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Very recently, the following question was asked:

Often, we encounder the assumption that $(\Omega,\mathcal{F},\mathbb{F},\mathbb{P})$ is a stochastic base on which a Brownian motion is defined. Often times there can be many independent Brownian motions on such as space....

However, does there exist a stochastic base on which there exists exactly one Brownian motion? (By this I mean there do not exist two independent Brownian motions).

The question has received negative response and has been deleted by the owner. I think the question makes sense and should be answered, which is the purpose of this posting.

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  • $\begingroup$ 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. $\endgroup$ – Dan Sep 5 '19 at 14:11
  • $\begingroup$ @Dan : It is not clear to me what you mean by "a Brownian motion in its own filtration". $\endgroup$ – Iosif Pinelis Sep 5 '19 at 21:56
  • $\begingroup$ 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 $B^i$? $\endgroup$ – Dan Sep 9 '19 at 12:44
  • $\begingroup$ @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. $\endgroup$ – Iosif Pinelis Sep 9 '19 at 16:49
  • $\begingroup$ 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. $\endgroup$ – Dan Sep 10 '19 at 11:38
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Upon request of Iosif Pinelis, here is my comment (slightly edited).


There cannot exist two independent Brownian motions adapted to the standard Brownian filtration. Indeed, suppose that $B_t$ is a Brownian motion. By the martingale representation theorem, every $L^2$ martingale $X_t$ adapted to the filtration generated by $B_t$ is an Itô integral with respect to $B_t$: $$X_t = \int C_t dB_t$$ for some predictable $C_t$. If $X_t$ is a Brownian motion itself, the integrand $C_t$ can only take values $\pm 1$ (almost everywhere). If $X_t$ and $Y_t$ are two such Brownian motions, and $C_t$, $D_t$ the corresponding integrands, then $C_t D_t \ne 0$ almost everywhere, and hence $X_t$ and $Y_t$ are dependent (their co-variation is non-zero)

Interestingly, the situation is quite different for random walks ("discrete-time Brownian motions") with jumps following a continuous distribution $\mu$. 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 $B_n$. Indeed: simply define $$(X_{n+1} - X_n, Y_{n+1} - Y_n) = \phi(B_{n+1} - B_n),$$ where $\phi : \mathbb{R} \to \mathbb{R}^2$ is a measure-theoretic isomorphism between $(\mathbb{R}, \mu)$ and $(\mathbb{R}^2, \mu \times \mu)$.

However, if the jumps of $B_n$ have atoms (say, $B_n$ is a standard simple random walk), the answer is again negative.

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A partial answer to this question is as follows. Take any Brownian motion $B=(B_t)$ with a base $\mathcal B=(\Omega,\mathcal{F},\mathbb{F},\mathbb{P})$. Consider the Karhunen–Loève expansion $$B_t=\sqrt2\sum_{k=1}^\infty Z_k \frac{\sin((k-1/2)t)}{(k-1/2)\pi} $$ for $t\in[0,1]$, where the $Z_k$'s are certain iid standard normal random variables, defined on the same probability space $\mathcal P:=(\Omega,\mathcal{F},\mathbb{P})$. For $t\in[0,1]$, let $$C_t=\sqrt2\sum_{k=1}^\infty Z_{2k-1} \frac{\sin((k-1/2)t)}{(k-1/2)\pi},\quad D_t=\sqrt2\sum_{k=1}^\infty Z_{2k} \frac{\sin((k-1/2)t)}{(k-1/2)\pi}. $$ Then $C$ and $D$ are independent Brownian motions on $[0,1]$, defined on the same probability space $\mathcal P$, but apparently not on the same base $\mathcal B$ -- which is why the answer is only partial.

Now it easy to use shifting in $t$ and gluing to construct two (or even countably many) independent Brownian motions on $[0,\infty)$, again on the same probability space. Here we need to partition the set of all natural numbers into countably many countable subsets (rather than into two subsets, of all odd natural numbers and all even natural numbers, as was done above). Here the term "countable" is of course used in the sense of "countably infinite".


On the other hand, take any Brownian motion $B=(B_t)$ and let $\mathcal B=(\Omega,\mathcal{F},\mathbb{F},\mathbb{P})$ be the (smallest) base generated by $B$, with $\Omega$ being the set of all paths of $B$. Then there is no Brownian motion $C$ on base $\mathcal B$ that is independent of $B$.

Indeed, otherwise the sigma-algebra $\mathcal G$ generated by $C$ would be a sub-sigma-algebra of $\mathcal F$ that is independent of $\mathcal F$. So, the sigma-algebra $\mathcal G$ would be independent of itself. That is only possible is all members of $\mathcal G$ have probability $0$ or $1$. But then $\mathcal G$ cannot be the sigma-algebra generated by a Brownian motion.

However, as noted by Robert Israel, there still might exist, on the same base, two independent Brownian motions $C_1$ and $C_2$.


So, the problem remains open.

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    $\begingroup$ 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 $B$. $\endgroup$ – Robert Israel Aug 28 '19 at 18:42
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    $\begingroup$ @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. $\endgroup$ – Iosif Pinelis Aug 28 '19 at 19:56
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    $\begingroup$ 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). $\endgroup$ – Mateusz Kwaśnicki Aug 28 '19 at 20:29
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    $\begingroup$ 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. $\endgroup$ – Mateusz Kwaśnicki Aug 28 '19 at 20:32
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    $\begingroup$ @MateuszKwaśnicki : Thank you! Can you please post your comments as a formal answer, so that I could accept it and the question be settled? $\endgroup$ – Iosif Pinelis Aug 28 '19 at 21:37

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