Level-Hitting Times and Exit-Time Events

Hitting and Exit Times as Pathwise Events

In exit problems, an event such as “the process reaches a boundary by time $T$” is not the same as an event about its value at the single time $T$. A path may hit a level and then move away from it, or it may leave an interval through one boundary and later visit the other. Continuity lets us describe these events exactly, but it is important to use the path up to the relevant first-hitting or first-exit time.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$, adapted to $(\mathcal{F}_t)$, with continuous paths and $B_0=0$. For clarity, the pathwise definitions below also apply to any continuous adapted real-valued process $X_t$; when a general starting point $x$ is needed, take $X_t=x+B_t$. The class $\mathcal{L}^2$ consists of progressively measurable integrands $H_t$ such that $\mathbb{E}\int_0^T H_t^2\,dt<\infty$. No stochastic integral is needed here.

Definition: For a real-valued process $X_t$ and a level $c\in\mathbb{R}$, its first hitting time of $c$ is $$ \tau_c=\inf\{t\ge0:X_t=c\}, $$ where the infimum of the empty set is $+\infty$. For an open interval $(a,b)$, its first exit time is $$ \tau_{(a,b)}=\inf\{t\ge0:X_t\notin(a,b)\}. $$ A random time $\tau$ is a stopping time if $\{\tau\le t\}\in\mathcal{F}_t$ for every $t\ge0$.

If $X_0$ is already at the level $c$, then $\tau_c=0$. If $X_0\notin(a,b)$, then $\tau_{(a,b)}=0$. The most informative interval-exit statements below assume $X_0\in(a,b)$; this is the usual setting when a process starts inside a region and is stopped at its boundary.

Hitting Is Not the Same as Crossing

A path hits $c$ on $[0,T]$ if $X_s=c$ for at least one $s\in[0,T]$. It strictly crosses $c$ from below to above on $[0,T]$ if there are times $s<u$ in that interval with $X_s<c<X_u$. A crossing from above to below is defined by reversing the inequalities. Continuity and the intermediate value theorem imply that a strict crossing includes a hit. A hit alone need not be a crossing: a path can touch a level and turn back without taking values on the other side.

Worked Example: Touching a Level Without Crossing It

On $[0,2]$, take the continuous path $x(t)=-(t-1)^2$ and the level $c=0$. The path hits $0$ at $t=1$, since $x(1)=-(1-1)^2=0$. For every $t\in[0,2]$, $x(t)\le0$, so it never takes a value strictly above $0$. Thus it does not strictly cross from below to above. It also does not cross from above to below, since it never takes a value strictly above $0$.

Check: its values at $t=0,1,2$ are $-1,0,-1$, respectively. The path reaches the level and returns to the same side; the hit at $t=1$ is not a crossing.

This distinction is pathwise and does not itself assert how likely either behavior is for Brownian motion. For a first-hitting problem, the defining event is equality with the level at some time, not a change of sign across it.

Stopping-Time Property

Proposition: If $X_t$ is continuous and adapted, then its first hitting time of a closed set $A\subseteq\mathbb{R}$, defined by $\tau_A=\inf\{s\ge0:X_s\in A\}$, is a stopping time. In particular, $\tau_c$ and $\tau_{(a,b)}$ are stopping times.

Proof. Fix $t\ge0$. On the compact interval $[0,t]$, continuity ensures that the function $s\mapsto\operatorname{dist}(X_s,A)$ attains its minimum. The path hits $A$ by time $t$ exactly when this minimum is zero. Let $D_t$ consist of the rational numbers in $[0,t]$, together with $t$. This is a countable dense subset of $[0,t]$. By continuity, the minimum is zero exactly when $\inf_{s\in D_t}\operatorname{dist}(X_s,A)=0$. Each $X_s$ for $s\le t$ is $\mathcal{F}_t$-measurable, because $X$ is adapted and $\mathcal{F}_s\subseteq\mathcal{F}_t$. Therefore the countable infimum is $\mathcal{F}_t$-measurable, and $$ \{\tau_A\le t\} = \left\{\inf_{s\in D_t}\operatorname{dist}(X_s,A)=0\right\} \in\mathcal{F}_t. $$ A singleton $\{c\}$ is closed, so this proves the claim for $\tau_c$. The complement $\mathbb{R}\setminus(a,b)$ is closed, so applying the same argument to that set proves the claim for $\tau_{(a,b)}$. $\square$

The closed-set hypothesis matters in this argument: distance zero means the path is in a closed set, whereas for a nonclosed set a point can have distance zero without belonging to it. The interval exit is expressed using the closed complement of the open interval for precisely this reason.

Characterizing a Level Hit by the Path Range

Proposition: For a continuous real-valued path on $[0,T]$, let $m_T=\min_{0\le s\le T}X_s$ and $M_T=\max_{0\le s\le T}X_s$. Then, for every $c\in\mathbb{R}$, $$ \{\tau_c\le T\}=\{m_T\le c\le M_T\}. $$

Proof. The extrema exist because a continuous function on a compact interval attains its minimum and maximum. If $\tau_c\le T$, then $X_{\tau_c}=c$, so $m_T\le c\le M_T$. Conversely, suppose $m_T\le c\le M_T$. There are times $s_-,s_+\in[0,T]$ with $X_{s_-}=m_T$ and $X_{s_+}=M_T$. If $m_T=M_T$, the path is constant and that value must equal $c$. If $m_T<M_T$, continuity and the intermediate value theorem show that the path takes every value between its minimum and maximum between $s_-$ and $s_+$, regardless of which of those times comes first. In particular, it takes the value $c$ by time $T$. Hence $\tau_c\le T$. $\square$

The inequalities in this characterization are non-strict because touching the level counts as hitting it. Replacing either inequality by a strict one would omit paths whose minimum or maximum equals $c$.

Worked Example: Testing a Level-Hitting Event

Consider the path $x(t)=t-1$ on $[0,3]$, the level $c=0$, and the horizon $T=3$. Its minimum is $m_3=x(0)=-1$, and its maximum is $M_3=x(3)=2$. Since $-1\le0\le2$, the range criterion gives $\tau_0\le3$. Solving $x(t)=0$ gives $t-1=0$, so $\tau_0=1$.

For comparison, on $[0,\tfrac12]$ the minimum is $-1$ and the maximum is $-\tfrac12$, so $0\le M_{1/2}$ fails and the level has not been hit. At $T=1$, the maximum equals $0$, and the non-strict condition $c\le M_1$ holds: the path hits the level exactly at the endpoint $t=1$.

Check: the direct solution $x(t)=0$ gives $t=1$, agreeing with both range tests: no hit by time $\tfrac12$, but a hit by time $1$.

First Exit and the Boundary Reached

Assume now that $X_0\in(a,b)$, where $a<b$, and write $\tau=\tau_{(a,b)}$. The exit time is the first time the path is outside the open interval. For a continuous path starting inside, it cannot jump over a boundary: if the exit time is finite, the path reaches one of the two boundary values at that time.

Proposition: For a continuous path with $X_0\in(a,b)$, if $\tau<\infty$, then $X_\tau\in\{a,b\}$. Moreover, $$ \tau=\tau_a\wedge\tau_b, \qquad \{\tau\le T\}=\{m_T\le a\}\cup\{M_T\ge b\}. $$

Proof. Continuity at time $0$ and $X_0\in(a,b)$ imply that the path remains inside the interval for some positive time, so $\tau>0$. For every $s<\tau$, the definition of the infimum gives $X_s\in(a,b)$. If $\tau<\infty$, there are times outside $(a,b)$ arbitrarily close to $\tau$ from at or after $\tau$; otherwise the infimum of the outside times would be strictly greater than $\tau$. Continuity then implies $X_\tau\notin(a,b)$, because the complement is closed. Also, taking times $s<\tau$ tending to $\tau$ gives $X_\tau\in[a,b]$. Therefore $X_\tau\in\{a,b\}$.

The first time outside $(a,b)$ is consequently the first time at either boundary: before $\tau_a\wedge\tau_b$, neither boundary nor the exterior has been reached; at the first boundary hit, the path is outside the open interval. Thus $\tau=\tau_a\wedge\tau_b$. Finally, if $m_T\le a$, continuity and $X_0>a$ force a hit of $a$ by time $T$, so $\tau\le T$. If $M_T\ge b$, continuity and $X_0<b$ force a hit of $b$ by time $T$, also giving $\tau\le T$. Conversely, if $\tau\le T$, the boundary property just proved gives $X_\tau=a$ or $X_\tau=b$, implying $m_T\le a$ or $M_T\ge b$. This proves the event identity. $\square$

The exit-side events require attention to which boundary is reached first. Since $a\ne b$, a continuous path starting inside cannot hit both at the same finite time. With the convention that an infinite time is larger than every finite time, the precise descriptions are $$ \{\tau\le T,\ X_\tau=a\}=\{\tau_a\le T,\ \tau_a<\tau_b\}, \qquad \{\tau\le T,\ X_\tau=b\}=\{\tau_b\le T,\ \tau_b<\tau_a\}. $$ Each event records both a time bound and the ordering of the two boundary hits.

Worked Example: The First Boundary Matters

Take the continuous path $x(t)=t$ for $0\le t\le2$, and $x(t)=4-t$ for $2\le t\le5$. It starts at $0$ inside $(-1,2)$. The path first hits $2$ at $t=2$, so $\tau_2=2$. It later hits $-1$ at $t=5$, so $\tau_{-1}=5$. Therefore the first exit time is $\tau=\min(2,5)=2$, and the exit boundary is $2$.

By time $T=5$, the path minimum is $-1$ and maximum is $2$. The condition $m_5\le-1$ is true, but it does not mean the path exited through $-1$: it exited through $2$ at time $2$. The side formula confirms this, since $\tau_2=2<5=\tau_{-1}$.

Check: at the computed first exit time, $x(2)=2$, the upper boundary; $x(5)=-1$ is a later visit and cannot change the first exit side.

Strict and Non-Strict Time Bounds

For a fixed $T$, the event $\{\tau\le T\}$ includes an exit exactly at time $T$, while $\{\tau<T\}$ excludes it. For a continuous path starting inside $(a,b)$, the exact path-range descriptions are $$ \{\tau\le T\}=\{m_T\le a\}\cup\{M_T\ge b\}, \qquad \{\tau>T\}=\{m_T>a,\ M_T<b\}. $$ The second identity follows because $\tau>T$ means the path remains strictly inside throughout $[0,T]$. Conversely, if both extrema are strictly inside the interval, the path has a positive distance from both boundaries on the compact time interval, so continuity at $T$ ensures it does not exit immediately after $T$; hence $\tau>T$. In particular, replacing the strict inequalities in the second event by non-strict ones would incorrectly include a path that reaches a boundary at $T$.

Worked Example: Exit Exactly at the Horizon

Let $x(t)=t$ on $[0,1]$, starting at $0$, and consider the interval $(-1,1)$. The first exit time is $\tau=1$, because $x(t)<1$ for $0\le t<1$ and $x(1)=1$. At $T=1$, the extrema are $m_1=0$ and $M_1=1$. Thus $m_1\le-1$ is false and $M_1\ge1$ is true, giving $\tau\le1$. But $M_1<1$ is false, correctly showing that $\tau>1$ does not hold.

At $T=\tfrac12$, $m_{1/2}=0$ and $M_{1/2}=\tfrac12$. Both strict inequalities $m_{1/2}>-1$ and $M_{1/2}<1$ hold, so $\tau>\tfrac12$. These calculations distinguish exit at the horizon from remaining inside beyond it.

Check: solving $x(t)=1$ gives $t=1$, consistent with $\tau\le1$ and inconsistent with $\tau<1$.

Common Pitfall: Using the Whole Horizon to Name the Exit Side

A minimum or maximum over $[0,T]$ can show that an exit happened by time $T$, but after the first exit the path is no longer constrained to stay inside the interval. The later path may visit the other boundary. In the preceding piecewise-linear example, $m_5=-1$ and $M_5=2$; the incorrect calculation “$m_5=-1$, therefore the path exited at the lower boundary” ignores that the upper boundary was hit first at $t=2$. The correction is to compare first-hitting times: $\tau_2=2<\tau_{-1}=5$, so the exit side is the upper boundary.

These pathwise descriptions are useful whenever a payoff or boundary condition is evaluated at an exit time. On the event $\{\tau<\infty\}$, continuity identifies the exit value as $X_\tau=a$ or $X_\tau=b$; any expression involving that value must still account for which boundary was reached first. The statements here identify the relevant events without assigning probabilities to them.

Check Your Understanding

Use the pathwise definitions and keep endpoint equalities explicit.

For $x(t)=2-t$ on $[0,3]$, find the first hitting time of $1$. Does the path strictly cross $1$ from above to below by time $3$?
Explain why $\{\tau_c\le T\}$ uses $m_T\le c\le M_T$, rather than strict inequalities.
A path starts inside $(a,b)$, hits $b$ at time $2$, and hits $a$ at time $4$. What is its first exit time and exit side? Why does its minimum over $[0,4]$ not answer the exit-side question by itself?
For an interval exit time $\tau$, what does $M_T=b$ imply about $\tau\le T$, and why does it not imply $\tau<T$?
State the stopping-time event $\{\tau_c\le t\}$ in terms of the minimum distance of the path from $c$ on $[0,t]$.