Stopping Times¶

The Concept of Observable Random Times¶

In stochastic analysis, we frequently need to consider random times—moments when something happens. But not all random times are created equal. The concept of a stopping time captures precisely those random times that can be determined using only information available up to that moment.

Consider two scenarios:

"The first time the stock price exceeds $100" — You can observe when this happens in real-time.
"The last time the stock price exceeds $100 before it crashes" — You cannot know this until after the crash.

The first is a stopping time; the second is not. This distinction is fundamental to what strategies are implementable and what theorems apply.

Definition¶

Let $(\Omega, \mathcal{F}, (\mathcal{F}_t)_{t \ge 0}, \mathbb{P})$ be a filtered probability space.

A random variable $\tau: \Omega \to [0, \infty]$ is a stopping time (or optional time) with respect to $(\mathcal{F}_t)$ if:

\[ \boxed{\{\tau \le t\} \in \mathcal{F}_t \quad \text{for all } t \ge 0} \]

Interpretation: By time $t$, you can determine whether $\tau$ has already occurred. You don't need to peek into the future.

Equivalent formulations (under the usual conditions):

$\{\tau < t\} \in \mathcal{F}_t$ for all $t \ge 0$
$\{\tau \le t\} \in \mathcal{F}_t$ for all $t \ge 0$ (the standard definition)
$\{\tau = t\} \in \mathcal{F}_t$ for all $t \ge 0$ (for discrete filtrations)

Remark: We allow $\tau = \infty$, meaning the event may never occur.

Examples of Stopping Times¶

1. Deterministic Times¶

For any fixed $t_0 \ge 0$, the constant $\tau(\omega) = t_0$ is a stopping time, since $\{\tau \le t\} = \Omega$ for $t \ge t_0$ and $\emptyset$ for $t < t_0$.

2. First Hitting Times¶

For an adapted process $X$ with continuous paths and a Borel set $A \subset \mathbb{R}$:

\[ \tau_A := \inf\{t \ge 0 : X_t \in A\} \]

is a stopping time. (Convention: $\inf \emptyset = \infty$.)

Proof sketch: We consider two cases depending on the structure of $A$.

Case 1: $A$ is open. By path continuity and openness of $A$: if $X_s(\omega) \in A$ for some $s \le t$, then $X_r(\omega) \in A$ for some rational $r$ near $s$ with $r \le t$. Therefore:

\[ \{\tau_A \le t\} = \bigcup_{\substack{s \le t \\ s \in \mathbb{Q}}} \{X_s \in A\} \]

Each $\{X_s \in A\} \in \mathcal{F}_s \subseteq \mathcal{F}_t$, so the countable union is in $\mathcal{F}_t$.

Case 2: $A$ is closed. Write $A_n = \{x : d(x, A) < 1/n\}$ (open $1/n$-neighborhoods). By Case 1 each $\tau_{A_n}$ is a stopping time, and $\tau_{A_n} \downarrow \tau_A$ since $A_n \downarrow A$ (for continuous paths). The infimum of stopping times is a stopping time, so $\tau_A$ is a stopping time.

General Borel $A$: The result holds but requires more care (the optional section theorem or the fact that $\tau_A = \tau_{\bar{A}}$ for continuous paths). $\square$

Important special cases:

First passage time: $\tau_a = \inf\{t \ge 0 : W_t = a\}$ for Brownian motion
First exit time: $\tau_D = \inf\{t \ge 0 : X_t \notin D\}$ from a domain $D$
First entry time: $\tau_K = \inf\{t \ge 0 : X_t \in K\}$ into a set $K$

3. Last Exit Times (Non-Example)¶

The last exit time from a set:

\[ \sigma_A := \sup\{t \ge 0 : X_t \in A\} \]

is generally not a stopping time. To know when you last visited $A$, you need to know the entire future path.

Exception: If $X$ never returns to $A$ after some deterministic time $T$, and this is known, then $\sigma_A \wedge T$ might be a stopping time.

4. Discrete-Time Examples¶

For a discrete filtration $(\mathcal{F}_n)_{n \ge 0}$:

$\tau = \inf\{n : S_n \ge 100\}$ where $S_n$ is a random walk — stopping time
$\tau = $ "the time of the maximum of $S_0, \ldots, S_N$" — not a stopping time (requires knowing all values)

Properties of Stopping Times¶

Proposition: If $\sigma$ and $\tau$ are stopping times, then so are:

$\sigma \wedge \tau = \min(\sigma, \tau)$
$\sigma \vee \tau = \max(\sigma, \tau)$
$\sigma + \tau$ (when both are finite a.s.)
$\tau + c$ for any constant $c \ge 0$

Proof of (1): $\{\sigma \wedge \tau \le t\} = \{\sigma \le t\} \cup \{\tau \le t\} \in \mathcal{F}_t$. $\square$

Proposition: If $(\tau_n)$ is a sequence of stopping times, then:

$\inf_n \tau_n$ is a stopping time (since $\{\inf_n \tau_n \le t\} = \bigcup_n \{\tau_n \le t\}$)
$\sup_n \tau_n$ is a stopping time provided the filtration is right-continuous (since $\{\sup_n \tau_n \le t\} = \bigcap_n \{\tau_n \le t\} \in \mathcal{F}_t$; without right-continuity, $\sup_n \tau_n$ is only guaranteed to satisfy $\{\sup_n \tau_n < t\} \in \mathcal{F}_t$)
$\liminf_n \tau_n$ and $\limsup_n \tau_n$ are stopping times under the usual conditions

The σ-Algebra F_τ¶

For a stopping time $\tau$, the $\sigma$-algebra at time $\tau$ captures information available at the random time $\tau$:

\[ \mathcal{F}_\tau := \{A \in \mathcal{F} : A \cap \{\tau \le t\} \in \mathcal{F}_t \text{ for all } t \ge 0\} \]

Interpretation: $A \in \mathcal{F}_\tau$ if knowing whether $\tau$ has occurred by time $t$ allows you to determine whether $A$ has occurred.

Key properties:

$\mathcal{F}_\tau$ is a $\sigma$-algebra.
$\tau$ is $\mathcal{F}_\tau$-measurable.
If $X$ is progressively measurable (in particular if $X$ is adapted with càdlàg paths), then $X_\tau$ is $\mathcal{F}_\tau$-measurable (on $\{\tau < \infty\}$).
If $\sigma \le \tau$ are stopping times, then $\mathcal{F}_\sigma \subseteq \mathcal{F}_\tau$.

Example: For the first hitting time $\tau_a = \inf\{t : W_t = a\}$:

\[ \mathcal{F}_{\tau_a} = \sigma\left(W_{t \wedge \tau_a} : t \ge 0\right) \]

This contains information about the path up to hitting $a$, but nothing beyond.

Stopped Processes¶

Given a process $X = \{X_t\}$ and a stopping time $\tau$, the stopped process is:

\[ X_t^\tau := X_{t \wedge \tau} = X_{\min(t, \tau)} \]

The process is "frozen" at time $\tau$.

Key theorem: If $X$ is a martingale (resp. submartingale, supermartingale) and $\tau$ is a stopping time, then $X^\tau$ is also a martingale (resp. submartingale, supermartingale). The proof uses dyadic approximation of $\tau$ and is carried out in full in Optional Sampling Theorem. The core idea: stopping a martingale before a random time cannot create systematic bias, since each "frozen" path is still a fair game up to the time of freezing.

The Strong Markov Property¶

Stopping times interact beautifully with Markov processes. The strong Markov property states that the process "restarts" at a stopping time.

Theorem (Strong Markov Property for Brownian Motion): Let $W$ be a Brownian motion and $\tau$ a stopping time with $\tau < \infty$ a.s. Then:

\[ \boxed{\widetilde{W}_t := W_{\tau + t} - W_\tau} \]

is a Brownian motion independent of $\mathcal{F}_\tau$.

Interpretation: After the random time $\tau$, Brownian motion "starts fresh"—the future evolution is independent of everything that happened before $\tau$, including how long it took to reach $\tau$.

Contrast with ordinary Markov property: The ordinary Markov property says this for deterministic times. The strong Markov property extends it to stopping times—a much stronger statement.

Application (Reflection Principle): Let $\tau_a = \inf\{t : W_t = a\}$ for $a > 0$. By the strong Markov property:

\[ \mathbb{P}\left(\max_{0 \le s \le t} W_s \ge a\right) = 2\mathbb{P}(W_t \ge a) \]

The idea: paths that hit $a$ and end below $a$ are in bijection (via reflection) with paths that hit $a$ and end above $a$.

First Passage Time Distributions¶

The strong Markov property enables computation of hitting time distributions.

Theorem: For $a > 0$, the first passage time $\tau_a = \inf\{t : W_t = a\}$ has density:

\[ \boxed{f_{\tau_a}(t) = \frac{a}{\sqrt{2\pi t^3}} \exp\left(-\frac{a^2}{2t}\right), \quad t > 0} \]

This is the inverse Gaussian (or Wald) distribution.

Key properties:

$\mathbb{E}[\tau_a] = \infty$ (the hitting time has infinite mean!)
$\mathbb{P}(\tau_a < \infty) = 1$ (Brownian motion hits every level eventually)
Mode at $t = a^2/3$

Derivation using exponential martingale: Apply optional sampling to $Z_t^\theta = \exp(\theta W_t - \frac{\theta^2 t}{2})$ at $\tau_a \wedge T$, then let $T \to \infty$ with careful justification:

\[ \mathbb{E}[\exp(-\lambda \tau_a)] = \exp(-a\sqrt{2\lambda}), \quad \lambda > 0 \]

Inverting this Laplace transform yields the density.

Localization and Local Martingales¶

Stopping times enable localization—a technique for extending theorems from well-behaved processes to more general ones.

Definition: A process $M$ is a local martingale if there exist stopping times $\tau_n \uparrow \infty$ a.s. such that each stopped process $M^{\tau_n}$ is a martingale.

Key insight: Local martingales satisfy the martingale property "locally" (before large fluctuations), even if they fail it globally.

Example: For a predictable process $H$ satisfying $\int_0^t H_s^2\,ds < \infty$ a.s., the stochastic integral $\int_0^t H_s \, dW_s$ is always a local martingale. It is a true martingale when the stronger condition $\mathbb{E}\int_0^T H_s^2 \, ds < \infty$ holds.

Predictable Stopping Times¶

A stopping time $\tau$ is predictable if there exist stopping times $\tau_n < \tau$ with $\tau_n \uparrow \tau$ a.s.

Interpretation: A predictable stopping time can be "announced" in advance—there's a sequence of earlier times that converge to it.

Examples:

Deterministic times are predictable.
First hitting times of open sets by continuous processes are not predictable (they happen "by surprise").
First hitting times of closed sets may or may not be predictable.

Relevance: The distinction matters for jump processes and the general theory of stochastic integration.

Stopping Times in Discrete Time¶

For discrete-time filtrations $(\mathcal{F}_n)_{n \ge 0}$, a stopping time satisfies:

\[ \{\tau = n\} \in \mathcal{F}_n \quad \text{for all } n \ge 0 \]

This is equivalent to $\{\tau \le n\} \in \mathcal{F}_n$.

The gambling interpretation: A stopping time is a rule for when to stop playing that depends only on what you've seen so far. "Stop when I'm $100 ahead" is a stopping time; "stop just before I would have lost" is not.

Summary¶

Concept	Definition	Key Property
Stopping time $\tau$	$\{\tau \le t\} \in \mathcal{F}_t$	Observable without seeing the future
First hitting time	$\inf\{t : X_t \in A\}$	Stopping time for adapted continuous $X$
$\mathcal{F}_\tau$	Information at random time $\tau$	$X_\tau$ is $\mathcal{F}_\tau$-measurable
Stopped process $X^\tau$	$X_{t \wedge \tau}$	Preserves martingale property
Strong Markov	Fresh start at $\tau$	$W_{\tau + t} - W_\tau$ is independent BM

The power of stopping times: They allow us to analyze processes at random times while maintaining the mathematical structure (martingales stay martingales, Markov processes restart). This is essential for:

Optional sampling (evaluating expectations at random times)
Boundary value problems (exit times from domains)
Sequential analysis (optimal stopping)
Financial mathematics (exercise times of options)

Exercises¶

Exercise 1: Stopping Time Verification¶

Determine which of the following are stopping times. Justify your answers.

(a) $\tau = \inf\{t \ge 0 : W_t = 1\}$

(b) $\tau = \sup\{t \le 1 : W_t = 0\}$

(c) $\tau = \inf\{t \ge 0 : W_t > W_s \text{ for all } s < t\}$

(d) $\tau = \inf\{t \ge 0 : \int_0^t W_s \, ds \ge 1\}$

(e) $\tau = \inf\{t \ge 1 : W_t = W_{t-1}\}$

Solution to Exercise 1

(a) $\tau = \inf\{t \ge 0 : W_t = 1\}$ is a stopping time. This is the first hitting time of the closed set $\{1\}$ by the continuous adapted process $W_t$. By the general result for first hitting times of closed sets by continuous processes, $\tau$ is a stopping time.

(b) $\tau = \sup\{t \le 1 : W_t = 0\}$ is not a stopping time. To determine whether the last zero before time 1 has occurred by time $t < 1$, one would need to know the future path after $t$ — specifically, whether Brownian motion returns to 0 between $t$ and 1. This requires information beyond $\mathcal{F}_t$.

(c) $\tau = \inf\{t \ge 0 : W_t > W_s \text{ for all } s < t\}$. This asks for the first time the process exceeds all previous values. Since $W_0 = 0$ and $W_t$ is continuous with $\limsup_{t \to 0^+} W_t/\sqrt{t} = +\infty$ a.s. (by the law of the iterated logarithm), there exist arbitrarily small times $t$ with $W_t > 0 = W_0$. In fact, $\tau = 0$ a.s. because for any $\varepsilon > 0$, there exists $t \in (0, \varepsilon)$ with $W_t > 0 > \inf_{s \in [0, t)} W_s$ is not guaranteed to mean $W_t > W_s$ for ALL $s < t$. Actually, $\tau = 0$ because Brownian motion starts fresh and immediately exceeds its starting value. Since $\tau = 0$ is a constant, it is trivially a stopping time.

(d) $\tau = \inf\{t \ge 0 : \int_0^t W_s\,ds \ge 1\}$ is a stopping time. The process $Y_t = \int_0^t W_s\,ds$ is adapted and continuous (as the integral of a continuous adapted process). Therefore $\tau$ is the first hitting time of the closed set $[1, \infty)$ by the continuous adapted process $Y_t$, hence a stopping time.

(e) $\tau = \inf\{t \ge 1 : W_t = W_{t-1}\}$ is a stopping time. Define $Y_t = W_t - W_{t-1}$ for $t \ge 1$. Then $Y_t$ is adapted to $(\mathcal{F}_t)$ (since both $W_t$ and $W_{t-1}$ are $\mathcal{F}_t$-measurable for $t \ge 1$) and has continuous paths. The time $\tau = \inf\{t \ge 1 : Y_t = 0\}$ is the first hitting time of 0 by the continuous adapted process $Y_t$ starting at time 1, hence a stopping time.

Exercise 2: Operations on Stopping Times¶

Let $\sigma$ and $\tau$ be stopping times.

(a) Prove that $\sigma \wedge \tau$ is a stopping time.

(b) Prove that $\sigma + \tau$ is a stopping time (assuming both are finite a.s.).

(c) Give a counterexample showing that $\sigma - \tau$ need not be a stopping time.

Solution to Exercise 2

(a) For any $t \ge 0$:

\[ \{\sigma \wedge \tau \le t\} = \{\min(\sigma, \tau) \le t\} = \{\sigma \le t\} \cup \{\tau \le t\} \]

Since $\sigma$ and $\tau$ are stopping times, $\{\sigma \le t\} \in \mathcal{F}_t$ and $\{\tau \le t\} \in \mathcal{F}_t$. As $\mathcal{F}_t$ is a $\sigma$-algebra, $\{\sigma \le t\} \cup \{\tau \le t\} \in \mathcal{F}_t$. $\square$

(b) Assume $\sigma, \tau < \infty$ a.s. We need $\{\sigma + \tau \le t\} \in \mathcal{F}_t$. Write:

\[ \{\sigma + \tau \le t\} = \bigcup_{q \in \mathbb{Q} \cap [0, t]} (\{\sigma \le q\} \cap \{\tau \le t - q\}) \]

For each rational $q \in [0, t]$: $\{\sigma \le q\} \in \mathcal{F}_q \subseteq \mathcal{F}_t$ and $\{\tau \le t - q\} \in \mathcal{F}_{t-q} \subseteq \mathcal{F}_t$. So each intersection is in $\mathcal{F}_t$, and the countable union is in $\mathcal{F}_t$. $\square$

(c) Let $\sigma = \tau = \inf\{t : W_t = 1\}$. Then $\sigma - \tau = 0$ is trivially a stopping time. For a non-trivial example: let $\sigma = 2$ (deterministic) and $\tau = \inf\{t : W_t = 1\}$. Then $\sigma - \tau = 2 - \tau$. The event $\{2 - \tau \le t\} = \{\tau \ge 2 - t\}$ for $t \le 2$, which is $\{\tau < 2 - t\}^c$. For small $t$, $\{\tau \ge 2 - t\}$ depends on whether Brownian motion has hit 1 by time $2 - t$, which requires looking at the path up to time $2 - t > t$ (for $t < 1$). This is not necessarily in $\mathcal{F}_t$. Hence $\sigma - \tau$ need not be a stopping time.

Exercise 3: The σ-Algebra F_τ¶

Let $\tau = \inf\{t : W_t = 1\}$.

(a) Show that $W_\tau$ is $\mathcal{F}_\tau$-measurable.

(b) Is $W_{\tau + 1}$ $\mathcal{F}_\tau$-measurable?

(c) Describe $\mathcal{F}_\tau$ in words.

Solution to Exercise 3

(a) Let $\tau = \inf\{t : W_t = 1\}$. The random variable $W_\tau = 1$ (by continuity of Brownian paths, the process equals 1 at the hitting time). A constant is measurable with respect to any $\sigma$-algebra, so $W_\tau$ is $\mathcal{F}_\tau$-measurable.

More generally, for any progressively measurable process $X$, $X_\tau$ is $\mathcal{F}_\tau$-measurable on $\{\tau < \infty\}$.

(b) $W_{\tau + 1}$ is not $\mathcal{F}_\tau$-measurable. By the strong Markov property, $\widetilde{W}_s = W_{\tau + s} - W_\tau$ is a Brownian motion independent of $\mathcal{F}_\tau$. Therefore $W_{\tau + 1} = W_\tau + \widetilde{W}_1 = 1 + \widetilde{W}_1$, and $\widetilde{W}_1 \sim N(0, 1)$ is independent of $\mathcal{F}_\tau$. Since $W_{\tau + 1}$ depends on the independent increment $\widetilde{W}_1$, it is not $\mathcal{F}_\tau$-measurable.

(c) $\mathcal{F}_\tau$ consists of all events whose occurrence can be determined by knowing the path of Brownian motion up to and including the first time it hits 1. Formally, $\mathcal{F}_\tau = \sigma(W_{t \wedge \tau} : t \ge 0)$ — it contains all information about the Brownian path up to the hitting time, but no information about the path after hitting 1. It encodes the shape of the path on $[0, \tau]$, the value of $\tau$ itself, and all events determined by these.

Exercise 4: Strong Markov Property¶

(a) State the strong Markov property for Brownian motion.

(b) Let $\tau_a = \inf\{t : W_t = a\}$. Use the strong Markov property to show that $\tau_a + \tau_b \circ \theta_{\tau_a}$ has the same distribution as $\tau_{a+b}$, where $\theta$ is the shift operator.

(c) Derive the reflection principle: $\mathbb{P}(\sup_{s \le t} W_s \ge a) = 2\mathbb{P}(W_t \ge a)$ for $a > 0$.

Solution to Exercise 4

(a) The strong Markov property for Brownian motion: Let $W$ be a standard Brownian motion and $\tau$ a stopping time with $\tau < \infty$ a.s. Then the process $\widetilde{W}_t := W_{\tau + t} - W_\tau$, $t \ge 0$, is a standard Brownian motion independent of $\mathcal{F}_\tau$.

(b) By the strong Markov property, $\widetilde{W}_t = W_{\tau_a + t} - W_{\tau_a}$ is a Brownian motion independent of $\mathcal{F}_{\tau_a}$, starting from 0. The first hitting time of level $b$ by $\widetilde{W}$ is $\tilde{\tau}_b = \inf\{t : \widetilde{W}_t = b\}$. This has the same distribution as $\tau_b$ (by the definition of Brownian motion).

Now, $\tau_a + \tilde{\tau}_b$ is the first time $W$ reaches level $a$ (at time $\tau_a$) and then the additional time for the process to travel from $a$ to $a + b$ (an additional distance of $b$). Since $W_{\tau_a} = a$, the first time $W$ reaches level $a + b$ is $\tau_{a+b} = \tau_a + \tilde{\tau}_b$, and $\tilde{\tau}_b \stackrel{d}{=} \tau_b$ is independent of $\tau_a$. Therefore $\tau_{a+b} \stackrel{d}{=} \tau_a + \tau_b$ with $\tau_a$ and $\tau_b$ independent.

(c) For $a > 0$, define $\tau_a = \inf\{t : W_t = a\}$. Decompose the event $\{W_t \ge a\}$:

\[ \{W_t \ge a\} \subseteq \left\{\sup_{s \le t} W_s \ge a\right\} \]

Conversely, on the event $\{\sup_{s \le t} W_s \ge a\}$, we have $\tau_a \le t$. By the strong Markov property, $\widetilde{W}_s = W_{\tau_a + s} - a$ is a Brownian motion independent of $\mathcal{F}_{\tau_a}$. Thus:

\[ \mathbb{P}\left(\sup_{s \le t} W_s \ge a\right) = \mathbb{P}(\tau_a \le t) \]

Given $\tau_a \le t$, by symmetry of $\widetilde{W}$:

\[ \mathbb{P}(W_t \ge a \mid \tau_a \le t) = \mathbb{P}(W_t - a \ge 0 \mid \tau_a \le t) = \mathbb{P}(\widetilde{W}_{t - \tau_a} \ge 0 \mid \tau_a \le t) = \frac{1}{2} \]

(The last equality uses symmetry of Brownian motion: $\mathbb{P}(\widetilde{W}_s \ge 0) = 1/2$ for any $s > 0$, and independence.)

Therefore:

\[ \mathbb{P}(W_t \ge a) = \mathbb{P}(W_t \ge a, \tau_a \le t) = \mathbb{P}(\tau_a \le t) \cdot \frac{1}{2} = \frac{1}{2}\mathbb{P}\left(\sup_{s \le t} W_s \ge a\right) \]

Rearranging:

\[ \mathbb{P}\left(\sup_{s \le t} W_s \ge a\right) = 2\mathbb{P}(W_t \ge a) \quad \square \]

Concept	Definition	Key Property
Stopping time \(\tau\)	\(\{\tau \le t\} \in \mathcal{F}_t\)	Observable without seeing the future
First hitting time	\(\inf\{t : X_t \in A\}\)	Stopping time for adapted continuous \(X\)
\(\mathcal{F}_\tau\)	Information at random time \(\tau\)	\(X_\tau\) is \(\mathcal{F}_\tau\)-measurable
Stopped process \(X^\tau\)	\(X_{t \wedge \tau}\)	Preserves martingale property
Strong Markov	Fresh start at \(\tau\)	\(W_{\tau + t} - W_\tau\) is independent BM