Probability space with exactly one Brownian motion

Question

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.

Answer 1

Upon request of Iosif Pinelis, here is my comment (slightly edited).

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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.

Answer 2

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".

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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 $C_2$.

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So, the problem remains open.