Consider a filtered probability space in which there exist two independent Brownian motions $W$ and $B$. For every progressively measurable process $u=(u_t)_{t\ge 0}$ taking values in $[0,1]$, define two processes $(X^u, Y^u)$
$$X^u_t :=1 +\int_0^t u_s ds +W_t,\quad Y^u_t :=1 +\int_0^t (1-u_s) ds +B_t,\qquad \forall t\ge 0.$$ If $u$ is given by the Atlas-model, we denote $(X^*, Y^*)$, i.e., $$X^*_t :=1 +\int_0^t {\bf 1}_{\{X^*_t\le Y^*_t\}} ds +W_t,\quad Y^*_t :=1 +\int_0^t{\bf 1}_{\{X^*_t>Y^*_t\}} ds +B_t,\qquad \forall t\ge 0.$$ As a SDE, the above one is well defined. My question is whether one has the following path-wise inequality : for every progressively measurable $u$ taking values in $[0,1]$
$$|X^*_t-Y^*_t| \le |X^u_t-Y^u_t|, \qquad \forall t\le \tau?$$
Here $\tau:=\min(\tau^u,\tau^*)$ is the stopping time, where
$$\tau^u:=\inf\{t\ge 0: \min(X^u_t,Y^u_t)\le 0\},\quad \tau^*:=\inf\{t\ge 0: \min(X^*_t,Y^*_t)\le 0\}.$$
Note that $X^*_t+Y^*_t=X^u_t+Y^u_t$ for all $t\ge 0$. Many thanks for your help!
