Exist $\alpha < \infty$, $\beta > 0$ such that $\mathbb{P}\{T_\lambda > t\} \le \alpha e^{-\beta t}?$

Question

Let $B_t$ be a standard one-dimensional Brownian motion. Suppose $\lambda > 0$ and let$$T_\lambda = \min\{t : |B_t| = \lambda\}.$$Do there exist $\alpha < \infty$ and $\beta > 0$ (which may depend on $\lambda$) such that for all $t$, we have$$\mathbb{P}\{T_\lambda > t\} \le \alpha e^{-\beta t}?$$

Answer 1

Using that $$M_t := \exp(i \xi B_t + \frac{t}{2} \xi^2)$$ is a martingale, one can show that for any $\lambda>0$ there exists $\beta>0$ such that $$\mathbb{E}e^{\beta T_{\lambda}}<\infty.$$ By Markov's inequality, this implies in particular that

$$\mathbb{P}(T_{\lambda} \geq t) \leq e^{-\beta t} \underbrace{\mathbb{E}(e^{\beta T_{\lambda}})}_{\alpha}.$$

(See e.g. René Schilling, Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Problem 5.17(i) for a proof.)

Remark: As @zhoraster pointed out, one can calculate $\mathbb{E}e^{\beta T_{\lambda}}$ explicitly and find that the expectation is finite for any $\beta \in (0,\frac{\pi^2}{8\lambda})$.