Minimum And Maximum Of Stopping Times

Two Ways to Combine Stopping Times

A stopping time records when an event can first be known from the information available so far. Given two stopping times, their minimum is the time when at least one of the two has occurred, while their maximum is the time by which both have occurred. These descriptions suggest different set operations: “at least one” leads to a union, and “both” leads to an intersection. The distinction is essential for proving that the combined times are still stopping times.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$, where $(\mathcal{F}_t)$ is increasing, and suppose the space carries a Brownian motion $B_t$ adapted to this filtration. No Brownian-motion property is needed in the results below. The class $\mathcal{L}^2$ consists of progressively measurable processes $H$ such that $\mathbb{E}\int_0^T H_s^2\,ds<\infty$ for each finite $T$; no stochastic integrals are used in this tutorial.

Definition: A stopping time is a map $\tau:\Omega\to[0,\infty]$ such that $\{\tau\le t\}\in\mathcal{F}_t$ for every $t\ge0$. For two stopping times $\tau$ and $\sigma$, define their pointwise minimum and maximum by $(\tau\wedge\sigma)(\omega)=\min\{\tau(\omega),\sigma(\omega)\}$ and $(\tau\vee\sigma)(\omega)=\max\{\tau(\omega),\sigma(\omega)\}$. The value $+\infty$ is allowed.

The definition asks for information about a time threshold: by time $t$, has the stopping time occurred? Consequently, the proof needs to test the events $\{\tau\wedge\sigma\le t\}$ and $\{\tau\vee\sigma\le t\}$ for every deterministic $t\ge0$. It does not require that the filtration be complete or right-continuous. In particular, these closure properties hold before any usual-condition augmentation as well as after one.

The Threshold-Event Identities

Theorem: For any maps $\tau,\sigma:\Omega\to[0,\infty]$ and every $t\ge0$, $$ \{\tau\wedge\sigma\le t\}=\{\tau\le t\}\cup\{\sigma\le t\}, \qquad \{\tau\vee\sigma\le t\}=\{\tau\le t\}\cap\{\sigma\le t\}. $$ If $\tau$ and $\sigma$ are stopping times for $(\mathcal{F}_t)$, then $\tau\wedge\sigma$ and $\tau\vee\sigma$ are stopping times for that same filtration.

Proof. Fix $t\ge0$ and $\omega\in\Omega$. The inequality $\min\{\tau(\omega),\sigma(\omega)\}\le t$ holds exactly when at least one of $\tau(\omega)$ and $\sigma(\omega)$ is at most $t$. This proves the first event identity, including when one or both values are $+\infty$. The inequality $\max\{\tau(\omega),\sigma(\omega)\}\le t$ holds exactly when both values are at most $t$, proving the second identity. If $\tau,\sigma$ are stopping times, then $\{\tau\le t\}$ and $\{\sigma\le t\}$ belong to $\mathcal{F}_t$. A sigma-field is closed under finite unions and finite intersections, so both threshold events for the minimum and maximum belong to $\mathcal{F}_t$. This holds for every $t\ge0$, which is precisely the stopping-time condition. $\square$

The identities explain the different set operations. The minimum has already occurred by $t$ if either component has occurred by $t$; the maximum has occurred by $t$ only if both have occurred by $t$. Neither statement requires independence, a probability calculation, or a relation between the two stopping times.

Worked Example: Discrete-Valued Stopping Times

Let $A\in\mathcal{F}_1$ and $C\in\mathcal{F}_2$, and define $\tau=1$ on $A$, $\tau=3$ on $A^c$, $\sigma=2$ on $C$, and $\sigma=4$ on $C^c$. First, $\{\tau\le t\}$ is empty for $t<1$, equals $A$ for $1\le t<3$, and equals $\Omega$ for $t\ge3$. These events are in $\mathcal{F}_t$ in their respective ranges because the filtration is increasing. Similarly, $\{\sigma\le t\}$ is empty for $t<2$, equals $C$ for $2\le t<4$, and equals $\Omega$ for $t\ge4$. Thus both maps are stopping times.

On $A\cap C$, the pair of values is $(1,2)$, so the minimum is $1$ and the maximum is $2$. On $A\cap C^c$, the values are $(1,4)$, giving minimum $1$ and maximum $4$. On $A^c\cap C$, the values are $(3,2)$, giving minimum $2$ and maximum $3$. On $A^c\cap C^c$, the values are $(3,4)$, giving minimum $3$ and maximum $4$.

Therefore, for $0\le t<1$, both threshold events are empty. For $1\le t<2$, the minimum event is $A$ and the maximum event is empty. For $2\le t<3$, the minimum event is $A\cup C$, while the maximum event is $A\cap C$. For $3\le t<4$, the minimum event is $\Omega$, while the maximum event is $C$. For $t\ge4$, both events are $\Omega$.

Check: At $t=2$, the minimum is at most $2$ on $A\cup C$, because either $\tau=1$ or $\sigma=2$. The maximum is at most $2$ only on $A\cap C$, where $\tau=1$ and $\sigma=2$. The formulas give exactly these events.

Complementary Tail Events

Proposition: For every $t\ge0$, $$ \{\tau\wedge\sigma>t\}=\{\tau>t\}\cap\{\sigma>t\}, \qquad \{\tau\vee\sigma>t\}=\{\tau>t\}\cup\{\sigma>t\}. $$ These identities hold for arbitrary $[0,\infty]$-valued maps.

Proof. The event $\{\tau\wedge\sigma>t\}$ is the complement of $\{\tau\wedge\sigma\le t\}$. By the first identity in the Theorem, it is the complement of $\{\tau\le t\}\cup\{\sigma\le t\}$. De Morgan’s law makes this $\{\tau>t\}\cap\{\sigma>t\}$. Likewise, $\{\tau\vee\sigma>t\}$ is the complement of $\{\tau\vee\sigma\le t\}$, which by the second identity is the complement of $\{\tau\le t\}\cap\{\sigma\le t\}$. De Morgan’s law gives $\{\tau>t\}\cup\{\sigma>t\}$. Complements include the cases where either time equals $+\infty$, so no finiteness assumption is needed. $\square$

These tail identities are useful when a problem is phrased in terms of survival beyond a threshold rather than occurrence by a threshold. They are set identities; they do not assert that the two events on the right are independent. Without independence, their probabilities generally cannot be multiplied to find the probability of an intersection.

Worked Example: The First of Two Brownian Hitting Times

Let $B_t$ be the Brownian motion in the stated filtered space, and define $\tau_+=\inf\{t\ge0:B_t=1\}$ and $\tau_-=\inf\{t\ge0:B_t=-1\}$, with the infimum of the empty set defined as $+\infty$. The earlier result on first hitting times of continuous adapted processes gives that each is a stopping time. Apply the threshold identities to $\rho=\tau_+\wedge\tau_-$ and $\eta=\tau_+\vee\tau_-$: for every $t\ge0$,

\{\rho\le t\}=\{\tau_+\le t\}\cup\{\tau_-\le t\}, \qquad \{\eta\le t\}=\{\tau_+\le t\}\cap\{\tau_-\le t\}.

Thus $\rho\le t$ means that Brownian motion has hit at least one of the two levels by time $t$; $\eta\le t$ means it has hit both levels by time $t$. For example, at $t=0$, both events are empty because $B_0=0$, so neither level has been hit. If one level is never hit, its hitting time is $+\infty$: then $\rho$ is the other hitting time if that time is finite, while $\eta=+\infty$.

Check: The union in the formula for $\rho$ includes outcomes on which exactly one level has been hit. The intersection for $\eta$ excludes those outcomes. This agrees with the meanings of “first of the two” and “both have occurred.”

Finite Minima and Maxima

The same reasoning extends to any fixed finite collection. This is a finite closure property, and its proof uses only induction and closure of a sigma-field under finite unions and intersections.

Corollary: If $\tau_1,\ldots,\tau_n$ are stopping times, where $n\ge1$ is a fixed integer, then their pointwise minimum and pointwise maximum are stopping times. For each $t\ge0$, $$ \left\{\min_{1\le j\le n}\tau_j\le t\right\} =\bigcup_{j=1}^{n}\{\tau_j\le t\}, \qquad \left\{\max_{1\le j\le n}\tau_j\le t\right\} =\bigcap_{j=1}^{n}\{\tau_j\le t\}. $$

Proof. Fix $t\ge0$ and $\omega\in\Omega$. The minimum of the $n$ values is at most $t$ if and only if at least one value is at most $t$, giving the union identity. Their maximum is at most $t$ if and only if every value is at most $t$, giving the intersection identity. These equivalences hold also when some or all values are $+\infty$. Each event $\{\tau_j\le t\}$ is in $\mathcal{F}_t$, so the finite union and finite intersection are in $\mathcal{F}_t$. Since this holds for every $t\ge0$, both combined times are stopping times. The assumption $n\ge1$ ensures that the minimum and maximum are defined. $\square$

The word finite matters: this argument invokes only finite closure of a sigma-field. It makes no claim about a countably infinite family, for which further questions about limits and the stopping-time definition must be addressed separately.

Common Mistake: Reversing the Operations

A common error is to use an intersection for the minimum and a union for the maximum. That reverses the meanings of the threshold events. For instance, if $\tau=1$ and $\sigma=3$, then at $t=2$ the minimum has occurred, but the maximum has not. The incorrect calculation would give $\{\tau\wedge\sigma\le2\}=\{\tau\le2\}\cap\{\sigma\le2\}$, which is empty, and $\{\tau\vee\sigma\le2\}=\{\tau\le2\}\cup\{\sigma\le2\}$, which is the whole sample space. The correct calculation gives the whole sample space for the minimum event and the empty set for the maximum event.

A second pitfall is to confuse “the maximum is at most $t$” with “at least one time is at most $t$.” The latter describes the minimum. For the maximum to be at most $t$, neither time may exceed $t$, so both must be at most $t$. Keeping the question “what does this threshold event mean?” in view is a reliable way to select the correct set operation.

Combined time	Meaning of being at most $t$	Threshold event
$\tau\wedge\sigma$	At least one time is at most $t$	$\{\tau\le t\}\cup\{\sigma\le t\}$
$\tau\vee\sigma$	Both times are at most $t$	$\{\tau\le t\}\cap\{\sigma\le t\}$
$\tau\wedge\sigma>t$	Both times exceed $t$	$\{\tau>t\}\cap\{\sigma>t\}$
$\tau\vee\sigma>t$	At least one time exceeds $t$	$\{\tau>t\}\cup\{\sigma>t\}$

Check Your Understanding

For each question, use threshold events and state the set operation that applies.

If $\tau$ and $\sigma$ are stopping times, write $\{\tau\wedge\sigma\le t\}$ and explain why it belongs to $\mathcal{F}_t$.
Write $\{\tau\vee\sigma>t\}$ using $\{\tau>t\}$ and $\{\sigma>t\}$.
Suppose $\tau=2$ and $\sigma=5$ deterministically. For $t=3$, which of $\{\tau\wedge\sigma\le t\}$ and $\{\tau\vee\sigma\le t\}$ is the whole sample space?
If three stopping times are given, express the threshold event for their maximum and state the finite closure property used to prove it is in $\mathcal{F}_t$.
Why does the finite-family argument alone not establish a claim about a countably infinite family?

Combined time	Meaning of being at most \(t\)	Threshold event
\(\tau\wedge\sigma\)	At least one time is at most \(t\)	\(\{\tau\le t\}\cup\{\sigma\le t\}\)
\(\tau\vee\sigma\)	Both times are at most \(t\)	\(\{\tau\le t\}\cap\{\sigma\le t\}\)
\(\tau\wedge\sigma>t\)	Both times exceed \(t\)	\(\{\tau>t\}\cap\{\sigma>t\}\)
\(\tau\vee\sigma>t\)	At least one time exceeds \(t\)	\(\{\tau>t\}\cup\{\sigma>t\}\)

Minimum And Maximum Of Stopping Times

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