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.
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,
$$\mathbb{E}B_{\tau_a}=a$$
This question is a bit old, but here's another simple answer. $M_t = B_t^2 - t$ is a martingale, and let's stop it when it first hits $a>0$ or $-b<0$. Let $\tau=\tau_{a,-b}$ be this stopping time. Then, applying the optional stopping theorem, we get that
$$ E[M_{\tau \wedge N}] = E[M_0] = 0, $$
where $N>0$. This implies that $E[B_{\tau \wedge N}^2] = E[\tau \wedge N]$, and we can let $N \to \infty$. Applying the DCT on the left, and the MCT on the right, we get $E[B_{\tau}^2] = E[\tau]$. Now, what is $E[B_{\tau}^2]$? By using the fact that $B_t$ is a martingale, I'll leave it as an exercise that $E[B_{\tau}^2]=ab$, and thus $E[\tau]=ab$. Now let $b \to +\infty$ and we see that the expected time to hit $a$ is infinite.
$$\mathbb{P}(t(a)=t)= \mathbb{P}(s(a)=t\cap W_t=a) + \mathbb{P}\left(s(a)=h\cap W_h=-a \cap (r(a)+t(a))\leq (t-h):h<t \right) $$
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 $t=\infty$.
– Jan Stuller Jul 20 '20 at 11:13