Exponential decay of tail probability for Brownian stopping time A previous question, Exit Time of an Interval Brownian Motion - Distribution, asked about the tail probabilities for exit times from a region $(-a,b)$. In particular, the question was about exponential decay of the tail probability:

Let $W_t$ be a Brownian motion, fix $-a<0<b$ and let $\tau_x=\mathrm{inf}(t\ge0:W_t=x)$.


Show there is an $\alpha<1$: $P(\tau_{-a} \wedge \tau_b>n )\le \alpha^n$ for all $n \in \mathbb{N}$.

The accepted answer showed that this was indeed the case, and the resulting $\alpha$ depended implicitly on $a$ and $b$. However, I would like a more explicit form for $\alpha$'s dependence on $a$ and $b$. In particular, my question is:
Does $P(\tau_{-a} \wedge \tau_b>t )\le e^{-ct/(a+b)}$ for some constant $c$? If no, is there another way to express a simple bound showing both exponential decay in $t$ and some dependence on $a$ and $b$?
 A: You can actually compute the moment generating function of that random variable with some cleverness using an appropriate martingale. This computation is one I saw in "Continuous Exponential Martingales and BMO" by Kazamaki. It is Lemma 1.3 in Chapter 1.2 and is currently available on Springer's website. The idea is recorded there as being due to Leipngle. I'll just comment that the relevant local martingale is
$$
M_t^{\theta} = e^{\frac{1}{2}\theta^2t}\cos\left(\theta B_t - \frac{\theta(b-a)}{2}\right).
$$
One has to do a little work to show that optional stopping can be applied, but I'll skip that and refer you to the book for the relevant estimates if you need them. The end result is that for $0 \leq \theta < \frac{\pi}{a+b}$,
$$
\mathbb{E}[e^{\frac{1}{2}\theta^2 \tau_{-a}\wedge \tau_b}] = \frac{\cos\left(\frac{a-b}{2} \theta \right)}{\cos\left(\frac{a+b}{2} \theta\right)}
$$
Now we can use Markov's inequality:
$$
\mathbb{P}(\tau_{-a}\wedge \tau_b > t) \leq \frac{\mathbb{E}\left[e^{\theta^2 \tau_{-a} \wedge \tau_b}\right]}{e^{\frac{1}{2}\theta^2 t}} = \frac{\cos\left(\frac{a-b}{2} \theta \right)}{\cos\left(\frac{a+b}{2} \theta\right)} e^{-\frac{1}{2}\theta^2 t}
$$
Choosing for example $\theta = \frac{\pi}{2(a+b)}$, and noticing that the numerator is non-negative and less than or equal to $1$ we obtain
$$
\mathbb{P}(\tau_{-a}\wedge \tau_b > t) \leq \sqrt{2}e^{-\frac{\pi^2 t}{8(a+b)^2}}.
$$
