research36 min studyLesson 686 in Mathematics

Optimal stopping and Snell’s envelope

Construct the smallest dominating supermartingale and separate stopping rules from hindsight.

On this page

What you’ll understand

  • Compute a finite-horizon Snell envelope by backward induction.
  • Verify optimality through contact with the reward and a stopped martingale.
  • Explain why uniform integrability, path regularity, and the horizon matter in continuous time.

In optimal stopping, the action is the decision to stop observing and collect the current reward. The decision must use current information, so choosing the largest reward after seeing an entire path is usually inadmissible. The Snell envelope records the best achievable conditional reward and simultaneously gives a supermartingale bound on every stopping rule. A finite tree makes its construction exact, while continuous and infinite horizons reveal the assumptions hidden by a short backward recursion.

The finite recursion creates the smallest bound

For a finite discrete horizon and integrable adapted rewards, define the envelope at the final time to equal the reward. Move backward by taking the maximum of stopping now and the conditional expected envelope next period. This process dominates the reward and is a supermartingale. If another supermartingale dominates the reward, backward induction shows it also dominates this envelope. Thus the recursion constructs the smallest supermartingale with the required domination property.

All stopping times here take values in the finite decision grid. The first time the envelope equals the reward is a stopping time, since equality is observable at that date. Before contact, the recursion chooses the conditional continuation value, so the stopped envelope is a martingale. Optional sampling and contact then give expected reward equal to the initial envelope, while any other stopping rule is bounded by it.

UN=ZN,Un=max⁡{Zn,E[Un+1∣Fn]},τ∗=min⁡{n:Un=Zn}.U_N=Z_N,\qquad U_n=\max\{Z_n,\mathbb E[U_{n+1}\mid\mathcal F_n]\},\qquad \tau^*=\min\{n:U_n=Z_n\}.
The maximum is reward maximization, in contrast to the cost-minimization convention of the controlled-diffusion lessons.↔ Scroll wide equations horizontally.

Contact alone does not replace martingale verification

A dominating supermartingale bounds every allowed stopped reward through optional sampling. To attain that bound, a rule must collect the envelope at contact and avoid losing expected envelope value before it stops. Hence contact together with the martingale property of the stopped envelope certifies optimality. A rule that contacts only after the envelope has already fallen in expectation may be suboptimal. The martingale condition keeps track of that otherwise invisible loss.

In continuous time, define the value family using an essential supremum over stopping times, rather than a pointwise supremum over sample paths. Its aggregation into a right-continuous Snell process and the existence of an optimal first-contact rule require regularity. For example, a bounded continuous adapted reward on a finite horizon with the usual Brownian filtration has the standard optimal-contact result. Boundedness ensures uniform integrability of stopped rewards; more general theorems replace it by class D and appropriate regularity along stopping times.

Uθ=ess sup⁡θ≤τ≤TE[Zτ∣Fθ],{Zτ:0≤τ≤T} uniformly integrable.U_\theta=\operatorname*{ess\,sup}_{\theta\le\tau\le T}\mathbb E[Z_\tau\mid\mathcal F_\theta],\qquad \{Z_\tau:0\le\tau\le T\}\text{ uniformly integrable}.
Uniform integrability of the stopped reward family is the class-D condition; additional path regularity is needed for the stated continuous-time conclusions.↔ Scroll wide equations horizontally.

A finite horizon prevents one form of nonattainment

An infinite horizon requires additional terminal or transversality control when limits of finite-horizon envelopes and stopping rules are taken. Uniform integrability alone does not force an optimal finite stopping time to exist. Consider deterministic rewards increasing to one while remaining strictly below it at every finite date. Waiting longer approaches the supremum, but every finite stopping rule misses it. If stopping at infinity is allowed, its payoff must be defined explicitly, producing a different problem.

Hindsight creates a separate issue even on a bounded finite tree. A pathwise rule that chooses whichever earlier date later turns out best need not be a stopping time: its decision to stop now can depend on a future branch. Its expected payoff is a useful upper bound, often called a prophet value, but it is not automatically attainable. The filtration-based recursion computes what can actually be achieved under the allowed information.

Zn=1−1n+1⟹sup⁡τ<∞E[Zτ]=1 with no finite optimizer.Z_n=1-\frac1{n+1}\quad\Longrightarrow\quad \sup_{\tau<\infty}\mathbb E[Z_\tau]=1\text{ with no finite optimizer}.
These bounded deterministic rewards demonstrate nonattainment despite uniform integrability when only finite stopping times are permitted.↔ Scroll wide equations horizontally.

An early-stop branch and a continuation branch

At time zero the reward is three. A fair first branch leads to reward seven on U and one on D at time one. From U, fair terminal rewards are four and eight; from D, they are zero and four. Compute the Snell envelope, an optimal stopping rule, and the hindsight upper bound.

  1. At the U node, terminal continuation has mean six, below the current seven, so the envelope is seven and stopping is optimal. At D, continuation has mean two, above the current one, so the envelope is two and waiting is optimal.

  2. The root continuation value is (7+2)/2=4.5, exceeding its current reward three. Therefore wait initially, stop at U, and continue to the terminal date at D. The expected collected reward is 7/2+0/4+4/4=4.5.

  3. With full hindsight, the four equally likely pathwise maxima are seven, eight, three, and four, averaging 5.5. The choice of time zero only on a future low branch would use information unavailable at time zero, so this larger value does not contradict optimality.

U1(U)=7,U1(D)=2,U0=max⁡{3,12(7+2)}=92.U_1(U)=7,\quad U_1(D)=2,\quad U_0=\max\{3,\tfrac12(7+2)\}=\frac92.
The stopping rule uses information available at each node, rather than the final path maximum.↔ Scroll wide equations horizontally.
The result

The optimal stopping value is 4.5. Hindsight gives 5.5, but its pathwise date choices are not an admissible stopping rule.

Explore the model

Interactive lab

Interactive lab · markov

Compare early stopping with terminal-only collection

Use a separate two-state stopping model with reward zero in state zero and four in state one, and two updates remaining. Pause and Reset, then set p to 0.5 and q to 1. Predict state-one probability 0.5 after one Step and 0.25 after two. Terminal-only expected reward is one. On paper, stopping at state one after the first update and otherwise waiting gives value three. The instrument evolves the unstopped chain; it does not compute a Snell envelope. Reset and set both controls before fresh comparisons because edits retain elapsed time.

Compare early stopping with terminal-only collectionExact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. Current readings: P(state 1) at step 0: 0%; Stationary P(state 1): 60%; Chain behavior: Monotone convergence. All parameters are adjustable with the labeled controls below.Two-state Markov distribution after 0 steps: state zero probability 1, state one probability 0.Initial distribution (1, 0) · step 0p = 0.3q = 0.201100%0%1−p1−qP(0)P(1)Monotone convergence
P(state 0)P(state 1)Stationary distribution
P(state 1) at step 0
0%
Stationary P(state 1)
60%
Chain behavior
Monotone convergence
t = 0 s

Exact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. 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 forAn early stopping rule changes collected reward without changing the transition law of the underlying unstopped process.

Check your understanding

For the finite-horizon Snell envelope, what certifies that a stopping rule attains its initial bound?

Try it yourself

Keep the worked tree unchanged after time zero, but raise the initial reward to 4.8. Find the new envelope value and the smallest optimal stopping time. Explain why future branch policies need not be recomputed.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The values at time one depend only on later rewards, so compare the new initial reward with the same continuation expectation.

Compare with the solution

The time-one envelope remains seven at U and two at D. Its initial conditional expectation is still 4.5. The initial maximum is now 4.8, so immediate stopping is optimal and the smallest optimal time is zero. The higher initial reward changes only the root comparison, not the continuation optimization on the unchanged subtrees.

Take it with you

  • The Snell envelope is the smallest supermartingale dominating the reward.
  • Optimal contact requires preservation of expected envelope value before stopping.
  • Infinite-horizon attainment and clairvoyant pathwise maxima require separate scrutiny.

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