Let $T_x=\inf\{t>0:B(t)=x\}$, i.e $T_x$ is the first hitting time the Brownian motion $B(t)$ hits the point $x$.

With $B(0)=0$, the first time a standard Brownian motion escapes from strip $[a,b]$ is $T_{ab}=\min\{T_a,T_b\}$.

However I don't understand this definition. This is something necessary for the Brownian motion to escape, not a sufficient one, since $B(t)$ could hit $a$ or $b$ at some time, and then still remain in the strip...

Any help would be appreciated.

Edit: Based on Did's comment,

We want $P(\tau= 0) = 1$, where $\tau=\inf\{t>0:B(t)>0\}$ and his hint is to use $T_x=\inf \{t>0:B(t)=x\}$.


Knowing that $P(\tau = 0) = 1$ for a Standard Brownian Motion (SBM),and without loss of generality assuming that $T_a < T_b$, consider the process $B^*(s)=\{B(T_a+s)-B(T_a): s \geq 0\}$, which is also a SBM .

So $P(\tau=0)=1$ which is equivalent to $P(\inf\{t>0:B(T_a+t)>B(T_a)\}=0)=1$, i.e. exit time of $[b,a]$ which, with our assumptions, is $\inf\{ t>0: B(t)>a \}$ is a.s. $T_a$, which is also the exit time of $(b,a)$.

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    $\begingroup$ First, the hitting time of $\{a,b\}$ is also the escape time from the open interval $(a,b)$ (this is pure logic). Second, the escape time from $(a,b)$ is almost surely equal to the escape time from the closed interval $[a,b]$ (this is a result). $\endgroup$ – Did Jan 2 '18 at 20:05
  • $\begingroup$ @Did Thanks for the comment. I still have a doubt though... How can you say that those two escape times are a.s. equal? Any hints on how to prove that second assertion? $\endgroup$ – An old man in the sea. Jan 2 '18 at 21:50
  • $\begingroup$ This is based on the well known fact that, if $B(0)=0$, then $\tau=\inf\{t>0:B(t)>0\}$ is such that $P(\tau=0)=1$. To show this, one can use the explicit distribution of $T_x$ for $x>0$ and consider the limit $x\to0^+$ to show that $\lim\limits_{x\to0^+}T_x=0$ almost surely. $\endgroup$ – Did Jan 2 '18 at 23:03
  • $\begingroup$ @Did Thanks for the comments. I've edited my question based on your comments. Is the edit correct? Also, I'm using Resnick's Adventures in Stochastic Processes, and that fact is stated nowhere. What bibliography would you recommend for this? $\endgroup$ – An old man in the sea. Jan 3 '18 at 11:10
  • $\begingroup$ Of course, $P(T_x=0)=1$ is wrong for every $x\ne0$. $\endgroup$ – Did Jan 3 '18 at 14:16

Since Brownian motion has continuous sample paths, we know that the hitting time of $\{a,b\}$ equals the first exit time from the open interval $(a,b)$. It remains to show that the first exit time from $(a,b)$ equals almost surely the first exit time from $[a,b]$, and for this it is enough to prove that the stopping time

$$\tau := \inf\{t>0; B_t >0\}$$


$$\mathbb{P}(\tau=0)=1. \tag{1}$$

To this end, note that

$$\mathbb{P}(\tau>0) = \lim_{n \to \infty} \mathbb{P}(\tau>1/n).$$

By the very definition of $\tau$, we have $B_t(\omega) \leq 0$ for all $t \leq \tau(\omega)$ and so

$$\mathbb{P}(\tau>1/n) \leq \mathbb{P} \left( \sup_{t \leq 1/n} B_t = 0 \right).$$

The reflection principle implies that

$$\sup_{t \leq T} B_t \sim |B_T|$$

(see e.g. Brownian motion by Schilling & Partzsch, Chapter 6), and therefore

$$\mathbb{P}(\tau>1/n) \leq \mathbb{P}(|B_{1/n}|=0)=0$$

for all $n \in \mathbb{N}$. Thus,

$$\mathbb{P}(\tau>0) = \lim_{n \to \infty} \mathbb{P}(\tau>1/n)=0,$$

and this proves $(1)$.

  • $\begingroup$ Many thanks. I have to leave now, but as soon as I get home, and study your answer, I'll give you feedback. $\endgroup$ – An old man in the sea. Jan 3 '18 at 15:51
  • $\begingroup$ @Anoldmaninthesea. You are welcome. Don't worry; I'm not in a hurry. $\endgroup$ – saz Jan 3 '18 at 15:51
  • $\begingroup$ Saz, after studying your answer, I have two doubts. 1) Shouldn't we have $P(\tau>1/n)\leq P(\sup_{t\leq 1/n}B_t \leq 0)$ instead? 2) After proving that $P(\tau>0)=1$, how do we go about proving that the exit time for $[b,a]$ is the same as for $(b,a)$? Meaning, what I wrote in the question concerning this part, is it correct? $\endgroup$ – An old man in the sea. Jan 4 '18 at 14:13
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    $\begingroup$ @Anoldmaninthesea. 1) Note that $B_0=0$ and therefore $\sup_{t \leq T} B_t \geq B_0 = 0$ for any $T \geq 0$. In particular, $$\mathbb{P} \left( \sup_{t \leq T} B_t = 0 \right) = \mathbb{P} \left( \sup_{t \leq T} B_t \leq 0 \right).$$ 2) Yes, your reasoning concerning that part is correct. $\endgroup$ – saz Jan 4 '18 at 14:54
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    $\begingroup$ @Anoldmaninthesea. Yes, that's true. What do you mean by "[...] contradict the whole effort"? I don't see a contradiction... $\endgroup$ – saz Jan 4 '18 at 20:53

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