Case study: replicating a payoff, and complete markets

A market with three states

Take a market that lasts one year and can end in one of three states: down, middle or up. The numbers are invented to keep the arithmetic small. Real markets have far more than three outcomes; the desk notes at the end come back to that.

• The bond pays $1 in every state and costs 0.97 today.
• The stock ends at 80, 100 or 125 and costs 98.75 today.
• The call with strike $K = 100$ pays $\max(S - 100, 0)$, where $S$ is the stock's final price (case study 1.1.5). That is 0, 0 and 25. It costs 6.75.
• The put with the same strike pays $\max(100 - S, 0)$: 20, 0 and 0. Its price is the question.

Both options expire at the end of the year, when the bond pays.

Payoffs as a matrix

Put each asset's payoffs in a column, one row per state. This is the payoff matrix $D$, with the bond, the stock and the call as its columns:

A portfolio holds $h_1$ bonds, $h_2$ shares and $h_3$ calls: the holdings vector $\mathbf{h}$. A negative holding is a short position: you sell the asset today, receive its price, and owe its payoff at the end.

In each state the portfolio pays the sum of its holdings times their payoffs. That is the column picture of lesson 3.3.1: $D\mathbf{h}$ is a combination of the asset columns. Its price today is $\mathbf{p} \cdot \mathbf{h}$, the dot product (lesson 3.1.2) with the prices $\mathbf{p} = (0.97, 98.75, 6.75)$.

Copying the put

To replicate the put, find holdings whose payoff equals the put's in every state: $D\mathbf{h} = (20, 0, 0)$. That is a linear system of three equations in three unknowns, and elimination (lesson 3.3.1) solves it.

Back substitution gives $h_3 = 25 \div 25 = 1$, then $h_2 = -20 \div 20 = -1$, then $h_1 = 20 - 80h_2 = 100$. So: buy 100 bonds, sell one share short, buy one call.

Check it in each state: down $100 - 80 + 0 = 20$, middle $100 - 100 + 0 = 0$ and up $100 - 125 + 25 = 0$. The portfolio pays exactly what the put pays.

One payoff, one price

The replicating portfolio costs $100 \times 0.97 - 98.75 + 6.75 = 5.00$ today. The law of one price says two positions with the same payoff in every state must cost the same. So the put is worth $5.00.

Suppose it traded at $5.40. Sell the put for 5.40 and buy the portfolio for 5.00, keeping 0.40 today. At the end, the portfolio pays exactly what you owe on the put, in every state.

That is an arbitrage: a profit today with nothing to pay later. Traders would keep doing it until the price came back to 5.00.

At $4.60 the trade runs the other way. Buy the put and take the opposite holdings: sell 100 bonds (borrowing 97), buy a share and sell a call.

Put–call parity

The holdings were $K = 100$ bonds, $-1$ share and $+1$ call. Write $B = 0.97$ for the bond price, $S_0 = 98.75$ for the stock price today, and $C$ and $P$ for the call and put prices. Then the price of the put is $P = KB - S_0 + C$.

It does not depend on the three states. For every final price $S$, $\max(S - K, 0) - S + K = \max(K - S, 0)$: above the strike the call cancels the falling line, below it the call pays nothing. So parity holds in any market where the bond, the stock and the two options trade, whatever the stock can do.

State prices

A state price $\psi_s$ (psi) is today's price of $1 paid in state $s$ only, and nothing in the others. If you know the three, any payoff $\mathbf{x}$ costs $\boldsymbol{\psi} \cdot \mathbf{x}$: so much in each state, at that state's price.

Each traded asset must obey this, so its price gives one equation. The bond gives $\psi_1 + \psi_2 + \psi_3 = 0.97$; the stock gives $80\psi_1 + 100\psi_2 + 125\psi_3 = 98.75$; the call gives $25\psi_3 = 6.75$. These are the columns of $D$ dotted with $\boldsymbol{\psi}$, so the system uses the transpose of lesson 3.2.3.

The state prices add up to 0.97, the price of $1 in every state: the bond. Divided by 0.97 they become 0.2577, 0.4639 and 0.2784, positive numbers that add up to 1.

The market prices assets as if these were the chances of the states. They are called risk-neutral probabilities, and the Derivatives Modelling course builds option pricing on them.

All three state prices are positive here. A set of prices allows no arbitrage exactly when it has positive state prices; the Derivatives Modelling course proves it. A negative state price is a warning, as the desk notes below show.

Complete markets

A market is complete when every payoff can be replicated: $D\mathbf{h} = \mathbf{x}$ has a solution for every $\mathbf{x}$. By lesson 3.3.2, that needs a pivot in every row of $D$: the rank of $D$ must equal the number of states.

Here $D$ has rank 3 and there are 3 states, so the market is complete. A straddle, a call plus a put (case study 1.1.5), pays 20, 0 and 25. Solving gives 100 bonds, $-1$ share and 2 calls, at a price of 11.75: the put's 5.00 plus the call's 6.75.

Without the call

Now suppose only the bond and the stock trade. Their payoff matrix $D_2$ has 3 rows and 2 columns, so its rank is at most 2: a tall system (lesson 3.3.2). To copy the call, eliminate on $[D_2 \mid \text{call}]$:

The steps $R_2 - R_1$, $R_3 - R_1$ and $R_3 - \frac{9}{4}R_2$, as before, give

The last row reads $0 = 25$, so there is no solution. The rank of $D_2$ is 2 and the rank of $[D_2 \mid \text{call}]$ is 3.

The second panel of the figure above shows why. A portfolio of $a$ bonds and $b$ shares pays $a + bS$, a straight line in $S$. The call's three payoffs, 0, 0 and 25, do not lie on one line: the line through two of them misses the third.

What the bond and the stock do fix

Bond and stock still give two pricing equations for three state prices. By lesson 3.3.2 that leaves one free variable: take $\psi_3 = t$. Elimination gives the line of solutions

All three are positive only for $t$ between 0.07 and 0.47. The call costs $25t$, so any call price between 1.75 and 11.75 is free of arbitrage, and the market's 6.75 is one of them. The bond and the stock bound the call's price; they do not fix it.

The lower bound has a meaning. 1.75 is the price of the payoff $S - 100$, which $-100$ bonds and 1 share replicate, and the call never pays less than that. Case study 3.4.3 adds a fund to the bond and the stock, and asks whether it fills the gap.

import numpy as np

# columns: bond, stock, call with strike 100; rows: the states down, middle, up
D = [[1, 80, 0], [1, 100, 0], [1, 125, 25]]
PRICES = [0.97, 98.75, 6.75]
PUT = [20, 0, 0]


def test_put_holdings():
    """The put is copied by 100 bonds, −1 share and 1 call"""
    h, _ = replicate(D, PRICES, PUT)
    expect_close(h, [100, -1, 1], what="holdings that copy the put")


def test_put_price():
    """The put costs 5.00"""
    _, price = replicate(D, PRICES, PUT)
    expect_close(price, 5.0, what="price of the put")


def test_straddle():
    """The straddle, paying 20, 0 and 25, needs 2 calls and costs 11.75"""
    h, price = replicate(D, PRICES, [20, 0, 25])
    expect_close(h, [100, -1, 2], what="holdings that copy the straddle")
    expect_close(price, 11.75, what="price of the straddle")


def test_bond_copies_itself():
    """The bond's own payoff gives holdings (1, 0, 0) and its price, 0.97"""
    h, price = replicate(D, PRICES, [1, 1, 1])
    expect_close(h, [1, 0, 0], what="holdings that copy the bond", abs_tol=1e-12)
    expect_close(price, 0.97, what="price of the bond")


def test_state_prices():
    """The state prices are 0.25, 0.45 and 0.27"""
    expect_close(state_prices(D, PRICES), [0.25, 0.45, 0.27], what="state prices")


def test_two_states():
    """Two states (stock 80 or 125): the call is 0.5556 shares and −44.44 bonds, and costs 11.75"""
    h, price = replicate([[1, 80], [1, 125]], [0.97, 98.75], [0, 25])
    expect_close(h, [-80 * 25 / 45, 25 / 45], what="holdings that copy the call")
    expect_close(price, 11.75, what="price of the call")


def test_call_too_cheap():
    """A call quoted at 1.50 makes the down state price negative: ValueError"""
    try:
        state_prices(D, [0.97, 98.75, 1.50])
    except ValueError:
        return
    raise AssertionError("expected ValueError: the state prices are -0.0125, 0.9225 and 0.06")


def test_call_too_dear():
    """A call quoted at 12.00 makes the middle state price negative: ValueError"""
    try:
        state_prices(D, [0.97, 98.75, 12.0])
    except ValueError:
        return
    raise AssertionError("expected ValueError: the state prices are 0.5125, -0.0225 and 0.48")


def test_incomplete():
    """With only the bond and the stock, the call cannot be copied: ValueError"""
    try:
        replicate([[1, 80], [1, 100], [1, 125]], [0.97, 98.75], [0, 0, 25])
    except ValueError:
        return
    raise AssertionError("expected ValueError: two assets cannot cover three states")


def test_repeated_asset():
    """A second bond, paying 2 in every state, adds nothing new: ValueError"""
    try:
        replicate([[1, 80, 2], [1, 100, 2], [1, 125, 2]], [0.97, 98.75, 1.94], PUT)
    except ValueError:
        return
    raise AssertionError("expected ValueError: the payoff matrix has rank 2, below 3 states")