How replication prices a one-period option

If a stock can only go up or down, what is an option on it worth today?

Build a portfolio of shares and cash that pays exactly what the option pays in both cases. The option must cost what that portfolio costs. Any other price lets a trader lock in a profit.

Before you start · 3 quick checks on payoffs, linear systems and discounting, about 2 min

Setup: the one-period model

Today (time 0) a stock trades at $S_0 \gt 0$. At time 1 it is worth either $uS_0$ (the up state) or $dS_0$ (the down state), with $0 \lt d \lt u$.
A riskless bond grows by the factor $1+r$ over the period, with $r \gt -1$: $\$1$ today is $\$(1+r)$ at time 1 in both states.
An option is a European claim: it pays $C_u$ in the up state and $C_d$ in the down state at time 1, and nothing before. A call with strike $K$ pays $(S - K)^+ = \max(S-K, 0)$ and a put pays $(K - S)^+$.
A portfolio holds $\Delta$ shares and $B$ dollars in the bond today. $\Delta \lt 0$ is a short sale and $B \lt 0$ is borrowing.

Fine print: assumptions and the rate convention

The market is frictionless: no transaction costs, no dividends, no taxes. Any real number of shares can be bought or sold short, and anyone can borrow or lend any amount at the rate $r$.
The option itself can be bought or sold (written) in any amount at a single market price $P$, with no bid-ask spread. It cannot be exercised early and has no cash flows before time 1.
The real probability of the up state is some $p$ with $0 \lt p \lt 1$, so both states can happen.
This page uses simple growth $1+r$ per period. If a rate is quoted with continuous compounding, $\rho$ per year for a period of length $\Delta t$, replace every $1+r$ by $e^{\rho\Delta t}$. Nothing else changes.
An arbitrage is a portfolio that costs nothing today, never loses at time 1, and gains in at least one state. Receiving cash today and owing nothing later is the same thing: put the cash in the bond.

1 See

Copy the option with shares and a bond

A hypothetical stock trades at $100. At the end of one period it will be $120 or $80. Money in the bond grows by 5% per period. Pick an option, then choose the number of shares and the bond amount so that your portfolio pays what the option pays in both states.

Key point Two states give two equations, and two holdings, $\Delta$ shares and $B$ dollars in the bond, solve them. What that portfolio costs today is the option's price.

2 Try

What the price does and does not depend on

Trade against a wrong price

The $100-strike call from stage 1 now trades in the market at a price you set.

This part uses the exact copy of the $100-strike call, so it opens once you have copied that call in stage 1.

When the stock and bond alone give an arbitrage

Key point The price depends on $u$, $d$, $r$ and the payoffs. The real probability $p$ changes the expected payoff but not the cost of copying it. If $d \lt 1+r \lt u$ fails, the stock and bond alone give an arbitrage.

3 Prove

The replicating portfolio and the risk-neutral formula

The idea in three steps

Write one equation per state: the portfolio's value at time 1 equals the option's payoff.
Subtracting the equations removes the bond and gives $\Delta$. Either equation then gives $B$.
The cost $\Delta S_0 + B$ rearranges into a weighted average of the payoffs, discounted at $r$. A trader can lock in a profit at any other price.

Lemma (no arbitrage from the stock and bond)

Under the setup above, a portfolio of the stock and the bond alone can be an arbitrage if and only if $d \lt 1+r \lt u$ fails.

Theorem (one-period replication)

Assumptions. The setup and fine print above: $S_0 \gt 0$, $0 \lt d \lt u$, $r \gt -1$, real probability $0 \lt p \lt 1$, a frictionless market for the stock and bond, and a European option paying any real amounts $C_u, C_d$ at time 1 that trades at one price $P$ in both directions. In addition, $d \lt 1+r \lt u$.

Conclusion. Every payoff $(C_u, C_d)$ is replicated by exactly one portfolio,

\[ \Delta = \frac{C_u - C_d}{(u-d)S_0}, \qquad B = \frac{uC_d - dC_u}{(u-d)(1+r)}. \]

With $q = \dfrac{(1+r)-d}{u-d}$, which lies strictly between 0 and 1, its cost is

\[ C_0 = \Delta S_0 + B = \frac{qC_u + (1-q)C_d}{1+r}, \]

and $C_0$ is the only price of the option that does not allow an arbitrage. The formulas hold for every payoff pair, including constant payoffs $C_u = C_d$, where $\Delta = 0$; lines 3 to 6 of the proof state them in that general form.

Proof of the theorem (8 lines, build it or read it)

Scratch work, not part of the proof. For line 5, put the two parts of the cost over $(u-d)(1+r)$: \[ \frac{C_u - C_d}{u-d} + \frac{uC_d - dC_u}{(u-d)(1+r)} = \frac{(1+r)C_u - (1+r)C_d + uC_d - dC_u}{(u-d)(1+r)} = \frac{\big((1+r)-d\big)C_u + \big(u-(1+r)\big)C_d}{(u-d)(1+r)}. \] For line 7, $qu + (1-q)d = d + q(u-d) = d + (1+r) - d = 1+r$.

What $q$ is and is not

The weights $q$ and $1-q$ are positive and add to 1, so they can be read as probabilities. Under them the stock's average price at time 1 is $quS_0 + (1-q)dS_0 = (1+r)S_0$: the stock grows on average at the riskless rate. That is why $q$ is called the risk-neutral probability.

$q$ is built from $u$, $d$ and $r$ only. It is not a forecast of the up move, and it need not equal the real probability $p$. The formula $C_0 = \big(qC_u + (1-q)C_d\big)/(1+r)$ is a rewriting of the replication cost, so it gives the price whatever the value of $p$.

Key point Matching both states gives $\Delta = (C_u - C_d)/((u-d)S_0)$ and $B = (uC_d - dC_u)/((u-d)(1+r))$. Their cost is $\big(qC_u + (1-q)C_d\big)/(1+r)$, and any other price can be traded for a riskless profit.

4 Apply

Problems

All prices are hypothetical. Rates are simple rates per period.