Let $$\tau_{a} = \inf\{t>0 : W_{t} + at = 5\}.$$ Prove that $\mathbb{P}(\tau_{a}<\infty) = 1$ for $a\ge0.$
My solution:
We know that $W_{0} +a*0 < 5$. Furthermore, because $W_{t} \sim \sqrt{2tlnlnt}$, we can say that $W_{t} + at \xrightarrow{t \rightarrow\infty}\infty$. And that is why $\mathbb{P}(\tau_{a}<\infty) = 1.$
My question is whether it can be solved like this. I'm not sure about using $W_{t} \sim \sqrt{2tlnlnt}$.
On
Brownian motion is recurrent, so almost surely there exists $t$ with $B_t \ge 5$. That is, for almost every $\omega$, there is $t < \infty$, depending on $\omega$, such that $B_t(\omega) \ge 5$. Since $a \ge 0$, we have $B_t(\omega) + at \ge B_t(\omega) \ge 5$. So by the intermediate value theorem, there is $s \le t$ with $B_t(\omega) + at =5$. Then $\tau_a(\omega) \le s$. So $\tau_a(\omega) < \infty$, and this is true for almost every $\omega$.
Intuitively, $B_t$ is a.s. going to cross the level 5, and with the positive drift, $B_t + at$ is going to cross it even sooner.
LIL says $\lim \sup _{t \to \infty} \frac {W_t} {\sqrt {2t\ln\, \ln\, t}}=1$ almost surely. This implies that $\lim \sup_{t \to \infty} W_t= \infty$ almost surely. Hence, $\lim \sup_{t \to \infty} (W_t+at)= \infty$ almost surely. From this and IVP you get $P\{\tau_a <\infty\}=1$.