I've been looking at this for some time now and still have no sensible solutions, can somebody help me out please.

Say I define the stopping time of a Brownian motion as followed: $$\tau(a) = \min (t \geq 0 : W(t) \geq a)$$ (first time the random process hits level $a$)

Now, how do I go about compute $E[\tau(a)]$ - the expected stopping time?

Can someone please give me some clues? Thanks!


2 Answers 2


The expected hitting time of $a$ by a Brownian motion starting from $0$ is infinite.

Here is an elementary proof. Let $t(a)$ and $s(a)$ denote the expected hitting times of $a$ and of $\{-a,+a\}$ by a Brownian motion starting from $0$.

At the first hitting time of $\{-a,+a\}$, the Brownian motion is uniformly distributed on $\{-a,a\}$. That one can hit $\{-a,+a\}$ at $-a$ rather than $a$ (with probability $\frac12$) is the reason why $t(a)\gt s(a)$. Which amount of time should one add to reach $a$ in this case? Let $r(a)$ denote the expected hitting time of $0$ by a Brownian motion starting from $-a$. Starting from $-a$, the expected hitting time of $a$ is the sum of $r(a)$ (to hit $0$ again) and $t(a)$ (to hit $a$ starting from $0$). Thus, $$ t(a)=s(a)+\tfrac12(r(a)+t(a)). $$ By space homogeneity, $r(a)=t(a)$ hence $t(a)=s(a)+t(a)$. Since $s(a)\gt0$, this equation has exactly one solution in $[0,+\infty]$, which is $t(a)=+\infty$.

This uses the strong Markov property of Brownian motion (several times) and its invariance by the translations $x\mapsto x+c$ and by the symmetry $x\mapsto-x$.

This approach can be adapted to every Brownian motion with drift since one looses only the invariance by the symmetry $x\mapsto-x$. Considering $p=P_0[\text{hits}\ a\ \text{before}\ -a]$, one gets $$t(a)=s(a)+(1-p)(r(a)+t(a))=s(a)+2(1-p)t(a). $$ If the drift is positive, then $p\gt\frac12$ hence $t(a)=s(a)/(2p-1)$ is finite. If the drift is nonpositive, then $p\leqslant\frac12$ hence $t(a)$ is infinite.

  • $\begingroup$ Thanks Did, that was very helpful. I just started studying Brownian Motion and I think I need to spend some time getting familiar with all the terminologies and concepts. Your answer is appreciated!:) $\endgroup$
    – Vol_Smile
    Oct 20, 2013 at 8:26
  • $\begingroup$ @Did: I do not fully comprehend the equation $t(a)=s(a)+\tfrac12(r(a)+t(a))$, since there is 0.5 probability that $-a$ gets hit before $a$. Wouldn't it be easier to state your claim as: $$\mathbb{P}(t(a)=t)= \mathbb{P}(s(a)=t\cap W_t=a) + \mathbb{P}\left(s(a)=h\cap W_h=-a \cap (r(a)+t(a))\leq (t-h):h<t \right) $$ Above, we can then say that by symmetry: $r(a) = t(a)$, so: $$\mathbb{P}(t(a)=t)= \mathbb{P}(s(a)=t\cap W_t=a) + \mathbb{P}\left(s(a)=h\cap W_h=-a \cap 2t(a)\leq (t-h):h<t \right) $$ The above statement is a logical contradiction unless $t=\infty$. $\endgroup$ Jul 20, 2020 at 11:13

Let $a \neq 0$ and define

$$\tau_a := \inf\{t>0; W(t) \geq a\} $$

First of all, we note that $\tau_a<\infty$ almost surely, since the Brownian motion has continuous sample paths and satisfies $$\limsup_{t \to \infty} W_t = \infty \qquad \qquad \liminf_{t \to \infty} W_t = -\infty$$

On the other hand, $\tau_a$ is not integrable, i.e. $\mathbb{E}\tau_a = \infty$. This is a direct consequence of Wald's identities (see e.g. René L. Schilling/Lothar Partzsch: Brownian motion - An Introduction to Stochastic Processes, pp. 55). They state in particular that for any integrable stopping time $\tau$,


Obviously, this is not satisfied for $\tau_a$ since, by the continuity of the sample paths,


  • $\begingroup$ Thanks Saz, let me mull over your answer, I'm new to Brownian Motion and need a little time to get up to the speed. $\endgroup$
    – Vol_Smile
    Oct 20, 2013 at 8:24
  • $\begingroup$ @Vol_Smile You are welcome. Don't hesitate to ask if you don't get along with my answer. $\endgroup$
    – saz
    Oct 20, 2013 at 8:36

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