Finite-State Completeness and Replication

Replicating Claims in a Finite-State Market

In a finite-state market, each asset’s terminal payoff is a vector: one entry for each possible state. A portfolio is a linear combination of these payoff vectors. Replication is therefore a linear-algebra problem: can the claim’s payoff vector be expressed as such a combination? The rank of the payoff matrix answers whether this is possible for every claim.

Consider a one-period market, with trading at time \(0\) and claims paid at time \(1\). Let \((\Omega,\mathcal{F},(\mathcal{F}_t),\mathbb{P})\) be the probability space, with \(\mathcal{F}_0\) trivial and \(\mathcal{F}_1=\mathcal{F}\). Suppose the terminal information consists of \(n\) states \(\omega_1,\ldots,\omega_n\), each with positive probability. Every time-\(1\) claim is then represented by its values in those states. This finite-state model does not use a Brownian motion or stochastic integrals.

Assume there are \(m\) traded assets, with time-\(0\) prices \(p_1,\ldots,p_m\). Write \(a_{ij}\) for the time-\(1\) payoff of asset \(j\) in state \(\omega_i\). The payoff matrix \(A\) has \(n\) rows and \(m\) columns. A portfolio \(\theta=(\theta_1,\ldots,\theta_m)^\mathsf{T}\), where short positions are allowed, has terminal payoff \(A\theta\) and initial cost \(p^\mathsf{T}\theta\). We impose no portfolio constraints beyond these unrestricted positions.

The Rank Criterion

Proof. The columns of \(A\) are the traded assets’ payoff vectors. Thus \(x\) is replicable precisely when it is a linear combination of those columns, or equivalently when adjoining \(x\) as an extra column does not enlarge their span. The rank is the dimension of the column span, so this is exactly \(\operatorname{rank}(A)=\operatorname{rank}([A\mid x])\). This argument includes \(x=0\), which is always in the span.

For completeness, if every \(x\in\mathbb{R}^n\) is replicable, the columns of \(A\) span \(\mathbb{R}^n\), whose dimension is \(n\); hence \(\operatorname{rank}(A)=n\). Conversely, if \(\operatorname{rank}(A)=n\), its column span has dimension \(n\) and is a subspace of \(\mathbb{R}^n\), so it equals \(\mathbb{R}^n\). Every claim is then in that span and is replicable. In particular, completeness requires \(m\ge n\), since a matrix with \(m\) columns has rank at most \(m\). \(\square\)

To find a portfolio, solve the linear system \(A\theta=x\). Gaussian elimination preserves the solution set: swapping rows, multiplying a row by a nonzero number, or adding a multiple of one row to another produces an equivalent system. A contradiction row, such as \(0=1\), shows that the claim is not replicable; otherwise, the pivot equations give a solution, with any free variables marking non-uniqueness.

Full Rank and Portfolio Construction

When \(A\) has full row rank, the theorem guarantees a solution for every claim. A square invertible payoff matrix gives a unique solution; a full-row-rank matrix with more columns than rows may give multiple solutions. For a particular claim, the augmented-rank test determines whether a solution exists even when the market is incomplete.

When a market is complete, any claim can be replicated, but that fact alone does not say that the replicating portfolio is unique. The next result makes the distinction precise.

Proof. Two portfolios \(\theta\) and \(\eta\) have the same payoff exactly when \(A\theta=A\eta\), or \(A(\theta-\eta)=0\). Thus a nonzero vector in the null space of \(A\) gives two distinct portfolios with the same payoff: if \(Az=0\) and \(z\ne0\), then \(\theta\) and \(\theta+z\) have equal payoffs. Conversely, if two distinct portfolios have equal payoffs, their difference is a nonzero vector in that null space. Therefore every replicable claim has a unique portfolio exactly when the null space contains only zero. By the rank-nullity identity, this is equivalent to \(\operatorname{rank}(A)=m\). For completeness as well, the rank must equal \(n\); both conditions together require \(m=n\) and rank \(n\). \(\square\)

Prices, Interpretation, and a Common Pitfall

Replication is a statement about terminal payoffs, whereas pricing also uses initial asset prices. If a vector \(q\in\mathbb{R}^n\) of state prices satisfies \(p=A^\mathsf{T}q\), then any replicating portfolio has cost \(p^\mathsf{T}\theta=q^\mathsf{T}A\theta=q^\mathsf{T}x\). This identity follows by matrix multiplication. It does not require that \(q\) be a probability vector. Under no-arbitrage in a finite-state model with unrestricted portfolios and all states possible, positive state prices are the relevant pricing objects; the rank criterion itself does not require choosing or deriving them.

A common mistake is to infer completeness from the number of assets alone. Three assets for three states do not guarantee rank three. For example, if the payoff matrix is

the first and second columns are identical. Hence the rank is at most \(2\), despite having three assets and three states. The claim \((1,0,0)^\mathsf{T}\) cannot be replicated: the first two payoff equations would require \(\theta_1+\theta_2=1\) and \(\theta_1+\theta_2=0\) simultaneously. The correction is to calculate rank, not merely count columns. In a multi-period finite-state tree, the same linear-algebra test applies locally: at each node, use the possible next states and the assets’ next-step payoffs there.