Timeline for Can a diffusion have negative minimum or achieve large value at a given time?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| S Jul 12, 2022 at 13:39 | history | suggested | Christophe Leuridan | CC BY-SA 4.0 |
Correction of an error continous martingale -> Brownian motion
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| Jul 7, 2022 at 21:14 | comment | added | Christophe Leuridan | Thank you. I believe more on your second argument. $X-X_0$ is not independent of $X_0$ in general since $d\langle X \rangle_t/dt$ equals $\sigma(0,X_0)^2$ at time $0$. A more complete argument could be that the quadratic variation computed under $P$ is still the quadratic variation computed or under $P[\cdot|A]$ when $A \in \sigma(X_0)$ has positive probability. Calling $\tau_\cdot$ the inverse of the quadratic variation, we derive that $B := X_{\tau_\cdot}$ is a Brownian motion under $P$ and also under $P[\cdot|A]$ when $A \in \sigma(X_0)$ has positive probability. | |
| Jul 7, 2022 at 20:28 | comment | added | GJC20 | @ChristopheLeuridan Otherwise, we may use D-S theorem to X_t-X_0 conditioning on the event $\{X_0=x\}$ | |
| Jul 7, 2022 at 20:27 | comment | added | GJC20 | @ChristopheLeuridan That's a good point. Indeed, the D-S theorem is applied to $X_t-X_0$ which is independent to $X_0$. Therefore the independence between $B$ and $X_0$ still holds | |
| Jul 7, 2022 at 19:50 | review | Suggested edits | |||
| S Jul 12, 2022 at 13:39 | |||||
| Jul 7, 2022 at 19:49 | comment | added | Christophe Leuridan | I agree with that, provided $B$ independent of $X_0$. You use that twice, and you should mention it. That $B$ is independent of $X_0$ is not so obvious, although it looks true. | |
| Jul 7, 2022 at 18:08 | history | edited | GJC20 | CC BY-SA 4.0 |
added 2 characters in body
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| Jul 7, 2022 at 17:59 | history | answered | GJC20 | CC BY-SA 4.0 |