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!
 A: 
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.
A: 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,
$$\mathbb{E}B_{\tau_a}=a$$
