Pricing Rainbow Options

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1 Prcng Ranbow Opons Peer Ouwehand, Deparmen of Mahemacs and Appled Mahemacs, Unversy of Cape Town, Souh Afrca E-mal address: Graeme Wes, School of Compuaonal & Appled Mahemacs, Unversy of he Wwaersrand, Souh Afrca. Fnancal Modellng Agency, Souh Afrca. E-mal address: Absrac A prevous paper (Wes 2005 ackled he ssue of calculang accurae un-, b- and rvarae normal probables. Ths has mporan applcaons n he prcng of mulasse opons, e.g. ranbow opons. In hs paper, we derve he Black Scholes prces of several syles of (mul-asse ranbow opons usng change-of-numerare machnery. Hedgng ssues and devaons from he Black-Scholes prcng model are also brefly consdered. Keywords exoc opon, Black-Scholes model, exchange opon, ranbow opon, equvalen marngale measure, change of numerare, rvarae normal. 1. Defnon of a Ranbow Opon Ranbow Opons refer o all opons whose payoff depends on more han one underlyng rsky asse; each asse s referred o as a colour of he ranbow. Examples of hese nclude: Bes of asses or cash opon, delverng he maxmum of wo rsky asses and cash a expry (Sulz 1982, (Johnson 1987, (Rubnsen 1991 Call on max opon, gvng he holder he rgh o purchase he maxmum asse a he srke prce a exprry, (Sulz 1982, (Johnson 1987 Call on mn opon, gvng he holder he rgh o purchase he mnmum asse a he srke prce a expry (Sulz 1982, (Johnson 1987 Pu on max opon, gvng he holder he rgh o sell he maxmum of he rsky asses a he srke prce a expry, (Margrabe 1978, (Sulz 1982, (Johnson 1987 Pu on mn opon, gvng he holder he rgh o sell he mnmum of he rsky asses a he srke a expry (Sulz 1982, (Johnson 1987 Pu 2 and call 1, an exchange opon o pu a predefned rsky asse and call he oher rsky asse, (Margrabe Thus, asse 1 s called wh he srke beng asse 2. Thus, he payoffs a expry for ranbow European opons are: Bes of asses or cash max(s 1, S 2,...,S n, K Call on max max(max(s 1, S 2,...,S n K, 0 Call on mn max(mn(s 1, S 2,...,S n K, 0 Pu on max max(k max(s 1, S 2,...,S n, 0 Pu on mn max(k mn(s 1, S 2,...,S n, 0 Pu 2 and Call 1 max(s 1 S 2, 0 To be rue o hsory, we deal wh he las case frs. 2. Noaon and Seng Defne he followng varables: S = Spo prce of asse, K = Srke prce of he ranbow opon, 74 Wlmo magazne

2 σ = volalyof asse, q = dvdend yeld of asse, ρ j = correlaon coeffcen of reurn on asses and j, r = he rsk-free rae (NACC, τ = he erm o expry of he ranbow opon. Our sysem for he asse dynamcs wll be ds/s = (r qd + AdW (1 where he Brownan moons are ndependen. A s a square roo of he covarance marx, ha s AA =. As such, A s no unquely deermned, bu would be ypcal o ake A o be he Cholesk decomposon marx of (ha s, A s lower rangular. Under such a condon, A s unquely deermned. Le he h row of A be a. We wll say ha a s he volaly vecor for asse S. Noe ha f we were o wre hngs where S had a sngle volaly σ hen σ 2 = n j=1 a2 j, so σ = a, where he norm s he usual Eucldean norm. Also, he correlaon beween he reurns of S and S j s gven by a a j a a j. 3. The Resul of Margrabe The heory of ranbow opons sars wh (Margrabe 1978 and has s mos sgnfcan oher developmen n (Sulz (Margrabe 1978 began by evaluang he opon o exchange one asse for he oher a expry. Ths s jusfably one of he mos famous early opon prcng papers. Ths s concepually lke a call on he asse we are gong o receve, bu where he srke s self sochasc, and s n fac he second asse. The payoff a expry for hs European opon s: whch can be valued as: where max(s 1 S 2, 0, V M = S 1 e q1 τ N(d + S 2 e q2 τ N(d, (2 d ± = ln f 1 f 2 ± 1 2 σ 2 τ σ τ (3 f = S e (r qτ (4 σ 2 = σ σ 2 2 2ρσ 1σ 2 (5 Margrabe derves hs formula by developng and hen solvng a Black-Scholes ype dfferenal equaon. Bu he also gves anoher argumen, whch he creds o Sephen Ross, whch wh he hndsgh of modern echnology, would be consdered o be he mos approprae approach o he problem. Le asse 2 be he numerare n he marke. In oher words, asse 2 forms a new currency, and asse one coss S1 n ha S2 currency. The rsk free rae n hs marke s q 2. Thus we have he opon o buy asse one for a srke of 1. Ths has a Black-Scholes prce of V = S 1 S 2 e q1 τ N(d + e q2 τ N(d ln d ± = S 1 S (q 2 q 1 ± 12 σ 2 τ σ τ where σ s he volaly of S1. To ge from a prce n he new asse 2 currency o a prce n he orgnal economy, we mulply by S 2 : he exchange S2 rae, whch gves us (2. So, wha s σ? We show ha (5 s he correc answer o hs queson n Change of Numerare Suppose ha X s a European syle dervave wh expry dae T. Snce (Harrson & Plska 1981 has been known ha f X can be perfecly hedged (.e. f here s a self fnancng porfolo of underlyng nsrumens whch perfecly replcaes he payoff of he dervave a expry, hen he me value of he dervave s gven by he followng rsk neural valuaon formula: X = e r(t E Q [X T ] where r s he rskless rae, and he symbol E Q denoes he expecaon a me under a rsk neural measure Q. A measure Q s sad o be rsk neural f all dscouned asse prces S = e r S are marngales under he measure Q,.e. f he expeced value of each S a an earler me u s s curren value S u : E Q u [ S ] = S u whenever 0 u (Here we assume for he momen ha S pays no dvdends. Now le A = e r denoe he bank accoun. Then he above can be rewren as [ ] X = E Q XT.e. X = E Q [ ] X T A A T Thus X s a Q marngale. In an mporan paper, (Geman, El Karou & Roche 1995 was shown ha here s nohng specal abou he bank accoun: gven an asse (1995 Â, we can dscoun each underlyng asse usng Â: Ŝ = S  Thus Ŝ s he prce of S measured no n money, bu n uns of Â. The asse  s referred o as a numérare, and mgh be a porfolo or a dervave he only resrcon s ha s value  s srcly posve durng he me perod under consderaon. I can be shown (cf. (Geman, e al ha n he absence of arbrage, and modulo some echncal condons, here s for each numérare (1995  a measure ˆQ wh he propery ha each numérare deflaed ^ Wlmo magazne 75

3 asse prce process Ŝ s a ˆQ marngale,.e. E ˆQ u [Ŝ ] = Ŝ u whenever 0 u (Agan, we assume ha S pays no dvdends. We call ˆQ he equvalen marngale measure (EMM assocaed wh he numérare Â. I hen follows easly ha f a European syle dervave X can be perfecly hedged, hen ˆX = E ˆQ [ ˆX T ] and so X =  E ˆQ Indeed, f V s he value of a replcang porfolo, hen (1 X = V by he law of one prce, and (2 ˆV = V s a ˆQ marngale. Thus  [ ] ˆX = ˆV = E ˆQ ˆV T [ ] = E ˆQ ˆX T usng he fac ha V T = X T by defnon of replcang porfolo. I follows ha f N 1, N 2 are numérares, wh assocaed EMM s Q 1, Q 2, hen ] ] N 1 (E Q1 [ XT N 1 (T = N 2 (E Q2 [ XT N 2 (T Indeed, boh sdes of he above equaon are equal o he me prce of he dervave. To ge slghly more echncal, he EMM ˆQ assocaed wh numérare  s obaned from he rsk neural measure Q va a Grsanov ransformaon (whose kernel s he volaly vecor of he numérare. In parcular, he volaly vecors of all asses are he same under boh Q and ˆQ. A mnor modfcaon of he above reasonng s necessary n case he asses pay dvdends. Suppose ha S s a share wh dvdend yeld q. If we buy one share a me = 0, and f we renves he dvdends n he share, we wll have e q shares a me, wh value S(e q. If  s he new numérare, wh dvdend yeld ˆq, hen s he rao S(e q Â(eˆq ha s a ˆQ marngale, and no he rao S(. Â( Suppose now ha we have n asses S 1, S 2,...,S n, and ha we model he asse dynamcs usng an n dmensonal sandard Brownan moon. If a s he volaly vecor of S, hen, under he rsk neural measure Q, he dynamcs of S are gven by ds S = (r q d + a dw where q s he dvdend yeld of S, and W s an n dmensonal sandard Q Brownan moon. When we work wh asse S j as numérare, we wll be neresed n he dynamcs of he asse rao processes S /j ( = S ( S j ( under he assocaed EMM Q j. Now by Io s formula he rsk neural dynamcs of S /j are gven by [ XT ds /j = (q j q + a S j 2 a a j d + (a dw /j  T ] However, when we change o measure Q j, we know ha Y( = S /j ( e (q q j s a Q j marngale. Applyng Io s formula agan, we see ha he rsk neural dynamcs of Y are gven by dy ( Y = a j 2 a a j d + (a dw Snce Y( s a Q j marngale, s drf under Q j s zero, and s volaly remans unchanged. Thus he Q j dynamcs of Y( are dy Y = (a dw j where W j s a sandard n dmensonal Q j Brownan moon. Applyng Io s formula once agan o S /j ( = Y(e (q q j, follows easly ha he Q j dynamcs of S /j ( are gven by ds /j S /j = (q j q d + (a dw j Reurnng o 3, we have σ 2 = a 1 a 2 2 = a a 2 2 2ρ a 1 a 2 =σ σ 2 2 2ρσ 1σ 2, as requred. 5. The Resuls of Sulz (Sulz 1982 derves he value of wha are now called wo asse ranbow opons. Frs he value of he call on he mnmum of he wo asses s derved, by evaluang he (raher unpleasan bvarae negral. Then a mn-max pary argumen s nvoked: havng a wo asse ranbow maxmum call and he correspondng wo asse ranbow mnmum call s jus he same as havng wo vanlla calls on he wo asses. Fnally pu-call pary resuls are derved, enablng evaluaon of he pu on he mnmum and he pu on he maxmum. Raher han gong no any deals we mmedaely proceed o he more general case where we derve far more pleasan ways of mmedaely fndng any such valuaon. 6. Many Asse Ranbow Opons In (Johnson 1987 exensons of he resuls of (Sulz 1982 are clamed o any number of underlyngs. However, he formulae n he paper are acually que dffcul o nerpre whou ambguy: hey are presened nducvely, and he formula (even for n = 3 s dffcul o nerpre wh cerany. Moreover, he formulae are no proved only nuons are provded nor s any numercal work underaken o provde some comfor n he resuls. The argumens bascally nvolve nung wha he dela s of he opon n each of he n underlyngs should be, and exrapolang from here o he prce. So one can say bravo gven ha s possble o acually formally derve proofs for hese many asse prcng formulae. Wha we do s consruc general Marngale-syle argumens for all cases n 2 whch are n he syle of he proof frs found by Margrabe and Ross. Johnson s resuls are saed for any number of asses. A ranbow opon wh n asses wll requre he n-varae cumulave normal funcon for applcaon of hs formulae. As n ncreases, so he compuaonal effor and execuon me for havng such an approxmaon wll ncrease 76 Wlmo magazne

4 fg-1.eps Le σ /j = a. We know ha under Q j we have ds/j = (q S/j j q d+ (a dw j, so ln S /j (T φ(ln S /j ( + (q j q 1 σ 2 2 /j τ, σ /j τ. Noe ha, and defne σ 2 /j = σ 2 + σ 2 j ln S ( d /j S j ( + ± = d ± = ln S ( K + 2ρ j σ σ j ( q j q ± 1 2 σ 2 /j σ /j τ ( r q ± 1 2 σ 2 σ τ (τ (τ Hence Q j [S /j (T <>1] = N( d /j. Noe ha d /j ± = d j/. Also, he correlaon beween S /k (T and S j/k (T s Fgure 1: Mone Carlo for call on mnmum on 3 asses. On he horzonal axs: number of expermens n 1000 s, usng ndependen Sobol sequences, on he vercal axs: prce. The exac opon value usng he formula presened here s ρ j,k := (a a k (a j a k a a k a j a k a a j a a k a k a j + σ 2 k = (σ 2 + σ 2 k 2a a k (σ 2 j + σ 2 k 2a j a k (7 dramacally. In (Wes 2005 we have vb and c++ code for n 3 based upon he Forran of (Genz 2004, so here we apply hs code o European ranbow opons wh hree sock underlyngs, S 1, S 2 and S 3. Code for n > 3 does no seem o be avalable (n any language, a leas n a form ha would make he compuaonal me beer han drec Mone Carlo valuaon of he orgnal opon. Usng ha code, for he case n = 3 we can compare Mone Carlo smulaon o he prces n (Johnson 1987; see for example Fgure Maxmum Payoffs We wll frs prce he dervave ha has payoff max(s 1, S 2,...,S n, where he S sasfy he usual properes. In fac, hs s noaonally que cumbersome, and all he deas are encapsulaed n any reasonably small value of n, so we choose n = 4 (as we wll see laer, he fourh asse wll be he srke. Frsly, he value of he dervave s he sum of he value of 4 oher dervaves, he h of whch pays S (T f S (T >S j (T for j =, and 0 oherwse. Le us value he frs of hese, he ohers wll have smlar values jus by cyclng he coeffcens. We are consderng he asse ha pays S 1 (T f S 1 (T s he larges prce. Now le S 1 be he numerare asse wh assocaed marngale measure Q 1. We see ha he value of he dervave s V 1 ( = S 1 (e q1 τ E Q1 [1; S 2/1 (T <1, S 3/1 (T <1, S 4/1 (T <1] = S 1 (e q1 τ Q 1 [S 2/1 (T <1, S 3/1 (T <1, S 4/1 (T <1] = S 1 (e q1 τ Q 1 [ln S 2/1 (T <0, ln S 3/1 (T <0, ln S 4/1 (T <0] where S /j (T = S(T Sj(T. (6 ρ j σ σ j ρ k σ σ k ρ kj σ k σ j + σ 2 k = (σ 2 + σ 2 k 2ρ kσ σ k (σ 2 j + σ 2 k 2ρ jkσ j σ k Hence Q 1 [ln S 2/1 (T <0, ln S 3/1 (T <0, ln S 4/1 (T <0] = N 3 ( d 2/1,- d 4/1, 1 where 1, 2, 3 and 4 are 3 3 marces; he smples way o hnk of hem s ha hey are nally 4 4 marces, wh k havng ρ j,k n he (, j h poson, and hen he k h row and k h column are removed. Thus, he value of he dervave ha pays off he larges asse s V max ( = S 1 (e q1 τ N 3 ( d 2/1, 1, d 4/2, 2 + S 3 (e q3 τ N 3 ( d 1/3, d 4/3, 3 + S 4 (e q4 τ N 3 ( d 1/4, d 2/4, d 3/4, 4 = S 1 (e q1 τ N 3 ( d 2/1,ρ 23,1,ρ 24,1,ρ 34,1, d 4/2,ρ 13,2,ρ 14,2,ρ 34,2 + S 3 (e q3 τ N 3 ( d 1/3, d 4/3,ρ 12,3,ρ 14,3,ρ 24,3 + S 4 (e q4 τ N 3 ( d 1/4, d 2/4, d 3/4,ρ 12,4,ρ 13,4,ρ 23,4 6.2 Bes and Wors of Call Opons Le us sar wh he case where he payoff s he bes of asses or cash. The payoff a expry s max(s 1, S 2, S 3, K. If we consder hs o be he bes of four asses, where he fourh asse sasfes S 4 ( = Ke rτ and has zero volaly, hen we recover he value of hs opon from 6.1. Ths fourh asse no only has no volaly bu also s ndependen of he oher hree asses. Thus, a 4 = 0, ρ j,4 = ρ j, σ /4 = σ = σ 4/, d /4 ± = d ±, d4/ ± = d. Thus (8 ^ Wlmo magazne 77

5 V max ( = S 1 (e q1 τ N 3 ( d 2/1, d 1 +,ρ 23,1,ρ 24,1,ρ 34,1, d 2 +,ρ 13,2,ρ 14,2,ρ 34,2 + S 3 (e q3 τ N 3 ( d 1/3, d 3 +,ρ 12,3,ρ 14,3,ρ 24,3 + Ke rτ N 3 ( d 1, d2, d3,ρ 12,ρ 13,ρ 23 (9 V cmn ( = S 1 (e q1 τ N 3 (d 2/1, d 1 +,ρ 23,1, ρ 24,1, ρ 34,1 + S 2 (e q2 τ N 3 (d 1/2, d 3/2, d 2 +,ρ 13,2, ρ 14,2, ρ 34,2 + S 3 (e q3 τ N 3 (d 1/3, d 2/3, d 3 +,ρ 12,3, ρ 14,3, ρ 24,3 Ke rτ N 3 (d 1, d2, d3,ρ 12,ρ 13,ρ 23 (13 Now le us consder he ranbow call on he max opon. Recall, hs has payoff max(max(s 1, S 2, S 3 K, 0. Noe ha and so max(max(s 1, S 2, S 3 K, 0 = max(max(s 1, S 2, S 3, K K = max(s 1, S 2, S 3, K K V cmax ( = S 1 (e q1 τ N 3 ( d 2/1, d 1 +,ρ 23,1,ρ 24,1,ρ 34,1, d 2 +,ρ 13,2,ρ 14,2,ρ 34,2 + S 3 (e q3 τ N 3 ( d 1/3, d 3 +,ρ 12,3,ρ 14,3,ρ 24,3 Ke rτ [1 N 3 ( d 1, d2, d3,ρ 12,ρ 13,ρ 23 ] Fnally, we have he ranbow call on he mn opon. (Recall, hs has payoff max(mn(s 1, S 2, S 3 K, 0. Because of he presence of boh a maxmum and mnmum funcon, new deas are needed. As before we frs value he dervave whose payoff s max(mn(s 1, S 2, S 3, S 4. If S 4 s he wors performng asse, hen he payoff s he second wors performng asse. For 1 3 he value of hs payoff can be found by usng asse S as he numerare. For example, he value of he dervave ha pays S 1, f S 4 s he wors and S 1 he second wors performng asse, s S 1 (e q1 τ N 3 (d 2/1,ρ 23,1, ρ 24,1, ρ 34,1 If S 4 s no he wors performng asse, hen he payoff s S 4. Now he probably ha S 4 s he wors performng asse s N 3 (d 1/4, d 2/4, d 3/4,ρ 12,4,ρ 13,4,ρ 23,4 and so he value of he dervave ha pays S 4, f S 4 s no he wors performng asse, s S 4 (e q4 τ [1 N 3 (d 1/4, d 2/4, d 3/4,ρ 12,4,ρ 13,4,ρ 23,4 ] Thus, he value of he dervave whose payoff s max(mn(s 1, S 2, S 3, S 4 s V( = S 1 (e q1 τ N 3 (d 2/1,ρ 23,1, ρ 24,1, ρ 34,1 + S 2 (e q2 τ N 3 (d 1/2, d 3/2, d 4/2,ρ 13,2, ρ 14,2, ρ 34,2 + S 3 (e q3 τ N 3 (d 1/3, d 2/3, d 4/3,ρ 12,3, ρ 14,3, ρ 24,3 + S 4 (e q4 τ [1 N 3 (d 1/4, d 2/4, d 3/4,ρ 12,4,ρ 13,4,ρ 23,4 ] Hence he dervave wh payoff max(mn(s 1, S 2, S 3, K has value V( = S 1 (e q1 τ N 3 (d 2/1, d 1 +,ρ 23,1, ρ 24,1, ρ 34,1 + S 2 (e q2 τ N 3 (d 1/2, d 3/2, d 2 +,ρ 13,2, ρ 14,2, ρ 34,2 + S 3 (e q3 τ N 3 (d 1/3, d 2/3, d 3 +,ρ 12,3, ρ 14,3, ρ 24,3 + Ke rτ [1 N 3 (d 1, d2, d3,ρ 12,ρ 13,ρ 23 ] and he call on he mnmum has value (10 (11 (12 7. Fndng he Value of Pus Ths s easy, because pu-call pary akes on a parcularly useful role. I s always he case ha V c (K + Ke rτ = V p (K + V c (0 (14 where he parenheses denoes srke. V could be an opon on he mnmum, he maxmum, or ndeed any ordnal of he baske. If we have a formula for V c (K, as esablshed n one of he prevous secons, hen we can evaluae V c (0 by akng a lm as K 0, eher formally (usng facs of he manner N 2 (x,,ρ= N 1 (x and N 3 (x, y,, = N 2 (x, y,ρ xy or nformally (by forcng our code o execue wh a value of K whch s very close o, bu no equal o, 0 - hus avodng dvson by 0 problems bu mplcly mplemenng he above-menoned fac. By rearrangng, we have he pu value. 8. Delas of Ranbow Opons By nspecng (9 one mgh expec ha V max S 1 = e q1 τ N 3 ( d 2/1, d 1 +,ρ 23,1,ρ 24,1,ρ 34,1 wh smlar resuls holdng for Vmax and Vmax, and ndeed for he dual S2 S3 dela Vmax. K Thus urns ou o be rue n hs case, bu o clam as an obvous fac would be erroneous. Recall Euler s Homogeneous Funcon Theorem, whch we wll cas n our case of a funcon of four varables V(x 1, x 2, x 3, x 4. The heorem saes ha f V(λx 1,λx 2,λx 3,λx 4 = λv(x 1, x 2, x 3, x 4 for any V V V V consan λ hen V(x 1, x 2, x 3, x 4 = x 1 + x x1 2 + x x2 3 + x x3 4. x4 The argumen of (Johnson 1987 s essenally an applcaon of hs heorem: he nus wha V s and hen reassembles V usng hs resul. S However, o clam a converse of he form ha f V(x 1, x 2, x 3, x 4 = x x x x 4 4 for some nce hen of necessy = V x s false. (6 provdes a counerexample; because ceranly s no he case ha V = 0 for > 1. To jump a he clam ha s an obvous fac ha S he above s he formula for dela s probably an applcaon of hs false converse. However, hese clams are rue n he case of V max, as s for V cmax, V cmn, V pmax and V pmn. 9 Fndng he Capal Guaranee on he Bes of Asses or Cash Opon We wsh o deermne he srke K of he bes of asses or cash opon so ha a ncepon he valuaon of he opon s equal o K. Denong he 78 Wlmo magazne

6 value of such an opon as V(K mplcly fxng all oher varables besdes he srke we wsh o solve V(K = K. To do so usng Newon s mehod s forunaely que manageable, for he same reasonng ha we have already seen. As prevously promsed we have from (9 ha V K = e rτ N 3 ( d 1, d2, d3,ρ 12,ρ 13,ρ 23 Hence he approprae Newon mehod eraon s K n+1 = K n V(K n K n V K K=Kn 1 (15 and hs s eraed o some desred level of accuracy. An alernave would be o erae K n+1 = V(K n, our dfferenaon shows ha he funcon V s a conracon, and so hs eraon wll converge o he fxed pon V(K = K by he conracon mappng heorem. I s mporan o noe ha he process of fndng he far heorecal srke s no jus a curosy. In he frs place, s aracve for he buyer of he opon ha hey wll ge a leas her premum back. (There s a floor on he reurn of 0%. Moreover, f K s hs far srke, he rader wll srke he opon a an K, where K > K, n order o expec fa n he deal. To see hs, we can consruc n a complee marke a smple arbrage sraegy: magne ha he dealer sells he clen for K an opon sruck a K, and hedges hs wh he far dealer by payng K for an opon sruck a K. 1 The dfference K K s nvesed n a rsk free accoun for he expry dae. Three cases hen arse: If max(s 1, S 2, S 3 K hen we owe K. The far rader pays K and we oban K K from savng, and prof from he me value of K K. If K < max(s 1, S 2, S 3 K, hen he far rader pays S 1 say. We sell hs, and oban he balance o K from savng. If K < max(s 1, S 2, S 3, hen he far rader pays S 1 say and we delver hs. 10 Prcng Ranbow Opons n Realy The model ha has been developed here les whn he classcal Black- Scholes framework. As s well known, he assumpons of ha framework do no hold n realy; varous sylsed facs argue agans ha model. For vanlla opons, he model s adjused by means of he skew hs skew exacly ensures ha he prce of he opon n he marke s exacly capured by he model. Models whch exrac nformaon from ha skew and of how ha skew wll evolve are of paramoun mporance n modern mahemacal fnance. Afer a momen s hough one wll realse wha a dffcul ask we are faced wh n applyng hese skews here. Le us sar by beng compleely naïve: we wsh o mark our ranbow opon o marke by usng he skews of he varous underlyngs. Frsly, wha srke do we use for he underlyng? How does he srke of he ranbow ranslae no an approprae srke for an opon on a sngle underlyng? Secondly, suppose we somehow resolved hs problem, and for a raded opon, wshed o know s mpled volaly? A famlar problem arses: ofen he opon wll have wo, somemes even hree dfferen volales of one of he asses whch recover he prce (all oher npus beng fxed. To be more mahemacal, he map from volaly o prce s no njecve, so he concep of mpled volaly s ll defned. See Fgure 2. To see he sensvy o he npus, suppose o he seup n Fgure 2 we add a hrd asse as elaboraed n Fgure 3. Of course he general level fg-2.eps Fgure 2: The prce for a call on he mnmum of wo asses. S 1 = 2, S 2 = 1, K = 1, τ = 1, r = 10%, ρ = 70%, 20% σ 1 60%, σ 2 100%. fg-3.eps Fgure 3: The prce for a call on he mnmum of hree asses. As above, n addon S 3 = 1, σ 3 = 30% fxed, correlaon srucure ρ 12 = 70%, ρ 13 = 30%, ρ 23 = 20% ^ Wlmo magazne 79

7 of he value of he asse changes, bu so does he enre geomery of he prce surface. Anoher ssue s ha of he assumed correlaon srucure: agan, correlaon s dffcul o measure; f here s mpled daa, hen wll have a srke aached. Fnally, he jon normaly hypohess of reurns of prces wll ypcally be rejeced. A popular approach s o use skews from he vanlla marke o nfer he margnal dsrbuon of reurns for each of he ndvdual asses and hen glue hem ogeher by means of a copula funcon. Gven a mulvarae dsrbuon of reurns, ranbow opons can hen be prced by Mone Carlo mehods. FOOTNOTES & REFERENCES 1. The far dealer s he perfec hedger, whose replcang porfolo ends up wh exacly he payoff. Geman, H., El Karou, N. & Roche, J. (1995 Changes of numerare, changes of probably measure and opon prcng, Journal of Appled Probably 32. Genz, A. (2004 Numercal compuaon of recangular bvarae and rvarae normal and probables, Sascs and Compung 14, *hp:// Harrson, J. M. & Plska, S. R. (1981, Marngales and sochasc negrals n he heory of connuous radng, Sochasc Processes and her Applcaon 11, Johnson, H. (1987 Opons on he maxmum or he mnmum of several asses, Journal of Fnancal and Quanave Analyss 22, Margrabe, W. (1978, The value of an opon o exchange one asse for anoher, The Journal of Fnance 23, Rubnsen, M. (1991 Somewhere over he ranbow, Rsk 4, Sulz, R. M. (1982, Opons on he mnmum or he maxmum of wo rsky asses, Journal of Fnancal Economcs XXXIII, No. 1, Wes, G. (2005, Beer approxmaons o cumulave normal funcons, WILMOTT Magazne May, ACKNOWLEDGEMENTS Thanks o he 2005 Mahemacs of Fnance class a he Unversy of he Wwaersrand for aler feedback n lecures. W