Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 32 / 1

The positive state-price density alternative characterizes one-period arbitrage. Interpret $A_0$ as the vector of current asset prices and $A_1$ as their terminal payments. A fixed vector of holdings produces initial cost $x^{\top }{A}_0$ and terminal payment $x^{\top }{A}_1$. The first alternative is a nonpositive-cost investment with a nonnegative payment and strict improvement either initially or on an event of positive probability. A strictly positive state-price density prices every such payment and excludes both forms of arbitrage, as proved below.

If there is a traded bank account costing one and paying a deterministic $R>0$, the density satisfies $1=R\mathbb{E}\nu$. Hence $dQ/dP=R\nu$ defines an equivalent martingale measure andConversely this pricing relation under an equivalent martingale measure gives $\nu =R^{-1}dQ/dP$. Without such an account, $\nu$ is a pricing density, not necessarily a probability density or even an integrable random variable by itself; the stated requirement is integrability of $\nu A_1$. This distinction keeps the abstract alternative valid without an unstated asset paying one.

Suppose a strictly positive pricing density exists. Its weighted payment for any fixed holdings is integrable, andIf the terminal payment is nonnegative and strictly positive on an event of positive probability, positivity of $\nu$ makes this expected value strictly positive. It cannot therefore have nonpositive initial cost. Thus the first form of arbitrage is incompatible with a positive pricing density. The existence argument in part (b) proves that this form occurs whenever separation within the payment span obstructs a density.

The same pricing relation gives $x^{\top }{A}_0\ge 0$ whenever the terminal payment is nonnegative. Therefore negative initial cost also contradicts a strictly positive state-price density, even if the terminal payment is identically zero.

This second form is needed for directions invisible in the terminal payments. Let $L$ be the linear span of the essential values of $A_1$, equivalentlyIf $A_0\notin L$, let $v$ be its nonzero orthogonal projection on $L^{\perp }$ and set $x=-v$. Then $x^{\top }{A}_1=0$ almost surely while $x^{\top }{A}_0=-\Vert v{\Vert }^2<0$. Hence the second form exists in this case. Part (b) handles all remaining cases.

We prove exhaustiveness, including existence of the positive density when neither form of arbitrage occurs. Use the subspace $L$ from part (a)(ii), and considerThis is a nonempty convex cone closed under positive scaling: ${\nu }_0=(1+\Vert A_1\Vert {)}^{-1}$ supplies one element and guarantees weighted integrability even when $A_1$ itself has no first moment. Crucially, $\mathcal{K}$ is relatively open in $L$; it need not be closed.

To prove openness at a point $z=\mathbb{E}[\nu A_1]$, put $Z=\nu A_1$. The vectors $\mathbb{E}[hZ]$, for bounded real measurable $h$, span $L$. Indeed, a vector orthogonal to all of them would satisfy $\mathbb{E}[h{u}^{\top }Z]=0$ for every such $h$, forcing $u^{\top }Z=0$ almost surely. Since $\nu >0$, this says $u\in L^{\perp }$. Choose finitely many bounded $h_j$ for which $\mathbb{E}[h_{j}Z]$ form a basis of $L$. For coefficients $t_j$ sufficiently small,This gives a relative neighbourhood of $z$. It also proves the openness step behind the strictly positive barycentre cone lemma without assuming a closed moment cone.

If $A_0\notin L$, the second arbitrage form was already constructed. If $A_0\in L\setminus \mathcal{K}$, the separation of a point and an open convex set, applied inside the finite-dimensional space $L$, gives a nonzero $x\in L$ with $x^{\top }{A}_0\le {\inf }_{z\in \mathcal{K}}{x}^{\top }z$. Positive scaling in the cone forcesWe claim $x^{\top }{A}_1\ge 0$ almost surely. Otherwise $E=\{x^{\top }{A}_1<0\}$ has positive probability, and the admissible positive weights ${\nu }_m={\nu }_0(1+m{1}_E)$ givecontradicting the separation inequality. Since $x\in L$ is nonzero, $x^{\top }{A}_1$ cannot vanish almost surely; thus it is strictly positive on an event of positive probability. This is the first arbitrage form.

Consequently absence of both forms forces $A_0\in \mathcal{K}$, which supplies the required $\nu$. Conversely either form excludes every such density by part (a). We have proved exactly one of arbitrage or a strictly positive pricing density holds. If $L=\{0\}$, the same argument reduces to the elementary cases $A_0=0$ and $A_0\ne 0$, so no nondegeneracy assumption has been hidden.