The fundamental theorem of asset pricing

Definition A finite market model consists of a finite sample space $\Omega = \{\omega_1, \dots, \omega_m\}$ with a physical probability $\mathbb{P}$, trading dates $t = 0, 1, \dots, T$ with a filtration $(\mathcal{F}_t)$ in which $\mathcal{F}_0$ is trivial, and finitely many traded assets with adapted price processes $S^0, S^1, \dots, S^d$, where $S^0 = B$ is a strictly positive numeraire, here the money-market account. Trading is frictionless. There are no transaction costs, taxes or margin requirements, assets are perfectly divisible, and every predictable self-financing strategy with real holdings is allowed, including short positions and borrowing. In the one-period case $T = 1$, with $B_0 = 1$ and a deterministic $B_1 = R > 0$, let $S_0^i$ be the price of asset $i$ today and $S_1^i(\omega_j)$ its price in state $\omega_j$. The discounted gain of asset $i$ is the vector in $\mathbb{R}^m$

and the gain space is the linear subspace $G = \operatorname{span}\{g^1, \dots, g^d\}$ of $\mathbb{R}^m$. A portfolio holding $\theta_i$ units of each risky asset $i$, financed entirely through the account, costs nothing at date $0$, and its value at date $1$ in account units is $\sum_i \theta_i g^i$. So $G$ is the set of discounted payoffs of zero-cost portfolios.

Definition An arbitrage in a finite market model is a self-financing predictable strategy whose value process $V$ satisfies

where $V_T \ge 0$ holds in every state. The model satisfies the no-arbitrage condition, written NA, if no arbitrage exists. A strategy with $V_0 < 0$ and $V_T \ge 0$ counts as an arbitrage, since placing its initial surplus $-V_0$ in the account makes the final value positive in every state. Under the full-support condition of the next card, the one-period model satisfies NA exactly when

where $\mathbb{R}^m_{+}$ is the set of vectors with nonnegative coordinates. To see this, note that a portfolio with risky holdings $\theta$ and initial value $V_0$ has $V_1/R = V_0 + g$ with $g = \sum_i \theta_i g^i$. If it is an arbitrage, then $g = V_1/R - V_0 \ge 0$, and $g \ne 0$ because either $V_0 < 0$ or $V_1$ is positive in some state. Conversely, a nonzero $g \ge 0$ in $G$ is the discounted payoff of a zero-cost portfolio that is positive in some state, and that state has positive probability. The law of one price, which says that portfolios with equal payoffs have equal costs, follows from NA but is weaker.

Definition The full-support condition requires $\mathbb{P}(\omega) > 0$ for every $\omega \in \Omega$, so every listed state is possible. In a tree it means that every node can be reached and every branch has positive conditional probability. A probability $\mathbb{Q}$ on $\Omega$ is equivalent to $\mathbb{P}$ if $\mathbb{Q}(A) = 0$ exactly when $\mathbb{P}(A) = 0$, and under full support this holds exactly when $\mathbb{Q}(\omega) > 0$ for every $\omega$. The condition enters twice. It turns “positive in some state” into “positive with positive probability” in the definition of arbitrage, and through equivalence it forbids a pricing measure from giving zero weight to a possible state, where an arbitrage could hide. A state that the model regards as impossible is removed from $\Omega$ before any price is computed, never because a pricing calculation makes it inconvenient.

Definition In a finite market model with numeraire $B$, an equivalent martingale measure, or EMM, is a probability $\mathbb{Q}$ on $\Omega$ equivalent to $\mathbb{P}$ under which the discounted price process $S^i/B$ of every traded asset is a $\mathbb{Q}$-martingale with respect to $(\mathcal{F}_t)$, with the discounted cumulative gain process in its place for an asset that pays dividends. In the one-period model with $B_0 = 1$, $B_1 = R$ and full support, an EMM is a vector $q \in \mathbb{R}^m$ with

where the last equation holds for every risky asset $i$ and says $q^{\top} g^i = 0$. So $q$ is orthogonal to $G$. Its state prices $\pi_j = q_j/R$ are strictly positive and price every traded asset, and conversely strictly positive state prices that price every traded asset, the account included, give an EMM through $q_j = R\,\pi_j$. An EMM belongs to its numeraire, and it need not be unique.

Definition Fundamental theorem of asset pricing, finite version. In a finite market model with full support, finitely many dates, frictionless trading, unrestricted predictable self-financing strategies and a strictly positive numeraire $B$, the no-arbitrage condition holds if and only if an equivalent martingale measure for $B$ exists. In the one-period case NA holds if and only if some $q \in \mathbb{R}^m$ satisfies

The “if” direction needs only linearity. Each $g \in G$ is a combination of the $g^i$, so $q^{\top} g = 0$. A nonzero $g \ge 0$ would have $q^{\top} g = \sum_j q_j g_j > 0$, since some $g_j > 0$ and every $q_j > 0$, so $G$ contains no such vector and NA holds. With several dates the argument runs through the martingale property. The discounted value of a self-financing predictable strategy is its initial value plus a sum of predictable holdings times martingale increments, so it is a $\mathbb{Q}$-martingale and $\mathbb{E}^{\mathbb{Q}}[V_T/B_T] = V_0/B_0 \le 0$. A final value with $V_T \ge 0$ and $\mathbb{P}(V_T > 0) > 0$ also has $\mathbb{Q}(V_T > 0) > 0$ by equivalence, which would make that expectation positive. The “only if” direction needs the separating hyperplane of the next card. In a finite tree, NA holds if and only if the one-period model at every reachable node is arbitrage-free. A local arbitrage at such a node can be entered only after the node is reached, which is a predictable strategy and a global arbitrage, and positive local weights multiplied along each path give a global EMM. The theorem is a statement about finite models. Infinite state spaces, infinite horizons and continuous trading need admissibility conditions and the no-free-lunch forms of the theorem, which this proof does not cover.

Definition A hyperplane in $\mathbb{R}^m$ is a set $\{x : a^{\top} x = b\}$ with $a \ne 0$. The converse of the theorem uses the following strong separation result. Let $C \subseteq \mathbb{R}^m$ be nonempty, closed and convex with $0 \notin C$, and let $a$ be the point of $C$ nearest to $0$, which exists because $C$ is closed and is unique because $C$ is convex. Then

for every $c \in C$. Indeed, for $c \in C$ and $0 < \lambda \le 1$ the point $a + \lambda(c - a)$ lies in $C$, so

Expanding and dividing by $\lambda$ gives $2a^{\top}(c - a) + \lambda \lVert c - a \rVert^2 \ge 0$, and letting $\lambda \to 0$ gives the inequality.

For the proof of the theorem, let $\Delta$ be the unit simplex, the set of vectors in $\mathbb{R}^m$ with nonnegative coordinates that add up to $1$. Under NA the sets $G$ and $\Delta$ are disjoint, because every point of $\Delta$ is nonzero and nonnegative. The set $C = \Delta - G$ of all differences $x - g$ with $x \in \Delta$ and $g \in G$ is convex, it is closed because $\Delta$ is compact and the subspace $G$ is closed, and it does not contain $0$. The result gives a vector $a$ with

for all $x \in \Delta$ and $g \in G$. Since $G$ is a subspace, $g$ can be replaced by $s g$ for every real $s$, which forces $a^{\top} g = 0$ on $G$. Then $a^{\top} e_j > 0$ for every unit vector $e_j$, which lies in $\Delta$, so every $a_j > 0$, and

is a strictly positive probability vector orthogonal to $G$, an EMM. The hyperplane $\{x : a^{\top} x = 0\}$ contains $G$ and leaves all of $\Delta$ strictly on its positive side. Compactness of $\Delta$ matters, since two disjoint closed convex sets that are both unbounded need not be strictly separated.