First-Passage Probabilities and Optimal Barriers for Brownian Motion with Drift
Let $X_t = \mu t + W_t$ where $W_t$ is a standard Brownian motion and $\mu$ is a constant drift. Two absorbing barriers sit at $-L < 0$ and $U > 0$. Define $\tau = \inf\{t \geq 0 : X_t \in \{-L, U\}\}$ as the first time the process hits either barrier.
- Compute $P(X_\tau = U)$, the probability of hitting the upper barrier before the lower.
- Compute $E[\tau]$, the expected time to absorption.
- Describe what happens in the limiting cases $\mu \to 0$ and $U, L \to \infty$.
- Suppose you can choose a single symmetric barrier $b = U = L > 0$ before trading begins. How should the optimal $b$ change with $|\mu|$ to balance hit probability against expected duration? Discuss the trade-off qualitatively and describe what the optimal $b$ looks like as $|\mu| \to 0$ and $|\mu| \to \infty$.
Hints
- Think about what function of $X_t$ kills the drift and produces a martingale -- the exponential family $e^{\lambda X_t}$ is the right place to start. Choose $\lambda$ so the $dt$ terms cancel.
- For $E[\tau]$, set up the ODE $\mu u'(x) + \frac{1}{2}u''(x) = -1$ from Dynkin's formula with boundary conditions $u(-L) = u(U) = 0$. The particular solution takes the form $-x/\mu$, and you solve for two constants using the boundary conditions.
- For the symmetric barrier trade-off, write the profit rate as $b(2p(b) - 1)/E_b[\tau]$ and check the limiting behavior as $b \to 0$ (coin-flip, zero edge) and $b \to \infty$ (drift dominates, profit rate approaches $\mu$). The interior maximum scales as $b^{*} \sim 1/|\mu|$ for small drift.
Worked Solution
How to Think About It: Think of $X_t$ as a position that drifts steadily in one direction while also jiggling randomly. You want to know: does the jiggle carry you to $U$ before the drift pulls you down to $-L$? The key tool is the exponential martingale $e^{-2\mu X_t}$ (when $\mu \neq 0$) -- it is a martingale under the original measure, so optional stopping pins down the exit probabilities exactly. For the expected duration, Dynkin's formula applied to $X_t^{2} - t$ gives you a clean closed form via an ODE. For the symmetric barrier question, think of it as a trade-off: a tight barrier ($b$ small) gets hit quickly but pays out a coin-flip; a wide barrier ($b$ large) gives the drift time to work but takes forever to resolve.
Quick Estimate: Take $\mu = 0.5$, $U = 2$, $L = 1$. With no randomness, the drift alone would carry you to $U = 2$ in 4 units of time and to $-L = -1$ in 2 units, so the drift clearly favors the upper barrier. Using the exit probability formula $P(\text{hit } U \text{ first}) = (1 - e^{2\mu L}) / (e^{-2\mu U} - e^{2\mu L})$ with $\mu = 0.5$, $U = 2$, $L = 1$: numerator $= 1 - e^{1} \approx -1.718$, denominator $= e^{-2} - e^{1} \approx 0.135 - 2.718 = -2.583$, so $P \approx 0.665$. That is about two-thirds, which feels right given positive drift favoring the upper barrier.
Approach: Use two martingales. First, $M_t = e^{-2\mu X_t}$ is a martingale (verify by Ito's lemma), giving the exit probabilities via optional stopping. Second, the generator $\mathcal{L} = \mu \partial_x + \frac{1}{2} \partial_{xx}$ applied to Dynkin's formula yields an ODE for $E[\tau]$.
Formal Solution:
*Part 1: Exit Probabilities*
When $\mu \neq 0$, apply Ito's lemma to $f(x) = e^{-2\mu x}$: $$df = -2\mu e^{-2\mu X_t} dX_t + \frac{1}{2}(4\mu^{2}) e^{-2\mu X_t} dt = e^{-2\mu X_t}(-2\mu \, dX_t + 2\mu^{2} \, dt)$$
Substituting $dX_t = \mu \, dt + dW_t$: $$df = e^{-2\mu X_t}(-2\mu(\mu \, dt + dW_t) + 2\mu^{2} \, dt) = e^{-2\mu X_t}(-2\mu \, dW_t)$$
So $M_t = e^{-2\mu X_t}$ is indeed a local martingale (and a true martingale on a bounded domain). By optional stopping at $\tau$: $$E[M_0] = E[M_\tau]$$ $$e^{0} = P(\text{hit } U) \cdot e^{-2\mu U} + P(\text{hit } -L) \cdot e^{2\mu L}$$
Let $p = P(X_\tau = U)$. Then $1-p = P(X_\tau = -L)$, so: $$1 = p \, e^{-2\mu U} + (1-p) e^{2\mu L}$$ $$\boxed{p = \frac{1 - e^{2\mu L}}{e^{-2\mu U} - e^{2\mu L}}}$$
When $\mu > 0$, the numerator is negative and the denominator is negative (since $e^{-2\mu U} < 1 < e^{2\mu L}$), so $p > 0$ as expected. When $\mu < 0$, drift pulls toward $-L$ and $p < 1/2$ for symmetric barriers.
*Part 2: Expected Hitting Time*
Using Dynkin's formula for the generator $\mathcal{L} = \mu \partial_x + \frac{1}{2} \partial_{xx}$, let $u(x) = E_x[\tau]$ where $E_x$ means starting from $x \in (-L, U)$. Then $u$ satisfies the ODE: $$\mu u'(x) + \frac{1}{2} u''(x) = -1, \quad u(-L) = u(U) = 0$$
The homogeneous solution is $C_1 + C_2 e^{-2\mu x}$. A particular solution: try $u_p = ax$. Then $\mu a = -1$, so $a = -1/\mu$. Thus: $$u(x) = C_1 + C_2 e^{-2\mu x} - \frac{x}{\mu}$$
Apply boundary conditions $u(-L) = 0$ and $u(U) = 0$: $$C_1 + C_2 e^{2\mu L} + \frac{L}{\mu} = 0$$ $$C_1 + C_2 e^{-2\mu U} - \frac{U}{\mu} = 0$$
Subtracting the second from the first: $$C_2(e^{2\mu L} - e^{-2\mu U}) = -\frac{L+U}{\mu}$$ $$C_2 = \frac{-(L+U)/\mu}{e^{2\mu L} - e^{-2\mu U}}$$
And $C_1 = \frac{U}{\mu} - C_2 e^{-2\mu U}$. The expected duration starting from $x = 0$ is: $$E[\tau] = u(0) = C_1 + C_2$$
Substituting and simplifying: $$\boxed{E[\tau] = \frac{1}{\mu}\left[U \cdot \frac{e^{2\mu L} - 1}{e^{2\mu L} - e^{-2\mu U}} - L \cdot \frac{1 - e^{-2\mu U}}{e^{2\mu L} - e^{-2\mu U}}\right]}$$
*Part 3: Limiting Cases*
Case $\mu \to 0$: Apply L'Hopital's rule (or Taylor-expand the exponentials) to recover the pure BM results: $$p \to \frac{L}{L+U}, \quad E[\tau] \to LU$$
This is the standard gambler's ruin for Brownian motion: you hit $U$ with probability $L/(L+U)$ and the expected time is $LU$. Both are purely geometric -- drift plays no role at the margin.
Case $U, L \to \infty$ with $\mu > 0$: As the barriers widen, drift dominates. With $\mu > 0$ and $L$ fixed, as $U \to \infty$: $p \to 1$ (you reach $+\infty$ before $-L$ almost surely when drift is positive and the upper barrier is arbitrarily far). The expected duration scales like $U/\mu$ (time to walk up to $U$ at drift rate $\mu$).
*Part 4: Optimal Symmetric Barrier*
With $U = L = b$, define $\alpha = 2\mu b$. Then: $$p(b) = \frac{1 - e^{\alpha}}{e^{-\alpha} - e^{\alpha}} = \frac{e^{\alpha} - 1}{2\sinh(\alpha)}$$
For $\mu > 0$: $p(b) > 1/2$ and increases toward 1 as $b \to \infty$ (drift wins if you wait long enough). Meanwhile $E[\tau]$ grows with $b$ (wider barriers take longer). The trade-off depends on what you are optimizing.
If you maximize the expected profit per unit time (e.g., for a trading strategy where profit $= b$ if you hit $+b$ and loss $= b$ if you hit $-b$), you want to maximize: $$\frac{b(2p(b) - 1)}{E_b[\tau]}$$
For $\mu > 0$, as $b \to 0$: $p \to 1/2$, so the numerator vanishes and profit rate goes to 0. As $b \to \infty$: $p \to 1$ and $E[\tau] \sim b/\mu$, so profit rate approaches $\mu$ (just the drift). There is an intermediate maximum.
The key qualitative result: when $|\mu|$ is large relative to the diffusion, barriers should be wide -- the drift works quickly and reliably, so you can afford to wait. When $|\mu| \approx 0$, barriers should be tight -- there is little drift working in your favor, so capture a quick profit before diffusion carries you the wrong way. Formally, the optimal $b^{*}$ scales as $1/|\mu|$ in the limit $|\mu| \to 0$, reflecting that the signal is weak and you need the diffusion to do most of the work.
Answer:
- $P(X_\tau = U) = \dfrac{1 - e^{2\mu L}}{e^{-2\mu U} - e^{2\mu L}}$
- $E[\tau] = \dfrac{1}{\mu}\left[U \cdot \dfrac{e^{2\mu L} - 1}{e^{2\mu L} - e^{-2\mu U}} - L \cdot \dfrac{1 - e^{-2\mu U}}{e^{2\mu L} - e^{-2\mu U}}\right]$
- As $\mu \to 0$: $P(\text{hit } U) \to L/(L+U)$, $E[\tau] \to LU$ (gambler's ruin).
- Optimal symmetric $b^{*}$: wide when $|\mu|$ is large, tight (scales as $1/|\mu|$) when $|\mu|$ is small, reflecting the balance between drift reliability and diffusion speed.
Intuition
The exponential martingale trick -- turning a drifted process into a martingale by composing it with $x \mapsto e^{-2\mu x}$ -- is a specific instance of a much broader idea: the Girsanov change of measure. When you apply optional stopping to $e^{-2\mu X_t}$, you are essentially computing how much the drift tilts the exit probabilities away from the purely diffusive case. The $\mu \to 0$ limit ($p \to L/(L+U)$) is the gambler's ruin result for pure BM -- a beautiful sanity check, because the exit probability for pure BM is linear in starting position.
The symmetric barrier optimization problem is a stylized version of a real trading decision: how wide should your take-profit and stop-loss be? The answer depends entirely on your signal-to-noise ratio $|\mu|/\sigma$. When you have a strong edge ($|\mu|$ large), run your winners -- use wide targets because the drift will reliably carry you there, and cutting profits short wastes the edge. When your edge is weak or uncertain ($|\mu| \approx 0$), use tight targets -- diffusion is indifferent to your target width, but a tight target gets you out faster before the random walk can undo your temporary gain. This is why stat arb strategies with small but reliable edges often use tight profit targets and fast recycling, while directional bets with high-conviction views use wider stops.