Very recently, the following question [was asked](https://mathoverflow.net/questions/339366/probability-space-with-exactly-one-brownian-motion):



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> 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....
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> 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).
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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.



Probability space with exactly one Brownian motion
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.


