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Jun 15, 2020 at 7:27 history edited CommunityBot
Commonmark migration
Sep 10, 2019 at 11:41 comment added Dan For an example, suppose $(\Omega,\mathcal{F},\mathbb{P})$ supports two independent Brownian motions $B^1$ and $B^2$. Define a filtration $\mathbb{F}=(\mathcal{F}_t)_{t \ge 0}$ by setting $\mathcal{F}_t = \sigma(B^1_s,B^2_r : s \le t, r \ge 0)$. That is, the whole path of $B^2$ is known at time zero. Then $B^1$ is an $\mathbb{F}$-Brownian motion, but $B^2$ is not because it does not satisfy the independent increment property with respect to $\mathbb{F}$. However, $B^2$ is $\mathbb{F}$-adapted and is a Brownian motion with respect to its own filtration.
Sep 10, 2019 at 11:38 comment added Dan For a process $B$ to be Brownian with respect to a filtration $(\mathcal{F}_t)_{t \ge 0}$ it must be the case, among these things, that $B_t-B_s$ is independent of $\mathcal{F}_s$ for each $t > s \ge 0$. So for a process to be "a Brownian motion on $(\Omega,\mathcal{F},\mathbb{F},\mathbb{P})$" is a stronger requirement than for it to be $\mathbb{F}$-adapted and Brownian in its own filtration.
Sep 9, 2019 at 16:49 comment added Iosif Pinelis @Dan : Alas, I still don't understand what your question is. For one thing, any process is adapted to its natural filtration; so, why mention it as a condition? Maybe, it will help if we avoid terms such as "base" and "supports" and speak only in terms of being adapted to a filtration.
Sep 9, 2019 at 12:44 comment added Dan I suppose we agree on what it means for a process $B$ to be a Brownian motion on a stochastic base $(\Omega,\mathcal{F},\mathbb{F},\mathbb{P})$. The answers provided say that there exists a stochastic base on which there are not two independent Brownian motions. But suppose we ask instead for a stochastic base which supports two independent $\mathbb{F}$-adapted processes $B^1$ and $B^2$, such that, for $i=1,2$, $B^i$ is a Brownian motion on the stochastic base $(\Omega,\mathcal{F},\mathbb{F}^{B^i},\mathbb{P})$, where $\mathbb{F}^{B^i}$ is the natural filtration of $B^i$?
Sep 5, 2019 at 21:56 comment added Iosif Pinelis @Dan : It is not clear to me what you mean by "a Brownian motion in its own filtration".
Sep 5, 2019 at 14:11 comment added Dan An alternative and intriguing (though admittedly less natural) interpretation of the question remains unanswered: Does there exist a stochastic basis $(\Omega,\mathcal{F},\mathbb{F},\mathbb{P})$ on which there do not exist two independent $\mathbb{F}$-adapted processes, each of which is a Brownian motion in its own filtration? The given answers address the case where both processes are required to be $\mathbb{F}$-Brownian.
Aug 29, 2019 at 17:14 vote accept Iosif Pinelis
Aug 29, 2019 at 16:58 answer added Mateusz Kwaśnicki timeline score: 4
Aug 28, 2019 at 22:33 history edited YCor CC BY-SA 4.0
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Aug 28, 2019 at 19:37 history edited Iosif Pinelis CC BY-SA 4.0
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Aug 28, 2019 at 17:54 answer added Iosif Pinelis timeline score: 2
Aug 28, 2019 at 17:52 history asked Iosif Pinelis CC BY-SA 4.0