Multi-Period Binomial Option Valuation by Backward Induction

From One-Step Replication to a Multi-Period Tree

A multi-period option tree is a sequence of linked one-period pricing problems. At each node, the option’s possible values one step later can be replicated with shares and a risk-free asset. That replication gives the node’s continuation value. For a European option, continue this calculation back from the maturity payoff. For an American option, compare the continuation value at each node with the value of exercising there.

Fix a horizon of $N$ periods. Use the finite sample space $\Omega=\{u,d\}^N$, with its power-set sigma-field and any probability measure $\mathbb{P}$ that assigns positive probability to every path. Let $\mathcal{F}_n$ record the first $n$ moves. The stock price $S_n$ is determined by the path up to time $n$: from a node with price $S$, its next price is $uS$ after an up move and $dS$ after a down move. The risk-free asset is worth $R^n$ at time $n$, where $R>0$ is its gross return per period. Assume throughout that $0<d<R<u$.

The inequalities ensure that the two next-period stock prices lie on opposite sides of the risk-free growth of the current price. Define the pricing weight $q=(R-d)/(u-d)$. It satisfies $0<q<1$, since $R-d>0$, $u-R>0$, and $u-d>0$; moreover $1-q=(u-R)/(u-d)$. These are weights derived from replication. They are not, by definition, the probabilities under the underlying measure $\mathbb{P}$.

Definition: A node at time $n$ is a particular history of the first $n$ moves. Its continuation value is the current cost of a portfolio that replicates the option’s values in the two child nodes. Backward induction means calculating these values from time $N$ toward time $0$.

The One-Step Replication Formula

Suppose the required values at the up and down child nodes are $V^u$ and $V^d$. A portfolio holding $\Delta$ shares and $b$ units of the risk-free asset has value $\Delta S+b$ now and values $\Delta uS+Rb$ and $\Delta dS+Rb$ next period. Replication requires those two future values to equal $V^u$ and $V^d$.

\Delta=\frac{V^u-V^d}{S(u-d)},\qquad b=\frac{uV^d-dV^u}{R(u-d)}.

Substitution into the portfolio’s current cost gives the one-step pricing formula. The result holds at every node, provided the same $u,d,R$ apply there and $0<d<R<u$.

Theorem (One-step replication): For $S>0$, $0<d<R<u$, and finite child values $V^u,V^d$, there is a unique portfolio of shares and risk-free asset that produces those child values. Its current value is $\big(qV^u+(1-q)V^d\big)/R$, where $q=(R-d)/(u-d)$.

Proof. The two replication equations are $\Delta uS+Rb=V^u$ and $\Delta dS+Rb=V^d$. Subtracting the second from the first gives $\Delta S(u-d)=V^u-V^d$, which determines $\Delta$ uniquely because $S(u-d)>0$. Substitution into either equation gives $b=(uV^d-dV^u)/(R(u-d))$, so $b$ is also unique. The current cost is therefore

\Delta S+b =\frac{R(V^u-V^d)+uV^d-dV^u}{R(u-d)} =\frac{(R-d)V^u+(u-R)V^d}{R(u-d)} =\frac{qV^u+(1-q)V^d}{R}.

The last equality uses $q=(R-d)/(u-d)$ and $1-q=(u-R)/(u-d)$. Thus the stated portfolio exists, is unique, and has the stated cost. $\square$

European Options: Work Back from Maturity

A European option can be exercised only at maturity. Let its payoff at time $N$ be $g(S_N)$, such as $(S_N-K)^+$ for a call or $(K-S_N)^+$ for a put, where $x^+=\max(x,0)$. Set $V_N=g(S_N)$. At a node at time $n<N$, let $V^u$ and $V^d$ be the already calculated values at its up and down children. The one-step theorem gives the recursion below.

Theorem (European backward induction): Under the one-step assumptions at every node, the value at maturity is $V_N=g(S_N)$, and at every earlier node $$ V_n=\frac{qV^u+(1-q)V^d}{R}. $$ These values are obtained by a self-financing replicating strategy, constructed one step at a time from maturity to time $0$.

Proof. At maturity the claim pays $g(S_N)$, so assigning that value at every terminal node is correct. Suppose the values at all nodes at time $n+1$ have been calculated. At each time-$n$ node, apply the one-step replication theorem to the two child values $V^u,V^d$. It supplies a portfolio whose next-period values equal those child values and whose current cost is $\big(qV^u+(1-q)V^d\big)/R$. Therefore the portfolio replicates the claim from that node onward. Starting at $n=N-1$ and repeating this argument down to $n=0$ constructs replication at every node. Uniqueness at each step follows from the one-step theorem, so the recursion determines the replicating value uniquely. $\square$

Worked Example: A Two-Period European Call

Take $S_0=4$, $u=\tfrac32$, $d=\tfrac12$, $R=\tfrac54$, and a call strike $K=4$. The pricing weight is $q=(\tfrac54-\tfrac12)/(\tfrac32-\tfrac12)=\tfrac34$, so $1-q=\tfrac14$. At each node the up and down weights sum to $\tfrac34+\tfrac14=1$, and the parameter condition holds: $0<\tfrac12<\tfrac54<\tfrac32$.

The stock prices at time $1$ are $6$ and $2$. At time $2$, the prices following the up node are $9$ and $3$, and following the down node they are $3$ and $1$. The call payoffs at those four terminal nodes are respectively $5,0,0,0$, because $(9-4)^+=5$, $(3-4)^+=0$, and $(1-4)^+=0$.

At the time-1 up node, the continuation value is $\big(\tfrac34\cdot5+\tfrac14\cdot0\big)/(\tfrac54)=3$. At the time-1 down node, both terminal payoffs are zero, so its value is $\big(\tfrac34\cdot0+\tfrac14\cdot0\big)/(\tfrac54)=0$. Hence at time $0$ the value is $\big(\tfrac34\cdot3+\tfrac14\cdot0\big)/(\tfrac54)=\tfrac95$.

Check by direct terminal weighting: The only positive payoff is reached by two up moves, whose product pricing weight is $(\tfrac34)^2=\tfrac9{16}$. Discounting two periods gives $(\tfrac9{16}\cdot5)/(\tfrac54)^2=\tfrac{45}{16}\cdot\tfrac{16}{25}=\tfrac95$, as above.

Check by replication: At the up node, $\Delta=5/(6( \tfrac32-\tfrac12))=\tfrac56$ and $b=(\tfrac32\cdot0-\tfrac12\cdot5)/(\tfrac54\cdot1)=-2$. The cost is $\tfrac56\cdot6-2=3$. At the root, $\Delta=(3-0)/(4\cdot1)=\tfrac34$ and $b=(\tfrac32\cdot0-\tfrac12\cdot3)/(\tfrac54\cdot1)=-\tfrac65$; the cost is $\tfrac34\cdot4-\tfrac65=\tfrac95$.

American Options: Compare Exercise with Continuation

An American option may be exercised at any allowed time up to and including maturity. At a node with stock price $S$, immediate exercise pays $g(S)$. Continuing means holding the option for at least one more period; its value is found by replicating the values in the next two nodes. Define $A_N=g(S_N)$. At earlier nodes, calculate the continuation value $C_n$ from the children’s American values and choose the larger of exercise and continuation.

C_n=\frac{qA^u+(1-q)A^d}{R},\qquad A_n=\max\{g(S_n),C_n\}.

Theorem (American backward induction): On a finite binomial tree satisfying $0<d<R<u$, the recursion $A_N=g(S_N)$ and $A_n=\max\{g(S_n\},C_n\}$, with $C_n$ as above, gives the value at every node when exercise is permitted at any time through $N$. Choosing exercise when the first term is at least the second, and otherwise continuing, attains this value.

Proof. At time $N$, exercise is required, so the value is $g(S_N)$. Assume the stated values are optimal at all nodes at time $n+1$. At a node at time $n$, any permitted strategy either exercises immediately, receiving $g(S_n)$, or continues for one step. If it continues, its remaining value in the two child nodes cannot exceed $A^u,A^d$, by the induction assumption. The one-step replication formula and the positive weights $q,1-q$ show that the cost of continuing with those optimal child values is $C_n$, and no continuation using smaller child values has a larger value. Thus no strategy at the current node exceeds $\max\{g(S_n),C_n\}$. Exercising now attains the first candidate. Continuing and using the optimal choices in each child attains the second candidate. Selecting the larger therefore attains the maximum. Induction from $N$ back to $0$ proves the recursion and the stated exercise rule. $\square$

Worked Example: An American Put with Early Exercise

Use $S_0=4$, $u=\tfrac32$, $d=\tfrac12$, $R=\tfrac54$, and a two-period American put with strike $K=4$. Here $q=\tfrac34$, $1-q=\tfrac14$, and each node’s pricing weights sum to $1$. The parameter condition is $0<\tfrac12<\tfrac54<\tfrac32$.

The terminal stock prices are $9,3,3,1$, and the corresponding put payoffs are $0,1,1,3$. At the time-1 up node, $S_1=6$, so immediate exercise pays $(4-6)^+=0$. The continuation value is $\big(\tfrac34\cdot0+\tfrac14\cdot1\big)/(\tfrac54)=\tfrac15$. The American value at this node is therefore $\max\{0,\tfrac15\}=\tfrac15$.

At the time-1 down node, $S_1=2$, so immediate exercise pays $(4-2)^+=2$. Its continuation value is $\big(\tfrac34\cdot1+\tfrac14\cdot3\big)/(\tfrac54)=(\tfrac34+\tfrac34)/(\tfrac54)=\tfrac65$. Thus the node value is $\max\{2,\tfrac65\}=2$, and exercise is optimal there.

At the root, immediate exercise pays $(4-4)^+=0$. Continuing gives $\big(\tfrac34\cdot\tfrac15+\tfrac14\cdot2\big)/(\tfrac54)=(\tfrac3{20}+\tfrac12)/(\tfrac54)=\tfrac{13}{25}$. Therefore the American put value is $\max\{0,\tfrac{13}{25}\}=\tfrac{13}{25}$. The decision at the root is to continue, even though exercise is optimal at the later down node.

Check against the European put: If exercise is allowed only at maturity, the time-1 continuation values are $\tfrac15$ and $\tfrac65$. The root value is $\big(\tfrac34\cdot\tfrac15+\tfrac14\cdot\tfrac65\big)/(\tfrac54)=(\tfrac3{20}+\tfrac3{10})/(\tfrac54)=\tfrac9{25}$. The American value $\tfrac{13}{25}$ is larger, as expected because the holder can choose the European strategy or exercise early.

A Three-Period Calculation and a Common Error

Worked Example: A Three-Period European Put

Let $S_0=8$, $u=\tfrac32$, $d=\tfrac12$, $R=1$, and $K=8$, with maturity after three periods. Then $q=(1-\tfrac12)/(\tfrac32-\tfrac12)=\tfrac12$ and $1-q=\tfrac12$; the weights at every node sum to $1$, and $0<\tfrac12<1<\tfrac32$.

At time $3$, the stock prices with zero, one, two, and three up moves are $1,3,9,27$. The put payoffs are $7,5,0,0$, respectively. At time $2$, the nodes with zero, one, and two up moves have stock prices $2,6,18$. Their values are $(7+5)/2=6$, $(5+0)/2=\tfrac52$, and $(0+0)/2=0$.

At time $1$, the down node has stock price $4$, and its children have values $6$ and $\tfrac52$, giving $(6+\tfrac52)/2=\tfrac{17}{4}$. The up node has stock price $12$, and its children have values $\tfrac52$ and $0$, giving $(\tfrac52+0)/2=\tfrac54$. The root value is $(\tfrac{17}{4}+\tfrac54)/2=\tfrac{11}{4}$.

Check by terminal weighting: With $q=\tfrac12$, the probabilities generated by the pricing weights for zero, one, two, and three up moves are $\tfrac18,\tfrac38,\tfrac38,\tfrac18$. The weighted terminal payoff is $\tfrac18\cdot7+\tfrac38\cdot5+\tfrac38\cdot0+\tfrac18\cdot0=\tfrac{22}{8}=\tfrac{11}{4}$. Since $R=1$, discounting three periods does not change this value.

Common mistake: discounting only once. In the two-period call example, the time-1 up-node value is $3$, but it is not yet the time-0 price. A wrong calculation omits the root-period discount and reports $\tfrac34\cdot3=\tfrac94$. The root calculation must discount the next-period value by $R=\tfrac54$: $(\tfrac34\cdot3+\tfrac14\cdot0)/(\tfrac54)=\tfrac95$. The direct terminal check also discounts twice, by $R^2=(\tfrac54)^2$, and confirms $\tfrac95$.

Backward induction is useful because it reduces a large tree to local calculations at its nodes. The terminal payoffs are the only initial data for a European option; for an American option, each node also requires the immediate exercise payoff. The replication formula then gives the continuation value, and the exercise comparison makes the optimal stopping decision explicit. A reliable calculation records both quantities at American nodes, keeps the discount factor at every step, and checks the root value by a second route whenever the tree is small enough.

Check Your Understanding

Use the one-step replication formula and show the intermediate values in each calculation.

For $u=\tfrac43$, $d=\tfrac23$, and $R=1$, calculate $q$ and verify that both pricing weights are positive and sum to $1$.
At a node with $S=10$, suppose the two child values are $V^u=6$ and $V^d=2$. With $u=\tfrac32$, $d=\tfrac12$, and $R=\tfrac54$, find the replicating holdings and current value.
For a European option, explain why its maturity payoffs are the starting values for backward induction.
At an American-option node, the immediate exercise value is $3$ and the continuation value is $\tfrac{11}{4}$. Which action is optimal, and what is the node value?
In the three-period put example, verify the time-1 down-node value by substituting its two child values into the recursion.