0

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 $W_{t} \sim \sqrt{2tlnlnt}$.

Guesttt
  • 183
  • Well, it's not true that $W_t \approx \sqrt{2t\log(\log(t))}$, that holds only for the limes superior, assuming that $W_t$ is standard Brownian motion, which you should explain in your question. – user159517 Jan 08 '19 at 22:51
  • Also, the title of your question asks to show that the stopping time is bounded, while in the body of your question you ask to show that the stopping time is almost surely finite. – user159517 Jan 08 '19 at 22:55

2 Answers2

1

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.

Nate Eldredge
  • 101,664
0

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 $P\{\tau_a <\infty\}=1$.

Kavi Rama Murthy
  • 359,332