Is it possible to find a coupling of two Wiener processes $W^0, W^x$ (i.e. two Wiener processes defined on a common probability space). One starting from $0$ and the other from $x$ such that

$W_t^0 - W_t^x \rightarrow_t 0$ almost surely and in $L^1$.

Using some random-walks considerations I suspect that it is *not* possible to have convergence in $L^1$ but I do not know how to prove it.

The **answer** is: $W_t^0 - W_t^x$ *cannot* converge to $0$ in $L^1$

The proof is as follow (it is a slightly extended version of the proof by smalldeviations). Obviously

$|W_t^0 - W_t^x|_1\geq \inf _{\gamma\in \Gamma } \{\int _{\mathbf{R}^d\times \mathbf{R}^d} |x-y| \text{d}\gamma(x,y)\},$

where $\Gamma$ is the set of all couplings of $\mathcal{L} (W_t^0)$ and $\mathcal{L} (W_t^x)$ ($\mathcal{L}$ is the law of given variable). By the duality formula (see http://en.wikipedia.org/wiki/Transportation_theory) the right hand side is equal to

$\sup \{ \int_{\mathbf{R}^d} \phi (x) \mathrm{d} \mu (x) + \int_{\mathbf{R}^d} \psi (y) \mathrm{d} \nu (y) \},$

where the supremum runs over all pairs of bounded and continuous functions such that $\phi(x)+\psi(y)\leq |x-y|$ and $\mu =\mathcal{L}(W_t^x)$ and $\nu = \mathcal{L} (W_t^0)$. In our case it is sufficent to take $\phi(x) = x, \psi(y)=y$. Then the expression under the sup is equal to:

$\mathbf{E}(W^x_t) - \mathbf{E}(W^0_t) = x-0 =x$. Therefore $|W_t^0 - W_t^x|_1\geq x$.