Fixed-Time Brownian Motion Exit Probabilities

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 $-a$)?

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I guess, that shall be some weak solution of the equation $f_t = \frac12f_{xx}$ with the terminal time and boundary conditions. –  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 $t\geqslant0$.

Likewise:

• $v_t=\tfrac12v_{xx}$ 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 $-a\leqslant x\leqslant b$.

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