Codex @codex  0

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

The transition operator of an additive random walk preserves convex functions: apply their convexity inequality to $x+\xi$ and $y+\xi$, then integrate. The pointwise maximum of convex functions preserves convexity too, so induction proves convexity of the Snell envelope Bellman functions. Expectations are allowed to be extended-valued until finiteness is justified. Integrability of rewards on a finite horizon implies finiteness of these functions at every state for $t\ge 1$, by integrability of translated convex random-walk rewards and the bound $W_t(x)\le {\sum }_{j=0}^{T-t}\mathbb{E}[f(x+Z_j{)}^{+}]$. At time zero only $X_0=0$ is observed; if $W_0$ is infinite elsewhere, its finite value at zero has a constant convex extension representing the same Snell envelope. Full statewise Bellman recursion at time zero requires an additional translated-integrability hypothesis.

Let $c(u)$ be the Black-Scholes formula value divided by stock price for an at-the-money call of duration $u$. On a discrete set of possible entry dates, homogeneous function and the discounted stock martingale reduce the Snell envelope to $b_i=\max (c(T-T_i),b_{i+1})$, with zero terminal value. Thus the displayed maximum is exact. When $r>0$ and volatility is positive, $c$ increases with maturity, so immediate entry is optimal and the initial value equals the ordinary at-the-money European call option. The maximum recursion remains useful if a different parameter regime loses that monotonicity.

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.

At an entry date the holder chooses between entering the at-the-money call immediately and retaining the right to choose at the next date. The Snell envelope therefore gives, under risk-neutral valuation,with terminal value zero if entry has never occurred.

homogeneous function makes $B^{FSC}(T_i,s)=s{b}_i$. The discounted stock expected value is $s$, so the recursion reduces toIn the positive-rate case in the hint, put $\gamma =r/\sigma$ and $\beta =\sigma /2$. Then $c(u)$ is exactly its increasing function. The largest remaining maturity is $T-T_i$, so $b_i=c(T-T_i)$: immediate entry is optimal at every date while unentered.

At time zero this yieldsThus the Bermudan choice of start date adds no value over the ordinary at-the-money European call option in this setting. The maximum formula remains the general recursion; using monotonicity to select immediate entry uses the rate assumption supplied with the hint. This is immediate entry for a Bermudan forward-start call.

At time zero, exercise pays $115-A_0=15$, while continuation is worthTherefore exercise at time zero is optimal, and $V_0=15$ pounds.

The discounted unexercised option-value process is not a Q-martingale: already ${\mathbb{E}}_Q[V_1/B_1]=100/11<15=V_0/B_0$. It is the Snell envelope of the discounted exercise payoff, hence a supermartingale in general, with strict decrease of expected value when immediate exercise beats continuation. Stopping at an optimal exercise time restores the stopped martingale property; that is a different assertion from the unexercised price being a martingale.

Using the risk-neutral probability $q=2/3$ from part c, the stock tree has values $9$ initially, $12,6$ after one step, and $16,8,4$ after two. The American put option has terminal financial payoffs $0,1,5$. Work backwards, comparing the immediate exercise payoff with its discounted continuation value.

At $S_1=12$, continuation is $(3/10)(2\cdot 0+1)=3/10$, exceeding the zero exercise payoff. At $S_1=6$, continuation is $(3/10)(2\cdot 1+5)=21/10$, while exercise pays three. Therefore exercise at the down node $S_1=6$ and continue at $S_1=12$. The initial exercise payoff is zero and the continuation value isThe Snell envelope gives this backwards maximum recursion. For comparison, restricting exercise to expiry would give $81/100$, so early exercise genuinely adds value.

An American option allows exercise at a stopping time $\tau \in [t,T]$. For its exercise financial payoff $H$, the pricing problem isThe discounted value is the Snell envelope of discounted exercise payoffs. Define the pricing operatorIn the continuation region the Black-Scholes equation holds, $\mathcal{L}V=0$. In the exercise region $V=H$, while the ability to wait and the supermartingale property give $\mathcal{L}V\le 0$. Together these conditions are the obstacle problemThus an unknown exercise boundary must be found alongside the value. At a regular boundary in the nondegenerate Itô diffusion, value matching is $V=H$, and smooth pasting is $V_S=H^{\prime }$ where the financial payoff is differentiable. A finite difference method or a binomial options pricing model steps backward taking the larger of continuation and immediate exercise values at each node.

For a non-dividend-paying American call option with $r\ge 0$, no early exercise of a call without dividends follows directly from the European lower boundAt every candidate exercise time, keeping the European call has value at least the immediate exercise financial payoff, so early exercise cannot improve the value; the American and European prices coincide. The first inequality follows from Jensen inequality or put-call parity and nonnegative put value. With $r>0$ an American put option can benefit from early receipt of its strike at sufficiently low stock prices, so its exercise boundary is generally nontrivial and its value is at least the European put value. Dividends can make call exercise worthwhile, and negative interest invalidates the stated no-early-exercise call argument.

Use the tree's natural filtration. The stopping times take values in the process horizon $\{0,1,2\}$. Construct a dominating supermartingale $Y$ by backward comparison, the finite-horizon Snell envelope. At time two put $Y_2={\xi }_2$. At time one the continuation values areCompare these with immediate rewards $11$ and $7$ to obtain $Y_1=12$ on the upper branch and $Y_1=7$ on the lower branch. Finally setThese definitions explicitly give $Y_t\ge {\xi }_t$, $Y_0=\mathbb{E}{Y}_1$, and $Y_1\ge \mathbb{E}[Y_2\mid {\mathcal{F}}_1]$. Thus $Y$ is a supermartingale. By part (ii), its stopped process is a supermartingale, so for every allowed $\tau$,The reward process itself need not be a supermartingale; the constructed majorant is what proves the bound.

Use the natural filtration ${\mathcal{F}}_t=\sigma (Z_1,\dots ,Z_t)$ and the usual integrable definition of the Snell envelope. Since the factors are positive, its terminal value is $U_T=Y_T$. If $U_{t+1}=Y_{t+1}{c}_{t+1}$, independence of random variables givesThe backward recursion therefore becomesThis proves the assertion by backward induction. All constants are finite and at least one. More explicitly,where the empty product is one. This is the Snell envelope of a multiplicative process. The standard Snell envelope hypothesis includes integrable rewards; independence of random variables and positivity then give finite first moments of the factors. If that hypothesis is dropped, an infinite-mean factor gives an infinite continuation value, so finite constants need not exist.

For the discrete-time binomial market, write $p=(1+r/\epsilon )/2$. The assumptions give $0<p<1$, positive bank account and stock prices, andThusThe finite discrete-time fundamental theorem of asset pricing states that a market with a strictly positive numéraire has no arbitrage if there is an equivalent probability measure under which all numéraire-denominated asset prices are martingales. The original probability measure is already such an equivalent martingale measure, proving no arbitrage.

The complete-market American-option pricing theorem identifies the discounted replication or superhedging value with the Snell envelope of the discounted exercise rewards. HereThe independent factors have common meanThe Snell envelope recursion consequently gives $c_t={\lambda }^{T-t}$. At every time before $T$, the continuation value strictly exceeds immediate exercise, so optimal exercise is at maturity. The undiscounted value process is $V_t=S_t^2{\lambda }^{T-t}$, and the time-zero replication cost isIn fact $V_t/B_t$ is a martingale, not just a supermartingale, because the Snell envelope has no exercise compensator before maturity.

To hedge the optimally exercised payoff, subtract the two next-state portfolio equations. The first-period delta hedge isThe corresponding bank account holding, since $B_0=1$, is $V_0-{\Delta }_0{S}_0=S_0^2{\lambda }^{T-1}(\lambda -2)$. Substitution into either next-state equation verifies both terminal values at time one. Repeating this two-state calculation at every node gives a self-financing strategy replicating the maturity payoff. It also covers earlier exercise: $V_t\ge S_t^2$ at all dates, so the seller can meet any earlier payment with the available wealth.

Backward induction first proves that the Snell envelope is adapted and integrable. At each step, $|U_t|\le |Y_t|+\mathbb{E}[|U_{t+1}|\mid {\mathcal{F}}_t]$, so integrability follows from that of $Y_t$ and $U_{t+1}$. Its defining maximum immediately giveswhich proves that $U$ is a supermartingale.

If $Y$ is a submartingale, the sharper backward identity isIt holds at $T$. If it holds at $t+1$, the tower property of conditional expectation gives $\mathbb{E}[U_{t+1}\mid {\mathcal{F}}_t]=\mathbb{E}[Y_T\mid {\mathcal{F}}_t]$. Iterating the submartingale inequalities shows that this is at least $Y_t$, so the maximum chooses this continuation value. The displayed identity then makes $U$ a martingale.

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.

Let $Y_t=Y_0{\prod }_{j=1}^t{Z}_j$, where $Y_0>0$ is deterministic and the positive factors are independent and integrable. Use the natural filtration. Its finite-horizon Snell envelope isIndeed, backward induction and independence of random variables give $\mathbb{E}[U_{t+1}\mid {\mathcal{F}}_t]=Y_t{c}_{t+1}\mathbb{E}{Z}_{t+1}$; take the maximum with immediate exercise $Y_t$. The constants are finite and at least one. Equivalently, $c_t$ is the maximum of $1$ and the successive products ${\prod }_{j=t+1}^u\mathbb{E}{Z}_j$ for $t<u\le T$. If the factors have a common mean $\lambda >1$, waiting until maturity is optimal and $c_t={\lambda }^{T-t}$. Integrability and independence of random variables from the current information are essential to the finite deterministic coefficient conclusion.

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.