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$B_t$ 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.

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  • $\begingroup$ 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. $\endgroup$ – Nate Eldredge Apr 24 '17 at 14:32
  • $\begingroup$ @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? $\endgroup$ – Siskaa Apr 24 '17 at 15:59
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    $\begingroup$ 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. $\endgroup$ – Nate Eldredge Apr 24 '17 at 17:23
  • $\begingroup$ @NateEldredge That's right. Thank you very much! $\endgroup$ – Siskaa Apr 24 '17 at 17:46

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