Phase 14 - Lesson 14.1

No-Arbitrage and One-Period Pricing

The single principle underneath all of derivative pricing, made concrete in a one-period market.

⏱ 50 min● Intermediate🔗 Prereqs: Probability (Phase 7), linear algebra (Phase 3)

↖ Phase 14 hub

Leads to: Replication (14.2) and the fundamental theorems (14.3) generalize this one-period picture.

Learning Objectives

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Define an arbitrage opportunity precisely and explain why prices must exclude it.
Model a one-period market with a bond and a stock and identify its payoff vectors.
Derive the no-arbitrage bounds $d\lt R\lt u$ and interpret their failure as free money.
Compute the price of a claim by the law of one price in a complete one-period market.
Solve a one-period replication problem for the hedge $(\Delta,B)$ and the implied price.

Key Vocabulary

Arbitrage: A self-financing strategy with zero cost today and, tomorrow, a nonnegative payoff that is strictly positive with positive probability - risk-free profit.

One-period market: Two dates $t=0,1$ and finitely many states; assets are a risk-free bond and a risky stock with known state payoffs.

Gross return $R$: One plus the risk-free rate over the period; $\$1$ in the bond becomes $\$R$ tomorrow.

Law of one price: Two portfolios with identical payoffs in every state must have identical prices today, or an arbitrage exists.

Replicating portfolio: Holdings $(\Delta,B)$ in stock and bond whose date-1 payoff equals the claim’s in every state.

State prices: Nonnegative numbers $\psi_s$ pricing an $\$1$-in-state-$s$ security; a claim’s price is $\sum_s\psi_s\,\text{payoff}_s$.

Intuition & Motivation

Intuition

One idea generates almost all of pricing theory: there is no free lunch. If two packages of cash flows are identical in every future state, they must cost the same today - otherwise buy the cheap one, sell the dear one, and pocket the difference at zero risk. A market that permits such a trade is instantly exploited and cannot persist. Turning this into arithmetic in the simplest possible market - one period, one stock, one bond, two states - already gives the whole logic of option pricing: find a stock-plus-bond portfolio that replicates the option, and the option’s price is forced to equal that portfolio’s cost. Probabilities of the states never enter.

The one-period market

Two dates, $t=0$ (now) and $t=1$ (period end), and two states of the world, up ($\omega_u$) and down ($\omega_d$). Two traded assets:

A bond: costs $\$1$ now, pays $\$R$ in both states, where $R=1+r\gt 0$ is the gross risk-free return.
A stock: costs $S_0$ now, pays $S_0u$ up and $S_0d$ down, with $0\lt d\lt u$.

A portfolio holds $\Delta$ shares of stock and $B$ dollars in the bond; its cost is $\Delta S_0+B$ and its date-1 payoff is $\Delta S_0u+BR$ (up) or $\Delta S_0d+BR$ (down).

What arbitrage is

Definition - Arbitrage opportunity

A portfolio with cost $\Delta S_0+B\le0$ today whose date-1 payoff is $\ge0$ in every state and $\gt 0$ in at least one state. Equivalently: something for nothing, with no possibility of loss.

The pricing axiom is that observed prices admit no arbitrage. This is not an empirical law of markets so much as a consistency condition: any violation is competed away almost instantly by traders, so any sensible pricing model must be arbitrage-free.

The no-arbitrage bound $d\lt R\lt u$

Theorem - One-period no-arbitrage condition

The one-period stock–bond market is arbitrage-free if and only if $d\lt R\lt u$; equivalently, iff there is a probability $q\in(0,1)$ with $R=qu+(1-q)d$.

Why: if $R\le d$ the stock dominates the bond (it returns at least $R$ in every state and more in the up state), so borrow at $R$, buy stock, and profit risk-free. If $R\ge u$, short the stock and lend at $R$. Only $d\lt R\lt u$ prevents both. Solving $R=qu+(1-q)d$ gives the risk-neutral probability

\[q=\frac{R-d}{u-d}\in(0,1),\qquad 1-q=\frac{u-R}{u-d}.\] (14.1)

Key Idea

The number $q$ is not the real-world probability of an up move. It is a pricing weight - the unique probability under which the stock’s expected gross return equals the risk-free $R$. Real-world beliefs $p$ never appear in the price.

Replication prices the claim

A claim (e.g. a call) pays $C_u$ up and $C_d$ down. Find $(\Delta,B)$ matching both states:

\[\Delta S_0u+BR=C_u,\qquad \Delta S_0 d+BR=C_d.\] (14.2)

Subtracting gives the delta (the hedge ratio), then back-substitute for $B$:

\[\Delta=\frac{C_u-C_d}{S_0(u-d)},\qquad B=\frac{1}{R}\,\frac{u\,C_d-d\,C_u}{u-d}.\] (14.3)

By the law of one price the claim must cost exactly the replicating portfolio’s price $V_0=\Delta S_0+B$. Remarkably this equals a discounted risk-neutral expectation, $V_0=\tfrac1R\big(qC_u+(1-q)C_d\big)$ - the seed of everything in 14.2.

Worked Example - Price a one-period call by replication

Data: $S_0=100,\ u=1.2,\ d=0.9,\ R=1.05$; call strike $K=105$. So $S_0u=120,\ S_0d=90$.

Payoffs: $C_u=(120-105)^+=15,\ C_d=(90-105)^+=0$.

Delta: $\Delta=(15-0)/(100\cdot0.3)=15/30=0.5$ shares.

Bond: match up state, $0.5\cdot120+1.05B=15\Rightarrow B=(15-60)/1.05=-42.857$ (borrow \$42.857).

Price: $V_0=\Delta S_0+B=0.5\cdot100-42.857=7.143$. Check via $q=(1.05-0.9)/0.3=0.5$: $V_0=\tfrac1{1.05}(0.5\cdot15)=7.143$. Identical.

Interactive: solve the replication linear system

Common Mistakes to Avoid

Discounting the claim with the real-world probability $p$ of an up move; the price uses $q$, which depends only on $u,d,R$, not on beliefs.
Forgetting that $d\lt R\lt u$ is required; if it fails, a static stock–bond arbitrage exists and no consistent price can be assigned.
Sign errors on the bond position $B$; a negative $B$ means borrowing - part of the hedge, not an inconsistency.
Believing the option’s price reflects its ‘expected payoff’ under real probabilities; it reflects the cost of the replicating hedge.

Quant Practitioner Tips

First compute $\Delta$ (the hedge), then $B$, then the price - this order works in every one-period problem.
Cross-check any one-period price two ways: replication cost and $\tfrac1R(qC_u+(1-q)C_d)$. They must match exactly.
State prices $\psi_u=q/R,\ \psi_d=(1-q)/R$ price any claim by $\psi_uC_u+\psi_dC_d$; a compact bookkeeping device.

Knowledge Check

Q1 Medium

An arbitrage opportunity in the one-period market is a portfolio that:

Has positive expected return

Costs $\le0$ today and pays $\ge0$ in every state, $\gt 0$ in some state

Always beats the stock

Has zero variance

Q2 Easy

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

$R\lt d$

$d\lt R\lt u$

$R\gt u$

$u=d$

Q3 Medium

In the replication price, the real-world probability $p$ of an up move:

Equals $q$

Determines the price

Does not enter the price at all

Must exceed $q$

Practical Exercise

A one-period market has $S_0=50,\ u=1.3,\ d=0.8,\ R=1.10$. A put has strike $K=55$. (a) Find the put payoffs $(P_u,P_d)$. (b) Compute $q$, the hedge $(\Delta,B)$, and the put price. (c) Verify no-arbitrage holds.

▶ Show full solution

(a) $S_0u=65,\ S_0d=40$, so $P_u=(55-65)^+=0,\ P_d=(55-40)^+=15$.

(b) $q=(R-d)/(u-d)=(1.1-0.8)/0.5=0.6$. Delta: $\Delta=(P_u-P_d)/(S_0(u-d))=(0-15)/(50\cdot0.5)=-0.6$ (short 0.6 shares - a put hedge). Bond: from $\Delta S_0d+BR=P_d$, $-0.6\cdot40+1.1B=15\Rightarrow B=(15+24)/1.1=35.4545$. Price: $V_0=\Delta S_0+B=-30+35.4545=5.4545$. Check: $\tfrac1{1.1}(0.6\cdot0+0.4\cdot15)=6/1.1=5.4545$. Agreed.

After the reveal, answer for yourself: Why does the put hedge hold a negative number of shares, and what does that say about a put’s delta?

Lesson Summary

No-arbitrage - no riskless profit from a zero-cost portfolio - is the axiom from which pricing follows. In the one-period stock–bond market it holds iff $d\lt R\lt u$, equivalently iff a risk-neutral probability $q=(R-d)/(u-d)$ exists. Any claim is priced by building a replicating portfolio $(\Delta,B)$; its cost equals the discounted risk-neutral expectation $\tfrac1R(qC_u+(1-q)C_d)$, and real-world probabilities never enter.

Retrieval Practice

Close the lesson and answer from memory before checking. This is deliberate, effortful recall - the single highest-yield study action.

▶ Show retrieval prompts & answers

Q: State the one-period no-arbitrage condition and the resulting risk-neutral probability.
A: Arbitrage-free iff $d\lt R\lt u$; then $q=(R-d)/(u-d)\in(0,1)$ is the unique probability making the stock’s expected gross return equal $R$.

Q: How is a one-period claim priced, and does the real probability $p$ matter?
A: Replicate it with $(\Delta,B)$ and charge the portfolio cost, equal to $\tfrac1R(qC_u+(1-q)C_d)$. The physical $p$ does not enter the price.

Completion Checklist

I can explain the core ideas in my own words
I worked the derivations/examples by hand
I completed the interactive workbench(es)
I passed the knowledge check

Confidence / mastery rating

Personal notes

Source References

This lesson synthesizes and paraphrases concepts from the sources below. No copyrighted text, problem sets, or solutions are reproduced. Return to the originals for full depth.

Stochastic Calculus for Finance II (Steven Shreve, 2004) foundational - Ch. 1 - Vol. I Ch. 1 (adapted): single-period binomial no-arbitrage, replication, and risk-neutral probabilities.
Methods of Mathematical Finance (Karatzas & Shreve, 1998) foundational - Ch. 1 - Ch. 1: arbitrage, the law of one price, and state-price/contingent-claim valuation in finite markets.

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No-Arbitrage and One-Period Pricing

Learning Objectives

Key Vocabulary

Intuition & Motivation

The one-period market

What arbitrage is

The no-arbitrage bound \(d\lt R\lt u\)

Replication prices the claim

Interactive: solve the replication linear system

Knowledge Check

Practical Exercise

Lesson Summary

Retrieval Practice

Completion Checklist

Source References