Timeline for Probability space with exactly one Brownian motion
Current License: CC BY-SA 4.0
10 events
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| Aug 28, 2019 at 21:37 | comment | added | Iosif Pinelis | @MateuszKwaśnicki : Thank you! Can you please post your comments as a formal answer, so that I could accept it and the question be settled? | |
| Aug 28, 2019 at 20:32 | comment | added | Mateusz Kwaśnicki | Interestingly, the situation is quite different for random walks ("discrete-time Brownian motions") with jumps following a continuous distribution: in this case it is easy to construct two independent copies $X_n$, $Y_n$ of a random walk $B_n$, adapted to the filtration generated by $B_n$. However, if the jumps of $B_n$ have atoms (say, $B_n$ is a standard simple random walk), the answer is again negative. | |
| Aug 28, 2019 at 20:29 | comment | added | Mateusz Kwaśnicki | There cannot exist two independent Brownian motions adapted to the standard Brownian filtration: by the martingale representation theorem, every $L^2$ martingale $X_t$ adapted to the filtration generated by a Brownian motion $B_t$ is an Itô integral with respect to $B_t$, and if $X_t$ is a Brownian motion itself, the integrand can only take values $\pm 1$ (almost everywhere). If $X_t$ and $Y_t$ are two such Brownian motions, and $A_t$, $B_t$ the corresponding integrands, then $A_t B_t \ne 0$ almost everywhere, and hence $X_t$ and $Y_t$ are dependent (their co-variation is non-zero). | |
| Aug 28, 2019 at 19:56 | comment | added | Iosif Pinelis | @RobertIsrael : Thank you for this very good point. I have now added another construction to the answer. However, I do not have a complete answer yet. | |
| Aug 28, 2019 at 19:52 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Aug 28, 2019 at 19:28 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Aug 28, 2019 at 19:19 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Aug 28, 2019 at 19:10 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Aug 28, 2019 at 18:42 | comment | added | Robert Israel | There is no Brownian motion $C$ that is independent of $B$, but there might be two Brownian motions $C_1$ and $C_2$ that are independent of each other, while neither is independent of $B$. | |
| Aug 28, 2019 at 17:54 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |