Non replication of options

Christos Kountzakis∗, Ioannis A. Polyrakis†and Foivos Xanthos‡

June 30, 2008

Abstract

In this paper we study the scarcity of replication of options in the two periodmodel of financial markets with a finite set of states. Especially we study this prob-lem in financial markets without binary vectors and in strongly resolving markets.We start our study by proving that a financial market does not have binary vectorsif and only if for any portfolio, at lest one non trivial option is replicated. After thischaracterization we prove that in these markets, for any portfolio, at mostm − 3options can be replicated wherem is the number of states, therefore for any port-folio, the number of the replicated options is between the natural numbers1 andm− 3. Note that by the existing result of Baptista (2007), the set of non replicatedoptions is of measure zero, and as it is known there are infinite sets with measurezero.

In the sequel we generalize the definition of strongly resolving markets to amore general class of financial markets by considering the payoff matrix of prim-itive securities, not with respect to the usual basis ofRm, but with respect to thepositive basis of the financial completion of the market. This allows us to general-ize the result of Aliprantis-Tourky (2002) about the non-replication of options in abigger class of financial markets.

In this study, the theory of positive bases developed in [5] and [6] plays acentral role. This theory simplifies and unifies the theory of options.

JEL Classification : C600, D520, G190Keywords: Replicated options; Completion by options; Strongly resolving mar-

kets; Positive bases.

1 The economic model and some essential notions

In this article we study a two-period security market with a finite number of statesΩ = 1, 2, ..., m during the date1, a finite number of primitive securities (assets)with payoffs the linearly independent vectorsx1, x2, ..., xn of the payoff spaceRm.

∗Department of Statistics and Actuarial- Financial Mathematics, University of the Aegean, 83200 Samos,Greece†Department of Mathematics, National Technical University of Athens, Zographou 157 80, Athens,

Greece, e-mail: ypoly@math.ntua.gr‡Department of Mathematics, National Technical University of Athens, Zographou 157 80, Athens,

Greece.

1

A portfolio is a vectorθ = (θ1, θ2, ...θn) of Rn whereθi is the number of the unitsof thei security. ThenT (θ) =

∑ni=1 θixi ∈ Rm is thepayoff of θ. Since the operator

T is one-to-one, it identifies portfolios with their payoffs. So the vectorsx1, x2, ..., xn

will be mentioned asprimitive securities, the subspace

X = [x1, x2, ..., xn],

of E, generated by the vectorsxi as thespace of marketed securitiesor theasset spanand the vectors ofX will be also referred as portfolios. A vectorx ∈ Rm is marketedor x is replicated if it is the payoff of some portfolioθ, or equivalently ifx ∈ X.

Recall that the vector spaceRm = x = (x(1), x(2), ..., x(m))|x(i) ∈ R for each i is ordered by the pointwise ordering i.e. for anyx, y ∈ Rm we have: x ≥ y ifx(i) ≥ y(i) for eachi. Rm

+ = x ∈ Rm|x(i) ≥ 0 for each i is the positive coneof Rm. For anyx, y ∈ Rm x ∨ y =

(x(1) ∨ y(1), x(2) ∨ y(2), ..., x(m) ∨ y(m)

)is the supremum andx ∧ y =

(x(1) ∧ y(1), x(2) ∧ y(2), ..., x(m) ∧ y(m)

)is the

infimum of x, y in Rm. x+ = x ∨ 0 =(x(1) ∨ 0, x(2) ∨ 0, ..., x(m) ∨ 0

)and

x− = (−x) ∨ 0 are the positive and the negative part ofx. Note also that for any tworeal numbersa, b, a ∨ b is the supremum anda ∧ b is the infimum ofa, b. A linearsubspaceZ of Rm is a sublattice or a Riesz subspaceof Rm if for any x, y ∈ Z,x ∨ y andx ∧ y belong toZ. Also for anyx = (x(1), x(2), ..., x(m)) ∈ Rm, the setsupp(x) = i = 1, 2, ..., m|x(i) 6= 0 is the support ofx.

For any subsetB of Rm, thesublatticeS(B) of Rm generated byB is the inter-section of the sublattices ofRm which containB. Theriskless bond1 is the vector ofRm whose every coordinate is equal to 1. Thecall option written on the vectorx ∈ Rm

with exercise pricea is the vectorc(x, a) = (x− a1)+ of Rm and theput option of xwith exercise pricea is p(x, a) = (a1− x)+. We havex− a1 = c(x, a)− p(x, a).

If both c(x, a) > 0 andp(x, a) > 0, we say that the call optionc(x, a) is nontrivial and also we say that the put optionp(x, a) is non trivial. In this case we say thata is anon trivial exercise price of x. We denote byKx the set of non trivial exerciseprices ofx. If c(x, a) ∈ X we say thata is acall-replicated exercise price ofx andif p(x, a) ∈ X, we say thata is aput-replicated exercise price ofx. If both c(x, a),p(x, a) are inX we say thata is a replicated exercise price ofx. If 1 ∈ X, we have:c(x, a) ∈ X if and only if p(x, a) ∈ X. If the riskless bond is not contained inX itis possible only one of the call and put options to be replicated. In this paper we donot suppose always that the riskless bond1 belongs toX. Of course1 belongs to thecompletion by options ofX.

Thecompletion by optionsof X is the subspace ofRm which arises inductivelyby adding in the market the call and put options of the marketed securities and bytaking again call and put options which are added again in the market. In [3] a math-ematical definition of the completion by options in infinite securities markets is given.Especially in the above article, a more general study of the completion by options ofthe market is presented by Kountzakis and Polyrakis (2006) were the options are nottaken with respect to the riskless bond1 but with respect to some risky vectors from astandard subspaceU of Rm and the completion by options ofX is denoted byFU (X).This study is very general and also includes the case of exotic options. In the classicalcase where the options are taken with respect to the riskless bond1, the completion

2

by options ofX is denoted in [3] byF1(X) and we will preserve this notation in thepresent article. In the above article it is proved that if the payoff space is a generalvector latticeE thenFU (X) is the sublattice ofE generated by the setX ∪ U . In ourcase where the payoff space is the spaceRm and the call and put options are taken withrespect to the riskless bond1, the completion by optionsF1(X) of X is the sublatticeof Rm generated by the setX ∪1. In the case where1 ∈ X, F1(X) is the sublatticeof Rm generated by the setX. In the above results and also in the results of the presentarticle, the theory of lattice-subspaces and positive bases developed by Polyrakis in thepapers of the bibliography is very important. This theory simplifies and unifies thetheory of options.

Ross (1976) proved in [7] that the existence of an efficient fund inX is equivalentto the fact that the span of call and put options written on the elements ofX is equalto the entire spaceRm. In Aliprantis and Tourky (2002), a scarcity result for replicatedoptions for strongly resolving markets is proved. Especially it is proved that if1 ∈ X,n ≤ m+1

2 and the asset span isstrongly resolving, i.e. anyn×n submatrix of the payoffmatrix of the primitive securities is non singular, then any non-trivial option written onsome element ofX is not replicated.

In Baptista (2007), the replication of options is studied in the case where the assetspanX doesn’t contain binary vectors and1 ∈ X. Note thatx ∈ Rm is a binary vectorif x 6= 0, x 6= 1 andx(i) = 0 or x(i) = 1 for anyi. Especially in [2] it is proved thatfor anyx ∈ X the set of non-trivial and non replicated exercise prices is a subset offull measure ofKx, or equivalently the set of the non-trivial, replicated exercise pricesis a subset ofKx of measure zero.

In this article, we start our study by a characterization of the markets without binaryvectors based on the replication of options.

Especially, if1 ∈ X, we show thatX doesn’t contain binary vectors if and only iffor any non-constant vectorx ∈ X at least one non-trivial option ofx is non-replicated,Theorem 5. In the sequel, Theorem 6, we prove that ifX doesn’t contain binary vec-tors, then for anyx ∈ X the set of replicated exercise prices has at mostk−3 elementswherek is a real number smaller thanm and we also determine disjoint intervals ofKx each of which contains at most one replicated exercise price. Since there are in-finite subsets ofKx of measure zero, our theorem is an essential improvement of theexcellent result of Baptista.

In this article we also generalize the definition and the results of strongly resolvingmarkets of Aliprantis-Tourky. Especially we consider the payoff matrix of primitivesecuritiesxi with respect to the positive basisbi of F1(X) and we define the notionof strongly resolving markets with respect to the basisbi. If F1(X) = Rm, thetwo definitions coincide. In Theorem 10 we prove that ifF1(X) is a proper subset ofRm the market cannot be strongly resolving. The marketX of Example 13 is stronglyresolving with respect to the basisbi but X is not strongly resolving, therefore ourdefinition is a generalization of the existing one.

In Theorem 11 we extent the result of [1] for strongly resolving markets withrespect to the basisbi. Our proof is the analogous with the excellent proof ofAliprantis-Tourky.

For a study of the two-periods security markets we refer to the book of LeRoy andWerner (2001), [4].

3

2 Determination of the completionF1(X) of X

In this section we describe the method of determination of the completion by optionsof X as it is presented in [3]. According to this method we consider the set

A = x+1 , x−1 , x+

2 , x−2 , ..., x+n , x−n ,1.

Any maximal subsety1, y2, ..., yr of linearly independent vectors ofA is abasicsetof the market, wherex+

i , x−i are the positive and negative parts of the vectorsxi.Note that a basic set is not necessarily unique. In general it is possible to find differentbasic sets of the market but all these sets have the same cardinal numberr. Especiallyr is the dimension of the linear subspace ofRm generated byA and a basic set is abasis of it.

Theorem 1 ( [3], Theorem11).F1(X) is the sublattice ofRm generated by a basicsety1, y2, . . . , yr of the market.

After this result we use the theory of lattice-subspaces and positive bases devel-oped by Polyrakis in [5] and [6] for the determination ofF1(X). SinceF1(X)is a sublattice ofRm which contains1, we have thatF1(X) has a positive basisb1, b2, ..., bµ which is apartition of the unit , i.e. the vectorsbi have disjoint sup-ports and

∑µi=1 bi = 1, see in Theorem 16 of the Appendix. This basis is unique. So

we have:

Theorem 2. F1(X) has a positive basisb1, b2, ..., bµ which is a partition of the unit.

For the determination of the positive basisbi of F1(X) which is a partition ofthe unit we follow the steps of Polyrakis algorithm, see Theorem 18 in the appendix,where a positive basis of the sublattice ofRm generated by a finite set of positive andlinearly independent vectors is determined. We start by the determination of a basicfunction set ofy1, y2, . . . , yr of the market. In the sequel we determine thebasicfunction of y1, y2, . . . , yr which is very important for the theory of lattice-subspacesand positive bases. This function has been defined in [5] and is the following:

β(i) =(y1(i)

y(i),y2(i)y(i)

, ...,yr(i)y(i)

), for each i = 1, 2, ...,m, with y(i) > 0,

wherey = y1 + y2 + ... + yr. This function takes values in the simplex∆r = ξ ∈Rr

+|∑r

i=1 ξi = 1 of Rr+.

Denote byR(β) is the range (i.e. the set of values) ofβ and bycardR(β) thecardinal number ofR(β) (i.e. the number of the different values ofβ).

We continue the algorithm and we obtain a positive basisd1, d2, ..., dµ of F1(X).The elements of this basis have disjoint support and eachbi is constant on its support.So by a normalization of the basisdiwe obtain the basis positive basisbi of F1(X)which is also a partition of the unit.

Except the determination of the positive basisbi of F1(X), by the basic functionwe can ask very easy to two important questions of the theory of options. Especially we

4

can check directly wether any option is replicated and also if the completion by optionsof X is the whole spaceRm. Recall that ifX = F1(X), any option is replicatedand we say then thatX is complete by options(with respect to1). As it is provedin Polyrakis (1999), Theorem 3.7 (see also in the Appendix) the dimension of thesublattice generated byy1, y2, . . . , yr, i.e. the dimension ofF1(X), is equal to thecardinal number ofR(β). So if R(β) hasn elements we have thatF1(X) = X and ifR(β) hasm elements we have thatF1(X) = Rm. so we have:

Theorem 3. The dimension ofF1(X) is equal to the cardinal number of the rangeR(β), therefore we have:

(i) F1(X) = X if and only if cardR(β) = n,

(ii) F1(X) = Rm if and only if cardR(β) = m,

(iii) F1(X) ( Rm if and only if cardR(β) < m.

3 Markets without binary vectors

Throughout this paper we will denote byb1, b2, ..., bµ, or for simplicity bybi, thepositive basis ofF1(X) which is a partition of the unit. For anyx =

∑µi=1 λibi ∈

F1(X), λ1, λ2, ..., λµ are the coefficients ofx in the basisbi. We put

a1 = minλi|i = 1, 2, ..., µ andΦ1 = i|λi = a1,

a2 = minλi|λi > a1 andΦ2 = i|λi = a2,and by continuing this process we take the real numbersa1, a2, ..., ak and the subsetsΦ1, Φ2, ..., Φk of 1, 2, ..., µ. The numbersa1, a2, ..., ak will be referred as thees-sential coefficientsand the setsΦ1, Φ2, ..., Φk as theessential sets of statesof x, withrespect to the basisbi.

The essential coefficients are in increasing order, i.e.ai < aj for any i < j. Ofcourse for the numberk of the essential coefficients ofx we have:k ≤ µ ≤ m.

If for exampleµ = 5 andx = 2b1 − 3b2 + 2b3 + b4 + b5, thena1 = −3, a2 =1, a3 = 2 are the essential coefficients andΦ1 = 2, Φ2 = 4, 5, Φ3 = 1, 3 theessential sets of states ofx. We havex = a1

∑i∈Φ1

bi + a2

∑i∈Φ2

bi + a3

∑i∈Φ1

bi.

Proposition 4. For anyx =∑µ

i=1 λibi ∈ F1(X) we have:

(i) c(x, a) =∑µ

i=1(λi − a)+bi and p(x, a) =∑µ

i=1(a− λi)+bi,

(ii) if a1, a2, ..., ak are the essential coefficients ofx, the intervalKx = (a1, ak) isthe set of non-trivial exercise prices ofx.

Proof. (i): The basisbi is a partition of the unit therefore the vectorsbi have dis-joint supports and

∑µi=1 bi = 1. Therefore, the vectorsbi have disjoint supports and

bi(j) = 1 for any j ∈ supp(bi). So we havec(x, a) = (x − a1)+ = (∑µ

i=1 λibi −a

∑µi=1 bi)+ = (

∑µi=1(λi − a)bi)+. Since the basisbi is a partition of the unit, for

any j ∈ supp(bi) we have that thej-coordinate ofc(x, a) in the usual basisei of

5

Rm is (λj − a)+, therefore it is easy to show thatc(x, a) =∑µ

i=1(λi − a)+bi. Theproof for the put option is analogous.

(ii): If a ≤ a1 then(a − λi)+ = 0 for any i, thereforep(x, a) = 0 becausea1 isthe minimum of the coefficientsλi of x. Thereforea is a trivial exercise price ofx. Ifa ≥ ak, similarly we have that(λi − a)+ = 0 for any i, thereforec(x, a) = 0 andais a trivial exercise price. For anya ∈ (a1, ak) we have that(λi − a)+ > 0 for atleast onei and also(a− λj)+ > 0 for at least onej. Since the elementsbi of the basisare positive we have thatc(x, a) > 0 andp(x, a) > 0, hencea is a non-trivial exerciseprice ofx. ¥

We give below a characterization of the markets without binary vectors. We saythat a vectorx ∈ Rm is anon-constant vectorif x is not a multiple of1 i.e. x 6= λ1for someλ ∈ R. According to this definition,0 is a constant vector ofRm.

Theorem 5. If 1 ∈ X, we have:X does not contain binary vectors if and only if forany non-constant vectorx ∈ X at least one non-trivial option ofx is non-replicated.

Proof. Suppose thatX does not contain binary vectors. Suppose also that there existx ∈ X so thatc(x, α) ∈ X for eachα ∈ Kx. Also for anyα /∈ Kx we havethat c(x, α) = 0 or p(x, α) = 0, thereforec(x, α) andp(x, α) are elements ofXbecausex − α1 = c(x, α) − p(x, α). Therefore ifL = [x] is the one-dimensionalsubspace generated byx, then by [3], Theorem 21, the completion by optionsF1(L)of L is the subspace generated by the set of call options written on the elements ofthe subspaceY of Rm generated by the setL ∪ 1, i.e. the completion is takenduring the first steep of the process. Each vectory of Y is of the formy = λx + ξ1thereforec(y, a) = (y − a1)+ = (λx− (a− ξ)1) = c(λx, (a− ξ)). So we have thatany call option written on an element ofY is a call option written on an element ofL, therefore it belongs toX as we have proved before. So we have thatF1(L) ⊂ X.Since1 ∈ F1(L) we have thatF1(L) is an at lest two-dimensional sublattice, thereforeF1(L) has a positive basis which is also a partition of the unit. The elements of thisbasis are binary vectors, and these elements belong toX, contradiction. So for anyx ∈ X at least one non-trivial option ofx is non-replicated.

For the converse suppose for anyx ∈ X at least one non-trivial option ofx isnon-replicated. If we suppose thatx is a binary vector ofX, then it is easy to showthat the essential coefficients ofx in the basisbi of F1(X) area1 = 0, a2 = 1,thereforeKx = (0, 1) andx =

∑i∈Φ2

bi = b2 ∈ X. For anyα ∈ (0, 1) we havethatp(x, α) = (1− α)b2 ∈ X which is a contradiction. ThereforeX does not containbinary vectors. ¥

Theorem 6. Suppose that the asset spanX doesn’t contain binary vectors andx isa non-constant vector ofX. If a1, a2, ..., ak are the essential coefficients ofx withrespect to the basisbi, then:

(i) If k = 2, each non-trivial call option ofx is non-replicated. Ifk > 2, each of theintervals(a1, a2), [a2, a3), ..., [ak−2, ak−1) contains at most one call-replicatedexercise price, therefore there are at mostk − 2 call-replicated exercise pricesof x.

6

(ii) If k = 2, each non-trivial put option ofx is non-replicated. Ifk > 2, eachof the intervals(a2, a3], ..., (ak−2, ak−1], (ak−1, ak) contains at most one put-replicated exercise price, therefore there are at mostk−2 put-replicated exerciseprices ofx.

(iii) If we suppose moreover that1 ∈ X, we have: Ifk = 3, each non-trivial option ofx is non-replicated. Ifk > 3, each of the intervals(a2, a3), [a3, a4), ..., [ak−2, ak−1)contains at most one replicated exercise price, therefore there are at mostk − 3replicated exercise prices ofx.

Proof. Suppose thatx ∈ X, x 6= λ1, x =∑µ

i=1 λibi is the expansion ofx in the basisbi and suppose thata1, a2, ..., ak are as the essential coefficients andΦ1, Φ2, ..., Φk

are the essential sets of states ofx with respect to the basisbi. Since we have sup-posed thatx is not a positive multiple of the riskless bond we have thatx has at leasttwo essential coefficients, hencek ≥ 2.

Thenx = a1

∑

i∈Φ1

bi + a2

∑

i∈Φ2

bi + ... + ak

∑

i∈Φk

bi. (1)

We putbj =∑

i∈Φjbi, j = 1, 2, ..., k and we remark that every such vector is a binary

vector. Also, we have

x =k∑

j=1

ajbj .

The set of non trivial exercise prices is the intervalKx = (a1, ak). For anya ∈(a1, ak) we have

c(x, a) =k∑

j=r+1

(aj − a)bj ,

wherer = 1 if a ∈ (a1, a2) andr = ν if a ∈ [aν , aν+1) for ν = 2, 3, ..., k − 1.If a ∈ [ak−1, ak) then c(x, a) = (ak − a)bk is a positive multiple of a binary

vector, thereforec(x, a) /∈ X, So for anya ∈ [ak−1, ak), c(x, a) is non-replicated.This means also that ifk = 2, i.e. if a1, a2 are the essential coefficients ofx, then anycall option ofx is non-replicated.

Suppose now thata, a′ are different exercise prices belonging to the same subinter-val of (a1, ak), i.e. a, a′ ∈ (a1, a2) or a, a′ ∈ [ar, ar+1) for somer = 2, 3, ..., k − 2.Then we have

c(x, a)− c(x, a′) =k∑

j=r+1

((aj − a)− (aj − a′))bj = (a′ − a)k∑

j=r+1

bj .

If we suppose thatc(x, a) andc(x, a′) belong toX we have that

(a′ − a)k∑

j=r+1

bj ∈ X,

7

which is a contradiction because∑k

j=r+1 bj is a binary vector. This implies that atmost one ofc(x, a), c(x, a′) belongs toX.

So any of the subintervals(a1, a2), [a2, a3), ..., [ak−2, ak−1) of (a1, ak) containsat most one call-replicated exercise price, therefore there are at mostk−2 call-replicatedexercise prices.

Similarly for the case of put options we write(a1, ak) = (a1, a2] ∪ (a2, a3]... ∪(ak−1, ak) and we have: For anya ∈ (a1, a2] the put optionp(x, a) is a binary vec-tor, therefore the interval(a1, a2] does not contain put-replicated exercise prices. Forany twoa, a′ different exercise prices belonging in the same subinterval of(a1, ak),p(x, a) − p(x, a′) is a multiple of a binary vector and as above we have that any ofthe intervals(a2, a3], ..., (ak−2, ak−1], (ak−1, ak) contains at most one put-replicatedexercise price, therefore there are at mostk − 2 put-replicated exercise prices.

If we suppose that1 ∈ X and at least one ofc(x, a), p(x, a) is replicated, then bothof them are replicated because

x− a1 = c(x, a)− p(x, a),

therefore an exercise pricea is call-replicated if and only ifa is put-replicated. So, if1 ∈ X, by (i) and(ii) we have that there are at mostk − 3 replicated exercise pricesbecause strike prices in the intervals(a1, a2] and[ak−1, ak) are excluded. So in thecase where1 ∈ X, each interval(a2, a3), [a3, a4), ..., [ak−2, ak−1) has at most onereplicated exercise price forx, therefore there are at mostk − 3 replicated exerciseprices.

¥

If the dimension ofF1(X) is at most three, then the basisbi of X has at mostthree elements and for anyx ∈ X the essential coefficients ofx are at most three realnumbersa1, a2, a3 and the next corollary is obvious:

Corollary 7. Suppose that the asset spanX doesn’t contain binary vectors.

(i) If dimF1(X) = 2, then any non-trivial option written on some element ofX isnon-replicated.

(ii) If dimF1(X) = 3 and1 ∈ X, any non-trivial option written on some elementof X is non-replicated.

Example 8. Suppose thatx1 = (1, 1, 2, 2, 0, 0, 0, 0), x2 = (0, 0, 0, 0, 3, 3, 4, 4), x3 =(1, 1, 1, 1, 1, 1, 1, 1) are the primitive securities andX = [x1, x2, x3] is the marketedspace. It is easy to show thatX does not contain binary vectors.

According to the methodology of the determination ofF1(X) we start by the de-termination of a basic set and we find thaty1, y2, y3 = x1, x2, x3 is a basic set ofthe market. In order to determine a positive basis ofF1(X) we follow the algorithm ofTheorem 18. So we determine the basic functionβ of y1, y2, y3. β = 1

y (y1, y2, y3),wherey is the sum ofyi and we find that

β(1) = β(2) =12(1, 0, 1) = P1, β(3) = β(4) =

13(2, 0, 1) = P2

8

β(5) = β(6) =14(0, 3, 1) = P3, β(7) = β(8) =

15(0, 5, 1) = P4.

So we have thatcard(R(β)) = 4 therefore the completionF1(X) is a four-dimensionalsublattice ofR8.

The three first vectorsP1, P2, P3 of R(β) are linearly independent, so we preservethe enumeration ofR(β). According to the algorithm,I4 = β−1(P4) = 7, 8 andwe define the new vectory4 = (0, 0, 0, 0, 0, 0, 5, 5). We determine the basic functionγ = 1

y′ (y1, y2, y3, y4) wherey′ is the sum of these vectors. We find that

γ(1) = γ(2) =12(1, 0, 1, 0) = P ′1, γ(3) = γ(4) =

13(2, 0, 1, 0) = P ′2

γ(5) = γ(6) =14(0, 3, 1, 0) = P ′3, γ(7) = γ(8) =

110

(0, 4, 1, 5) = P ′4

A positive basis ofF1(X) is given by the formula(d1, d2, d3, d4)T = A−1(y1, y2, y3, y4)T

where A is the matrix whose columns are the vectorsP ′i , i = 1, ..., 4. We find that thevectors

d1 = (2, 2, 0, 0, 0, 0, 0, 0), d2 = (0, 0, 3, 3, 0, 0, 0, 0),

d3 = (0, 0, 0, 0, 4, 4, 0, 0), d4 = (0, 0, 0, 0, 0, 0, 10, 10),

define a positive basis ofF1(X). By a normalization of this basis we have that thevectors

b1 = (1, 1, 0, 0, 0, 0, 0, 0), b2 = (0, 0, 1, 1, 0, 0, 0, 0),

b3 = (0, 0, 0, 0, 1, 1, 0, 0), b4 = (0, 0, 0, 0, 0, 0, 1, 1),

define the positive basis ofF1(X) which is a partition of the unit.Consider the portfoliox = −x1 + x2 = (−1,−1,−2,−2, 3, 3, 4, 4). The expan-

sion ofx in the basisbi is x = −b1 − 2b2 + 3b3 + 4b4 and according to the abovetheorema1 = −1, a2 = −2, a3 = 3, a4 = 4 are the essential coefficients ofx. For anyα ∈ (−1, 2] or α ∈ [3, 4), any option is non replicated. By the previous resultsx hasexactly one replicated exerciseα ∈ (2, 3). We can determine it as follows:

c(x, α) = (3− α)b3 + (4− α)b4 ∈ X,

therefore

c(x, α) = ρ1x1 +ρ2x2 +ρ3x3 = ρ1(b1 +2b2)+ρ2(3b3 +4b4)+ρ3(b1 +b2 +b3 +b4).

We find thatρ1 = ρ3 = 0, 3ρ2 = 3− α, 4ρ2 = 4− α, thereforeα = 0.Indeed,c(x, 0) = 3b3 + 4b4 = x2 ∈ X andp(x, 0) = b1 + 2b2 = x1 ∈ X.

9

4 Strongly resolving markets

The replication of options in strongly resolving markets has been studied by Aliprantis-Tourky (2002). If we expand the vectorsxi in the usual basise1, ..., em of Rm wehave them× n matrix

A(xi, ei) =

x1(1) x2(1) ... xn(1)x1(2) x2(2) ... xn(2)

. . .

. . .

. . .x1(m) x2(m) ... xn(m)

,

which is thepayoff matrix of vectorsxi. The notion of strongly resolving market isdefined in the above article as follows: If anyn × n submatrix ofA(xi, ei) is non-singular, the asset spanX = [x1, ..., xn] (or the market) is called strongly resolving.As we have noted in the introduction, Aliprantis and Tourky prove that if1 ∈ X,n ≤ m+1

2 and the asset span is strongly resolving, any non-trivial call and put optionwritten on some element ofX is not replicated.

In this paper we extend the definition of strongly resolving markets by taking thepayoff matrix of the payoff vectors in the basisbi of F1(X). So if b1, ..., bµ is thepositive basis ofF1(X) which is a partition of the unit, we expand eachxi in this basisand suppose thatxi =

∑µj=1 xb

i (j)bj . Theµ× n matrix

A(xi, bi) =

xb1(1) xb

2(1) ... xbn(1)

xb1(2) xb

2(2) ... xbn(2)

. . .

. . .

. . .xb

1(µ) xb2(µ) ... xb

n(µ)

,

is the payoff matrix of the basic securitiesxi in the basisbi.Definition 9. If any n × n submatrix ofA(xi, bi) is non-singular, the marketX isstrongly resolving with respect to the basisbi.

In the next theorem we prove that ifF1(X) 6= Rm, the market cannot be stronglyresolving. So if the market is strongly resolving, thenF1(X) = Rm and the twodefinitions coincide because the basisbi of F1(X) is the usual basisei of Rm andthereforeA(xi, bi) = A(xi, ei). Also the market of Example 13 is strongly resolvingwith respect to the basisbi but not strongly resolving, therefore our definition ofstrongly resolving markets is a generalization of the existing one.

Theorem 10. If n ≥ 2 and the completion by optionsF1(X) of X is a proper subspaceof Rm, then the market is not strongly resolving.

Proof. The assumption thatF1(X) 6= Rm, impliesµ < m, whereb1, b2, ..., bµ isthe positive basis ofF1(X) which is also a partition of the unit. Sinceµ < m, thesupport of at least one of the elements of the basis is not a singleton. So we may

10

suppose thati1, i2 ∈ supp(br) for somer. For anyxi we havexi =∑µ

j=1 xbi (j)bj ,

thereforexi(i1) = xbi (r)br(i1) = xb

i (r) becausebr(i1) = 1. Similarlyxi(i2) = xbi (r),

therefore for any vectorxi we havexi(i1) = xi(i2). This implies that thei1 andi2-rowof the matrixA(xi, ei) coincide and the theorem is true. ¥

For anyx ∈ F1(X) we expandx in the basisbi of F1(X) which is a partition ofthe unit and suppose thatx =

∑µi=1 λibi. Thensuppb(x) = i|λi 6= 0 is the support

of x andzerosb(x) = i|λi = 0 is the set of zeros ofx with respect to the basisbi. Also #suppb(x) and#zerosb(x) is the cardinal number of the setssuppb(x)andzerosb(x).

Theorem 11. Suppose that the riskless bond1 is contained in X. If the market isstrongly resolving with respect to the basisbi and n ≤ µ+1

2 then any nontrivialoption written on some element ofX is non replicated.

Proof. Let x =∑µ

i=1 λibi ∈ X and suppose thaty = c(x, α) =∑µ

i=1(λi − α)+bi

is a nontrivial call option. Theny > 0 and also the corresponding put optionz =p(x, α) =

∑µi=1(α− λi)+bi is also greater of zero,z > 0. Let

#suppb(y) = β, #zerosb(y) = γ, #suppb(z) = β′,#zerosb(z) = γ′.

We shall show thatmaxγ, γ′ ≥ µ

2.

It is clear thati ∈ suppb(y) ⇒ i ∈ zerosb(z) andi ∈ suppb(z) ⇒ i ∈ zerosb(y),thereforeγ′ ≥ β andγ ≥ β′.

Also β + γ = β′ + γ′ = µ. If β ≥ γ thenβ ≥ µ2 thereforeγ′ ≥ µ

2 . If γ ≥ βthenγ ≥ µ

2 and the assertion is true. Since the risklesss bond belongs toX we havethat bothy, z are replicated or not. Suppose thaty, z are replicated. Then as we haveproved above at lest one of them, for example the call optiony has a number of zerocoordinates in the basisbi greater or equal toµ2 , i.e. γ ≥ µ

2 .Sincey ∈ X, it can be expanded in the basisx1, ..., xn of X and suppose that

y =∑n

i=1 ρixi. Then we have

(λ1 − α)+

(λ2 − α)+

.

.

.(λµ − α)+

=

xb1(1) xb

2(1) ... xbn(1)

xb1(2) xb

2(2) ... xbn(2)

. . .

. . .

. . .xb

1(µ) xb2(µ) ... xb

n(µ)

ρ1

ρ2

.

.

.ρn

. (2)

By our assumption thatn ≤ µ+12 we have thatn ≤ µ

2 + 12 ≤ γ + 1

2 , therefore wehave thatn ≤ γ becausen, γ are natural numbers. Therefore that at leastn coordinatesof y in the basisbi are equal to zero and suppose that(λi1 − α)+ = (λi2 − α)+ =

11

... = (λin − α)+ = 0. Then

xb1(i1) xb

2(i1) ... xbn(i1)

xb1(i2) xb

2(i2) ... xbn(i2)

. . .

. . .

. . .xb

1(in) xb2(in) ... xb

n(in)

ρ1

ρ2

.

.

.ρn

=

00...0

, (3)

where(ρ1, ρ2, ..., ρn) 6= (0, 0, ..., 0) becauseρi are the coordinates ofy in thebasisxi andy > 0. This is a contradiction because the matrix of the system isnon-singular. So none ofy, z belong to X and the theorem is true. ¥

Suppose thatx ∈ X and suppose thata1, a2, ..., ak and Φ1, Φ2, ..., Φk are theessential coefficients and the essential sets of states ofx in the basisbi. For anyr = 1, 2, ..., k we definecx(r) = card(Φ1∪...∪Φr) andpx(r) = card(Φr+1∪...∪Φk),i.e. cx(r) andpx(r) are the cardinal numbers ofΦ1 ∪ ... ∪ Φr andΦr+1 ∪ ... ∪ Φk.

Proposition 12. Suppose that the security marketX is strongly resolving with resectto the basisbi of F1(X), x ∈ X anda1, a2, ..., ak are the essential coefficients ofxwith respect to the basisbi.

(i) If cx(r) ≥ n, the interval[ar, ar+1) does not contain call-replicated exerciseprices ofx,

(ii) If px(r) ≥ n, the interval(ar, ar+1] does not contain put-replicated exerciseprices ofx,

(iii) if 1 ∈ X andmaxcx(r), px(r) ≥ n, the interval[ar, ar+1] does not containreplicated exercise prices ofx.

Proof. Suppose thatcx(r) ≥ n and thata ∈ [ar, ar+1) is a call-replicated exerciseprice. Then

y = c(x, a) =k∑

j=r+1

(aj − a)bj ∈ X,

wherebj =∑

i∈Φjbi for anyj = 1, 2, ..., k.

We expandedy in the basisx1, ..., xn of X and suppose thaty =∑n

i=1 λixi andsuppose also thatxi =

∑µj=1 xb

i (j)bj . By our hypothesis we have thatzerosb(y) =cx(r) ≥ n, therefore at leastn of the coordinatesξi of y in the basisbi are equal tozero and suppose thatξi1 , ξi2 , ..., ξin aren such coordinates. This leads to the system

xb1(i1) xb

2(i1) ... xbn(i1)

xb1(i2) xb

2(i2) ... xbn(i2)

. . .

. . .

. . .xb

1(in) xb2(in) ... xb

n(in)

λ1

λ2

.

.

.λn

=

00...0

.

12

The fact thatX is strongly resolving implies that the system has the unique solutionλ1 = λ2 = ... = λn = 0 thereforey = c(x, a) = 0, which is a contradiction becausea is a non trivial exercise price ofx. Hence[ar, ar+1) does not contain call-replicatedexercise prices and statement(i) is true.

The proof of statement(ii) is analogous and(iii) follows by (i) and(ii) and bythe fact that1 ∈ X. ¥

Example 13. Let x1 = (4, 4, 3, 3, 2, 2, 1, 1), x2 = (1, 1, 2, 2, 3, 3, 4, 4), andX =[x1, x2]. 1 is contained inX becausex1 + x2 = 51. The payoff matrix

A(xi, ei) =

4 14 13 23 22 32 31 41 4

,

has singular2× 2 submatrices, therefore the market is not strongly resolving.In order to apply our theorem, we determine the positive basis ofF1(X) which is

a partition of the unit and we find that the vectors

b1 = (1, 1, 0, 0, 0, 0, 0, 0), b2 = (0, 0, 1, 1, 0, 0, 0, 0), b3 = (0, 0, 0, 0, 1, 1, 0, 0), b4 = (0, 0, 0, 0, 0, 0, 1, 1),

define this basis. We expand the vectorsxi in the basisbi and we find that

A(xi, bi) =

4 13 22 31 4

,

is the payoff matrix with respect to this basis and we remark that the market isstrongly resolving with respect to the basisbi. Sincen = 2 ≤ µ+1

2 = 52 we have

that any option written on elements ofX is non-replicated.

5 Appendix: Lattice-subspaces and positive bases inC(Ω)

In this section we give the basic mathematical notions and results, which are needed forthis article.C(Ω) is the space of real valued functions defined on a compact Hausdorfftopological spaceΩ. C(Ω) is ordered by the pointwise ordering, i.e for anyx, y ∈C(Ω) we have:x ≥ y if and only if x(t) ≥ y(t) for eacht ∈ Ω. C+(Ω) = x ∈C(Ω)

∣∣x(t) ≥ 0 for eacht ∈ Ω is the positive cone ofC(Ω). Recall that if the setΩis finite, for example ifΩ = 1, 2, ...,m, thenC(Ω) is the vector spaceRm, thereforethe results presented below hold also for the spaceRm which we use in this paper. But

13

we present the results inC(Ω) as they are formulated in [5] and [6]. The results ofthese articles are presented below.

The spaceC(Ω), ordered by the pointwise ordering is a vector lattice i.e. for anyx, y ∈ C(Ω) the supremumx ∨ y and the infimumx ∧ y of x, y in C(Ω) exists.Suppose thatL is anordered subspaceof C(Ω), i.e. L is a linear subspace ofC(Ω)ordered again by the pointwise ordering. ThenL+ = C+(Ω) ∩ L is the positive coneof L. If L is a vector lattice, i.e. if for anyx, y ∈ L the supremumsupLx, y and theinfimum infLx, y of x, y in L exist, thenL is a lattice-subspaceof C(Ω). Thenwe have

supLx, y ≥ x ∨ y ≥ x ∧ y ≥ infLx, y.If for any x, y ∈ L, x∨ y ∈ L andx∧ y ∈ L, L is asublattice of C(Ω). It is clear

that any sublattice ofC(Ω) is a lattice-subspace but the converse is not true. In generalan ordered subspaceL of C(Ω) is not a lattice-subspace and also a lattice-subspace isnot always a sublattice.

For any subsetB of C(Ω), the intersection of all sublattices ofC(Ω) which containB is a sublattice ofC(Ω) and it is the minimum sublattice ofC(Ω) which containsB.This subspace is thesublattice ofC(Ω) generated byB.

Suppose thatL is finite dimensional. A basisb1, b2, ..., br of L is apositive basisof L if L+ = x =

∑ri=1 λibi | λi ∈ IR+ for eachi. In other words, the basisbi

of L is positive if for anyx ∈ L we have:x(t) ≥ 0 for any t ∈ Ω if and only thecoefficientsλi of x in the basisbi are positive. AlthoughL has infinitely many basesthe existence of a positive basis ofL is not always ensured.

Suppose thatb1, b2, ..., br is a positive basis ofL. Then it is easy to show thatfor anyx =

∑ri=1 λibi, y =

∑ri=1 µibi ∈ L we havex ≥ y if and only if λi ≥ µi for

eachi. This property implies thatsupLx, y =∑r

i=1(λi ∨ µi)bi andinfLx, y =∑ri=1(λi ∧ µi)bi, thereforeL is a lattice-subspace. The converse is also true. One can

prove it directly or by using the Choquet-Kenthal theorem, see in [6], Proposition 1.1.So we have the following:

Theorem 14. A finite-dimensional ordered subspaceL of C(Ω) is a lattice-subspaceif and only ifL has a positive basis.

Also each vectorbi of the positive basis ofL is an extremal point ofL+. ( A vectorx0 ∈ L+, x0 6= 0 is an extremal point ofL+ if for any x ∈ L, 0 ≤ x ≤ x0 impliesx = λx0 for some real numberλ). This property implies that a positive basis ofL isunique in the sense of positive multiples.

Theorem 15. [ [6], Proposition2.2] A finite-dimensional ordered subspaceL of C(Ω)is a sublattice ofC(Ω) if and only if L has a positive basisb1, b2, ..., br with theproperty: b−1

i (0, +∞) ∩ b−1j (0, +∞) = ∅ for anyi 6= j.

As an application of the above result we have:

Theorem 16. Suppose thatL is a sublattice ofRm. If the constant vector1 =(1, 1, ..., 1) is an element ofL, thenL has a positive basisb1, b2, ..., br which isa partition of the unit , i. e. the vectorsbi have disjoint supports and1 =

∑ri=1 bi.

This basis is unique.

14

Indeed, by the previous propositionL has a positive basisd1, d2, ..., dr withdisjoint supports. Since1 ∈ L we have1 =

∑ri=1 λidi and for eachj ∈ supp(di) we

have1 = 1(j) = λidi(j), thereforedi(j) = 1λi

for anyj ∈ supp(di). So eachdi isconstant on its support, therefore the basisbi = λidi is a positive basis ofL whichis a partition of the unit.

We suppose now thatz1, z2, ..., zr are fixed, linearly independent, positive vectorsof C(Ω) and that

L = [z1, z2, ..., zr],

is the subspace ofC(Ω) generated by the vectorszi. We study the problem: underwhat conditionsL is a lattice-subspace or a sublattice ofC(Ω)? In the case whereLfails to be a lattice-subspace we study ifL is contained in a finite-dimensional minimallattice-subspace ofC(Ω) or if the sublattice generated byL is finite-dimensional.

The function

β(t) =(z1(t)

z(t),z2(t)z(t)

, ...,zr(t)z(t)

), for each t ∈ Ω, with z(t) > 0,

wherez = z1+z2+...+zr, is thebasic function of z1, z2, ..., zr. This function is veryimportant for the study of lattice-subspaces and positive bases and has been defined in[5]. The setR(β) = β(t)

∣∣t ∈ Ω with z(t) > 0, is the range ofβ and the cardinalnumbercardR(β) of R(β) is the number of the (different) elements ofR(β). Underthe above notations we have, see in [6] Theorem 3.6:

Theorem 17 (Polyrakis ). L is a sublattice ofC(Ω) if and only ifcardR(β) = r.If R(β) = P1, P2, . . . , Pr, a positive basisb1, b2, ..., br of L is given by the

formula:(b1, b2, ..., br)T = A−1(z1, z2, ..., zr)T , (4)

whereA is the r × r matrix whose theith column is the vectorPi, for eachi =1, 2, ..., r, and(b1, b2, ..., br)T , (z1, z2, ..., zr)T are the matrices with rows the vectorsb1, b2, ..., br, z1, z2, ..., zr.

5.1 The algorithm for the sublattice generated byL

The next result gives an algorithm for the construction of the sublatticeZ of C(Ω)generated by a finite setz1, z2, ..., zr of linearly independent and positive vectors, inthe case whereZ is finite-dimensional. In this case a positive basis ofZ is determined.As in the previous theorem,β is the basic function ofz1, z2, ..., zr. Statement(d)determines the positive basis ofZ. In fact (d) is an application of the previous theoremfor the determination of a positive basis ofZ. For more details, see in [6], Theorem3.7.

Theorem 18 (Polyrakis ). LetZ be the sublattice ofC(Ω) generated byz1, z2, ..., zrand letµ ∈ N. Then the statements(i) and(ii) are equivalent:

(i) dim(Z) = µ.

(ii) R(β) = P1, P2, ..., Pµ.

15

If statement(ii) is true thenZ is constructed as follows:

(a) EnumerateR(β) so that itsr first vectors to be linearly independent (such anenumeration always exists). Denote again byPi, i = 1, 2, . . . , µ the new enu-meration and we putIr+k = t ∈ Ω

∣∣β(t) = Pr+k, for eachk = 1, 2, ..., µ− r.

(b) Define the vectorszr+k, k = 1, 2, ..., µ− r as follows:

zr+k(i) = z(i) if i ∈ Ir+k and zr+k(i) = 0 if i 6∈ Ir+k,

wherez = z1 + z2 + ... + zr is the sum of the vectorszi.

(c) Z = [z1, z2, . . . , zr, zr+1, . . . , zµ].

(d) A positive basisb1, b2, ..., bµ of Z is constructed as follows:

Consider the basic functionγ of z1, z2, . . . , zr, zr+1, . . . , zµ and suppose thatP ′1, P ′2, ..., P ′µ is the range ofγ (the range ofγ has exactlyµ points). Then

(b1, b2, ..., bµ)T = D−1(z1, z2, ..., zµ)T ,

whereD is theµ× µ matrix with columns the vectorsP ′1, P′2, ..., P

′µ.

References

[1] C. D. Aliprantis and R. Tourky,Markets that don’t replicate any option, EconomicLetters76 (2002), 443–447.

[2] A. M. Baptista,On the non-existence of redundant options, Economic Theory31(2007), 205–212.

[3] C. Kountzakis and I. A. Polyrakis,The completion of security markets, Decisionsin Economics and Finance29 (2006), 1–21.

[4] S. F. LeRoy and J. Werner,Principles of Financial Economics, Cambridge Univer-sity Press, 2001.

[5] I. A. Polyrakis,Finite-dimensional lattice-subspaces ofC(Ω) and curves ofIRn,Transactions of the American Mathematical Society384(1996), 2793–2810.

[6] , Minimal lattice-subspaces, Transactions of the American MathematicalSociety351(1999), 4183–4203.

[7] S. A. Ross,Options and efficiency, Quartely Journal of Economics90 (1976), 75–89.

16