In this link p3, I read that the mirror coupling of Brownian motion is actually a maximal coupling. However, they do not prove this and I was wondering if anyone might have a reference to a proof of this fact.
To be more specific, I am asking for a proof that: $$P\{T(X1, X2) ≥ t\} = \phi_t(|x1 − x2|)$$ where $T(X1, X2)$ is the earliest time at which the two Brownian motions from the mirror coupling coincide and where $$\phi_t(r)=\frac{2}{\sqrt{2\pi t}} \int_0^{r/2}e^{-\rho^2/2}d\rho$$ Thank you in advance!
$\phi_t(r)$ gives the tail probability of the first passage time of a 1D Brownian motion as they say on the first page (and whose proof is a direct application of reflection principle), i.e. $P(\tau_{r/2} \geq t) = \phi_t (r)$
You can reduce the hitting time of an n-dimensional Brownian motion to an n-1 dimensional hyperplane to the hitting time of a one-dimensional Brownian motion by spherical symmetry and essentially discarding all the directions orthogonal to the normal vector of the hyperplane. In particular, if an n-dimensional Brownian motion starts $r = |x_1 - x_2|/2$ away from the hyperplane, the distribution of its hitting time $\tau$ will be given by $P(\tau \geq t) = \phi_t (|x_1 - x_2|)$
$T(X_1, X_2) = \tau$, since as soon as $X_1$ and $X_2$ coincide on the hyperplane, they will stick together, so we just need to look at the first time $X_1$ hits the hyperplane, which is given by $\tau$, and thus they share the same CDF.