Brownian Motion With Unit Drift: Probability of Ever Reaching -1
A process follows
$$dX(t) = dt + dW(t), \qquad X(0) = 0,$$
where $W$ is a standard Brownian motion (so $X$ has drift $+1$ and unit volatility). What is the probability that $X$ ever reaches the level $-1$?
Hints
- The drift is positive, so the process escapes to $+\infty$ almost surely; the question is how often it dips to $-1$ first. Think of an upper barrier at $+N$ and let $N \to \infty$.
- $e^{-2X(t)}$ is a martingale for $dX = dt + dW$ (Ito: the drift of $e^{\lambda X}$ is $\lambda + \lambda^2/2$, zero at $\lambda = -2$).
- Optional stopping between $-1$ and $+N$: $p\,e^{2} + (1 - p)\,e^{-2N} = 1$. Let $N \to \infty$.
Worked Solution
How to Think About It: With positive drift the process drifts off to $+\infty$, so "ever reaching $-1$" is a genuine event with probability strictly between 0 and 1. Set up the standard two-barrier problem with barriers at $-1$ and $+N$, solve it with the exponential martingale, and let the upper barrier go to infinity.
Quick Estimate: In the first unit of time the process moves by $N(1, 1)$, so it is below $-1$ at $t = 1$ with probability $\Phi(-2) \approx 2.3\%$; the running minimum dips lower than the endpoint, and integrating over all times pushes the total up to about $13.5\%$.
Formal Solution:
*Step 1 -- Exponential martingale.* For $M(t) = e^{\lambda X(t)}$, Ito's lemma gives
$$dM = M\left[\left(\lambda \cdot 1 + \tfrac{1}{2}\lambda^2\right)dt + \lambda\,dW\right],$$
so $M$ is a driftless (and, when stopped between bounded barriers, bounded) martingale for $\lambda = -2$: $M(t) = e^{-2X(t)}$, $M(0) = 1$.
*Step 2 -- Two-barrier problem.* Fix $N > 0$ and let $\tau_N$ be the first time $X$ hits $-1$ or $+N$; let $p_N = P(X \text{ hits } -1 \text{ before } +N)$. Optional stopping ($E[\tau_N] < \infty$ and $M(t \wedge \tau_N)$ is bounded by $e^{2}$):
$$p_N\,e^{2} + (1 - p_N)\,e^{-2N} = 1 \quad\Longrightarrow\quad p_N = \frac{1 - e^{-2N}}{e^{2} - e^{-2N}}.$$
*Step 3 -- Remove the upper barrier.* The events $\{X \text{ hits } -1 \text{ before } +N\}$ increase to $\{X \text{ ever hits } -1\}$ as $N \to \infty$ (a path that reaches $-1$ does so at a finite time, before which it is bounded, hence below some $N$). By continuity of probability,
$$P(X \text{ ever reaches } -1) = \lim_{N \to \infty} p_N = \frac{1}{e^{2}} = e^{-2} \approx 0.1353.$$
*Step 4 -- General formula.* For $dX = \mu\,dt + \sigma\,dW$ with $\mu > 0$, the same argument with $\lambda = -2\mu/\sigma^2$ gives $P(\text{ever reach } -b) = e^{-2\mu b/\sigma^2}$; the running minimum of the process is exponentially distributed with rate $2\mu/\sigma^2$.
Answer: $P(X \text{ ever reaches } -1) = e^{-2} \approx 0.135$.
Intuition
A Brownian motion with drift $\mu > 0$ that starts at $0$ reaches level $-b$ with probability $e^{-2\mu b}$: here $e^{-2}$, about $13.5\%$. The exponential martingale $e^{-2\mu X}$ turns the question into a one-line optional-stopping calculation once you push the upper barrier to infinity. This is the continuous-time "probability a trending P&L ever hits the stop-loss" formula: the maximum drawdown of a Brownian motion with drift is exponential with rate $2\mu/\sigma^2$, which is why doubling the edge squares the odds of avoiding any given stop.