Early exercise of American options

When is early exercise worth more than waiting?

At each node of a tree, compare two numbers: what exercising pays now and what holding the option is worth. The option is worth the larger one, and the best exercise rule exercises where exercising is worth at least as much. Where the two tie, either choice is best. Which nodes those are depends on the option, the rate and the dividends.

Before you start · 3 quick checks on backward induction, exercise values and put-call parity, about 2 min

Setup: American options on a binomial tree

A share trades today at $S_0 \gt 0$. The time to expiry $T$ is split into $n$ steps of length $\Delta t = T/n$. In each step the share price is multiplied by $u$ or by $d$, with $0 \lt d \lt u$. After $k$ steps with $j$ up moves it is $S_0u^jd^{k-j}$ at node $(k, j)$, so the tree recombines.
A bond pays a constant, continuously compounded rate $r$: $\$1$ grows to $\$e^{r\Delta t}$ over one step. The share and the bond allow no arbitrage: $d \lt e^{r\Delta t} \lt u$, so $q = (e^{r\Delta t} - d)/(u - d)$ lies strictly between 0 and 1.
An option with strike $K \gt 0$ expires at step $n$. Exercising it at a node with share price $S$ pays $g = (S - K)^+$ for a call and $g = (K - S)^+$ for a put. A European option can be exercised only at expiry; an American option at any step $0, 1, \dots, n$.
Values come from backward induction. At expiry $V = g$. Before expiry the holding value is $H = e^{-r\Delta t}\big(qV_u + (1-q)V_d\big)$, from the values $V_u, V_d$ at the two next nodes. The European value is $V = H$; the American value is $V = \max(g, H)$.
An exercise rule says, at each node before expiry, whether to exercise if the option is still held there. Its value is found by the same backward induction, with the rule's choice in place of the maximum. The best rule exercises where $g \gt 0$ and $g \ge H$. At a tie, $g = H$, holding is just as good; this page then marks the node as exercise when $g \gt 0$.

Fine print: assumptions and conventions

The market is frictionless: no transaction costs and no taxes. Any real number of shares can be bought or sold short, and anyone can borrow or lend any amount at the rate $r$.
The share pays no dividends, except in the dividend part of stage 2 and in problem P5, which state their own conventions.
Each option can be bought or sold (written) in any amount at a single price. The European option has no cash flows before expiry.
In this model the no-arbitrage price of an American option is its backward-induction value $V_0$. A seller who receives $V_0$ can trade the share and the bond so as to cover the payoff whenever the holder exercises, and a holder who follows the best rule receives payments worth exactly $V_0$. This page takes that result as given; see Shreve, chapter 4.
Exercising where $g = 0$ gives the option up for nothing. The best rule never does that before expiry.
Rates are continuously compounded. With a simple rate $r$ per step instead, replace every $e^{r\Delta t}$ by $1 + r$. Nothing else changes.

1 See

Choose where to exercise

A call on the same share

Key point A rule's value comes from backward induction with the rule's own choices, and no rule beats the one that takes the larger of exercising and holding at every node. On this tree the put is exercised early at low share prices, while exercising the call early never raises its value.

2 Try

What changes the answer

The interest rate

A dividend

Key point On this tree, raising the rate never removes a node where the best rule exercises the put, and it adds nodes, because the strike received early becomes worth more. A dividend breaks the argument for the call: exercising just before the payment can be the best choice.

3 Prove

The best rule, and why the call waits

The idea in three steps

Backward induction shows that no exercise rule is worth more than the value $V_0$, and the best rule reaches it.
A call plus a bond paying $K$ never pays less than a share at expiry, so the call is worth at least $S - Ke^{-r\tau}$.
With $r \ge 0$ that bound is at least $S - K$. A call price is never negative, so together they give at least the exercise value $(S - K)^+$, and holding never loses to exercising.

Lemma (the best rule)

Under the setup, every exercise rule has value $W_0 \le V_0$, and the best rule has value $V_0$.

Theorem (with no dividends, early exercise of a call never beats holding)

Assumptions.

The tree of the setup: $S_0 \gt 0$, $0 \lt d \lt u$, $n \ge 1$ steps of length $\Delta t \gt 0$, and a constant continuously compounded rate $r$ with $d \lt e^{r\Delta t} \lt u$. In addition, $r \ge 0$.
The share pays no dividends before expiry.
Frictionless market as in the fine print: shares and the bond trade in any amount, short sales are allowed, and anyone can borrow and lend at $r$.
A European call and an American call, both with strike $K \gt 0$ and expiry $T = n\Delta t$, each trade at one price, bought or written. The European call has no cash flows before $T$. The American call pays $(S - K)^+$ when exercised.
No arbitrage: no trade brings in money today and never pays out later, and no trade costs nothing, never loses and gains in some state.

Notation. At a node at step $k \lt n$, $S$ is the share price, $\tau = (n-k)\Delta t \gt 0$ the time left, $C^E$ the European call's price there, which in this model is its backward-induction value, and $V^A$ the American call's backward-induction value. Portfolio A is one European call and a bond paying $K$ at expiry; portfolio B is one share.

Conclusion. At every node before expiry, $C^E \ge S - Ke^{-r\tau}$ and

\[ V^A = C^E \ge (S - K)^+ , \]

so holding the American call is worth at least as much as exercising it: early exercise never beats holding. If $r \gt 0$, then at every node before expiry with $S \gt K$, holding is worth strictly more: $V^A - (S - K) \ge K(1 - e^{-r\tau}) \gt 0$.

Proof of the theorem (6 lines, build it or read it)

Scratch work, not part of the proof. For line 1, split into two cases. If $S_T \ge K$: $(S_T - K) + K = S_T = \max(S_T, K)$. If $S_T \lt K$: $0 + K = K = \max(S_T, K)$. For line 6: $(S - Ke^{-r\tau}) - (S - K) = K - Ke^{-r\tau} = K(1 - e^{-r\tau})$.

Boundary cases and exceptions: $r = 0$, $r \lt 0$ and dividends

With a dividend. Line 2 sells the share short and holds the position to expiry. If the share pays a dividend before then, the short seller owes it, and the bound drops by the value at the node of the dividends paid before expiry. Stage 2 showed a tree where exercising just before the dividend is the best choice. The put needs no dividend to be exercised early: the parity bound $P^E \ge Ke^{-r\tau} - S$ lies below the put's exercise value $(K - S)^+$ when $r \gt 0$, at every share price: $Ke^{-r\tau} - S \lt K - S \le (K - S)^+$. So it does not rule exercise out.

Key point The call plus a bond paying $K$ dominates the share, so $C^E \ge S - Ke^{-r\tau}$. With $r \ge 0$ this is at least $S - K$, so an American call on a share with no dividends is worth the same as the European call, and exercising it early never beats holding.

4 Apply

Problems

All prices are hypothetical. Rates are continuously compounded.