How backward induction prices a binomial tree

If a stock moves up or down in each of several periods, what is an option on it worth today?

Start at expiration, where the option is worth its payoff. Each earlier node is a one-period problem: its value is the one-period price of the two values after it. Working back node by node to today gives the price.

Before you start · 3 quick checks on one-period pricing, conditional expectation and discounting, about 3 min

Setup: the n-period binomial model

Time runs over \(n\) periods: \(t = 0, 1, \dots, n\). The stock starts at \(S_0 \gt 0\), and each period its price is multiplied by \(u\) (an up move) or by \(d\) (a down move), with \(0 \lt d \lt u\).
An up move followed by a down move ends at the same price as a down move followed by an up move, so the tree recombines: after \(t\) periods with \(j\) up moves the stock is at \(S_0u^jd^{t-j}\), one of \(t+1\) prices.
A riskless bond grows by the factor \(1+r\) each period, with \(r \gt -1\) and \(d \lt 1+r \lt u\). As in the one-period model, \(q = \dfrac{(1+r)-d}{u-d}\) then lies strictly between 0 and 1.
A European option pays \(f(S_n)\) at time \(n\) and nothing before. A call with strike \(K\) pays \((S_n - K)^+\) and a put pays \((K - S_n)^+\).
At each node before time \(n\), a portfolio holds \(\Delta\) shares and \(B\) dollars in the bond. It can be changed at any node. \(\Delta \lt 0\) is a short sale and \(B \lt 0\) is borrowing.

Fine print: assumptions and the rate convention

The market is frictionless: no transaction costs, no dividends, no taxes. Any real number of shares can be bought or sold short at every node, and anyone can borrow or lend any amount at the rate \(r\) per period.
The option can be bought or sold (written) at time 0 in any amount at a single price \(P\), with no bid-ask spread, and is then held to time \(n\). It cannot be exercised early; American options are the subject of step 7.
Every path through the tree has a positive real probability. For example, the stock may go up with some probability \(p\), \(0 \lt p \lt 1\), each period.
This page uses simple growth \(1+r\) per period. If a rate is quoted with continuous compounding, \(\rho\) per year for periods of length \(\Delta t\), replace every \(1+r\) by \(e^{\rho\Delta t}\). Nothing else changes.
This is the binomial model of Cox, Ross and Rubinstein. One common way to tie \(u\) and \(d\) to a volatility \(\sigma\) is \(u = e^{\sigma\sqrt{\Delta t}}\) and \(d = 1/u\). The examples on this page pick \(u\) and \(d\) directly.

1 See

Fill the tree from the end

The last column of the tree holds the option's payoffs. Work backward: pick a node whose two next nodes are filled, enter its value \(V\) and the number of shares \(\Delta\) that copies the option over the next period, and check it. Give \(V\) to the cent and \(\Delta\) to two decimals: an entry within 1 cent of the exact value of \(V\), and within 0.01 of the exact \(\Delta\), counts as right, and the page then shows the exact values. A wrong node is marked, with a hint about what went wrong.

Nodes ready to fill (their two next nodes are done). You can also click a dashed node in the tree.

Key point Each node before expiration is a one-period problem whose payoffs are the values of the two nodes after it: \(V = \big(qV_u + (1-q)V_d\big)/(1+r)\). The values fill in from the end, and the value at the root is the option's price.

2 Try

Follow the copy, then count the paths

Paying for the changes

The real probability

Counting paths

Key point The copy may change its share count at a node, but each trade is paid for inside the portfolio: the net cash flow is exactly 0, so the copy needs only its cost at time 0. Its price does not involve the real probability, and summing over final nodes, with \(\binom{n}{k}\) paths to each, gives the same number as backward induction.

3 Prove

Backward induction, self-financing and the closed form

The idea in three steps

At each node, the one-period theorem gives a portfolio that is worth the two next values one period later.
At the next node, the old portfolio is worth exactly what the new one costs, so every trade nets to zero, and by induction the strategy ends with the payoff on every path.
Unrolling the recursion gives the closed form; trading the option against the strategy rules out every other price.

Theorem (backward induction on an n-period tree)

Assumptions. The setup and fine print above: \(S_0 \gt 0\), \(0 \lt d \lt u\), \(r \gt -1\) with \(d \lt 1+r \lt u\), an integer number of periods \(n \ge 2\), every path with positive real probability, a frictionless market for the stock and the bond, and a European option paying any real amount \(f(S_n)\) at time \(n\) that trades at time 0 at one price \(P\) in both directions and is then held. (One period is the one-period page, step 4.)

Definitions. Let \(S_t(j) = S_0u^jd^{t-j}\), \(V_n(j) = f\big(S_n(j)\big)\), and for \(t \lt n\)

\[ V_t(j) = \frac{qV_{t+1}(j+1) + (1-q)V_{t+1}(j)}{1+r}, \qquad \Delta_t(j) = \frac{V_{t+1}(j+1) - V_{t+1}(j)}{(u-d)S_t(j)}, \qquad B_t(j) = V_t(j) - \Delta_t(j)S_t(j). \]

Conclusion. (a) The strategy that starts with \(V_0(0)\) dollars and holds \(\Delta_t(j)\) shares and \(B_t(j)\) dollars in the bond at each node before time \(n\) never adds or removes cash after time 0, and it is worth \(f(S_n)\) at time \(n\) on every path. (b) Its starting value is

\[ V_0(0) = (1+r)^{-n}\sum_{k=0}^{n}\binom{n}{k}q^k(1-q)^{n-k}f(S_0u^kd^{n-k}). \]

The payoff \(f\) is arbitrary, so lines 1 to 4 of the proof are stated for every payoff \(f\): each is one claim about all payoffs at once, and a candidate line that fails for some payoff is false. Some false lines happen to hold for one particular payoff; a constant payoff, for example, has \(\Delta = 0\) everywhere.

In the proof, at a node before time \(n\), the symbols \(S\), \(V\), \(\Delta\), \(B\) stand for \(S_t(j)\), \(V_t(j)\), \(\Delta_t(j)\), \(B_t(j)\), and \(V_u\), \(V_d\) for \(V_{t+1}(j+1)\), \(V_{t+1}(j)\), the values of the two nodes after it. \(\Delta_u, B_u\) and \(\Delta_d, B_d\) are the holdings at those two nodes. At the root the symbols are \(S_0\), \(V_0\), \(\Delta_0\), \(B_0\), and at a final node \(S_n\) and \(V_n\).

What counts as an arbitrage here

An arbitrage is a strategy in the stock, the bond and the option that costs nothing at time 0, never adds or removes cash after time 0, is never worth less than zero at time \(n\), and is worth more than zero on at least one path. Since every path has positive real probability, that last condition means a gain with positive probability.

Receiving cash at time 0 and owing nothing later is the same thing: put the cash in the bond. Cash taken out before time \(n\) can instead be left in the bond until time \(n\).

Proof of (a), (b) and the first half of (c) (5 lines, build it or read it)

Scratch work, not part of the proof. For line 1, with \(B = V - \Delta S\), \(\Delta S = (V_u - V_d)/(u-d)\) and \(V = \big(qV_u + (1-q)V_d\big)/(1+r)\):

The last step uses \(q + \big(u - (1+r)\big)/(u-d) = 1\).

Lemma (the second half of (c): no arbitrage at \(P = V_0\))

Under the assumptions of the theorem, no strategy is an arbitrage when the option trades at \(P = V_0\).

Why the real probability does not enter

No line of the proof uses the real probabilities. The strategy pays the option's payoff on every path, so its cost is the price whatever those probabilities are, as long as every path can happen. The real probabilities do change how likely each final node is, and so they can change the option's expected payoff.

The weights \(q^k(1-q)^{n-k}\) are positive and, summed over all \(2^n\) paths, add to 1, so they can be read as path probabilities. They are built from \(u\), \(d\) and \(r\) only and are not a forecast.

What happens with many periods

A finer tree uses more, shorter periods for the same time to expiration, with \(u\) and \(d\) tied to a volatility as in the fine print. How the tree price behaves as the number of periods grows, and how it relates to the Black–Scholes formula, is explained on a later page (step 18). This page uses only finite trees.

Key point At every node the one-period copy is worth the two next values, and moving into the next node's copy costs exactly 0. So the strategy started with \(V_0\) pays the option on every path, \(V_0\) is the sum over final nodes with weights \(\binom{n}{k}q^k(1-q)^{n-k}/(1+r)^n\), and no other price is free of arbitrage.

4 Apply

Problems

All prices are hypothetical. Growth factors are per period.