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.