Probability & Statistics

Will the Drifted Path Reach Minus One

X is a generalized Wiener process with unit upward drift, dX = dt + dW. How likely is it ever to fall as low as -1?

solvedmedium1 min

Let XX be the generalized Wiener process dX=dt+dW(t)dX = dt + dW(t), started at 00, where WW is a standard Brownian motion. What is the probability that XX ever reaches −1-1?

Reveal solutionHide solution

#A drifted walk that climbs

The process Xt=t+WtX_t = t + W_t has positive drift, so it marches off toward +∞+\infty. Ever reaching −1-1 means beating that drift downward, which only grows harder the deeper the level.

#A martingale gives the answer

With unit drift and unit variance, e−2Xte^{-2X_t} is a martingale, starting at e0=1e^{0} = 1. Let τ\tau be the first time the path touches −1-1. Stopping the martingale, the path either reaches −1-1, where e−2Xτ=e2e^{-2X_\tau} = e^{2}, or escapes to +∞+\infty, where e−2Xt→0e^{-2X_t} \to 0. Matching the mean to its starting value,

P(hit −1)⋅e2+P(never)⋅0=1⟹P(hit −1)=e−2.(1)\PP(\text{hit } -1)\cdot e^{2} + \PP(\text{never})\cdot 0 = 1 \quad\Longrightarrow\quad \PP(\text{hit } -1) = e^{-2}. \tag{1}
0-1drift up
The positive drift carries the path upward, so a dip all the way to -1 means fighting the trend. Because exp(-2 X_t) is a martingale, that downward escape happens with probability exactly e to the minus 2.

#Read it off

P(X ever reaches −1)=e−2≈0.135.(2)\PP(X \text{ ever reaches } -1) = e^{-2} \approx 0.135. \tag{2}

The drift turns a dip to −1-1 into a roughly one-in-seven event, and each additional unit of depth costs another factor of e−2e^{-2}.