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Exit Probabilities for Brownian Motion Between Two Barriers

Article Quant Q&A · Author: oxheron
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Summary

This note derives the probabilities that a standard Brownian motion starting at zero first reaches an upper or lower barrier. With the lower barrier below zero and the upper barrier above zero, the process stops when it hits either boundary. Applying Wald’s identity to the stopped process gives an expected terminal value of zero, which relates the two possible exit values to their probabilities.

Solving that relation yields the probability of an upper-boundary exit as the distance from the lower boundary to the starting point divided by the full distance between the barriers; the lower-exit probability is the complement. The result concerns which boundary is hit, not the distribution of the stopping time or the path before exit. It assumes standard Brownian motion and barriers on opposite sides of the starting point.

Key ideas

  • The setup assumes standard Brownian motion starts at zero with one barrier on each side.
  • The process terminates at the first hit of either boundary.
  • Wald’s identity links the boundary values to their respective hitting probabilities.
  • The upper-exit probability depends on the relative distances between the start and barriers.

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# Probability distribution function for boundaries on brownian motion


# Probability distribution function for boundaries on brownian motion












What would be the probability distribution function for Brownian motion with two boundaries, i.e. a stop loss and take profit. The process is a standard Brownian motion. However at values $a$ and $b$ any random walk past these points is stopped. What would be the distribution for this function.

## Answer by Wei (score 1)

https://quant.stackexchange.com/a/80610

Assume that the standard Brownian motion $W$ starts at $0$, and that $a<0<b$. Let $\tau$ be the first hitting time of $W$ at $a$ or $b$. By Wald's identity, $0=\mathbb E W_\tau = b\mathbb P[W_\tau=b] + a(1-\mathbb P[W_\tau=b])$, so $$P[W_\tau=b] = \frac{-a}{b-a}.$$ And so $$P[W_\tau=a] = 1-\mathbb P[W_\tau=b] = \frac{b}{b-a}.$$

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.