I have two different standard one-dimensional Brownian motions on different filtered spaces, $\langle\Omega,\mathcal F, (\mathcal F_t)_{t\geq 0}, \mathbb P, (W_t)_{t\geq0}\rangle$ and $\langle\hat\Omega,\hat{\mathcal F}, (\hat{\mathcal F}_t)_{t\geq 0}, \hat{\mathbb P}, (\hat{W}_t)_{t\geq0}\rangle$. Assume each filtration is right-continuous and each measure complete.)

Question: Does there always exist a probability measure $\mathbb P^*$ on $\langle \Omega\times\hat\Omega, \mathcal F \otimes \hat{\mathcal F}\rangle$ with marginals $\mathbb P$ and $\hat{\mathbb P}$ such that $\mathbb P^*\{(\omega,\hat\omega)\in\Omega\times\hat\Omega:\ W_t(\omega)= \hat W_t(\hat\omega) \ \forall t\geq0\}=1$?

Of course, the answer would immediately be "yes" if I knew that $\langle\Omega,\mathcal F, (\mathcal F_t)_{t\geq 0}, \mathbb P, (W_t)_{t\geq0}\rangle = \langle\hat\Omega,\hat{\mathcal F}, (\hat{\mathcal F}_t)_{t\geq 0}, \hat{\mathbb P}, (\hat{W}_t)_{t\geq0}\rangle$, and in particular if both Brownian motions lived on the canonical probabilistic basis. But I don't want to assume that.


(Too long for a comment.)

As this does not work even for simple random variables (see below), I am rather sure it will fail for Brownian motions, too.

Let $\Omega$, $\hat\Omega$ be disjoint non-measurable subsets of $[0, 1]$ of full outer Lebesgue measure. Let $\mathbb{P}$ and $\hat{\mathbb{P}}$ be "restrictions" of the Lebesgue measure on $[0, 1]$ to $\Omega$ and $\hat\Omega$, respectively. Finally, let $X(\omega) = \omega$, $\hat X(\hat\omega) = \hat\omega$. Then both $X$ and $\hat X$ are uniformly distributed on $[0, 1]$, but they take values in disjoint subsets, and therefore no strong coupling of $X$ and $\hat X$ is possible.


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