Trinomial Model

Introduction

The binomial model restricts the stock price to two possible outcomes at each time step: up or down. While this yields a clean, complete market in which every contingent claim can be uniquely replicated, real markets offer a richer set of possibilities. The trinomial model generalizes the binomial framework by allowing three possible price movements per period: up, middle, and down.

This seemingly minor extension has a profound structural consequence. With three possible states but only two traded assets (stock and bond), the payoff space of the market is two-dimensional while the state space is three-dimensional. As a result, not every contingent claim can be replicated, and the market becomes incomplete. The risk-neutral measure is no longer unique---there is an entire family of equivalent martingale measures, each producing a different arbitrage-free price for the same derivative.

The trinomial model thus serves a dual pedagogical purpose:

• Computational: the Boyle (1988) trinomial tree is a practical numerical method that often converges faster than the binomial tree
• Conceptual: it provides the simplest concrete example of market incompleteness, motivating the Second Fundamental Theorem of Asset Pricing

Prerequisites

• Binomial Model: one-period setup, no-arbitrage condition, risk-neutral probability
• Replicating Portfolio: portfolio construction and the notion of replication
• Risk-Neutral Measure: the measure \(\mathbb{Q}\) and expectation pricing

Learning Objectives

By the end of this section, you will be able to:

1. Define the one-period trinomial model and identify its three possible stock price outcomes
2. Derive the no-arbitrage condition for three states
3. Characterize the set of risk-neutral measures and explain why it is not a singleton
4. Prove that the trinomial market is incomplete by showing that replication fails for generic claims
5. Compute the interval of no-arbitrage prices for a European contingent claim
6. Describe the Boyle (1988) parameterization and its connection to matching moments

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One-Period Trinomial Model

Setup

We work on a single period \(t \in \{0, \Delta t\}\) with two traded assets.

Risk-free asset (bank account). Starting from \(B_0 = 1\), the bank account grows deterministically:

\[ B_{\Delta t} = e^{r \Delta t} \]

where \(r \geq 0\) is the continuously compounded risk-free rate.

Risky asset (stock). Starting from \(S_0 > 0\), the stock price at time \(\Delta t\) takes one of three values:

\[ S_{\Delta t} = \begin{cases} u \, S_0 & \text{with probability } p_u \quad \text{(up state)} \\[6pt] m \, S_0 & \text{with probability } p_m \quad \text{(middle state)} \\[6pt] d \, S_0 & \text{with probability } p_d \quad \text{(down state)} \end{cases} \]

where the multiplicative factors satisfy \(u > m > d > 0\) and the physical probabilities satisfy \(p_u, p_m, p_d > 0\) with \(p_u + p_m + p_d = 1\).

Notation Convention

The middle factor \(m\) is often chosen so that \(m = 1\) (the stock price stays flat) or \(m = e^{r \Delta t}\) (the stock earns the risk-free rate in the middle state). Neither convention is required---the theory works for any \(m\) with \(d < m < u\).

The Sample Space

The one-period trinomial model lives on a finite probability space \((\Omega, \mathcal{F}, \mathbb{P})\) with three states:

\[ \Omega = \{\omega_u, \omega_m, \omega_d\} \]

The filtration is \(\mathcal{F}_0 = \{\emptyset, \Omega\}\) (no information at time \(0\)) and \(\mathcal{F}_{\Delta t} = 2^\Omega\) (full information at time \(\Delta t\)). A contingent claim is any \(\mathcal{F}_{\Delta t}\)-measurable random variable, which in this finite setting is simply a vector \(H = (H_u, H_m, H_d)\) specifying the payoff in each state.

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No-Arbitrage Condition

Portfolios

A portfolio \((\Delta, \beta)\) consists of \(\Delta\) shares of stock and \(\beta\) units of the bank account. Its value at each time is:

\[ V_0 = \Delta \, S_0 + \beta, \qquad V_{\Delta t} = \Delta \, S_{\Delta t} + \beta \, e^{r \Delta t} \]

An arbitrage is a portfolio with \(V_0 \leq 0\), \(V_{\Delta t} \geq 0\) in all states, and \(V_{\Delta t} > 0\) in at least one state.

Derivation

The same logic as in the binomial model applies, but now the extreme states are \(u\) and \(d\).

If \(e^{r \Delta t} \geq u\): short the stock, invest in the bank. Since \(u > m > d\), the payoff \(S_0 e^{r \Delta t} - S_{\Delta t}\) is non-negative in all three states and strictly positive in the middle and down states. This is an arbitrage.

If \(e^{r \Delta t} \leq d\): buy the stock, borrow from the bank. The payoff \(S_{\Delta t} - S_0 e^{r \Delta t}\) is non-negative in all three states and strictly positive in the middle and up states. This is an arbitrage.

Conversely, if \(d < e^{r \Delta t} < u\), one can show that no portfolio \((\Delta, \beta)\) with \(V_0 \leq 0\) achieves \(V_{\Delta t} \geq 0\) in all three states with strict inequality somewhere (this follows from the First FTAP, or can be verified directly by solving the resulting linear system).

No-Arbitrage Condition (Trinomial Model)

The one-period trinomial market is arbitrage-free if and only if:

\[ \boxed{d < e^{r \Delta t} < u} \]

Interpretation: Exactly as in the binomial case, the risk-free return must lie strictly between the worst and best possible stock returns. The middle factor \(m\) does not appear in the condition---it is the extreme states that determine whether arbitrage exists.

Why \(m\) Does Not Matter

The no-arbitrage condition depends only on \(d\) and \(u\) because a portfolio that exploits a deterministic dominance of one asset over another must do so in every state. Only the extremes \(d\) and \(u\) are relevant for bounding the risk-free rate.

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Risk-Neutral Measures

The Martingale Condition

A probability measure \(\mathbb{Q}\) on \(\Omega = \{\omega_u, \omega_m, \omega_d\}\) is a risk-neutral measure (equivalent martingale measure) if:

1. Equivalence: \(\mathbb{Q}(\omega) > 0\) for every \(\omega \in \Omega\), i.e., \(q_u, q_m, q_d > 0\)
2. Normalization: \(q_u + q_m + q_d = 1\)
3. Martingale property: The discounted stock price is a \(\mathbb{Q}\)-martingale:

\[ S_0 = e^{-r \Delta t} \, \mathbb{E}^{\mathbb{Q}}[S_{\Delta t}] \]

Expanding the expectation and dividing by \(S_0\):

\[ e^{r \Delta t} = q_u \, u + q_m \, m + q_d \, d \]

Characterizing the Set of Risk-Neutral Measures

We now have three unknowns \((q_u, q_m, q_d)\) subject to:

\[ q_u + q_m + q_d = 1 \]

\[ q_u \, u + q_m \, m + q_d \, d = e^{r \Delta t} \]

\[ q_u > 0, \quad q_m > 0, \quad q_d > 0 \]

This is a system of two linear equations in three unknowns with positivity constraints. The solution set is a one-parameter family. Using the normalization constraint to eliminate \(q_m = 1 - q_u - q_d\) and substituting into the martingale equation:

\[ q_u \, u + (1 - q_u - q_d) \, m + q_d \, d = e^{r \Delta t} \]

\[ q_u (u - m) + q_d (d - m) = e^{r \Delta t} - m \]

\[ q_u (u - m) - q_d (m - d) = e^{r \Delta t} - m \]

We can parameterize the family by a free parameter. Let \(q_d = \lambda\) where \(\lambda > 0\). Then:

\[ q_u = \frac{e^{r \Delta t} - m + \lambda(m - d)}{u - m} \]

\[ q_m = 1 - q_u - \lambda \]

The constraints \(q_u > 0\), \(q_m > 0\), \(q_d = \lambda > 0\) restrict \(\lambda\) to an open interval \((\lambda_{\min}, \lambda_{\max})\).

Family of Risk-Neutral Measures

Under the no-arbitrage condition \(d < e^{r \Delta t} < u\), the set of risk-neutral measures for the trinomial model is a one-parameter family \(\{(q_u(\lambda), q_m(\lambda), q_d(\lambda)) : \lambda \in (\lambda_{\min}, \lambda_{\max})\}\), where each member satisfies the martingale condition and strict positivity.

In contrast, the binomial model has a unique risk-neutral measure \(q = (e^{r\Delta t} - d)/(u - d)\).

Geometric Interpretation

The martingale condition defines a plane (actually a line, after intersecting with the probability simplex) in the simplex \(\{(q_u, q_m, q_d) : q_i \geq 0, \sum q_i = 1\}\). In the binomial model, two constraints on two unknowns yield a single point. In the trinomial model, two constraints on three unknowns yield a line segment in the interior of the simplex.

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Market Incompleteness

Replication Failure

In the binomial model, any contingent claim \(H = (H_u, H_d)\) can be replicated by choosing \((\Delta, \beta)\) to solve two equations in two unknowns. In the trinomial model, replication requires:

\[ \Delta \, u \, S_0 + \beta \, e^{r \Delta t} = H_u \]

\[ \Delta \, m \, S_0 + \beta \, e^{r \Delta t} = H_m \]

\[ \Delta \, d \, S_0 + \beta \, e^{r \Delta t} = H_d \]

This is a system of three equations in two unknowns \((\Delta, \beta)\). Generically, such a system has no solution.

Incompleteness Theorem

Proposition. In the one-period trinomial model with two traded assets, a contingent claim \(H = (H_u, H_m, H_d)\) is replicable if and only if:

\[ \frac{H_u - H_m}{(u - m) \, S_0} = \frac{H_m - H_d}{(m - d) \, S_0} \]

That is, replication is possible only when the payoff differences are proportional to the stock price differences---the claim must be affine in \(S_{\Delta t}\).

Proof sketch. The first two equations determine \(\Delta\) and \(\beta\) uniquely. Substituting into the third equation yields the compatibility condition above. \(\square\)

Claims that fail this condition---such as a generic European call option---cannot be replicated. The market is incomplete.

Connection to the Second Fundamental Theorem

The First FTAP states: no-arbitrage if and only if there exists at least one risk-neutral measure. The Second FTAP adds:

The market is complete if and only if the risk-neutral measure is unique.

In the trinomial model:

• Risk-neutral measures exist (by no-arbitrage), so the First FTAP is satisfied
• Multiple risk-neutral measures exist, so by the Second FTAP, the market is incomplete

This connection is developed rigorously in Complete Markets and Uniqueness.

Completing the Market

To make the trinomial market complete, one could introduce a third traded asset (e.g., an option on the stock). With three assets and three states, the payoff matrix becomes \(3 \times 3\) and generically invertible, restoring uniqueness of the risk-neutral measure. This is a common strategy in practice: liquid options serve as additional hedging instruments.

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No-Arbitrage Price Bounds

The Pricing Interval

Since the risk-neutral measure is not unique, the risk-neutral pricing formula \(V_0 = e^{-r \Delta t} \, \mathbb{E}^{\mathbb{Q}}[H]\) gives a different price for each choice of \(\mathbb{Q}\). Every such price is consistent with no-arbitrage. The set of all no-arbitrage prices forms an interval:

\[ \boxed{V_0 \in \left( \inf_{\mathbb{Q} \in \mathcal{Q}} e^{-r \Delta t} \, \mathbb{E}^{\mathbb{Q}}[H], \; \sup_{\mathbb{Q} \in \mathcal{Q}} e^{-r \Delta t} \, \mathbb{E}^{\mathbb{Q}}[H] \right)} \]

where \(\mathcal{Q}\) is the set of all risk-neutral measures. The infimum and supremum are taken over the one-parameter family derived above.

Open vs Closed Interval

For a non-replicable claim, the interval is open: the extreme prices correspond to degenerate measures where one of \(q_u, q_m, q_d\) equals zero, violating the equivalence requirement. For a replicable claim, all measures agree and the interval collapses to a single point.

Super-Replication Interpretation

The upper bound \(\sup_{\mathbb{Q}} e^{-r \Delta t} \, \mathbb{E}^{\mathbb{Q}}[H]\) equals the cost of the cheapest portfolio that super-replicates \(H\) (i.e., \(V_{\Delta t} \geq H\) in every state). Similarly, the lower bound equals the negative of the super-replication cost of \(-H\). These bounds have direct financial meaning:

• A seller of the claim needs at most the upper bound to hedge
• A buyer would pay at most the lower bound to avoid being overcharged relative to the market

Any price within the open interval is consistent with no-arbitrage, but the market does not determine a unique fair value. Additional criteria---such as utility maximization, model calibration, or risk preferences---are needed to select a specific price.

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The Boyle Parameterization

Motivation

For computational purposes, we need a specific choice of \(u\), \(m\), \(d\) (and often a specific risk-neutral measure) for the trinomial tree. Boyle (1988) proposed choosing parameters to match the first two moments (mean and variance) of the stock's log-return under the risk-neutral measure, plus an additional symmetry condition.

Construction

Fix the time step \(\Delta t\) and the volatility \(\sigma > 0\). Boyle's parameterization sets:

\[ u = e^{\sigma \sqrt{2 \Delta t}}, \qquad m = 1, \qquad d = e^{-\sigma \sqrt{2 \Delta t}} \]

Note that \(u \, d = 1\) and \(m = 1\) (the stock is unchanged in the middle state). The risk-neutral probabilities are chosen to match the mean and variance of \(\ln(S_{\Delta t}/S_0)\):

\[ q_u = \left( \frac{e^{r \Delta t / 2} - e^{-\sigma \sqrt{\Delta t / 2}}}{e^{\sigma \sqrt{\Delta t / 2}} - e^{-\sigma \sqrt{\Delta t / 2}}} \right)^2 \]

\[ q_d = \left( \frac{e^{\sigma \sqrt{\Delta t / 2}} - e^{r \Delta t / 2}}{e^{\sigma \sqrt{\Delta t / 2}} - e^{-\sigma \sqrt{\Delta t / 2}}} \right)^2 \]

\[ q_m = 1 - q_u - q_d \]

Moment-Matching Property

By construction, \(\mathbb{E}^{\mathbb{Q}}[S_{\Delta t}] = S_0 \, e^{r \Delta t}\) (martingale condition) and \(\text{Var}^{\mathbb{Q}}[\ln(S_{\Delta t}/S_0)] = \sigma^2 \Delta t + O((\Delta t)^2)\). The two free parameters in the risk-neutral family are pinned down by matching the variance and imposing the symmetry \(u \, d = 1\).

Advantages of the Trinomial Tree

Compared to the Cox-Ross-Rubinstein binomial tree:

• Three branches per node produce a more refined lattice, often giving faster convergence to the continuous-time (Black-Scholes) price
• The lattice naturally recombines: an up-then-down path, a middle-then-middle path, and a down-then-up path can all reach the same node
• The extra degree of freedom allows better moment matching and more flexibility in fitting dividend yields or time-varying parameters
• Trinomial trees are especially natural for interest rate models (e.g., the Hull-White trinomial tree in Chapter 20)

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Worked Example: European Call in the Trinomial Model

We now compute the interval of no-arbitrage prices for a European call option in a concrete one-period trinomial model.

Parameters

\[ S_0 = 100, \quad u = 1.2, \quad m = 1.0, \quad d = 0.8, \quad r = 0.05, \quad \Delta t = 1 \text{ year} \]

The risk-free growth factor is \(e^{r \Delta t} = e^{0.05} \approx 1.05127\).

Verify no-arbitrage: \(d = 0.8 < 1.05127 < 1.2 = u\). The condition holds.

Stock and Call Payoffs

Consider a European call with strike \(K = 100\):

State	\(S_{\Delta t}\)	Call payoff \(H = (S_{\Delta t} - K)^+\)
Up	\(120\)	\(20\)
Middle	\(100\)	\(0\)
Down	\(80\)	\(0\)

Is the Call Replicable?

Check the replication condition:

\[ \frac{H_u - H_m}{(u - m) \, S_0} = \frac{20 - 0}{(1.2 - 1.0)(100)} = \frac{20}{20} = 1 \]

\[ \frac{H_m - H_d}{(m - d) \, S_0} = \frac{0 - 0}{(1.0 - 0.8)(100)} = 0 \]

Since \(1 \neq 0\), the call is not replicable. The market is incomplete for this claim.

Computing the Price Interval

The risk-neutral probabilities must satisfy:

\[ q_u + q_m + q_d = 1 \]

\[ 1.2 \, q_u + 1.0 \, q_m + 0.8 \, q_d = 1.05127 \]

Parameterize by \(q_d = \lambda > 0\). From the two equations:

\[ q_u = \frac{1.05127 - 1.0 + \lambda(1.0 - 0.8)}{1.2 - 1.0} = \frac{0.05127 + 0.2\lambda}{0.2} = 0.25634 + \lambda \]

\[ q_m = 1 - q_u - \lambda = 1 - 0.25634 - 2\lambda = 0.74366 - 2\lambda \]

Positivity constraints:

• \(q_d = \lambda > 0\): requires \(\lambda > 0\)
• \(q_u = 0.25634 + \lambda > 0\): automatically satisfied for \(\lambda > 0\)
• \(q_m = 0.74366 - 2\lambda > 0\): requires \(\lambda < 0.37183\)

So \(\lambda \in (0, \, 0.37183)\).

Call price as a function of \(\lambda\):

\[ V_0(\lambda) = e^{-0.05} \bigl( q_u \cdot 20 + q_m \cdot 0 + q_d \cdot 0 \bigr) = e^{-0.05} \cdot 20 \, q_u \]

\[ V_0(\lambda) = e^{-0.05} \cdot 20 \, (0.25634 + \lambda) = 19.0246 \cdot (0.25634 + \lambda) \]

Bounds:

• As \(\lambda \to 0^+\): \(V_0 \to 19.0246 \times 0.25634 \approx 4.877\)
• As \(\lambda \to 0.37183^-\): \(V_0 \to 19.0246 \times (0.25634 + 0.37183) \approx 19.0246 \times 0.62817 \approx 11.949\)

Pricing Interval for the European Call

The no-arbitrage price interval for the European call with \(K = 100\) is:

\[ V_0 \in (4.88, \; 11.95) \]

Any price in this open interval is consistent with no-arbitrage. The market alone does not determine a unique fair value.

For comparison, the binomial model with the same \(u = 1.2\), \(d = 0.8\) gives the unique price:

\[ V_0^{\text{bin}} = e^{-0.05} \cdot \frac{e^{0.05} - 0.8}{1.2 - 0.8} \cdot 20 = e^{-0.05} \cdot 0.62817 \cdot 20 \approx 11.95 \]

The binomial price coincides with the upper bound of the trinomial interval. This is because the binomial model is a special case where the middle state is absent, collapsing the family of measures to a single point at the upper end.

Interpretation

The wide interval \((4.88, \, 11.95)\) reflects the fundamental pricing ambiguity in incomplete markets. In practice, a trader would narrow this interval by:

1. Adding traded instruments: if a second option is liquidly traded, its market price pins down additional constraints on \(\mathbb{Q}\), shrinking the interval (or eliminating it entirely if the market becomes complete)
2. Imposing a model: choosing a specific parameterization (e.g., Boyle's) selects one \(\mathbb{Q}\) from the family
3. Utility-based pricing: an agent's risk preferences select a unique price within the interval

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Summary

Concept	Binomial Model	Trinomial Model
States per period	2	3
Traded assets	2 (stock + bond)	2 (stock + bond)
No-arbitrage condition	\(d < e^{r\Delta t} < u\)	\(d < e^{r\Delta t} < u\)
Risk-neutral measures	Unique	One-parameter family
Market completeness	Complete	Incomplete
Derivative pricing	Unique price	Price interval

Key Takeaways

1. The trinomial model extends the binomial framework by adding a middle state \(m\) with \(d < m < u\), yielding three possible stock outcomes per period.

2. The no-arbitrage condition \(d < e^{r\Delta t} < u\) is identical in form to the binomial case---only the extreme factors matter.

3. With two assets and three states, the system of martingale equations is underdetermined, producing a one-parameter family of risk-neutral measures. By the Second FTAP, this means the market is incomplete.

4. For non-replicable claims, no-arbitrage determines only a price interval \((\inf_{\mathbb{Q}} \mathbb{E}^{\mathbb{Q}}[e^{-r\Delta t} H], \; \sup_{\mathbb{Q}} \mathbb{E}^{\mathbb{Q}}[e^{-r\Delta t} H])\), not a unique price.

5. The Boyle (1988) parameterization pins down specific values of \(u\), \(m\), \(d\) and selects a risk-neutral measure by matching the mean and variance of log-returns, producing a practical trinomial tree for numerical pricing.

6. The trinomial model provides the simplest concrete example of incomplete markets, motivating the general theory in the FTAP section.

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What's Next

Section	Topic
Binomial to Black-Scholes	Continuous-time limit of the binomial tree
Complete Markets and Uniqueness	Second FTAP: uniqueness of \(\mathbb{Q}\) and completeness

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Exercises

Exercise 1. Consider the trinomial model with \(S_0 = 100\), \(u = 1.3\), \(m = 1.05\), \(d = 0.7\), \(r = 5\%\), and \(\Delta t = 1\). Verify the no-arbitrage condition. Then parameterize the family of risk-neutral measures by \(q_d = \lambda\) and determine the admissible range of \(\lambda\).

Solution to Exercise 1

Given \(S_0 = 100\), \(u = 1.3\), \(m = 1.05\), \(d = 0.7\), \(r = 5\%\), \(\Delta t = 1\).

No-arbitrage condition: We need \(d < e^{r\Delta t} < u\).

\[ e^{0.05} = 1.05127 \]

\[ 0.7 < 1.05127 < 1.3 \quad \checkmark \]

Parameterizing the risk-neutral family: With \(q_d = \lambda\), the two constraints are \(q_u + q_m + q_d = 1\) and \(1.3q_u + 1.05q_m + 0.7q_d = 1.05127\).

Substituting \(q_m = 1 - q_u - \lambda\):

\[ 1.3q_u + 1.05(1 - q_u - \lambda) + 0.7\lambda = 1.05127 \]

\[ 1.3q_u + 1.05 - 1.05q_u - 1.05\lambda + 0.7\lambda = 1.05127 \]

\[ 0.25q_u - 0.35\lambda = 0.00127 \]

\[ q_u = \frac{0.00127 + 0.35\lambda}{0.25} = 0.00508 + 1.4\lambda \]

\[ q_m = 1 - (0.00508 + 1.4\lambda) - \lambda = 0.99492 - 2.4\lambda \]

Positivity constraints:

• \(q_d = \lambda > 0\): requires \(\lambda > 0\)
• \(q_u = 0.00508 + 1.4\lambda > 0\): automatically satisfied for \(\lambda > 0\)
• \(q_m = 0.99492 - 2.4\lambda > 0\): requires \(\lambda < 0.41455\)

Admissible range: \(\lambda \in (0, \, 0.41455)\).

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Exercise 2. Using the trinomial model from Exercise 1, compute the no-arbitrage price interval for a European put with strike \(K = 100\). Express \(V_0(\lambda)\) as a function of \(\lambda\) and find the supremum and infimum over the admissible range. Compare the width of the interval to that of a call with the same strike.

Solution to Exercise 2

Using the trinomial model from Exercise 1, the European put with \(K = 100\) has payoffs:

State	\(S_{\Delta t}\)	Put payoff \(H = (K - S_{\Delta t})^+\)
Up	\(130\)	\(0\)
Middle	\(105\)	\(0\)
Down	\(70\)	\(30\)

Put price as a function of \(\lambda\):

\[ V_0(\lambda) = e^{-0.05}(q_u \times 0 + q_m \times 0 + q_d \times 30) = e^{-0.05} \times 30\lambda = 28.537\lambda \]

Bounds:

• As \(\lambda \to 0^+\): \(V_0 \to 0\)
• As \(\lambda \to 0.41455^-\): \(V_0 \to 28.537 \times 0.41455 = 11.83\)

Price interval for the put: \(V_0 \in (0, \; 11.83)\).

Width comparison:

• Put interval width: \(11.83 - 0 = 11.83\)
• Call interval width (from text): \(11.95 - 4.88 = 7.07\)

The put interval is wider. This is because the put payoff is concentrated entirely in the down state, so varying \(q_d\) (which is our free parameter \(\lambda\)) has maximum impact on the put price. The call payoff is concentrated in the up state, and \(q_u = 0.00508 + 1.4\lambda\) has a baseline component \(0.00508\) that limits how small the call price can become.

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Exercise 3. Prove that a contingent claim \(H = (H_u, H_m, H_d)\) in the one-period trinomial model is replicable if and only if the payoff is affine in \(S_{\Delta t}\), i.e., \(H = a \cdot S_{\Delta t} + b\) for some constants \(a, b\). Show that this is equivalent to the condition:

\[ \frac{H_u - H_m}{u - m} = \frac{H_m - H_d}{m - d} \]

Solution to Exercise 3

A claim \(H = (H_u, H_m, H_d)\) is replicable if there exist \(\Delta, \beta\) satisfying:

\[ \Delta \cdot u S_0 + \beta e^{r\Delta t} = H_u \qquad (1) \]

\[ \Delta \cdot m S_0 + \beta e^{r\Delta t} = H_m \qquad (2) \]

\[ \Delta \cdot d S_0 + \beta e^{r\Delta t} = H_d \qquad (3) \]

From \((1) - (2)\): \(\Delta S_0(u - m) = H_u - H_m\), so \(\Delta = \frac{H_u - H_m}{(u-m)S_0}\).

From \((2) - (3)\): \(\Delta S_0(m - d) = H_m - H_d\), so \(\Delta = \frac{H_m - H_d}{(m-d)S_0}\).

For both to hold simultaneously, we need:

\[ \frac{H_u - H_m}{(u - m)S_0} = \frac{H_m - H_d}{(m - d)S_0} \]

which simplifies to:

\[ \frac{H_u - H_m}{u - m} = \frac{H_m - H_d}{m - d} \]

Equivalence to affine payoff: This condition says the "slope" of the payoff is constant between adjacent states. Setting \(a = \frac{H_u - H_m}{(u-m)S_0}\) and solving for \(\beta\) from equation (1):

\[ \beta = e^{-r\Delta t}(H_u - a \cdot uS_0) \]

Then \(H = a \cdot S_{\Delta t} + b\) where \(b = \beta e^{r\Delta t}\) is the bond component. The payoff is affine in \(S_{\Delta t}\). Conversely, any affine payoff \(H = aS_{\Delta t} + b\) automatically satisfies the replication conditions with \(\Delta = a\) and \(\beta = be^{-r\Delta t}\). \(\square\)

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Exercise 4. In the worked example from the text (\(S_0 = 100\), \(u = 1.2\), \(m = 1.0\), \(d = 0.8\), \(r = 5\%\)), the binomial price of the call equals the upper bound of the trinomial interval. Prove this is not a coincidence: show that for any claim with \(H_m = H_d\) (the middle and down payoffs coincide), the binomial price always equals the upper bound of the trinomial price interval.

Solution to Exercise 4

We prove that when \(H_m = H_d\), the binomial price (with states \(u\) and \(d\) only) equals \(\sup_{\mathbb{Q}} e^{-r\Delta t}\mathbb{E}^{\mathbb{Q}}[H]\) over the trinomial risk-neutral family.

The trinomial price as a function of \(\lambda\) is:

\[ V_0(\lambda) = e^{-r\Delta t}(q_u H_u + q_m H_m + q_d H_d) \]

Since \(H_m = H_d\):

\[ V_0(\lambda) = e^{-r\Delta t}(q_u H_u + (q_m + q_d) H_d) = e^{-r\Delta t}(q_u H_u + (1 - q_u) H_d) \]

This depends only on \(q_u\). From the parameterization, \(q_u\) increases as \(\lambda\) increases (since \(q_u = c_1 + c_2\lambda\) with \(c_2 > 0\), where \(c_1, c_2\) depend on the specific model).

If \(H_u > H_d\) (as for a call), then \(V_0\) is increasing in \(q_u\), hence increasing in \(\lambda\). The supremum is reached as \(\lambda \to \lambda_{\max}\), which is when \(q_m \to 0^+\).

When \(q_m = 0\), the trinomial model collapses to the binomial model (middle state has zero probability). In this limit, \(q_u + q_d = 1\) and the martingale condition becomes \(q_u u + q_d d = e^{r\Delta t}\), giving:

\[ q_u = \frac{e^{r\Delta t} - d}{u - d} \]

This is exactly the binomial risk-neutral probability, and the corresponding price is the binomial price. Since this is the limiting value as \(\lambda \to \lambda_{\max}\) (approached but not reached), the supremum of the trinomial prices equals the binomial price. \(\square\)

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Exercise 5. Suppose the trinomial market is "completed" by adding a traded European call option with strike \(K = 100\) and observed market price \(C_0 = 8.50\) (using the same parameters as the worked example in the text). With three assets (stock, bond, call) and three states, the risk-neutral measure becomes unique. Find the unique risk-neutral measure \((q_u, q_m, q_d)\) and use it to price a European put with strike \(K = 110\).

Solution to Exercise 5

With the call market price \(C_0 = 8.50\) and call payoffs \(H^C = (20, 0, 0)\), we have three traded assets: stock payoffs \((120, 100, 80)\), bond payoffs \((e^{0.05}, e^{0.05}, e^{0.05})\), and call payoffs \((20, 0, 0)\).

The unique risk-neutral measure must satisfy:

\[ q_u + q_m + q_d = 1 \qquad (1) \]

\[ 120q_u + 100q_m + 80q_d = 100 \times e^{0.05} = 105.127 \qquad (2) \]

\[ 20q_u \times e^{-0.05} = 8.50, \quad \text{so } q_u = \frac{8.50 \times e^{0.05}}{20} = \frac{8.50 \times 1.05127}{20} = 0.44679 \qquad (3) \]

From (1): \(q_m + q_d = 1 - 0.44679 = 0.55321\).

From (2): \(120 \times 0.44679 + 100q_m + 80q_d = 105.127\)

\[ 53.615 + 100q_m + 80q_d = 105.127 \]

\[ 100q_m + 80q_d = 51.512 \]

Using \(q_m = 0.55321 - q_d\):

\[ 100(0.55321 - q_d) + 80q_d = 51.512 \]

\[ 55.321 - 20q_d = 51.512 \]

\[ q_d = \frac{3.809}{20} = 0.19045 \]

\[ q_m = 0.55321 - 0.19045 = 0.36276 \]

Unique risk-neutral measure: \((q_u, q_m, q_d) = (0.4468, \, 0.3628, \, 0.1905)\).

Pricing a European put with \(K = 110\): Put payoffs \(H^P = (110 - S_{\Delta t})^+\):

• Up: \((110 - 120)^+ = 0\)
• Middle: \((110 - 100)^+ = 10\)
• Down: \((110 - 80)^+ = 30\)

\[ P_0 = e^{-0.05}(0.4468 \times 0 + 0.3628 \times 10 + 0.1905 \times 30) \]

\[ = 0.9512 \times (0 + 3.628 + 5.715) = 0.9512 \times 9.343 = 8.89 \]

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Exercise 6. For the Boyle (1988) parameterization with \(\sigma = 0.20\), \(r = 0.05\), and \(\Delta t = 0.25\), compute \(u\), \(m\), \(d\), \(q_u\), \(q_d\), and \(q_m\). Verify that (a) \(q_u + q_m + q_d = 1\), (b) the martingale condition \(q_u u + q_m m + q_d d = e^{r\Delta t}\) holds, and (c) \(\text{Var}^{\mathbb{Q}}[\ln(S_{\Delta t}/S_0)] \approx \sigma^2 \Delta t\).

Solution to Exercise 6

Given \(\sigma = 0.20\), \(r = 0.05\), \(\Delta t = 0.25\).

Boyle parameters:

\[ u = e^{\sigma\sqrt{2\Delta t}} = e^{0.20\sqrt{0.5}} = e^{0.20 \times 0.7071} = e^{0.14142} = 1.15191 \]

\[ m = 1, \qquad d = e^{-\sigma\sqrt{2\Delta t}} = e^{-0.14142} = 0.86817 \]

Risk-neutral probabilities:

\[ q_u = \left(\frac{e^{r\Delta t/2} - e^{-\sigma\sqrt{\Delta t/2}}}{e^{\sigma\sqrt{\Delta t/2}} - e^{-\sigma\sqrt{\Delta t/2}}}\right)^2 \]

First compute the intermediate quantities: \(\sqrt{\Delta t/2} = \sqrt{0.125} = 0.35355\).

\[ e^{\sigma\sqrt{\Delta t/2}} = e^{0.20 \times 0.35355} = e^{0.07071} = 1.07327 \]

\[ e^{-\sigma\sqrt{\Delta t/2}} = e^{-0.07071} = 0.93174 \]

\[ e^{r\Delta t/2} = e^{0.05 \times 0.125} = e^{0.00625} = 1.00627 \]

\[ q_u = \left(\frac{1.00627 - 0.93174}{1.07327 - 0.93174}\right)^2 = \left(\frac{0.07453}{0.14153}\right)^2 = (0.5266)^2 = 0.2773 \]

\[ q_d = \left(\frac{1.07327 - 1.00627}{1.07327 - 0.93174}\right)^2 = \left(\frac{0.06700}{0.14153}\right)^2 = (0.4734)^2 = 0.2241 \]

\[ q_m = 1 - 0.2773 - 0.2241 = 0.4986 \]

(a) Normalization: \(q_u + q_m + q_d = 0.2773 + 0.4986 + 0.2241 = 1.0000\) \(\checkmark\)

(b) Martingale condition:

\[ q_u u + q_m m + q_d d = 0.2773 \times 1.15191 + 0.4986 \times 1 + 0.2241 \times 0.86817 \]

\[ = 0.31943 + 0.49860 + 0.19455 = 1.01258 \]

\[ e^{r\Delta t} = e^{0.0125} = 1.01258 \quad \checkmark \]

(c) Variance of log-return:

\[ \ln u = 0.14142, \quad \ln m = 0, \quad \ln d = -0.14142 \]

\[ \mathbb{E}^{\mathbb{Q}}[\ln(S_{\Delta t}/S_0)] = 0.2773 \times 0.14142 + 0.4986 \times 0 + 0.2241 \times (-0.14142) = 0.03922 - 0.03169 = 0.00753 \]

\[ \mathbb{E}^{\mathbb{Q}}[(\ln(S_{\Delta t}/S_0))^2] = 0.2773 \times 0.14142^2 + 0.4986 \times 0 + 0.2241 \times 0.14142^2 \]

\[ = (0.2773 + 0.2241) \times 0.02000 = 0.5014 \times 0.02000 = 0.01003 \]

\[ \text{Var}^{\mathbb{Q}}[\ln(S_{\Delta t}/S_0)] = 0.01003 - 0.00753^2 = 0.01003 - 0.00006 = 0.00997 \]

\[ \sigma^2 \Delta t = 0.04 \times 0.25 = 0.01000 \]

The computed variance \(0.00997 \approx 0.01000 = \sigma^2 \Delta t\) \(\checkmark\)