advanced35 min studyLesson 664 in Mathematics

Optional sampling and uniform integrability

Identify when stopping preserves conditional means and why an almost-surely finite stopping time alone is insufficient.

On this page

What you’ll understand

  • Prove the bounded-stopping expectation identity.
  • State an unbounded-stopping conclusion with uniform integrability or closedness.
  • Calculate finite-barrier hitting probabilities and expected durations without an unjustified limit exchange.

A martingale has no predictable one-step drift, but choosing when to inspect it introduces a new limit question. Bounded stopping times preserve conditional fairness because only finitely many integrable increments are involved. For an unbounded stopping time, even one that is finite almost surely, rare stopped paths can retain expectation while disappearing in an almost-sure limit. Uniform integrability supplies the missing control. Optional sampling is therefore a theorem with hypotheses, not a permission to replace deterministic times by arbitrary random ones.

Bounded stopping uses predictable indicators

Write a martingale stopped at T as its initial value plus its increments multiplied by the indicators that T has not occurred before the corresponding step. Those indicators are predictable, so each retained increment has mean zero. If T is bounded by a fixed integer N, the finite sum is integrable and taking expectations gives the original mean. More generally, for stopping times S at most T at most N, conditional expectation of the stopped terminal value given the stopped information at S equals the value at S.

The stopped sigma-algebra consists of events whose intersection with stopping by each time is measurable by that time. It records what is known when the rule actually stops, rather than information at one predetermined index. The bounded conditional sampling statement can be checked by testing these events and splitting them over the possible stopping values. This formulation makes explicit both the information constraint on the times and the integrability needed for the conditional expectation to exist.

MT∧n=M0+∑k=1n1{T≥k}ΔMk,S≤T≤N⟹E[MT∣FS]=MSM_{T\wedge n}=M_0+\sum_{k=1}^n\mathbf1_{\{T\ge k\}}\Delta M_k,\qquad S\le T\le N\Longrightarrow\mathbb E[M_T\mid\mathcal F_S]=M_S
M is an integrable discrete-time martingale and S,T are stopping times. A deterministic finite N bounds both times in the conditional statement.↔ Scroll wide equations horizontally.

The limit needs more than almost-sure stopping

For an almost-surely finite T, the stopped variables converge almost surely to the value at T, but their expectations need not converge. A sufficient additional condition is uniform integrability of the family of stopped variables indexed by deterministic n. This gives convergence in L1, so the bounded expectation identities pass to the limit. Bounded stopped values provide an easy sufficient condition. Another sufficient setting is a martingale closed by one integrable terminal variable: each value is its conditional expectation under the current information.

A closed martingale is uniformly integrable, and at finite stopping times it remains the conditional expectation of that same closing variable given the stopped information. This yields conditional optional sampling between almost-surely finite stopping times. Other hypotheses are possible, such as bounded increments together with an integrable stopping time for the expectation identity, but they cannot be silently exchanged for almost-sure finiteness. The applicable sufficient condition should be named before removing a deterministic truncation.

lim⁡K→∞sup⁡nE ⁣[∣MT∧n∣1{∣MT∧n∣>K}]=0⟹MT∧n→MT in L1,EMT=EM0\lim_{K\to\infty}\sup_n\mathbb E\!\left[|M_{T\wedge n}|\mathbf1_{\{|M_{T\wedge n}|>K\}}\right]=0\Longrightarrow M_{T\wedge n}\to M_T\text{ in }L^1,\quad \mathbb E M_T=\mathbb E M_0
Assume T is finite almost surely and M is an integrable martingale. The displayed tail condition is uniform integrability of the stopped family, not merely bounded first moments.↔ Scroll wide equations horizontally.

Barrier stopping supplies its own bounds

For a fair nearest-neighbor walk stopped upon reaching either of two finite barriers, the stopped positions stay between those barriers. Absorption occurs almost surely: from any interior site, a fixed finite sequence of steps has a positive probability of reaching a barrier, giving a geometric bound on survival over blocks. Thus bounded positions justify optional sampling for the walk. The terminal values then convert the preserved mean into a hitting probability. No independence of the stopping time and the walk is required.

To compute the expected duration, apply the bounded theorem to the squared walk minus time at the truncated stopping time. The squared stopped positions are uniformly bounded and converge to the squared terminal position. The truncated times increase to the stopping time, so monotone convergence identifies their expected limit. This proves the duration is integrable instead of assuming that fact to justify the very identity used to calculate it. A one-sided barrier has no analogous uniform bound on stopped positions and requires a separate argument.

Two asymmetric barriers

A fair walk starts at zero and stops at T=inf⁡{n:Sn∈{−2,3}}T=\inf\{n:S_n\in\{-2,3\}\}. Find the probability of exiting at three and the expected stopping time, with the limit justifications stated.

  1. Absorption is almost surely finite by the finite-barrier argument. The stopped walk lies in [−2,3][-2,3], so its family is uniformly integrable and EST=0\mathbb E S_T=0.

  2. Writing q for the upper-exit probability gives 3q−2(1−q)=03q-2(1-q)=0, so q=2/5q=2/5. The terminal squared position has expectation 9(2/5)+4(3/5)=69(2/5)+4(3/5)=6.

  3. For each n, bounded stopping of Sk2−kS_k^2-k gives E(T∧n)=EST∧n2\mathbb E(T\wedge n)=\mathbb E S_{T\wedge n}^2. Dominated convergence applies to the right side and monotone convergence to the left, yielding ET=6\mathbb E T=6.

The result

The upper-exit probability is two fifths and the expected duration is six steps. Both conclusions follow from justified stopped limits, with no assumption that the stopping time is independent.

Explore the model

Interactive lab

Interactive lab · probability

An endpoint event differs from a stopping event

Pause and Reset, set both p=0.5,n=3p=0.5,n=3, and predict P(S3≥2)=1/8P(S_3\ge2)=1/8 using S3=2K−3S_3=2K-3. On paper compare the event of having reached +2+2 by time three, whose probability is 1/41/4 from the first two heads. Reset and set both p=0.5,n=2p=0.5,n=2 to inspect that endpoint probability directly. The lab provides exact binomial endpoints, not stopped paths. Control edits retain time; Reset clears both controls and clock.

An endpoint event differs from a stopping eventThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

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 forThe time of observation changes which event is being measured, even for the same fair walk.

Check your understanding

Which condition is sufficient to pass from bounded stopping to E[MT] = E[M0] when T is finite almost surely?

Try it yourself

Repeat the barrier calculation for a fair walk stopped at −1-1 or 44. Explain why stopping only at +1+1 does not permit the same preserved-mean conclusion from almost-sure finiteness alone.

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 zero terminal mean and then the terminal squared mean for the two-sided calculation. For the one-sided rule inspect the terminal position itself.

Compare with the solution

For the two barriers, 4q−(1−q)=04q-(1-q)=0 gives q=1/5q=1/5. The expected squared exit position is 16/5+4/5=416/5+4/5=4, so the truncation argument yields expected duration four. The fair one-dimensional walk hits +1+1 almost surely, but at that stopping time its value is identically one, not mean zero. Its stopped family is not uniformly integrable; values arbitrarily far below zero prevent transferring the finite-time expectation identity.

Take it with you

  • Bounded stopping preserves conditional martingale means.
  • Almost-surely finite stopping still needs a justified integrable limit.
  • Uniform integrability or an integrable closing variable supplies sufficient control.

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