Timeline for Is a martingale constant on the event that its quadratic variation is zero?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Nov 4, 2022 at 3:28 | vote | accept | Nate River | ||
| Nov 2, 2022 at 8:27 | answer | added | JGWang | timeline score: 2 | |
| Oct 23, 2022 at 22:35 | comment | added | Thomas Kojar | @JohnDawkins I agree. Perhaps some conditional version might do the trick i.e. $H(t):=E[\chi_E|Ft]$. | |
| Oct 22, 2022 at 16:48 | comment | added | John Dawkins | @Thomas Kojar: Such an $H$ is not likely to be adapted. | |
| Oct 20, 2022 at 22:58 | comment | added | Nate River | Oh damn, that works indeed.. | |
| Oct 20, 2022 at 22:11 | comment | added | Thomas Kojar | how about we use the Ito isometry for martingales $$\operatorname{E}\left(\left(\int_0^t H_{s}\,dM_{s}\right)^2\right) = \operatorname{E}\left(\int_0^tH_{s}^2\,d[M]_{s}\right).$$ but for $H$ constantly equal to the indicator of $E$ at least in the case of E being measurable with respect to the filtration of M? | |
| Oct 20, 2022 at 21:28 | history | edited | Nate River | CC BY-SA 4.0 |
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| Oct 20, 2022 at 21:23 | history | edited | Nate River | CC BY-SA 4.0 |
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| Oct 20, 2022 at 21:17 | history | asked | Nate River | CC BY-SA 4.0 |