Completeness and Uniqueness of Martingale Measures

From Replication to Risk-Neutral Probabilities

In a finite-state market, both replication and risk-neutral pricing can be expressed using the same matrix of discounted asset payoffs. Replication asks whether the columns of that matrix can produce every state-contingent payoff. An equivalent martingale measure asks whether a strictly positive probability vector solves the pricing equations. The key connection is that the payoff matrix has full row rank exactly when those pricing equations have at most one solution.

Consider a one-period market with dates \(0\) and \(1\), and a finite state space \(\Omega=\{\omega_1,\ldots,\omega_n\}\). Let \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t=0,1},\mathbb{P})\) be its filtered probability space, with \(\mathcal{F}_0=\{\varnothing,\Omega\}\), \(\mathcal{F}_1=2^\Omega\), and \(\mathbb{P}(\{\omega_i\})>0\) for every state. There are \(m\) traded assets, including a riskless asset. Use discounted prices, so the riskless asset has price \(1\) at both dates. Let \(A_{ij}\) be asset \(j\)'s discounted payoff at date \(1\) in state \(\omega_i\), and let \(p_j\) be its discounted price at date \(0\). Write \(A\) for the \(n\)-by-\(m\) payoff matrix and \(p\) for the vector of initial prices.

The riskless asset's payoff column consists of ones and its initial price is \(1\), so its equation in \(A^{\mathsf T}q=p\) is precisely the probability normalization \(\sum_i q_i=1\). For each other asset, the equation says its initial discounted price is its expected discounted payoff under \(\mathbb{Q}\). In this finite-state setting, the martingale condition at date \(0\) is this expectation identity because \(\mathcal{F}_0\) is trivial.

We assume throughout the main result that at least one equivalent martingale measure exists. In a finite-state market this is the no-arbitrage case covered by the First Fundamental Theorem. The issue here is not existence, but whether that measure is unique.

Completeness Is a Rank Condition

Proof. By definition, the payoff vectors attainable from portfolios are exactly the vectors \(Ah\) as \(h\) ranges over \(\mathbb{R}^m\). These vectors form the column space of \(A\). The market is complete precisely when this column space is all of \(\mathbb{R}^n\), which holds precisely when its dimension, \(\operatorname{rank}(A)\), is \(n\). This also covers the case of no nonzero attainable payoffs: if the column space is \(\{0\}\) and \(n>0\), it is not all of \(\mathbb{R}^n\), and the rank is less than \(n\). \(\square\)

The number of traded assets alone does not decide completeness. What matters is the number of linearly independent payoff patterns across states. An asset whose payoff column is a linear combination of other columns adds no new direction for replication.

The Completeness–Uniqueness Theorem

Proof. By the proposition, completeness is equivalent to \(\operatorname{rank}(A)=n\). The martingale-measure equations are \(A^{\mathsf T}q=p\). If \(\operatorname{rank}(A)=n\), then the null space of \(A^{\mathsf T}\) is \(\{0\}\). If \(q\) and \(\widehat q\) are two equivalent martingale measures, subtracting their equations gives \(A^{\mathsf T}(q-\widehat q)=0\), so \(q-\widehat q=0\). Thus they are equal, proving uniqueness.

For the converse, suppose \(\operatorname{rank}(A)<n\). Then the null space of \(A^{\mathsf T}\) contains a nonzero vector \(v\), so \(A^{\mathsf T}v=0\). Because the riskless asset's payoff column is the all-ones vector, this equation also gives \(\sum_i v_i=0\). Let \(q\) be an existing equivalent martingale measure, so every \(q_i>0\). Choose \(\varepsilon>0\) small enough that \(q_i+\varepsilon v_i>0\) and \(q_i-\varepsilon v_i>0\) for every \(i\). Such a choice exists: for each \(i\) with \(v_i\ne0\), it is enough to take \(\varepsilon<q_i/|v_i|\), and there are only finitely many such indices. Then \(q+\varepsilon v\) and \(q-\varepsilon v\) are distinct, strictly positive vectors. They both sum to one because \(\sum_i v_i=0\), and both satisfy the pricing equations because \(A^{\mathsf T}v=0\). They are therefore two distinct equivalent martingale measures. Hence uniqueness fails when the market is incomplete. \(\square\)

The positivity assumption is essential in the converse. It lets us perturb an existing probability vector in both directions while keeping every state probability positive. The proof does not say that every solution of the pricing equations is an equivalent measure; a solution with a zero or negative component is not one.

Worked Payoff Matrices

A Common Pitfall: Counting Equations Is Not Enough

In the three-state incomplete example, one might set \(q_3=0\), solve the equations to obtain \(q_2=1\) and \(q_1=0\), and then call this the unique martingale measure. This is wrong twice over: the choice \(q_3=0\) was imposed rather than deduced, and the resulting vector is not strictly positive, so it does not define a measure equivalent to \(\mathbb{P}\). The actual equivalent measures have \(q=(t,1-2t,t)\) for every \(0<t<\tfrac12\). Uniqueness concerns all strictly positive solutions, not a solution selected by arbitrarily fixing one coordinate or a boundary solution that assigns zero probability to a state.

The equivalence also gives a useful diagnostic. With an existing equivalent martingale measure, a missing payoff direction creates a nonzero vector \(v\) invisible to every traded payoff: \(A^{\mathsf T}v=0\). Small positive and negative perturbations along \(v\) preserve all asset prices but change state probabilities. Full rank removes every such direction, leaving at most one pricing probability vector.