Course 7: Derivatives Pricing

Options Pricing: Binomial Model

Master the binomial tree framework for option pricing: replication, risk-neutral valuation, and the path to continuous-time models

Learning Objectives

1.Understand option fundamentals: calls, puts, payoffs, and put-call parity
2.Derive option prices via replicating portfolios in single-period binomial models
3.Master risk-neutral valuation and state prices
4.Apply backward induction for multi-period binomial trees
5.Price American options and analyze early exercise boundaries
6.Understand convergence to Black-Scholes via Cox-Ross-Rubinstein parameterization

1. Option Fundamentals and Put-Call Parity

Definition 1.1: Call and Put Options

Consider a stock with current price $S_0$ and uncertain future price $S_T$ at time $T$ .

A European call option with strike price $K$ gives the holder the right (not obligation) to buy the stock at price $K$ at maturity $T$ . Its payoff is:

C_T = \max(S_T - K, 0) = (S_T - K)^+

A European put option with strike price $K$ gives the holder the right to sell the stock at price $K$ at maturity $T$ . Its payoff is:

P_T = \max(K - S_T, 0) = (K - S_T)^+

The option price today ( $C_0$ for call, $P_0$ for put) is what we seek to determine.

Example 1.1: Option Payoff at Maturity

Stock: $S_0 = \$100$ , Strike: $K = \$105$ , Maturity: $T = 1$ year

At maturity, suppose $S_T$ can be $90, $110, or $130.

Call payoffs:

If $S_T = \$90$ : $C_T = \max(90-105, 0) = \$0$
If $S_T = \$110$ : $C_T = \max(110-105, 0) = \$5$
If $S_T = \$130$ : $C_T = \max(130-105, 0) = \$25$

Put payoffs:

If $S_T = \$90$ : $P_T = \max(105-90, 0) = \$15$
If $S_T = \$110$ : $P_T = \max(105-110, 0) = \$0$
If $S_T = \$130$ : $P_T = \max(105-130, 0) = \$0$

Theorem 1.1: Put-Call Parity

For European options on a non-dividend-paying stock, the following relationship holds:

C_0 - P_0 = S_0 - Ke^{-rT}

where $r$ is the risk-free rate (continuous compounding).

Proof:

Consider two portfolios at time 0:

Portfolio A: Long call + $Ke^{-rT}$ in cash
Value at $t=0$ : $C_0 + Ke^{-rT}$

Portfolio B: Long put + Long stock
Value at $t=0$ : $P_0 + S_0$

At maturity $T$ , the cash grows to $K$ . Consider two cases:

Case 1: $S_T > K$

Portfolio A: Exercise call, pay $K$ , receive stock. Value = $S_T$
Portfolio B: Put expires worthless, keep stock. Value = $S_T$

Case 2: $S_T \leq K$

Portfolio A: Call expires worthless, keep cash $K$ . Value = $K$
Portfolio B: Exercise put, sell stock for $K$ . Value = $K$

Both portfolios have identical payoffs at $T$ in all states. By no-arbitrage:

C_0 + Ke^{-rT} = P_0 + S_0

Rearranging yields put-call parity.

∎

Remark 1.1: Implications of Put-Call Parity

Put-call parity has important consequences:

It's a model-free relationship—holds regardless of stock price dynamics
Can synthesize any position: e.g., $C_0 = P_0 + S_0 - Ke^{-rT}$
Arbitrage opportunities arise if parity is violated in liquid markets
Does not hold for American options (inequality instead)

2. Single-Period Binomial Model

Definition 2.1: Single-Period Binomial Tree

At time $t=0$ , stock price is $S_0$ . At time $t=T$ , stock can move to:

Up state: $S_u = uS_0$ with $u > 1$
Down state: $S_d = dS_0$ with $d < 1$

The risk-free asset grows from $1 to $R = e^{rT}$ , where $r$ is the continuously compounded rate.

No-arbitrage condition: $d < R < u$

Remark 2.1: Why d < R < u is Necessary

If $R \geq u$ : Risk-free dominates stock (sell stock, buy bonds = arbitrage)

If $R \leq d$ : Stock dominates risk-free (borrow, buy stock = arbitrage)

Thus, $d < R < u$ ensures no arbitrage opportunities.

Theorem 2.1: Replication Portfolio

Consider a European option with payoffs $C_u$ (up state) and $C_d$ (down state).

A portfolio holding $\Delta$ shares of stock and $B$ dollars in bonds replicates the option if:

\begin{cases} \Delta S_u + BR = C_u \\ \Delta S_d + BR = C_d \end{cases}

Solving this system yields:

\Delta = \frac{C_u - C_d}{S_u - S_d} = \frac{C_u - C_d}{S_0(u-d)}

B = \frac{uC_d - dC_u}{(u-d)R}

The option price is the replication cost:

C_0 = \Delta S_0 + B

Proof:

From the two equations:

\Delta S_u + BR = C_u \quad (1)

\Delta S_d + BR = C_d \quad (2)

Subtracting (2) from (1):

\Delta(S_u - S_d) = C_u - C_d

\Delta = \frac{C_u - C_d}{S_u - S_d}

Substituting $\Delta$ into equation (1):

\frac{C_u - C_d}{S_u - S_d} \cdot S_u + BR = C_u

BR = C_u - S_u \cdot \frac{C_u - C_d}{S_u - S_d}

BR = \frac{C_u(S_u - S_d) - S_u(C_u - C_d)}{S_u - S_d} = \frac{C_uS_d - C_dS_u}{S_u - S_d}

Using $S_u = uS_0$ and $S_d = dS_0$ :

B = \frac{uC_d - dC_u}{(u-d)R}

∎

Example 2.1: Call Option Pricing in Single-Period Model

Setup: $S_0 = \$100$ , $u = 1.2$ , $d = 0.9$ , $r = 0.05$ , $T = 1$ year, $K = \$105$

Risk-free growth: $R = e^{0.05} \approx 1.0513$

Stock prices at T:

$S_u = 1.2 \times 100 = \$120$
$S_d = 0.9 \times 100 = \$90$

Call payoffs:

$C_u = \max(120 - 105, 0) = \$15$
$C_d = \max(90 - 105, 0) = \$0$

Replication portfolio:

\Delta = \frac{15 - 0}{120 - 90} = \frac{15}{30} = 0.5

B = \frac{1.2 \times 0 - 0.9 \times 15}{(1.2 - 0.9) \times 1.0513} = \frac{-13.5}{0.3154} \approx -\$42.80

Call price:

C_0 = 0.5 \times 100 - 42.80 = \$7.20

Interpretation: Buy 0.5 shares (=$50), borrow $42.80 at risk-free rate. This portfolio replicates the call option.

Theorem 2.2: Risk-Neutral Valuation

Define the risk-neutral probability:

q = \frac{R - d}{u - d}

Under no-arbitrage ( $d < R < u$ ), we have $0 < q < 1$ .

The option price can be expressed as:

C_0 = \frac{1}{R}[qC_u + (1-q)C_d] = e^{-rT}\mathbb{E}^Q[C_T]

where $\mathbb{E}^Q$ denotes expectation under risk-neutral probabilities.

Proof:

From the replication formulas:

C_0 = \Delta S_0 + B = \frac{C_u - C_d}{u - d} + \frac{uC_d - dC_u}{(u-d)R}

Multiply the first term by $\frac{R}{R}$ :

C_0 = \frac{R(C_u - C_d) + uC_d - dC_u}{(u-d)R}

C_0 = \frac{RC_u - RC_d + uC_d - dC_u}{(u-d)R}

C_0 = \frac{C_u(R-d) + C_d(u-R)}{(u-d)R}

Let $q = \frac{R-d}{u-d}$ , then $1-q = \frac{u-R}{u-d}$ :

C_0 = \frac{1}{R}[qC_u + (1-q)C_d]

This is the discounted expected payoff under the risk-neutral measure $Q$ .

∎

Example 2.2: Risk-Neutral Valuation (Continued)

Using data from Example 2.1:

q = \frac{1.0513 - 0.9}{1.2 - 0.9} = \frac{0.1513}{0.3} \approx 0.5043

1 - q \approx 0.4957

Call price via risk-neutral valuation:

C_0 = \frac{1}{1.0513}[0.5043 \times 15 + 0.4957 \times 0]

C_0 = \frac{7.565}{1.0513} \approx \$7.20

Matches the replication approach! Risk-neutral valuation is computationally simpler.

Proposition 2.1: State Prices

Define state prices $\psi_u$ and $\psi_d$ as the current prices of Arrow-Debreu securities paying $1 in the up and down states, respectively.

\psi_u = \frac{q}{R}, \quad \psi_d = \frac{1-q}{R}

Any derivative with payoffs $C_u, C_d$ has price:

C_0 = \psi_u C_u + \psi_d C_d

State prices provide a unified framework for pricing all derivatives in the binomial model.

3. Multi-Period Binomial Trees and Backward Induction

Definition 3.1: N-Period Binomial Tree

Divide time interval $[0, T]$ into $N$ periods of length $\Delta t = T/N$ .

At each step, stock multiplies by $u$ (up) or $d$ (down) with risk-free rate $r\Delta t$ per period.

After $N$ steps with $j$ up moves:

S_{N,j} = S_0 u^j d^{N-j}, \quad j = 0, 1, \ldots, N

Risk-neutral probability per step:

q = \frac{e^{r\Delta t} - d}{u - d}

Theorem 3.1: Backward Induction Algorithm

For a European option, work backward from maturity:

Step 1 (Terminal): At $t = T$ , compute payoffs:

C_{N,j} = \max(S_{N,j} - K, 0) \text{ (call)} \quad \text{or} \quad P_{N,j} = \max(K - S_{N,j}, 0) \text{ (put)}

Step 2 (Recursion): For $n = N-1, N-2, \ldots, 0$ :

C_{n,j} = e^{-r\Delta t}[qC_{n+1,j+1} + (1-q)C_{n+1,j}]

Step 3: The option price today is $C_{0,0}$ .

Example 3.1: Two-Period Call Option

Setup: $S_0 = \$100$ , $u = 1.1$ , $d = 0.95$ , $r = 0.04$ , $\Delta t = 0.5$ years, $K = \$105$ , $N = 2$

Risk-neutral probability:

q = \frac{e^{0.04 \times 0.5} - 0.95}{1.1 - 0.95} = \frac{1.0202 - 0.95}{0.15} \approx 0.4680

Stock tree:

               S₂,₂ = 100 × 1.1² = 121
              /
    S₁,₁ = 110
   /          \
S₀ = 100      S₂,₁ = 100 × 1.1 × 0.95 = 104.5
   \          /
    S₁,₀ = 95
              \
               S₂,₀ = 100 × 0.95² = 90.25

Terminal payoffs (t=1 year):

$C_{2,2} = \max(121 - 105, 0) = \$16$
$C_{2,1} = \max(104.5 - 105, 0) = \$0$
$C_{2,0} = \max(90.25 - 105, 0) = \$0$

Backward induction (t=0.5 years):

C_{1,1} = e^{-0.04 \times 0.5}[0.468 \times 16 + 0.532 \times 0] = 0.9802 \times 7.488 \approx \$7.34

C_{1,0} = e^{-0.04 \times 0.5}[0.468 \times 0 + 0.532 \times 0] = \$0

Initial price (t=0):

C_{0,0} = e^{-0.04 \times 0.5}[0.468 \times 7.34 + 0.532 \times 0] \approx \$3.37

Proposition 3.1: Closed-Form for European Options

For a European call maturing at $T = N\Delta t$ :

C_0 = e^{-rT} \sum_{j=0}^{N} \binom{N}{j} q^j (1-q)^{N-j} \max(S_0 u^j d^{N-j} - K, 0)

This can be simplified to:

C_0 = S_0 \Phi(a; N, q') - Ke^{-rT} \Phi(a; N, q)

where $\Phi(a; N, p)$ is the complementary binomial CDF, $a$ is the minimum number of up moves for the call to be in-the-money, and:

q' = \frac{qu e^{r\Delta t}}{e^{r\Delta t}} = qu

Remark 3.1: American Options

For American options, the holder can exercise at any time. Modify the backward induction:

C^{\text{Am}}_{n,j} = \max\left(\text{Intrinsic}, e^{-r\Delta t}[qC^{\text{Am}}_{n+1,j+1} + (1-q)C^{\text{Am}}_{n+1,j}]\right)

For a call: Intrinsic value = $\max(S_{n,j} - K, 0)$

For a put: Intrinsic value = $\max(K - S_{n,j}, 0)$

Key insight: American puts may be exercised early (e.g., deep in-the-money), while American calls on non-dividend stocks should never be exercised early.

Example 3.2: American Put with Early Exercise

Setup: $S_0 = \$100$ , $u = 1.1$ , $d = 0.8$ , $r = 0.05$ , $\Delta t = 0.5$ , $K = \$100$ , $N = 2$

$q = \frac{e^{0.025} - 0.8}{1.1 - 0.8} \approx 0.7508$

Stock tree: $S_{2,2} = 121$ , $S_{2,1} = 88$ , $S_{2,0} = 64$

Terminal put payoffs: $P_{2,2} = 0$ , $P_{2,1} = 12$ , $P_{2,0} = 36$

At t=0.5, node (1,0) where S=80:

\text{Hold value} = e^{-0.025}[0.7508 \times 12 + 0.2492 \times 36] \approx \$17.63

\text{Exercise value} = 100 - 80 = \$20

Since $20 > $17.63, exercise early: $P^{\text{Am}}_{1,0} = \$20$

This demonstrates the value of the early exercise feature for American puts.

4. Cox-Ross-Rubinstein Model and Convergence to Black-Scholes

Definition 4.1: Cox-Ross-Rubinstein Parameterization

To ensure convergence to the Black-Scholes model as $N \to \infty$ , Cox, Ross, and Rubinstein (1979) proposed:

u = e^{\sigma\sqrt{\Delta t}}, \quad d = e^{-\sigma\sqrt{\Delta t}} = \frac{1}{u}

where $\sigma$ is the volatility of the stock and $\Delta t = T/N$ .

Risk-neutral probability:

q = \frac{e^{r\Delta t} - d}{u - d} = \frac{e^{r\Delta t} - e^{-\sigma\sqrt{\Delta t}}}{e^{\sigma\sqrt{\Delta t}} - e^{-\sigma\sqrt{\Delta t}}}

Theorem 4.1: Convergence to Black-Scholes

As $N \to \infty$ (equivalently, $\Delta t \to 0$ ), the binomial option price with CRR parameterization converges to the Black-Scholes formula:

\lim_{N \to \infty} C_0^{\text{binom}}(N) = C_0^{\text{BS}}

where

C_0^{\text{BS}} = S_0 \Phi(d_1) - Ke^{-rT}\Phi(d_2)

d_1 = \frac{\ln(S_0/K) + (r + \sigma^2/2)T}{\sigma\sqrt{T}}, \quad d_2 = d_1 - \sigma\sqrt{T}

and $\Phi$ is the standard normal cumulative distribution function.

Proof:

(Sketch) With CRR parameterization, the log stock return over one period:

\ln(S_{n+1}/S_n) = \begin{cases} \sigma\sqrt{\Delta t} & \text{with prob } q \\ -\sigma\sqrt{\Delta t} & \text{with prob } 1-q \end{cases}

Over $N$ periods, $\ln(S_T/S_0)$ is the sum of $N$ independent random variables.

By the Central Limit Theorem, as $N \to \infty$ :

\ln(S_T/S_0) \to N\left((r - \sigma^2/2)T, \sigma^2 T\right)

This matches the log-normal distribution assumption in the Black-Scholes model:

S_T = S_0 e^{(r - \sigma^2/2)T + \sigma\sqrt{T}Z}, \quad Z \sim N(0,1)

Taking expectations under the risk-neutral measure yields the Black-Scholes formula.

∎

Example 4.1: Numerical Convergence

Setup: $S_0 = \$100$ , $K = \$100$ , $r = 0.05$ , $\sigma = 0.2$ , $T = 1$ year

Black-Scholes price: $C_0^{\text{BS}} \approx \$10.45$

Binomial convergence:

$N = 10$ : $C_0 \approx \$10.38$ (error ≈ 0.67%)
$N = 50$ : $C_0 \approx \$10.44$ (error ≈ 0.10%)
$N = 100$ : $C_0 \approx \$10.45$ (error ≈ 0.05%)
$N = 500$ : $C_0 \approx \$10.450$ (error ≈ 0.01%)

Convergence is rapid. Even with $N=50$ steps, the binomial price is within 0.1% of Black-Scholes.

Remark 4.1: Practical Implementation

For European options, Black-Scholes is more efficient. However, binomial trees are essential for:

American options: No closed-form formula; numerical methods required
Path-dependent options: Barriers, lookbacks, Asian options
Dividends: Easy to incorporate discrete dividends at specific nodes
Changing parameters: Time-varying volatility or interest rates

Typical practice: Use $N = 100$ to $N = 500$ steps for accurate pricing.

Key Takeaways

1.Options can be priced by replicating their payoffs with stock and bonds (complete market)
2.Risk-neutral valuation: Price = discounted expected payoff under risk-neutral probabilities (not real-world probabilities!)
3.Multi-period binomial trees use backward induction to price options recursively
4.American options require comparing continuation value vs. immediate exercise value at each node
5.Cox-Ross-Rubinstein parameterization ensures convergence to Black-Scholes as N → ∞
6.Binomial model is the foundation for understanding option pricing and provides intuition for more advanced continuous-time models

Practice Problems

Options Binomial Model

Questions

Correct

Accuracy

A stock is currently priced at

50. In one year, it will be worth either

60 or $45. The risk-free rate is 4% (continuous). What is the risk-neutral probability q of the up move?

Not attempted

Using the data from Q1, what is the price of a European call option with strike K = $52?

Not attempted

In a single-period binomial model, Δ (delta) represents:

Not attempted

Put-call parity states that C0 - P0 equals:

Not attempted

What is the no-arbitrage condition in the binomial model?

Not attempted

In a 2-period binomial tree, how many different terminal stock prices are possible?

Not attempted

For American options in the binomial model, we use:

Not attempted

The Cox-Ross-Rubinstein parameterization sets u = ?

Not attempted

Which statement about risk-neutral probabilities is TRUE?

Not attempted

An American call on a non-dividend stock should:

Not attempted

In a 3-period binomial tree (N=3), how many nodes are there at time step 2?

Not attempted

State prices ψ_u and ψ_d satisfy which property?

Not attempted

As the number of periods N → ∞ in the CRR model, the binomial distribution converges to:

Not attempted

For a put option, if K =

100 and S_T =

110, the payoff is:

Not attempted

The main advantage of binomial trees over Black-Scholes for American options is:

Not attempted

Frequently Asked Questions

What is the key insight of the binomial model?

Options can be priced by constructing a replicating portfolio using the underlying stock and risk-free bonds. In a complete market, this replication is unique and yields a no-arbitrage price.

Why do we use risk-neutral probabilities?

Risk-neutral probabilities simplify pricing: the option value equals the discounted expected payoff under these probabilities. They're not real-world probabilities but mathematical tools ensuring no-arbitrage pricing.

What's the difference between European and American options?

European options can only be exercised at maturity, while American options can be exercised any time before maturity. American options are worth at least as much as European options.

How does the binomial model relate to Black-Scholes?

As the number of periods increases and the time step shrinks, the binomial model converges to the Black-Scholes formula. The Cox-Ross-Rubinstein parameterization ensures this convergence.

Can we price path-dependent options with binomial trees?

Yes! Binomial trees naturally handle path-dependent features like early exercise (American options), barriers, and lookback features by tracking the state at each node.