Chapter 13: Binomial Trees
6 min readThe One-Step Binomial Model
A simple yet powerful approach: assume the stock price can move either up or down over a single time step of length Δt\Delta t.
- Stock price S0S_0 goes up to S0⋅uS_0 \cdot u with probability pp
- Stock price S0S_0 goes down to S0⋅dS_0 \cdot d with probability 1−p1-p
Example: S0=20S_0 = 20, u=1.1u = 1.1, d=0.9d = 0.9, T=3T = 3 months, r=12%r = 12\%, K=21K = 21.
Up state: Su=22S_u = 22, call payoff = max(22−21,0)=1\max(22 - 21, 0) = 1 Down state: Sd=18S_d = 18, call payoff = max(18−21,0)=0\max(18 - 21, 0) = 0
No-Arbitrage Argument
Create a riskless portfolio: long Δ\Delta shares of stock and short 1 call:
Δ⋅Su−fu=Δ⋅Sd−fd\Delta \cdot S_u - f_u = \Delta \cdot S_d - f_d
Solving:
Δ=fu−fdS0u−S0d\Delta = \frac{f_u - f_d}{S_0 u - S_0 d}
The portfolio earns the risk-free rate, so:
S0Δ−f=(S0uΔ−fu)⋅e−rTS_0 \Delta - f = (S_0 u \Delta - f_u) \cdot e^{-rT}
Solving for ff, the call price:
f=S0Δ−(S0uΔ−fu)e−rT=0.633f = S_0 \Delta - (S_0 u \Delta - f_u) e^{-rT} = 0.633
Risk-Neutral Valuation
The same result emerges if we define risk-neutral probabilities:
p=erΔt−du−dp = \frac{e^{r \Delta t} - d}{u - d}
Then the option value is the expected payoff discounted at the risk-free rate:
f=e−rΔt⋅[p⋅fu+(1−p)⋅fd]f = e^{-r \Delta t} \cdot [p \cdot f_u + (1 - p) \cdot f_d]
Key insight: In a risk-neutral world, all investors are indifferent to risk. The expected return on all assets is the risk-free rate. The option price is the same in the real world and the risk-neutral world.
For our example: p=(e0.12×0.25−0.9)/(1.1−0.9)=0.6523p = (e^{0.12 \times 0.25} - 0.9) / (1.1 - 0.9) = 0.6523
f=e−0.12×0.25×[0.6523×1+0.3477×0]=0.633f = e^{-0.12 \times 0.25} \times [0.6523 \times 1 + 0.3477 \times 0] = 0.633
Two-Step Binomial Trees
Extend to two steps. At each node:
- Compute the option values at the final nodes (maturity)
- Work backwards using risk-neutral valuation
European options: One pass backward through the tree American options: At each node, compare the value of immediate exercise with the computed value (take the maximum)
Matching Volatility
The parameters uu and dd are chosen to match the stock's volatility σ\sigma:
u=eσΔt,d=e−σΔt=1uu = e^{\sigma \sqrt{\Delta t}}, \quad d = e^{-\sigma \sqrt{\Delta t}} = \frac{1}{u}
The risk-neutral probability:
p=erΔt−du−dp = \frac{e^{r \Delta t} - d}{u - d}
Delta (Δ\Delta)
Delta is the ratio of the change in option price to the change in stock price:
Δ=fu−fdS0u−S0d\Delta = \frac{f_u - f_d}{S_0 u - S_0 d}
For the one-step example: Δ=(1−0)/(22−18)=0.25\Delta = (1 - 0)/(22 - 18) = 0.25
Delta is the number of shares to hold to create a riskless hedge. It changes at each node in the tree.
Increasing the Number of Steps
As the number of time steps nn increases:
- The binomial pricing converges to the Black–Scholes–Merton price
- A tree with 30 steps usually gives adequate accuracy
- DerivaGem can handle up to 500 steps
Cox-Ross-Rubinstein Formulas
The binomial tree is consistent with Black–Scholes–Merton as Δt→0\Delta t \to 0 when:
u=eσΔt,d=e−σΔtu = e^{\sigma \sqrt{\Delta t}}, \quad d = e^{-\sigma \sqrt{\Delta t}}
For a European call with nn steps, the price is:
c=e−rnΔt∑j=0n(nj)pj(1−p)n−jmax(S0ujdn−j−K,0)c = e^{-rn \Delta t} \sum_{j=0}^{n} \binom{n}{j} p^j (1-p)^{n-j} \max(S_0 u^j d^{n-j} - K, 0)
Options on Other Assets
The binomial tree adapts to other underlyings by adjusting the risk-neutral probability pp:
| Underlying | pp |
|---|---|
| Stock index (yield qq) | (e(r−q)Δt−d)/(u−d)(e^{(r-q)\Delta t} - d)/(u - d) |
| Currency (rfr_f = foreign rate) | (e(r−rf)Δt−d)/(u−d)(e^{(r-r_f)\Delta t} - d)/(u - d) |
| Futures contract | (1−d)/(u−d)(1 - d)/(u - d) |
Appendix: From Binomial to Black–Scholes–Merton
As n→∞n \to \infty and Δt→0\Delta t \to 0, the binomial model converges to the lognormal stock price distribution assumed by Black–Scholes–Merton. The binomial sum converges to the Black–Scholes–Merton pricing formula.