Strictly positive state prices and no arbitrage

When do strictly positive state prices exist?

Exactly when the market has no arbitrage. A state price is a value today for one dollar paid in a single state and nothing in the others. If each asset's price today equals the sum of its state payoffs weighted by strictly positive state prices, no trade makes money from nothing; if no such weights exist, some trade does.

Before you start · 3 quick checks on replication, linear systems and separating lines, about 2 min

Setup: a one-period market with finitely many states

Today is time 0. At time 1 exactly one of \(m\) states happens, \(\omega = 1, \ldots, m\), with \(m \ge 2\).
There are \(n+1\) assets, \(j = 0, 1, \ldots, n\). Asset \(j\) costs \(p_j\) today and pays \(D_{\omega j}\) in state \(\omega\) at time 1, with no other cash flows. Asset 0 is a riskless bond: \(p_0 = 1\), and it pays \(1+r\) in every state, with \(r \gt -1\).
A portfolio \(\theta\) holds \(\theta_j\) units of asset \(j\); a negative \(\theta_j\) is a short sale, and a negative \(\theta_0\) is borrowing. It costs \(V_0 = \sum_{j=0}^{n} \theta_j p_j\) today and pays \(V_\omega = \sum_{j=0}^{n} \theta_j D_{\omega j}\) in state \(\omega\).
An arbitrage is a portfolio that costs at most 0 today, pays at least 0 in every state, and either costs strictly less than 0 or pays strictly more than 0 in at least one state.
A state-price vector is a list \(\psi = (\psi_1, \ldots, \psi_m)\) that prices every asset: \(p_j = \sum_{\omega=1}^{m} \psi_\omega D_{\omega j}\) for every \(j\). It is strictly positive when every \(\psi_\omega\) is greater than 0. Having every \(\psi_\omega\) at least 0 is weaker, and this page shows that it is not enough.
For the bond the price equation reads \((1+r)(\psi_1 + \cdots + \psi_m) = 1\), so the weights \(q_\omega = (1+r)\psi_\omega\) add to 1, and every price is \(p_j = \frac{1}{1+r}\sum_{\omega=1}^{m} q_\omega D_{\omega j}\).

Fine print: trading assumptions and the two kinds of arbitrage

The market is frictionless: no transaction costs, taxes or bid-ask spreads. Any real amount of any asset can be bought or sold short at its single price, and anyone can lend or borrow any amount through the bond.
Every state can happen: each has a real-world probability greater than 0. Nothing else about the probabilities is used.
An arbitrage that brings in money today and never loses can be turned into one that costs exactly 0: put the money received in the bond. The arbitrage portfolios on this page all cost exactly 0 today.
The market is complete when every list of payoffs \((x_1, \ldots, x_m)\) is the payoff of some portfolio, that is, when the payoff matrix \(D\) has rank \(m\). Otherwise it is incomplete.
This page uses simple growth \(1+r\) per period, and the markets in stages 1 and 2 use \(r = 5\%\). All prices are hypothetical.

1 See

Move prices until the market breaks

Each market below has the bond and one or two stocks with fixed payoffs. Pick a market and move today's prices. For every price the page solves the price equations for \(\psi\), checks the signs of the solutions, and separately searches for an arbitrage portfolio. Work through the goals, then try anything you like.

Key point As long as today's prices can be written with strictly positive state prices, the page finds no arbitrage. The moment they cannot, it finds one, even when the only state-price vector left has a zero entry and none negative. With fewer assets than states, many state-price vectors fit the same prices, and the market is incomplete.

2 Try

More assets than states, and a claim no portfolio copies

More assets than states

Quoting a claim in an incomplete market

Key point When the price equations contradict each other, no state-price vector exists and an arbitrage does. When they leave room and the original market is arbitrage-free, a claim that no portfolio copies has a whole range of no-arbitrage prices, and the ends of that range are excluded. If the original market already has an arbitrage, no price of a new claim removes it.

3 Prove

The fundamental theorem for finite markets

The idea in three steps

With state prices, the cost of any portfolio is its payoffs weighted by the state prices.
If every weight is strictly positive, a portfolio that never loses costs something, unless it pays nothing at all. So no arbitrage exists.
In two states, with a stock that costs \(S_0\) and ends at \(uS_0\) (up) or \(dS_0\) (down), the price equations can be solved by hand, and the state prices are positive exactly when \(d \lt 1+r \lt u\).

Theorem (fundamental theorem of asset pricing, one period, finitely many states)

Assumptions. The setup and fine print above: \(m \ge 2\) states, each possible; assets \(j = 0, \ldots, n\) with asset 0 the bond (\(p_0 = 1\), paying \(1+r\) in every state, \(r \gt -1\)); frictionless trading of any real amounts at one price per asset; arbitrage as defined in the setup.

Conclusions.

The market has no arbitrage if and only if it has a state-price vector \(\psi\) with \(\psi_\omega \gt 0\) for every \(\omega\).
If the market has no arbitrage, its strictly positive state-price vector is unique if and only if the market is complete.
For such a \(\psi\), the weights \(q_\omega = (1+r)\psi_\omega\) are positive and add to 1, and every price is \(p_j = \frac{1}{1+r}\sum_{\omega=1}^{m} q_\omega D_{\omega j}\): the risk-neutral probabilities.

The proof below has two parts. Lines 1 to 4 prove the easy direction of conclusion 1: if a strictly positive state-price vector \(\psi\) exists, there is no arbitrage. Lines 5 to 7 compute the state prices of a two-state market with a stock price \(S_0 \gt 0\) and \(0 \lt d \lt u\), and show when they are positive. The other direction is sketched under "Going further".

Proof (7 lines, build it or read it)

Part 1 assumes a state-price vector \(\psi\) with \(\psi_\omega \gt 0\) for every \(\omega\). Part 2 starts at line 5 and assumes nothing about signs.

Scratch work, not part of the proof. For line 6: from \((u-d)\psi_u = 1 - \frac{d}{1+r} = \frac{(1+r)-d}{1+r}\) we get \(\psi_u = \frac{(1+r)-d}{(1+r)(u-d)}\). Then \[ \psi_d = \frac{1}{1+r} - \psi_u = \frac{(u-d) - \big((1+r)-d\big)}{(1+r)(u-d)} = \frac{u-(1+r)}{(1+r)(u-d)}. \]

Why uniqueness is the same as completeness

The price equations say that \(\psi\) solves a linear system with one equation per asset and one unknown per state; its coefficient matrix is the transpose of the payoff matrix \(D\). If \(D\) has rank \(m\), that system has at most one solution. If the rank is less than \(m\), there is a nonzero vector \(k\) with \(\sum_{\omega=1}^{m} k_\omega D_{\omega j} = 0\) for every asset \(j\). Then \(\psi + tk\) prices every asset for every real \(t\), and when \(\psi\) is strictly positive, so is \(\psi + tk\) for every \(t\) close enough to 0. So a strictly positive state-price vector is unique exactly when \(D\) has rank \(m\), which is the definition of completeness.

Corollary (pricing a new claim)

Add to an arbitrage-free market a European claim that pays \(X_\omega\) in state \(\omega\) at time 1 and nothing before, traded at price \(c\) on the same frictionless terms as the other assets (any amount, bought or sold at that one price). The larger market has no arbitrage if and only if \(c = \sum_{\omega=1}^{m} \psi_\omega X_\omega\) for some strictly positive state-price vector \(\psi\) of the original market.

This is conclusion 1 applied to the larger market: its strictly positive state-price vectors are the original ones that also price the claim at \(c\). If a portfolio copies the claim, every such \(\psi\) gives the same value, the cost of the copy. If none does, the strictly positive state-price vectors form a bounded, relatively open convex set on which the value is not constant, so the no-arbitrage prices form an open interval. In stage 2 that interval runs between the values at the two end vectors, and the ends are excluded because each end vector has a zero entry.

Going further: the hard direction, by a separating hyperplane

Claim. Under the setup's assumptions, if the market has no arbitrage, then a strictly positive state-price vector exists.

Sketch. For each portfolio form the vector \(z(\theta) = (-V_0, V_1, \ldots, V_m)\) in \(\mathbb{R}^{m+1}\): the money received today, then the payoff in each state. Because prices and payoffs are linear in the holdings, these vectors form a linear subspace \(M\). No arbitrage says exactly that \(M\) meets the set of vectors with every entry at least 0 only at the zero vector. So \(M\) does not meet the set \(K\) of vectors with every entry at least 0 and entries adding to 1, which is convex, closed and bounded.

The separating hyperplane theorem (a closed convex set and a disjoint, closed, bounded convex set can be strictly separated) gives a vector \(\lambda\) and a number \(c\) with \(\lambda \cdot z \lt c\) for every \(z\) in \(M\) and \(\lambda \cdot k \gt c\) for every \(k\) in \(K\). Since \(M\) is a subspace and contains \(0\), the first inequality forces \(\lambda \cdot z = 0\) on \(M\) and \(c \gt 0\). Taking \(k\) to be each unit vector shows every entry of \(\lambda\) is greater than \(c\), so every entry is positive. Then \(\lambda_0 V_0 = \sum_{\omega=1}^{m} \lambda_\omega V_\omega\) for every portfolio, and \(\psi_\omega = \lambda_\omega / \lambda_0\) is a strictly positive state-price vector: apply the identity to one unit of each asset.

What it uses. Finitely many states and assets, so everything lives in \(\mathbb{R}^{m+1}\); prices that are linear in the holdings (no frictions, short sales allowed); and the strict separation theorem, which needs one set closed and the other closed and bounded. With infinitely many states the vectors no longer live in a finite-dimensional space, and the argument needs measure theory and more care; this page stays with finitely many.

Key point With a strictly positive \(\psi\), every portfolio costs \(\sum_{\omega=1}^{m} \psi_\omega V_\omega\). So a portfolio that never loses costs something unless it pays nothing at all, and no arbitrage exists. A zero state price breaks this step. In two states, both state prices are positive exactly when \(d \lt 1+r \lt u\).

4 Apply

Problems

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