Dahl, Geir

Dahl, Kristina Rognlien

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Dept. of Math./CMA University of OsloPure Mathematics no 05ISSN 0806–2439 August 2012Linear optimization and mathematical financeGeir DahlKristina Rognlien DahlDepartment of Mathematics, University of Oslo.1 IntroductionA goal of this short note is to explain how linear optimization, also calledlinear programming (LP), may be used to solve two basic problems in math-ematical finance concerning arbitrage and dominant trading strategies, re-spectively. The note is an extension of a previous version [1] where we wantto make the presentation self-contained, with complete proofs of the dualitytheorem of linear optimization.We focus on the discrete one-step model in mathematical finance. Whatwe obtain are new proofs based on LP duality, and an efficient computationalapproach for these problems. The idea here is to start with some “natural”LP problems describing the investors problem. This note supplements thepresentation in [3]. (The proofs in [3], see (1.9) and (1.16), are related, butdifferent.)However, as many students of mathematical finance are not familiar withLP, a brief introduction to this topic is in order. In particular, a proofof the important linear programming duality theorem will be given. Ourproof is based on convex analysis, in particular the separating hyperplanetheorem and Farkas’ lemma. For an introduction to linear optimization,including efficient algorithms, we recommend [4] (here a constructive proofof the duality theorem based on the simplex algorithm is given).We treat vectors as column vectors and O denotes the zero vector (ormatrix).12 A preparation: distances and convex setsWe start with some preparations concerning nearest points of sets.Recall from the course MAT1110 that a set S ⊆ Rn is closed if it containsthe limit point of each convergent sequence of points in S. For a nonempty,closed set S ⊆ Rn and a point z ∈ Rn we define dS(z) = inf{‖s− z‖ : s ∈ S}and call this the distance of x to S. We here use the Euclidean norm. Wesay that s0 ∈ S is a nearest point of S to z if ‖s0 − z‖ = dS(z). Onecan show that a nearest point always exists. In fact, dS(z) = dS′(z) whereS ′ = {s ∈ S : ‖s− z‖ ≤ dS + 1} and the minimum distance from z to S′ isattained as S ′ is closed and bounded (i.e., compact) and the Euclidean normis continuous; recall here the extreme value theorem.Thus, closedness of (a nonmpty set) S assures that a nearest point of agiven point exists. But such a point may not be unique. However, for a VERYinteresting class of sets there is a unique nearest point! A set C ⊆ Rn is calledconvex if λx+(1−λ)y ∈ C for all x, y ∈ C and λ ∈ [0, 1]. Geometrically, thisproperty means that whenever we choose two points in the set, say x, y ∈ C,then all points on the line segment between x and y also lie in C.Lemma 1 (Unique nearest point for convex sets) Let C ⊆ Rn be anon-empty closed convex set. Then, for every z ∈ Rn, there is a uniquenearest point c to z in CProof. Assume that both c0 and c1 are nearest points to z in C, and letd = dC(z) = ‖z − c0‖ = ‖z − c1‖. Then c0 and c1 both lie on the boundaryof the closed ball B = {y ∈ Rn : ‖y − z‖ ≤ d} with radius d and center z.But the midpoint c∗ = (1/2)c0 + (1/2)c1 lies in C, as C is convex, and c∗also lies in the interior of B. Therefore ‖c∗ − z‖ < d, a contradiction. Thisproves that a nearest point must be unique.3 The separating hyperplane theoremA hyperplane is a ”generalized plane”. More formally, it is a set H ⊆ Rnof the form H = {x ∈ Rn : aTx = α} for some nonzero vector a and areal number α. The vector a is a normal vector of the hyperplane. In R2, ahyperplane is a line, in R3 it is a plane. We denote Ha,α = H = {x ∈ Rn :2aTx = α}. Also, we define the halfspacesH−a,α := {x ∈ Rn : aTx ≤ α};H+a,α := {x ∈ Rn : aTx ≥ α}.(1)These halfspaces represent the two sides of the hyperplane as we can convinceourselves in R2 or R3.Now, consider two sets S and T in Rn. We say that the hyperplane Ha,αstrongly separates S and T if there is an ǫ > 0 such that S ⊆ H−a,α−ǫ andT ⊆ H+a,α+ǫ or vice versa. In R2, this means that the sets S and T are onseparate sides of the line Ha,α, and neither of the sets intersect the line.CH hyperplanezFigure 1: Strong separationTheorem 2 (Separating hyperplane theorem) Let C ⊆ Rn be a nonemptyclosed convex set. Let z ∈ Rn and assume that z 6∈ C. Then C and z can bestrongly separated.Proof. Consider C and z as indicated above, and let p be the uniquenearest point to x in C (see Lemma 1). Let x ∈ C and let 0 < λ < 1.Since C is convex, (1 − λ)p + λx ∈ C and since p is a nearest point wehave that ‖(1 − λ)p + λx − z‖ ≥ ‖p − z‖, i.e., ‖(p − z) + λ(x − p)‖ ≥‖p−z‖. By squaring both sides and calculating the inner products we obtain‖p − z‖2 + 2λ(p − z)T (x − p) + λ2‖x − p‖2 ≥ ‖p − z‖2. We now subtract3‖p− z‖2 on both sides, divide by λ, let λ → 0+ and finally multiply by −1.This gives the inequality(z − p)T (x− p) ≤ 0 for all x ∈ C. (2)Consider the hyperplane H containing p and having normal vector a := z−p,i.e., H = {x ∈ Rn : aTx = α} where α = aTp. The inequality (2) shows thatC ⊆ H−a,α. Moreover, z 6∈ H−a,α as z 6= p (because z 6∈ C). Now, consider thehyperplane H∗ which is parallel to H (i.e., having the same normal vector)and contains the point (1/2)(z + p). Then it is easy to see that H∗ stronglyseparates z and C as desired.An illustration is in Figure 1. There are several separation theorems forconvex sets, see e.g. [2] and the references given there.4 Farkas’ lemmaUsing the separating hyperplane theorem (Theorem 2) we will prove Farkas’lemma which characterizes when a system of linear inequalities has a solution.It is central in optimization, as we shall see in the next section.Theorem 3 (Farkas’ lemma) Let A be a real m×n matrix and let b ∈ Rm.Then there exists an x ≥ O satisfying Ax = b if and only if for each y ∈ Rmwith yTA ≥ O it also holds that yT b ≥ 0.Proof. Let a1, a2, . . . , an denote the column vectors of the matrix A. Con-sider the set C = {∑nj=1 λjaj : λj ≥ 0 for j = 1, . . . , n} ⊆ Rm; this set iscalled the convex cone generated by a1, a2, . . . , an, see [2]. Then C is closed(this can be checked from the definition of C, but is also follows from Propo-sition 2.5.5 in [2]). We observe that Ax = b has a nonnegative solution x ifand only if b ∈ C.Assume now that x satisfies Ax = b and x ≥ O. If yTA ≥ O, thenyT b = yT (Ax) = (yTA)x ≥ 0 as the inner product of two nonnegative vectors.Conversely, if Ax = b has no nonnegative solution, then b 6∈ C. But then,by the separating hyperplane theorem (Theorem 2), C and b can be stronglyseparated, so there is a nonzero vector y ∈ Rn and α ∈ R with yTx ≥ αfor each x ∈ C and yT b < α. As O ∈ C, we have α ≤ 0. We claim thatyTx ≥ 0 for each x ∈ C: for if yTx < 0 for some x ∈ C, there would be apoint λx ∈ C with λ > 0 such that yT (λx) < α, a contradiction. Therefore4(as aj ∈ C) yTaj ≥ 0 so yTA ≥ O. Since yT b < 0 we have proved the otherdirection of Farkas’ lemma.Farkas’ lemma can be understood geometrically: b lies in the cone Cdefined in the proof above if and only if there is no hyperplane H = {x ∈Rn : yTx = 0} (having normal vector y) that separates b and C, i.e., yT b < 0and yTaj ≥ 0 for each j.5 Linear programming dualityLinear programming (LP), or linear optimization, is to maximize a linearfunction in n variables subject to a finite number of linear constraints. Theseconstraints are linear equations and/or linear inequalities. A recommendedbook in LP is [4]. Hence, a standard linear programming problem is of theformsup{cTx : Ax ≤ b} (3)where c ∈ Rn, b ∈ Rm and the m × n matrix A are given, and the variablevector is x ∈ Rn. Here, ≤ means componentwise inequality (i.e., it holds foreach component).We call problem (3) the primal problem. The primal problem is feasibleif there is an x satisfying Ax ≤ b, and such an x is called a feasible solution.A feasible solution x0 is optimal if cTx0 = sup{cTx : Ax ≤ b}, so then thesupremum is attained and we may write max in stead of sup. We define thesupremum in (3) to be −∞ if the problem is not feasible, and it is +∞ ifthe problem is unbounded, meaning that there is a sequence (xk) of feasiblesolutions such that cTxk →∞ as k →∞.Associated to each (primal) LP problem there is another LP problem,called its dual problem. The dual problem associated with problem (3) isinf{bT y : ATy = c, y ≥ O}. (4)As for the primal problem we use the terms feasible problem, feasible solutionand optimal solution for the dual problem. We define the infimum in (4) tobe∞ if the problem is not feasible, and it is −∞ if the problem is unbounded,i.e., there is a sequence (yk) of feasible solutions such that bT yk → −∞ ask →∞.One of the main theorems in optimization is the LP duality theorem (see[2], [4]).5Theorem 4 (Linear programming duality theorem) (i) Assume thatthe primal problem (3) has an optimal solution. Then the dual problem (4)also has an optimal solution andmax{cTx : Ax ≤ b} = min{bTy : ATy = c, y ≥ O}. (5)(ii) If one of the problems is unbounded, then the other problem is not feasible.Thus, when at least one problem is feasible, sup{cTx : Ax ≤ b} = inf{bTy :ATy = c, y ≥ O}.Proof. (i) Let x be feasible in the primal problem and y feasible in thedual problem, so Ax ≤ b and ATy = c, y ≥ O. ThencTx = (ATy)Tx = yTAx ≤ yT b = bT ywhere the inequality follows from Ax ≤ b as y ≥ O. By first taking the supre-mum over feasible x and then the infimum over feasible y in this inequalitywe obtain(∗) sup{cTx : Ax ≤ b} ≤ inf{bTy : AT y = c, y ≥ O}.Let x0 be an optimal solution of the primal problem. Let aTi denote theith row in the matrix A (so ai is a column vector). Define I = {i ≤ m :aTi x0 = bi} which corresponds to the indices of inequalities from Ax ≤ b thathold with equality for x = x0.Claim: For each z ∈ Rn satisfying aTi z ≤ 0 for all i ∈ I, the inequalitycT z ≤ 0 also holds.Otherwise, there is a z ∈ Rn with aTi z ≤ 0 for all i ∈ I and cT z > 0.Then, for suitably small ǫ > 0, the point x′ = x0 + ǫz satisfies Ax′ ≤ bbecause (i) for each i ∈ I we have aTi x′ = aTi x0 + ǫaTi z = bi+ ǫaTi z ≤ bi , and(ii) for each i ≤ m with i 6∈ I we have aTi x′ = aTi x0 + ǫaTi z < bi + ǫaTi z, soaTi x′ ≤ bi for ǫ small. But cTx′ = cTx0 + ǫcT z > cTx0 which contradicts thatx0 is an optimal solution. This proves the Claim.Next, the Claim makes it possible to apply Farkas’ lemma (Theorem 3)to the matrix A′ whose columns are the vectors ai for i ∈ I (so A′ plays therole of A in Theorem 3). As a result there must exist nonnegative numbersyi for i ∈ I such that∑i∈I yiai = c. Therefore ATy = c where y ∈ Rm isthe vector with components yi for i ∈ I (those we just found), and yi = 06otherwise. Since AT y = c and y ≥ O, y is a feasible solution in the dualproblem. Moreover, using that yi = 0 for i 6∈ I we getcTx0 = yTAx0 =∑i∈Iyi(aTi x0) =∑i∈Iyibi = bT y.This proves, due to the inequality (∗), that y is an optimal solution of the dualproblem and that the maximum in the primal problem equals the minimumin the dual problem, and (5) holds.(ii) Consider the inequality (∗). If the primal problem is unbounded,then the dual problem is not feasible (for, due to (∗), bT y would be an upperbound on cTx). So, in this case, both sides of (∗) are∞. Similarly, if the dualproblem is unbounded, both sides of (∗) are −∞. Finally, if the supremumin (∗) is finite, one can show that the supremum is attained and therefore, bythe first part of the theorem, ”sup=max=min=inf”. The same is true whenthe infimum is finite. (We omit the detailed argument here, it involves thestructure of polyhedra - the feasible sets of LP problems).Thus, for an LP problem, there are only three possible situations: (i) itis not feasible (i.e., no feasible solution exist), (ii) it is unbounded, or (iii) anoptimal solution exists; then both problems have an optimal solution, andthe corresponding optimal values are equal.Note the special case where b = O (in our primal and dual problems):then the function to be minimized in the dual is constant equal to 0. Thus,the problem is simply to determine if there are any feasible solutions in thedual. This special case will be useful later.6 The fundamental theorem of asset pricingvia linear programmingIn this section, we will apply LP theory in order to prove a version of the fun-damental theorem of asset pricing and also to find dominant trading strate-gies. See [3] for a discussion of these notions and mathematical finance.We shall use the following notation:• K: number of states (scenarios), n: number of assets• P = [pij ]: payoff matrix of size K × n where pij is payoff under state ifor asset j (this is ∆S∗j (ωi) in Pliska’s notation)7• h ∈ Rn: a trading strategy, we buy hj units of asset j• x ∈ RK : the payoff, i.e., outcome of some trading strategy under dif-ferent states• O, e, I: the zero vector (or zero matrix) O, the all ones vector e andthe identity matrix I (of suitable size). We treat vectors as columnvectors.• Nul(A), Col(A): nullspace and columnspace of a matrix AA risk-neutral probability measure is a vector y with positive componentsthat sum to 1 such that the dot product of y and each column of P is zero(meaning that expected payoff of each asset is zero).The first LP model: find an arbitrageConsider the LP problemmax∑Ki=1 xisubject tox = Phx ≥ O(6)Here x = Ph relates payoff x and trading strategy h; the linear equation saysthat x is a linear combination of the columns in P . The nonnegativity of xis very desirable: we will not loose money under any state. The objective(goal) is to maximize the sum of the payoffs, where we sum over all states.It should be a reasonable goal and can (if you like) be given a probabilisticinterpretation. The main point is that it reflects that we look for positivepayoffs for at least one scenario, i.e., an arbitrage possibility.So: an arbitrage exists if and only if the optimal value of the LP problem(6) is positive. Since LP problems can be solved very fast using differentalgorithms (e.g. the simplex algorithm), we can find an arbitrage, or provethat it does not exist, efficiently, even if K and n are very large. For instance,even with some thousands of assets and several hundreds of states it shouldonly take a couple of seconds to solve the problem assuming you have a goodLP code. Moreover, we may obtain an important theoretical result from thisLP viewpoint.8Theorem 5 (The arbitrage theorem) There is no arbitrage if and only ifthere is a risk-neutral probability measure.Proof. We prove the arbitrage theorem by applying duality theory to ourLP above, and doing some matrix calculations for partitioned matrices.The LP problem (6) may be written in the form of the primal problemin (3) as follows:max {[Oe]T [hx]: P −I−P IO −I[hx]≤ OOO}Here we used that Ph = x is equivalent to Ph− x ≤ O, −Ph+ x ≤ O. Thedual of this problem (confer (4)) ismin { OOOT  y1y2y3 :[P T −P T O−I I −I] y1y2y3 =[Oe], y1, y2, y3 ≥ O}Note that the objective function is simply 0! By using the substitutiony = y2 − y1, z = y3 the dual problem becomesmin {0 : P Ty = O, y = z + e, z ≥ O}.Check this! This problem has a feasible solution y if and only if there isa vector y ∈ Nul(P T ) = Col(P )⊥ such that y ≥ e. This, again, mustbe equivalent to the existence of an y ∈ Nul(P T ) with yi > 0 (i ≤ K) and∑i yi = 1; this follows by suitable scaling of y. Summing up, we have verifiedthat the following statements are equivalent:• there is no arbitrage• the optimal value in the LP problem (6) is zero• there is a strictly positive vector y ∈ Nul(P T ) with∑i yi = 1; this isprecisely a risk-neutral probability measure.So the proof is complete.We now turn to the second theorem; it concerns dominant trading strate-gies.9The second LP model: find a dominant trading strategyThis problem is very similar to (6) but it contains an extra variable ǫ ∈ R:max ǫsubject tox = Phx ≥ ǫe(7)The final constraints means that xj ≥ ǫ for each j ≤ n, and the goal is tofind a trading strategy which maximizes the minimum outcome (the optimalǫ)! Note that (7) has feasible solutions (e.g., the zero vector).So: a dominant trading strategy exists if and only if the optimal value ofthe LP problem (7) is positive.Recall that a linear pricing measure is just like a risk-neutral probabilitymeasure, except that some probabilities may be zero.Theorem 6 (Dominant trading strategy/linear pricing measure) There is nodominant dominant trading strategy if and only if there is a linear pricingmeasure.Proof. The proof is very similar to the previous one. First, we write theLP problem (7) in the form of the primal problem in (∗):max { OO1T  hxǫ : P −I O−P I OO −I e hxǫ ≤ OOO}The dual of this problem is (see (∗) and do a calculation as before):min{0 : P T (y1 − y2) = 0, −y1 + y2 − y3 = 0, eT y3 = 1, y1, y2, y3 ≥ O}.With the substitution y = y2 − y1 and π = y3 we see that we can eliminatey, and the problem simplifies tomin{0 : P Tπ = 0,∑jπj = 1, π ≥ O}.A feasible solution in this problem is precisely a linear pricing measure. Thisproves that the following statements are equivalent:10• there exists a linear pricing measure• the optimal value in the LP problem (7) is zero• there is no dominant trading strategyThat’s it!Final comments:1. Using LP problems (6) and (7) you may solve efficiently arbitrage anddominant trading strategy problems. Note that any (good) LP solversolves both the primal and the dual. So, e.g., if a linear pricing measureexist, you will get it!2. The LP approach above may be extended in different ways. For in-stance, in (6), you may add the constraint h ≥ O (no short-selling),and add an upper bound on the components of h, or further linearconstraints on the payoff vector x. Well, the problem then gets morecomplex, but it is still an LP problem, and should be easy to solvecomputationally.3. Perhaps this motivates you to learn more about LP: take a look at thecourse page for INF-MAT3370 Linear optimizationhttp://www.uio.no/studier/emner/matnat/ifi/INF-MAT3370/There is also a Master course (INF-MAT5360) in optimization andconvexity (see [2]), both useful areas for mathematical finance.References[1] G. Dahl. Mathematical finance and linear programming (optimization).Notes, University of Oslo, 2009.[2] G. Dahl. An Introduction to Convexity. Lecture notes, University ofOslo, 2000 (see http://folk.uio.no/geird/)[3] S.R. Pliska. Introduction to Mathematical Finance: Discrete Time Mod-els. Blackwell Publishers, 1997.[4] R. Vanderbei. Linear programming: foundations and extensions.Springer, Third edition, 2008.11

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