Binomial Pricing

Options, Futures and Derivative Securities

Lorenzo Naranjo

Spring 2026

Introduction

Binomial Trees

• One of the easiest ways to describe the evolution over time of a stock price is to use what in finance we call a binomial tree.
• At each point there are only two possibilities for the future stock price happening with probability p and 1 - p, respectively.
• We will see that these probabilities play no role in the price of an option.

• It is useful to define S_{u} = S \times u and S_{d} = S \times d, where u and d are the gross percentage increase and decrease of the stock price over the next period, respectively.

Example: One-Period Binomial Tree

• The current stock price is $100.
• Next period, the asset can go up or down by 10% with probability p and 1 - p, respectively.
• In this example u = 1.10 and d = 0.90.

The Replicating Portfolio Approach

Pricing Options

• Consider a non-dividend paying stock that currently trades for $100.
• Over the next 6-months the stock can go up or down by 10%.

• The interest rate is 6% per year with continuous compounding.
• What should be the price of a European call option with maturity 6 months and strike price $100?

The Binomial Tree of a Bond

• Consider a risk-free bond with maturity 6 months and face value equal to the strike price of the call, i.e., $100.
• The price of the bond is: Z = 100 e^{-0.06 \times 6/12} = \$97.04
• The binomial tree for the bond is:

The Replicating Portfolio Approach

• We will price the option by using the stock and bonds to replicate the payoffs of the call.

Step 1: Setting the Objective

• Say we purchase N_S units of the stock and N_Z units of the bond.
• Such a portfolio would pay \text{Payoff} = \begin{cases} 110 N_S + 100 N_Z & \text{if $S = 110$} \\ 90N_S + 100 N_Z & \text{if $S = 90$} \end{cases}

Step 2: Replicating the Call

• Furthermore, say we choose N_S and N_Z such that payoff of the portfolio matches the payoff of the call, i.e. \begin{align*} 110 N_S + 100 N_Z & = 10 \\ 90N_S + 100 N_Z & = 0 \end{align*}
• We can solve for N_S and N_Z to find: \begin{align*} N_S & = \frac{10 - 0}{110 - 90} = 0.50 \\ N_Z & = - \frac{90}{100} N_S = (-0.90)(0.5) = -0.45 \end{align*}

Step 3: Pricing the Call

• Therefore, by buying 0.50 units of the stock and shorting 0.45 units of the bond we can exactly match the payoffs of the call.
• The price of the call must then match the price of the portfolio, otherwise there would be an arbitrage opportunity: C = 0.5 \times 100 - 0.45 \times 97.04 = \$6.33

Example: Call Arbitrage

• What would happen in the previous analysis if the call was trading for $6?
• Buy the call and sell the synthetic call: short 0.5 shares and buy 0.45 bonds.

Position	Today	S_{T} = 110	S_{T} = 90
Long Call	-6.00	10	0
Short Synthetic Call	6.33	-10	0
Total	0.33	0	0

Call Leverage

• The replication analysis reveals that the call can be seen as a levered position on the stock.

• A 1% fall in the stock reduces the call by about 0.5/6.33 = 7.9\%.

Replicating a Put Option

• Consider now a European put option with the same maturity and strike price as the call.

• As before, we replicate the payoffs of the put option by trading the stock and the bond: \begin{align*} 110 N_S + 100 N_Z & = 0 \\ 90N_S + 100 N_Z & = 10 \end{align*}

Pricing the Put

• We can solve for N_S and N_Z to find: \begin{align*} N_S & = \frac{0 - 10}{110 - 90} = -0.50 \\ N_Z & = - \frac{110}{100} N_S = (-1.1)(-0.5) = 0.55 \end{align*}
• Therefore, by shorting 0.50 units of the stock and going long 0.55 units of the bond we can exactly match the payoffs of the put.
• The price of the put must then match the price of the portfolio: P = -0.5 \times 100 + 0.55 \times 97.04 = \$3.37

Put Leverage

• Where the call is a levered long position in the stock, the put is an investment in the risk-free bond financed in part by shorting stocks.

• A 1% rise in the stock reduces the put by about 0.5/3.37 = 14.8\%.

Replicating A Generic Derivative

• The analysis so far suggests that we can generalize the replicating approach to price any derivative.

• The bond has face value M and price Z = M e^{-r T}.
• As before, we start by replicating the payoffs of the derivative by trading the stock and the bond: \begin{align*} S_{u} N_{S} + M N_{Z} & = X_{u} \\ S_{d} N_{S} + M N_{Z} & = X_{d} \end{align*}

Pricing the Derivative

• We can solve for N_S and N_Z to find: \begin{align*} N_{S} & = \frac{X_{u} - X_{d}}{S_{u} - S_{d}} \\ N_{Z} & = \frac{X_{u} - S_{u} N_{S}}{M} = \frac{X_{d} - S_{d} N_{S}}{M} \end{align*}
• The price of the derivative must then match the price of the portfolio, otherwise there would be an arbitrage opportunity: X = N_{S} S + N_{Z} Z
• The number of shares N_{S} needed to replicate the derivative is called the delta of the instrument.

Pricing a Derivative in the Binomial Model

• In a one-period binomial model, the price X of a European call or put option with strike K and maturity T takes the form: X = N_S S + N_Z Z where

  • S is the current stock price
  • Z is the price of a risk-free zero-coupon bond with face value K and maturity T
  • N_{S} and N_{Z} are the number of shares and risk-free bonds, respectively, needed to replicate the derivative

The Risk-Neutral Approach

The Risk-Neutral Approach

• In replicating the payoffs of the option, we never used the actual probabilities.
• As a matter of fact, these probabilities might even change based on who’s thinking about the asset.
• Since the previous reasoning is silent about the probabilities and the type of investor pricing the asset, we can assume in our reasoning that all investors are risk neutral.
• Even if this is not true in real markets, such assumption would not affect the logic of the replicating-portfolio argument.

The Parable of the Bookmaker

• A bookmaker takes bets on a two-horse race (Baxter and Rennie 1996).
  • Careful study gives the horses a 25% and a 75% chance of winning.
  • The public bets $5,000 on the first horse and $10,000 on the second.
• Odds set from the true probabilities: zero expected profit, but a $5,000 loss if the outsider wins.
• Odds set from the money wagered, as if the chances were 1/3 and 2/3: the stakes cover every payout, and the bookmaker breaks even whichever horse wins.
• In the binomial model, the price of the stock plays the role of the bets, and the probability q it implies is the risk-neutral probability.

The Real Probabilities Are Irrelevant

• In a world populated by risk-neutral investors, all expected payoffs should be discounted at the risk-free rate, regardless of their riskiness.
• Hence, the price of the stock in this world, which is $100, should be equal to the expected payoff discounted at the risk-free rate: 100 = (110 q + 90 (1 - q)) e^{-0.06 \times 6/12}
• We can reverse-engineer the probability of the stock going up that makes consistent valuations in this world: q = \frac{100 e^{0.06 \times 6/12} - 90}{110 - 90} = 0.6522

The Risk-Neutral Probability

• In general, the risk-neutral probability of an up-move is q = \frac{S e^{r T} - S_{d}}{S_{u} - S_{d}} and the price of any derivative is X = (q X_{u} + (1 - q) X_{d}) e^{-r T}.
• For q to be a probability we need S_{d} < S e^{r T} < S_{u}.
• This is also what rules out arbitrage between the stock and the bond:
  • If S e^{r T} \geq S_{u}, short the stock and invest in the bond.
  • If S e^{r T} \leq S_{d}, borrow to buy the stock.

Pricing the Call and Put Again

• The price of the call is also equal to the expected payoff under this risk-neutral probability, discounted at the risk-free rate: C = (10 q + 0 (1 - q)) e^{-0.06 \times 6/12} = \$6.33
• Similarly, for the put we have that: P = (0 q + 10 (1 - q)) e^{-0.06 \times 6/12} = \$3.37
• Of course, the prices are the same as before since both approaches are consistent with each other.

Example 1

• A non-dividend paying stock trades at $50 and over the next 6-months can go up to $60 or down $40.
• The risk-free rate is 6% per year with continuous compounding.
• Compute the price of a European call option expiring in 6 months with strike price $48.

Example 1: Solution

• The risk-neutral probability of the stock moving up is: q = \frac{50 e^{0.06 \times 6/12} - 40}{60 - 40}
• The price of the call is: C = (12 q + 0 (1 - q)) e^{-0.06 \times 6/12} = 6.71

Example 2

• A non-dividend paying stock trades at $120 and over the next 3-months can increase or decrease by 10%.
• The risk-free rate is 5% per year with continuous compounding.
• Compute the price of an asset that pays in 3 months $100 if the stock increases in price and $200 otherwise.

Example 2: Solution

• The stock can move up to 132 or down to 108.
• The risk-neutral probability of the stock moving up is: q = \frac{120 e^{0.05 \times 3/12} - 108}{132 - 108}
• The price of the asset is: X = (100 q + 200 (1 - q)) e^{-0.05 \times 3/12} = 141.93

State Prices

• The risk-neutral probabilities are intimately related to the so-called Arrow-Debreu securities depicted below.

• The price of each security is then the expected payoff using the risk-neutral probabilities, discounted at the risk-free rate: \begin{align*} \pi_{u} & = (1 q + 0 (1 - q)) e^{-r T} \\ \pi_{d} & = (0 q + 1 (1 - q)) e^{-r T} \end{align*}

Using State Prices

• The risk-neutral probabilities are state prices grown at the risk-free rate: \pi_{u} = q e^{-r T} and \pi_{d} = (1 - q) e^{-r T}.
• Holding both securities pays $1 for sure, so \pi_{u} + \pi_{d} = e^{-r T}.
• Any derivative is a bundle of X_{u} up-securities and X_{d} down-securities: X = X_{u} \pi_{u} + X_{d} \pi_{d}
• In our example, \pi_{u} = 0.6329 and \pi_{d} = 0.3375, so C = 10 \pi_{u} = \$6.33 and P = 10 \pi_{d} = \$3.37.

Example 3

• A non-dividend paying stock trades at $120 and over the next 3-months can increase or decrease by 10%.
• The risk-free rate is 5% per year with continuous compounding.
  1. Compute the price of an asset that pays in 3 months $1 if the stock price increases and $0 otherwise.
  2. Compute the price of an asset that pays in 3 months $0 if the stock price increases and $1 otherwise.
  3. Using the previous results, compute the price of an asset that pays in 3 months $100 if the stock increases in price and $200 otherwise.

Example 3: Solution

• Using the same risk-neutral probabilities as in Example 2, we have that:
  1. \pi_{u} = (1 q + 0 (1 - q)) e^{-0.05 \times 3/12} = 0.5559
  2. \pi_{d} = (0 q + 1 (1 - q)) e^{-0.05 \times 3/12} = 0.4317
  3. X = 100 \pi_{u} + 200 \pi_{d} = 141.93

References

Baxter, Martin, and Andrew Rennie. 1996. Financial Calculus: An Introduction to Derivative Pricing. Cambridge University Press.