Martingales and stopping times
Express fairness relative to information, distinguish legitimate stopping rules from future selection, and apply optional stopping with its hypotheses.
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The question for today
What you’ll understand
- Check adaptedness, integrability, and conditional fairness.
- Determine whether a rule is a stopping time.
- Use a bounded or otherwise justified optional-stopping argument.
A fair process can wander far from its starting value, and selecting an attractive stopping rule does not automatically produce a profitable strategy. Martingales make fairness conditional on everything available now. Stopping times restrict decisions to information already revealed, while optional-stopping theorems specify when stopping preserves expectation. These qualifications are the substance of the theory. Ignoring them creates convincing false proofs, from guaranteed betting profits to unjustified first-passage calculations for noisy dynamical systems.
Fairness is conditional and information dependent
A filtration is an increasing family of sigma algebras representing information revealed through time. A martingale Mₙ is adapted, so its current value is known at time n; each value is integrable; and its next conditional expectation equals its present value. A symmetric random walk with independent increments of plus or minus one is the basic example. Fairness refers to expected increments given the past, not to the path remaining close to zero or alternating wins and losses.
The filtration matters. Revealing tomorrow's increment today would destroy the martingale property of a nontrivial symmetric walk relative to that enlarged information. Conversely, an integrable terminal payoff Z generates a martingale Mₙ=E[Z|Fₙ] by the tower property. This is a model for updating predictions as evidence arrives: realized predictions change, but their conditional future mean equals the current prediction. Zero unconditional mean increments alone do not establish this stronger information-dependent property.
A decision can use the past but cannot inspect the future
A nonnegative integer-valued random time T is a stopping time when the event T≤n is determined by information available at n, for every n. The first time a walk hits a threshold qualifies: by time n, the observed path reveals whether a hit has occurred. The last time a walk visits a threshold before an unknown future endpoint generally does not qualify, because deciding that a visit was the last one requires inspecting future observations.
For a bounded stopping time and an integrable discrete-time martingale, optional stopping gives E[M_T]=E[M₀]. For unbounded stopping times, further hypotheses are essential. One common route is uniform integrability of the stopped family; another uses bounded increments together with finite expected stopping time. Merely knowing T is finite almost surely is insufficient. Always identify the theorem version and show its hypotheses rather than appending an informal phrase about waiting until the favorable event eventually occurs.
Localization is an argument, not an automatic limit
An unbounded stopping time can be truncated to T∧n. Each truncated time is bounded, so the corresponding expectation identity is valid. Passing n to infinity requires control of the stopped variables. Bounded convergence works when their magnitudes share a deterministic bound; uniform integrability is a more general tool. Almost-sure convergence alone does not preserve expectations, because probability mass can escape through rarer and larger values. This is the same obstruction that appears in disappearing-spike examples from integration theory.
For a symmetric walk started at zero, the first hit of +1 is finite almost surely, and the stopped value is exactly one. If unrestricted optional stopping applied, its expectation would simultaneously be zero and one. The contradiction identifies a failed hypothesis: this one-sided hitting time has infinite mean, and the truncated stopped walk is not uniformly integrable. In contrast, a walk stopped on leaving a finite interval stays bounded, allowing a controlled gambler's-ruin probability calculation. Keeping these two exit rules side by side helps isolate the hypothesis: both eventually stop, but only the bounded-interval stopped values admit the simple bounded-convergence argument.
Work it through
A fair walk between two absorbing boundaries
A symmetric unit-step walk starts at 1 and stops at its first hit of 0 or 4. Find its probability of hitting 4 first and its expected stopping time.
The stopped walk always lies between 0 and 4. Exit occurs almost surely with finite mean: from any interior state, a run of four consecutive downward steps exits within four moves, an event with conditional probability at least 1/16. Repeated four-step blocks give a geometric tail bound.
Apply optional stopping to the bounded stopped walk. If q is the probability of exit at 4, then E[S_T]=4q=E[S₀]=1. Thus q=1/4. The expected squared exit position is E[S_T²]=16q=4.
For a symmetric unit-step walk, Sₙ²−n is a martingale. Apply it at T∧n, then use bounded convergence for S² and monotone convergence for T∧n, with the finite-mean control just established. This gives E[S_T²]−E[T]=S₀²=1, so E[T]=4−1=3 moves.
The upper-boundary probability is 1/4 and the expected exit time is 3 moves. Bounded state values and a controlled stopping-time tail justify the limiting steps.
The living notebook
Explore the model
Interactive lab · probability
A fixed horizon is different from a stopping rule
Set p to 1/2 and vary n. The exact binomial count K translates to the endpoint of a fair walk by Sₙ=2K−n, with mean zero. The bars describe a fixed deterministic horizon; they do not stop paths at barriers or display hitting probabilities. Compare that fixed-horizon distribution with the boundary argument in the worked example.
- Mean · np
- 5
- Variance · np(1−p)
- 2.5
- Total probability
- 1
These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forChanging a deterministic horizon and choosing a path-dependent stopping time are different operations, each requiring its own argument.
Pause & reason
Check your understanding
Your turn
Try it yourself
For the same symmetric walk stopped at 0 or 4, start at 3 instead of 1. Compute the upper-exit probability and mean exit time.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Use E[S_T]=3 and E[S_T²−T]=9 after the same stopping justification.
Compare with the solution
Let q be the probability of reaching 4 first. Then 4q=3, so q=3/4. Also E[S_T²]=16q=12. The squared-walk martingale gives E[T]=12−9=3. The reflected starting positions 1 and 3 have complementary upper-exit probabilities and equal mean exit times, which checks the symmetry.
Keep the idea
Take it with you
- A martingale is fair relative to a specified filtration, with adaptedness and integrability included.
- A stopping rule must be decidable from current information without looking ahead.
- Unbounded optional stopping needs explicit control; almost-sure finiteness alone is not enough.
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