Unit 1, lesson 2

Arbitrage and the law of one price

What an arbitrage is in a finite one-period market, why no arbitrage implies the law of one price and linear pricing by replication, how payoff dominance orders prices, and what a short-selling constraint changes.

  • Video 2
  • Contracts, arbitrage and returns

A call on a stock is quoted at 14, while a portfolio of half a share and some borrowing pays exactly what the call pays in every state and costs about 11.90. Can a trader sell the call, buy the portfolio and keep the difference without any risk, and what permissions does that trade need? This lesson defines arbitrage as a precise pattern of cash flows, proves that no arbitrage forces equal prices for equal payoffs and ordered prices for ordered payoffs, and solves a two-state replication. It also shows how a short-selling constraint turns these equalities into inequalities.

Key terms

Arbitrage

Definition Consider a one-period market with finite state space Ω={ω1,…,ωm}\Omega = \{\omega_1, \ldots, \omega_m\} in which every state has positive physical probability, P(ωj)>0\mathbb{P}(\omega_j) > 0. There are NN traded instruments with time-00 price vector c∈RNc \in \mathbb{R}^N and terminal payoff matrix A∈Rm×NA \in \mathbb{R}^{m \times N}, whose entry AjiA_{ji} is the payment of instrument ii in state ωj\omega_j, so states are rows and instruments are columns. A holdings vector θ∈RN\theta \in \mathbb{R}^N costs c⊤θc^{\top}\theta at time 00 and pays AθA\theta at TT. Unless a constraint is stated, every real θ\theta is admissible, so borrowing and short selling are allowed at the quoted prices, with no fees, spreads or counterparty default. An arbitrage is an admissible θ\theta with

c⊤θ≤0,Aθ≥0,c^{\top}\theta \le 0, \qquad A\theta \ge 0,

and in addition

c⊤θ<0orAθ≠0,c^{\top}\theta < 0 \quad \text{or} \quad A\theta \ne 0,

where inequalities between vectors hold entry by entry. So an arbitrage costs at most zero, never requires a payment at TT, and either brings in money today or pays a positive amount in some state, which has positive probability. When a cash instrument with a positive price and a positive payment in every state is traded, investing the money received today in it turns a strategy of the first kind into one with zero cost and a payoff that is nonnegative in every state and positive in some state. The market is free of arbitrage when no admissible θ\theta is an arbitrage.

In plain words An arbitrage never asks for money, today or in any state later, and it provides money today or in at least one possible state. A high expected return is not enough, because a single state with a negative payment, however unlikely, rules the trade out.

Example In a constructed market, a stock costs 100 and pays 80 or 120 at TT, and a cash instrument costs 100 and pays 105 in both states. Buying one share with 100 borrowed by selling one cash instrument costs 0 and pays 80−105=−2580 - 105 = -25 in the down state and 120−105=15120 - 105 = 15 in the up state. It is not an arbitrage, whatever the probability of the up state, because the down state has positive probability and requires a payment. A strategy that receives 2 today and pays 0 in both states is an arbitrage, and investing the 2 in the cash instrument gives one that costs 0 and pays 2.12.1 in both states.

Law of one price

Definition A market satisfies the law of one price if any two admissible portfolios with the same payoff in every state have the same cost,

Aθ=Aϕ  ⟹  c⊤θ=c⊤ϕ.A\theta = A\phi \;\Longrightarrow\; c^{\top}\theta = c^{\top}\phi.

In the frictionless one-period market of the previous card, no arbitrage implies the law of one price. To prove it, suppose Aθ=AϕA\theta = A\phi and c⊤θ<c⊤ϕc^{\top}\theta < c^{\top}\phi. The strategy θ−ϕ\theta - \phi buys the cheaper portfolio and sells the dearer one. It costs c⊤θ−c⊤ϕ<0c^{\top}\theta - c^{\top}\phi < 0 and pays Aθ−Aϕ=0A\theta - A\phi = 0 in every state, which is an arbitrage. The proof needs θ−ϕ\theta - \phi to be admissible, so it uses the freedom to sell. The converse fails, since the law of one price only compares portfolios with equal payoffs and says nothing about a portfolio that costs nothing and has a nonnegative, nonzero payoff.

In plain words Two recipes that produce the same payments in every state must cost the same. Otherwise one buys the cheap recipe, sells the expensive one, keeps the difference today and owes nothing later.

Example Change the constructed stock so that it costs 100 and pays 110 or 120, and keep the cash instrument at 100 paying 105. The payoff matrix is

A=[110105120105].A = \begin{bmatrix} 110 & 105 \\ 120 & 105 \end{bmatrix}.

Its columns are linearly independent, so Aθ=AϕA\theta = A\phi forces θ=ϕ\theta = \phi, and the law of one price holds trivially. Yet buying one share and selling one cash instrument costs 0 and pays 5 in the down state and 15 in the up state, an arbitrage. The law of one price is weaker than no arbitrage.

Replication

Definition A payoff X∈RmX \in \mathbb{R}^m is attainable, or replicable, if some admissible θ\theta satisfies Aθ=XA\theta = X, and any such θ\theta is a replicating portfolio for XX. Replication must match XX in every state, not only on average or in one scenario. If the law of one price holds, all replicating portfolios of XX have the same cost, so the price

Π(X)=c⊤θfor any θ with Aθ=X,\begin{gathered} \Pi(X) = c^{\top}\theta \\[0.4em] \text{for any } \theta \text{ with } A\theta = X, \end{gathered}

belongs to the payoff and not to a particular portfolio. If a claim with payoff XX is also quoted at a price other than Π(X)\Pi(X), buying the cheaper of the claim and its replicating portfolio and selling the other is an arbitrage, provided both trades are admissible.

In plain words To price a claim, build it from instruments whose prices are known and add up their costs. The probabilities of the states never enter, because the recipe pays the claim exactly in every state.

Example Return to the market where the stock costs 100 and pays 80 or 120, and the cash instrument costs 100 and pays 105. Let Δ\Delta shares and bb cash instruments replicate the call with strike 100, whose payoff vector is (0,20)(0, 20). The state equations are

80Δ+105b=0,120Δ+105b=20.\begin{aligned} 80\Delta + 105b &= 0, \\[0.4em] 120\Delta + 105b &= 20. \end{aligned}

Subtracting the first from the second removes bb, because cash pays the same in both states, so 40Δ=2040\Delta = 20 and Δ=1/2\Delta = 1/2. Then 105b=−40105b = -40, so b=−8/21b = -8/21. The negative holding is borrowing, and bb counts cash instruments, each costing 100, so its value today is −800/21-800/21. The cost of the portfolio is

V0=100Δ+100b=50−80021=25021≈11.9048.\begin{aligned} V_0 &= 100\Delta + 100b \\[0.4em] &= 50 - \tfrac{800}{21} = \tfrac{250}{21} \\[0.4em] &\approx 11.9048 . \end{aligned}

As a check, half a share pays 40 or 60, the borrowing requires 821⋅105=40\tfrac{8}{21} \cdot 105 = 40 in both states, and the net payments are 0 and 20. If the call is quoted at 14, selling it and buying the replicating portfolio receives 14−250/21=44/21≈2.095214 - 250/21 = 44/21 \approx 2.0952 today, and the payments at TT cancel in both states.

Linear pricing

Definition The attainable payoffs form the subspace M={Aθ:θ∈RN}M = \{A\theta : \theta \in \mathbb{R}^N\} of Rm\mathbb{R}^m, the column space of AA. Under the law of one price the price map Π\Pi on MM is linear. If Aθ=XA\theta = X and Aϕ=YA\phi = Y, then A(aθ+bϕ)=aX+bYA(a\theta + b\phi) = aX + bY for all real aa and bb, so

Π(aX+bY)=a Π(X)+b Π(Y).\begin{aligned} &\Pi(aX + bY) \\[0.4em] &\quad = a\,\Pi(X) + b\,\Pi(Y). \end{aligned}

Linearity uses that holdings can be scaled by any real number and added, and that each instrument is bought and sold at one price. A stronger statement is the finite version of the fundamental theorem of asset pricing, stated here and proved later in the course. In this market, with every state of positive probability, there is no arbitrage if and only if there is a state-price vector π∈Rm\pi \in \mathbb{R}^m with

A⊤π=c,πj>0 for every j.\begin{gathered} A^{\top}\pi = c, \\[0.4em] \pi_j > 0 \text{ for every } j. \end{gathered}

Then Π(X)=π⊤X\Pi(X) = \pi^{\top}X for every attainable XX, the strictly positive linear valuation operator of Lesson 1. When M=RmM = \mathbb{R}^m the state prices are unique. Otherwise several state-price vectors price the traded instruments, and all of them give the same price to each attainable payoff.

In plain words Once one recipe prices a payoff, scaling the recipe scales the price and combining recipes adds their prices. Without arbitrage these prices come from strictly positive prices of the unit claims, one for each state.

Example In the market with the stock paying 80 or 120 and the cash instrument paying 105, the conditions A⊤π=cA^{\top}\pi = c read

80πd+120πu=100,105πd+105πu=100,\begin{aligned} 80\pi_d + 120\pi_u &= 100, \\[0.4em] 105\pi_d + 105\pi_u &= 100, \end{aligned}

with the solution πd=5/14≈0.3571\pi_d = 5/14 \approx 0.3571 and πu=25/42≈0.5952\pi_u = 25/42 \approx 0.5952. Both are positive, so this market has no arbitrage, and the call is worth 20⋅25/42=250/2120 \cdot 25/42 = 250/21, the replication cost. Multiplying by the gross cash return 1.051.05 gives the weights 5/85/8 for the up state and 3/83/8 for the down state, which add to 1, and every price equals the expected payoff under these weights divided by 1.051.05. The stock’s gross returns are 0.80.8 and 1.21.2, and the up weight equals (1.05−0.8)/(1.2−0.8)(1.05 - 0.8)/(1.2 - 0.8), a ratio that lies strictly between 0 and 1 because 0.8<1.05<1.20.8 < 1.05 < 1.2. These are risk-neutral probabilities, pricing weights rather than forecasts of the states. In the market with the stock paying 110 or 120, the same equations give πu=−10/21<0\pi_u = -10/21 < 0, which matches the arbitrage found there.

Payoff dominance

Definition A payoff XX dominates a payoff YY if X(ωj)≥Y(ωj)X(\omega_j) \ge Y(\omega_j) in every state, and strictly dominates it if in addition X(ωj)>Y(ωj)X(\omega_j) > Y(\omega_j) in at least one state. In the frictionless one-period market without arbitrage, for attainable XX and YY,

X≥Y  ⟹  Π(X)≥Π(Y),X \ge Y \;\Longrightarrow\; \Pi(X) \ge \Pi(Y),

and strict dominance gives Π(X)>Π(Y)\Pi(X) > \Pi(Y). To prove it, suppose Π(X)<Π(Y)\Pi(X) < \Pi(Y). Buying a replicating portfolio of XX and selling one of YY costs Π(X)−Π(Y)<0\Pi(X) - \Pi(Y) < 0 and pays X−Y≥0X - Y \ge 0, an arbitrage. If XX strictly dominates YY and Π(X)≤Π(Y)\Pi(X) \le \Pi(Y), the same trade costs at most zero and pays X−YX - Y, which is nonnegative and positive in a state of positive probability, again an arbitrage. Taking Y=0Y = 0 shows that an attainable payoff that is nonnegative and not zero has a positive price, the order property that strictly positive state prices express.

In plain words A payoff that is never worse in any state cannot cost less, and one that is strictly better in some state must cost strictly more. The comparison is state by state, so a payoff that is better on average but worse in one state is not ordered by this argument.

Example In the market with the stock paying 80 or 120, the call with strike 100 pays (0,20)(0, 20) and the call with strike 110 pays (0,10)(0, 10). The first strictly dominates the second, and indeed 250/21≈11.90250/21 \approx 11.90 exceeds 125/21≈5.95125/21 \approx 5.95. Dominance also bounds the price CC of the call with strike 100 without solving for it. Since (ST−100)+≤ST(S_T - 100)^{+} \le S_T, the call costs at most the share, 100. The call’s payoff is also at least 0 and at least ST−100S_T - 100, the payoff of one share less 100 sure units at TT. Those sure units cost 100⋅100/105=2000/21100 \cdot 100/105 = 2000/21, so

C≥max⁡(0, 100−200021)=10021≈4.76.\begin{aligned} C &\ge \max\bigl(0,\, 100 - \tfrac{2000}{21}\bigr) \\[0.4em] &= \tfrac{100}{21} \approx 4.76 . \end{aligned}

The replication price 250/21250/21 lies between these bounds.

Short-selling constraint

Definition A short-selling constraint forbids negative holdings in some instruments, so the admissible set becomes {θ:θi≥0 for each constrained i}\{\theta : \theta_i \ge 0 \text{ for each constrained } i\} instead of all of RN\mathbb{R}^N, and the definition of arbitrage applies only to admissible strategies. Borrowing is a negative holding of the cash instrument, so a borrowing limit is a constraint of the same kind. The proofs of the law of one price, of linear pricing and of dominance all sell a portfolio, so under a constraint their conclusions can fail in one direction, and no arbitrage then gives inequalities instead of equalities. Fees and bid-ask spreads have a similar effect, because they make the buying price and the selling price differ. The arithmetic of a strategy stays correct under a constraint. What changes is whether the strategy is allowed.

In plain words A price gap is an arbitrage only if the trades that exploit it are permitted. Forbidding a short sale removes the trade that would push an overpriced claim back toward its replication cost.

Example Suppose the call with strike 100 is quoted at 14 and cannot be sold short, while the stock and the cash instrument trade freely. The arbitrage of the replication card sells the call, so it is not admissible. In fact no admissible arbitrage exists. Add the call as a third column of AA and take the state prices π=(5/14,25/42)\pi = (5/14, 25/42), which price the stock and the cash instrument exactly and value the call’s payoff at 250/21250/21. Every holdings vector with call holding θC\theta_C then satisfies

c⊤θ−π⊤Aθ=(14−25021)θC=4421 θC.\begin{aligned} &c^{\top}\theta - \pi^{\top}A\theta \\[0.4em] &\quad = \bigl(14 - \tfrac{250}{21}\bigr)\theta_C \\[0.4em] &\quad = \tfrac{44}{21}\,\theta_C . \end{aligned}

An admissible strategy has θC≥0\theta_C \ge 0, so its cost is at least π⊤Aθ\pi^{\top}A\theta, which is nonnegative when the payoff is nonnegative. A cost of zero or less then forces θC=0\theta_C = 0 and a zero payoff, so no admissible strategy is an arbitrage. Under this constraint the market is free of arbitrage exactly when the call price is at least 250/21250/21, since a lower quote can still be exploited by buying the call and selling its replicating portfolio, which needs no short sale of the call.

Check yourself

  1. In the market where the stock costs 100 and pays 80 or 120 and the cash instrument costs 100 and pays 105, find the replicating portfolio and the price of a European put with strike 100. Check the price with the state prices 5/145/14 and 25/4225/42.
Show the answer

The put pays (20,0)(20, 0). The state equations are 80Δ+105b=2080\Delta + 105b = 20 and 120Δ+105b=0120\Delta + 105b = 0. Subtracting gives 40Δ=−2040\Delta = -20, so Δ=−1/2\Delta = -1/2, and then 105b=20+40=60105b = 20 + 40 = 60, so b=4/7b = 4/7. The portfolio is short half a share and long 4/74/7 of a cash instrument, and it costs

100⋅(−12)+100⋅47=507≈7.1429.\begin{aligned} &100 \cdot \bigl(-\tfrac12\bigr) + 100 \cdot \tfrac47 \\[0.4em] &\quad = \tfrac{50}{7} \approx 7.1429 . \end{aligned}

It pays −40+60=20-40 + 60 = 20 in the down state and −60+60=0-60 + 60 = 0 in the up state. With state prices the put is worth 20⋅5/14=50/720 \cdot 5/14 = 50/7, the same number. The call minus the put pays ST−100S_T - 100, worth 100−2000/21=100/21100 - 2000/21 = 100/21, and indeed 250/21−150/21=100/21250/21 - 150/21 = 100/21.

  1. Every state of a two-state market has positive probability, and the up state has probability 0.99. Strategy A costs 0 and pays 30 in the up state and −1-1 in the down state. Strategy B receives 1 today and pays 0 in both states. Which of them is an arbitrage, and why does the definition require every listed state to have positive probability?
Show the answer

Strategy A is not an arbitrage. Its expected payoff is 0.99⋅30−0.01⋅1=29.690.99 \cdot 30 - 0.01 \cdot 1 = 29.69, but it requires a payment in the down state, which has positive probability. Strategy B is an arbitrage, since it costs −1<0-1 < 0 and its payoff is nonnegative in every state. The definition asks for a positive payment in some state, and that gain is real only if the state can occur. With every listed state of positive probability, a payoff that is nonnegative and not zero is positive with positive probability, which is the gain an arbitrage must offer.

  1. In the same market, the call with strike 100 is quoted at 10. Construct an arbitrage, find the amount received today and the payoff after that amount is invested in the cash instrument, and name the short sale the strategy needs. Is there still an arbitrage if the stock cannot be sold short?
Show the answer

The call is cheap relative to its replication cost 250/21250/21, so buy it and sell its replicating portfolio. Selling that portfolio means selling half a share short and buying 8/218/21 of a cash instrument, which is lending. Today the strategy receives

−10+50−80021=4021≈1.9048.\begin{aligned} &-10 + 50 - \tfrac{800}{21} \\[0.4em] &\quad = \tfrac{40}{21} \approx 1.9048 . \end{aligned}

At TT the call pays 0 or 20, and the sold portfolio requires exactly 0 or 20, so the net payment is 0 in both states. Investing 40/2140/21 in the cash instrument buys 2/1052/105 of a unit, which pays 2 in both states. The strategy needs a short sale of half a share. If the stock cannot be sold short, this strategy is not admissible, and no other one works. The state prices 19/4219/42 and 1/21/2 price the cash instrument exactly, since 105⋅(19/42+21/42)=100105 \cdot (19/42 + 21/42) = 100, and the call exactly, since 20⋅1/2=1020 \cdot 1/2 = 10, while they value the stock’s payoff at 2020/21≈96.192020/21 \approx 96.19, below its price 100. So every strategy in the stock, the cash instrument and the call satisfies c⊤θ−π⊤Aθ=(80/21) θSc^{\top}\theta - \pi^{\top}A\theta = (80/21)\,\theta_S, which is nonnegative when the stock holding θS\theta_S is nonnegative. As in the short-selling card, an admissible strategy with a nonnegative payoff then has a nonnegative cost, and a cost of zero forces a zero payoff, so no admissible arbitrage exists.

How this lesson connects