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Nov 21, 2022 at 17:07 vote accept Nate River
Nov 21, 2022 at 17:07 vote accept Nate River
Nov 21, 2022 at 17:07
S Nov 21, 2022 at 16:28 history bounty ended Nate River
S Nov 21, 2022 at 16:28 history notice removed Nate River
Nov 20, 2022 at 17:41 answer added Christophe Leuridan timeline score: 1
S Nov 19, 2022 at 14:19 history bounty started Nate River
S Nov 19, 2022 at 14:19 history notice added Nate River Draw attention
S Jan 25, 2022 at 10:00 history bounty ended CommunityBot
S Jan 25, 2022 at 10:00 history notice removed CommunityBot
Jan 23, 2022 at 15:16 comment added Tobsn Intuitively I would expect that by Martingale representation theorem all such processes can be realised as Ito integrals and are therefore indeed not differentiable a.s.
S Jan 17, 2022 at 8:23 history bounty started Nate River
S Jan 17, 2022 at 8:23 history notice added Nate River Draw attention
Jan 4, 2022 at 22:29 comment added jlewk I stand corrected @StephanSturm. The question should link to mathoverflow.net/questions/397639/…
Jan 4, 2022 at 14:37 comment added Stephan Sturm @jlewk: While the argument is formally correct, I do not think it is useful here. A Lipschitz martingale is of bounded total variation, hence of zero quadratic variation. So the only martingales satisfying this conditions are constants anyway.
Jan 4, 2022 at 3:43 comment added jlewk $E[M_{t+h}-M_t|F_t] =0$ by the martingale property, but on the other hand, $E[h^{-1}(M_{t+h}-M_t)|F_t]\to^{h\to 0} M_t'$ by dominated convergence if you additionally assume that $M_t$ is $C$-Lipschitz almost surely. So $M_t'=0$ necessarily in this case.
Jan 3, 2022 at 23:52 history asked Nate River CC BY-SA 4.0