# Existence of strong couplings for Brownian motion

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.

## 1 Answer

(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.