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Let $W$ be the standard Wiener process. The task is to establish for which $a$ and $b$ the stopping moment is finite almost surely. $$\sigma_{a,b}=inf\{t \geq 0\space W_t+at=b\}$$ I established that for a=0 and all b it is finite almost surely, but I am not sure on how to approach this drift. Thank you for your hints. Should I use some martingale (in the case of a=0 I used the fact that $W_t$ is a Martingale)?

Ryszard Eggink
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    Because of the symmetry of Brownian motion, we have $$\mathbb{P}(\sigma_{a,b} < \infty) = \mathbb{P}(\sigma_{-a,b}<\infty),$$ and therefore it is enough to determine $\mathbb{P}(\sigma_{a,b}<\infty)$ for $a>0$, $b \in \mathbb{R}$. For $b \geq 0$ you can use the inequality $$W_t + at \geq W_t$$ to obtain that $\sigma_{a,b} \leq \sigma_{0,b}$, and therefore $\sigma_{a,b}$ is finite almost surely. For $b<0$ see this question. – saz May 13 '19 at 10:49
  • Hi, thank you for your answer, one little question shouldn't it be for $a \in R$, $b>0$ instead of $a>0,b \in R$? – Ryszard Eggink May 13 '19 at 11:26
  • Well, actually it doesn't matter. But you are right that something in my previous comment was off; the first equation should read $$\mathbb{P}(\sigma_{a,b} < \infty) = \mathbb{P}(\sigma_{-a,-b} < \infty).$$ – saz May 13 '19 at 13:09

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