A standard computation using martingale techniques allows us to compute probability that a Brownian motion started at zero exits the interval $[-a,b]$ ($a, b > 0$) at $-a$ or $b$. It appears to me that if we replace $B_t$ by $B_{t \wedge \delta}$ for small $\delta$, then the argument used in that computation breaks down because we no longer know that the new martingale exits this region almost surely. This feels like a classical problem that should have been analyzed in detail at some point and I am curious if anyone happens to know the answer off-hand.

Formally, there are two related questions I am interested in: Are there explicit formulas for $P(B_{t\wedge \delta}$ exits $(-a,b))$ or $P(B_{t \wedge \delta}$ exits $(-a,b)$ at $-a$)?

  • $\begingroup$ I guess, that shall be some weak solution of the equation $f_t = \frac12f_{xx}$ with the terminal time and boundary conditions. $\endgroup$ – Ilya Jan 19 '13 at 22:49

The main idea is to consider these quantities simultaneously for every starting point $x$ and every time $t$. Let $T=\inf\{t\geqslant0\mid B_t\in\{-a,b\}\}$.

The first quantity is $u(\delta,0)$ where, for every $-a\leqslant x\leqslant b$ and every $t\geqslant0$, $$ u(x,t)=\mathbb P_x(T\leqslant t). $$ Likewise, the second quantity is $v(\delta,0)$ where, for every $x$ in $[-a,b]$ and $t\geqslant0$, $$ v(x,t)=\mathbb P_x(T\leqslant t,B_T=-a). $$ Each function $u$ and $v$ is the unique solution of some partial differential equation with boundary conditions on the domain $-a\leqslant x\leqslant b$, $t\geqslant0$. For example:

  • $u_t=\tfrac12u_{xx}$ for every $-a\lt x\lt b$ and every $t\gt0$,
  • $u(x,0)=0$ for every $-a\lt x\lt b$,
  • $u(-a,t)=1$ for every $t\geqslant0$,
  • $u(b,t)=1$ for every $t\geqslant0$.


  • $v_t=\tfrac12v_{xx}$ for every $-a\lt x\lt b$ and every $t\gt0$,
  • $v(x,0)=0$ for every $-a\lt x\lt b$,
  • $v(-a,t)=1$ for every $t\geqslant0$,
  • $v(b,t)=0$ for every $t\geqslant0$.

These differential equations guarantee that, when $t\to\infty$, $u(t,x)\to1$ and $v(t,x)\to\frac{b-x}{b+a}$ for every $-a\leqslant x\leqslant b$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.