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An American option may be exercised at any time up to maturity. Its value is the Snell envelope of its discounted exercise payoff.

Choosing a stopping time to maximize an expected reward. For an adapted integrable reward process over a finite discrete horizon, the Snell envelope computes the conditional optimal reward by comparing immediate exercise with continuation.

An optimal stopping time attains the supremum of the expected reward over an admissible class of stopping times. For a finite-horizon Snell envelope, the first time at which its value equals the current reward is optimal.

Backward induction makes the Snell envelope integrable and adapted: $|U_t|\le |Y_t|+\mathbb{E}(|U_{t+1}|\mid {\mathcal{F}}_t)$. Its definition gives $U_t\ge Y_t$ and $U_t\ge \mathbb{E}(U_{t+1}\mid {\mathcal{F}}_t)$, so it is a supermartingale dominating the reward.

For a stopping time $\tau$ taking values in $\{0,\dots ,T\}$, expand its stopped value asThe indicators are ${\mathcal{F}}_t$-measurable. Taking conditional expectations in each summand makes its expectation nonpositive by the supermartingale property. Since ${\mathcal{F}}_0$ is trivial, $U_0$ is deterministic andThis proves the finite-horizon optional sampling theorem directly in the instance needed here, without assuming nonnegative rewards.

The first-contact time ${\tau }_{\ast }\le T$ is a stopping time: the event $\{{\tau }_{\ast }\le t\}$ is the finite union of events $\{U_s=Y_s\}$ for $s\le t$. It is finite because $U_T=Y_T$. On $\{{\tau }_{\ast }>t\}$ we have $U_t>Y_t$, so the maximum defining the Snell envelope selects continuation andThe stopped process $U_{t\wedge {\tau }_{\ast }}$ is therefore a martingale: before stopping its conditional increment is zero and after stopping its increment vanishes. Applying the stopped-sum argument with zero conditional increments givesThe upper bound from part (a) now proves that first contact is an optimal stopping time.

Let $v_j$ be the optimal expected value before seeing the next offer when $j$ rounds remain. With one round left the offer must be accepted, so $v_1=1/2$. For $j>1$, observing $x$ gives a choice between $x$ now and the continuation value $v_{j-1}$. Independence of future uniform distributions makes that continuation value independent of past offers. Thus the uniform-offer stopping recursion isIt gives $v_2=5/8$ and $v_3=89/128$. The Snell envelope rule therefore yields the explicit strategyOnly rounds actually reached are played. At a threshold the two actions have identical continuation expected value; either convention is optimal, and exact equality has probability zero. The optimal expected payout before the first offer is

Backward induction makes every $U_t$ adapted and integrable: the conditional expectation of an integrable random variable is integrable, and $|\max (a,b)|\le |a|+|b|$. The Snell envelope recursion directly gives $U_t\ge \mathbb{E}[U_{t+1}\mid {\mathcal{F}}_t]$, so $U$ is a supermartingale.

If $Z$ is a submartingale, iterating its defining inequality gives $\mathbb{E}[Z_T\mid {\mathcal{F}}_t]\ge Z_t$. Starting at $U_T=Z_T$ and applying the tower property of conditional expectation backwards showsThis last conditional expectation is a true martingale. The conclusions areIn the second case waiting until $T$ attains the optimal stopping value. It is not generally true that $U_t=Z_t$: that further equality holds when $Z$ itself is a martingale.

For the American put option, compare exercise with continuation at time one. At $S_1=6$, both values are zero. At $S_1=4$, immediate exercise pays $1$, whereas the continuation value is the European value $4/9$, so exercise is optimal. At time zero, immediate exercise pays zero and continuation isThe optimal policy is therefore to continue at time zero, exercise at time one if $S_1=4$, and otherwise continue to maturity. This nodewise maximum is the Snell envelope value.

With independent uniform distributions on $[0,1]$ as sequential offers and a finite number of rounds, the continuation value with $j-1$ offers remaining is $v_{j-1}$. Accept the current offer when it reaches that threshold. Integrating the maximum of the offer and continuation value gives the recursion, by the Snell envelope principle.