Let $\{B_t\}_{t \ge 0}$ a Brownian Motion, prove that for all $a \gt 0$: $P(B_t \ge a$ for some $t \ge 0 )=1$

Question

Let $\{B_t\}_{t \ge 0}$ a standard Brownian Motion, prove that for all $a \gt 0$:

$P(B_t \ge a$ for some $t \ge 0 )=1$

Use the principle of reflection

Thoughts:

The reflection principle is:

Let $\{ B_t^a: t \gt 0\}$ a standard Brownian Motion which starts in a, if $a \lt b$ then for all $t \ge0:$

$P( B_s^a \ge b$ for some $s \in [0,t])=2P(B_t^a \ge b)$

I do not know how to start, I was thinking take $s \in [0, \infty)$ but I don't think it's correct because s must be in a closed interval.

Any ideas?

Answer 1

Do not get confused between the $a$ in the title and the $a$ in the reflection principle. Take $a=0$ in the reflection principle. Note that $2P(B_t \geq b)=2P(X \geq b/\sqrt t) \to 1$ as $ t \to \infty$ (where $X$ is a standard normal variable). Hence $P(B_s \geq b \, \text{for some} \, s\leq t) \to 1$ as $t \to \infty$ which is what we want to prove.