Quantitative Analysis.
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I.Basic math.
II.Pricing and Hedging.
1.Basics of derivative pricing I.
2.Change of numeraire.
3.Basics of derivative pricing II.
4.Market model.
5.Currency Exchange.
6.Credit risk.
7.Incomplete markets.
A.Single time period discrete price incomplete market.
a.Existence of pricing vector.
b.Uniqueness of pricing vector.
c.Bid and ask.
B.Coherent measure.
C.Incomplete market with multiple participants.
D.Example: uncertain local volatility.
III.Explicit techniques.
IV.Data Analysis.
V.Implementation tools.
VI.Basic Math II.
VII.Implementation tools II.
Bibliography.
Forum Notation Index Contents

Existence of pricing vector.

efinition

Given the market structure MATH the set of acceptable opportunities MATH is defined byMATH

The MATH is matrix multiplication. The MATH is k-th component.

Remark

The assumption that the market MATH without acceptable opportunities is equivalent to the assumption that for any $x$ such that MATH eitherMATH orMATH orMATH Observe that if the last inequality is the only one that prevents $x$ from being as acceptable opportunity then the $x$ may be simply scaled down (remember that the $f_{k}$ is negative):MATH Such $y$ taken in place of $x$ becomes an acceptable opportunity. Hence, we introduce the following definition.

Definition

Given the market structure MATH the assumption of "no acceptable opportunities" is the requirement that the setMATH is empty. Note the abesence of stress measures in the present definition (see the last remark).

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

Proposition

There are no acceptable opportunities iff there exists a pricing vector.

In other words MATH iff there exists w s.t.MATH

Proof

(If there exists w then there are no strictly acceptable x). We assumeMATH orMATH It follows from MATH that we assumeMATH Since w is striclty positive the components MATH cannot be all non negative and one of them striclty positive.

Proof

(If there are no strictly acceptable x then there exists w). We assume that the conditionsMATH are never satisfied.

We introduce the transformation MATH. LetMATH The starting assumption is equivalent to MATH We would like to conclude thatMATH If there exists MATH then MATH and MATH s.t. $Ax_{0}=y$ and $Ax_{1}=y$. Hence, it is enough to show thatMATH If that is not true then a portfolio $x_{1}$ with the property math may be turned into MATH in all components by subtracting a portfolio $x_{0}$ that has zero initial cost. This contradicts the assumption because MATH and, consequently, $x_{0}$ is an acceptable opportunity:MATH We concludeMATH

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

Note that the vector $w$ is defined as a normal vector to some separation hyperplane. Set MATH and consider the consequence of (*)MATH

If we assume existence of a riskless asset then there is no choice but to conclude $C=1$. Hence, the $w^{T}P$ represents some convex combination of the original measures and the equalityMATH means that $w^{T}P$ is a risk neutral measure.

Definition

Given the set of available intruments MATH we introduce the set of risk neutral probability measures MATH s.t.MATH

Corollary

If there are no acceptable opportunities and there is a riskless asset thenMATH

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

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

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

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

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

Remark

Note that the last proof expands to the case of continuous random variable x if we have weak compactness of the set of probability measures $\left\{ P\right\} $.





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