Suppose $W_t$ is Brownian motion and consider the following two stopping times: $$\tau_a \equiv \inf \{t \ge 0 : W_t + at \ge b(t) \} \wedge T$$ and $$\tau_{-a} \equiv \inf\{t \ge 0: W_t - at \ge b(t)\} \wedge T$$ for some $T > 0$, $a > 0$, and an arbitrary (strictly positive) boundary $b(\cdot)$. Obviously $\tau_a \leq \tau_{-a}$ by construction. Is it true that $$\frac{E(\tau_{-a})}{E(\tau_{a})} \leq C$$ for some constant $C$?
I have tried to prove it in the following way:
We convert to a measure (denoted by $\widetilde{E}$) where $B_t \equiv W_t - 2at$ is Brownian motion via Girsanov's theorem. In particular, we have, denoting the stochastic exponential by $\mathcal{E}$, $$E(\tau_{-a}) = \widetilde{E}(\tau_{-a} \mathcal{E}(-2aB_{\tau_{-a}}))$$
Noting that $\tau_{-a} = \inf\{t \ge 0 : B_t + at \ge b(t)\}$, we see that $(B_{\tau_{-a}},\tau_{-a})$ has the same law under the measure denoted $\widetilde{E}$ as $(W_{\tau_a}, \tau_{a})$ has under the original measure denoted $E$. Hence, it suffices to bound the following: $$\frac{E(\mathcal{E}(-2aW_{\tau_a}) \tau_{a})}{E(\tau_{a})}.$$ Is this possible?