Market completeness and nonunique prices

Definition Consider a one-period market with $m$ states, each with positive physical probability, and $n$ traded instruments. Their prices today form the vector $c \in \mathbb{R}^n$, and their payoffs form the payoff matrix $A \in \mathbb{R}^{m \times n}$, whose entry $A_{ji}$ is the payoff of instrument $i$ in state $j$. Trading is frictionless, so a portfolio $\theta \in \mathbb{R}^n$ with entries of either sign costs $c^{\top}\theta$ and pays $A\theta$. A claim, a payoff vector $X \in \mathbb{R}^m$, is attainable if $X = A\theta$ for some $\theta$, so the attainable claims form the column space $\operatorname{col}(A)$. Under NA every portfolio that replicates an attainable claim has the same cost, and this cost equals $\pi^{\top} X$ for every strictly positive state-price vector $\pi$, that is every $\pi > 0$ with $A^{\top}\pi = c$, because

When the account, which costs $1$ and pays $R > 0$ in every state, is traded, this common value is $\mathbb{E}^{\mathbb{Q}}[X]/R$ for every EMM $\mathbb{Q}$.

Definition The market is complete if every claim $X \in \mathbb{R}^m$ is attainable, that is if $\operatorname{col}(A) = \mathbb{R}^m$, or equivalently $\operatorname{rank}(A) = m$. A finite one-period market that satisfies NA and trades the account is complete if and only if its state-price vector is unique, and then its EMM for the account is unique too. If $\operatorname{rank}(A) = m$, then $A^{\top}$ has trivial null space, so $A^{\top}\pi = c$ has at most one solution. If $\operatorname{rank}(A) < m$, there is a $z \ne 0$ with $A^{\top} z = 0$. Starting from a strictly positive solution $\pi$, which exists under NA, the vectors $\pi + \varepsilon z$ and $\pi - \varepsilon z$ are strictly positive for small $\varepsilon > 0$, both solve $A^{\top}\pi = c$, and they differ. Because the account is traded, every solution satisfies $R \sum_j \pi_j = 1$, so $q_j = R\,\pi_j$ turns distinct state-price vectors into distinct EMMs. Completeness is relative to the modeled states and information. Splitting one state into two makes a complete market incomplete unless an instrument is added whose payoff tells the two apart.

Definition The rank of the payoff matrix $A$ is the number of linearly independent payoff columns. It equals the number of linearly independent rows and the dimension of the space $\operatorname{col}(A)$ of attainable claims, and $\operatorname{rank}(A) \le \min(m, n)$. A new instrument raises the rank only when its payoff is not already a combination of the existing payoffs, and under NA an instrument whose payoff is $A\theta$ must cost $c^{\top}\theta$. Under NA the solutions of $A^{\top}\pi = c$ form an affine set of dimension $m - \operatorname{rank}(A)$, and the strictly positive ones form a nonempty, relatively open subset of it with the same dimension. So $m - \operatorname{rank}(A)$ counts the free directions left in the pricing measure.

Definition A finite market is incomplete if $\operatorname{rank}(A) < m$, so some claims are not attainable. Under NA it then has infinitely many EMMs, by the completeness result. Every attainable claim gets the same value under all of them, while every unattainable claim gets different values under some pair of them, since some $z$ with $A^{\top} z = 0$ has $z^{\top} X \ne 0$ exactly when $X$ is not in $\operatorname{col}(A)$. Incompleteness is not arbitrage. Each EMM gives a consistent arbitrage-free system of prices, and the no-arbitrage condition alone does not choose among them.

Definition Let a finite one-period market with full support and frictionless trading trade the account and satisfy NA, and let $X$ be a claim. A price $v$ for $X$ is arbitrage-free if the market extended by $X$ as a new instrument with price $v$ still satisfies NA. By the fundamental theorem applied to the extended market, the arbitrage-free prices are exactly the numbers

which are the values $\mathbb{E}^{\mathbb{Q}}[X]/R$ over all EMMs $\mathbb{Q}$. This set is an interval, because the strictly positive state-price vectors form a convex set. If $X$ is attainable it is a single point, the replication cost. If $X$ is not attainable it is an open interval

with $\underline{\pi}(X) < \overline{\pi}(X)$, whose endpoints are the subhedging and superhedging bounds of the next card, and neither endpoint is an arbitrage-free price.

Definition In the market of the previous card, the superhedging bound of a claim $X$ is the smallest cost of a portfolio that pays at least $X$ in every state, and the subhedging bound is the largest cost of a portfolio that pays at most $X$ in every state,

with the inequalities taken coordinate by coordinate. A seller who receives $\overline{\pi}(X)$ can buy an optimal superhedge and meet the claim in every state, so $\overline{\pi}(X)$ is the seller’s bound, and $\underline{\pi}(X)$ is the buyer’s bound in the same way. Because the account pays $R > 0$ in every state, both problems are feasible, and under NA linear programming duality gives

where $\Pi$ is the set of nonnegative state-price vectors, all $\pi \ge 0$ with $A^{\top}\pi = c$. All four optima are attained. The set $\Pi$ contains boundary vectors with zero entries, which do not give measures equivalent to $\mathbb{P}$. Under NA every point of $\Pi$ is a limit of strictly positive state-price vectors, so the bounds are also the supremum and the infimum of $\mathbb{E}^{\mathbb{Q}}[X]/R$ over the EMMs. For an unattainable claim an optimal superhedge $\theta^*$ has $A\theta^* \ge X$ and $A\theta^* \ne X$, so selling $X$ at $\overline{\pi}(X)$ and buying $\theta^*$ costs nothing and pays the nonnegative nonzero vector $A\theta^* - X$, an arbitrage. The same argument with an optimal subhedge excludes $\underline{\pi}(X)$. The closed bounds are therefore the ends of the open interval of arbitrage-free prices, not members of it.