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!
Best Answer
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.