1$ and $\sigma(s,x)=0$ if $|x|=1$? | |
| Jul 31, 2021 at 16:49 | history | asked | GJC20 | CC BY-SA 4.0 |