Replicating Portfolio¶

Introduction¶

In a complete market, every contingent claim can be replicated—there exists a portfolio of traded assets that produces exactly the same payoff in every state of the world. The cost of this replicating portfolio is the unique no-arbitrage price of the claim.

This section develops replication from two complementary perspectives:

Trading Basis: Replicate derivatives using stock and bond

— the practical approach that can be directly implemented in real markets.
State Price Basis: Replicate everything using Arrow-Debreu securities

— hypothetical assets that pay $1 in exactly one state of the world and nothing otherwise

— revealing the deeper structure of arbitrage-free pricing.

Both approaches yield the same prices.

Prerequisites

Binomial Model (market setup, no-arbitrage condition)

Learning Objectives

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

Replicate any derivative using stock and bond
Understand Arrow-Debreu securities and state prices
Express stock and bond in terms of state prices
See why all pricing approaches are equivalent
Apply replication to price calls, puts, digitals, and forwards

The Payoff Space¶

One-Period Binomial Model¶

In the one-period binomial model, there are exactly two possible states at time $\Delta t$:

Up state: Stock price is $uS_0$
Down state: Stock price is $dS_0$

Any derivative has a payoff that depends on which state occurs:

\[ H = \begin{cases} H_u & \text{(up state)} \\[4pt] H_d & \text{(down state)} \end{cases} \]

We can represent this as a vector in $\mathbb{R}^2$:

\[ H = (H_u, H_d) \]

The Payoff Space Is Two-Dimensional¶

The set of all possible payoffs forms a two-dimensional vector space. Any payoff is just a pair of numbers—what you receive in the up state and what you receive in the down state.

To replicate arbitrary payoffs, we need two linearly independent assets. The binomial model provides exactly this:

Stock: payoff $(uS_0, \, dS_0)$
Bond: payoff $(e^{r\Delta t}, \, e^{r\Delta t})$

These vectors are linearly independent (the stock payoff varies across states; the bond payoff is constant), so they span $\mathbb{R}^2$.

Market Completeness

The one-period binomial market is complete: any payoff $(H_u, H_d)$ can be replicated by some portfolio of stock and bond.

Completeness implies:

Every derivative has a unique no-arbitrage price
There is exactly one risk-neutral measure $\mathbb{Q}$

Method 1: Replication with Stock and Bond¶

The Replication Problem¶

Given a derivative with payoff $(H_u, H_d)$, find a portfolio $(\Delta, B)$ where:

$\Delta$ = number of shares of stock
$B$ = units of bond (cash position)

such that the portfolio replicates the derivative in both states.

The Replication Equations¶

Up state: Portfolio payoff must equal $H_u$

\[ \Delta \cdot uS_0 + B \cdot e^{r\Delta t} = H_u \]

Down state: Portfolio payoff must equal $H_d$

\[ \Delta \cdot dS_0 + B \cdot e^{r\Delta t} = H_d \]

This is a system of 2 linear equations in 2 unknowns.

Solution¶

Solving for $\Delta$ (subtract down equation from up equation):

\[ \Delta \cdot S_0(u - d) = H_u - H_d \]

\[ \boxed{\Delta = \frac{H_u - H_d}{(u - d)S_0}} \]

Solving for $B$ (substitute $\Delta$ into up equation):

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

\[ \boxed{B = e^{-r\Delta t}\left(H_u - \frac{(H_u - H_d)u}{u - d}\right) = e^{-r\Delta t}\left(\frac{uH_d - dH_u}{u - d}\right)} \]

The Replication Price¶

The cost of the replicating portfolio at time $0$ is:

\[ \boxed{V_0 = \Delta \cdot S_0 + B} \]

By the law of one price, this must equal the derivative price. If the derivative traded at any other price, arbitrage would be possible.

Examples: Stock-Bond Replication¶

We use these parameters throughout:

Parameter	Value
$S_0$	$100$
$u$	$1.2$
$d$	$0.9$
$r$	$5\%$
$\Delta t$	$1$ year
$e^{r\Delta t}$	$1.0513$

Example 1: European Call (Strike K = 105)¶

Payoffs:

\[ H_u = (120 - 105)^+ = 15, \qquad H_d = (90 - 105)^+ = 0 \]

Replicating portfolio:

\[ \Delta = \frac{15 - 0}{(1.2 - 0.9) \times 100} = \frac{15}{30} = 0.5 \]

\[ B = e^{-0.05}\left(\frac{1.2 \times 0 - 0.9 \times 15}{1.2 - 0.9}\right) = 0.9512 \times (-45) = -42.80 \]

Price:

\[ C_0 = 0.5 \times 100 + (-42.80) = 7.20 \]

Interpretation: Buy 0.5 shares, borrow $42.80.

Example 2: European Put (Strike K = 105)¶

Payoffs:

\[ H_u = (105 - 120)^+ = 0, \qquad H_d = (105 - 90)^+ = 15 \]

Replicating portfolio:

\[ \Delta = \frac{0 - 15}{30} = -0.5, \qquad B = 0.9512 \times \frac{1.2 \times 15 - 0.9 \times 0}{0.3} = 57.07 \]

Price:

\[ P_0 = -0.5 \times 100 + 57.07 = 7.07 \]

Interpretation: Short 0.5 shares, lend $57.07.

Example 3: Forward Contract (Forward Price F)¶

Payoffs:

\[ H_u = uS_0 - F = 120 - F, \qquad H_d = dS_0 - F = 90 - F \]

Replicating portfolio:

\[ \Delta = \frac{(120 - F) - (90 - F)}{30} = \frac{30}{30} = 1 \]

\[ B = e^{-0.05}(120 - F - 1 \times 120) = -Fe^{-0.05} \]

Price:

\[ V_0 = 1 \times 100 - Fe^{-0.05} = S_0 - Fe^{-r\Delta t} \]

Setting $V_0 = 0$ (forward has zero initial cost):

\[ F = S_0 e^{r\Delta t} = 105.13 \]

Method 2: Arrow-Debreu Securities and State Prices¶

Arrow-Debreu Securities¶

An Arrow-Debreu security (or state-contingent claim) pays $1 in exactly one state and $0 in all other states.

In the binomial model, there are two Arrow-Debreu securities:

Up-state digital $(\mathbf{1}_u)$:

\[ \mathbf{1}_u = \begin{cases} 1 & \text{(up state)} \\ 0 & \text{(down state)} \end{cases} = (1, 0) \]

Down-state digital $(\mathbf{1}_d)$:

\[ \mathbf{1}_d = \begin{cases} 0 & \text{(up state)} \\ 1 & \text{(down state)} \end{cases} = (0, 1) \]

The State Price Basis¶

The Arrow-Debreu securities form a basis for the payoff space. Any payoff $(H_u, H_d)$ can be written as:

\[ \boxed{H = H_u \cdot \mathbf{1}_u + H_d \cdot \mathbf{1}_d} \]

This is trivially true—you just hold $H_u$ units of the up-state digital and $H_d$ units of the down-state digital.

State Prices¶

Let $\psi_u$ and $\psi_d$ denote the prices of the Arrow-Debreu securities:

$\psi_u$ = price of up-state digital (price today of receiving $1 if up)
$\psi_d$ = price of down-state digital (price today of receiving $1 if down)

These are called state prices (or Arrow-Debreu prices).

Universal Pricing Formula¶

Since any payoff is a linear combination of Arrow-Debreu securities:

\[ \boxed{V_0 = \psi_u H_u + \psi_d H_d} \]

This is the fundamental pricing formula. Once you know the state prices, you can price any derivative by simple multiplication.

Deriving State Prices¶

The state prices must be consistent with the traded asset prices (stock and bond). We derive them by replicating the Arrow-Debreu securities.

Replicating the Up-State Digital¶

The up-state digital has payoff $(1, 0)$. Using the stock-bond replication formulas:

\[ \Delta_u = \frac{1 - 0}{(u-d)S_0} = \frac{1}{(u-d)S_0} \]

\[ B_u = e^{-r\Delta t}\left(\frac{u \cdot 0 - d \cdot 1}{u-d}\right) = e^{-r\Delta t}\left(\frac{-d}{u-d}\right) \]

Price of up-state digital:

\[ \psi_u = \Delta_u S_0 + B_u = \frac{1}{u-d} + e^{-r\Delta t}\left(\frac{-d}{u-d}\right) = \frac{1 - de^{-r\Delta t}}{u-d} \]

Simplifying:

\[ \boxed{\psi_u = e^{-r\Delta t} \cdot \frac{e^{r\Delta t} - d}{u - d} = e^{-r\Delta t} q} \]

where $q = \frac{e^{r\Delta t} - d}{u-d}$ is the risk-neutral probability.

Replicating the Down-State Digital¶

The down-state digital has payoff $(0, 1)$:

\[ \Delta_d = \frac{0 - 1}{(u-d)S_0} = \frac{-1}{(u-d)S_0} \]

\[ B_d = e^{-r\Delta t}\left(\frac{u \cdot 1 - d \cdot 0}{u-d}\right) = e^{-r\Delta t}\left(\frac{u}{u-d}\right) \]

Price of down-state digital:

\[ \psi_d = \Delta_d S_0 + B_d = \frac{-1}{u-d} + e^{-r\Delta t}\left(\frac{u}{u-d}\right) = \frac{ue^{-r\Delta t} - 1}{u-d} \]

Simplifying:

\[ \boxed{\psi_d = e^{-r\Delta t} \cdot \frac{u - e^{r\Delta t}}{u - d} = e^{-r\Delta t}(1-q)} \]

State Prices Summary¶

Security	Payoff	Price
Up-state digital	$(1, 0)$	$\psi_u = e^{-r\Delta t} q$
Down-state digital	$(0, 1)$	$\psi_d = e^{-r\Delta t}(1-q)$

With our parameters ($q = 0.5043$):

\[ \psi_u = 0.9512 \times 0.5043 = 0.4797 \]

\[ \psi_d = 0.9512 \times 0.4957 = 0.4716 \]

Replicating Stock and Bond with Arrow-Debreu Securities¶

The Arrow-Debreu basis allows us to express the traded assets as linear combinations:

Stock Replication¶

The stock payoff is $(uS_0, dS_0)$:

\[ S_{\Delta t} = uS_0 \cdot \mathbf{1}_u + dS_0 \cdot \mathbf{1}_d \]

Stock price using state prices:

\[ S_0 = \psi_u \cdot uS_0 + \psi_d \cdot dS_0 = S_0(\psi_u u + \psi_d d) \]

This implies:

\[ \boxed{\psi_u u + \psi_d d = 1} \]

Bond Replication¶

The bond payoff is $(e^{r\Delta t}, e^{r\Delta t})$:

\[ B_{\Delta t} = e^{r\Delta t} \cdot \mathbf{1}_u + e^{r\Delta t} \cdot \mathbf{1}_d = e^{r\Delta t}(\mathbf{1}_u + \mathbf{1}_d) \]

Bond price using state prices:

\[ B_0 = 1 = \psi_u \cdot e^{r\Delta t} + \psi_d \cdot e^{r\Delta t} = e^{r\Delta t}(\psi_u + \psi_d) \]

This implies:

\[ \boxed{\psi_u + \psi_d = e^{-r\Delta t}} \]

Interpretation: The sum of state prices equals the discount factor. Receiving $1 in every state is equivalent to holding a zero-coupon bond.

Connection to Risk-Neutral Probability¶

State Prices and Probability¶

The state prices can be written as:

\[ \psi_u = e^{-r\Delta t} q, \qquad \psi_d = e^{-r\Delta t}(1-q) \]

where $q = \frac{e^{r\Delta t} - d}{u-d} \in (0,1)$.

The pricing formula becomes:

\[ V_0 = \psi_u H_u + \psi_d H_d = e^{-r\Delta t}(qH_u + (1-q)H_d) = e^{-r\Delta t}\mathbb{E}^{\mathbb{Q}}[H] \]

State Prices and Risk-Neutral Measure

\[ \psi_u = e^{-r\Delta t} q, \qquad \psi_d = e^{-r\Delta t}(1-q) \]

State prices are discounted risk-neutral probabilities. The risk-neutral measure $\mathbb{Q}$ encodes the state prices in probability form.

Why q Is Called "Risk-Neutral"¶

Under the measure $\mathbb{Q}$ with $\mathbb{Q}(up) = q$:

\[ \mathbb{E}^{\mathbb{Q}}[S_{\Delta t}] = q \cdot uS_0 + (1-q) \cdot dS_0 = S_0 e^{r\Delta t} \]

The stock's expected return equals the risk-free rate—as if investors were neutral to risk.

Examples: State Price Method¶

Using state prices $\psi_u = 0.4797$ and $\psi_d = 0.4716$:

European Call (Strike K = 105)¶

\[ C_0 = \psi_u \cdot 15 + \psi_d \cdot 0 = 0.4797 \times 15 = 7.20 \text{ ✓} \]

European Put (Strike K = 105)¶

\[ P_0 = \psi_u \cdot 0 + \psi_d \cdot 15 = 0.4716 \times 15 = 7.07 \text{ ✓} \]

Forward Contract¶

\[\begin{array}{lll} V_0 &=&\displaystyle \psi_u(120 - F) + \psi_d(90 - F)\\ &=&\displaystyle 0.4797 \times 120 + 0.4716 \times 90 - F(\psi_u + \psi_d)\\ &=&\displaystyle 57.56 + 42.44 - F \times 0.9512\\ &=&\displaystyle 100 - 0.9512F \end{array}\]

Setting $V_0 = 0$: $F = 100/0.9512 = 105.13$ ✓

Comparison of Methods¶

Aspect	Stock-Bond Replication	State Price Method
Basis	$(S, B)$	$(\mathbf{1}_u, \mathbf{1}_d)$
Payoffs	$(uS_0, dS_0)$ and $(e^{r\Delta t}, e^{r\Delta t})$	$(1, 0)$ and $(0, 1)$
To price $H$	Solve 2×2 system for $(\Delta, B)$	Compute $\psi_u H_u + \psi_d H_d$
Computation	Requires solving equations each time	Direct multiplication once $\psi$ known
Insight	How to construct the hedge	Structure of arbitrage-free pricing

Both methods give identical prices because they span the same payoff space.

The Fundamental Theorems (Preview)¶

The replication results connect to the Fundamental Theorems of Asset Pricing:

First Fundamental Theorem

The market is arbitrage-free if and only if there exist positive state prices $(\psi_u, \psi_d) > 0$.

Equivalently: there exists a risk-neutral measure $\mathbb{Q}$.

Second Fundamental Theorem

The market is complete if and only if the state prices (equivalently, $\mathbb{Q}$) are unique.

In the one-period binomial model: - No-arbitrage ($d < e^{r\Delta t} < u$) guarantees positive state prices - Two assets for two states guarantees uniqueness

See FTAP for the general theory.

Summary¶

Concept	Formula
Stock-bond replication	$\Delta = \dfrac{H_u - H_d}{(u-d)S_0}$, $B = e^{-r\Delta t}\dfrac{uH_d - dH_u}{u-d}$
Replication price	$V_0 = \Delta S_0 + B$
Up-state price	$\psi_u = e^{-r\Delta t} q$
Down-state price	$\psi_d = e^{-r\Delta t}(1-q)$
State price formula	$V_0 = \psi_u H_u + \psi_d H_d$
Sum of state prices	$\psi_u + \psi_d = e^{-r\Delta t}$

Key Takeaways

Two equivalent bases: Stock-bond and Arrow-Debreu both span the payoff space.
Replication determines price: The cost of the replicating portfolio is the unique no-arbitrage price.
State prices are fundamental: $\psi_u$ and $\psi_d$ encode all pricing information.
State prices = discounted $\mathbb{Q}$-probabilities: $\psi = e^{-r\Delta t} \times$ risk-neutral probability.
Completeness = unique prices: Two independent assets for two states means every payoff has a unique price.

What's Next¶

Section	Topic
Delta Hedging	Pricing via risk elimination
Risk-Neutral Measure	The measure $\mathbb{Q}$ and expectation pricing
Multi-Period Model	Extending to multiple time steps

Exercises¶

Exercise 1. In the one-period binomial model with $S_0 = 80$, $u = 1.25$, $d = 0.85$, $r = 4\%$, and $\Delta t = 1$, find the replicating portfolio $(\Delta, B)$ and the price of a European put with strike $K = 85$. Verify the replication by checking that the portfolio payoff matches the put payoff in both states.

Solution to Exercise 1

Given $S_0 = 80$, $u = 1.25$, $d = 0.85$, $r = 4\%$, $\Delta t = 1$, and $K = 85$.

Stock prices at $\Delta t$:

Up: $S_u = 1.25 \times 80 = 100$
Down: $S_d = 0.85 \times 80 = 68$

Put payoffs:

\[ H_u = (85 - 100)^+ = 0, \qquad H_d = (85 - 68)^+ = 17 \]

Replicating portfolio:

\[ \Delta = \frac{H_u - H_d}{(u - d)S_0} = \frac{0 - 17}{(1.25 - 0.85) \times 80} = \frac{-17}{32} = -0.53125 \]

\[ B = e^{-r\Delta t}\left(\frac{uH_d - dH_u}{u - d}\right) = e^{-0.04}\left(\frac{1.25 \times 17 - 0.85 \times 0}{0.40}\right) = 0.96079 \times 53.125 = 51.04 \]

Price:

\[ P_0 = \Delta S_0 + B = -0.53125 \times 80 + 51.04 = -42.50 + 51.04 = 8.54 \]

Verification:

Up state: $\Delta \cdot uS_0 + B \cdot e^{r\Delta t} = -0.53125 \times 100 + 51.04 \times 1.04081 = -53.125 + 53.125 = 0 = H_u$ $\checkmark$
Down state: $\Delta \cdot dS_0 + B \cdot e^{r\Delta t} = -0.53125 \times 68 + 51.04 \times 1.04081 = -36.125 + 53.125 = 17 = H_d$ $\checkmark$

Exercise 2. Prove that in the one-period binomial model, the stock and bond payoff vectors are linearly independent if and only if $u \neq d$. What would happen to the replication problem if $u = d$?

Solution to Exercise 2

The stock payoff vector is $(uS_0, dS_0)$ and the bond payoff vector is $(e^{r\Delta t}, e^{r\Delta t})$.

These are linearly independent if and only if one is not a scalar multiple of the other. Suppose $(uS_0, dS_0) = c(e^{r\Delta t}, e^{r\Delta t})$ for some scalar $c$. Then $uS_0 = ce^{r\Delta t}$ and $dS_0 = ce^{r\Delta t}$, which implies $uS_0 = dS_0$, hence $u = d$.

Conversely, if $u \neq d$, then $uS_0 \neq dS_0$ while $e^{r\Delta t} = e^{r\Delta t}$, so the stock payoff vector cannot be a multiple of the bond payoff vector. The two vectors are linearly independent.

If $u = d$: The stock payoff is the same in both states, so $S_{\Delta t} = uS_0$ deterministically. The stock becomes equivalent to a scaled bond, and the two assets span only a one-dimensional subspace of $\mathbb{R}^2$. Replication of a general payoff $(H_u, H_d)$ with $H_u \neq H_d$ is impossible because no portfolio can distinguish between the two states.

Exercise 3. Using the parameters from the text ($S_0 = 100$, $u = 1.2$, $d = 0.9$, $r = 5\%$, $\Delta t = 1$), compute the state prices $\psi_u$ and $\psi_d$. Verify that (a) $\psi_u u + \psi_d d = 1$, (b) $\psi_u + \psi_d = e^{-r\Delta t}$, and (c) a European call with strike $K = 105$ priced via state prices gives the same result as the stock-bond replication method.

Solution to Exercise 3

With $S_0 = 100$, $u = 1.2$, $d = 0.9$, $r = 5\%$, $\Delta t = 1$, we have $q = \frac{e^{0.05} - 0.9}{1.2 - 0.9} = \frac{0.15127}{0.3} = 0.5043$.

State prices:

\[ \psi_u = e^{-0.05} \times 0.5043 = 0.9512 \times 0.5043 = 0.4797 \]

\[ \psi_d = e^{-0.05} \times 0.4957 = 0.9512 \times 0.4957 = 0.4716 \]

(a) $\psi_u u + \psi_d d = 0.4797 \times 1.2 + 0.4716 \times 0.9 = 0.5756 + 0.4244 = 1.0000$ $\checkmark$

(b) $\psi_u + \psi_d = 0.4797 + 0.4716 = 0.9513 \approx e^{-0.05} = 0.9512$ $\checkmark$ (rounding)

(c) European call with $K = 105$: $H_u = 15$, $H_d = 0$.

State price method:

\[ C_0 = \psi_u \times 15 + \psi_d \times 0 = 0.4797 \times 15 = 7.20 \]

Replication method: $\Delta = 0.5$, $B = -42.80$, $V_0 = 50 - 42.80 = 7.20$ $\checkmark$

Exercise 4. A bear spread consists of a long put with strike $K_2$ and a short put with strike $K_1$, where $K_1 < K_2$. Using the state price method with $S_0 = 100$, $u = 1.2$, $d = 0.9$, $r = 5\%$, $\Delta t = 1$, $K_1 = 95$, and $K_2 = 110$, compute the bear spread price. Then verify your answer by pricing each put separately and taking the difference.

Solution to Exercise 4

With the standard parameters, $\psi_u = 0.4797$ and $\psi_d = 0.4716$.

Put with $K_1 = 95$:

$H_u = (95 - 120)^+ = 0$, $H_d = (95 - 90)^+ = 5$
$P_1 = 0.4797 \times 0 + 0.4716 \times 5 = 2.36$

Put with $K_2 = 110$:

$H_u = (110 - 120)^+ = 0$, $H_d = (110 - 90)^+ = 20$
$P_2 = 0.4797 \times 0 + 0.4716 \times 20 = 9.43$

Bear spread (long $K_2$ put, short $K_1$ put):

\[ V_0^{\text{bear}} = P_2 - P_1 = 9.43 - 2.36 = 7.07 \]

Direct calculation: Bear spread payoff is $H_u = 0 - 0 = 0$, $H_d = 20 - 5 = 15$.

\[ V_0 = \psi_u \times 0 + \psi_d \times 15 = 0.4716 \times 15 = 7.07 \quad \checkmark \]

Exercise 5. Suppose a derivative pays $H_u = 10$ in the up state and $H_d = 25$ in the down state. Using the standard parameters ($S_0 = 100$, $u = 1.2$, $d = 0.9$, $r = 5\%$, $\Delta t = 1$), compute the replicating portfolio. Interpret the sign of $\Delta$ and explain why the replicating portfolio is long or short the stock.

Solution to Exercise 5

Given $H_u = 10$, $H_d = 25$, with $S_0 = 100$, $u = 1.2$, $d = 0.9$, $r = 5\%$, $\Delta t = 1$.

Replicating portfolio:

\[ \Delta = \frac{H_u - H_d}{(u - d)S_0} = \frac{10 - 25}{0.3 \times 100} = \frac{-15}{30} = -0.5 \]

\[ B = e^{-0.05}\left(\frac{1.2 \times 25 - 0.9 \times 10}{0.3}\right) = 0.9512 \times \frac{30 - 9}{0.3} = 0.9512 \times 70 = 66.58 \]

Price:

\[ V_0 = -0.5 \times 100 + 66.58 = -50 + 66.58 = 16.58 \]

Interpretation of $\Delta < 0$: The derivative pays more in the down state ($H_d = 25$) than in the up state ($H_u = 10$). It behaves like a put-like instrument whose value increases when the stock falls. To replicate it, we must short 0.5 shares of stock (so the stock leg gains when the stock falls) and lend $66.58 (the positive cash position provides the base payoff level).

Exercise 6. Consider a one-period binomial model where a third asset (a call option with known market price $C_0^{\text{mkt}}$) is also traded. Show that if $C_0^{\text{mkt}}$ differs from the no-arbitrage price $C_0 = \psi_u H_u + \psi_d H_d$, then an arbitrage portfolio can be constructed using the stock, bond, and call. Describe the arbitrage strategy explicitly for $C_0^{\text{mkt}} > C_0$.

Solution to Exercise 6

The no-arbitrage price of the call is $C_0 = \psi_u H_u + \psi_d H_d$. If $C_0^{\text{mkt}} \neq C_0$, we can construct an arbitrage.

Case $C_0^{\text{mkt}} > C_0$ (call is overpriced):

The strategy is to sell the overpriced call and buy the replicating portfolio:

Sell 1 call at price $C_0^{\text{mkt}}$
Buy the replicating portfolio $(\Delta_C, B_C)$ at cost $C_0 = \Delta_C S_0 + B_C$

Initial cash flow: $C_0^{\text{mkt}} - C_0 > 0$ (profit at inception).

Terminal payoff in each state:

Up state: Replicating portfolio pays $H_u$, call obligation is $H_u$. Net = $0$.
Down state: Replicating portfolio pays $H_d$, call obligation is $H_d$. Net = $0$.

The replicating portfolio exactly offsets the call obligation in every state, so the terminal net payoff is zero. The initial profit $C_0^{\text{mkt}} - C_0 > 0$ is a riskless arbitrage.

Investing this profit in the bank account yields $(C_0^{\text{mkt}} - C_0)e^{r\Delta t} > 0$ at maturity with zero risk.

Exercise 7. Derive the state price formulas $\psi_u = e^{-r\Delta t} q$ and $\psi_d = e^{-r\Delta t}(1-q)$ directly from the two consistency conditions $\psi_u u + \psi_d d = 1$ and $\psi_u + \psi_d = e^{-r\Delta t}$, without using the replication argument. That is, solve the $2 \times 2$ linear system for $\psi_u$ and $\psi_d$.

Solution to Exercise 7

We solve the system:

\[ \psi_u u + \psi_d d = 1 \qquad (1) \]

\[ \psi_u + \psi_d = e^{-r\Delta t} \qquad (2) \]

From (2): $\psi_d = e^{-r\Delta t} - \psi_u$. Substituting into (1):

\[ \psi_u u + (e^{-r\Delta t} - \psi_u)d = 1 \]

\[ \psi_u(u - d) + de^{-r\Delta t} = 1 \]

\[ \psi_u = \frac{1 - de^{-r\Delta t}}{u - d} = \frac{e^{-r\Delta t}(e^{r\Delta t} - d)}{u - d} = e^{-r\Delta t} \cdot \frac{e^{r\Delta t} - d}{u - d} = e^{-r\Delta t} q \]

Then:

\[ \psi_d = e^{-r\Delta t} - \psi_u = e^{-r\Delta t} - e^{-r\Delta t} q = e^{-r\Delta t}(1 - q) \]

This confirms $\psi_u = e^{-r\Delta t} q$ and $\psi_d = e^{-r\Delta t}(1-q)$ directly from the consistency conditions.

Parameter	Value
\(S_0\)	\(100\)
\(u\)	\(1.2\)
\(d\)	\(0.9\)
\(r\)	\(5\%\)
\(\Delta t\)	\(1\) year
\(e^{r\Delta t}\)	\(1.0513\)

Security	Payoff	Price
Up-state digital	\((1, 0)\)	\(\psi_u = e^{-r\Delta t} q\)
Down-state digital	\((0, 1)\)	\(\psi_d = e^{-r\Delta t}(1-q)\)

Aspect	Stock-Bond Replication	State Price Method
Basis	\((S, B)\)	\((\mathbf{1}_u, \mathbf{1}_d)\)
Payoffs	\((uS_0, dS_0)\) and \((e^{r\Delta t}, e^{r\Delta t})\)	\((1, 0)\) and \((0, 1)\)
To price \(H\)	Solve 2×2 system for \((\Delta, B)\)	Compute \(\psi_u H_u + \psi_d H_d\)
Computation	Requires solving equations each time	Direct multiplication once \(\psi\) known
Insight	How to construct the hedge	Structure of arbitrage-free pricing

Concept	Formula
Stock-bond replication	\(\Delta = \dfrac{H_u - H_d}{(u-d)S_0}\), \(B = e^{-r\Delta t}\dfrac{uH_d - dH_u}{u-d}\)
Replication price	\(V_0 = \Delta S_0 + B\)
Up-state price	\(\psi_u = e^{-r\Delta t} q\)
Down-state price	\(\psi_d = e^{-r\Delta t}(1-q)\)
State price formula	\(V_0 = \psi_u H_u + \psi_d H_d\)
Sum of state prices	\(\psi_u + \psi_d = e^{-r\Delta t}\)