Binomial Pricing

Options and Futures

Lorenzo Naranjo

Spring 2024

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.

• 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 put 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 put, i.e., $100.
• The price of the bond is: \[ B = 100 e^{-0.06 \times 6/12} = \$97.04 \]
• 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 put.

Step 1: Setting the Objective

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

Step 2: Replicating the Put

• Furthermore, say we choose \(N_S\) and \(N_B\) such that payoff of the portfolio matches the payoff of the put, i.e. \[ \begin{align*} 110 N_S + 100 N_B & = 0 \\ 90N_S + 100 N_B & = 10 \end{align*} \]
• We can solve for \(N_S\) and \(N_B\) to find: \[ \begin{align*} N_S & = \frac{0 - 10}{110 - 90} = -0.50 \\ N_B & = - \frac{110}{100} N_S = (-1.1)(-0.5) = 0.55 \end{align*} \]

Step 2: Pricing The Put

• 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, otherwise there would be an arbitrage opportunity: \[ P = -0.5 \times 100 + 0.55 \times 97.04 = \$3.38 \]

Example: Put Arbitrage

• What would happen in the previous analysis if the put was trading for $3?

Put Leverage

• The replication analysis shows that the put can be seen as an investment in the risk-free bond that is financed in part by shorting stocks.

Replicating A Call Option

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

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

Pricing the Call

• We can solve for \(N_S\) and \(N_B\) to find: \[ \begin{align*} N_S & = \frac{10 - 0}{110 - 90} = 0.50 \\ N_B & = - \frac{90}{100} N_S = (-0.90)(0.5) = -0.45 \end{align*} \]
• 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: \[ C = 0.5 \times 100 - 0.45 \times 97.04 = \$6.33 \]

Call Leverage

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

Replicating A Generic Derivative

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

• As before, we start by replicating the payoffs of the derivative by trading the stock and the bond: \[ \begin{align*} S_{u} N_{S} + F N_{B} & = X_{u} \\ S_{d} N_{S} + F N_{B} & = X_{d} \end{align*} \]

Pricing the Derivative

• We can solve for \(N_S\) and \(N_B\) to find: \[ \begin{align*} N_{S} & = \frac{X_{u} - X_{d}}{S^{u}- S^{d}} \\ N_{B} & = \frac{X_{u} - S_{u} N_{S}}{F} = \frac{X^{d} - S^{d} N_{S}}{F} \end{align*} \]
• The price of the derivative must then match the price of the portfolio, otherwise there would be an arbitrage opportunity: \[ X = N_{S} S + N_{B} B \]
• 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: \[ X = N_S S + N_B B \] where

  • \(S\) is the current stock price
  • \(B\) is the price of a risk-free zero-coupon bond with face value \(K\) and maturity \(T\)
  • \(N_{S}\) and \(N_{B}\) 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 whose 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 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: \[ 100 = (110 p + 90 (1 - p)) e^{-0.06 \times 6/12} \]
• We can reverse-engineer the probability of the stock going up that makes consistent valuations in this world: \[ p = \frac{100 e^{0.06 \times 6/12} - 90}{110 - 90} = 0.6522 \]

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: \[ C = (10 p + 0 (1 - p)) e^{-0.06 \times 6/12} = \$6.33 \]
• Similarly, for the put we have that: \[ P = (0 p + 10 (1 - p)) e^{-0.06 \times 6/12} = \$3.38 \]
• 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: \[ p = \frac{50 e^{0.06 \times 6/12} - 40}{60 - 40} \]
• The price of the call is: \[ C = (12 p + 0 (1 - p)) e^{-0.06 \times 6/12} = 6.71 \]

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: \[ p = \frac{120 e^{0.05 \times 3/12} - 108}{132 - 108} \]
• The price of the asset is: \[ X = (100 p + 200 (1 - p)) e^{-0.05 \times 3/12} = 141.93 \]

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: \[ \begin{align*} \pi_{u} & = (1 p + 0 (1 - p)) e^{-r T} \\ \pi_{d} & = (0 p + 1 (1 - p)) e^{-r T} \end{align*} \]

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. \(\pi_{u} = (1 p + 0 (1 - p)) e^{-0.05 \times 3/12} = 0.5559\)
  2. \(\pi_{d} = (0 p + 1 (1 - p)) e^{-0.05 \times 3/12} = 0.4317\)
  3. \(X = 100 \pi_{u} + 200 \pi_{d} = 141.93\)