The Fundamental Theorem of Asset Pricing with d Risky Assets
In the stochastic discount factor notebook we proved the SDF existence theorem in a finite state space \mathbb{R}^{S} using a direct separation argument. That proof exploits two finiteness assumptions simultaneously: a finite number of states (|\Omega| = S) and a finite number of risky assets (d assets with strategies in \mathbb{R}^{d}).
This notebook removes the finite-state assumption. We work on an arbitrary probability space (\Omega, \mathcal{F}, P) with payoffs in L^{p} = L^{p}(\Omega, \mathcal{F}, P) for any p \in [1, \infty). The strategy space remains finite-dimensional: d risky assets with deterministic portfolio weights \theta \in \mathbb{R}^{d}. This finite dimensionality is the key structural feature: the Bolzano-Weierstrass theorem implies that simple no-arbitrage (NA) is sufficient for the existence of a risk-neutral measure, without strengthening to no free lunch (NFL). The companion notebook FTAP: Infinite-Dimensional Strategies treats the harder case where the strategy space itself is infinite-dimensional, and NA must be strengthened to NFL.
The L^{p} Setting
Let p \in [1, \infty) and let q denote the conjugate exponent defined by 1/p + 1/q = 1 (with q = \infty when p = 1). We work in L^{p} = L^{p}(\Omega, \mathcal{F}, P), the space of random variables with finite p-th moment, equipped with the norm \lVert{x}\rVert_{p} = \operatorname{E}(|x|^{p})^{1/p}. By the Riesz representation theorem, the dual space satisfies (L^{p})^{*} = L^{q}.
The positive cone L^{p}_{+} = \{x \in L^{p} : x \geq 0 \text{ a.s.}\} has empty interior for all p \in [1, \infty) on a non-atomic probability space. This rules out a global separation argument of the type used in \mathbb{R}^{S}, but the finite dimensionality of \mathbb{R}^{d} provides an alternative route.
The One-Period Model
There are d risky assets with price changes \Delta S = (\Delta S^{1}, \ldots, \Delta S^{d}) \in (L^{p})^{d} and a risk-free numeraire with return normalized to zero. A trading strategy is a vector \theta \in \mathbb{R}^{d} of portfolio weights, yielding terminal payoff \theta \cdot \Delta S := \sum_{i=1}^{d} \theta^{i} \Delta S^{i} \in L^{p}.
Define the cone of claims super-replicable at non-positive cost: A = \{y \in L^{p} : y \leq \theta \cdot \Delta S \;\text{a.s. for some}\; \theta \in \mathbb{R}^{d}\}, and let \bar{A} denote its closure in L^{p}. Then A = R - L^{p}_{+} where R = \{\theta \cdot \Delta S : \theta \in \mathbb{R}^{d}\}, and -L^{p}_{+} \subseteq A (take \theta = 0).
No-Arbitrage Conditions
Assumption 1 (No Arbitrage) A \cap L^{p}_{+} = \{0\}.
Assumption 2 (No Free Lunch) \bar{A} \cap L^{p}_{+} = \{0\}.
NFL is in general stronger than NA. In the companion FTAP: Infinite-Dimensional Strategies notebook, the two conditions are genuinely distinct when the strategy space is infinite-dimensional. Here they coincide:
Property 1 (No Arbitrage Implies Closedness) In the one-period model with d risky assets and payoffs in L^{p}, NA implies A is closed in L^{p}, so NA \Leftrightarrow NFL.
This is the key structural fact: the strategy space \mathbb{R}^{d} is finite-dimensional, so bounded sequences of strategies have convergent subsequences. The proof is contained in Property 2 below.
The Fundamental Theorem
Property 2 (Fundamental Theorem of Asset Pricing) The following conditions are equivalent:
- A \cap L^{p}_{+} = \{0\} (no arbitrage),
- A \cap L^{p}_{+} = \{0\} and A = \bar{A} (no arbitrage and A is closed),
- \bar{A} \cap L^{p}_{+} = \{0\} (no free lunch),
- there exists \tilde{P} \sim P with d\tilde{P}/dP \in L^{q} such that \operatorname{E}_{\tilde{P}}[\Delta S] = 0.
Condition (d) says there is an equivalent martingale measure \tilde{P} under which each risky asset has zero expected return, with Radon-Nikodym density in the dual space L^{q}. The implications (b) \Rightarrow (c) \Rightarrow (a) are trivial; the substance is (a) \Rightarrow (b) and (c) \Rightarrow (d).
Two special cases are worth noting. When p = 1, the density d\tilde{P}/dP \in L^{\infty} is bounded, recovering the result of Kabanov and Stricker (2001). When p = 2, the density is square-integrable: d\tilde{P}/dP \in L^{2}.
Proof Tools
The proof uses two results.
Lemma 1 (Hahn-Banach Separation in L^{p}) Let C \subseteq L^{p} be a non-empty closed convex cone containing 0, and let x \in L^{p}_{+} with x \notin C. Then there exists z \in L^{q} such that \operatorname{E}(z\, y) \leq 0 < \operatorname{E}(z\, x) \quad \text{for all } y \in C.
This follows from the Hahn-Banach separation theorem in the locally convex space L^{p}, combined with the duality (L^{p})^{*} = L^{q}. When p = 1, the dual (L^{1})^{*} = L^{\infty} gives z \in L^{\infty} normalizable to z \leq 1 a.s. When p = 2, the Hilbert-space self-duality (L^{2})^{*} = L^{2} means the separator lies in L^{2} itself.
Lemma 2 (Halmos-Savage Theorem) Let \{\mu_{\alpha} : \alpha \in I\} be a family of measures on (\Omega, \mathcal{F}), each absolutely continuous with respect to P. If for every measurable A with P(A) > 0 there exists some \alpha with \mu_{\alpha}(A) > 0, then there exists a countable subfamily with the same null sets as the full family.
Proof of Property 2
(a) \Rightarrow (b): NA implies A is closed. Suppose y^{(n)} \in A with y^{(n)} \to y in L^{p}. Each y^{(n)} \leq \theta^{(n)} \cdot \Delta S a.s. for some \theta^{(n)} \in \mathbb{R}^{d}. We show \theta^{(n)} is bounded.
Let N = \ker(\theta \mapsto \theta \cdot \Delta S) \subseteq \mathbb{R}^{d} and write \theta^{(n)} = \phi^{(n)} + \eta^{(n)} with \phi^{(n)} \in N^{\mathrel\bot} and \eta^{(n)} \in N. Since \eta^{(n)} \cdot \Delta S = 0 a.s., the payoff satisfies \theta^{(n)} \cdot \Delta S = \phi^{(n)} \cdot \Delta S, so it suffices to show \phi^{(n)} is bounded in N^{\mathrel\bot}.
Suppose for contradiction that \|\phi^{(n)}\| \to \infty along some subsequence. Normalize: \hat\phi^{(n)} = \phi^{(n)}/\|\phi^{(n)}\| lies on the unit sphere of N^{\mathrel\bot}, which is compact. Pass to a subsequence with \hat\phi^{(n)} \to \hat\phi \in N^{\mathrel\bot}, \|\hat\phi\| = 1. Since \hat\phi^{(n)} \to \hat\phi in \mathbb{R}^{d} and \Delta S \in (L^{p})^{d}, we have \hat\phi^{(n)} \cdot \Delta S \to \hat\phi \cdot \Delta S in L^{p}. Since y^{(n)}/\|\phi^{(n)}\| \to 0 in L^{p}, passing to an a.s.-convergent subsequence of each gives \hat\phi \cdot \Delta S \;\geq\; \lim_{n} \frac{y^{(n)}}{\|\phi^{(n)}\|} \;=\; 0 \quad \text{a.s.} By NA, \hat\phi \cdot \Delta S \in A \cap L^{p}_{+} = \{0\}, so \hat\phi \in N. But \hat\phi \in N^{\mathrel\bot} and \|\hat\phi\| = 1, giving \hat\phi \in N \cap N^{\mathrel\bot} = \{0\}, a contradiction.
Hence \phi^{(n)} is bounded in N^{\mathrel\bot}, so \theta^{(n)} is effectively bounded. Pass to a subsequence with \phi^{(n)} \to \phi. Then \phi^{(n)} \cdot \Delta S \to \phi \cdot \Delta S in L^{p}, so passing to an a.s.-convergent subsequence, y \leq \phi \cdot \Delta S a.s. and y \in A.
(b) \Rightarrow (c): Trivial since A = \bar{A} under (b).
(c) \Rightarrow (d): Since (c) implies (a) and (a) implies (b), A is closed. The cone A is a closed convex cone in L^{p} satisfying -L^{p}_{+} \subseteq A and A \cap L^{p}_{+} = \{0\}.
Step 1: Point-wise separation. Fix any x \in L^{p}_{+} with x \neq 0; by NFL, x \notin A. By Lemma 1, there exists z_{x} \in L^{q} such that \operatorname{E}(z_{x}\, y) \leq 0 \quad \text{for all } y \in A, \qquad \operatorname{E}(z_{x}\, x) > 0. To show z_{x} \geq 0 a.s.: since -\mathbf{1}_{z_{x} < 0} \in -L^{p}_{+} \subseteq A, we have \operatorname{E}(z_{x}\, \mathbf{1}_{z_{x} < 0}) \geq 0. But z_{x} < 0 on \{z_{x} < 0\}, so z_{x}\, \mathbf{1}_{z_{x} < 0} \leq 0 a.s., forcing P(z_{x} < 0) = 0. Normalize so that \lVert{z_{x}}\rVert_{q} = 1, and define \mu_{x}(B) := \operatorname{E}(z_{x}\, \mathbf{1}_{B}) for B \in \mathcal{F}. In particular, for any B with P(B) > 0, taking x = \mathbf{1}_{B} gives \mu_{\mathbf{1}_{B}}(B) = \operatorname{E}(z_{\mathbf{1}_{B}}\, \mathbf{1}_{B}) > 0, so the family \{\mu_{x} : x \in L^{p}_{+},\, x \neq 0\} charges every set of positive P-measure.
Step 2: Extract a countable equivalent subfamily. By Lemma 2, there exists a countable sequence \{z_{x_{n}}\}_{n \geq 1} with the same null sets as the full family \{z_{x}\}.
Step 3: Construct the density. Define \rho = \sum_{n=1}^{\infty} 2^{-n}\, z_{x_{n}}. Since \lVert{z_{x_{n}}}\rVert_{q} = 1, the triangle inequality gives \lVert{\rho}\rVert_{q} \leq \sum_{n=1}^{\infty} 2^{-n} = 1, so \rho \in L^{q}. Moreover \rho > 0 a.s.: if P(\rho = 0) > 0, let B = \{\rho = 0\}; since each z_{x_{n}} \geq 0 and \rho = \sum 2^{-n} z_{x_{n}}, we get z_{x_{n}} = 0 a.s. on B for all n, so \mu_{x_{n}}(B) = 0 for all n. But P(B) > 0, so the family \{\mu_{x}\} charges B (Step 1), contradicting the null-set equivalence of Step 2. Since \rho > 0 a.s. we have \operatorname{E}(\rho) > 0; set d\tilde{P}/dP = \rho/\operatorname{E}(\rho) \in L^{q}. Then \tilde{P} \sim P.
Step 4: \tilde{P} is a martingale measure. Each z_{x_{n}} satisfies \operatorname{E}(z_{x_{n}}\, y) \leq 0 for all y \in A. Since \pm\theta \cdot \Delta S \in R \subseteq A for every \theta \in \mathbb{R}^{d}, applying this to both signs gives \operatorname{E}(z_{x_{n}}\, \theta \cdot \Delta S) = 0 for all \theta. By Hölder’s inequality, |\operatorname{E}(z_{x_{n}}\, \theta \cdot \Delta S)| \leq \lVert{z_{x_{n}}}\rVert_{q}\lVert{\theta \cdot \Delta S}\rVert_{p} = \lVert{\theta \cdot \Delta S}\rVert_{p}, so the series \sum 2^{-n} \operatorname{E}(z_{x_{n}}\, \theta \cdot \Delta S) converges absolutely and by linearity of expectation \operatorname{E}(\rho\, \theta \cdot \Delta S) = 0 for all \theta, hence \operatorname{E}_{\tilde{P}}[\Delta S] = 0.
(d) \Rightarrow (a): Let \xi \in A \cap L^{p}_{+}, so 0 \leq \xi \leq \theta \cdot \Delta S a.s. for some \theta. Then 0 \leq \operatorname{E}_{\tilde{P}}[\xi] \leq \operatorname{E}_{\tilde{P}}[\theta \cdot \Delta S] = \theta \cdot \operatorname{E}_{\tilde{P}}[\Delta S] = 0. Since \xi \geq 0 and \tilde{P} \sim P, we get \xi = 0 a.s.
Why Infinite Strategies Are Harder
The simplicity of this proof rests on one structural feature: the strategy space \mathbb{R}^{d} is finite-dimensional.
NA implies closedness via compactness. The closedness argument works by contradiction. If the norms of effective strategies (those in N^{\mathrel\bot}) grow without bound, normalizing them produces a sequence on the unit sphere of N^{\mathrel\bot}. That sphere is compact, so Bolzano-Weierstrass supplies a convergent subsequence. The limit direction then yields a nonnegative payoff, an arbitrage, contradicting NA. Compactness is what converts NA into closedness of A.
In the companion notebook, the strategy space is an infinite-dimensional subspace X \subseteq L^{p}: bounded sets in infinite-dimensional Banach spaces are not compact, Bolzano-Weierstrass fails, and one can exhibit markets where NA holds but A is not closed. This is why NFL must be imposed as an explicit hypothesis rather than derived from NA.
The same issue appears in dynamic and continuous-time models. Dybvig and Huang (1988) show that a nonnegative wealth constraint (W_{t} \geq 0 a.s. for all t) rules out all arbitrage — including doubling strategies — precisely because it ensures closedness of the attainable payoff cone. Like finite-dimensionality in the static model, it is a structural condition that does the work NFL must do explicitly when the strategy space is unconstrained.