Binomial Pricing
Options, Futures and Derivative Securities
Spring 2025
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.
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- 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%.
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- 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:
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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?
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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.
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Replicating A Call Option
- Consider now a European call option with the same maturity and strike price as the put.
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- 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.
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Replicating A Generic Derivative
- The analysis so far suggests that we can generalize the replicating approach to price any derivative.
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- 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
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 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
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} = \$3.38
- 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:
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.
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- 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*}
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.
- Compute the price of an asset that pays in 3 months $1 if the stock price increases and $0 otherwise.
- Compute the price of an asset that pays in 3 months $0 if the stock price increases and $1 otherwise.
- 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:
- \pi_{u} = (1 q + 0 (1 - q)) e^{-0.05 \times 3/12} = 0.5559
- \pi_{d} = (0 q + 1 (1 - q)) e^{-0.05 \times 3/12} = 0.4317
- X = 100 \pi_{u} + 200 \pi_{d} = 141.93
