Suppose we have 2 geometric brownian motions, $(X^{(1)}_t)_{t \geq 0}$ and $(X^{(2)}_t)_{t \geq 0}$ driven by the same Brownian motion and started at the same position. Suppose they have the same volatility $\sigma$ but the drifts are such that $\mu_2 \geq \mu_1$. Then it is not hard to show that $\mathbb{P}[X^{(1)}_t \leq X^{(2)}_t, \forall t \geq 0] = 1.$ So one process dominates the other.
Then, does that imply for any stopping time (w.r.t. the filtration generated by the Brownian motion) $\tau$:
a)$\mathbb{P}[X^{(1)}_\tau \leq X^{(2)}_{\tau}] = 1$?
b)$\mathbb{E}[X^{(1)}_\tau] \leq \mathbb{E}[X^{(2)}_{\tau}]$?
Of course a) implies b) but I do not know if a) is true.