# first hitting time of drift brownian motion is stopping time?

$B_t$ be an ($\mathscr{F}_t$)-Brownian motion started from $0$. $a>0$ and define $$\sigma_a = \inf \{t \geq 0 : B_t \leq t-a \}.$$

I want to show that $\sigma_a$ is a stopping time and that $\sigma_a < \infty$ a.s.

For prove second part, I think that I can use $\frac{B_t -t}{t} \to -1$ a.s. is it right?

Also I need some help for first part... Thanks.

• Do you know how to show that the hitting time of an ordinary Brownian motion is a stopping time? The proof is almost exactly the same. – Nate Eldredge Apr 24 '17 at 14:32
• @NateEldredge $B_t − t$ is measurable wrt $F_t$ , right? Then {σa≤t} = min{ $B_s − s ≤a :0≤s≤t$} , so it should be a stopping time. Is this correct? – Siskaa Apr 24 '17 at 15:59
• Yes, that's basically the idea - except you should write $\{\min\{B_s - s : 0 \le s \le t\} \le a\}$, and you should justify that the min is also measurable with respect to $F_t$. You can also write the event as a countable intersection, taking advantage of the fact that Brownian motion is continuous and rationals are dense. – Nate Eldredge Apr 24 '17 at 17:23
• @NateEldredge That's right. Thank you very much! – Siskaa Apr 24 '17 at 17:46