{VERSION 2 3 "IBM INTEL NT" "2.3" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 }{CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 256 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 1 10 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 1 8 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 260 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Output" 0 11 1 {CSTYLE "" -1 -1 " " 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 3 0 -1 -1 -1 0 0 0 0 0 0 -1 0 } {PSTYLE "" 11 12 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 } 1 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 11 256 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }1 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 } {PSTYLE "" 0 257 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 } 3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 258 1 {CSTYLE "" -1 -1 " " 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 } {PSTYLE "" 0 259 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 } 3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 12 260 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 } {PSTYLE "" 0 261 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 } 3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 262 1 {CSTYLE "" -1 -1 " " 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 257 "" 0 "" {TEXT -1 10 "IMAFA-ESSI" }}{PARA 257 "" 0 "" {TEXT -1 52 "Processus Stochastiques Finis de la Finance - TD \+ n\2602" }}{PARA 259 "" 0 "" {TEXT -1 28 "Prix d'une Option Europ\351en ne" }}{PARA 258 "" 0 "" {TEXT 258 30 "(c)1997 M. Diener - UNSA, CNRS" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 8 " Bonjour!" }}{PARA 0 "" 0 "" {TEXT -1 37 "Nous reprenons les donn\351es du TD n\2601:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "S0:=140;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "S:= proc(n,j)options remember;" } }{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "if n=0 then S0 " }}{PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 34 "elif j=0 then evalf(S(n-1,0)*down)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "else evalf(S(n-1,j-1)*up) fi" }}{PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 4 "end;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "d:=6; " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "delta_t:=evalf(1.0/d):" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "sigma:=0.4;R:=ln(1+.1247); mu:=R*1. 3;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "up:=1+sigma*sqrt(delta_t)+mu* delta_t;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "down:=1-sigma*sqrt(delt a_t)+mu*delta_t;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 32 "Corrig\351 de l'exercice du TD n\2601:" }}{PARA 256 "" 1 "" {TEXT -1 106 "Il fallai t essayer de dessiner l'arbre binaire complet (c'est-\340-dire toutes \+ les trajectoires possibles). " }}{PARA 0 "" 0 "" {TEXT -1 133 "Voici \+ une solution, inspir\351e par celle envoy\351e par St\351phane MARTIN \+ (nous avons veill\351 \340 donner l'option \"remember\" a la proc\351d ure S):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "with(plottools): " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "branche:=proc(i,j)" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 " local dessin;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "dessin:=NULL:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "if i " 0 "" {MPLTEXT 1 0 57 " \+ dessin:=line([i,S(i,j)],[i+1,S(i+1,j)]),branche(i+1,j):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 68 " dessin:=dessin,line([i,S(i,j)],[i+1,S(i+1,j +1)]),branche(i+1,j+1) " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "elif i=d -1 then " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 42 " dessin:=line([i,S(i ,j)],[i+1,S(i+1,j)]):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 51 " dessin:= dessin,line([i,S(i,j)],[i+1,S(i+1,j+1)]) " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 1 " " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "fi:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "dessin" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 4 " end;" }}{PARA 12 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "A pr\351sent, calculons et dessinons :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "arbre:=branche(0,0):" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 22 "plots[display](arbre);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{EXCHG {PARA 260 "" 1 "" {TEXT 259 22 "Une repr\351sentation 3d: " }}{PARA 12 "" 0 "" {TEXT -1 68 "Du fait que up*down=down*up , le pr ocessus S a la propri\351t\351 d'\352tre " }{TEXT 256 11 "recombinant " }{TEXT -1 39 " et de ce fait l'arbre constitu\351 des 2^" }{TEXT 257 1 "d" }{TEXT -1 214 " chemins possibles apparait comme un graphe \+ \340 mailles quadrangulaires. Pour mieux comprendre le processus S, no us avons ajout\351 ci-dessous une dimension de profondeur, premettant \+ de dissocier les diverses branches. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "branche3d:=proc(i,k,j) " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 " local dessin ,decalage; " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 36 " dessin:=NULL:decalage:=2^(d-2-i):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "if i " 0 "" {MPLTEXT 1 0 61 " dessin:= line([i,k,S(i,j)],[i+1,k-decalage,S(i+1,j)]), \+ " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 45 " branche3d(i+1,k-deca lage,j): " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 24 " dessin:= dessi n, " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 64 " line([i,k,S( i,j)],[i+1,k+decalage,S(i+1,j+1)]), " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 41 " branche3d(i+1,k+decalage,j+1)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "elif i=d-1 then " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 55 " dessin:=line([i,k,S(i,j)],[i+1,k-decalage,S(i+1,j)] ):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 " dessin:=dessin, " }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 " line([i,k,S(i,j)],[i+1,k+ decalage,S(i+1,j+1)]) " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "fi:" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "dessin" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "end;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "arbre3d:=branche3d(0,0,0): " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 218 "En modifiant l'orientation ( ou angle de vue) [theta,phi]=[-90+90], on retrouve l'image usuelle. [s \351lectionner et r\351orienter l'image au moyen du bouton gauche de l a souris, puis \"redraw\" du menu associ\351 au bouton droit]" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "theta:=-64:phi:=58:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "plots[display](arbre3d,axes= boxed,orientation=[theta,phi]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 218 "Ci-dessous, nous r \351\351crit une proc\351dure b3 intervertissant les variables d (=nom bre de up) et k. Cela permet de mieux visualiser l'arbre (et pr\351par e une proc\351dure dans laquelle on remplacera k par une autre informa tion)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "b3:=proc(i,j,k) " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 " local dessin ,decalage; " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 36 " dessin:=NULL:decalage:=2^(d -2-i):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "if i " 0 "" {MPLTEXT 1 0 61 " dessin:= line([i,S(i,j),k],[i+1,S(i+1,j ),k-decalage]), " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 38 " \+ b3(i+1,j,k-decalage): " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 24 " de ssin:= dessin, " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 64 " \+ line([i,S(i,j),k],[i+1,S(i+1,j+1),k+decalage]), " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 " b3(i+1,j+1,k+decalage)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "elif i=d-1 then " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 55 " dessin:=line([i,S(i,j),k],[i+1,S(i+1,j),k-decalage] ):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 " dessin:=dessin, " }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 " line([i,S(i,j),k],[i+1,S( i+1,j+1),k+decalage]) " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "fi:" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "dessin" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "end;" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%#b3G:6%%\"iG %\"jG%\"kG6$%'dessinG%)decalageG6\"F-C&>8$%%NULLG>8%)\"\"#,(%\"dG\"\" \"!\"#F89$!\"\"@&2F:,&F7F8F;F8C$>F06$-%%lineG6$7%F:-%\"SG6$F:9%9&7%,&F :F8F8F8-FG6$FLFI,&FJF8F3F;-F$6%FLFIFO>F06%F0-FC6$FE7%FL-FG6$FL,&FIF8F8 F8,&FJF8F3F8-F$6%FLFYFZ/F:F>C$>F0FB>F06$F0FTF0F-F-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "arbre3d:=b3(0,0,0);" }}{PARA 12 "> " 1 "" {MPLTEXT 1 0 19 "theta:= -64:phi:=58:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "plots[displ ay](arbre3d,axes=boxed,orientation=[theta,phi]);" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 83 "Ici, avec [theta,phi]=[-90+90] montre pr\351cis \351ment l'arbre soutandant le processus S" }}}{EXCHG {PARA 12 "" 0 " " {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 127 "Nous en venons maintenant au th \350me du jour: prix d'une option Call europ\351enne, calcul\351e au m oyen du mod\350le discret S ci-dessus." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 42 "Reprenons la d\351finition de nos constantes:" }}{PARA 0 "" 0 "" {TEXT -1 77 "On divise l'unit\351 de temps macroscopique en d intervalles de temps deta_t " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "d:=100;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "delta_t:=evalf(1. 0/d);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 33 "Choisissons une volatili t\351 sigma" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "sigma:=0.4; " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 102 "Choisissons un taux annuel r ho; R=\"taux annuel continu\"; r=\"taux annuel discret\" (1+r delta_t) ^d=1+rho" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "rho:=.1247;R:=l n(1+rho); r:=d*(exp(R/d)-1); " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 179 "Choisissons mu; a noter que la th\351orie pr\351voit que la valeur ma croscopique du prix de l'option europ\351enne ne d\351pends pas de la \+ valeur de mu; on aurait de ce fait pu choisir mu=r" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "mu:=R*1.3; " }}}{EXCHG {PARA 261 "" 0 " " {TEXT -1 67 "Valeur (du portefeuille de couverture) d'une option Cal l europ\351enne" }}{PARA 0 "" 0 "" {TEXT -1 358 "La th\351orie de Cox- Ross-Rubinstein \351tabli qu'\340 chaque instent la valeur (du portefe uille de couverture) de l'option est egale \340 la valeur actualis\351 e de l'esp\351rence de la valeur du sous-jacent \340 la p\351riode sui vante, pour la probabilit\351 dite \"risque-neutre\". Nous notons act u le facteur d'actualisation, p_up et p_down les valeurs de la prob a risque-neutre:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "actu:=1 /(1+r*delta_t);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "up:=1+sigma*sqrt (delta_t)+mu*delta_t;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "down:=1-si gma*sqrt(delta_t)+mu*delta_t; " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "p _down:=0.5 +(mu-r)*sqrt(delta_t)/(2*sigma);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "p_up:=0.5 -(mu-r)*sqrt(delta_t)/(2*sigma);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%actuG$\"+q_D))**!#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#upG$\"+7x_T5!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# >%%downG$\"+BrF:'*!#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'p_downG$\" +#H#)R/&!#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%p_upG$\"+3x,c\\!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 174 "Red\351finissons la procedure donnant la valeur du sou s-jacent en fonction du nombre n d'intervalles de dur\351e delta_t ( ou \"tics\") \351coul\351s et le nombre d de up intervenus, " } {TEXT 260 64 "en introduisant la valeur initiale S0 comme troisi\350m e param\350tre" }{TEXT -1 1 ":" }}{PARA 0 "" 0 "" {TEXT -1 61 "Nous en profitons pour d\351r\351cursiver la d\351finition de S(ouja)." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "SouJa := proc(n,j,S0)options remember;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "#if n=0 then S0 " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "#elif j=0 t hen evalf(SouJa(n-1,0,S0)*down)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 36 " #else evalf(SouJa(n-1,j-1,S0)*up) fi" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "evalf(S0*up^j*down^(n-j))" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "e nd;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&SouJaG:6%%\"nG%\"jG%#S0G6\"6 #%)rememberGF*-%&evalfG6#*(9&\"\"\")%#upG9%F2)%%downG,&9$F2F5!\"\"F2F* F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "SouJa(d,49,S00); " }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 88 "Introduisons la valeur CE du po rtefeuille de couverture du Call Europ\351en en fonction " }}{PARA 0 "" 0 "" {TEXT -1 37 " - du nombre n de tics \351coul\351s " }} {PARA 0 "" 0 "" {TEXT -1 34 " - nombre d de up intervenus" }} {PARA 0 "" 0 "" {TEXT -1 46 " - de la valeur initiale S0 du sous-j acent" }}{PARA 0 "" 0 "" {TEXT -1 53 " - de la valeur d'exercice (st rike) K de l'option" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "CE :=proc (n,j,S0,K) option remember;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "if n=d then " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 51 " if SouJa(d,j, S0)>K then SouJa(d,j,S0)-K else 0 fi" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "else (p_down*CE(n+1,j,S0,K)+p_up*CE(n+1,j+1,S0,K))*actu" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "fi" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "end;" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%#CEG:6&%\"nG%\"jG%#S0G% \"KG6\"6#%)rememberGF+@%/9$%\"dG@%29'-%&SouJaG6%F19%9&,&F5\"\"\"F4!\" \"\"\"!*&,&*&%'p_downGF;-F$6&,&F0F;F;F;F8F9F4F;F;*&%%p_upGF;-F$6&FD,&F 8F;F;F;F9F4F;F;F;%%actuGF;F+F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "Exemples pour S0:=140<160=:K" }}{PARA 0 "" 0 "" {TEXT -1 62 "Valeur f inale (n=d) si le titre n'a fait que descendre (j=0)" }}{PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 16 "CE(d,0,140,160);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 87 "Valeur du sous-jacent puis de l'option, \340 l'instant fi nal, apr\350s 51 up; S0=140, K=160:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "SouJa(d,51,140);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "CE(d,51,140,160);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 58 "Valeur \340 l'instant initial: n=0=j: S0=140, K=160, puis 180" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "K:=160;S0:=140;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "CE(0,0,S0,K);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "CE(0,0,140,180);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 377 "La biblioth\350que (library) \"finance\" de Ma ple donne la valeur de cette option selon une formule donn\351e par MM . Black et Scholes; jetez un coup d'oeil \340 cette biblioth\350que et cette fonction, puis comparez. On peut montrer (Cutland, Kopp, Willin ger, puis Van den Berg [http://math.unice.fr/~rgr]) que BS et CRR donn ent des r\351sultat infiniment proches lorsque d et infiniment grand" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "?finance" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "with(finance);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "?blackscholes" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 103 "Syntaxe, avec nos notations: blackscholes( S0, K, R, T, sigma , ' hedge' ); comparez avec vos r\351sultats!" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 43 "blackscholes(140,160,R,1,sigma , 'hedge' );" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(\");" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "evalf(blackscholes(S0,180,R,1,sigma,'hedg e' ));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "?erf" }}}{EXCHG {PARA 262 "" 0 "" {TEXT -1 156 "Probl\350me: programmer une proc\351du re CRR( S0, K, R, T, sigma ) analogue, mais fond\351e sur le mod\350le Cox-Ross-Rubinstein, utilisant une \"constante\" locale d=100" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}}{MARK "63 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 }