A Simple Computational Approach to the Fundamental Theorem - Hikari

Applied Mathematical Sciences, Vol. 6, 2012, no. 72, 3555 - 3562

A Simple Computational Approach to the Fundamental Theorem of Asset Pricing Cherng-tiao Perng Department of Mathematics Norfolk State University 700 Park Avenue, Norfolk, VA 23504 [email protected] Abstract In this short note, we will establish the fundamental theorem of asset pricing (FTAP) by basic linear algebra. More precisely, we prove Stiemke’s Theorem, which is equivalent to FTAP. For comparison purpose, many existing proofs rely on linear programming, the separating hyperplane theorem, or some technical algorithm. In deriving our proof, we used a version of separating hyperplane theorem in dimension two, which is easy and can be shown directly.

Mathematics Subject Classification: 15A39, 91B02, 91B25 Keywords: the fundamental theorem of asset pricing, Stiemke’s theorem

1

Introduction

Relating the existence of an equivalent martingale measure to the no-arbitrage condition, the Fundamental Theorem of Asset Pricing plays a central role in Mathematical Finance. For example, the seminal papers by Black, Scholes and Merton (Nobel Prize for Economics, 1997) on the pricing of options were based on No Arbitrage principle. The pioneering work related to the Fundamental Theorem of Asset Pricing was done around the late eighties; we merely list some names with years without mentioning the precise works, since we will not need them in this paper: Cox and Ross’ method of risk neutral valuation (1976), Harrison and Kreps (1979) and Harrison and Pliska (1981, 1983). For the purpose of this paper, we don’t need the martingale measure or any probability theory (which involves basically a translation using different terminologies; see [9] or any textbook on financial economics). We note that,

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mathematically, the Fundamental Theorem of Asset Pricing in the single period model is equivalent to Stiemke’s Theorem ([11]), which follows directly from a version of separating hyperplane theorem. There are various approaches to proving this: some involves non-constructive method such as Hahn-Banach Theorem (this requires additional work, as the versions of this theorem from any standard textbook do not directly apply). Other proofs include the ones by applying Farkas’s Lemma ([5],[8]), which appears to be stronger (i.e. it implies Stiemke, but not the other way around) ([10]), by careful study of polyhedral cones ([4],[6]), by linear programming, or by Fredhlom’s “alternative theorem” ([2]). In this note, we will use a constructive method based on basic linear algebra. To see the simplicity of our method, one may wish to compare the work [1], which used a technical algorithm devised by A. Dax ([3]). Our method resembles the approach via Arrow-Debreu securities in the manner of extending the securities to make the market complete (cf. [9]), but our proof can be done explicitly by simple geometric consideration without those terminologies and we have avoided the use of linear programming.

2

Preliminaries

We briefly recall some definitions. A large part of notations follow those of [9]. A single-period market model consists of a finite number of n securities (or assets) S1 , S2 , · · · , Sn , and their values at time 0 and time 1. At time 0, the prices are known, but at time 1, the prices are random variables defined on a sample space Ω = {ω1 , ω2 , · · · , ωm } consisting of states. We assume that the probability P(ωi ) > 0 for each i = 1, 2, · · · , m, but as far as this article is concerned, we won’t need any other notions of probability. More precisely, we have the following Definition 2.1 Denote by Sj (t, ωi) the security price at time t and the state ωj . Here we are concerned with only t = 0 and t = 1, and we assume that j = 1, 2, · · · , n and i = 1, 2, · · · , m. We denote the vector [S1 (0, ωi), · · · , Sn (0, ωi)] by S(0), which represents the time 0 prices of securities, since we assume that initially the prices do not depend on ωi . And we denote by S(1, Ω) the matrix (aij ) where aij = Sj (1, ωi ). A trading strategy (or portfolio) represented by a column vector θ = [θ1 , θ2 , · · · , θn ]T is a buy and sell consideration of the securities, thus the corresponding price of the portfolio at time 0 is given by S(0)θ = S1 (0)θ1 + S2 (0)θ2 + · · · + Sn (0)θn ,

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where Sj (0) = Sj (0, ωi) for 1 ≤ i ≤ m and 1 ≤ j ≤ n. Notations. X ≥ 0 means that all the entries of X are nonnegative, X > 0 means that all the entries are nonnegative and there exists at least one positive entry, and X  0 means that all the entries are positive (similarly for 0 (ii) S(0)θ < 0 and S(1, Ω)θ = 0 Note. The condition (ii) was not mentioned in [9] but tacitly assumed in their proof of the Fundamental Theorem of Asset Pricing. The author of [7] defines (i) to be type B Arbitrage, and (ii) as type A Arbitrage. Definition 2.3 A state price vector ψ is a strictly positive vector [ψ(ω1 ), ψ(ω2 ), · · · , ψ(ωm )] for which S(0) = ψS(1, Ω).

2.1

The Fundamental Theorem of Asset Pricing

We are now ready to state the following Theorem 2.4 (Fundamental Theorem of Asset Pricing) Keeping the above notations, the model is arbitrage free if and only if there exists a state price vector ψ  0 such that S(0) = ψS(1, Ω).

2.2

Stiemke’s Theorem

Stiemke’s Theorem ([11]) concerns the solutions of system of linear inequalities, which is one of the famous theorems of alternative (for these formulations, see eg. [8]) in the theory of linear programming or optimization. For our purpose, it is convenient to state it in the following form. Theorem 2.5 (Stiemke’s Theorem) Let A be an m × n matrix. Then there does not exist any column vector x such that Ax > 0, if and only if there exists a row vector y  0 such that yA = 0.

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Remark. Geometrically, the above theorem says that the linear span V of m the column vectors of A intersects Rm + trivially (i.e. V ∩ R+ = {0}) if and only T if there exists y ∈ Rm ++ such that y A = 0.

2.3

Proof of the Fundamental Theorem of Asset Pricing Using Stiemke’s Theorem

Proof. (Sufficiency) If there exists a state price vector ψ with S(0) = ψS(1, Ω), then S(1, Ω)θ > 0 ⇒ ψS(1, Ω)θ > 0 ⇒ S(0)θ > 0; similarly S(1, Ω)θ = 0 ⇒ ψS(1, Ω)θ = 0 ⇒ S(0)θ = 0. Therefore, by Definition 2.2, there are no arbitrage opportunities. (Necessity) If there are no arbitrage opportunities, then there does not exist any portfolio θ satisfying (i) or (ii) of Definition 2.2, which is equivalent to the following condition: There does not exist any θ such that   S(1, Ω) θ > 0. −S(0) Applying Stiemke’s Theorem to the matrix   S(1, Ω) A= , −S(0) we see that there exists a vector y  0 such that yA = 0. If we normalize the last component of y and write y = [ψ, 1], then we get yA = 0 ⇒ S(0) = ψS(1, Ω), as required.



Remark. By essentially reversing the above argument, we can show that FTAP implies Stiemke’s Theorem. Hence the two theorems are equivalent.

3

Constructive Proof of Stiemke’s Theorem

As the main result of this paper, we provide a constructive proof of Stiemke’s Theorem. To facilitate the proof we make precise some definitions, and establish some lemmas.

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Definition 3.1 A set W is called a linear polyhedral cone in R if there exist column vectors v1 , v2 , · · · , vk in R such that    αi vi , αi ≥ 0, ∀i . W = v|v= i

Remark. If  ≤ 2, then W is easily seen to be a closed subset of R . It is true in general that W is a closed subset. Since we will only need the case for  = 2, we do not include here the proof of this general fact (for an elementary account of this, see [10]). Lemma 3.2 Let W be a linear polyhedral cone generated by a set of vectors Λ in R2 , and V be the linear polyhedral cone generated by the standard basis {e1 , e2 }, i.e. V = R2+ . If W ∩ V = {0}, then there exists a nonzero vector w ∈ R2 such that the extended linear polyhedral cone spanned by Λ ∪ {w, −w} intersects V at {0}. Proof. It is convenient to possibly enlarge W be adding two more generators: −e1 and −e2 ; let’s call this new linear polyhedral cone W  . Then it is easy to see that W  is bounded by two boundary rays enclosing an angle between 90◦ and 180◦ . We may assume that the two boundary rays have slope mi , with −∞ ≤ mi ≤ 0, i = 1, 2, where by −∞ we mean the boundary ray is generated by −e2 . Using this convention, the choice of w is easy to describe: if the two boundary rays have the same slope m, then W  must be a half-space, and we can choose w to be one of the directional vectors of a line with slope m. If the two boundary rays have different slopes m1 and m2 with m1 < m2 , then we can choose w to be one of the directional vectors of a line with slope m such that m1 < m < m2 .  Lemma 3.3 Let V be a linear polyhedral cone in Rn generated with nonnegative coefficients by a finite set of vectors {vi }, i ∈ Λ, |Λ| < ∞. Furthermore let n·x ≥ 0 be a half-space constraint, where n·x  = 0 is a homogeneous equation of n  a hyperplane in R . Then V = {v | v = i αi vi , αi ≥ 0, i ∈ Λ and n · v ≥ 0} is again a linear polyhedral cone, i.e. generated with nonnegative coefficients by a finite set of vectors. Proof. We partition the generators for V into 3 types, according as whether − } (a priori, n · vi > 0, = 0, or < 0 : {v1+ , · · · , vk+ } ∪ {v10 , · · · , v0 } ∪ {v1− , · · · , vm either of the sets could be empty). By multiplying by a positive constant, we may assume that their inner products with n are either 1, 0 or −1. It suffices to prove the

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Claim. The vectors v1+ , · · · , vk+ , v10 , · · · , v0 , and vi+ + vj− , where 1 ≤ i ≤ k and 1 ≤ j ≤ m, form a set of generators for V  . Proof of Claim. Any nonnegative linear combination of the above vectors is − + clearly in V  . Conversely, if v = α1 v1+ + · · · + αk vk+ + β1 v1− + · · · + βm vm 0 0  γ1 v1 + · · · + γ v (all coefficients being nonnegative) is in V , then we have α1 + · · · + αk − β1 − · · · − βm ≥ 0, using n · v ≥ 0. By subtracting suitable combination of v·+ and all the combination of v·0 , we easily reduce to the case − v = α1 v1+ +· · ·+αk vk+ +β1 v1− +· · ·+βm vm with α1 +· · ·+αk −β1 −· · ·−βm = 0 and it suffices to prove that this can be generated by vectors of the form  + − wij := vi + vj , i.e. v = ij αij wij , where αij ≥ 0. This can be done by rearranging terms: for example, if α1 ≥ β1 , we can let α11 = β1 , α1 = α1 − β1 − and write v = α11 (v1+ +v1− )+α1 v1+ +· · ·+αk vk+ +β2 v2− +· · ·+βm vm ; if α1 < β1 , + −  we can let α11 = α1 , β1 = α1 + β1 , and write v = α11 (v1 + v1 ) + α2 v2+ + · · · + − αk vk+ +β1 v1− +β2 v2− +· · ·+βm vm . Dropping the first term α11 (v1+ +v2− ), we have less terms in the combination, while keeping α1 +α2 +· · ·+αk −β2 −· · ·−βm = 0 (resp. α2 + · · · + αk − β1 − β2 − · · · − βm = 0). Hence by induction, we are able to replace the terms v·+ or v·− by vi+ + vj− with suitable i and j and αij ≥ 0.  Proof of Theorem 2.5. Given an m × n matrix A, suppose that there does not exist any column vector x such that Ax > 0. We assert that there exists a row vector y  0 such that yA = 0. Clearly we may assume that the column vectors are linearly independent. Furthermore, our assumption implies that m > n: since the rank of A equals n, we have necessarily m ≥ n by rank consideration; if m = n, then A is invertible, and there always exists x such that Ax > 0. We will prove our assertion by induction on d = m − n. Case 1. d = 1. Choose a permutation matrix  P such that the first n rows of B , where B is an n × n invertible P A is linearly independent, say P A = C matrix. Let Q be the inverse of B. Then P AQ is of the form   I P AQ = , C where C  = CQ is a row vector. It follows that all the entries of C  must be negative, otherwise there will be a column vector of the form Ax in AQ satisfying Ax > 0 contradicting our assumption. Letting C  = [−a1 , · · · , −an ] with ai > 0, ∀i, and y0 = [a1 , · · · , an , 1], it is straightforward to check that y0 P AQ = 0. Therefore y := y0 P satisfies yA = 0 and y  0, thus Case 1 is proven.

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Case 2. d = 2. Choose P and Q as in Case 1 such that P AQ is of the form   I  A := P AQ = , C where C  is a 2 × n matrix. Thinking of C  as given by n column vectors in R2 , it is clear by assumption that the vector formed by any nontrivial linear combination with nonnegative coefficients must contain a negative entry. Geometrically this says the linear polyhedral cone determined by the column vectors of C  intersects the first quadrant (including the boundary) of R2 at 0. By Lemma 3.2, there exists a nonzero vector w such that the linear polyhedral cone W positively spanned (i.e. the coefficients are nonnegative) by the column vectors of C  and {w, −w} satisfies W ∩ R2+ = {0}.   I 0 . It is easy to see Now define an m × (n + 1) matrix A1 by A1 = C w that rank(A1 ) = n + 1 and that A1 x1 > 0 is false for any x1 ∈ Rn+1 , using the property of w constructed above. Therefore by Case 1, there exists y0  0 such that y0 A1 = 0. But y0 A1 = 0 ⇒ y0 A = y0 P AQ = 0. Therefore, as in Case 1, the vector y := y0 P satisfies y  0 and yA = 0. This establishes Case 2, i.e. when d = 2. To prove the theorem, it suffices to show that for any d > 2, we can describe a procedure analogous to Case 2, so at to reduce to the case of d − 1, then the result follows by induction. Starting with A with d = m − n > 2, we can find a permutation matrix P and an invertible matrix Q such that P AQ is of the following form   I P AQ = , C where C  is an (m − n) × n matrix. Let V be the linear polyhedral cone formed by taking nonnegative linear combination of the column vectors of P AQ. Let W be the subset of V such that the last m−(n+2) components are nonnegative. By Lemma 3.3, W is another linear polyhedral cone, since it is obtained from V by taking successive halfspace constraints. Necessarily either the (n + 1)-th or (n + 2)-th component of any nonzero vector in W must be negative. Put in a different way, by restricting W to the (n + 1)-th and (n + 2)-th components, we obtain a 2 × n matrix, satisfying the same condition as in Case 2, if W = 0. Now we just need to consider two subcases: Subcase 1. If W = 0, we can use the argument in Case 2 to choose w so that adding a column vector of the form [0, wT , 0]T to P AQ reduces to the case d − 1. Here and below, the column vector is of size n + 2 + (m − n − 2). Subcase 2. If W = 0, this would mean that for any nonzero vector in V , the last m − n − 2 entries contain at least one negative number. In this case, we

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simply define w = [1, −1]T and add the column vector [0, wT , 0]T to P AQ. Hence by induction, we have shown the direction: there does not exist x such that Ax > 0 implies there exists y  0 such that yA = 0. To prove the other direction, we need to show that if there exists y  0 such that yA = 0, then there does not exist x such that Ax > 0. This can be seen as follows: if y  0 such that yA = 0 and Ax > 0 are both true, then Ax > 0 ⇒ 0 = yAx > 0, a contradiction. 

References [1] F. Acedo, F. Benito, A. Falc´o, A. Rubia and J. Torres, A computational approach to the fundamental theorem of asset pricing in a single-period market, Computational Economics, 18, number 3, (2002), 233 - 249. [2] A. Ben-Israel, Notes on linear inequalities, I: the intersection of the nonnegative orthant with complementary orthogonal subspaces, (1963) O. N. R. Research Memorandum, No. 78, Northwestern University [3] A. Dax, An elementary proof of Farkas’s lemma, SIAM Rev., 39 (1997), 503 - 507. [4] D. Gale, Convex polyhedral cones and linear inequalities, Cowles Commission for Research in Economics, Monograph No. 13, New York, 1951, 287 - 297 [5] D. Gale, The theory of linear economic models, McGraw-Hill, New York, 1960. [6] M. Gerstenhaber, Theory of convex polyhedral cones, Cowles Commission for Research in Economics, Monograph No. 13, New York, 1951, 298 - 316 [7] D. Luenberger, Investment Science, Oxford University Press, 1998, 240 241. [8] O. Mangasarian, Nonlinear programming, McGraw-Hill, New York, 1969. [9] H. Panjer, et al., Financial Economics: With Applications to Investments, Insurance and Pensions, The Actuarial Foundation, The Society of Actuaries, Illinois, 2001. [10] C. Perng, A note on Farkas’s lemma and related theorems, submitted. ¨ [11] E. Stiemke, Uber positive L¨osungen homogener linearer Gleichungen, Math. Ann., 76 (1915), 340 - 342. Received: March, 2012