Let $ B=(B_t)_{t\geq 0}$ be a Brownian Motion and $a \in \mathbb R$, and
$\tau_a:=\inf \{ t\geq 0 :B_t =a \} $ is a finite stopping time.
How can we prove that :
$$ \mathbb E(\tau_a)=\infty \;\;\text{ and } \;\;\limsup_{t \to \infty}B_t=+\infty\; a.s $$ $ \;\;\liminf_{t \to \infty}B_t=-\infty\;$ a.s
Thanks in advance for any help!
Best Answer
In this question you find two proofs that $\mathbb{E}(\tau_a)=\infty$. For the limiting behaviour of Brownian motion, note that $(-B_t)_{t \geq 0}$ is also a Brownian motion, and therefore it suffices to show that $$\limsup_{t \to \infty} B_t = \infty \quad \text{a.s.}\tag{1}$$ This is proved here using martingale methods. Alternatively, you can also use the reflection principle. Since Brownian motion has continuous sample paths (hence bounded on compact time intervals), it follows that $(1)$ is equivalent to saying that $\mathbb{P}(\tau_a<\infty)=1$ for all $a>0$. By the reflection principle,
$$\mathbb{P}(\tau_a \leq t) = \mathbb{P} \left( \sup_{s \leq t} B_s \geq a \right) = \mathbb{P}(|B_t| \geq a) = \mathbb{P}(\sqrt{t} |B_1| \geq a),$$
and so
$$\mathbb{P}(\tau_a < \infty) = \lim_{t \to \infty} \mathbb{P}(\tau_a \leq t) = \lim_{t \to \infty} \mathbb{P}(|B_1| \geq a/\sqrt{t})=1$$ for all $a>0$.