Unit 4, lesson 25

One period replication and state prices

How one-period replication becomes a linear system in the payoff matrix, how state prices solve the dual pricing system, and how a pricing kernel writes the same prices with physical probabilities.

  • Video 25
  • Replication, state prices and binomial trees

A stock trades at 100100 today and will be worth 8080 or 120120 after one period, and a call with strike 100100 pays 00 or 2020. What should the call cost today when two analysts disagree about how likely the up move is? If a portfolio of the stock and a cash account pays exactly what the call pays in each state, any other price for the call would allow an arbitrage, and no probability is needed. This lesson writes replication as a linear system, solves its transpose for state prices, and expresses the same prices through a pricing kernel and physical probabilities.

Key terms

One-period model

Definition A one-period model has two dates, 00 and 11, and a finite set of states Ω={ω1,…,ωK}\Omega = \{\omega_1, \dots, \omega_K\}, exactly one of which is revealed at date 11. There are NN traded assets. Asset nn costs cnc_n at date 00 and pays AjnA_{jn} at date 11 in state ωj\omega_j, both in currency units. The market is frictionless, which means that any real holdings, including short and fractional ones, can be traded at these prices with no transaction costs, taxes or limits on borrowing. A physical probability measure P\mathbb{P} gives every state a probability pj=P({ωj})>0p_j = \mathbb{P}(\{\omega_j\}) > 0, so every listed state can occur. In this lesson one traded asset is a cash account with price B0=1B_0 = 1 that pays B1=RB_1 = R in every state, where R>0R > 0 is the gross one-period return. A claim is a payoff vector X∈RKX \in \mathbb{R}^K, with XjX_j paid in state ωj\omega_j.

In plain words The model lists everything that can happen by date 11 and what each traded asset pays in each case. Probabilities say how likely the states are, and prices say what the assets cost today. They are separate inputs, and replication uses only the second.

Example The constructed market of this lesson has two states, down and up, always listed in that order. A stock costs S0=100S_0 = 100 and pays Sd=80S_d = 80 or Su=120S_u = 120. The cash account costs 11 and pays R=1.05R = 1.05 in both states. The call with strike 100100 pays X=(0,20)X = (0, 20). One physical model gives the up state probability 0.40.4, so p=(0.6,0.4)p = (0.6, 0.4), and under it the stock’s expected payoff is 0.6⋅80+0.4⋅120=960.6 \cdot 80 + 0.4 \cdot 120 = 96.

Payoff matrix

Definition The payoff matrix of a one-period model is the K×NK \times N matrix AA whose row jj lists the payoffs of all assets in state ωj\omega_j and whose column nn is the payoff vector of asset nn. A portfolio is a vector θ∈RN\theta \in \mathbb{R}^N of holdings, in units of each asset. With the price vector c∈RNc \in \mathbb{R}^N, its payoff and its cost are

Aθ∈RK,c⊤θ=∑n=1Ncnθn.\begin{gathered} A\theta \in \mathbb{R}^{K}, \\[0.4em] c^{\top}\theta = \sum_{n=1}^{N} c_n \theta_n . \end{gathered}

A claim XX is attainable when Aθ=XA\theta = X has a solution, which means that XX lies in the column space Col⁡A\operatorname{Col} A. Every claim is attainable exactly when rank⁡A=K\operatorname{rank} A = K, and an attainable claim has only one replicating portfolio exactly when rank⁡A=N\operatorname{rank} A = N, that is, when the columns are linearly independent.

In plain words The rows are the states and the columns are the assets. Whether a payoff can be built depends only on the columns, while what it costs also depends on the prices.

Example For the stock and the cash account, with the price vector c=(100,1)c = (100, 1) written as a column,

A=[801.051201.05].A = \begin{bmatrix} 80 & 1.05 \\ 120 & 1.05 \end{bmatrix}.

Its determinant is 80⋅1.05−1.05⋅120=−42≠080 \cdot 1.05 - 1.05 \cdot 120 = -42 \ne 0, so its rank is 2=K=N2 = K = N. Every claim on the two states is attainable, and in exactly one way. If the stock were replaced by an asset paying 2.12.1 in both states, its column would be twice the cash column, the rank would drop to 11, and only claims that pay the same amount in both states could be built.

Replicating portfolio

Definition A replicating portfolio for a claim XX is a portfolio θ\theta with Aθ=XA\theta = X, so that it pays exactly XjX_j in every state ωj\omega_j. Its replication cost is V0=c⊤θV_0 = c^{\top}\theta. Suppose two portfolios θ\theta and θ′\theta' both replicate XX but have different costs. Buying the cheaper one, selling the dearer one and placing the difference in the cash account costs nothing today, and since the two payoffs cancel, it pays RR times that difference in every state, which is an arbitrage. So in a market without arbitrage that trades the cash account, every attainable claim has one replication cost, whichever replicating portfolio is used. This is the law of one price for attainable claims. It is weaker than the absence of arbitrage, because a market can give equal payoffs equal prices and still offer an arbitrage.

In plain words A replicating portfolio is a hedge that matches the claim state by state, not only on average. If the claim also traded at a price other than the hedge’s cost, buying the cheaper of the two and selling the dearer one would lock in the difference.

Example For the call X=(0,20)X = (0, 20), write Δ\Delta for the number of shares and bb for the currency amount placed in the cash account at date 00, which grows to 1.05 b1.05\,b. The replication equations, one for each state, are

80Δ+1.05 b=0,120Δ+1.05 b=20.\begin{aligned} 80\Delta + 1.05\,b &= 0, \\[0.4em] 120\Delta + 1.05\,b &= 20. \end{aligned}

Subtracting the first from the second gives 40Δ=2040\Delta = 20, so Δ=12\Delta = \tfrac12, and then 1.05 b=−401.05\,b = -40, so b=−80021≈−38.095b = -\tfrac{800}{21} \approx -38.095. The portfolio holds half a share and borrows about 38.09538.095, owing 4040 at date 11 in both states. It pays 40−40=040 - 40 = 0 in the down state and 60−40=2060 - 40 = 20 in the up state, and it costs

V0=12⋅100−80021=25021≈11.905.\begin{aligned} V_0 &= \tfrac12 \cdot 100 - \tfrac{800}{21} \\[0.4em] &= \tfrac{250}{21} \approx 11.905 . \end{aligned}

No probability entered the calculation. The amount bb counts currency, which equals the number of cash-account units only because each unit costs 11. With a bond priced at 100100 that pays 105105, the same borrowing would be a position of −821-\tfrac{8}{21} bonds.

State price

Definition A state-price vector is a vector π∈RK\pi \in \mathbb{R}^K that prices every traded asset as the sum of its state payoffs weighted by π\pi,

A⊤π=c,A^{\top}\pi = c ,

which says cn=∑jπjAjnc_n = \sum_{j} \pi_j A_{jn} for each asset nn. The number πj\pi_j is the state price of ωj\omega_j, in date-00 currency per unit of currency paid in state ωj\omega_j. When the Arrow security that pays 11 in state ωj\omega_j and 00 in every other state is attainable, every state-price vector gives it the value πj\pi_j, which is its replication cost. The cash account forces R∑jπj=1R \sum_j \pi_j = 1, so the state prices add up to the discount factor 1/R1/R, not to 11. A result proved in The fundamental theorem of asset pricing says that this finite market, which trades the cash account, has no arbitrage if and only if it has a state-price vector with πj>0\pi_j > 0 for every jj.

In plain words A state price is the cost today of receiving one unit of currency in one particular state and nothing otherwise. Adding the state prices gives the cost of one unit received for sure, which is less than 11 when R>1R > 1.

Example For the stock and the cash account the state-price equations are

1.05(πd+πu)=1,80πd+120πu=100.\begin{aligned} 1.05(\pi_d + \pi_u) &= 1, \\[0.4em] 80\pi_d + 120\pi_u &= 100. \end{aligned}

The first gives πd+πu=2021\pi_d + \pi_u = \tfrac{20}{21}. Putting πd=2021−πu\pi_d = \tfrac{20}{21} - \pi_u into the second leaves 40πu=100−160021=5002140\pi_u = 100 - \tfrac{1600}{21} = \tfrac{500}{21}, so πu=2542≈0.5952\pi_u = \tfrac{25}{42} \approx 0.5952 and πd=514≈0.3571\pi_d = \tfrac{5}{14} \approx 0.3571. Both are positive. For a general stock price S0S_0 and payoffs Sd<SuS_d < S_u, the same two equations give

πu=RS0−SdR(Su−Sd),πd=Su−RS0R(Su−Sd),\begin{aligned} \pi_u &= \frac{R S_0 - S_d}{R(S_u - S_d)}, \\[0.4em] \pi_d &= \frac{S_u - R S_0}{R(S_u - S_d)}, \end{aligned}

so both state prices are positive exactly when Sd<RS0<SuS_d < R S_0 < S_u, which here reads 80<105<12080 < 105 < 120.

Dual pricing system

Definition The replication system Aθ=XA\theta = X and the state-price system A⊤π=cA^{\top}\pi = c form a dual pricing system. The first combines the columns of AA to build a payoff, and the second weights the rows of AA to reproduce the prices. For any solutions of the two systems,

π⊤X=π⊤Aθ=(A⊤π)⊤θ=c⊤θ.\begin{aligned} \pi^{\top}X &= \pi^{\top}A\theta \\[0.4em] &= (A^{\top}\pi)^{\top}\theta \\[0.4em] &= c^{\top}\theta . \end{aligned}

So every state-price vector values every attainable claim at its replication cost. When AA is square and invertible, both systems have unique solutions, θ=A−1X\theta = A^{-1}X and π=(A⊤)−1c\pi = (A^{\top})^{-1}c, and every claim gets the single value π⊤X\pi^{\top}X. When rank⁡A<K\operatorname{rank} A < K and the dual system has a solution, it has infinitely many, because A⊤A^{\top} then has a nonzero null space. These solutions agree on every attainable claim, but they do not all agree on a claim outside Col⁡A\operatorname{Col} A, which therefore has no single price fixed by replication. That case is treated in Market completeness and nonunique prices.

In plain words Building the claim from assets and pricing the states are two views of one linear algebra problem. Transposing the payoff matrix moves from one view to the other, and both give the same number for any payoff that can be built.

Example The call has the state-price value 20πu=20⋅2542=2502120\pi_u = 20 \cdot \tfrac{25}{42} = \tfrac{250}{21}, the replication cost found above. A sure payment of 1010 is replicated by 101.05=20021≈9.524\tfrac{10}{1.05} = \tfrac{200}{21} \approx 9.524 in the cash account and no stock, and its state-price value is 10(πd+πu)=2002110(\pi_d + \pi_u) = \tfrac{200}{21} as well. Now replace the stock by an asset that costs 22 and pays 2.12.1 in both states. Its price equation 2.1(πd+πu)=22.1(\pi_d + \pi_u) = 2 repeats the cash equation, so the dual system reduces to πd+πu=2021\pi_d + \pi_u = \tfrac{20}{21}, and every pair of positive numbers with this sum is a positive state-price vector. All these pairs give the sure payment the value 20021\tfrac{200}{21}, but they value the call, which is no longer attainable, anywhere in the open interval from 00 to 40021\tfrac{400}{21}.

Pricing kernel

Definition Let π\pi be a state-price vector and let P\mathbb{P} give every state a probability pj>0p_j > 0. The pricing kernel, also called the stochastic discount factor, is the random variable mm with m(ωj)=πj/pjm(\omega_j) = \pi_j / p_j. For every claim XX,

EP[mX]=∑j=1Kpj πjpj Xj=π⊤X,\begin{aligned} \mathbb{E}^{\mathbb{P}}[mX] &= \sum_{j=1}^{K} p_j \, \frac{\pi_j}{p_j} \, X_j \\[0.4em] &= \pi^{\top}X , \end{aligned}

so V0=EP[mX]V_0 = \mathbb{E}^{\mathbb{P}}[mX] for every attainable claim. Applied to the cash account this gives EP[m]=1/R\mathbb{E}^{\mathbb{P}}[m] = 1/R, and therefore

V0=EP[X]R+Cov⁡P(m,X).\begin{aligned} V_0 &= \frac{\mathbb{E}^{\mathbb{P}}[X]}{R} \\[0.4em] &\quad + \operatorname{Cov}^{\mathbb{P}}(m, X). \end{aligned}

The kernel depends on the market prices and on the chosen P\mathbb{P}, it is positive in every state exactly when the state prices are, and the division needs pj>0p_j > 0 in every state. Discounting the physical mean payoff at the cash return gives the price of an attainable claim exactly when its covariance with mm is zero.

In plain words The kernel rewrites each state price as a probability times a discount that depends on the state. Averaging kernel times payoff with the physical probabilities gives back the market price, because the probabilities cancel. It does not say that a physical expected payoff should be discounted at one fixed rate.

Example With p=(0.6,0.4)p = (0.6, 0.4),

md=5/140.6=2542≈0.5952,mu=25/420.4=12584≈1.4881.\begin{aligned} m_d &= \frac{5/14}{0.6} = \frac{25}{42} \approx 0.5952, \\[0.4em] m_u &= \frac{25/42}{0.4} = \frac{125}{84} \approx 1.4881. \end{aligned}

Then EP[m]=0.6⋅2542+0.4⋅12584=2021\mathbb{E}^{\mathbb{P}}[m] = 0.6 \cdot \tfrac{25}{42} + 0.4 \cdot \tfrac{125}{84} = \tfrac{20}{21}, the discount factor, and EP[mX]=0.4⋅12584⋅20=25021\mathbb{E}^{\mathbb{P}}[mX] = 0.4 \cdot \tfrac{125}{84} \cdot 20 = \tfrac{250}{21}, the call price. Discounting the physical mean payoff instead gives 0.4⋅20/1.05=16021≈7.6190.4 \cdot 20 / 1.05 = \tfrac{160}{21} \approx 7.619, which misses the price by the covariance term 9021\tfrac{90}{21}. Under this P\mathbb{P} the stock’s expected payoff 9696 is below 1.05⋅100=1051.05 \cdot 100 = 105, so the covariance of mm with the stock payoff is positive and the kernel is larger in the up state. That ordering follows from these constructed inputs and is not a general statement about investors.

Check yourself

  1. In the same market, a put with strike 100100 pays 2020 in the down state and 00 in the up state. Find its price from the state prices, find its replicating portfolio, and check put-call parity against the call price 25021\tfrac{250}{21}.
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The state prices give the put the value 20πd=20⋅514=507≈7.14320\pi_d = 20 \cdot \tfrac{5}{14} = \tfrac{50}{7} \approx 7.143. The replication equations are

80Δ+1.05 b=20,120Δ+1.05 b=0,\begin{aligned} 80\Delta + 1.05\,b &= 20, \\[0.4em] 120\Delta + 1.05\,b &= 0, \end{aligned}

so 40Δ=−2040\Delta = -20, which gives Δ=−12\Delta = -\tfrac12, and then 1.05 b=601.05\,b = 60, which gives b=4007≈57.143b = \tfrac{400}{7} \approx 57.143. The portfolio is short half a share and lends about 57.14357.143, and it costs −50+4007=507-50 + \tfrac{400}{7} = \tfrac{50}{7}, the same value. For parity, the call minus the put pays S1−100S_1 - 100 in both states, which one share and a loan of 1001.05\tfrac{100}{1.05} replicate. So the call price minus the put price must be 100−1001.05=10021100 - \tfrac{100}{1.05} = \tfrac{100}{21}, and indeed 25021−15021=10021\tfrac{250}{21} - \tfrac{150}{21} = \tfrac{100}{21}.

  1. A classmate prices the call as its physical mean payoff discounted at the cash return, using p=(0.6,0.4)p = (0.6, 0.4), and gets 16021≈7.619\tfrac{160}{21} \approx 7.619. Suppose the call traded at that price. Build an arbitrage and find what it pays.
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The call would be cheaper than its replication cost 25021\tfrac{250}{21}. Buy the call for 16021\tfrac{160}{21} and sell the replicating portfolio, that is, short half a share and place 80021\tfrac{800}{21} in the cash account, which brings in 50−80021=2502150 - \tfrac{800}{21} = \tfrac{250}{21}. The net receipt today is 9021=307≈4.286\tfrac{90}{21} = \tfrac{30}{7} \approx 4.286, and it also goes into the cash account, so the whole position costs nothing. At date 11 the call pays XX and the sold portfolio owes exactly XX, so these cancel, and the cash account pays 307⋅1.05=4.5\tfrac{30}{7} \cdot 1.05 = 4.5 in both states. The mistake is the formula itself. The price is EP[mX]\mathbb{E}^{\mathbb{P}}[mX], which exceeds EP[X]/R\mathbb{E}^{\mathbb{P}}[X]/R here by Cov⁡P(m,X)=9021\operatorname{Cov}^{\mathbb{P}}(m, X) = \tfrac{90}{21}.

  1. Keep the stock at 100100 with payoffs 8080 and 120120, but let the cash account pay R=1.25R = 1.25. Solve for the state prices and decide whether the market is free of arbitrage. What does the replication cost of the call mean in this market?
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The formulas give

πu=125−801.25⋅40=0.9,πd=120−1251.25⋅40=−0.1.\begin{aligned} \pi_u &= \frac{125 - 80}{1.25 \cdot 40} = 0.9, \\[0.4em] \pi_d &= \frac{120 - 125}{1.25 \cdot 40} = -0.1. \end{aligned}

The down-state price is negative, because the condition Sd<RS0<SuS_d < R S_0 < S_u fails with RS0=125>120R S_0 = 125 > 120. The market has an arbitrage. Short one share for 100100 and put the 100100 in the cash account. The position costs nothing and pays 125−80=45125 - 80 = 45 in the down state and 125−120=5125 - 120 = 5 in the up state. The payoff matrix is still invertible, so the call can still be replicated, with Δ=12\Delta = \tfrac12 and b=−32b = -32, at the cost 18=20⋅0.918 = 20 \cdot 0.9. That number is only the cost of one portfolio in a market that already offers a riskless profit, not an arbitrage-free price. The Arrow security for the down state, whose payoff is nonnegative, would cost −0.1-0.1. Solving the linear systems is not the test, and the positivity of the state prices is.

How this lesson connects