Separating Hyperplanes and State-Price Vectors

Market Data as a Geometric Object

In a finite-state market, a portfolio has one initial cost and a vector of terminal payoffs, one for each possible state. No-arbitrage says that no portfolio can have nonpositive cost and nonnegative payoffs in every state, with some strict gain. The separating-hyperplane argument turns this condition into a positive linear valuation of terminal payoffs: each state receives a strictly positive price weight.

Consider a one-period market with \(n\) possible terminal states and \(d\) traded assets. Write \(p\in\mathbb{R}^d\) for the vector of initial asset prices and \(A\in\mathbb{R}^{n\times d}\) for the payoff matrix, where \(A_{ij}\) is the payoff of asset \(j\) in state \(i\). A portfolio \(\theta\in\mathbb{R}^d\) costs \(p\cdot\theta\) and has terminal payoff vector \(A\theta\). Short positions are allowed. Assume each of the \(n\) states is possible and that portfolios are unrestricted.

The state-price equation means that the cost of every portfolio equals the state-price-weighted sum of its terminal payoffs: \(p\cdot\theta=q\cdot(A\theta)\). State prices are measured in units of current currency per unit of terminal payoff. They are not, in general, probabilities.

The Separation Lemma

The geometric step is a finite-dimensional fact about a linear subspace and the nonnegative orthant. The orthant \(\mathbb{R}^N_+\) consists of vectors whose coordinates are all nonnegative; its interior consists of vectors whose coordinates are all strictly positive.

Proof. First suppose \(z\in L^\perp\) and \(z_i>0\) for all \(i\). If \(y\in L\cap\mathbb{R}^N_+\) is nonzero, then at least one coordinate of \(y\) is positive, so \(z\cdot y>0\). But \(y\in L\) and \(z\in L^\perp\) imply \(z\cdot y=0\), a contradiction. Thus \(L\cap\mathbb{R}^N_+=\{0\}\).

Conversely, suppose \(L\cap\mathbb{R}^N_+=\{0\}\), and let \(\Delta=\{x\in\mathbb{R}^N_+: \sum_{i=1}^N x_i=1\}\). This set is compact and disjoint from the closed subspace \(L\). The distance from a point to \(L\) is continuous, so it attains a minimum on \(\Delta\). This minimum is positive: if it were zero, a minimizing point in \(\Delta\) would belong to the closed set \(L\), contradicting disjointness. Choose \(x_0\in\Delta\) minimizing the distance, and let \(l_0\in L\) be a closest point in \(L\) to \(x_0\). Put \(z=x_0-l_0\), so \(z\ne0\).

For any \(l\in L\), the squared distance from \(x_0\) to \(l_0+tl\) is minimized at \(t=0\). Expanding this square for real \(t\) gives \(z\cdot l=0\); hence \(z\in L^\perp\). For any \(x\in\Delta\), the point \(x_0+t(x-x_0)\) belongs to \(\Delta\) for \(0\le t\le1\). Since \(x_0\) minimizes the distance to \(L\), its squared distance to \(l_0\) cannot decrease in this direction. Expanding at \(t=0\) gives \(z\cdot(x-x_0)\ge0\). Also, \(z\cdot l_0=0\), so \(z\cdot x_0=z\cdot(l_0+z)=\|z\|^2>0\). Therefore \(z\cdot x\ge\|z\|^2>0\) for every \(x\in\Delta\). Each coordinate unit vector \(e_i\) belongs to \(\Delta\), so \(z_i=z\cdot e_i>0\) for every \(i\). This proves the converse. \(\square\)

Geometrically, the lemma gives a hyperplane through the origin, with normal \(z\), that contains \(L\). Every nonzero vector in the nonnegative orthant lies strictly on the positive side of this hyperplane because its dot product with \(z\) is positive.

No-Arbitrage and Strictly Positive State Prices

Proof. Map each portfolio to the vector consisting of its terminal payoffs and the negative of its initial cost: \(T\theta=(A\theta,-p\cdot\theta)\in\mathbb{R}^{n+1}\). The set \(L=T(\mathbb{R}^d)\) is a linear subspace. A vector \(T\theta\) belongs to \(\mathbb{R}^{n+1}_+\) exactly when the portfolio has nonnegative payoffs in every state and nonpositive initial cost. It is nonzero exactly when at least one of those inequalities is strict. Thus the absence of arbitrage is equivalent to \(L\cap\mathbb{R}^{n+1}_+=\{0\}\).

By the lemma, this intersection condition holds exactly when there is a vector \(z=(u,\lambda)\in L^\perp\) with \(u_i>0\) for \(i=1,\dots,n\) and \(\lambda>0\). Orthogonality to every \(T\theta\) says \(u\cdot A\theta-\lambda p\cdot\theta=0\) for every \(\theta\). Taking portfolios that hold one asset at a time gives \(A^{\mathsf T}u=\lambda p\). Since \(\lambda>0\), the vector \(q=u/\lambda\) is strictly positive and satisfies \(A^{\mathsf T}q=p\). In the reverse direction, if such a \(q\) exists and \(\theta\) is an arbitrage, then \(p\cdot\theta=q\cdot(A\theta)>0\): the dot product is positive because \(A\theta\ge0\), is nonzero, and every coordinate of \(q\) is positive. This contradicts \(p\cdot\theta\le0\). Therefore the market is arbitrage-free. \(\square\)

The theorem’s strict positivity is substantive. A zero state-price coordinate would fail to assign positive value to a payoff that occurs only in that state. Such a payoff can be a cost-free arbitrage, even though a nonnegative vector of state prices might still satisfy the pricing equations.

Worked State-Price Calculations

Interpretation and a Common Pitfall

The separating normal combines state-by-state payoffs with the initial-cost coordinate. In the proof, it has the form \((u,\lambda)\), where \(u\) assigns positive weights to terminal states and \(\lambda\) weights the negative-cost coordinate. Dividing by the positive number \(\lambda\) expresses the result in the familiar units of state prices. Every attainable portfolio vector has zero value under this normal; every nonzero vector representing nonnegative payoffs and nonpositive cost has positive value. The geometry therefore rules out arbitrage by placing attainable vectors on a hyperplane that cannot contain a nonzero arbitrage vector.

Common mistake: accepting nonnegative state prices. In the two-state arbitrage example, \(q=(0,1)\) satisfies both pricing equations. It is wrong to conclude from this that the market is arbitrage-free: the portfolio’s payoff \((1,0)\) has value \(q\cdot(1,0)=0\), while its cost is also zero. The correction is to require \(q_i>0\) in every possible state. Then any nonzero nonnegative payoff has strictly positive value, which contradicts the nonpositive cost required for arbitrage.

State prices should also not be confused with probabilities. Their defining role here is to price payoffs; their sum is determined by the price of a unit payoff in every state, if such an asset is present. The theorem does not require the state prices to sum to one. Normalizing them or interpreting them through a change of measure is a separate step.