Dry Markets and Statistical Arbitrage Bounds for European Derivatives

Transcription

1 Dry Markes and Sascal Arbrage Bounds for European Dervaves João Amaro de Maos Faculdade de Economa Unversdade Nova de Lsboa Campus de Campolde Lsboa, Porugal Ana Lacerda Deparmen of Sascs Columba Unversy 1255 Amserdam Avenue New York, NY , USA December 30, 2005 Absrac We derve sascal arbrage bounds for he buyng and sellng prce of European dervaves under ncomplee markes. In hs paper, ncompleeness s generaed due o he fac ha he marke s dry,.e., he underlyng asse canno be ransaced a ceran pons n me. In parcular, we re ne he noon of sascal arbrage n order o exend he procedure for he case where dryness s random,.e., a each pon n me he asse can be ransaced wh a gven probably. We analycally characerze several properes of he sascal arbragefree nerval, show ha s narrower han he super-replcaon nerval and domnaes somehow alernave nervals provded n he leraure. Moreover, we show ha, for su cenly ncomplee markes, he sascal arbrage nerval conans he reservaon prce of he dervave. 1 Inroducon In complee markes and under he absence of arbrage opporunes, he value of a European dervave mus be he same as he cheapes porfolo ha replcaes exacly s value a any gven pon n me. However, n he presence of some marke mperfecons, markes may become ncomplee, and s no possbleo exacly replcae he value of heeuropean dervave a all mes anymore. Neverheless, s possble o derve an arbrage-free range of varaon for he value of he dervave. hs nerval depends on 1

2 wo d eren facors. Frs, on he naure of marke ncompleeness; second, on he noon of arbrage opporunes. In wha follows we consder ha marke ncompleeness s generaed by he fac ha agens canno rade he underlyng asse on whch he dervave s wren whenever hey please. In fac, and as opposed o he radonal asse prcng assumpons, markes arevery rarelylqud and mmedacy s no always avalable. As Longsa (1995, 2001, 2004) recalls, he relevance of hs fac s pervasve hrough many nancal markes. he markes for many asses such as human capal, busness parnershps, penson plans, savng bonds, annues, russ, nherances and resdenal real esae, among ohers, are generally very llqud and long perods of me (monhs, somemes years) may be requred o sell an asse. hs pon becomes exremely relevan for he case of opon prcng when we consder ha s an ncreasngly common phenomenon even n well-esablshed secures markes, as llusraed by he 1998 Russan defaul crss, leadng many raders o be rapped n rsky posons hey could no unwnd. o address he mpac of hs ssue on dervaves prcng, we consder a dscree-me seng such ha ransacons are no possble whn a subse of pons n me. Alhough clearly very sylzed, he advanage of hs seng s ha ncorporaes n a very smple way he noon of marke llqudy as he absence of mmedacy. Under such llqudy we say ha markes are dry. In hs framework, dryness changes wha s oherwse a complee marke no a dynamcally ncomplee marke. hs was also he approach n Amaro de Maos and Anão (2001) when characerzng he spec c superreplcaon bounds for opons n such markes and s mplcaons. We furher exend hs seng by assumng haransacons occur a each pon n me wh a gven probably, re ecng a more realsc ex-ane uncerany abou he marke. As sressed above, here s no a unquearbrage-free value for a dervave under marke ncompleeness. However, for any gven dervave, porfoloscan be found ha havehesamepayo as he dervaven somesaes of naure, and hgher payo s n he oher saes. Such porfolos are sad o be superreplcang. Holdngsuch aporfoloshould be worh morehan he opon selfand herefore, hevalueof hecheapesof such porfolos should be seen as an upper bound on he sellng value of he opon. Smlarly, a lower bound for he buyng prce can be consruced. he naure of he superreplcang bounds s well characerzed n he conex of ncomplee markes n he papers by El Karou and Quenez (1991,1995), Edrsnghe, Nak and Uppal (1993) and Karazas and Kou (1996). he superreplcang bounds esablsh he lms of he nerval for arbrage-free value of he 2

3 opon. If he prce s ousde hs range, hen a posve pro s aanable wh probably one. herefore, he equlbrum prces a whch he dervave s ransaced should le beween hose bounds. Mos of he mes, however, hese superreplcang bounds are rval, n he sense ha hey are oo broad, no allowng a useful characerzaon of equlbrum prces vcny. As an alernave, Bernardo and Ledo (2000) propose a uly-based approach, resrcng he no-arbrage condon o rule ou nvesmen opporunes o erng hgh gan-loss raos, where gan (loss) s he expeced posve (negave) par of excess payo. In hs way, narrower bounds are obaned. Analogously, Cochrane and Saá-Requejo (2000) alsoresrc he no-arbrage condon by no allowng ransacons of good deals,.e. asses wh very hgh Sharpe rao. Followng Hansen and Jagannahan (1991), hey show ha hs resrcon mposes an upper bound on he prcng kernel volaly and leads o narrower prcng mplcaons when markes are ncomplee. Gven a se of prcng kernels compable wh he absence of arbrage opporunes, Cochrane and Sáa-Requejo exclude prcng kernels mplyng very hgh Sharpe raos, whereas Bernardo and Ledo exclude prcng kernels mplyngvery hgh gan-lossraos fora benchmark uly. Noceha, for a d eren uly, Bernardoand Ledo would exclude ad eren subse of prcng kernels, for he same levels of accepable gan-loss raos. Also noce ha he nerval desgned by Cochrane and Sáa-Requejo s no necessarly arbrage free, and herefore does no necessarly conan he equlbrum prce. In order o avod ad-hoc hresholds n eher Sharpe or gan-loss raos, or o make some paramerc assumpons abou a benchmark prcng kernel, as n Bernardo and Ledo (2000), he work of Bondarenko (2003)nroduces he noon of sascal arbrage opporuny, by mposng a weak assumpon on a funconal form of admssble prcng kernels, yeldng narrower prcng mplcaonsas compared ohe superreplcaon bounds. A sascal arbrage opporuny s characerzed as azero-cos radng sraegy for whch () he expeced payo s posve, and () he condonal expeced payo n each nal sae of he economy s nonnegave. Unlke a pure arbrage opporuny, a sascal arbrage opporuny may allow for negave payo s, provded ha he average payo n each nal sae s nonnegave. In parcular, rulng ou sascal arbrage opporunes mposes a novel marngale-ype resrcon on hedynamcs of secures prces. hemporan properes of he resrcon are ha s model-free, n he sense ha requres no paramerc assumpons abou he rue equlbrum model, and connues o hold when nvesors belefs are msaken. Alhough Bon- 3

4 darenko s nerval can be shown o be n he arbrage-free regon, does no necessarly conan he equlbrum value of he dervave. In hs paper we exend henoon ofsascal arbrageopporuny o he case where he underlyngasse can be ransaced a each pon n me wh a gven probably, and compare he sascal arbrage-free bounds wh he superreplcaon bounds. We show ha he sascal arbrage-free nerval s narrower han he pure arbrage bounds, and show also ha, for su cenly ncomplee markes (probably no oo close o1), hesascal arbrage nerval conans he reservaon prce of he dervave. We also provde examples ha allow comparson wh he resuls of Cochrane and Saá-Requejo (2000) and dscuss he comparson wh Bernardo and Ledo (2000). hs paper s organzed as follows. In secon 2, he model s presened and he pure arbrage resuls are derved. In secon 3 he noon of sascal arbrage n he spr of Bondarenko (2003) s de ned. In secon 4, he man resuls are presened. In Secon 5 we rs characerze he reservaon prces and hen show ha, n a su cenly dry marke, hey are conaned n he sascal arbrage nerval. In Secon 6 we llusrae how he sascal arbrage-free nerval somehow domnaes alernave nervals provded n he leraure. In secon 7 some numercal examples are presened n order ollusraesome mporan properes of he bounds. In he las secon several conclusons are presened. 2 he Model Consder a dscree-me economy wh perods, wh a rsky asse and a rskless asse. A each pon n me he prce of he rsky asse can be mulpled eher by U or by D o ge he prce of he nex pon n me. Equlbrum requreshau > R > D, wherer denoesoneplushersk-free neres rae. A me = 0 and = ransacons are ceranly possble. However, a = 1;:::; 1 here s uncerany abou he possbly of ransacon of he rsky asse. ransacons wll occur wh probably p a each of hese pons n me. A European Dervave wh maury s consdered. Consder he Bnomal ree process followed by he prce of he rsky asse. Le he se of nodes a dae be denoed by I ; and le each of he +1 elemens of I be denoed by = 1; :: : ; + 1. For any 0 <, le I 0 denoe he se of all he nodes a me 0 ha are predecessors of a gven node : A pah on he even ree s a se of nodes w = [ 2f0;1;:::;g such 4

5 ha each elemen n he unon sas es 1 2 I 1 : Le denoe he se of all pahs on he even ree: he payo s of a European dervave, a each ermnal node, wll be denoed G : A each node ; he sock prce s gven by S = U +1 D 1 S 0 : Moreover, a each node ; here s a number represenng he number of shares bough (or sold, f negave), and a number B denong he amoun nvesed (or borrowed, f negave) n he rsk-free asse. Hence, a here are + 1 values of, composng a vecor ( 1 ;:: : ; +1 ) 2 R +1 : Smlarly, we consruc he vecor B (B 1 ; :: : ;B+1 ) 2 R +1 : De non 1 A radng sraegy s a porfolo process µ = ( ; B ); composed of uns of he rsky asse and an amoun B nvesed n he rskless asse, such ha he porfolo s cos s S + B for = 0;1;:: : 1: In order o nd he upper (lower) bound of he arbrage-free range of varaon for he value of a European dervave we consder a nancal nsuon ha wshes obefully hedged when sellng (buyng) ha dervave. he objecve of he nsuon s o mnmze (maxmze) he cos of replcang he exercse value of he dervave a maury. he value deermned under such opmzaon procedure avods whas known asarbrage opporunes, re ecng he possbly of ceran pro s a zero cos. hs secon s organzed as follows. We rs characerze he upper bound, and hen he lower bound for he nerval of no-arbrage admssble prces. For each bound, we rs deal wh he complee marke case, and hen wh he fully ncomplee marke case, nally nroducng random ncompleeness. 2.1 he upper bound n he casep = 0 andp =1: Frs, we presen he well-known case where p = 1. he usual de non of an arbrage opporuny n our economy s as follows. De non 2 (Pure Arbrage n he case p = 1) In hs economy, an arbrage opporuny s a zero cos radng sraegy µ such ha 1. he value of he porfolo s posve a any nal node,.e., 1 1 S + RB 1 1 0; for any 1 2 I 1 and all 2 I ; and 2. he porfolo s self- nancng,.e., 1 1 S + RB 1 1 S +B for any 1 2 I 1 ; all 2 I and all 2 f0;:::; 1g: ; 5

6 he upper bound for he value of he European opon s he maxmum value for whch he dervave can be ransaced, whou allowng for arbrage opporunes. hs s he value of he cheapes porfolo ha he seller of he dervave can buy n order o compleely hedge hs poson agans he exercse a maury, whou he need of addonal nancng a any rebalancng daes. Hence, for p = 1; he upper bound s Cu 1 ; gven by Cu 1 = mn 0 S 0 + B 0 f ;B g =0;::::; 1 subjec o 1 1 S + RB 1 1 G ; wh 1 2 I 1 and all 2 I ; and he self- nancng consrans 1 1 S + RB 1 1 S + B ; for all 1 2 I 1 ; all 2 I and all 2 f0;:::; 1g; where he consrans re ec he absence of arbrage opporunes. hs problem leads o he famlar resul C 1 u = 1 R j=0 µ j µ R D U D j µ U R j G +1 j : U D Consder now he case where p = 0. In hs case, he noon of a radng srae# gy sasfyng he self- nancng consran s nnocuous, snce he porfolo µ canno be rebalanced durng he lfe of he opon. Under he absence of arbrage opporunes, he upper bound for he value s C 0 u sasfyng Cu 0 = mn 0S 0 +B 0 f 0;B0g subjec o 0 S + R 0 G ; for all 2 I : In hs case, he bound C 0 u can be shown o solve he maxmzaon problem on a se of posve consans f# g; wh P +1 =1 # = 1; subjec o Cu 0 = max 1 +1 # R # =1 G S 0 = 1 +1 R # =1 S ; For nsance, f a call opon wh exercse K s consdered, we have 1 Cu 0 = 1 µ R D U R U D S 0 K µ + U R D + U D S 0 K + : 1 See Amaro de Maos and Anão (2001). 6

7 2.2 he lower bound n he casep = 0 andp = 1: he lower bound for he value of an Amercan dervave s he mnmum value for whch he dervave can be ransaced whou allowng for arbrage opporunes. hs s he value of he mos expensve porfolo ha he buyer of he opon can sell n order o be fully hedged, and whou he need of addonal nancng a rebalancng daes. For p = 1, he lower bound for he value of he dervave under he absence of arbrage opporunes s hus Cl 1 ; gven by Cl 1 = max 0 S 0 + B 0 f ;Bg =0;::::; 1 subjec o 1 1 S + RB 1 1 G, wh 1 2 I 1 and all 2 I ; and he self- nancng consrans 1 1 S + RB 1 1 S + B ; for all 1 2 I 1 ; all 2 I and all 2 f0;:::; 1g; where he consrans re ec he absence of arbrage opporunes hs problem leads o he famlar resul C 1 l = 1 R µ j j=0 µ R D U D j µ U R j G +1 j U D ; (1) ha concdes wh he soluon obaned for C 1 u : In he case where p = 0; he lower bound for he value of he dervave s C 0 l ; sasfyng C 0 l = max f 0;B0g 0S 0 +B 0 subjec o 0 S +R B 0 G : As above, follows ha, for ase of posve consans f# g; wh P +1 =1 # = 1; hs bound s gven by subjec o Cu = mn # R # =1 G S 0 = 1 R +1 =1 # S : 7

8 In he case of a call opon wh exercse K,we have 2 Cl 0 = 1 Ã! R U (+1) D +1 U R U D U (+1) D +1 D S 0 K µ U D R U 1 R U D U (+1) D +1 D +1 S 0 K + : (2) where s de ned as he unque neger sasfyng U n (+1) D +1 < R n < U n D, and 0 n 1: 2.3 he Bounds on Probablsc Markes In he aforemenoned cases we consdered he cases where eher p = 0 or p = 1. However, f p s no equal o neher 0 nor 1, he formulaon has o be adjused. If he rsky asse can be ransaced wh a gven probably p 2 (0;1), hen he usual de non of arbrage opporuny reads as follows. De non 3 (Pure Arbrage for p 2 (0;1)) In hs economy, an arbrage opporuny s a zero cos radng sraegy such ha 1. he value of he porfolo s posve a any nal node,.e., S + R B 0; 2 I and all 2 I ; and he self- nancng consrans 2. he porfolo s self- nancng,.e., j j S + R j B j j S + B ; for all j 2 I j ; all 2 I and all 2 f0;:::; 1g: he upper bound C p u s he soluon of he followng problem: Cu p = mn 0 S 0 +B 0 f ;Bg =0;:::; 1 where ;B 2 R +1 ; = 0;:::; 1; subjec o he superreplcang condons S + R B G ; wh 2 I and all 2 I ; and he self- nancng consrans j j S +R j B j j S +B for all j 2 I j ; all 2 I and all 2 f0; :::; 1g: On he oher hand, he lower bound C p l solves he followng problem: 2 See Amaro de Maos and Anão (2001). 8

9 C p l = max 0 S 0 +B 0 f ;Bg =0;:::; 1 where ; B 2 R +1 ; = 0; :::; 1; subjec o he condons S + R B G ; wh 2 I and all 2 I ; and he self- nancng consrans j j S +R j B j j S +B for all j 2 I j ; all 2 I and all 2 f0;:::; 1g: Noce ha he consrans n he above opmzaon problems are mpled by he absence of arbrage opporunes and do no depend on he probably p: 3 herefore, neher Cu p nor C p l wll depend on p: We are now n condons o relae hese values o Cu 0 and C0 l as follows. heorem 4 For p 2 (0; 1) he upper and lower bound for he prces above do no depend on p: he opmzaon problems above lead o he same soluons as when p = 0: Proof. Consder rs he case of he upper bound. he consrans characerzng Cu p nclude all he consrans characerzng Cu 0: hus, Cp u Cu 0. Now, le 0 0 and B0 0 denoe he opmal values nvesed, a me = 0; when p = 0. he radng sraegy p = 0 0 and Bp = R B0 0, for all = 1;:::; 1, s an admssble sraegy for any gven p, hence Cu p = Cu 0: he case of he lower bound s analogous. henuon forhs resuls sraghforward. he upper (lower) bound of he European dervave remans he same as when p = 0; because wh probably 1 p would no be possble o ransac he sock a each pon n me. In order o be fully hedged, as requred by he absence of arbrage opporunes, he worse scenaro wll be resrcve n spe of s possbly low probably. he fac ha no nermedae ransacons may occur domnaes all oher possbles. he above resul s srongly drven by he de non of arbrage opporunes. Neverheless, f hs noon s relaxed n an economc sensble way, a narrower arbrage-free range of varaon for he value of he European dervavemaybeobaned, possblydependngnowon p: hsshesubjec of he res of he paper. 3 hs happens snce, n order o have an arbrage opporuny, we mus ensure ha, wheher marke exss or no a each me 2 f1;:::; 1g; he agen wll never lose wealh. herefore, he opmzaon problem canno depend onp: 9

10 3 Sascal Arbrage Opporuny Consder heeconomydescrbed n he prevous secon. Le p = f1;:::; 1g denoe he se ofpons n me. A each ofheseponshere smarke wh probably p; and here s no marke wh probably 1 p: he exsence (or no) of he marke a me corresponds o he realzaon of a random varable y ha assumes he value 0 (when here s no marke) and 1 (when here s marke). hs random varable s de ned for all 2 p and s assumed o be ndependen of he ordnary source of uncerany ha generaes he prce process. We can herefore alk abou a marke exsence process. In order o consruc one such process, le us sar wh he sae space. Le #( p ) denoe he number of pons n p : A each of hese pons, marke may eher exs or no exs, leadng o 2 #( p) possble saes of naure. We hen have he collecon of possble saes of naure denoed by ^ = fv g =1;:::;2 #( p) ; each v correspondng o a dsnc sae. Moreover, le ^F = ^F 1 ;: :: ; ^F 1 ; where ^F s he ¾-algebra generaed by he random varable y. Le p y be he probably assocaed wh he random varable y : For all 2 p ; we have p y (y = 1) = p and p y (y = 0) = 1 p: 3.1 he expeced value of a porfolo We now consruc a random varable ha allows o consruc he expeced fuure value of a porfolo n hs seng. For < 0 ; le x ; 0 be a random varable denfyng he las me ha ransacons ake place before dae 0 ; gven ha we are a me ; and ransacons n are currenly o possble. Le ^ be he subse of ^ such ha ^ = v 2 ^ : y (v ) = 1 : hen, x ; 0 : ^! ;: :: ; 0 1 ª Le p x; 0 be he probably assocaed wh x ; 0: hen, p x; 0 Moreover, for a gven s 2 (; 0 ); x; 0 = = (1 p) 0 1 : p x ; 0 x; 0 = s = p(1 p) 0 s 1 : Also noe ha 0 1 s= p x ; 0 x; 0 = s = (1 p) s= p(1 p)0 s 1 = 1;

11 as should. Consder a gven radng sraegy ( ; B ) =0;:::; ; where ( ; B ) ( ; B ) 2I s a ( + 1) dmensonal vecor. Consder a gven pah w and ( s ) s2f;:::;g ½ w. Suppose ha he agen s a a gven node ; where rebalancng s possble. As here s uncerany abou he exsence of marke a he fuure pons n me, here s also uncerany abou he porfolo ha he agen wll ³ be holdng ³ a a fuure node 0: In ³ fac, he porfolo a 0 may be any of ; B ; ;B ;: :: ; or ; B ; where ( s ) s2f;:::; 0 1g ½ w: Clearly, he expeced value of a gven radng sraegy a node 0, gven ha we are a node ; s E px ; 0 h x x S 0 + x R0 B x x = p s=;:::; 0 x 1 ; 0 x; 0 = s h s s S s 0 +R0 B s s ; where we use x o shor noaon for x ; 0: 3.2 Sascal versus pure arbrage A pure arbrage opporuny s azero-cos porfoloa me, such ha he value of each possble porfolo a node s posve,.e., +j +j S +R j B +j +j 0 for all +j such ha s a predecessor, j = 0;1;: :: ; 1 and E p h x ; x x S +R x B x x > 0; ogeher wh he self- nancng consrans S +j +j +Rj B +j +j S +j +j + B +j +j ; for all +j such ha s a predecessor, and j = 1; :: : ; 1: If sascal arbrage s consdered, however, an arbrage opporuny requres only ha, a node ; he expeced value of he porfolo a s posve, E px ; h x x S +R x B x x 0; ogeher wh weaker self- nancng condons. Le us regard hese laer condons n some deal. 11

12 Suppose ha we are a a gven node : If here ³ s marke a he nex pon n me we hen have, for sure, he porfolo ;B a me + 1. Hence, f node +1 s reached, he self- nancng condon s ³ S RB S B Consder now ha +2 s reached. A me here s uncerany abou he exsence of he marke ³ a me + 1: Hence, ³ a me + 2 we can eher have he porfolo ; B or he porfolo ;B : Under he concep of sascal arbrage, we wan o ensure ha, n expeced value, we are no gong o lose a node +2 : Hence, he self- nancng condon becomes s=;+1 p x;+2 (x ;+2 = s) ³ ³ s s S R+2 s Bs s S B More generally, for any a whch ransacon occurs and < 0 < ; he sascal self- nancng condon becomes h ³ x S 0 +R 0 0 x B x x 0 S 0 0 +B E px ; 0 De non 5 4 A sascal arbrage opporuny s a zero-cos radng sraegy for whch 1. A any node, he expeced value of he porfolo a any nal node s posve,.e., h x x ; S + R x Bx x ; 0 E px ; 4 hs noon of Arbrage Opporuny s n he spr of Bondarenko (2003). In hs de non 2, a Sascal Arbrage Opporuny (SAO) s de ned as a zero-cos radng sraegy wh a payo Z =Z(F), such ha ()E[ZjF0]>0; and ()E[ZjF0;» ] 0; for all» ; where» denoes he sae of he Naure a me; andf = (» 1 ;:::;» ) s he marke nformaon se, whf 0 =Á. Also, he second expecaon s aken a me = 0 and s condonal o he ermnal sae». However, noce ha elmnang SAO s a me =0 does no mply he absence of SAO s a fuure mes2[1; 1]. Hence, n order o ncorporae a dynamcally conssen absence of SAO s, we re ne he de non of a SAO as a zero-cos radng sraegy wh a payo Z =Z(F), such ha ()E[Z jf 0 ]>0; and ()E[ZjF;» ] 0; for all» and all2 [0; 1]: 12

13 for any 2 I and 2 f0;1; :: : ; 1g; and 2. he porfolo s sascally self- nancng,.e:, E p h ³ x ; 0 x x ; 0 S 0 + x R0 B x x ; 0 0 S 0 0 +B for any 2 I ; 0 >, 2 f0;1; :: : ; 2g and 0 2 f1;: :: ; 1g: he wo de nons of arbrage are relaed n he followng. heorem 6 If here are no sascal arbrage opporunes, hen here are no pure arbrage opporunes. Proof. If here s a pure arbrage opporuny hen he nequales presenn hede non ofarbrage opporuny, de non 3, arerespeced. Hence, as heseexpressons areheerms under expecaon n hede non of Sascal Arbrage opporuny, presened n de non 5, here s also a sascal arbrage opporuny. he se of porfolos ha represen a pure arbrage opporuny s a subse of heporfolos ha represen asascal arbrage opporuny,.e., here are porfolos ha, n spe of no beng a pure arbrage opporuny, represen a sascal arbrage opporuny. In order o have a sascal arbrage opporuny s no necessary (alhough s su cen) ha he value of he porfolo a he nal dae s posve. I s only necessary ha, for all, he expeced value of he porfolo a he nal dae s posve. Consder now he self- nancng condons under sascal arbrage. When rebalancng he porfolo s no necessary (alhough s su cen) ha he value of he new porfolo s smaller han he value of he old one. hs happens because fuure rebalancng s unceran, leadng o uncerany abou he porfolo ha he agen wll be holdng n any fuure momen. In order o avod a sascal arbrage opporuny s only necessary ha he expeced value of he porfolo a a gven pon n me s larger han he value of he rebalancng porfolo. Fnally, noce ha he concep of sascal arbrage opporuny reduces o he usual concep of arbrage opporuny n he lmng case p = 0: 3.3 Augmened measures For echncal reasons, we now de ne an augmened probably space Q on. In order o do ha, we de ne a sempah m from o 0, whch s a se 13

14 of nodes m = [ k2f;:::; 0 g k such ha k 2 I k+1 k : Le + ; 0 denoe he se of sempahs from o 0: De non n o 7 An augmened probably space n s a se of probables such ha 2 I ; m 2 + ;, = 0;: :: ; and q ( ;);m (;) 1 q ( ;);m (;) = 1; =0 m De non 8 A mod ed marngale probably measure s an augmened probably measure Q 2 Q whch sas es () S 0 = 1 R f 2I g q S where wh and () q = 1 =0 S 1 1 = 1 R n :2I o q ( ;);m ( n o ;) S ; m2 + ; n : 12I 1 o ¼ ¼ ( ;);m n : 1 2I o 1 ¼ ( ;);m (;) = 1 1 p x; (x ; = 1) =0 n ( ;);m (;) S (;) = 1 :2I 1 o n m2 + ;: 12m o q( ;);m (;) where 1 = p x; (x ; = 1) =0 n : 1 2I 1 o n :2I 1 o n m2 + ; : 12m o q( ;);m (;) ; 14

15 n () here exss ( 0;0 );m (;) o ;for all 0 2 I 0; 2 I ; m 2 + ; and 0 > for all = 0;:: : ; 1 such ha, for all 0 < k < ; S k k = 1 R k µ (;);m ( n : k 2I o ;) S + 0 >k k where and µ ( ;);m (;) = 1 " ( 0;0 );m (;) = 1 wh = n : k 2I 0 >k<k k o p x; 0 µ ( ;);m n : k 2I o k k =0 <k k =0 (;) + 0 >k p x; (x ; = k) n : 2I o k p x; 0 x; 0 = k n :2I k o p x; (x ; = k) n :2I k x; 0 = k n : 2I k o 1 R 0 k "( 0; 0 );m ( ;) S 0 0 ; " ( 0;0 );m (;) = 1 o n m2 + ;: k2m o q( ;);m (;) n o ( 0;0 );m (;) m2 + ; : k2m n o q( ;);m ( ;) + m2 + ; : k 2m n m2 + ; : k2m o ( 0;0 );m (;) : We denoe by Q S : he se of mod ed marngale probably measure. Such measures wll help wrng down he upper and lower bounds for he value of European dervaves under he absence of sascal arbrage opporunes. 4 Man Resuls 4.1 he upper bound he Problem he problem of deermnngheupper bound ofhesascal arbrage-free range of varaon for he value of a European dervave, can be saed as C u = mn 0 S 0 + B 0 f ;Bg =0;:::; 1 15

16 where ;B 2 R +1 ; = 0; :::; 1 subjec o he condons of a posve expeced payo x x S + R x Bx x G ; E px ; 0 E px ; for any 2 I and 2 f0;1;:: : ; 1g 5 ; and self- nancng condons h ³ x x S x 0 +R0 B x x 0 0 S B for any 2 I ; 0 >, 2 f0; 1;: :: ; 2g and 0 2 f1;: :: ; 1g 6 : Example 9 Illusraon of he opmzaon problem wh = 3. he evoluon of he prce underlyng asse can be represened by he ree n gure 1. =0 =1 =2 =3 Fgure 1: Evoluon of he undelyng asse prce. In wha concerns he evoluon of he prce process here are egh d eren saes,.e., = fw g =1;:::;8 : he problem ha mus be solved n order o nd he upper bound s he followng. where C u = mn f ;Bg =0;:::;2 0 S 0 + B 0 f 0 ; B 0 g = f( 0 ;B 0 )g f 1 ; B 1 g = ;B1 ; 2 1 ; B 2 ª 1 f 2 ; B 2 g = ;B1 ; 2 2 ; B 2 2 ; 3 2 ; B2 3 ª 5 For each here are 2 ( ) pahs, and as a resul, 2 ( ) (+1) resrcons a me. he oal number of resrcons s P 1 =0 2( ) (+1): 6 For each here are P 1 0 =+1 20 : Hence, for each here are (+1) P 1 0 =+1 20 : Hence, here are P 2 =0 (+1) P 1 0 =+1 20 resrcons. 16

17 subjec o he condons of a posve expeced payo h 2 2 S3 3 + RB2 2 G 3 3 ; for all 3 2 I 3 and 2 2 I 2 such ha 2 2 I2 3 h h p 2 2 S 3 3 +RB (1 p) 1 1 S R2 B 1 1 : (hese are 6 consrans); G 3 3 for all 3 2 I 3 ; 2 2 I 2 and 1 2 I 1 such ha 2 2 I 3 2 : and 1 2 I 2 1 ;.e., 1; 2 and 3 belong o he same pah (hese are 8 consrans) and p 2 + p(1 p) h h 2 2 S3 3 + RB2 2 +p(1 p) 1 1 S3 3 + R2 B1 1 h + +(1 p) 2 0 S3 3 + R3 B2 2 G 3 3 for all 3 2 I 3 ; 2 2 I 2 and 1 2 I 1 such ha 2 2 I2 3: and 1 2 I1 2 ;.e., 1; 2 and 3 belong o he same pah (hese are 8 consrans). Moreover, he self- nancng consrans mus also be consdered for any 1 2 I 1 (2 consrans), h (1 p) 0 S2 2 + R2 B2 2 0 S1 1 +RB S1 1 + B1 1 + p h 1 1 S2 2 +RB S2 2 + B2 2 for any 2 2 I 2 and 1 2 I 1 such ha 1 2 I 2 1 ;.e., 1;and 2 belong o he same pah (hese are 4 consrans) and, nally, 1 1 S RB S B 2 2 for any 2 2 I 2 and 1 2 I 1 such ha 1 2 I 2 1 ;.e., 1;and 2 belong o he same pah (hese are 4 consrans) Soluon heorem 10 here exss a mod ed marngale probably measure, q 2 Q S ; such ha he upper bound for arbrage-free value of a European opon can be wren as 1 C u = max q 2Q S R f 2I g q G : (3) 17

18 Proof. See proof n appendx A.1. Remark 11 If a Call Opon s consdered, he values for q ; n a model wh wo perods are explcly calculaed n appendx A.3. In ha case can be shown ha for a srcly posve p; he q 1 ; q 2 and q 3 are also srcly posve. In wha follows we characerze some relevan properes of C u : 1. C u C 0 u Proof. Le 0 0 and B0 0denoe he opmal values nvesed, a me = 0; n he sock and n he rsk-free asse respecvely, when p = 0. he radng sraegy ¹ = ¹ P=0 0 and B ¹ = R B ¹ 0 p=0, for = 1;:::; 1, s an admssble sraegy for any gven p. As a resul, he soluon of he problem for anypcanno be larger ha he value of hs porfolo a = 0 (whch s Cu 0). 2. C u Cu 1: Proof. Consder he radng sraegy ¹ ; ¹B ha solves =0;:::; 1 he maxmzaon problem ha characerzes he upper bound for a p 2 (0;1): hs s an admssble sraegy for he case p = 1, because s self- nancng,.e., 1 1 S + RB 1 1 S +B ; and superreplcaes he payo of he European dervave a maury,.e., 1 1 S + RB 1 1 G : Hence, he soluon of he problem for p = 1 canno be hgher han he value of hs porfolo a = 0 (whch s C u ). 3. lm p!0 C u = C 0 u and lm p!1 C u = C 1 u : Proof. See Appendx A.4 An example for a Call Opon and =2 s also show n appendx A C u s a decreasng funcon of p. Proof. See Appendx A.4 18

19 5. For a Call Opon and = 2, we can prove ha C u pc 1 u +(1 p)c0 u meanng ha he probablsc upper bound s a convex lnear combnaon of he perfecly lqud upper bound and he perfecly llqud upper bound. Proof. See appendx A he Lower Bound he organzaon of hs secon s analogous o he secon for he upper bound he Problem he problem of deermnng helower bound of he sascal arbrage-free range of varaon for he value of an European dervave, can be saed as where C l = max f ;Bg =0;:::; 1 0 S 0 + B 0 ;B 2 R +1 ; = 0; :::; 1 subjec o he condons of a posve expeced payo E px ; x x S + R x B x x G ; for any 2 I and 2 f0;1;:: : ; 1g 7 ;and self- nancng condons h ³ x x S x 0 +R0 B x x 0 S B E px ; 0 for any 2 I ; 0 >, 2 f0; 1;: :: ; 2g and 0 2 f1;: :: ; 1g 8 : 7 As n he upper bound case, for each here are 2 ( ) pahs, and as a resul, 2 ( ) (+1) resrcons a me. he oal number of resrcons s P 1 =0 2( ) (+1): 8 As n he upper bound case, for each here are P 1 0 =+1 20 : Hence, for each here are ( +1) P 1 0 =+1 20 : Hence, here are P 2 =0 (+1) P 1 0 =+1 20 resrcons. 19

20 4.2.2 Soluon heorem 12 here exss an mod ed marngale probably measure, q 2 Q S ; such ha he upper bound for arbrage-free value of an European opon can be wren as 1 C l = mn q 2Q S R f 2I g q G : Proof. he proof s analogous o he upper bound. Remark 13 If a call opon s consdered, he values for q ; n a model wh wo perods are explcly calculaed n appendx B.2. In ha case can be shown ha for a srcly posve p; he q 1 ; q 2 and q 3 are also srcly posve. In wha follows we characerze some relevan properes of C l : 1. C l C 0 l : 2. C l C 1 l : 3. lm p!0 C l = C 0 l and lm p!0 C l = C 1 l : An example for a call opon and = 2 s shown n appendx?? 4. C l s a ncreasng funcon of p. heproofs of heseproperes areanalogous o hose presened for he upper bound. 5 Uly Funcons and Reservaon Prces In hs secon we show ha he prce for whch any agen s nd eren beween ransacng or no ransacng he dervave, o be called he reservaon prce of he dervave, s conaned whn he sascal arbrage bounds derved above. Le u (:) denoe a uly funcon represenng he preferences of an agen a me. he argumen of he uly funcon s assumed o be he consumpon a me. Le y be he nal endowmen³ of he agen, and Z denoe he vecor of consumpon a me ;.e., Z = Z 2I : Le ½ be a dscoun facor. If an agen sells a European dervave by he amounc; 20

21 and ha dervave has a payo a maury gven G, he maxmum uly ha he or she can aan s u sell (C;p) = sup E G;P 0 f ;Bg =0 ½ u (Z ) =0;:::; 1 subjec o Z S 0 + B 0 C + y Z + S + B Z j j S j j S + R j B j +R j B j j G for all 2 I ; 2 I, j and = 1;: :: ; 1 where E G;P 0 denoes a bvarae expeced value, a = 0, wh respec o he probably P nduced by he marke exsence and he probably G underlyng he sochasc evoluon of he prce process. Smlarly, f he agen decdes no o nclude dervaves n hs or her porfolo, he maxmum uly ha he or she can aan s gven by subjec o j u (p) = sup E G;P 0 f ;Bg =0 ½ u (Z ) =0;:::; 1 Z S 0 +B 0 y Z + S + B Z j j S j j S +R j B j + R j B j j Lemma 14 In he case of random dryness, here s p > 0 such ha, for all p < p ; he uly aaned sellng he dervave by C u, s larger han he uly aaned f he dervave s no ncluded n he porfolo. u sell (C u; p) u (p): Proof. Le, for a gven p; f ; B g =0;:::; 1 denoe he soluon of he uly maxmzaon problem wh no dervave and f u ;Bu g =0;:::; 1 denoe he soluon of he mnmzaon problem ha mus be solved o nd he upper bound f sascal arbrage opporunes are consdered (see secon 4.1). Moreover, le sell ;B sell ª denoe an admssble soluon of he uly maxmzaon problem when he agen sells one un =0;:::; 1 of 21 j

22 he dervave. Now, consder he lm case, when p approaches zero. In ha case, he porfolo n o sell ;B sell f + u ;B + B ug =0;:::; 1 =0;:::; 1 s an admssble soluon of he uly maxmzaon problem when he agen sells one un of dervave by C u. he reason s as follows. he consran se of he problem ha mus be solved o nd he upper bound s connuous n p: Hence, when p! 0; he soluon of he problem s f u ;Bu g =1;:::; 1 = u 0 ;R B0 u ª where f u 0 ;B0 u g s he soluon of followng problem ½ ¾ mn S 0 + B s.a. S ;B + R B G ;8 : As C = C u = u 0 S 0 + B 0 ; he porfolo f u ;B u g =1;:::; 1 s an admssble soluon of he uly maxmzaon problem when one un of he dervave s beng sold. Moreover, guaranees a posve expeced uly. 9 Hence, he porfolo sell ; B sell ª s also admssble soluon for he =0;:::; 1 opmzaon problem, when one un of he dervave s beng sold, whch guaranees a hgher payo han he porfolo f ;B g =0;:::; 1 : Hence, u sell (C u; 0) u (0): Connuy on p of boh u sell and u ensure he resul. Remark 15 Noce ha he exsence of p follows from he connuy of he ules n p: Furhermore, s possble o have p = 1. Examples wh d eren values of p are gven n he end of hs paper. he range of values p < p characerzes wha was vaguely descrbed as su cenly ncomplee markes n he nroducon. he reservaon prce for an agen ha s sellng he opon s de ned as he value of C ha makes u sell (C;p) = u (p). Le R u denoe such reservaon prce. heorem 16 For all p < p we have R u C u : u sell Proof. he opmal uly value; (C; p) = sup C + y ( 0 S 0 +B 0 ) + E G;P 0 f ;B g =0;:::; 1 =1 ½ u (Z ); 9 I su ces o consderz 0 =y;z =0 andz = S +R B G 0: 22

23 s ncreasng n C: hs, ogeher wh lemma 14 leads o he resul. he same apples for he case when he agen s buyng a dervave. In ha case, f an agen s buyng he dervave by C, he maxmum uly ha he or she can aan s subjec o u buy (C;p) = sup E0 G;P f ;B g =0;:::; 1 Z S 0 +B 0 C +y Z + S + B j Z j j S j S + R j B j =0 ½ u (Z ) +R j B j j j +G Lemma 17 In he case of random dryness, here s p > 0 such ha, for all p < p ; he uly aaned buyng he dervave by C l, s larger han he uly aaned f he dervave s no ncluded n he porfolo. u buy (C l;p) u (p): Proof. he proof s analogous o he one n proposon 14 Le R l denoe he reservaon sellng prce,.e., he prce such ha u (p) = u buy (R l; p). heorem 18 For all p < p we have R l C p l : Proof. he proof s analogous o he one presened n heorem (16). However, n hs case he uly s a decreasng funcon of C; and we oban u buy (C l; p) u (p) ) R l C l : Several llusraons are presened n secon 7. 6 Comparsons wh he Leraure In wha follows we compare our mehodology wh ohers n he leraure, namely Bernardo and Ledo (2000) and Cochrane and Saá-Requejo (2000). Cochrane and Saá-Requejo (2000) nroduce he noon of good deals, or nvesmen opporunes wh hgh Sharpe raos. hey show ha rulng ou nvesmen opporunes wh hgh Sharpe raos, heycan oban narrower 23

24 bounds on secures prces. However, as sressed n Bondarenko (2003), no all pure arbrage opporunes qualfy as good deals. Moreover, for a gven se of parameers we found ou ha n order o conan he reservaon prces of a rsk neural agen he nerval s more broad han he one ha was obaned n our formulaon. We rs provde a smple example o compare our bounds wh pure arbrage bounds. Example 19 Consder a smple wo perods example, where ransacons are ceranly possble a mes = 0 and = 2: A me = 1 here are ransacons wh a gven probably p = 0:65: he nal sock prce s S 0 = 100 and may eher ncrease n each perod wh a probably 0:55; or decrease wh a probably 0:45: We ake and U = 1:2;D = 0:8 and R = 1:1: A call opon ha maures a me = 2 wh exercse prce K = 80 s consdered. Usng pure arbrage argumens we nd he followng range of varaon for he value of he call opon [33:88; 37:69] Usng he noon of sascal arbrage opporuny, he above range ges narrower and s gven by [34:31; 35:17]; clearly narrower ha he above nerval. If markes were complee (p = 1), he value of he opon would be 34:71. Also, he reservaon prce 10 for a rsk neural agen s equal o 35:09. Noce ha boh nervals nclude he complee marke value and he reservaon prce. We now use he same example o compare our mehodology wh he one presened by Cochrane and Saá-Requejo (2000). We show ha eher our nerval s conaned n hers, or else, her nerval do no conan he above menoned reservaon prce. Wh he Sharpe rao mehodology he lower bound s gven by C = mn fmg E m[s 2 K;0] +ª 10 In hs example, he reservaon prce for an agen who s buyng he dervave concdes wh ha of an agen who s sellng. 24

25 subjec o S 0 = E [ms 2 ]; m 0; ¾(m) h R 2; where S 0 s he nal prce of he rsky asse, and S 2 s he prce of he rsky asse a me = 2 11 : he upper bound s subjec o C = max fmg E m[s 2 K; 0] +ª p = E [ms 2 ] ;m 0; ¾(m) h R 2: Example 20 In order o compare he sascal arbrage nerval wh he Sharpe rao bounds, we mus choose he ad-hoc facor h so as o make one of he lmng bounds o concde. If we wan he upper bound of he Sharpe Rao mehodology o concde wh he upper bound obaned wh sascal arbrage, we mus ake h = 0:3173: In ha case, he lower bound wll be 33:88 and he range of varaon wll be [33:88; 35:17]; worse han he sascal arbrage nerval. Alernavely, f we wan he lower bound of he Sharpe Rao mehodology o concde wh he lower bound obaned wh sascal arbrage, we mus ake h = 0:28359: 12 In ha case, he upper bound wll be 34:49 and he range of varaon wll be [34:31; 34:49]: Alhough hs nerval s gher han he sascal arbrage nerval, does no conan he reservaon prce for a rsk neural agen. In a d eren paper Bernardo and Ledo (2000) preclude nvesmens o erng hgh gan-loss raos o a benchmark nvesor, somehow analogous 11 As sressed by Cochrane and Saá-Requejo, n a former paper Hansen and Jagannahan (1991) have shown haa consran on he dscoun facor volaly s equvalen o mpose an upper lm on he Sharpe rao of mean excess reurn o sandard devaon. 12 In order o ge a lower bound hgher han 33:88 s necessary o mpose adonally ham>0: If ha were no he case, hen he lower bound would only be de ned forh 0:2980 and would be equal o 33:88: 25

26 o he good deals of Cochrane and Sáa-Requejo. he creron, however, s d eren snce Bernardo and Ledo (2000) propose a uly-based approach, as sressed n he Inroducon. In hs way, he arbrage-free range of varaon for he value of he European dervave s narrower han n he case of pure arbrage. Le ~z + denoe he (random) gan and ~z denoe he (random) loss of a gven nvesmen opporuny. he uly of a benchmark agen characerzes a prcng kernel ha nduces a probably measure, accordng o whch he expeced gan-loss rao s bounded from above E (~z + ) E (~z ) ¹L: he far prce s he one ha makes he ne resul of he nvesmen o be null. In oher words, for a benchmark nvesor, would correspond o he prcng kernel ha would make E ~z + ~z = 0, E (~z+ ) E (~z ) = 1: hs las equaly characerzes he benchmark prcng kernel for a gven uly. Noce ha he far prce consruced n hs way concdes wh our de non of he reservaon prce. herefore, by choosng ¹L larger han one, he nerval bul by Bernardo and Ledo conans by consrucon he reservaon prce of he benchmark agen. On he oher hand, he arbrary hreshold ¹L can be chosen such ha her nerval s conaned n he sascal arbrage-free nerval. he dsadvanages, however, are clear. Frs, he hreshold s ad-hoc, jus as n he case of Cochrane and Sáa-Requejo; second, he consruced nerval depends on he benchmark nvesor; and nally, he only reservaon prce ha s conaned for sure n ha nerval, s he reservaon prce of he benchmark nvesor. In oher words, we canno guaranee ha he reservaon prce of an arbrary agen, d eren from he benchmark, s conaned n ha nerval. 7 Numercal Examples 7.1 Upper and Lower Bounds In hs secon several numercal examples are provded n order o llusrae he properes of he upper and lower bounds presened n he prevous 26

27 secons. Usng numercal examples we can conclude ha, for a call opon, C u pc 1 u + (1 p)c0 u : If he Call Opon s sold by he expeced value of he call, regardng he exsence (or no) of marke, here wll be an arbrage opporuny n sascal erms. he reason s ha he agen ha sells he call opon can buy a hedgng porfolo (n a sascal sense) by an amoun smaller han he expeced value of he call opon. As a resul, here s an arbrage opporuny, becausehes recevng more for he call opon han s payng for he hedgng porfolo. However, n whaconcerns helower bound, snopossbleoconclude wheher C l pcl 1+(1 p)c0 l or C l pcl 1+(1 p)c0 l : ha depends on he value of he parameers. Alhough n he wo-perod smulaon n Fgure 2 we seem o have he former case, he hree perod example n Fgure?? seems o sugges he laer, snce he lower bound behaves as a concave funcon ofpfor mos of s doman. Fnally, we may use Fgure 4 o llusrae several feaures. he Upper and Lower bounds of an European Call Opon for d eren values of p and K (K = 80, K = 100, K = 120, K = 140 e K = 160) n a hree perod model, wh U = 1:2; R = 1:1; D = 0:8 and S = 100. Frs, le us regard he suaons n hs Fgure ha are relaed o pure arbrage. hs ncludes he value of he dervave under perfecly lqud (p = 1) and perfecly llqud (p = 0) markes. In he former case, he unquevalueofhe dervave clearly decreases wh he exercse prce K; as should. In he laer, boh he upper bound Cu 0 and he lower bound C0 l also decrease wh K: More curously, however, he spread Cu 0 Cl 0 has a non-monoonc behavour. Obvously hs d erence s null for K less han he n mum value ofhe sock a maury and mus goozeroas he srke approaches he supremum of he sock s possble values a maury. In he mddle of hs range aans a maxmum. In our numercal example we observe ha he maxmum value of he spread s aaned for K close o 120. Regardng he sascal arbrage doman when p 2 (0; 1); we noce ha all he above remarks reman rue. he Fgure also suggess ha, for any gven p; he spread aans s maxmum for he same value of K as before. Noce ha he spread C u C l decreases wh p for xed srke K convergng o zero as p! 1: Hence, alhough somehow d eren from he radonal de non of arbrage, he noon of sascal arbrage seems 27

28 40 Upper Lower p Fgure 2: he Upper and Lower Bounds for a European Call Opon wh d eren value of p n a wo perod model, wh U = 1:2, R = 1:1, D = 0:8; S 0 = 100 and K = 80: o provde a very nce brdge, for 0 < p < 1, beween he wo exreme cases above (p = 0 and p = 1), where he orgnal concep of arbrage makes sense. hs can be seen n gure 5. A hrd ssue drven by he gure s he remark ha, for nermedae values of K; here s an overlappng of he d eren spreads Cu 0 C0 l : ake K = 120 and K = 140; for nsance, when p = 0. he upper bound for K = 140 s above he lower bound of hek = 120 dervave. he value 15 s n-beween. he spread s consruced n a way such ha f he K = 120 dervave s ransaced by 15; here are no arbrage opporunes. Bu, f he K = 140 dervave s ransaced by 15; here are also no arbrage opporunes. Hence, we may sell he K = 140 dervave by 15 and use he proceeds o buy he K = 120 dervave by 15. Snce he payo of he K = 120 dervave s always larger han he payo of he K = 140 dervave, we would have an arbrage opporuny... hs paradox has a smpleexplanaon. Our bounds are consruced under he assumpon ha here s a sngle dervave. In fac, he presence of more dervaves may 28

29 50 45 Upper Bound Lower Bound p Fgure 3: he Upper and Lower Bounds for a European Call Opon wh d eren value of p n a hree perod model, wh U = 1:3, R = 1, D = 0:6; S 0 = 100 and K = 100: help o complee he marke, makng he overlappng arbrage-free regons no vable. As markes become complee, he arbrage-free regons shrnk o a pon, correspondng o he unque value of he dervaves under complee markes. 7.2 Uly and Reservaon Prces In hs secon we llusrae several aspecs relaed o he deermnaon of he reservaon prce. In Fgure 6 we represen he uly of an agen n hree d eren suaons. Whou he dervave; a shor poson on he dervave, when he nsrumen s sold by he sascal arbrage upper bound; and a long poson on he dervave when he nsrumen s bough by he sascal arbrage lower bound. Noce ha for p = 0 he bes suaon s he shor poson on he dervave and he wors s whou radng he dervave. hs s conssen wh lemma 14 and lemma 17. Noce ha here s a value of p such ha, for larger probables, he uly whou radng he dervave s no longer he wors. ha crcal 29

30 50 45 K= K= K=120 K=140 K= p Fgure 4: he Upper and Lower bounds of an European Call Opon for d eren values of p and K (K = 80, K = 100, K = 120, K = 140 e K = 160) n a hree perod model, wh U = 1:2; R = 1:1; D = 0:8 and S = 100. value of p s wha we called p : Fgure 7 represens he sascal arbrage-free nerval ogeher wh he reservaon prce for a rsk neural agen. By consrucon, he probably assocaed o he pon where he reservaon prce concdes wh he upper bound, corresponds o he crcal probably p : Noce from Fgure 6ha he uly of he poson assocaed oa longposon on he dervave s always above he uly whou radng he dervave. hs mples ha he reservaon prce s always above he lower bound. Lkewse, he fac ha he uly of he shor poson on he dervave goes below he ulywhou radng he dervave, mplesha hereservaon prce goes above he upper bound. 30

31 Cu-Cl p= K p=1 Fgure 5: he Spread (C u C l ) of an European Call Opon for d eren values of K and p (p = 0; :: : ;1 wh ncremens of 0:1) n a hree perod model, wh U = 1:2; R = 1:1; D = 0:8 and S = Concluson In hs paper we have characerzed he sascal arbrage-free bounds for he value of an opon wren on an asse ha may no be ransaced. hs sascal arbrage-free nerval s by consrucon gher han he usual arbrage-free nerval, obaned under he superreplcaon sraegy. In ha sense, ourresul s close o he resuls of Bernardo and Ledo (2000) and Cochrane and Saá-Requejo (2000). By usng a concep of sascal arbrage, n hespr of Bondarenko s (2003), we were able o avod he arbrary hreshold ha led he former approaches oconsran he arbrage-free nerval. In a framework characerzed by he fac ha ransacons of he underlyng asse are possble wh a gven probably, we derved he range of varaon for he sascal arbrage-free value of an European dervave. If ransacons were possble a all pons n me here would be a unque arbrage-freevaluefor heeuropean dervave ha s conaned n hesascal arbrage-free range. Moreover, he sascal arbrage-free range s 31

32 Uly Value Whou Dervave Sellng Dervave Buyng Dervave p Fgure 6: Uly value for he followng parameers: U = 1:3, D = 0:6, R = 1, S 0 = 100, K = 100, ½ = 1=R, q = 0:5, y = 50 and = 3. conaned n he arbrage-free range of varaon f he marke s perfecly llqud. he upper bound s a decreasng funcon n he probably of exsence of he marke and he lower bound s a ncreasng funcon. hey are asympocally well behaved boh when p! 0 and when p! 1. Fnally, we could also prove ha, n he case of random llqudy, he reservaon prces (boh for sellng and buyng posons) are conaned n he sascal arbrage-freerange of varaon for he value ofhe European Opon. 32

33 Reservaon Sellng Prce Reservaon Buyng Prce Upper Bound Lower Bound p Fgure 7: Sascal Arbrage free bounds and reservaon prces for he followng parameers: U = 1:3, D = 0:6, R = 1, S 0 = 100, K = 100, ½ = 1=r, q = 0:5, y = 40 and = 3. 33

34 References [1] Amaro de Maos, J. and P. Anão, 2001, Super-Replcang Bounds on European Opon Prces when he Underlyng Asse s Illqud, Economcs Bullen, 7, 1-7. [2] Bernardo, A. and Ledo, O., 2000, Gan, Loss, and Asse Prcng, Journal of Polcal Economy, 108, 1, [3] Bondarenko, O., 2003, Sascal Arbrage and Secures Prces, Revew of Fnancal Sudes, 16, [4] Cochrane, J. and Saá-Requejo, J., 2000, Beyond Arbrage: Good-Deal Asse Prce Bounds n Incomplee Markes, Journal of Polcal Economy, 108, 1, [5] Edrsnghe, C., V. Nak and R. Uppal, 1993, Opmal Replcaon of Opons wh ransacon Coss and radng Resrcons, Journal of Fnancal and Quanave Analyss, 28, [6] El Karou, N. and M.C. Quenez, 1991, Programaon Dynamque e évaluaon des acfs conngens en marché ncomple, Compes Rendues de l Academy des Scences de Pars, Sére I, 313, [7] El Karou, N. and M.C. Quenez, 1995, Dynamc programmng and prcng of conngen clams n an ncomplee marke, SIAM Journal of Conrol and Opmzaon, 33, [8] Hansen. L.P. and R. Jagannahan. Implcaons of Secury Marke Daa for Models of Dynamc Economes. Journal of Polcal Economy, 99, , [9] Karazas, I. and S. G. Kou, 1996, On he Prcng of Conngen Clams wh Consraned Porfolos, Annals of Appled Probably, 6, [10] Longsa, F., 1995, How Much Can Markeably A ec Secury Values?, he Journal of Fnance, 50, [11] Longsa, F., 2001, Opmal Porfolo Choce and he Valuaon of Illqud Secures, Revew of Fnancal Sudes, 14, [12] Longsa, F., 2004, he Flgh-o-Lqudy Premum n U.S. reasury Bond Prces, Journal of Busness, 77, 3. [13] Mas-Colell, A., Whson, M. and Green, J., 1995, Mcroeconomc heory, Oxford Unversy Press. 34

35 A Some Proofs on he Soluon of he Upper Bound for Sascal Arbrage Opporunes A.1 Proof of heorem 10 Proof. For any gven pah m 2 + ; le ( ;);m ( ;) be he dual varable assocaed wh he superreplcaon consran E px ; h s=;:::; 0 1 p x ; 0 x; 0 = s h s s S s 0 +R0 B s s G wh k 2 I k+1 k and k = ;: :: ; 1: Le n be he number of nodes ha are predecessors of node a me ; where n s gven by n = mnf ( 1); 1; g+ 1 A each node ; ha s a predecessor of ; here are # ³ + ; pahs o reach : For any gven pah m 2 + ; le ( 0;0 );m be he dual varable 0 (;) assocaed wh he self- nancng consrans h ³ E px ; 0 x x ; 0 S 0 + x R0 Bx x ; 0 0 S 0 0 +B 0 0 0: 0 Consderng n 0 be he number of nodes ha are predecessors of node 0 a me we have n 0 = mn 0 ( 0 1); 0 1; 0 ª +1 ³ A each node ha s a predecessor of 0 here are # + ;. 0 he problem ha mus be solved n order o nd he upper bound of he range of varaon of he arbrage-free value of an European dervave s a lnear programmng problem. Is dual problem s +1 mn G j=1 where s he sum of he dual varables assocaed wh he posve expeced payo consrans ha have he rgh member equal o G ;.e., 1 = =0 f:2ig (;);m ( n o ;) S m2 + ; 35

36 he rs se of consrans s of nonnegavy of each dual varable,.e, ( ;);m ( ;) ; ( 0;0 );m ( ;) 0: he oher se of consrans consss of equaly consrans, one consran assocaed wh each varable of he prmal problem. As here are 2 1 =0 ( 1+ 1) ( +1) = 21+ = ( + 1) 2 prmal varables here are also ( + 1) consrans of he dual problem, whch are equaly consrans because he varables of he prmal problem are free. he consran for 0 s: P p x0; (x 0; = 0) + P ½ 1 Q 1 =2 2I Pm2 (;);m (0;0) +0; j=1 p y (y j = 0) P n:02i 0 S + (1;1) (0;0) S1 1 + (2;1) (0;0) S2 1 o P ¾ m2 + (;);m S 0 ; (0;0) = S 0 he consran for B 0 s: p x0; (x 0; = 0)R P P ³ 2I m2 ;m + 0 ; 0 + R (1;1) ( 0 ;0) + (2;1) ( 0 ;0) + P ½ 1 Q 1 =2 j=1 p y (y j = 0)R P n:02i 0 o P ¾ m2 + (;);m = 1 0 ; (0;0) For he consran ha concerns k he erm n s p x0; (x 0; = k) P P n m2 + o ( ;);m 0k ; : 0 2I ( 0 0 ;0) S + o P n p x1; (x 1; = k) P o n1:12i P k 1 nm2 + 1 ; : k 2m :::: p x; (x ; = k) P n:2i o P k nm2 + ; : k 2m P k=0 p x; (x ; = k) P n o P n o P n m2 + ; : k 2m :2I k : k 2I he erms ha nvolve are p x; 0 P <k P 0 >k o ( ;);m : k 2I ( k 1 ;1) S o P n : k 2I o ( ;);m (;) S = k o ( ;);m (;) S k x; 0 = k P n o P :2I k nm2 o + ; : ( 0;0 );m k2m (;) S 0 P P n <k :2I o k ( k;k);m (;) S k k = 0: 0 36