Let $B=(X,Y)$ be a correlated two-dimensional Brownian motion, that is, the components are standard Brownian motions and the covariance between $X_t$ and $Y_t$ is $t\rho$ for some constant $\rho \in [-1,1]$. The stopping time I am interested in is $$\tau = \inf \lbrace t>0: |X_t|=1 \text{ or } |Y_t|=1\rbrace.$$

Is there a closed formula for the expectation $E(\tau)$?

Without the absolute values in the definition of $\tau$ the distribution of $\tau$ has been calculated e.g. by Adam Metzler in
*Statistics and Probability Letters 80 (2010) 277–284* (for me, the formulas are terrifying).
However, the *modified* $\tau$ probably has expectation $\infty$.

## Edit

Having looked a bit at probability books, in paticular Kallenberg's *Foundations of Modern Probability* (2nd ed, chapter 24), I found the following:

Using a linear transformation $T:\mathbb R^2\to\mathbb R^2$ one reduces to an uncorrelated BM and a parallelogram $D=D_\rho$ instead of a square. If $g:D\times D\to\mathbb R$ is the Green function of $D$ and $\tau_x=\inf\lbrace t>0: x+ B_t\notin D\rbrace$ then (by Kallenberg, page 477 with $f=1$, originally this is probably due to Hunt) we get $$ E(\tau_x)= \int g(x,y) dy.$$ In principle, it would thus be enough (but it might be overkill) to know the Green function of a parallelogram. Perhaps this can be calculated using Schwarz-Christoffel formulas. Nevertheless, I do not see to what kind of formula for $E(\tau)=E(\tau_0)=f(\rho)$ this may lead.

A rather simple observation is $f(1)=f(-1)=2$ since a perfectly correlated two-dimensional BM is a one-dimensional BM "living on a diagonal".

modified$\tau$ (without absolute values). The reason for this is that for a one-dimensional Brownian motion $X$ one has $E(\inf\lbrace t>0: X_t=c\rbrace)=\infty$ for every $c\neq 0$. – Jochen Wengenroth Mar 15 '12 at 7:44