# Expected hitting time of given level by Brownian motion

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!

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.

• 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!:) Oct 20, 2013 at 8:26
• @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 $t=\infty$. 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 $\tau$,

$$\mathbb{E}B_{\tau}=0$$

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

$$\mathbb{E}B_{\tau_a}=a$$

• 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. Oct 20, 2013 at 8:24
• @Vol_Smile You are welcome. Don't hesitate to ask if you don't get along with my answer.
– saz
Oct 20, 2013 at 8:36