Existence of pricing vector.

We are going to drop the stress measures from further considerations of this chapter.

If the projections of vectors $\QTR{cal}{P}_{k}$ on the plane

are pointing in all directions then

may be replicated by a positive linear combination of $\QTR{cal}{P}_{k}$ . Otherwise, if the projections of $\QTR{cal}{P}_{k}$ are in the same half-plane with respect to any subdivision to half-planes in

then such replication is not possible because then

would have a non-zero projection on the plane.

It is now clear why we need the condition ( Basic coherence condition ): If all the $\QTR{cal}{P}_{k}$ lie in the plane then the replication is never possible.

Proof

(If there exists then there are no strictly acceptable x). We assume MATH Thus MATH Since $w_{k}>0$ this implies MATH Therefore MATH

Proof

(If there are no strictly acceptable then there exists ). We assume that the conditions MATH are never satisfied together.

We introduce the transformation $y=\QTR{cal}{P}x$ . Let MATH The starting assumption is equivalent to MATH We would like to conclude that MATH If there exists then and s.t. and . Hence, it suffices to show that MATH We assume existence of $x_{0},x_{1}$ with the properties $x_{0}\in C_{0}$ , $x_{1}\in C_{1}$ , and derive the contradiction: MATH We conclude MATH

The $\QTR{cal}{P}C_{0}$ is an affine set and $\QTR{cal}{P}C_{1}$ is a convex set. Hence, there is a hyperplane that strictly separates these two sets: $\exists w$ such that MATH By definition of $C_{1}$ and $\QTR{cal}{P}$ , for every , the components $y_{k}$ are non-negative. Hence, . Also, for every $x\in C_{0}$ MATH and $C_{0}$ has dimension N-1. Hence, MATH for some constant . This concludes the proof.

Note that the vector

is defined as a normal vector to some separating hyperplane. Set

and consider the consequence of

If we assume existence of a riskless asset then there is no choice but to conclude

. Hence, the $w^{T}P$ represents some convex combination of the original measures and the equality MATH

means that $w^{T}P$ is a risk neutral measure.

Here $conv_{k}P_{k}$ refers to the convex hull across various ways to model the market (across the index

Proof

Even though the corollary above clearly follows from considerations of the present section, such considerations are not easily extensible to more complex situations. Hence, we give a more generic proof.

For any portfolio with the property we introduce the set MATH Clearly, by assumption of no acceptable opportunities MATH The next task is to prove that MATH We argue by contradiction. Assume and introduce MATH By the assumption we have MATH By the theorem ( Separating hyperplane theorem ) there exists s.t. MATH From the latter inequality we derive that $u\geq0$ . Then from the former inequality we conclude MATH But is another portfolio with the property . Hence, we arrived to contradiction with the assumption of no acceptable opportunities. We conclude that any finite intersection is non empty.

We next want to prove that the intersection is not empty for any countable collection of . We introduce the linear subspace MATH The may be empty. Any x s.t. may be represented as a sum MATH where the $l\in L$ and the y has the property MATH Also, MATH Hence, we will restrict further proof to only (switch to the orthogonal complement of ). For the set always has an interior. Let $\mu$ be the Lebesgue measure acting in the space of : . By the structure of $C\left( y\right)$ the mapping is continuous with respect to y. We choose some sequence MATH where $B_{1}$ is the unit ball. By taking a subsequence we have MATH for some $y_{0}$ . We have MATH Since $C\left( y\right)$ always has an interior we also have MATH By taking further subsequence and using continuity of we have MATH where we use the notation for the difference of sets. Note, that has properties of distance between sets. Hence, cannot converge to zero (because and are $\mu-$ small and is at least as large as ). Consequently, MATH cannot converge to zero. Hence, we proved that is not empty for any countable collection .

To see that the same statement holds for any collection we note that there exists a countable everywhere dense set on and is continuous. Hence, any is infinitely close to some point $z\in D$ and we can always have MATH for however small $\varepsilon$ . Since we conclude MATH

It remains to note that the structure of the set $C\left( x\right)$ is such that the result extends from to general .