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Why a Brownian Path Must Cross a Lower Barrier First

Article Quant Q&A · Author: KACEFMA.
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Summary

The document explains an event equality for a continuous Brownian motion starting below a level u. It compares paths that reach a higher threshold, 2u−d, after the first time they hit u with paths that reach that higher threshold at any time before T. Given 0 < d < u, the higher threshold is above u, so continuity forces every path reaching it to have crossed u first. The event that the first hitting time is no later than T is therefore automatic whenever the higher threshold is reached.

The answers offer this pathwise argument and also express the relationship through the inclusion of maximum events. The result depends on continuity, the initial value being below u, and the higher threshold being strictly above u. The discussion is a proof of an event identity, not an analysis of trading data or a claim about market behavior; it does not extend the argument to discontinuous processes or other starting conditions.

Key ideas

  • A continuous path starting below u must pass through u before reaching any higher level.
  • The condition 0 < d < u makes 2u−d greater than u.
  • Reaching 2u−d by time T implies that the first hitting time of u is no later than T.
  • The identity relies on path continuity and the stated starting point.

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# How to show the equality of these two events?


# How to show the equality of these two events?












For $X_t$ a brownian motion defined for $t\in[0,T]$, How to show the equality of following events: $$ \{ \displaystyle \max_{[\tau,T]}(X_t)\ge 2u-d, \tau\leq T \}=\{\displaystyle \max_{[0,T]}(X_t)\ge 2u-d\} $$ where,

$\tau=min\{t:X_t=u\} $ and $0<d <u $

I Thank You in advance.

## Answer by LocalVolatility (score 3, accepted)

https://quant.stackexchange.com/a/30528

Let $\max_{[0, T]} \left( X_t \right) = M_T$. We have $2u - d > u$ and thus $M_T \geq 2u - d$ implies that $M_T > u$. Note that $X$ is a continuous process with initial value $X_0 = 0 < u$. Then, in order for a path of $X$ to start at $X_0 = 0$ and reach a point $M_T > 2u - d$ it has to cross the level $u$ first.

## Answer by KACEFMA. (score 1)

https://quant.stackexchange.com/a/30648

My idea to answer is:

Since $2u-d>u$, then

$$\{ \displaystyle\max_{[\tau,T]} (X_t)\ge 2u-d \}=\{\displaystyle\max_{[0,T]} (X_t)\ge 2u-d \}$$ because the random variable $\tau$ is the time of the first crossing of the barrier $u$ by the process $X$.

In addition, we have the following equality: $$\{\tau \le T\}=\{\displaystyle\max_{[0,T]} (X_t)\ge u \} $$ Finally, since $$\{ \displaystyle\max_{[0,T]} (X_t)\ge 2u-d \}\subset \{\displaystyle\max_{[0,T]} (X_t)\ge u \} $$ Therefore

$$\{ \displaystyle\max_{[\tau,T]} (X_t)\ge 2u-d, \tau \le T \}=\{\displaystyle\max_{[0,T]} (X_t)\ge 2u-d \}$$

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