A stopping time is a random time \tau at which
you decide to act — but the decision may use only the information available so far.
Relative to a
filtration
(\mathcal{F}_t), \tau is a stopping
time when, for every t,
\{\tau \le t\} \in \mathcal{F}_t.
Read it plainly: by time t you can already tell whether
\tau has happened. No peeking into the future —
the event "I have stopped by now" must be decidable from what you have observed up to now,
never from what is still to come.
One you can detect, one you cannot
The canonical stopping time is the first hitting time of a level
a,
\tau_a = \inf\{\, t \ge 0 : W_t = a \,\}.
The moment the path first touches a, you know it — no future
information required. Let us turn that intuition into a proof, line by line.
Step 1 — rewrite the event "hit by time t" without
mentioning \tau_a. Brownian paths are
continuous. A continuous path that starts below the level
a can only get above it by passing through it, so it has reached
a at some moment on [0,t] exactly when
its running maximum over [0,t] has reached
a:
\{\tau_a \le t\} = \Big\{\, \sup_{s \le t} W_s \ge a \,\Big\}.
(Continuity is doing real work here — for a jumpy path the two events can differ, since a
jump could clear a without ever equalling it.)
Step 2 — reduce the supremum to countably many observations. A supremum over
an uncountable set of times sounds hard to "see", but continuity rescues us again: a
continuous function attains its supremum over [0,t] as a limit
along the rationals, so
\sup_{s \le t} W_s = \sup_{\substack{q \in \mathbb{Q} \\ 0 \le q \le t}} W_q.
Step 3 — check each observation is known by time t.
For every rational q \le t the value
W_q is \mathcal{F}_t-measurable (it was
observed at or before t). A supremum of countably many
\mathcal{F}_t-measurable quantities is again
\mathcal{F}_t-measurable, so the running maximum is
\mathcal{F}_t-measurable, and therefore so is the event that it
clears a:
\Big\{\, \sup_{s \le t} W_s \ge a \,\Big\} \in \mathcal{F}_t.
Step 4 — conclude. By Step 1 that event is
\{\tau_a \le t\}, so
\{\tau_a \le t\} \in \mathcal{F}_t \quad \text{for every } t,
which is exactly the definition of a stopping time. The first hitting time passes the test:
you can always tell whether it has happened using only what you have already seen.
Let (W_t) be Brownian motion adapted to a filtration
(\mathcal{F}_t), and fix a level a. The
first hitting time \tau_a = \inf\{\, t \ge 0 : W_t = a \,\} is a
stopping time: \{\tau_a \le t\} \in \mathcal{F}_t for every
t. The argument needs only path continuity (to write
\{\tau_a \le t\} = \{\sup_{s\le t} W_s \ge a\}) and the reduction
of that supremum to the rationals.
The cautionary contrast. Compare the time at which a path attains its
maximum over [0,T],
\theta = \operatorname*{arg\,max}_{s \le T} W_s.
This is not a stopping time. To decide the event
\{\theta \le t\} — "the maximum has already occurred by
t" — you would have to be sure that nothing on the remaining
interval [t, T] ever climbs higher. That comparison is against the
future:
\{\theta \le t\} = \Big\{\, \sup_{s \le t} W_s \;\ge\; \sup_{t \le s \le T} W_s \,\Big\},
and the right-hand supremum is over [t,T], which is
not \mathcal{F}_t-measurable. So
\{\theta \le t\} \notin \mathcal{F}_t in general: a peak only
reveals itself as the peak once you have seen that nothing later beats it. Refresh
the figure for a fresh Brownian path; the level-hit marker is detectable live, the argmax
marker needs the whole future.
Why it matters
Stopping times are the rule of fair play in dynamic decisions. They power the
optional-stopping theorem
— which says a fair game stopped at a stopping time stays fair — and they model the
exercise time of an
American option:
the holder may exercise whenever they like, but the decision can only depend on information
seen so far, never on a peek at tomorrow's price. That "no clairvoyance" constraint is
exactly \{\tau \le t\}\in\mathcal{F}_t.
The headline result about stopping times is Doob's optional stopping
theorem. If (M_t) is a
martingale
— a fair game, with \mathbb{E}[M_t \mid \mathcal{F}_s] = M_s —
and \tau is a stopping time, then under suitable conditions the
expected value at the random stopping time is just the starting value:
\mathbb{E}[M_\tau] = \mathbb{E}[M_0].
Read it as: a fair game stays fair no matter what stopping rule you quit on. You
cannot manufacture an edge by cleverly choosing when to walk away, as long as your
decision uses only information seen so far — which is precisely what "stopping time" buys
you. The caveat matters. The conclusion can fail for an
unbounded stopping time: it needs a regularity condition such as
\tau being bounded (\tau \le T), or
(M_{t \wedge \tau}) being uniformly integrable. Without it the
classic counterexample is the doubling ("martingale") betting system on Brownian motion,
where \tau = \inf\{t : W_t = 1\} is finite almost surely yet
\mathbb{E}[W_\tau] = 1 \ne 0 = \mathbb{E}[W_0] — the unbounded
waiting time smuggles the apparent free money back out.
A fixed time t carries the information
\mathcal{F}_t. What is the analogue at a random stopping
time \tau? It is the σ-algebra
\mathcal{F}_\tau = \big\{\, A \in \mathcal{F} : A \cap \{\tau \le t\} \in \mathcal{F}_t \text{ for every } t \,\big\}.
The defining clause is the same "no peeking" idea applied to information itself: an event
A counts as known by time \tau when,
on the part of the sample space where \tau has already occurred
by t, you can decide A using only
\mathcal{F}_t. One can check this collection really is a
σ-algebra, that \tau itself is
\mathcal{F}_\tau-measurable, and that for a constant time
\tau \equiv t it reduces to the ordinary
\mathcal{F}_t — exactly what "information available up to the
moment you act" should mean. It is the natural conditioning σ-algebra in both optional
stopping and the strong Markov property.