Is this stopping time finite a.s ? bounded? Let $B$ and $W$ be two independent Brownian motions on $(\mathcal{F}_{t})$. Is $\tau$ defined as:
$$\tau = \inf \{ t \ge 0: B_{t} \ge W_{t} + e^{-t} \}$$ almost surely finite ? Bounded ?
I would say that it can not be finite due to Brownian motion and therefore can't be bounded. But I don't know how to prove it.
What do you think guys ?

The proof in the other answer shows that $\tau$ is not bounded, but we actually do have that $\tau < \infty$ almost surely.  To show that, first note that by Levy's characterization of Brownian motion, $\hat B_t := \frac{1}{\sqrt 2} (B_t - W_t)$ is a Brownian motion.  Then $\tau = \inf\{ t : B_t \ge W_t + e^{-t} \} = \inf\{ t : \hat B_t \ge \frac{1}{\sqrt 2} e^{-t}\}$, so we have
\begin{align*}
P(\tau = \infty) &= P(\hat B_t < \frac{1}{\sqrt{2}} e^{-t} \text{ for all }t \ge 0) \\
&\le P(\hat B_t < \frac{1}{\sqrt{2}} \text{ for all }t \ge 0). 
\end{align*}
By the recurrence property of Brownian motion, this final probability is $0$.
