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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".

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    $\begingroup$ Why would your $\tau$ have expectation $\infty$? Wouldn't it be dominated by the expected time until $X_t$ leaves $[-1,1]$? $\endgroup$
    – Paul Tupper
    Commented Mar 14, 2012 at 17:25
  • $\begingroup$ I meant the 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 $c\neq 0$. $\endgroup$
    – Jochen Wengenroth
    Commented Mar 15, 2012 at 7:44
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    $\begingroup$ As you must know, this is equivalent to computing $u(0,0)$ where $u$ solves the Poisson equation $\frac{1}{2} \Delta u(x,y)=-1$ inside a rhombus centred at the origin. $\endgroup$
    – Alekk
    Commented Mar 19, 2012 at 10:44
  • $\begingroup$ The "modified" stopping time has finite expected value for some angles (I believe angles less than $\frac{\pi}{2}$, or negative correlation) and infinite expected value for some angles including $\frac{\pi}{2}$. $\endgroup$
    – Douglas Zare
    Commented Sep 7, 2012 at 6:34

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