Hitting 3 Before -5: Brownian Motion With and Without Drift
(a) A standard Brownian motion $W(t)$ starts at $0$. What is the probability that it reaches $+3$ before it reaches $-5$?
(b) Now let the process have drift: $dX(t) = m\,dt + dW(t)$ with $X(0) = 0$ and constant $m$. What is the probability that $X$ reaches $+3$ before $-5$?
Hints
- Without drift, $W(t)$ is a martingale. Apply optional stopping at the exit time $\tau$ of $(-5, 3)$: $E[W(\tau)] = 0$.
- With drift, $X(t)$ is not a martingale, but $e^{-2mX(t)}$ is (check with Ito: the drift of $e^{\lambda X}$ is $\lambda m + \lambda^2/2$, which vanishes at $\lambda = -2m$).
- Optional stopping for $e^{-2mX}$: $p\,e^{-6m} + (1-p)\,e^{10m} = 1$. Solve for $p$, and check that $m \to 0$ recovers $5/8$.
Worked Solution
How to Think About It: Two-barrier hitting probabilities are optional-stopping exercises: find a martingale of the process, evaluate it at the exit time where the position is one of two known values, and solve one linear equation for the probability. For driftless Brownian motion the martingale is $W$ itself. With drift, the trick is the exponential martingale $e^{-2mX(t)}$.
Quick Estimate: With no drift, $-5$ is farther away than $3$, so hitting $3$ first is more likely than not: $5/8 = 62.5\%$. A positive drift $m$ pushes the probability up; with $m = 0.1$ it is already about $79\%$, and with $m = 1$ it is essentially $1$.
Formal Solution:
Part (a): No drift
*Step 1 -- Optional stopping.* Let $\tau = \inf\{t : W(t) \in \{3, -5\}\}$ and $p = P(W(\tau) = 3)$. Since $W$ is a martingale, $E[\tau] < \infty$ and $W(t \wedge \tau)$ is bounded, optional stopping gives $E[W(\tau)] = W(0) = 0$:
$$3p - 5(1 - p) = 0 \quad\Longrightarrow\quad p = \frac{5}{8}.$$
In general the probability of hitting $\alpha$ before $-\beta$ is $\beta/(\alpha + \beta)$.
Part (b): With drift $m$
*Step 2 -- Find a martingale.* For $Y(t) = e^{\lambda X(t)}$, Ito's lemma gives
$$dY = \lambda Y\,dX + \tfrac{1}{2}\lambda^2 Y\,(dX)^2 = Y\left[\left(\lambda m + \tfrac{1}{2}\lambda^2\right)dt + \lambda\,dW\right].$$
The drift vanishes for $\lambda = -2m$, so $M(t) = e^{-2mX(t)}$ is a martingale.
*Step 3 -- Optional stopping.* With $p = P(X \text{ hits } 3 \text{ before } -5)$ and $M(0) = 1$,
$$E[M(\tau)] = p\,e^{-6m} + (1 - p)\,e^{10m} = 1.$$
*Step 4 -- Solve.*
$$p\left(e^{-6m} - e^{10m}\right) = 1 - e^{10m} \quad\Longrightarrow\quad p = \frac{e^{10m} - 1}{e^{10m} - e^{-6m}}.$$
*Step 5 -- Checks.* As $m \to 0$, expand $e^{u} \approx 1 + u$: $p \to \dfrac{10m}{10m + 6m} = \dfrac{5}{8}$, matching part (a). As $m \to +\infty$, $p \to 1$; as $m \to -\infty$, $p \to 0$. For $m = 0.1$: $p = (e - 1)/(e - e^{-0.6}) \approx 0.792$.
*Alternative (ODE).* $p(x)$, the probability from start $x$, solves the generator equation $m\,p'(x) + \tfrac{1}{2}p''(x) = 0$ with $p(-5) = 0$, $p(3) = 1$, whose solution is $p(x) = \dfrac{e^{-2mx} - e^{10m}}{e^{-6m} - e^{10m}}$; at $x = 0$ this is the same formula.
Answer: (a) $P = \dfrac{5}{8}$. (b) $P = \dfrac{e^{10m} - 1}{e^{10m} - e^{-6m}}$ (reducing to $5/8$ as $m \to 0$).
Intuition
Without drift the hitting probability is pure gambler's ruin: $5/8$, the ratio of the distance to the far barrier over the total width. Drift breaks the martingale property of $X$ itself, but $e^{-2mX}$ restores it, and optional stopping yields $p = (e^{10m} - 1)/(e^{10m} - e^{-6m})$, which tends to $5/8$ as $m \to 0$ and to $1$ or $0$ as $m \to \pm\infty$. The exponential martingale is the continuous analogue of the biased-coin gambler's ruin ratio $((1-q)/q)^k$, and this exact formula is what prices one-touch and double-barrier claims and sets stop-loss versus take-profit odds for a trending position.