American options and optimal stopping

Definition Work on a finite set of states with a filtration $\mathcal{F}_0 \subseteq \mathcal{F}_1 \subseteq \dots \subseteq \mathcal{F}_T$, where $\mathcal{F}_0$ is trivial and $t = 0, 1, \dots, T$ indexes the allowed exercise dates. Let $B_t > 0$ be the cash account, an adapted numeraire, and let $\mathbb{Q}$ be a pricing measure for this numeraire that gives every state positive probability, in a complete model the unique one. An American-style option with exercise payoff $H_t \ge 0$, an adapted process so that $H_t$ is known at date $t$, lets its holder choose one admissible exercise time $\tau$, defined in the next card, and pays $H_{\tau}$ at date $\tau$. Its value at time $0$ is the value of an optimal stopping problem,

where $\mathcal{T}_{0,T}$ is the set of admissible exercise times. The states and the dates are finite, so there are finitely many such times, the maximum exists and every expectation is a finite sum. On a general probability space one assumes $\mathbb{E}^{\mathbb{Q}}[H_t/B_t] < \infty$ for every $t$ and takes a supremum. Exercise on a finite list of dates is called Bermudan in market language, and an option that can be exercised at every instant needs a continuous-time limit that this lesson does not treat. In a complete finite model the same number is also the smallest initial capital from which a self-financing portfolio can cover $H_t$ at every date and in every state, a result stated here without proof. The constant choices $\tau = 0$ and $\tau = T$ are admissible, so

where $V^{\mathrm{Eu}}_0$ is the value of the European claim that pays $H_T$ at $T$.

Definition A stopping time is a random date $\tau$ with values in $\{0, 1, \dots, T\}$ such that

for every date $t$. In discrete time this is equivalent to $\{\tau = t\} \in \mathcal{F}_t$ for every $t$, so whether exercise happens at date $t$ is decided by the information available at $t$. An admissible exercise time is a stopping time with values in the allowed exercise dates, and $\mathcal{T}_{t,T}$ denotes the admissible exercise times with values in $\{t, \dots, T\}$. Letting the option expire is the choice $\tau = T$, which pays $H_T$, and this is zero when the option ends out of the money. A rule that picks the date after seeing the whole path, such as the date with the largest discounted payoff, is in general not a stopping time and is not admissible.

Definition Let $Z_t = H_t/B_t$ be the exercise payoff in units of the cash account. Its Snell envelope under $\mathbb{Q}$ is defined backward from maturity by

for $t = T - 1, \dots, 0$, where $C_t$ is the continuation value of the next card. If the model also has dates at which exercise is not allowed, the maximum is taken only at the exercise dates and $Y_t = C_t$ at the others. In the finite model of the first card the following result holds. The process $Y$ dominates $Z$ and is a $\mathbb{Q}$-supermartingale, meaning $Y_t \ge \mathbb{E}^{\mathbb{Q}}[Y_{t+1} \mid \mathcal{F}_t]$ for every $t < T$, and it is the smallest such process, so every $\mathbb{Q}$-supermartingale $U$ with $U_t \ge Z_t$ for all $t$ satisfies $U_t \ge Y_t$ for all $t$. The first date at which exercise is at least as good as continuing,

is an admissible exercise time with $\tau^* \le T$, because $Y_T = Z_T$, and it is optimal,

so $V^{\mathrm{Am}}_0 = B_0 Y_0$. In the same way $V^{\mathrm{Am}}_t = B_t Y_t$ is the value at date $t$ of a right not yet exercised, since $\mathbb{E}^{\mathbb{Q}}[Z_{\tau} \mid \mathcal{F}_t] \le Y_t$ for every $\tau \in \mathcal{T}_{t,T}$, with equality for the first date from $t$ on at which $Y = Z$.

The proof uses only the finite horizon. Minimality follows by backward induction. At maturity $U_T \ge Z_T = Y_T$, and if $U_{t+1} \ge Y_{t+1}$, then $U_t \ge \mathbb{E}^{\mathbb{Q}}[U_{t+1} \mid \mathcal{F}_t] \ge C_t$ and $U_t \ge Z_t$, so $U_t \ge Y_t$. For optimality, stop $Y$ at an admissible time $\tau$. The increments of the stopped process are

and $\{\tau > t\} \in \mathcal{F}_t$, so their conditional expectation given $\mathcal{F}_t$ is $\mathbf{1}_{\{\tau > t\}}(C_t - Y_t) \le 0$. The stopped process is a supermartingale, hence $\mathbb{E}^{\mathbb{Q}}[Z_{\tau}] \le \mathbb{E}^{\mathbb{Q}}[Y_{\tau}] \le Y_0$. On the event $\{\tau^* > t\}$ we have $Y_t > Z_t$, hence $Y_t = C_t$, and the increments of the process stopped at $\tau^*$ have conditional mean zero. That stopped process is a martingale, so $\mathbb{E}^{\mathbb{Q}}[Z_{\tau^*}] = \mathbb{E}^{\mathbb{Q}}[Y_{\tau^*}] = Y_0$. Conditional expectations and martingales on a finite tree are set up in Martingales and discrete conditional pricing.

Definition For $t < T$ the continuation value $C_t = \mathbb{E}^{\mathbb{Q}}[Y_{t+1} \mid \mathcal{F}_t]$ is the value, in cash-account units, of keeping the right for at least one more date and then acting optimally. In currency it is $B_t C_t$. When the cash account grows by a deterministic factor $R$ per step, so that $B_{t+1} = R B_t$,

where $V_t = B_t Y_t$ is the American value at date $t$. The expectation uses the optimized values $V_{t+1}$, which already contain every later exercise decision, and not the value of holding the option to maturity. Since $Y$ is a supermartingale with $Y_T = Z_T$, it satisfies $Y_{t+1} \ge \mathbb{E}^{\mathbb{Q}}[Z_T \mid \mathcal{F}_{t+1}]$, so $C_t \ge \mathbb{E}^{\mathbb{Q}}[Z_T \mid \mathcal{F}_t]$, and in currency the continuation value is at least the date $t$ value of the European claim on $H_T$. It can still be smaller than the immediate payoff.

Definition At a date $t < T$ the exercise region is the set of states where immediate exercise attains the maximum in the recursion, the event $\{Z_t \ge C_t\}$, which is the same as $\{Y_t = Z_t\}$. The continuation region is the complementary event $\{Z_t < C_t\}$, and at maturity every state belongs to the exercise region. The rule $\tau^*$ stops the first time the state enters the exercise region. The exercise boundary is the border between the two regions. When the state is the stock price alone and the exercise region of a put at each date $t$ has the form $\{S_t \le b_t\}$, the boundary is described by critical prices $b_t$. In a tree the regions are sets of nodes, so the boundary is a classification of nodes, and a critical price is determined only up to the gap between neighboring node prices. The boundary depends on time, rates, dividends and the model state, and with several state variables a single price threshold need not describe it. At a node where $Z_t = C_t$ exercising and continuing are both optimal, and $\tau^*$ exercises. In continuous time the exercise boundary of an American put is a curve whose existence and regularity need further analysis, which is outside this lesson.

Definition The early-exercise premium is

where $V^{\mathrm{Eu}}_0$ values the European claim that pays $H_T$ at $T$. It is nonnegative because $\tau = T$ is admissible. In a finite model where $\mathbb{Q}$ gives every state positive probability, it is positive exactly when $Z_t > C_t$ at some date $t < T$ in some state. If that never happens, then $Y_t = C_t$ for every $t < T$, so $Y$ is a martingale and $Y_0 = \mathbb{E}^{\mathbb{Q}}[Z_T]$. If it happens on an event of $\mathcal{F}_t$, exercising there and holding to maturity otherwise beats the European claim, because $C_t$ is at least the conditional European value $\mathbb{E}^{\mathbb{Q}}[Z_T \mid \mathcal{F}_t]$.

One standard case has no premium. Suppose that the stock pays no dividends, so that $S_t/B_t$ is a $\mathbb{Q}$-martingale, that the cash account never decreases, so that $B_T \ge B_t$ for $t \le T$, which holds with nonnegative interest rates, and that exercise has no costs. For the call payoff $H_t = (S_t - K)^{+}$ the discounted European value $M_t = \mathbb{E}^{\mathbb{Q}}[(S_T - K)^{+}/B_T \mid \mathcal{F}_t]$ satisfies

and $M_t \ge 0$, so $M_t \ge Z_t$. A martingale is a supermartingale, so minimality gives $Y \le M$, and the American call is worth exactly the European call. Dividends, negative rates or costs of exercise break this argument, and with dividends or negative rates early exercise of a call can be optimal.