How do you price an option with a binomial tree? Worked example

Q: What is the risk-neutral probability in a binomial tree?

It is p = (e^(r dt) - d) / (u - d), the probability that makes the stock earn the risk-free rate over each step. With a 4 per cent rate, 24 per cent volatility and three steps over a year, u is 1.1486, d is 0.8706 and p is 0.5137. It is a pricing device, not the real-world chance that the stock rises.

Q: How many steps does a binomial tree need?

Enough that doubling them changes the answer by less than your tolerance. For an at-the-money one-year call the CRR error is 0.0237 at 100 steps, 0.0047 at 500 and 0.0024 at 1,000, so each doubling roughly halves it. Compare even step counts with even, because the error flips sign between odd and even counts at the money.

Q: Why use a binomial tree instead of Black-Scholes?

Because a tree can test for early exercise at every node, which the closed form cannot. On a one-year at-the-money put with a 4 per cent rate and 24 per cent volatility, 1,000 steps value the European put at 7.5306 and the American put at 7.9295, an early exercise premium of 0.3988. For a European option, the closed form is faster and exact.

A three-step Cox-Ross-Rubinstein tree worked by hand, its convergence to Black-Scholes, and the American put it can price and the formula cannot.

Julian R. Sterling · from Quantitative Finance

Split the option's life into steps, let the stock move up by u = e^σ√Δt or down by d = 1/u at each one, value the payoff at the end and discount it back with the risk-neutral probability p = (e^rΔt − d) / (u − d). For a one-year at-the-money call at spot 100, 4 per cent and 24 per cent volatility, a three-step tree gives 12.2102; at 1,000 steps it gives 11.4517, within 0.0024 of the Black-Scholes 11.4541.

Worked in full in Quantitative Finance by Julian R. Sterling, with every figure reproduced in a free workbook.See the book on Amazon →

The assumptions

The option is Kestrel 100, the European call that runs through Quantitative Finance, priced here on the Cox-Ross-Rubinstein tree. All inputs are illustrative.

Kestrel 100
Input	Symbol	Value
Spot price	S	100
Strike	K	100
Maturity, years	T	1
Risk-free rate, continuously compounded	r	4%
Volatility	σ	24%
Steps in the worked tree	n	3

Step 1: the tree's parameters

Δt = T ÷ n = 1 ÷ 3 = 0.3333 years; σ√Δt = 0.24 × 0.5774 = 0.1386

u = e^0.1386 = 1.1486; d = 1 ÷ u = 0.8706

Growth over one step e^rΔt = 1.0134, so p = (1.0134 − 0.8706) ÷ (1.1486 − 0.8706) = 0.5137

Discount per step e^−rΔt = 0.9868

Excel: =EXP(vol*SQRT(T/n)), =1/u, =(EXP(r*T/n)-d)/(u-d)

p is not a forecast that the stock goes up. It is the probability that makes the stock earn exactly the risk-free rate on the tree, which is what lets the option be priced without anyone's view on direction.

Step 2: the payoffs at expiry

After three steps the stock has made three, two, one or no up-moves. Because d = 1/u the tree recombines: up then down returns to 100.

Terminal nodes
Up-moves	Stock price	Call payoff
3	151.54	51.54
2	114.86	14.86
1	87.06	0.00
0	65.99	0.00

Step 3: roll back through the tree

Each node is worth the discounted probability-weighted average of the two nodes it leads to.

V = e^−rΔt × [p × V_up + (1 − p) × V_down]

Step 2, top node (stock 131.93): 0.9868 × (0.5137 × 51.54 + 0.4863 × 14.86) = 33.26

Step 2, middle node (stock 100.00): 0.9868 × (0.5137 × 14.86 + 0.4863 × 0) = 7.53

Step 2, bottom node (stock 75.80): 0.00

Step 1: 20.47 at 114.86 and 3.82 at 87.06

Step 0: 0.9868 × (0.5137 × 20.47 + 0.4863 × 3.82) = 12.2102

The tree also hands you the hedge. At the root, delta is the change in option value over the change in stock between the two step-1 nodes: (20.47 − 3.82) ÷ (114.86 − 87.06) = 0.5991 shares per option.

Step 4: add steps until it converges

Three steps overprice the option by 0.7561, 6.6 per cent. Each extra step brings the distribution of terminal prices closer to the lognormal one Black-Scholes assumes. The European price has a closed form in the binomial distribution, so even 1,000 steps need no tree to be laid out; the companion workbook computes each row of this table that way.

CRR call price by number of steps, against Black-Scholes 11.4541
Steps	Call	Error
3	12.2102	+0.7561
10	11.2202	−0.2339
25	11.5434	+0.0893
50	11.4067	−0.0473
100	11.4304	−0.0237
250	11.4446	−0.0095
500	11.4493	−0.0047
1,000	11.4517	−0.0024

Two patterns are worth knowing. The error flips sign between odd and even step counts at the money, because the strike sits on a node for even counts and between nodes for odd ones, so never judge convergence from two adjacent step counts. And past the oscillation, doubling the steps halves the error: 0.0047 at 500, 0.0024 at 1,000. That is first-order convergence, and it tells you how many steps a given accuracy costs.

What the tree does that Black-Scholes cannot: the American put

The reason to build a tree at all is early exercise. At each node, compare holding with exercising and keep the larger. In the three-step tree the bottom node at step 2, stock 75.80, is worth 24.20 exercised against 22.88 held, so it is exercised.

Put on Kestrel 100, European and American
Steps	European put	American put	Early exercise premium
3	8.2891	8.5941	0.3050
100	7.5093	7.9193	0.4099
1,000	7.5306	7.9295	0.3988

The European put converges on the Black-Scholes put of 7.5330; the American put is worth about 0.40 more. On a stock paying no dividend the American call is never exercised early, and the tree confirms it: 11.4517 either way.

The common mistakes

The most frequent is using the annual rate in p instead of the rate for one step. With e^r in place of e^rΔt, p becomes 0.6122 and the three-step call jumps to 17.5891. Check that 0 < p < 1 and that p moves towards one half as you add steps. The second is choosing a step count and stopping: a single price from a tree means little until you have priced it at a few step counts and watched the error shrink. The third is reading p as the real-world chance of a rise.

At 1,000 steps the tree and the closed form agree to 0.0024, about 0.02 per cent. One volatility point moves this option by roughly 0.38. As with every model in the book, the input deserves more scrutiny than the method.

Takeaway

u = e^σ√Δt, d = 1/u, p = (e^rΔt − d) / (u − d), discount e^−rΔt per step.
Three steps give 12.2102; 1,000 give 11.4517 against Black-Scholes 11.4541.
Doubling the steps halves the error once past the odd-even oscillation.
Use the tree where early exercise matters: the American put is worth about 0.40 more here.

The full convergence table recomputes for any volatility, strike or maturity in the free workbook for this case. For the closed form the tree converges to, see the Black-Scholes worked example.

Questions readers ask

What is the risk-neutral probability in a binomial tree?

It is p = (e^(r dt) - d) / (u - d), the probability that makes the stock earn the risk-free rate over each step. With a 4 per cent rate, 24 per cent volatility and three steps over a year, u is 1.1486, d is 0.8706 and p is 0.5137. It is a pricing device, not the real-world chance that the stock rises.

How many steps does a binomial tree need?

Enough that doubling them changes the answer by less than your tolerance. For an at-the-money one-year call the CRR error is 0.0237 at 100 steps, 0.0047 at 500 and 0.0024 at 1,000, so each doubling roughly halves it. Compare even step counts with even, because the error flips sign between odd and even counts at the money.

Why use a binomial tree instead of Black-Scholes?

Because a tree can test for early exercise at every node, which the closed form cannot. On a one-year at-the-money put with a 4 per cent rate and 24 per cent volatility, 1,000 steps value the European put at 7.5306 and the American put at 7.9295, an early exercise premium of 0.3988. For a European option, the closed form is faster and exact.

Read the whole case

This article is one calculation from Quantitative Finance. The book takes the same case from first principles to the decision, chapter by chapter, and every figure it prints is a live formula in the free companion workbooks.

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Also on this site

How do you calculate a Black-Scholes option price by hand?Call = S x N(d1) minus discounted K x N(d2). At spot 100, strike 100, 4% rate and 24% volatility, d1 is 0.2867 and the one-year call is worth 11.4541.
How do you calculate implied volatility with Newton's method?Step volatility by the pricing error over vega until the model matches the quote: a one-year at-the-money call quoted at 12.50 implies 26.7302% in two passes.
Does hedging monthly instead of daily lose money?Hedging monthly does not lose money on average. It raises one standard deviation of outcome from 0.5265 to 2.3031, or 20.1 per cent of the premium.
How do you calculate the gamma-theta break-even move?Square root of 2 x theta / gamma, with theta net of the hedge's financing: 1.2562 a day on a 24% vol at-the-money call, not the 1.5042 raw theta gives.
How many Monte Carlo paths do you need to price an option?Paths = (payoff sd / tolerance) squared. With a payoff sd of 17.43 on an 11.4541 call, a 0.01 standard error takes about 3 million paths, not one million.
How are a new bond's coupon and reoffer price calculated?Mid-swap 2.734% plus 118 bp gives a 3.914% reoffer yield; the coupon is rounded down to 3.875% and the price solves to 99.765. Worked example.

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