Joint Exit-Side and Exit-Time Transforms

What the Joint Transform Measures

When Brownian motion with drift leaves a bounded interval, two things matter: which endpoint it reaches, and how long it takes to get there. The exit probability records only the first piece of information. A discount factor $e^{-\lambda\tau}$ also assigns less weight to exits that occur later. The resulting quantity is a joint exit-side and exit-time transform.

Let $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carry a standard Brownian motion $B_t$. For a starting point $x\in(l,u)$, let $X_t=x+\mu t+\sigma B_t$, where $l<u$, $\mu\in\mathbb{R}$, and $\sigma>0$. Write $\mathbb{P}_x$ and $\mathbb{E}_x$ for probability and expectation when $X_0=x$. Define the first exit time $\tau=\inf\{t\ge0:X_t\notin(l,u)\}$. Since the paths are continuous, on $\{\tau<\infty\}$ the exit location is either $l$ or $u$. We take $\lambda\ge0$, so the discount factor is bounded by one.

Definition: The discounted upper-exit transform is $F(x)=\mathbb{E}_x[e^{-\lambda\tau}\mathbf{1}_{\{X_\tau=u\}}]$. Here $\mathbf{1}_{\{X_\tau=u\}}$ is the indicator of the event that the process exits at the upper endpoint. The exit time is finite almost surely, as shown below.

The interval is bounded, which ensures exit in finite time. To see this, fix $h>0$ and choose $c>u-l+|\mu|h$. If the process has not exited by time $nh$, then it is still inside the interval at time $nh$. An increment over the next time block satisfying $\sigma(B_{(n+1)h}-B_{nh})>c$ forces it above $u$ by time $(n+1)h$, unless it has exited earlier. This increment event has probability $p=\mathbb{P}(\sigma B_h>c)>0$, independent of $\mathcal{F}_{nh}$. Consequently, $\mathbb{P}_x(\tau>(n+1)h)\le(1-p)\mathbb{P}_x(\tau>nh)$, and hence $\mathbb{P}_x(\tau>nh)\le(1-p)^n$. This tends to zero.

The Boundary-Value Problem

In the interior, the drift coefficient of $X_t$ is $\mu$, and its diffusion coefficient is $\sigma$. Its generator on twice continuously differentiable functions is $\mathcal{L}f=\mu f'+\tfrac12\sigma^2f''$. The transform has payoff one at the upper endpoint and zero at the lower endpoint. Discounting at rate $\lambda$ leads to the classical boundary-value problem

\frac{\sigma^2}{2}F''(x)+\mu F'(x)-\lambda F(x)=0\quad(l<x<u), \qquad F(l)=0,\qquad F(u)=1.

The equation expresses the requirement that discounting balances the local drift of the candidate value. The boundary values express the exit payoff. More precisely, a twice continuously differentiable solution of this problem is the transform: the verification below proves that assertion directly, without assuming a stochastic representation theorem.

Theorem: For $\lambda\ge0$, $\sigma>0$, and $\mu\in\mathbb{R}$, the boundary-value problem above has the solution $$ F(x)= \begin{cases} \displaystyle e^{\mu(L-y)/\sigma^2} \frac{\sinh(qy/\sigma^2)}{\sinh(qL/\sigma^2)},&q>0,\\[1.2ex] \displaystyle \frac{y}{L},&q=0, \end{cases} \qquad y=x-l,\quad L=u-l,\quad q=\sqrt{\mu^2+2\lambda\sigma^2}. $$ This solution equals $\mathbb{E}_x[e^{-\lambda\tau}\mathbf{1}_{\{X_\tau=u\}}]$.

Proof. Set $y=x-l$, so $0<y<L$, and write $f(y)=F(l+y)$. The equation becomes $\tfrac12\sigma^2 f''+\mu f'-\lambda f=0$, with $f(0)=0$ and $f(L)=1$. Put $a=\mu/\sigma^2$ and set $f(y)=e^{-ay}g(y)$. Differentiating gives $f'=e^{-ay}(g'-ag)$ and $f''=e^{-ay}(g''-2ag'+a^2g)$. Substitution cancels the $g'$ terms and yields $g''=(q^2/\sigma^4)g$, where $q^2=\mu^2+2\lambda\sigma^2$.

If $q>0$, the solutions of this constant-coefficient equation are linear combinations of $\sinh(qy/\sigma^2)$ and $\cosh(qy/\sigma^2)$. The condition $g(0)=0$ removes the $\cosh$ term. Since $g(L)=e^{aL}$, the remaining constant is $e^{aL}/\sinh(qL/\sigma^2)$. Therefore $f(y)=e^{a(L-y)}\sinh(qy/\sigma^2)/\sinh(qL/\sigma^2)$. At $y=0$ this is zero; at $y=L$ it is one. If $q=0$, then $\mu=0$ and $\lambda=0$, and the equation reduces to $g''=0$. Its boundary conditions give $f(y)=g(y)=y/L$. These forms also show uniqueness: every solution of the differential equation has the stated general form, and each pair of boundary values fixes its constants.

It remains to verify the probabilistic representation. The candidate $F$ is smooth on the closed interval, so $F$ and $F'$ are bounded there. Apply Itô’s formula, including its second-derivative correction, to $e^{-\lambda t}F(X_t)$ stopped at $\tau$. For $f(t,z)=e^{-\lambda t}F(z)$, its derivatives are $f_t=-\lambda e^{-\lambda t}F(z)$, $f_z=e^{-\lambda t}F'(z)$, and $f_{zz}=e^{-\lambda t}F''(z)$. The correction term is $\tfrac12\sigma^2 e^{-\lambda t}F''(X_t)\,dt$, because $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$. Thus, for each finite $t$,

e^{-\lambda(t\wedge\tau)}F(X_{t\wedge\tau}) =F(x)+\int_0^{t\wedge\tau}e^{-\lambda s} \left(\mu F'(X_s)+\frac{\sigma^2}{2}F''(X_s)-\lambda F(X_s)\right)ds +\int_0^{t\wedge\tau}e^{-\lambda s}\sigma F'(X_s)\,dB_s.

The ordinary integral is zero by the boundary-value equation. For the stochastic integral, its integrand $H_s=e^{-\lambda s}\sigma F'(X_s)\mathbf{1}_{\{s\le\tau\}}$ is progressively measurable and satisfies $\mathbb{E}_x\int_0^t H_s^2\,ds\le \sigma^2\sup_{z\in[l,u]}|F'(z)|^2t<\infty$. It belongs to $\mathcal{L}^2$ on $[0,t]$. The Itô integral is therefore a true martingale: its increments have conditional expectation zero given the past, and it is integrable by the Itô isometry. Taking expectations gives $\mathbb{E}_x[e^{-\lambda(t\wedge\tau)}F(X_{t\wedge\tau})]=F(x)$.

As $t\to\infty$, $\tau<\infty$ almost surely, and the stopped expression converges to $e^{-\lambda\tau}F(X_\tau)$. Its absolute value is bounded by $\sup_{z\in[l,u]}|F(z)|$, so dominated convergence applies. Since $F(l)=0$ and $F(u)=1$, the limit is $e^{-\lambda\tau}\mathbf{1}_{\{X_\tau=u\}}$. Taking expectations proves the representation. $\square$

Worked Transforms and Checks

Worked Example: Zero Drift with Positive Discounting

Take $l=0$, $u=2$, $x=1$, $\mu=0$, $\sigma=1$, and $\lambda=\tfrac12$. Then $L=2$, $y=1$, and $q=\sqrt{0+2(\tfrac12)(1)}=1$. The formula gives

F(1)=\frac{\sinh(1)}{\sinh(2)} =\frac{\sinh(1)}{2\sinh(1)\cosh(1)} =\frac{1}{2\cosh(1)}.

Check: at $\lambda=0$, the $q=0$ formula gives the upper-exit probability $y/L=1/2$. For the stated positive discount, $\cosh(1)>1$, so $1/(2\cosh(1))<1/2$, as discounting requires.

Worked Example: An Upper-Exit Probability with Drift

Take $l=0$, $u=1$, $x=\tfrac12$, $\mu=1$, $\sigma=1$, and $\lambda=0$. Then $L=1$, $y=\tfrac12$, and $q=\sqrt{1}=1$. Hence

F\left(\frac12\right) =e^{1/2}\frac{\sinh(1/2)}{\sinh(1)} =e^{1/2}\frac{\sinh(1/2)}{2\sinh(1/2)\cosh(1/2)} =\frac{e^{1/2}}{2\cosh(1/2)} =\frac{e}{e+1}.

In the last equality, $2\cosh(1/2)=e^{1/2}+e^{-1/2}$, and multiplying numerator and denominator by $e^{1/2}$ gives $e/(e+1)$. Check: this is greater than $1/2$, since $e>1$, consistent with positive drift favouring the upper endpoint. The factor $e^{1/2}$ is essential: omitting it would give the wrong boundary value at $u=1$.

Worked Example: Drift and Discounting Together

Take $l=0$, $u=2$, $x=1$, $\mu=1$, $\sigma=2$, and $\lambda=\tfrac38$. Here $L=2$, $y=1$, $\sigma^2=4$, and $q=\sqrt{1+2(\tfrac38)(4)}=\sqrt4=2$. Therefore

F(1)=e^{(1)(2-1)/4} \frac{\sinh(2\cdot1/4)}{\sinh(2\cdot2/4)} =e^{1/4}\frac{\sinh(1/2)}{\sinh(1)}.

Check: replacing $y=1$ by $y=0$ makes the numerator $\sinh(0)=0$, while replacing $y$ by $L=2$ gives $e^0\sinh(1)/\sinh(1)=1$. Both boundary values are correct, including the exponential prefactor.

Common Mistakes and Interpretation

A frequent error is to solve the transformed equation for $g$ and then forget that $f(y)=e^{-ay}g(y)$ must equal one at $y=L$. Choosing $g(y)=\sinh(qy/\sigma^2)/\sinh(qL/\sigma^2)$ gives $f(L)=e^{-aL}$, not one. The correction is to impose $g(L)=e^{aL}$, producing the factor $e^{a(L-y)}$ in the formula. Checking both endpoints after solving is a simple way to catch this mistake.

The transform is not just an exit probability when $\lambda>0$. Because $0\le e^{-\lambda\tau}\le1$, it is no larger than the probability of upper exit; the inequality is strict whenever upper exit has positive probability and $\lambda>0$, since then $\tau>0$ almost surely. At $\lambda=0$, discounting disappears and the transform is exactly the upper-exit probability. The special case $\mu=\lambda=0$ must be handled by $F(x)=(x-l)/(u-l)$, rather than by substituting $q=0$ into a quotient of hyperbolic sines.

Feature	Role in the problem
$\mu$	Drift coefficient; it biases motion toward one endpoint.
$\sigma$	Diffusion coefficient; it controls the Brownian fluctuations.
$\lambda$	Discount rate; it penalizes later exits.
$F(l),F(u)$	Boundary payoffs: zero at the lower exit and one at the upper exit.

Check Your Understanding

Use the boundary-value equation and the transform formula to check your reasoning.

For $\mu=0$, $\sigma=1$, $\lambda=\tfrac12$, $l=0$, $u=4$, and $x=2$, write the transform in hyperbolic-function form.
What does the transform become when $\lambda=0$, $\mu=0$, and $x$ is the midpoint of the interval?
For positive drift and zero discount, explain from the formula why the upper-exit probability at the midpoint exceeds one half.
In the stopped Itô verification, why is the stochastic integral a true martingale on each finite time interval?
Which boundary condition fails if the exponential prefactor is omitted from the formula when $\mu\ne0$?

Feature	Role in the problem
\(\mu\)	Drift coefficient; it biases motion toward one endpoint.
\(\sigma\)	Diffusion coefficient; it controls the Brownian fluctuations.
\(\lambda\)	Discount rate; it penalizes later exits.
\(F(l),F(u)\)	Boundary payoffs: zero at the lower exit and one at the upper exit.

Joint Exit-Side and Exit-Time Transforms

What You'll Learn