Early exercise in continuous time

Definition Work under a pricing measure $\mathbb{Q}$ for the money-market account $B_u = e^{ru}$ with a constant continuously compounded rate $r$, on the filtration generated by a $\mathbb{Q}$-Brownian motion $W^{\mathbb{Q}}$, and let the stock follow

with a dividend yield $q \ge 0$ and $\sigma > 0$. An American claim with maturity $T$ and a continuous exercise payoff $g \ge 0$, such as the put payoff $g(s) = (K - s)^{+}$, pays $g(S_{\tau})$ at a time $\tau$ chosen by the holder. The admissible exercise times after $t$ form the set $\mathcal{T}_{t,T}$ of stopping times with values in $[t, T]$, so whether the holder has exercised by time $u$ must be decided by the information available at $u$. Write $\mathbb{E}^{\mathbb{Q}}_{t,s}$ for the expectation under $\mathbb{Q}$ given $S_t = s$, and assume $\mathbb{E}^{\mathbb{Q}}_{t,s}[\sup_{t \le u \le T} g(S_u)] < \infty$. The value of a right not yet exercised is

where

is the value of exercising at $\tau$. This is a continuous-time optimal stopping problem. The constant times $\tau = t$ and $\tau = T$ are admissible, so

where $V^{\mathrm{Eu}} = J_T$ is the value of the European claim on $g(S_T)$. The continuation region is the set of points $(t,s)$ with $V(t,s) > g(s)$, and the exercise region is the set where $V(t,s) = g(s)$. For the put and the call payoffs a standard result, stated here without proof, says that $V$ is continuous and that the first entry time into the exercise region,

is optimal. It is the continuous-time form of the Snell envelope rule of American options and optimal stopping, and it never exceeds $T$ because $V(T, s) = g(s)$.

Definition Let $\mathcal{L}$ be the generator of $S$ under $\mathbb{Q}$, which acts on twice differentiable functions by

and write $\mathcal{A}V = \partial_t V + \mathcal{L}V - rV$ for the operator of the Black-Scholes pricing PDE. The American value solves the obstacle problem, also called a variational inequality,

on $[0, T) \times (0, \infty)$. Equivalently, three conditions hold together,

The first holds because exercising at once is allowed. The second comes from waiting. Holding the right for a short time $h$ and then acting optimally is admissible, so

and where $V$ is smooth, Itô’s formula turns this into $\mathcal{A}V \le 0$ after dividing by $h$ and letting $h \to 0$. The third condition, called complementarity, says that at every point at least one of the inequalities is an equality, the pricing PDE $\mathcal{A}V = 0$ in the continuation region and $V = g$ in the exercise region. This classical form needs the derivatives to exist where they are used. For the American put in this model a standard result, stated here without proof, says that $V$ is continuous, that it is smooth in the continuation region, where $\mathcal{A}V = 0$, and that $\partial_s V$ is continuous across the boundary for $t < T$, while $\partial_{ss} V$ jumps there. The inequality is then read in each region separately, or in a weak or viscosity sense that states the same obstacle problem without assuming a second derivative at the boundary or at the kink of the payoff.

Definition The free boundary is the border between the continuation region and the exercise region. It is not given in advance, and finding it is part of the pricing problem, which is why the American problem is a free-boundary problem rather than a PDE on a fixed domain. For the put on a dividend-free stock with $r > 0$, a standard result, stated here without proof, says that for each $t < T$ the exercise region is $\{s \le b(t)\}$ for a critical price $b(t)$, where $b$ is continuous and nondecreasing with $b(t) \to K$ as $t \to T$. The value and the boundary then satisfy

• the pricing PDE $\mathcal{A}V = 0$ for $s > b(t)$ and $t < T$,
• the exercise condition $V = K - s$ for $s \le b(t)$,
• the terminal condition $V(T, s) = (K - s)^{+}$ and the decay $V(t, s) \to 0$ as $s \to \infty$,

together with value matching and smooth pasting at $s = b(t)$, described in the next two cards. Solving the PDE above an arbitrary curve, with $V = g$ on the curve, gives the value of the rule that exercises the first time the price reaches that curve. That value is at most $V$, so an arbitrary curve gives a lower bound and not the price.

For a call on a dividend-free stock with $r \ge 0$ there is no exercise region before $T$. Put-call parity gives

which is larger than $s - K$ because the European put is positive and $e^{-r(T - t)} \le 1$. The European call is also positive, so it lies strictly above the payoff $(s - K)^{+}$ before $T$. Its discounted value is a $\mathbb{Q}$-martingale, so stopping it at any exercise time gives at most its starting value, and the American call is worth exactly the European call. Dividends or negative rates break this argument.

Definition Value matching is the condition

at the free boundary. The value meets the payoff continuously, so approaching the boundary from the continuation region gives the exercise value. For the American put in the Black-Scholes model it follows from the continuity of $V$, and when a candidate boundary is chosen first and the continuation value is built on it, value matching holds by construction. In the perpetual put problem of the last card, with $q = 0$, $r > 0$ and $\sigma > 0$, fix a level $L$ in $(0, K)$ and let $\tau_L = \inf\{u \ge 0 : S_u \le L\}$. Write $\mathbb{E}^{\mathbb{Q}}_{s}$ for the expectation given $S_0 = s$. For $s \ge L$ the rule that exercises at $\tau_L$ is worth

where $\beta = -2r/\sigma^2$. The second equality holds because $\beta$ is a root of the characteristic equation of the last card, so that $e^{-ru}S_u^{\beta}$ is a $\mathbb{Q}$-martingale. Stopped at $\tau_L$ it stays below $L^{\beta}$, since $\beta < 0$ and $S_u \ge L$ before $\tau_L$, and it tends to $0$ on paths that never reach $L$, so optional stopping and bounded convergence give

Every level satisfies value matching, $f_L(L) = K - L$, so value matching alone describes a whole family of exercise rules and cannot pick out the best one.

Definition Smooth pasting, also called smooth fit, is the condition

at the free boundary, so that the value meets the payoff with the same slope. It is a result with hypotheses, not a law of every stopping problem. A standard result, stated here without proof, gives smooth pasting for a one-dimensional diffusion with continuous paths and a positive diffusion coefficient when the payoff is differentiable at the boundary, under further regularity conditions, and it covers the American put in the Black-Scholes model at every $t < T$. It can fail when the price can jump, when the diffusion coefficient vanishes, or when the payoff has a jump or a corner at the boundary. In the perpetual put problem it is exactly the first-order condition for the best level. For $s > L$,

which is positive for $L$ below $\beta K/(\beta - 1)$, negative above it, and zero exactly when $\beta(K - L)/L = -1$. The left side of this condition is the slope $f_L'(L)$ of the value curve at the level, and the right side is the payoff slope $g'(L) = -1$. So the best level is the one where the slopes match.

Definition A perpetual American put has no maturity. Let $q = 0$, $r > 0$, $\sigma > 0$ and $K > 0$, so that $dS_u = rS_u\,du + \sigma S_u\,dW^{\mathbb{Q}}_u$. Every stopping time with values in $[0, \infty]$ is admissible, and on the event $\{\tau = \infty\}$ the right is never exercised and pays nothing, so the value is

The problem looks the same at every date, so the value depends on the current price only, and the variational inequality becomes an ordinary differential inequality. The result, derived and verified below, is that with

the value is $V(s) = K - s$ for $s \le B$ and

for $s > B$, and that the first time the price falls to the boundary, $\tau_B = \inf\{u \ge 0 : S_u \le B\}$, is an optimal exercise time.

In the continuation region the value solves $\tfrac12\sigma^2 s^2 V'' + rsV' - rV = 0$. Trying $V = As^{\beta}$ and dividing by $As^{\beta}$ gives the characteristic equation

with roots $1$ and $-2r/\sigma^2$. The root $1$ gives a term that grows like $s$, while a put is worth at most $K$, so only the negative root stays. Value matching $AB^{\beta} = K - B$ and smooth pasting $\beta AB^{\beta - 1} = -1$ give, after dividing the second equation by the first, $\beta(K - B) = -B$, which is the formula for $B$. Since $\beta < 0$, the boundary lies strictly between $0$ and $K$.

The verification uses the piecewise smooth Itô formula. On $(B, \infty)$ the candidate is convex, because $\beta(\beta - 1) > 0$ and $K - B > 0$, and by value matching and smooth pasting its tangent at $B$ is the line $K - s$. So $V(s) \ge K - s$ there, and since $V > 0$, the candidate dominates the payoff $(K - s)^{+}$ everywhere. It satisfies $0 < V \le K$, its first derivative is continuous, and its second derivative is continuous on each side of $B$, so Itô’s formula applies with no extra term at $B$. The operator gives $\tfrac12\sigma^2 s^2 V'' + rsV' - rV = 0$ for $s > B$ and $-rK < 0$ for $s < B$, so the discounted candidate $e^{-ru}V(S_u)$ has a drift that is never positive. Its Brownian integrand $\sigma sV'(s)$ equals $-\sigma s$ for $s \le B$ and $\sigma\beta V(s)$ for $s > B$, so it is bounded, the stochastic integral is a martingale on every finite horizon, and the discounted candidate is a supermartingale.

For any stopping time $\tau$ and any $n$, the payoff lies below the candidate, so the expected discounted payoff at $\tau \wedge n$ is at most $V(s)$. As $n \to \infty$ the terms on $\{\tau < \infty\}$ converge, those on $\{\tau = \infty\}$ are at most $Ke^{-rn} \to 0$, and bounded convergence shows that no exercise rule is worth more than $V(s)$. For $s > B$ the drift vanishes before $\tau_B$, so the process stopped at $\tau_B \wedge n$ is a martingale with mean $V(s)$. On $\{\tau_B \le n\}$ it equals $e^{-r\tau_B}(K - B)$, the discounted payoff, and on $\{\tau_B > n\}$ it is at most $(K - B)e^{-rn}$, which tends to $0$ because $r > 0$. So exercising at $\tau_B$ is worth exactly $V(s)$. For $s \le B$ the rule exercises at once and $V(s) = K - s$.