1. Contents PrefaceIXAcknowledgmentsXlo.1 1 11 11Mathematical preliminaries 0.1 Solutions to Chapter 0 Exercises. . . . . . . . . . . . . . . .. 0.2 Supplemental Exercises. . . . . . . . . . . . . . . . . . . . .. 0.3 Solutions to Supplemental Exercises. . . . . . . . . . . . . ..1 Calculus review. Plain vanilla options. 17 1. 1 Solutions to Chapter 1 Exercises. . . . . . . . . . . . . . . .. 17 1. 2 Supplemental Exercises. . . . . . . . . . . . . . . . . . . . .. 32 1.3 Solutions to Supplemental Exercises . . . . . . . . . . . . . .. 33 2 Numerical integration. Interest Rates. Bonds. 45 2.1 Solutions to Chapter 2 Exercises. . . . . . . . . . . . . . . .. 45 2.2 Supplemental Exercises. . . . . . . . . . . . . . . . . . . . .. 57 2.3 Solutions to Supplemental Exercises. . . . . . . . . . . . . .. 58 3 Probability concepts. Black-Scholes formula. Hedging. 3.1 Solutions to Chapter 3 Exercises. . . . . . . . 3.2 Supplemental Exercises. . . . . . . . . . . . . 3.3 Solutions to Supplemental Exercises. . . . . .Greeks and . . . . . . . .. . . . . . . . .. . . . . . . . ..63 63 82 834 Lognormal variables. Risk-neutral pricing. 91 4.1 Solutions to Chapter 4 Exercises. . . . . . . . . . . . . . . .. 91 4.2 Supplemental Exercises. . . . . . . . . . . . . . . . . . . . . . 106 4.3 Solutions to Supplemental Exercises. . . . . . . . . . . . . . . 107 Vll

2. CONTENTSVl1l55.1 5.2 5.3 678 E 4 4 1 4 4 l 4Taylor formula.Taylor series. Solutions to Chapter 5 Exercises. . Supplemental Exercises. . . . Solutions to Supplemental Exercises. E 41 i 1 "3346Finite Differences. Black-Scholes PDE. 135 6.1 Solutions to Chapter 6 Exercises. . . . . . . . . . . . . . . . . 135 6.2 Supplemental Exercises . . . . . . . . . . . . . . . . . . . . . . 150 6.3 Solutions to Supplemental Exercises. . . . . . . . . . . . . . . 151 1ultivariable calculus: chain rule , integration tion , and extrema. 7.1 Solutions to Chapter 7 Exercises. . . . . . . . 7.2 Supplemental Exercises. . . . . . . . . . . . . 7.3 Solutions to Supplemental Exercises. . . . . .Lagrange multipliers. Newton's method. ity. Bootstrapping. 8.1 Solutions to Chapter 8 Exercises. . . . . . 8.2 Supplemental Exercises. . . . . . . . . . . 8.3 Solutions to Supplemental Exercises. . . .Bibliographyby substitu161 . . . . . . . . . 161 . . . . . . . . . 171 . . . . . . . . . 173Implied volatil179 . . . . . . . . . . . 179 . . . . . . . . . . . 197 . . . . . . . . . . . 198 203Preface The addition of this Solutions Ivlanual to "A Primer for the Mathematics of Financial Engineering" offers the reader the opportunity to undertake rigorous self-study of the mathematical topics presented in the Ivlath Primer , with the goal of achieving a deeper understanding of the financial applications therein. Every exercise from the Math Primer is solved in detail in the Solutions Manual. Over 50 new exercises are included , and complete solutions to these supplemental exercises are provided. 1any of these exercises are quite challenging and offer insight that promises to be most useful in further financial engineering studies as well as job interviews. Using the Solution Manual as a companion to the Ivlath Primer , the reader will be able to not only bridge any gaps in knowledge but will also glean a more advanced perspective on financial applications by studying the supplemental exercises and their solutions. The Solutions Ivlanual will be an important resource for prospective financial engineering graduate students. Studying the material from the Math Primer in tandem with the Solutions Manual would provide the solid mathematical background required for successful graduate studies. The author has been the Director of the Baruch College MFE Program 1 since its inception in 2002. Over 90 percent of the graduates of the Baruch IvIFE Program are currently employed i the financial industry. "A Primer for the Ivlathematics of Financial Engineering" and this Solutions Manual are the first books to appear in the Financial Engineering Advanced Backgroud Series. Books on Numerical Linear Algebra , on Probability, and on Differential Equations for financial engineering applications are forthcommg. Dan Stefanica New York , 2008 lBaruch MFE Program web page: http://www.baruch.cuny.edu/math/master tml QuantNetwork student forum web page: httr/ www.quantnet.org/forum/index lp IX 3. Acknowledgments "A Primer for the Mathematics of Financial Engineering" published in April 2008 , is based on material covered in a mathematics refresher course I taught to students entering the Financial Engineering Masters Program at Baruch College. Using the book as the primary text in the July 2008 refresher course was a challenging but exceptionally rewarding experience. The students from the 2008 cohort of the Baruch WIFE Program who took that course were driven to master the material and creatively incorporate ideas at an even deeper level than that of the WIath Primer book. WIany of the supplemental questions in the Solutions WIanual came about as a result of the remarkable interaction I had with is very talented group. I am grateful to all of them for their impressive efforts. Special thanks go to those who supported the proofreading effort: Barnett Feingold , Chuan Yuan-Huang , Aditya Chitral , Hao He , Weidong Huang , Eugene Krel , Shlomo Ben Shoshan , Mark Su , Shuwen Zhao , and Stefan Zota. The art for the book cover was again designed by Max Rumyantsev , and Andy Nguyen continued to lend his tremendous support to the fepress.org website and on QuantNet.org. I am indebted to them for all their help. This book is dedicated to my wonderful family. You give sense to my work and make everything worthwhile. Dan Stefanica New York , 2008Xl 4. XllACKN01iVLEDGJVIENTSChapter 0 Mathematical preliminaries. 0.1Solutions to Chapter 0 ExercisesProblem 1: Let f : IR IR be an odd function. (i) Show that xf(x) is a even function and x 2f (x) is an odd function. (ii) Show that the function gl : IR IR given by gl(X) = f(x22 is an even function and that the function g2 : IR IR given by ) = f(x :J) is an odd function. (iii) Let h : IR IR be dened as h(x) xtf(x J ) where i and j are positive integers. When is h(x) an odd function? Solutio Sincef(x) is an odd function , it follows that f( -x) = - f(x) ,(i) Let h ( = xf(x) andf2 (x) = x 2f(x).jxIR.(1)Using (1) , we nd thath(-x) = -xf(-x) = xf(x) = h(x) ,jxIR;(2)f2 (-x) = ( X)2 f( -x) = - x 2f(x) = - f2 (x) , j x IR. (3) We coclude from (2) that h (x) is an even function , and , from (3) , f2 (x) is an odd function. (ii) From (1) , it follows thatgl(-X) = f ((_X)2) = f(x 2) = gl(X) , j x IR; g2(-X) = f ((_x)3) = f(-x 3) = - f(x 3) = - g2(X) ,(4) jx IR. (5)We conclude from (4) that gl(X) is an even function , and , from (5) , that ) is an odd function. (iii) If j is a positive integer , it follows from (1) thatf ((- x )J)=(-1 )J f (x) , j x IR.(6) 5. 2JVIATHE1VIATICAL PRELIMINARIESThen , usi (6) we nd thath(-x) = (-x)if((-x)j) = (-1) ((-l)jf(x j )) ( - 1)i+ jxi f (x j ) = (-1)jh(x) V x JR. TTIerefore7if t +j is aIIeven iMeger7the fmctionfz(z)is an eveIl function , and , if i j is an odd integer , the function h(x) is a~ ~dd function. ?- (T(n 1, x)) ,3( 2 ( 1)3x n 1 (22 2-1)( 2)( l)x n2 n ) l -n (+ 3)( + 2)x 1 6(1 - x) 1 1/ ( 1)3( - 1)x n - 2 (22 2 -1)(n+2)( 1) x n 1Ii -n 2 ( 3)( 2)( 1)zn6/ 11 ( 2 O 2( 3)( 2 )( 1) Problem 2: Let S(n , 2) = I:~=1 k 2 and S( 3) = I:~=1 k 3 . (i) Let T(n , 2, x) = I:~=1 k 2 xk . Use the recursion formula T(n , 2 , x)0.1. SOLUTIONS TO CHAPTER 0 EXERCISES6 (+ 1) ( ( 1)2(- 1) + (22 2- 1)( 2) (+ 3)(+ 2))(7)6 (n+1)(2+and the fact that1)6 T( 1 x)Z ( 1)x n + 1 nx n + 2(1 - X)2(8)Therefore , S( 2)=(+ 1)(2+1)to show that T( 2 x)2 X +x ( +1)2 x n+1 (22 2 - 1)x n + 2 _ n 2x n + 3 (1 - x)3(ii) Note that S( 2) = T( 2" l);"Ys~ yHopital's rule to evaluate T( 2 1) and conclude that S ()= 71i(2) (iii) Compute T( 3 x) = I:~=1 k 3 x k using the recursion formulaJ)24m)) (iv) Note that S( 3) = T( 3 1). U~e I'Hopital's rule to evaluate T( 3 1) and conclude that S(n , 3) = ("(iii) Finding the value of T( 3 x) requires usi Quotient Rule to differen x). (iv) The solution follows similarly to that from part (ii) , albeit with more complicated computations. t iate T( 2 Problem 3: Compute S( 4) = I:~=1 k4 using the recursion formula c iFJn t u-MZH +.3 /Il--/ lI -l+ 3 A-.qb t A t 4lit-/ 1 1 I /QU -qJ /(9)for i = 4 , given that=(f1);~.aiution: (i) ~he res~lt follows from (7) and (8) by usi Quotient Rule to differentiate T ( 1 x).(ii) It is easy to see that T( 2 1) = I:~=1 S( 2). By using I'Hopital's rule we that T( 2 x) is equal to lim Z x 2 ( 1)2 x 1 (22 2 - 1)x n + 2 2 X n3 z 1 (1-x)3 lim 1 +2x ( 1)3 xn (22 2- 1)( 2)x n 1 _ n2( +3)x n + 2 -3(1 - x)21. =S()( Solution: For i = 4 , the recursion formula (9) becomes 4)=H 0)s(n ,j))n(+ 1)(2 1)6' 6. 4l!fATHENIATICAL PRELIMINARIES~()5 _ 1 -S(n , 0))))( 1)(63 92 -1)50.1. SOLUTIONS TO CHAPTER 0 EXERCISES (ii) We substitute 1 for n in (12) and obtain thatX n +3 -2x n 2 - Xn+l 2(+ 1) + 3.(14)By subtracti (12) from (14) , we find that X n +3 =Problem 4: It is easy to see that the sequence (x) 1 given by X n = I::~=l k2 satisfies the recursion X n ( +1? V 1 X n 1 -with Xl = 1. (i) By sstitutin+ 1 for/Fttin4 1 X n 1 2X n +l -Xn/, i 4 ' ' ' V ;13X n +2 - 3X n +l+ X n 2 V ; 1 with Xl = 1, X2 = 5, and = 14. (iii) Prove that the sequence (x n ) 1 satisfies the linear X n +4 - 4X n +3+ 6X n +2 -(13)P(z)Solve this recursion and show that ( = 1)(2+ 1)V > ' AI I J+ Xn-0, V n ~1.(16)= -4 + 6z 2 - 4z 1 =(z-1)4.The polynomial P (z) has root = 1 with multiplicity 4. We conclude that the there exist constants Ci , i = 1 : 4 , such that = 01 O2 032 04n3 V ;1.Since Xl = 1, X2 = 5, X3 = 14 , X4 = 30 , it follows that 0 1, O2 , 0 3 and 04 satisfy the following linear system (111 1 2 4 1 3 9 1 4 16recslOn0, V n 1.4x+1 X n/iEvhu The characteristic polynomial associated to the recursion (16) is /(12)1V n 1.By subtracting (13) from (15) , we find thatXnX n 3V n ~1. 2+3 3x n 1 3X n +2 Xn+l 2 X n +4 - 4X n +3 6X n +2 - 4X n +l+ ( 2?+3X n +l X n 2 with Xl 1 and X2 = 5. (ii) Similarly, show thatXnX n +4 =nuSubtract (10) from (11) to find that X n +2-(iii) We substitute 1 for n in (13) and obtain that(10) , obtain thatX n +2 -3x n 21 ) C 1 ) - /f 1 8I O25 27 I I 0 3 I - I 14 6 40430We obtain that 0 1 = 0 , O2 = ~ =i aBdq=i and therefore 23( + 1)(2n 1) n - = ,1.V n;1.Conclude that S( 2) k2(+ 1)(2+61)V> 1.Problem 5: Find the general form of the sequence (x n)n 2: 0 satisfyi the linear recursion X n +3 = 2X n +l X n , V n 0, withSolution: From (11) , we obtain that the first terms of the sequence (xn)n l are Xl = 1, X2 = 5, X3 = 14 , X4 = 30. (i) The recursion (12) follows immediately by subtracting (10) from (11).Xo= 1, Xl = -1 , andX2= 1.First Solution: By direct computation , we obtain X3 = -1 , X4 = 1 -1 = 1. It is nat al to conjecture that X n = (-l)n for a positive integer . This can be easily checked by induction. X6 7. 1VIATHE1VIATICAL PRELIMINARIES6SecondSolon: Alternati ythe linear recursion X n 3 P(z)-=we note that the sequece (x n ) o satises 0 , with characteristic polynomial+1 =(z 1)(z2 -Z -The roots of P(z) are 1 E aL. Thereore h re exist It t f e cst on S n /=01 (-1) O2I -~11 - vi 1 03 I U 2 I+I2/1 15 I=(-1) Vn o.X n +1, Vn O.3X n + 2, V ; 0 4x n 1 -with Xo = 1 and Xl = 5. (ii) Find the general forml i (17) -(18)=z2-4z 3=(z-1)(z-3) ,which has roots 1 and 3. Thus , X n 014X n+1 - 3x n 1 4x+2 - 3x 1(19)5X n+2 - 7X n 1=z3 - 5z2+ 7z -V ; o.V n o.(22)+ 1,V ; o.(23)+ 3x n, V n~ o.(24)3 = (z - 1) 2 (z - 3).Therefore , there exist constants 0 1 , O2 , 0 3 such that Xn=01 3 O2+ 0 3,V o.Since Xo = 1 = 5, and X2 = 18 , we find that 0 1 = ~ O2 = -~ =-i We conclude that 3+2 -+ 023n ,(21)(ii) The characteristic polynomial of the linear recursion (24) is P (x)~ o.3X n +1 +3 V o.Subtract (22) from (23) to find thatSltit1 (ii) The characteristic polynomial of the linear recsion (19) is P(z)Xn+2 -X n 30, V nV 0,Substitute + 1 for n in (22) to obtain that(17) from (18) and obtain that+ 3x n+ 3x r1with Xo = 1, Xl = 5, and X2 = 18. (ii) Find the general formula for X n , n o.Xn+3 -Xn+2 - 4x1(20)+2 V n 0,5x+2 7X n+1-Xn+2 -it follows thatSlractSubtract (20) from (21) to find that3x n , V 0,3x n 1 2.3x n By substituting n + 1 for i (20) we obtain thatfor X n o.Z+2 -WeZ+3(17)Solution: (i) Let n = 0 in (17) to find that Xl = 5. By-Solution: (i) The first three terms of the sequence can be computed from (20) and are Xo = 1, Xl = ?' and X~_~ 18.with Xo = 1. (i) Show that the sequence (x n ) satisfies the linear recursion X n 2 -with Xo = 1. (i) Show that the sequence (Xn)n:::::O satisfies the near 1i r ecSProblem 6: The sequence (x n )o satisfies the recursion X n 11, V O.-Problem 7: The sequence (x n ) satisfies the recursionBy solving the 3 x 3 linear system for 0 1 , O2 , and 0 3 obtained by requiring that Xo = 1, Xl = -1 , and X2 = 1 we find that 0 1 = 1, O2 = 0, and 0 3 = o. We conclude that Xn= 2 3nXn1).C1 , O2 , and 0 3 such that XnSince Xo = 1 and Xl = 5, we obtain that C 1 = -1 and C 2 = 2 and therefore2x - X n =z3 - 2z70.1. SOLUTIONS TO CHAPTER 0 EXERCISESXn -24~ V> o. 8. JVIATHEJVIATICAL PRELIMINARIES8Problem 8: Let P(z) = 2: iZi be the characteristic polynomial corresponding to the linear recursion2 i0, V n ~=(25)Assume that is a root of multiplicity 2 of P(z). Show that the sequece (Yn) given by Yn = c ; 0 where C is an arbitrary consta satisfies the recursion (25).It is easy to see thatIh(x) + h(x) I /' 1~ YY> """J h(x) I J.. lim ~;mn Ih(x) I 0 with ;Z( 7Zzb.Also , what is the largest possible value of b? 0.3Solutions to Supplemental ExercisesProblem 1: Let>0 be a positive number. Compute tt ion t hatlimit by l , ther1 1tfollows that l must satisfyj +j+ =d l (28)which can be solved for l to obtain that 1+ VI+ 2(29) 10. 12!/fATHEMATICALPRELIMINARIES.-.We now show that , for any> 0, the limit of /+ / +va-. does e whick i?quimieM to provig that the uenC e (X) o i s c on 0.3. SOLUTIONS TO SUPPLE1VIENTAL EXERCISESSolution: If we know that the continuous fraction (30) has a limit l, then l must satisfyl= +; l2 - al -where X 0= and vvaX n +1 -We will how that theI O.and thereforel creasm g.Let l be given by () i.e. , let l = vaX n +1va:+l = l ,The rstsince l is the positive O n of ( a d t he ref 28) 8 efl inductior1 we tllat X < l for all > O. fiIIld Xo(ii) The sequence (x r o is increasi sinceXn 0 for all n O. vVe conclude that2 1Xla3 + 24 32+1= X2= -:XQ = 1 33 222 l'1 11 4 2XnX2< l 07(33)with Yo = . Note that Yn = X2n for all O. Assume that Yn < l , where l is given by (32). Recall from (31) thatl2 -.al - 1 = 0(34)and thath t 2 t1 + a~...Yn+1< X n 1 for allProblem 2: Let > 0 be a positive number. Compute O.+17V n O. Xfew terms of the sequece (xn)no are< 0 ,> O. Note that 2:-zn -We note that the terms of the sequence are alternatively larger and smaller than the value of l given by (32) , i. e. ,It is easy to see that Xn(31)l= +va2.(i) The sequence (xn )o is bounded from above bv l. Note that Xo = < l. If we assume that Xn < l en va nand Yn+1 < l. Note that , by denition (33) , Yn > 0 for all O. Then , from (33) , it is easy to see thatYn+1 > Yn (2 I)Yn >d+U (; Yn - 1) < O. (36) From (35) , and using the assumptio that Yn < l , it follows thatYn+1 l , we show that Zn+1 < Zn and Zn > l. This proof is very similar to that given above for the sequece (Yro and is left to the reader as an exercise. We conclude that lim X2n1 = l./ d4-a Eom(36)and(37)pwe comiude tht7if yn(37) Um for anythe last equality can be derived as follows: 22 - al 1 = l l =2l - al l (l2 - al - 1) = 0 , where ttle last equality is tfle same as(34). We conclude from (38) that , if Yn < l ,' then Yn+1 < l for all O. III other words7we showed by induction that the sequece (Yn) giv by the recursion (33) with Yo = is increasi and bounded from above by l. Therefore , the sequece (Yn)o IS coergent. Denote by h = limn Yn the limit of the sequence (Yn)n '2 o. From (33)-and using (35)v, ;;e obtai~ th~tJ (2 l)h 1 h l (lt II - 1) = 0It - I) (It + -a) =0 Since h > 0 , it follows that h = l , i.e. , that limn Yn = l. Recall that Yn X2n for all; O. We showed that the subsequence made of the even terms of (x 7 1 by (32) , i.e. , that lim X2n l. (39) n 2lim Xn im Xn (2 + l)zn + Zn 1' V >0 , , .v:::-vl -+ v'zProblem 3: (i) Find x > 0 such thatXX"2.(41)(ii) Find the largest possible value of x > 0 with such that there exists a number b > 0 with XXX' = b. (42) Also , what is the largest possible value of b?Solution: (i) If there exists x such that (41) holds true , then x 2 = 2 l aI therefore x = y'2. We are left with proving that ~V2'-/2 Y";'= 2.Consider the sequence (x n )n '2 0 with Xo = y'2 and satisfying the following recurSIOn: +1 - -/2x n _ rz/27V 71 O. It is easy to see by induction that the sequence is increasing and bounded from above by 2 , since X n 1 > Xn 2Xn/2 > 2 xn - I/ 2 :> Xn > X n 1 ;Xn+1 < 2 ? 2xn / 2 < 2 ::} x n /2 < 1 ::} Xn < 2. We conclude that the sequece (x n)n'2 0 is convergent. If l = limn--t X n thenSimilarly, we define the sequence (zn)n '2 o by the recursion (40)From (39) and (40) , we find thatl (2 + I)Yn 0, xtz2- Xl + 3X3 , 3X2 - 2X3) ; (x) (x) (x) (x) ~(x) (x)Compute2~-u(x , t)u2(fff) (x) (x) ( (4X l - X2 ,D 2 f(x)I,f(l , -1 , 0) = 3 D f(l , ~1 0) = (5 , - 1 , - 3);(1. 6) (1. 7)(1 -1 (;-i e-;)(1. 8)at=1 2'J_t-u/~ V1-=y4e.r 2"4+x2.r 21=d 1Product ule 2.r e- 4i.r 2==e =e ; 4t 2 y4: 2t y4 uxax---1.---2t v41r t ~.r 2 4t,(x2(we find that 1I . I ~II 24t ))(1. 9) 16. 24CHAPTER 1. CALCULUS REVIE1iV. PLAIN VANILLA OPTIONS.1 1 x From (1. 9) and (1. 10) , we e c onc t i cll t hat 2 U/ t t 22 U t'nu 4t4ux 2 1 4ANote: This is a butrgy pead tMer mks a 10 Pi1 in a brfly pread if the price of the underWIngmset at maturity is expected to be h the K1 S(T) K3 range. SoJTLtiOTZJA butterny spread is &Il optiOIls portfolio made of a long positiOII IIe call option with strike kh a long position iIla call option with strike khazld a short position intwo calls with strike equal to the average of tlm strikes K 1 and K 3 , i.e. , with strike K2 ; all options have the same maturity and have the same underlying asse( The payof at maturity of a butterdy spread is aiways11OIIIlegativtand it is POSItive if ttle price of the mdeElying asset at maturity is betweeiltbe strikes K 1 and K 3 , i.e. , if K 1 < S(T) < K;. For our particular example?tfle values of the three call options at maturity e , respectively,C2 (T) C3 (T)max(S(T) - K 1 , 0) ax(S(T) - K 2 , 0) max(S(T) - K 3 , 0)max(S(T) - 30 , 0); max(S(T) - 35 , 0); max(S(T) - 40 , 0)and the value of the portfolio at maturity is V(T)= C 1 (T)S(T) E,,,,,,< 30 30 < S(T) < 35 35 < S T) < 40 40 < S(T)S(T) - 30O Omm 25O S(T) - 30S(T - 30 S(T - 35S(T) - 30sjTj35 SrT) - 4040 - S(T)L....JProblem 11: Cosider a portfolio with the following positions: long one call option with strike K 1 = 30; short two call options with strike K 2 = 35; long one call option with strike K 3 = 40. All options are on the same underlyi asset and have maturity T. Draw the payof dlagram at maturity of tile portfolio?i.e-7plot the value of the portfolio V (T ) at matu y ity as a funct ion of S (T )7t he price of t he m d e rl y Jrm at S se t time T.C1 (T)1.1. SOLUTIONS TO CHAPTER 1 EXERCISES2C2 (T)+ C 3 (T).Depend,ing o,n t~~ ~~lu~s_ of the spot S(T) of the underlying asset at maturity, the value V(T) of the portfolio at t'i~e T is given beiow~Problem 12: Draw the payoff diagram at maturity of a bull spread with a long position in a call with strike 30 and short a call with strike 35 , and of a bear spread with long a put of strike 20 and short a put of strike 15. Solution: The payoff of the bull spread at maturity T is (T)= max(S(T)30 , 0) -max(S(T) - 35 , 0).on the value of the spot price S(T) , the value of the bull spread at maturity TisDepedingI S(T) < 30 I 30 < S(T) < 35 I 35 < S(T) I 0 I S(T) 30 I 5 I vi (T) I The value of the bear spread at maturity T is (T)= max(20 - S(T) , 0) - max(15 - S(T) , 0) ,which can be writteni termsof the value of S(T) asA trader takes a long position in a bull spread if the underlying asset is expected to appreciate in value , and takes a long position in a bear spread if the value of the underlying asset is expected to depreciate. Problem 13: Which of the following two portfolios would you rather hold: Portfolio 1: Long one call option with strike K = X - 5 and long one call option with strike K = X 5; Portfolio 2: Long two call options with strike K = X? (All options are on the same asset and have the same maturity.) Solution: Note that being long Portfolio 1 and short Portfolio 2 is equivalent to bei long a butte 17. 26CHAPTER 1. CALCULUS REVIE1iV. PLAIN VANILLA OPTIONS.rather aomegative)payof M maturity-Ttmrefore7if you are to ss u n 7 POSItion ln either om of th portfolios ( ot to purchase the portfolios) you , are better of owniIlg Portfolio 17since its payofat maturity wiH always be at least as big as the payoff of Portfolio 2. More precisely, note thatV(T) (T)-1.1. SOLUTIONS TO CHAPTER 1 EXERCISESVeT) OK 1 < SeT) < K 2 K 2 < S(T ) < K 3 K 3 < SeT)- 5) 0) max(S(T) (X + 5) , 0) - 2max(S(T) - X , O).The value of the portfolio at timeTis detailed below:V(T) S(Tf ;0; 0; 0.For C 1 (0) = 8 , C2(0) = 5 , C3(0) = 3 and K 1 = 100 , K 2 = 120 , K 3 = 130 , the problem becomes finding Xl 0 and X2 and X3 such that8X1+ 5X2 + 3X3 30X1 + 10x2 Xl X2 + X3 > ;(1. 11) (1. 12) (1. 13)0; 0.(For these option prices , arbitrage will be possible since the middle option is overpriced relative to the other two options.) The easiest way to find values of Xl , X2 , and X3 satisfying the constraints above is to note that arbitrage can occur for a portfolio with long positions in the options with lowest and highest strikes , and with a short positio the option with middle strike (note the similarity to butterfly spreads). Then , choosi X3 = -Xl - X2 would be optimal; ef. (1. 13). The constraints (1. 11) and (1. 12) become5X1 + 2X2 < 3X1 + X2 >nunuThese constraints are satisfied , e.g. , for Xl corresponds to X3 = 2.1 and X2 --3 , which 18. 28CHAPTER 1. CALCULUS REVIEW. PLAIN VANILLA OPTIONS.Buying one option with strike 100 , selling three options with strike 120 , and buying two options with strike 130 will generate a positive cash flow of $1 , and will result in a portfolio that will not lose money, regardless of the value of the underlying asset at the maturity of the options. more shares of the underlying asset and usi the casg proceeds to make the dividend payments. Then , the short position in e- q1 shares at time 0 will become a short position in one share 1 at time T. The value of the portfolio at maturity isV(T) Problem 15: A stock with spot price 40 pays dividends continuously at a rate of 3%. The four months at-the-money put and call options on this asset are trading at $2 and $4 , respectively. The risk-free rate is constant and equal to 5% for all times. Show that the Put-Call parity is at satised and explain how would you take advantage of this arbitrage opportunity.Solution: The followi values are give : S = 40; K = 40; T = 1/3; r = 0.05;P(T) + S(T) - C(T)-C -39.5821> 39.3389 - Ke-= max(K - S(T) , 0) + S(T) - max(S(T) - K , 0) = K ,regardless of the value S(T) of the underlying asset at maturity. Therefore ,V(T) = (P(T) S(T) - C(T)) + 39.5821erT - - K + 39.5821e rT - - 40 + 40.2473 -The Put-Call parity is not satisfied , sinceP + Se-= - P(T) - S(T) + C(T) + 39.5821erT .Recall from the proof of the Put-Call parity thatq = 0.03; P = 2; C = 4. qTrT.(1. 14)Therefore , a riskless profit can be obtained by "buying low and selling by selling the portfolio a the left hand side of (1.14) and buying the portfolio on the right hand side of (1.14) (which is cash only). The riskless profit at maturity will be the future value at time T of the mispricing from the Put-Call parity, i.e. ,igh i.e. (39.5821 - 39.3389) = 0.2473(1. 15)To show this , start with no money and sell one put option , short e- qT shares , and buy one call option. This will generate the following cash amount:291.1. SOLUTIONS TO CHAPTER 1 EXERCISES0.2473.This value represents the risk-free profit made by exploiting the discrepancy from the Put-Call parity, and is the same as the future value at time T of the misprici from the Put-Call parity; d. (1. 15). Problem 16: The bid and ask prices for a six months European call option with strike 40 on a non-dividend-paying stock with spot price 42 are $5 and $5.5 , respectively. The bid and ask prices for a six months European put option with strike 40 on the same underlying asset are $2.75 and $3.25 , respectively. Assume that the risk free rate is equal to O. Is there an arbitrage opportunity present?V(O) = P(O) - S(O)e- qT C(O) + 39.5821 = O.Solution: For r = 0 , the Put-Call parity becomes P + S - C = K , which in this case can be written as C - P = 2. Thus , an arbitrage occurs if C - P can be "bought" for less than $2 (i. e. , if a call option is bought and a put option is sold for less than $2) , or if C - P can be "sold" for more than $2 (i.e. , if a call option can be sold and a put option can be bought for more than $2). From the bid and ask prices , we nd that the call can be bought for $5.5 and the put can be sold for $2.75. Then , C - P can be "bought" for $5. $2.75=$2.75 , which is more than $2. Therefore , no risk-free profit can be achieved this way. Also , a call can be sold for $5 and a put can be bought for $3.25. Therefore , C - P can be "sold" for $5-$3.25=$ 1. 75 , which is less than $2. Again , no risk-free profit can be achieved. Note that by shorting the shares you are responsible for paying the accrued dividends. Assume that the dividend prments are financed by shortinglThis is similar to converting a long pm;ition in c qT shares at time 0 into a long position share at time T: throh continus purchases of (fractions of) shares usi the dividend payments: which is a more intuitive process.P + Se- qT-C39.5821 ,since shorting the shares means that e- qT shares are borrowed and sold on the market for cash. (The short will be closed at maturity T by bi shares a the market and returning them to the borrower; see below for more details.) At time 0 , the portfolio consists of the following positions: short one put option with strike K and maturity T; short e- qT shares; long one call option with strike K and maturity T; cash: +$39.582 1. The initial value of the portfolio is zero , since no money were invested:i one 19. 30CHAPTER 1. CALCULUS REVIEVV. PLAIN VANILLA OPTIONS.Problem 17: You expect that an asset with spot price $35 will trade in the $40$45 range in one year. One year at-the-money calls on the asset can be bought for $4. To act on the expected stock price appreciation , you decide to either buy the asset , or to buy ATM calls. Which strategy is better , depending on where the asset price will be in a year?Solution: For every $1000 invested , the payoff i one year of the first strategy, i.e. , of buying the asset , isT) = 1~~0 S(T) , wereS(T) is the spot price of the asset in one year. For every $1000 invested , the payoff in one year of the second strategy, i.e. , of investing everything in buying call options , is1I2 (T)= max(S(T) - 35 0) ()-35)iijzjZZJIt is easy to see that , if S(T) is less than $35 , than the calls expire worthless and the speculative strategy of investing everything in call options will lose all the money invested in it , while the first strategy of buying the asset will not lose all its value. However , investing everything in the call options is very protable if the asset appreciates in value , i.e. , is S(T) is signicantly larger than $35. The breakeven point of the two strategies , i.e. , the spot price at maturity of the underlying asset where both strategies have the same payoff is $39.5161 , since 1000 _ ._, 351000:~ S(T) = (S(T)- 35) S(T)=39.516 1.If the price of the asset will , indeed , be i the $40$45 range in one year , then buying the call options is the more profitable strategy. Problem 18: The risk free rate is 8% compounded continuously and the dividend yield of a stock index is 3%. The index is at 12 ,000 and the futures price of a contract deliverable in three months is 12 ,100. Is there an arbitrage opportunity, and how do you take advantage of it?Solution: The arbitrage-free futures price of the futures contract is 12000er - q)T = 12000e(0.080.03)/412150.94 > 12100.Therefore , the futures contract is underpriced and should be bought while hedged statically by shorting e- qT = 0.9925 units of index for each futures contract that is sold.1.1. SOLUTIONS TO CHAPTER 1 EXERCISES31At maturity, the asset is bought for 12100 and the short is closed (the dividends paid on the short position increase the size of the short position to 1 unit of the index). The realized gain is the interest accrued on the cash resulting from the short position minus 12100 , i.e. , eO.08 / 4 (e- O.03 20. 32CHAPTER 1. CALCULUS REVIEW. PLAIN VANILLA OPTIONS.1.2Supplemental Exercises1. 3. SOLUTIONS TO SUPPLENIENTAL EXERCISES 8. Create a portfolio with the following payoff at time T:1. Compute(x) dx 2. ComputeV(T) =xne x dx(I n ( x) )n dxZ /it l-z /IIlt+< l l Ie ' -5. Letf(x)z 11I/ 1-z+ E ]V Z(= =exp( 0 S(T) < 20; 20 S(T) < 40; 40 S(T) 9. Call options on the same underlying asset and with the same maturity, with strikes K 1 < K 2 < K 3 , are trading for C1 , C2 and C3 , respectively (no Bid-Ask spread) , with C1 > C2 > C3 . Find necessary and sufficie conditions on the prices C 1 , C 2 and C 3 such that no-arbitrage exists corresponding to a portfolio made of positions in the three options.J4. Show that2S(T) , if ( < 60 - S(T) , if l S(T) - 20 , ifwhere S(T) is the spot price at time T of a given asset. Use plain vanilla options with maturity T as well as cash positions and positions in the asset itself. Assume , for simplicity, that the asset does not pay dividends and that interest rates are zero.J3. Compute33>- t 4C bid and Csk and by 1id and ~sk respectively, the bid and ask prices for a plain vanilla European call and for a plain vanilla European put option , both with the same strike K and maturity T , and 0e same underlying asset with spot price S and paying dividends continuously at rate q. Assume that the risk-free interest rates are constant equal to r. Find necessary and sufficient no-arbitrage conditions for Cbid , Cask , Pbid , and sk10. Denote by(x )2 22i)Assume that g : JR JR is a continuous function which is uniformly bounded , i.e. , there exists a constant C such that jg( x) I C for all x E JR. Then , show thatM: x=g()1.3Solutions to Supplemental ExercisesProblem 1: ComputeJn x ln(x) dx6. Let)CiSolution: If n -1 we use integration by parts and find thatCompute g' (y).f zdx = -Mz)-j zn+1-dz x n ln(x) n + l l f + 1 1)7. A derivative security pays a cash amount C if the spot price of the underlying asset at maturity is between K 1 and K 2 , where 0 < K 1 < K 2 , and expires worthless otherwise. How do you synthesize this derivative security (i.e. , how do you recreate its payoff almost exactly) using plain vanilla call options?nx 11n ( For n= 1xn 1A+ C.we obtain thatrn~x) dx WJ= (ln(x)?+C. 21. 34CHAPTER 1. CALCULUS REVIEW. PLAIN VANILLA OPTIONS.P VAO 'D emProblem 2: Compute''EA/ dx Solution:For every irger n 0 define the functionfn(x)hQU A I/ I i l UtliI/ al-Zz i- i(1. 16)(11 < In I 1 + -=. I < Vx> 1.Letx): +D;f(x)x(1. 17)+~)g(x)The 7= eX , the followi general forml fIJ4t-AZ 1By using integration by parts , it is easy to see thatG z neZ , 2 ' 4-j' (x) 414g'(x)1+. 2xIX1(x + 1) -Lx(x 1)I1(x + 1)2x2 (x + 1) < 0: ~ -, -1< 0x(x + 1)2We conclude that bo f (x) and g( x) are decreasing functions. Since Problem 3: Computelimj(1n(x)) dxX it follows thatSolution:f(x) =limX-cf(x) > 0 and g(x) > 0 forFor every integer 0 letg(x) = 0 ,allx > 0, andtherefore:>+:) ZLI7VZ>07 (x) j (In (x) )n dxwhich is what we wanted to show; cf. (1. 17).By using integration by parts , it is easy to see that , for any ; 1 Problem 5: Let(x)t dx = ) j (In(x))) =( f)2)and thereforefn(X ) = x(I n ( x)) fn-1()V 1.Since fo(x) = the following general formula can be obtained by iruction:j (1 (x))n dx = fn(x) = x L(k n ! .L ~nlI f,;(ln(x))k 0Vn 1.Assume that g : R ffi. is a continuous function which is uniformly bounded 2 , i.e. , there exists a constant 0 such that Ig(x)1 o for all x ffi.. Then , show that:g(x) dx = )2The uniform boundedness condition was chosen for simplicity: and it can be l'axed: e.g.: to functions which have polynomial growth at infinity. 22. CHAPTER 1. CALCULUS REVIE 1iV. PLAIN VANILLA OPTIONS.36Solution: Using the change of variables y = ?nd that f(x)g(x) dx g(x)exp(;)2)/ dy VkI/I1=dx (1. 19)since the function e-15is tue pability density function of the standard normal variable. From (1. 18) and (1. 19) we obtain that f1rf(x)g(x) dx = ) (g() - g( y)) e- dy. j d -/(1. 20)Our goal is to show that the right hand side of (1. 20) goes to 0 as O. Since g( x) is a continuous function , it follows that , for lY > 0 , there exists 61 () > 0 such that Ig( )-g(x)1 < forV Ix 1< 61() .LfdM)e4dy : ( dy < J !"L1fSince Ig( x) I C for all xr- 62()1=v L,1fIJ EIg() - g(+ )1 e dyJ 62() 0, the following condition must also be satisfied: X201O2(1. 39) 25. 42CHAPTER 1. CALCULUS REVIE1V. PLAIN VANILLA OPTIONS.Recall that C1 > C2 > C 3 . Using the value of that (1. 36) and (1. 37) hold true if and 0ly if C1 -X2Xl C2 -X3from (1. 38) , it followsC3(1. 40)-C3 'X 2 K 3 - K1 Xl-K3-(1. 41)K2Also , note that if (1 .40) holds true , then (1. 39) is satisfied as well , sinceC1 - C3 C2 - C3C1 C-C1 C3 C2 -C3 -(1 .4 2)Portfolio 2:Long the K 1-option , short the K 2-option , short the Arbitrage exists if we can d Xi > 0 , i = 1 : 3 , such that C1 - X2C2 - X3C3 Xl (80;- K 1 ) - x2(8 - K2 ) (8 - K3 ) 0 V 8 O.K3 option.(1. 43) (1. 44)The ineqlity (1. 44) holds if and only if the following two conditions are satisfied: Xl - X2 - X3 0; (1. 45) Xl (K3 -K 1) -x2(K3 -K2 ) O.(1. 46)However , (1. 43) and (1. 45) cannot be simultaneously satisfied. Since C 1 > C 2 > C3 , it is easy to see that 1=C? CC'}. C X1 +X C3 C3 Xl~~~since C1 > C2 > C3 . Therefore , no arbitrage can be obtained by being long the options with strikes K 1 and K 2 and short the optio with strike K 3 . We conclude that (1. 42) , i.e. ,2We conclude that arbitrage happens if and only if we can find Xl > 0 and X2 > 0 such tl(1.40) and (1. 41) are simultaneously satised. Therefore , no-arbitrage exists if and only if K 3 -K1 K 3 -K21.3. SOLUTIONS TO SUPPLE1VIENTAL EXERCISESIn other words , no arbitrage can be obtained by being long the option with strike K 1 and short the options with strikes K 2 and K 3 . Portfolio 3:Long the K 1-option , long the K2 option short the K 3-option. Arbitrage exists if we can find Xi > 0 , i = 1 : 3 , such that 0;C1 'K3 -K2 is the only condition required forC2 -C3no-arbitrage.Problem 10: Denote by Cbid and Cask , and by id and sk respectively, the bid and ask prices for a plain vanilla European call and for a plain vanilla European put option , both with the same strike K and maturity T , and on the same underlying asset with spot price 8 and paying dividends continuously at rate q. Assume that the risk-free interest rates are constant equal to 7'. Find necessary and sufficient no-arbitrage conditions for Cbid , Csk Pbid , and skSolution: Recall the Put-Call parity C-8e- qT -PKe- rT ,where the right hand represents the value of a forward contract on the underlying asset with strike K. An arbitrage would exist either if the purchase price of a long call short put portfolio , i.e. , C Pbid were less than the value 8e- qT - K e- rT of the forward contract , i.e. , if Csk idSe-qT- K e- rTis -arbitrage directlyCsk id 8e- qT-Kfollowing from the Put-e- r1 Cbid sk 26. Chapter 2 Improper integrals. Numerical integration. Interest rates. Bonds. 2.1Solutions to Chapter 2 ExercisesProblem 1: Compute the integral of the function f(x , y) = x 2 - 2y on the region bounded by the parabola y = (x + I? and the line y = 5x - 1. Solution: We first identify the integration domain D. Note that (x + I? = 5x -1 if and only if x = 1 and x =2 , and that (x + 1)2 5x - 1 if 1 < x < 2.Therefore , D = {(x , y) 11 Z 2 and (x+ 1)2 5x - I}.Then ,fin [( Z i;:( [(( )2)x 2 U|tcc) ) d U U::U; !: L J;I (5x-1 - ( (( 1)2 + Z 1 Z + 11'( x 1 1 ( 5 ( Z )Z =; l 3x -2 Z d Problem 2: Letf : (0 )lR denote the Gamma functio) x=V Z l 1 ~ 45i.e. le t. 27. CHAPTER 2. NUMERICAL INTEGRATION. BONDS.46(i) Show that f() is well defined for any > 0 , i.e. , show that both[x-1e- x dx[ ' x"1 e- xz-le-z 1. Show that f(l) = 1 and conclude that , for any 1 positive integer , f(n) = (-I)!.Solution: (i) Let > 0. Intuitively, note that , as x the order of x o:- 1 , since limx", 0 e- x = 1. Since~tIl}(1 x-1 t", 0 Jt0the functio Z-l e -x is 0Choose n() = max(m( N)). From (2.3) and (2 .4) we obtain thattoItx1e-x dx -lim I to Jt(2.5)We can then use XJim "~UVIJle- dx -and(2.6)< (2.6)to show that , for any s > () ,J/' X"-l ZX"-l Z 0 , there exists m(E) > such that 2e- m (f. )!2 < (2 .4)/,l(2.3)_-x/2 1m x- - e -,- =2e- n (f. )!2dx lim |zllim(l-t)Joe-z/27 Vx> N ,Ii -1 X-e- xsuch thatNote that there exists N >Slceand472.1. SOLUTIONS TO CHAPTER 2 EXERCISEStwhich is what we wanted to show; d. (2.2). (ii) It is easy to see thatm)=fev=JI tfzdz= (_e-t + 1) Assume that > 1. By integration by parts , we find that1,JiITl (1 - e- ) -f()"~UVit follows thatfzMe :Etlz-1xe- dx(2.1)exists and is finite. lIIaking these intuitive arguments precise is somewhat more subtle. We include a mathematically rigorous arguments for , e.g. , showing that the integral in (2.1) exists and is finite. By denition we need to prove that , for a > 0 , there exists() > such thatI-X dx < Vs> ()(2.2)= 1f l ' o k w h i i - ! U L 1 1 { 3 m ] e f ' 'z 1 1 J f n n U UiF IJ- i Z - e ffl'o -j d 0 0 / q& Z Z f 7 e K / z = 1 l t n1 A t 4ufbfqhl I / {Zsmce lim x o:- 1 e- xlimx-1z02Jlim t o:- 1 e- t t c-0,-1lim --:-t xeL= 0.for > 1;ZEB,m G dIil--tE 28. CHAPTER482.NUMERICAL INTEGRATION. BONDS.For any positive integer> 1, we nd that f(n) = f(l) = 1, it follows by induction that f() = (-1)!( -l)f(-1). 2.1. SOLUTIONS TO CHAPTER 2 EXERCISES No. Intervals NIidpoint Rule 0.17715737 4 0.17867774 8 0.17904866 16 0.17914062 32 0.17916354 64 0.17916926 128 0.17917070 256 0.17917105 512Since3Problem 3: Compute an approximate value of J1 -IX e-xdx using the NIidpoint rule , the Trapezoidal rule , and Simpson's rule. Start with = 4 intervals , and double the number of intervals until two consecutive approximations are within 10- 6 of each other.Solution: The approximate values of the integral found using the N dpoint , Ii Trapezoidal , and Simpson's rules can be found in the table below: No. Intervals NIidpoint Rule Trapezoidal Rule Simpson's Rule 4 0 .40715731 0 .4 1075744 0 .40835735 8 0 .40895737 0 .40807542 0 .40836940 16 0 .40829709 0 .40851639 0 .40837019 32 0 .40835199 0 .40840674 0.40837024 64 0 .40837937 0 .40837024 0 .40836569 128 0 .40836911 0 .4 0837253 0 .40837082 256 0 .40836996 512 0 .40837018 0 .4 0837039 0 .40837028 1024 0 .40837023(ii) Without computi f(4)(X) note tl the denominator 1 + x 2 of f (x) is bounded away from 0 , and that the fourth derivative of the numerator of f(x) is on the o-rder of X- 3 / 2 , which is not defined at 0 , and is unbounded in the limit as x o. (iii) Usi Simpsor rule the following approximate values of the integral are obtained: No. Intervals 4 8 16 32 64 128 256The approximate value of the integral is 0 .408370 , and is obtained for a 256 intervals partition using the NIidpoint rule , for a 512 intervals partition using the 'rapezoidal rule , and for a 16 intervals partition using Simpson's rule.49Simpson's Rule 0.179155099725056 0.179169815603871 0.179171055067087 0.179171162051226 0.179171171372681 0.179171172188741 0.179171172260393Problem 4: Let f : V R --+R gi (i) Use NIidpoint rule with tal = 10- 6 to compute an approximation of (1(1I 5/2= Jo f (x) dx = J ~ I I 1+ o OJ'--I----The approximate value of the integral is 0.17917117 , and is obtained for a partition of the interval [0 , 1] using 64 intervals. Problem 5: Let K , T g :R R asand rx~(ii) Show that f(4) (x) is not bounded on the interval (0 , 1). (iii) Apply Sipsor rule with n = 2 k = 2 : 8, intervals to compute the integral I. Conclude that Simpson's rule converges.Solution: (i) The approximate value of the integral is 0.179171 , and is obtained for a partition of the interval [0 , 1] using 512 intervals:g(x) where b(x) =(Inbe positive constants. Define the function= rhu?()() T) / (-IT). Compl. (x)Solution: Recall thatirf = f) 29. CHAPTER 2. NUMERICAL INTEGRATION. BONDS.50 Therefore ,2.1. SOLUTIONS TO CHAPTER 2 EXERCISES51it follows thatg"(t) = h(t) , Vib () Zg'(x)which is what we wanted to show.(lxO"-J2y(In() ( ) T)222TnProblem 7: The continuously compounded 6-month , 12-month , 18-month , and 24-month zero rates are 5% , 5.25% , 5.35% , and 5.5% , respectively. Find the price of a two year semiannual coupon bond with coupon rate 5%.~J'Solution: The value B of the semiannual coupon bond is Problem 6: Let 1) be a continuous fun0 such that Jo Ixh(x)ldx exists. Define g(t) byt)=[>O()h(x)2 (10 0 = h(t) .Solution: Recall that , if (t) and b(t) are differentiable functions and if f (x , t) is a continuous function such that ?It (x , t) exists and is continuous , then /-11iI Ill-/ M M P d-a PI rlu I VG , , Z z Z , 2 + rld vhu, , 'hu i'U rlu ' ' ' ' 1 V 1 V ' a A similar result can be derived for improper integrals , i.e. , /Ill- P M d-a ft rld YG I rld i'U z l Z Z z ' l 1 W 1 w Lcashow4 L ' L Uj?-U7} Tb// ' ' E / ' 1 Tb B J / i ' L U } /Tb ETbE J ' l t /I7luit// 1 1iTb ' U l ' /J Z a E "vcasow= [2.5 2.5 2.5 102.5] .The discount factors are (t)= t and f(x , t) = (x - t)h(x) ,(2.8)where h(x) is contius. Then , ~{ (x , t) = -h(x) is continuous. Note that f( (t) t)= f(t , t) = (t - t)h(t) = O.disc = [0.97530991 0.94885432 0.92288560 0.89583414], and the price of the bond is B =[' (x - t)h(x) dX)(2.9)Solution: Par yield is the coupon rate C that makes the value of the bond equal to its face value. For a 2-year semiannual coupon bond , the par yield can be found by solving 100 -~+= - [(X) dx T 28 U( 1: x) = 7Problem 8: The continuously compounded 6-month , 12-month , 18-month, and 24-month zero rates are 5% , 5.25% , 5.35% , and 5.5% , respectively. vVhat is the par yield for a 2-year semiannual coupon bond?From (2.7-2.9) , we conell that= (98.940623. iFor our problem ,Since= [0.5 1 1. 5 2];J ' Ug' (t) ; l (2 e- r Owhere C 0.05 and r(0 , 0.5) = 0.05 (0 1) = 0.0525 , r(0 ,1. 5) = 0.0535 , (0 2) = 0.055. The data below refers to the pseudocode from Table 2.5 of [2] for c om putir 19 the bond price given the zero rate curve. Input: n = 4- l ' /= 100 e- r (OO.5)O.5 100 e- r () 100 e- r ()1.5dand show that gil ( t)B100 e- r (O ,O.5)O.5 (100++ ~ 1C~100 e- r (O ,1)2+ 100 e- r (O 1. 52e- r (O2)QUC2(1 - e- r (O ,2)2) e- r (O O.5)O.5 e- r (O l) e- r (O 1. 5)L5 e- r (O 2)2 30. CHAPTER522.NUNIERICAL INTEGRATION. BONDS.For the zero rates given in this problem , the corresponding value of the par yield is C = 0.05566075 , i.e. , 5.566075%. 2.1. SOLUTIONS TO CHAPTER 2 EXERCISES (~ ~I 1002.5 disc(l) Problem 9: Assume that the continuously compounded instantaneous interest rate curve has the form r(t) = 0.05 0.0051 + t) , V t O.(i) Find the correspondi zero rate curve; (ii) Compute the 6-month , 12-moh 18-month , and 24-month discount factors; (iii) Find the price of a two year semiannual coupon bond with coupon rate 5%. Solution: (i) Recall that the zero rate curve r(O , t) can be obtained from the instantaneous interest rate curve r(t) as follows: l--tp''Ifo vv > nu nU , T T J I 7 TLVt' b/= ;+l (1) + t)e- r (O ,O.5)O.5_e- r (O ,l)0.94939392; -r(O 2)2 -0.97478242; 0.92408277 ;0.89899376.(iii) The price of the two year semiannual coupon bond with 5% coupon rate IS:0.05 ~({ ( ,,{ " 0.05 -100e-r(O?05)05 ~V 2+102.5 disc(4)Problem 10: The yield of a semiannual coupon bond with 6% coupon rate and 30 months to maturity is 9%. What are the price , duration and convexity of the bond? Solution: The price , duration , and convexity of the bond can be obtained from the yield y of the bond as follows:B D= 3 p(p( 1 f 3i I ): :-B t 2exp(i5 (5 + 103 ~exp -~y }I JI I 2 ~ ~ 2 2'-'I -~y I(2.10) ,, l 07 41A 1 4 1 ' /}1 (~9i VA"4(i2 5 (51 exp ( ~y) + 103 -A~ exp ( ~y ) I . 2 42'" } J 07 ~I{ 100 e- r (O ,1 I)}'--(2.12)-vcash_ow= [3 3 3 3 103] .Output: bond price B = 92.983915 , bond duration D = 2.352418 , and bond convexity C = 5.736739. (ii) The 6onth 12-month , 18-month , and 24-month discount factors are , respectively,e-r(O 1. 5) 1. 52.5 disc(3)Lcash_flow = [0.5 1 1. 5 2 2.5];0.005(1 t)-+The data below refers to the pseudocode from Table 2.7 of [2] for computing the price , duration and convexity of a bond given the yield of the bond. Input:= 5; y = 0.09;;(0 + 00 0 disc(l ) disc(2) disc(3) disc(4)2.5 disc(2)) ~ B t 005 fh t 0O 0 5 l ( ) n(0.04599.267508.= IThen ,OJ )+~({ t)e I VB'0.05. __ I IUU530.05+ 100 e-1. 5) 1. 5Problem 11: The yield of a 14 months quarterly coupo bond with 8% coupon rate is 7%. Compute the price , duration , and convexity of the bond. Solution: The quarterly bond will pay a cash flow of 1. 75 in 2, 5, 8, and 11 months , and will pay 10 1. 75 at maturity in 14 months. The formulas for the price , duration , and convexity of the bond in terms of the yield y of the bond are similar to those from (2.10-2.12). For example , the price of the bond can be computed as follows:B = 1. 75 exp(2( 5/ + 1. 75 exp I - ~-"y J + 1. 75 exp I - 12'-' J I ~-"y I 12'-'I - ~-"y I +1. 75 exp12'-' J( 1 1( 14 + 10 1. 75 exp I - -,,- y II ~~y I 12'" J/ 31. CHAPTER 2. NUMERICAL INTEGRATION. BONDS.54The data below refers to the pseudocode from Table 2.7 of [2] for computing the price , duration and convexity of a bond given the yield of the bond. Input: n = 5; y = 0.07; I 2 5 8 11 141 t_cow = 11-2 1~2 1~2 ~; ~;I;v_cash_flow = [2 2 2 2 Output: bond price B = 10 1. 704888 , bond duration D = 1. 118911 , and bond convexity C = 1. 285705. Problem 12: Compute the price , duration and convexity of a two year semiannual coupon bond with face value 100 and coupon rate 8% , if the zero rate curve is given by r(O , t) = 0.05 + O.Olln (1 + ~) Solution: The data below refers to the pseudocode from Table 2.5 of [2] for computing the price of a bond given the zero rate curve. Input:= 4; zero rate (0 t) = 0.05 + O.Olln (1 i); t_cashow= [0.5 1 1. 5 2] ; v_cashow = [4 4 4 104] .Discount factors: disc = [0.97422235 0.94738033 0.91998838 0.89238025]. Output: Bond price B = 104.17391 1. Note: To compute the duration and convexity of the bond , the yield would have to be known. The yield can be computed , e.g. , by using Newton's method , which is discussed in Chapter 8. We obtain that the yield of the bond is 0.056792 , i.e. , 5.6792% , and the duration and convexity of the bond are D = 1. 8901 and C = 3.6895 , respectively. 2.1. SOLUTIONS TO CHAPTER 2 EXERCISES55ac ou po np ay I rrrpayment.Problem 14: By how much would the price of a ten year zero-coupon bond change if the yield increases by ten basis points? (One perce age poi IS equal to 100 basis points. Thus , 10 basis points is equal to 0.00 1.) Solution: The duration of a zero-coupo bond is equal to the maturity of the bond , i.e. , D = T = 10. For small changes !:1 y in the yield , the percentage change in the value of a bond can be estimated as follows:!:1 B-!:1yD == - 0.0 1.0.001 . 10We conclude that the price of the bond decreases by 1%.Problem 15: A five year bond with duration 3~ years is worth 102. Find an approximate price of the bond if the yield decreases by fifty basis points. Solution: Note that , since the yield of the bond decreases , the value of the bond must increase. Recall that the percentage change in the price of the bond can be approximated by the duration of the bond multiplied by the parallel shift in the yield curve , with opposite sign , i.e. ,!:1 B- !:1 yDFor B = 102 , D = 3.5 and !:1 y = -0.005 (since 1% = 100 bp) , we nd that !:1B -!:1 y D B = 1. 785.The new value of the bond is Problem 13: If the coupon rate of a bond goes up , what can be said about the value of the bond and its duration? Give a financial argument. Check your answer mathematically, i.e. , by corgggand gg7and ShhOw these c are either always positive or always ive. fUl n ctiorIS 1legat Solution: If the coupon rate goes up , the coupon payments increase and therefore the value of the bond increases. The duration of the bond is the time weighted average of the cash flows , discounted with respect to the yield of the bond. If the coupo rate increases , the duration of the bond decreases. This is due to the fact that the earlier cash flows equal to the coupo payments become a higher fraction of the payment made at maturity, which is equal to the face value of the bond plusBnew-B +!:1 B -103.75.Problem 16: Establish the following relationship between duration and convexity: C _ D :t.Solution: Recall thatD n l-DU B-vu r can-D-=:.oy,dC12BBy2 32. CHAPTER56 Therefore ,2.NUMERICAL INTEGRATION. BONDS.2.2B = -DB.(2.13)Using Product Rl to differentiate (2.13) with respect to y , we nd that :-uu B B-2 B D DB -D-hu D B - BD n yB no-rC 1 1 1 1 / ' 'We conclude thatC12BB2.2. SUPPLEMENTAL EXERCISES2D2_ ~D. u057Supplemental Exercises1. Assume that the continuously compounded instantaneous rate curve (t) is given by (t) = 0.05 exp( (1+ t)2) .(i) Use Simpsor Rule to compute the 1-year and 2-year discount factors with six decimal digits accuracy, and compute the 3-year discount factor with eight decimal digits accuracy. (ii) Find the value of a three year yearly coupon bond with coupon rate 5% (and face value 100). 2. Consider a six months plain vanilla European put option with strike 50 on a lognormally distributed underlying asset paying dividends continuously at 2%. Assume that interest rates are constant at 4%. Use risk-neutral valuation to write the value of the put as an integral over a finite interval. Find the value of the put option with six decimal digits accuracy using the N dpoint Rule and using Simpson's Rule. Ii Also , compute the lack-Scholes value PBS of the put and report the approximation errors of the numerical integration approximations at each step. 3. The prices of three call options with strikes 45 , 50 , and 55 , on the same underlying asset and with the same maturity, are $4 , $6 , and $9 , respectively. Create a butterfly spread by going long a 45-call and a 55-call , and shorting two 50-calls. What are the payoff and the P &L at maturity of the butterfly spread? When would the butterfly spread be profitable? Assume , for simplicity, that interest rates are zero. 4. Dollar duration is defined asD= u and measures by how much the value of a bond portfolio changes for a small parallel shift in the yield curve. Similarly, dollar convexity is defined asC 2Bey2 33. 58CHAPTER 2. NUMERICAL INTEGRATION. BONDS.Note that , unlike classical duration and convexity, which can only be computed for individual bonds , dollar duration and dollar convexity can be estimated for any bond portfolio , assuming all bond yields change by the same amount. In particular , for a bond with value B , duration D , and convexity C , the dollar duration and the dollar convexity can be computed as D$ = BD and G$ = BG. You invest $1 million in a bond with duration 3.2 and convexity 16 and $2.5 million in a bond with duration 4 and convexity 24. (i) What are the dollardationand dollar convexity of your portfolio?(ii) If the yield goes up by ten basis points new approximate values for each of the bonds. What is the new value of the portfolio? (iii) You can buy or sell two other bos one with dion 1. 6 and convexity 12 and another one with duration 3.2 and convexity 20. What positions could you take in these bonds to immunize your portfolio (i. e. , to obtain a portfolio with zero dollar duration and dollar coexity)?2.3. SOLUTIONS TO SUPPLEJYIENTAL EXERCISES59(ii) The value of the three year yearly coupon bond is B= 5 disc(l) + 5 disc(2) + 105 disc(3) =100.236424.Problem 2: Consider a six months plai vanilla European put option with strike 50 on a lognormally distributed underlying asset paying dividends co tinuouslyat 2%. Assume that interest rates are constant at 4%. Use risk-neutral valuation to write the value of the put as an integral over a nite interval. Find the value of the put option with six decimal digits accuracy using the Midpoint Rule and using Simpson's Rule. Also , compute the Black-Scholes value PBS of the put and report the approximation errors of the numerical integration approximations at each step. Solution: If the underlying asset follows a lognormal distribution , the value S(T) of the underlying asset at maturity is a lognormal variable given by In(S(T))= In(S(O)) +/2 ( - q - v ) T + ITZ2where is the volatility of the underlying asset. Then , the probability density function h() of S(T) is2.3/Solutions to Supplemental Exercisesh(y)Problem 1: Assume that the continuously compounded instantaneous rate curve r(t) is given by (t) 0.05xp( (1+ t)2) .Solution: (i) Recall that the discount factor corresponding to time t is-fa' Using Simpson's Rule , we obtain that the 1-year, 2-year , and 3-year discount factors are= 0.956595;disc(2)= 0.910128;disc(3)~!= 0.86574100.LL "(2.14)if y > 0 , and h(y) = 0 if y O. Using risk-neutral valuation , we find that the value of the put is given b P -Use Simpson's Rule to compute the 1-year and 2-year discount factors with six decimal digits accuracy, and compute the 3-year discount factor with eight decimal digits accuracy. (ii) Find the value of a three year yearly coupon bond with coupon rate 5% (and face value 100).disc(l)=(l(0)) (r - q ) T)2 ie-rTER max(K - S(T) , 0)] I' K= e- rT / (K - y)h() dy ,(2.15)JQwhere h(y) is given by (2.14). The Black-Scholes value of the put is PBS = 4.863603. To compute a numerical appr()ximation of the integral (2.15) , we start with a partition of the interval [0 , K] into 4 intervals , and double the numbers of -intervals up to 8192 intervals. We report the Midpoint Rule and Simpson's Rule appronations to (2.15) and the correspondi approximation errors to the Black-Scholes value PBS in the table below: We rst note that the approximation error does not go below 6 . 10-6 . This is due to the fact that the Black-Scholes value of the put , which is given by PBS Ke-r(T-t) N( -d2 ) - Se-q(T-t) N( -d1) , 34. 60CHAPTER 2. NUIVIERICAL INTEGRATION. BONDS. No. Intervals Midpoint Rule 4 5.075832 8 4.922961 16 4.878220 32 4.867248 64 4.864518 128 4.863837 256 4.863666 512 4.863624 1024 4.863613 2048 4.863610 4096 4.863610 8192 4.863610Error Simpson's Rule 0.212228 4.855908 0.059357 4.863955 0.014616 4.863631 0.003644 4.863611 0.000914 4.863610 0.000233 4.863610 0.000020 4.863609 0.000009 4.863609 0.000006 4.863609 0.000006 4.863609 0.000006 4.863609 0.000006 4.863609Error 0.007696 0.000351 0.000027 0.000007 0.000006 0.000006 0.000006 0.000006 0.000006 0.000006 0.000006 0.000006is computed usi numerical approximations to estimate the terms N( -d1 ) and N( -d2 ) The approximation error of these approximations is on the order of 10- f Using numerical integration , the real value of the put option is computed , but the error of the Black-Scholes value will propagate to the approximation errors of the numerical integration. If we consider that convergence is achieved when the error is less than 10-5 , then convergence is achieved for 512 intervals for the Midpoint Rule and for 32 intervals for Simpson's Rule. This was to be expected given the quadratic convergence of the Midpoint Rule and the fourth order convergence of Simpson's Rule. Problem 3: The prices of three call options with strikes 45 , 50 , and 55 , 0 the same underlying asset and with the same maturity, are $4 , $6 , and $9 , respectively. Create a butterfly spread by going long a 45-call and a 55-call , and shorting two 50-calls. What are the payoff and the P&L at maturity of the butterfly spread? When would the butterfly spread be profitable? Assume , for simplicity, that interest rates are zero.Solution: The payoff V(T) of the butterfly spread at maturity is if S(T) 45; 45 , if 45 < S(T) 50; V(T)=i55-S(TL if50 0 F(x) = 0770Nis (iv) Show thatM t)Sol (i)E[(8T)2] var(8T )80u3 . p3 380 2d.p2(1_p)+ 380 2 . p(l _ p)2 80 d3 . (1 _ p)3 4 1. 7036; OU 3 )2 . p3 + 3(80u 2d)2. p2(1 - p) + 3(80ud2)2. p(l _ p)2 (80 d 3 )2 . (1 - p)3 = 1749.0762; E[(8T )2] (E[8T])2 9.8835. Problem 4: The density function of the exponential random variable X with parameter> 0 is J a e zif Z >f(x)=f z=Jt Z (-rz)lZ(1 - eZ1=(ii) By integration by parts we find thatJxez1xe x+r xUw~ e xxe --~_I e-ldx2r'+~ xe dxf Zlx2 ex2xe x2e z23T 'ME n W -P I - 1 10;)77if Z f x = Vt OIt is easy to see that fWe conclude thatE[8T]=Note: this result is used to show that the exponential variable is memoryless , i.e. , P(X t s I X t) = P(X s ).Note thatP(UUU) = p3; P(DDD) = (1 _ p)3; P(UUD) = P(UDU) = P(DUU) = p2(1_ p); P(UDD) = P(DUD) = P(DDU) = p(l- p?1(ii) Show tl the expected value and the variance of the expo rando'm variable X are E[X] = and var(X) = (iii) Show that the cumulative desity of X is8 = {UUU, UUD , UDU, UDD , DUU, DUD , DDU, DDD} , where U represents an "up" move and D represents a "down" move. The value 8T of the stock at time T is a random variable deed on 8 , and is given by 8T(UUU) S ; 8 T (DDD) = 8 0d3; 8 T (UUD) = 8 T (UDU) = 8T(DUU) = 8 0u 2d; 8 T (UDD) = 8 T (DUD) = 8 T (DDU) = 8 0ud2.673.1. SOLUTIONS TO CHAPTER 3 EXERCISES k z n / rJ l z i/ I t L z-e J J o d z--iii etnu = 1 1 1 /M / I l l h t z-dzfl04-g pud1-ded-/lil 38. 68CHAPTER 3. PROBABILITY. BLACK-SCHOLES FORJYIULA.E[X 2]LW)dz=due- dx Etjtpzdz3.1. SOLUTIONS TO CHAPTER 3 EXERCISES for any two continuous functionsSolution: Let _ 2~3ax) I:JRbe an arbitrary real number. Note that l21 g2(x)dx + )g 1 2 ii ( i) 1f x < 0 , then F(x) = J~ f(s) ds = O. 1f x 0 the(iv)1f t 0 =fa" f2= (-e- s) I~1-eXl 9" +bf(Z)g(Z)dz+ff2 0 if and only if , /lilf 1- P(X < t) = 1 ll' f d ( Z) = 1 (-e )VRbthen P(X t)V JR.Recall that a quadratic polynomial P(x) = Ax 2 Bx + C is nonnegative for all real values of x if and only if P(x) has at most one real double root , which happens if and only if B 2 - 4AC O. For our problem , it follows thatTherefore E[X [ ] ( Xbb~ 0,:sJR.b( + Uf ( Z )) J( -25ff-23+ )var ar( X = )f , g : JR 69t t t I rIdz/ny Z ' 4 ,Z G Ill-/ "LU /Il--F 1 1 A which is equivalent toI~P I ' ' 1 (3.6) rId "J'i rId Z '3' 15Z,2 G O. Compute E[ IXI ] and E[X 2].f(P) = 0.3;meanand5. Compute the expected value and variance of the Poisson distribution , i.e. , of a random variable X taking only positive integer values with probabilities3.3Solutions to Supplemental ExercisesProblem 1: What is the expected number of coin tosses of a fair coin in order to get two heads in a row? What if the coin is biased and the probability of getting heads is p?n kP(X=k) = V k 0 where>0 is a fixed positive number.6. Show that the values of a plain vanilla put option and of a plai vanilla call option with the same maturity and strike , and on the same underlying asset , are equal if and only if the strike is equal to the forward pnce. 7. You hold a portfolio made of a long position in 1000 put options with strike price 25 and maturity of six months , on a non-dividend-paying stock with lognormal distribution with volatility 30% , a long position in 400 shares of the same stock , which has spot price $20 , and $10 ,000 cash. Assume that the riskfree rate is constant at 4%. (i) How much is the portfolio worth? (ii) How do you adjust the stock position to make the portfolio Deltaneutral? (iii) A month later , the spot price of the underlying asset is $24. Wr is new value of your portfolio , and how do you adjust the stock position to make the portfolio Delta-neutral?Solution: If p is the probability of the coin toss resulting in heads , then the probability of the coin toss resulting in tails is 1 - p. The outcomes of the first two tosses are as follows: If the first toss is tails , which happens with probability 1 - p , then the process resets and the expected number of tosses increases by 1. If the st toss is heds and if the second toss is also heads , which happens with probability p then two consecutive heads were obtained after two tosses. If the first toss is heads?and if the second toss is tails?which happens with probability p(l - p) , then the process resets and the expected number of tosses increases by 2.If E[X] denotes the expected number of tosses in order to get two heads in a row , we conclude that E[X]= (1- p)(l + E[X]) 2p 2 p(lp)(2 + E[X]).(3.20)We solve (3.20) for E[X] and obtain that 1 E[X] = . p"For an u~biase_d coir:, i.e. , for p = ~ we find that E[X] = 6, and therefore the expected number of coin tosses to- obtain two heads i~ a row is 6. 46. CHAPTER 3. PROBABILITY. BLACK-SCHOLES FORIvIULA.84Problem 2: What is the expected number of tosses in order to get n heads in a row for a biased coin with probability of getting heads equal to p?3.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES var(U)= E [eu - E[U])2] = E [ (U jSolution: The probability that the first n throws are all heads is pn. If the k throws are heads and the (k + l)-th throw is tails , which happen wit h probability pk( p) then the process resets after the k 1 steps; here , l k = 0: (-1). Then , if x() denotes the expected number of tosses in order to get heads in a row , it follows that f(z-132)dz= ~(X_b;a)' rstZ n l PP 4 F k H V+kh Td o kkZK V -iJSSttPUA 4 1 1 d A 4P ARecall that Z Mp1- pn l-p7/' MVh LI rl l -P A+z nn 1 -PA J ' Kp-2 kpk/ F E 'hhUAp n (1 - p)21-p1 ( l)pn npn+11-p- x()(11-pand thereforeProblem 4: Let X be a normally distributed random variable with and standard deviation > O. Compute E[ IXI ] and E[X 2 ]. Solution: We compute E[+x(n)(l _ pn)vi l+ Ie dzE[IX ~ (/ ( ) e+L v IIV L/JrJ-xfJ /( + o- z) edz-tl:Vdz zlJNJdz1-p() =-I+ =e Tdz 2i J /-.ft7+ ,r.::= 1~ j/ TdIt is easy to see that1-duProblem 3: Calculate the mean and variance of the uniform distribution on the interval [].JUJul N(-~) = 1-N(~);(-e) I f= ); [~" e-'; dy N (;) ;f Solution: The probability density function of the uniform distribution U on the interval [a , b] is the cstant functi j(x) = b~a' for all x E [a , b]. Then ,rb ~" 1 rb b+ E[U] = I xj(x) dx = ~ I xdx = j b j 2IXI ] in terms of the cumulative distutionof the standard normal variable Z. Note that X = + Z. Then ,_ pn) ,pn (1- p)' We conclude that the expected number of tosses in order to get n headsin a row for a biased coin with p ability of gettileads al to pis 57 If the i were unbia i.e. , for p = ~ the expected number of tosses in order to get n heads in a row is 2+1 - 2. meanN(t) vi L znpn ( - l)pn1Then , we find that x()85(-e-~) J:/= ) 47. CHAPTER 3. PROBABILITY. BLACK-SCHOLES FORJVIULA.86873.3. SOLUTIONS TO SUPPLEMENTAL EXERCISESThen , it follows thatWe conclude that (l-N(;)) E[X] );) +) (2N (;) -1)+One way to compute E[X 2] would be to compute the E[X 2] =J-OOWhile this would provide the correct result , an easier way is to recall thatSince E[X] = and var(X)=2 e(3.22)e(3.23)From (3.21) and (3.22) , we nd that E[X] = Similarly, cE[X 2]= P(Xe ..=k) k2 =(k - 1)k(k-1)! x2 var(X) = E[X ] (E[X]?-ZJ-:-; -followi integral:E[(+ Z)2] = / ( M)2Jdz V' 1 and d < 1.ER( S(tt) ) J 8t S ( J = I / Recall tl if ln(Y) = Z is a lognormal random variable with parameters and the expected value and variance of Yaren(4.4)Solution 1: If the price S (t) of the non-dividend payi asset has lognormal distrib 0 ith drift nd at il r lity given by2 n(4.3)e(2r+2)8t S2(t).+ c5t)]ERN [S2(t c5t)]:Xi]932r8t=eTSt;(4.5)(2 (_ ~2) +)2) .(e()2- 1J4 . 6 ) ( -1)(4.7)Note that ,E[tft)] mr(tf)) E[S( 8t)] S(t)(4.8)-L-Var (S(t+60)-(4.9)S2(t)From (4.5) and (4.8) , and from (4.7) and (4.9) , respectively, we conclude that E[S(t 8t)]var (S(t + 8t))er8t S( t);e-1) S2(t)(4.10) 51. CHAPTER944.LOGNOR1VIAL VARIABLES.RNPRICING.4.1. SOLUTIONS TO CHAPTER 4 EXERCISESNote that var (S(t )) = E[S2(t=S2(t) =+ 8t)] (E[S(t )])2E[S2(t 6t)]V L-7r- e2r6ts2(t).(4.11)f(.L J- _2 exp 12 2 6t -J-xinI. 2) 6tFrom (4.10) and (4.11) it follows that95.2)6t / e-4dse 2r6t 26t S2(t) ,V L,7r J-x S2(t)e(2r+2)6t;which is what we wanted to show.Solution 2: Note that S(t ) can be written as a function of the standardthe substituti S -2v5t was used abovenormal variable Z as follows: IS(t )=I2 Problem 5: The results of the previous two exercises can be used to calibrate a binomial tree model to a lognormally distributed process. This means nding the up and down factors u and d , and the risk-neutral probability p (of goi up) such that the values of ERN[S(t 6t)] and ERN[S2(t+6t)] given by (4.1) and (4.2) coincide with the values (4.3) and (4 .4) for the lognormal model. In other words , we are looking for u , d , and p such thatrS(t) ~ ~ r ) 6t tZ)Then ,E[S(t + 6t)] f:S (t) i t m 6 +o-V5t t dx x;) 1pu (1 - p)dpu2r/ (x Vti? JS(t) Ot p +(1 _ p)d2 =viz -(e r6t - d) (u - er6t )1,ud -E[S2]= (00 S2(2 (_ ~2) 6t + 2vti x) e-~ I L,7rJ IS2(t) exp t 2r6t -Inm h a(4.14) / t t A UaTb t t IhJ(4.15)dx/ r-x-o-:.:I 6t + 2 yot x - 2) dxJ /1.Show that the solution can be written asSimiLr1377mobtaizltbMI(4.13)(ii) Derive the CoxRoSS s-Rubinsteir1 parametri zation for a binomial tree , by solving (4.14-4.15) with the additional condition thatsince eJ is the density function of the standard normal variable.= pwhere we used the substitution y = x vti and the fact thatfe(2r2)6tMl-ub t e-ed-r "S(t)eT5"t? t1(4.12)Since there are two constraints and three unknowns , the solution will not be umque. (i) Show that (4.12-4.13) are equivale to- 2-') )T )erI L,7r= er6t ;p= f; = u whereA =A+YA2; d=A-VA7~ ( e(r+e>2)6t).(4.16) 52. CHAPTER 4. LOGNORIvIAL VARIABLES. RN PRICING.96Solution: (i) Formula (4.14) can be obtained by solving the linear equation (4.12) for p. To obtain (4.15) , we rst square formula (4.12) to obtain4.1. SOLUTIONS TO CHAPTER 4 EXERCISESthe other solution of the quadratic equation (4.20) corresponds to the value of d, sinceL =A -1 A2 _ 1d = u 1e2r8tp2 U 2 + 2p(1 - p)ud + (1- p)2d2 -97A !v -and subtract this from (4.13). We d thatp(l - p)u 2 - 2p(1 - p)ud p(l - p)d2Prabl ill 6: Show that the series 1;l is conve while the series 2:%1i and E;2 are diverge i.e. equal to (2r2)8t _ e2r 8tNote: It is known thatwhich can be written asp(l - p)(u - d)2 = e 2r (eo-28t1)._Using formula (4.14) for p , it is easy to see thatp(l - p)=and( e r 8t (u-e r 8t ) (u - d)2(e r8td) (u - e-2 ) = e2r 8t (e o- 8t-1r26( ;-h())(4.18) whereFrom (4.17) and (4.18) , we conclude that r(4.17) 0.57721=~is called Euler's constant.Solution: Since all the terms of the series 2:%l are positive , it is enoh to show that the partial sums ~ 11)(ii) Bymtiplyi outue r8t(4.15) andusi thefact that ud = 1, we obtain that1 _ e 2r8t de r 8t = e(2r2) 8t-_ e 2( 4.19)After canceling out the term _e 2r8t , we divide (4.19) by e and obtain Ud- e- r8t- k2e(r 7are uniformly bounded , in order to conclude that the series is convergent. This can be seen as follows: 1 L-41 11 4 iwhich can be written asU 1 ( e- r8t + e(r2) z u/U++j K / / l i l 1+ (kll) 1 -il l i< /To show that the series 1;1+1-2A=0; U2,V n ~ 2.t is diverge we will prove that44d. (4.16) for the denition of A. In other words , u is a solution of the quadratic equationu 2 - 2Au 1 = 0, which has two solutions , conclude thatA + VAlandA-VA1. u=A JI;i() 1 , we)~2; -L 1. If x= -1. the series becomes ~ i:)!llce 'L ne 'L erms have S ~k=l kln(k)' Since the terms kln(k) alternating signs and decrease in absolute value to 0 , the series is covergent. If x = 1, the series becomes '2;1 kl~(k)' which was show to be divergent in Problem 6 of this chapter. R1Problem 8: Consider a put option with strike 55 and maturity 4 months on a non-dividend paying asset with spot price 60 which follows a lognormal 54. CHAPTER 4. LOGNORMAL VARIABLES. RN PRICING.100model with drift = 0.1 and volatility = 0.3. Assume that the risk-free rate is constant equal to 0.05. (i) Find the probability that the put will expire i the money (ii) Find the riskral probability that the put will expire in the money. (iii) Compute N( -d 2 ). Solution: (i) The probability that the put option will expire in the money is equal to the probability that the spot price at maturity is lower than the strike price , i.e. , to P(S(T) < K). Recall thatn/111 T 111/ 2 2/JIll 'qIll-/T 4.1. SOLUTIONS TO CHAPTER 4 EXERCISES The d 2101= 0.511983 , and N( -d 2 ) = 0.304331 ,which is the same as the risk-neutral probability that the put option will expire in the money; cf. (4.31) To understand this result , note that PRN(S(T)< K)=P(z< +(-q-4) 1 VTd z d. (4.31) and (4.32).N(-d 2 );=JThen , P(5(T)< K)P(;2< )=P(:2)< ))P( (-q-DT+dZ< )) )Problem 9: (i) Consider an ate-money call on a non-dividend paying asset; assume the Black-Scholes framewor k. Show that the Delta of the option is always greater than 0.5. (ii) If the underlyi asset pays dividends at the continuous rate q , when is the Delta of an at-the-money call1ess than 0.5? Note: For most cases , the Delta of an at-the-money call option is close to 0.5. Sol on:(i) Recall that the Delta of a call option is given by-qT =N( +( -q+ )T} 1For 5 = 60 , K = 55 , T = 1/3 = 0.1 , q = 0 = 0.3 , and 7' = 0.05 , we obtain that the probability that the put will expire in the money is 0.271525 , i.e. , 27.1525%. (ii) The risk-neutral probability that the put option will expire in the money is obtained just like the probability that the put expires i the money, by substituting the risk-free rate 7' for i.e. ,PRN (5(T) < K)P(z)- ( -q )1VT1For an at-th e-money call on a non-dividend paying asset , i.e. , for K = S and q = 0 , we find that=N ((r j) )(ii) If the underlyi asset pays dividends at the c or of ar1 ATM call iSJ (4.31)TN(d 1 )=e -q+ (iii) Recall that d2 (7' - q )T (4.32)For a fixed risk-free rateand xed maturity T , we conclude that ~(C) < 0.5 if and only if the dividend yield q and the volatility of the underlying asset 55. CHAPTER 4. LOGNORMAL VARIABLES. RN PRICING.102 satisT4.1. SOLUTIONS TO CHAPTER 4 EXERCISESWhen pricing a plain vanilla call using ris k-neutrality, we proved that n-rT r II _29CBS(O) = 8(0) = exp ( ( ) T+ dz-4:}dzthe following condition:V 7ir Ke e-T This happens , for example , if r = q and T is large enough , sice lim N (T00(jV! 1 ~1031andJlimT00IJ -d 2~L, )) /Lef ; d S (0 )N( d 1) Ke- T TN( ). d 2 0.5 =In other words , we showed that/ Adz=e (d2 );(4.34)V L,7i J -d2Problem 10: Use risk-neutral pricing to price a supershare , i.e. , an option that pays (max(8(T) - K , 0))2 at the maturity of the option. In other words , compute V(O) = e- rT E RN [(max(8(T) - K , 0))2 ], where the expected value is computed with respect to the risk-neutral distribution of the price 8(T) of the underlyi asset at maturity T , which is assumed to follow a lognormal process with drift r and volatility . Assume that the underlying asset pays no dividends , i.e. , q = O.Solution: Recall thatT C -cu nu exDA/ft-- /III- 'qr-2 T 111 d z/5 (r- ~2) T dz-2)dz= d1 )From (4.33) , (4.34) , and (4.35) , we conclude that V(O) = K 2e- rT N(d 2) - 2K8(0)N(d1 ) rT ? r / +8 2(0) (2 2 VTx - ~- dx. (4.36) )~ J-d2 The integral from (4.36) is computed by co pleti the square as follows:) 1 1 1 /~-rT eV/c/ r/( (2 - (j 2)T + 2dz-Z}dzL,7i J -d2L, )-rT(x r O)e~ / p ( ( L,7f v and note that- 2 VT)2z>) ( )ITTz V L,7i-d 28 2 (0) 2)T /2VT? } I( dy J LS ( e 0) ff7T'1 TTV K2 e- l; i 2 P( ( T+2 l: ((T2) Z x;) dxT (we used the substitution y = x - 2 -IT and the fact thatZ X Vi l:eS 0 d2( + (jL!"l7i J (d 2 +2 -IT)dxI 0) )T /{ (x 2 JThen ,m (2+ 2)T 1 L,J8(T) K (4.35)'rorm Fr (4.33)(4.36) and (4.37) , we conclude thatV(O -rT N(d2) -2K8(0)N(d1 )+ 8 2 (0) )TN(d2 2 ) . 56. CHAPTER 4. LOGNORMAL VARIABLES. RN PRICING.104Problem 11: If the price of an set follows a normal process , i.e. , d8 = dt+ dX then8(t2)8(tl) (t2 - t 1 ) + d Z V 0 < tl < t2=Assume that the risk free rate is constant and equal to r. (i) Use risk neutrality to nd the value of a call option with strike K and maturity T. (ii) Use the Put-Call parity to find the y~~ue of a put ~ption with s~rike K ~n'd maturity T , if the underlying asset follows a normal process as above. Solon: (i) Using risk-neutral prici it follows thatC(O) = e- rT ERmax(8(T) - K , 0)] ,= 8(0) +T VTZ.(4.38)> K iff Z>K - 8(0) - rT 1rJd-Ke- rT Jd e-hz y '21r f+d l tllatf f d d f I l d= e- rT ERN [8(T) - K] = e- rT (8(0)zc dz mJ!! 1-vi f: e = tm=it(-e I:d2e 2;1 N(d) = N(-d)where N(t) is the cumulative distribution of the standard normal variable. We conclude that rT rT 0(0) (8(0) + rT)e- N(-d) - K e- N(-d) + me- rT IT ~.L e .v+ T-K)since ER From (.39 and ( .4) we obtain that 4 ) 4 0 -rT vT e +J /2 since 1 - N( -d) Vi [(( O ( L l dx rTT O) ) T -r e- rNote0(0) - P(O)d2 ;[- dThen , 0(0)(ii) Regardless of the model used for describi the evolution of the price of the underlying asset , the Put-Call parity says that a portfolio made of a long position in a plain vanilla European call option and a short position in a plain vanilla European put option on the same asset and with the same strike and maturity as the call option has the same payoff at maturity as a long position in a unit of the underlying asset and a short cash position equal to the strike of the options. Using risk-neutral pricing , this can be written asKe-rTN(d) (8(0)Note that 8(T)105P(O) = 0(0) (8(0) T)e- rT Ke- rT K e -rT (1 - N ( -d)) (8(0) rT) -rT(l - N( -d))where the expected value is computed with respect to 8(T) given by 8(T)4.1. SOLUTIONS TO CHAPTER 4 EXERCISES(4.39)= N(d).+ rT)e- rT N(d)~-rTb_./ V .L e(4 .40) 57. CHAPTER 4. LOGNORMAL VARIABLES. RN PRICING.1064.24.3Supplemental Exercises=Solutions to Supplemental Exercises(- I()Xnis convergent to a limit between 0 and 1. Note: The limit of this sequence is 0.57721 the Euler's constant. 2. Assume that an asset with spot price 50 paying dividends continuously at rate q = 0.02 has lognormal distribution with mean = 0.08 and volatility = 0.3. Assume that the risk-free rates are constant and equal to r = 0.05.(ii) Find 95% and 99% risk-neutral condence intervals for the spot price of the asset in 15 days , 1 month , 2 months , 6 months , and 1 year , i. e. , assuming that the drift of the asset is equal to the risk-free rate. 3. If you play (American) 1 roulette 100 times , betting $100 on black each time , what is the probability of winning at least $1000 , and what is the probability of losing at least $10007 4. Use risk-neutral pricing to find the value of an option on a dividend-paying asset with lognormal distribution if the payoff of the option at maturity is equal to max((S(T))- K , O). Here > 0 is a fixed constant. 5. Find a binomial tree parame~rization for a risk-neutral probability (of going up or dow equal to ~. In other words , find the up and down factors u and d such that+(l-p)dpu2 (1- p)d2 u U=(- ln()is convergent to a limit between 0 and 1. Solution: Recall from (4.21) thatln()+j42bIt is easy to see thatXn+1 :-L-ln(+ l Therefore , X n 1 1000) =[ 100006000 /101 PI v Z> 1000 I 1919.-JP(Z > 1. 5284) = 0.0632.The probability of losing at least $1000 can be approximated as follows: -1000)[ 100006000 /101 PI v Z < -1000 I=P(vV

0 is a xed constant. 59. CHAPTER 4. LOGNORMAL VARIABLES. RN PRICING.110Solution: Using risk-neutral pricing , we find that the value of the option isV(O) = e- rT ERN[max((S(T)) - K , O) ], where)=4.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES note that the change of variables y = x JT was used above. We conclude thatv nu - s nuS(O (-DT VTZ)//ll1 ex D//I11 t 4Note that (S(T)) K is equivalent to S(T) K /. Using (4 .42) , we fin::::::}Z>d-a.The7 V(O)= 'where1S(T) K 1 /In 2lliI/ 111l/T N ( )+(-4)Tdku 'TN JT~~oblem 5: Find a binomia) tree parametrization for a risk-neutral proba r bility (of going up) equal to ~. In other word s 1 aI tL ld d such that ;/~((0)) ( S(+Recall from (.34) 4 t hat 4 'v ?p(1 - p)d 2 pu (1 - p)d2 pu5:e = e-rTN)111u" u7 ' 4) U ifp=i SolIt iseasy to see that , if p = ~ then U d =2eTdt;Therefore ,) =~1~ exp ((- 1) ~dz-Zhz -u2+ d2=2e(2r 2)dt7and thereforeKe-rTN()ud( d)27(u2+d2)- 2e2rM 2)MNote that u and d are the solutions of Z2 ( d)z ud = 0 , which is the same as Z2 _ 2erbt z 2e 2rM _ e(2r 2)M _ 0. (4 .43) Vi 1~ex((( p(We solve (4 .4 3) and find that p( ) 1)() ( ) 1:+Qd (-5) d 1 T) VIi ) ( +) T) 1: dy OUVT (r )T) N(a+QVT);ud since d 1(5.5) 62. CHAPTER 5. TAYLOR'S FORMULA. TAYLOR SERIES.116(x - t) n(-1 J I{X ~ (1 )7 n! d -/ . Jo n! t) {X ( -1)2(X - t)n A Jo (1 + t) (-x)nlln(l(5.6)!- < 7 0y ) .Note that pz1017D =157and C =25i (5.23) and (5.24). The following approximate values are obtained for b:.. y {0.001 0.005 , 0.01 , 0.02}: Bne DG 0.0010 100.8485 1B00.84D8 6 0.0001% 0.0050 100.2425 100.2457 0.0031% 0.01 99 .4850 99 .4976 0.0127% 0.02 97.9700 98.0205 0.0515% b:.. yBne D G - Bne D B ne DIPBS pproxl [ PBSr=Oq=b:.. B E - Db:.. y123The last column of the table represents the percent difference between the approximate value miIlg d11ratioa done?and ttle approximate value usiIIg both duration and convexity, i.e. ,Here , the Approximation Error is the relative approximation error defined asb:.. B Db:..y 5.1. SOLUTIONS TO CHAPTER 5 EXERCISES(5.24) 66. CHAPTER1245.2TAYLOR'S FORMULA. TAYLOR SERIES.5.Supplemental Exercises5.2. SUPPLEMENTAL EXERCISES 6. The goal of this exercise is to compute11In(1- x) In1. (i) Let g(x) be an initely differentiable function. Find the linear and q~adrati~ Taylor approximations of eg(x) around the point O.(ii) Use the result ~bove to compute the quadratic Taylor approximation around 0 of e(x+I)2 (iii) Compute the quadratic Taylor approxilation around 0 of by using Taylor approximations of eX and eX. 2. Show thate- X -L=0(as xl Ze(x+I)2(i) Show that :( - x) ln( ))l'PAVAOve/111 4i- =Z41&/ha4L0,(iii) Prove that4. Recall that z 111J/' 1-z= x/I ( - x) ln(x)) = lim-l1 1 1h ( ln ( Z around the point 0, and find its radius of convergence.(5.26)(ii) Use the Taylor series expansion of 1 1- x) for jxj < 1 to show that o;~) t Adxand conclude that the integral (5.26) can be regarded as a definite integral.3. Compute the Taylor series expansion of/JIl--1254EUto obtain that/Iit 4ti1111/ 1-z+ l-2 -b/fiI 1-z+ l-2< U< E Ll i f /Vv Z >- ln(x 5 45. (i) Find the radius of convergence of the series ~41 L2! 8 'L4!~12+. '6!(5.25)(ii) Show that the series from (5.25) is the Taylor series expansion of the function X2 ez2 e2 "2 67. Consider an ATM put option with strike 40 on an asset with volatility 30% and paying 2% dividends continuously. Assume that the interest rates are constant at 4.5%. Compute the relative approximation error to the Black-Scholes value of the option of the approximate value.IT ( 1 P oX r O q o - BV;1T (7' q)T (7' -FJ-if the put option expires in 1 , 3, 5, 10 , and 20 years.q)T 67. CHAPTER 5. TAYLOR'S FORMULA. TAYLOR SERIES.1265.35.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES Problem 2: Show thatSolutions to Supplemental ExercisesProblem 1: (i) Let g(x) be an infinitely differentiable function. Find the linear and quadratic Taylor approximations of eg(x) around the point O. (ii) Use the result above to compute the quadratic Taylor approximation around 0 of e(x+1? (iii) Compute the quadratic Taylor pproximation around 0 of e(x )2 by using Taylor approximations of eX and eX*.e-x -L=0(z2)7aS 1 Z= g'(x)eg(x)1 Z+ (g' (x))2)e g(X).+O(x 2) ,asxO.e- xj(O) _,x 2 _"+ x j' (O) f) +JO(x ) ,xas0,1 l+xasx 0; xareO.as2) xO. O(x3 ) In e g (Ix1 (L (2x)2 3 1 + 2x + O(x ) ,0,IxJ;52)=lO.=I0Solution: Note that the functionquadratic Taylor approximations(2X)2 " f ~3 J e""x = 1 + 2x + -; + O(x ) , x2 e _ 1 + x 2 O(x as x-taround the point 0 , and find its radius of convergence.as x O. (ii) By letting g(x) = (x I? in (5.27) , we find that asx(1 + 1-(5.27)e( )2 e + 2ex + 3ex 2 + 0 ()asProblem 3: Compute the Taylor series expansion ofbecomesUsi thel~Xas 3) l+xThe quadratic Taylor approximationeg(x) =0Note that we implicitly proved thateg(x) = eg(O) xe O)g'(O)(iii)1- x + x 2e- x -L= O()= j(O) x f' (O) + O(x 2) , as x 0,can be written asj(x)Therefore ,The linear Taylor approximation j(x)I-x m 1j"(x) = (g"(x)andZSolution: The qu atic Taylor approximations of e- x andSolution: (i) Let j(x) = eg(x). Thenf' (x)127asx0;is not defined for x = -lor x = 1. Therefore , the largest possible radius of convergence of its Taylor series expansion around 0 is 1. The Taylor series expansions of the functions In(l x) and In(l - x) are In (1+ x)it follows that e(x1)2=e -dz-et=e(l 2x 2x 2 )(1 x 2 ) O(x3 ) e 2ex + 3ex 2 + O(x 3 ) , as x O. In (1 - x)fz2z3z4(-OMJZZ +... k 2 '3 4 x kx2x3x4, V (-1 1];: = -x ... , V x [-1 1) fk 234 68. CHAPTER128TAYLOR'S FORMULA. TAYLOR SERIES.5.5.3. SOLUTIONS TO SUPPLEl1ENTAL EXERCISES Fr (5.31) we also find thatand have radius of convergence equal to 1. We conclude that the Taylor s~ries expansion of In (i~~) is/IIl- 1Z ;~)In(1 + x) l 1 - x)xk=, I.,12+liI/ /II11n-EAq G414+liI/ 1-z=1 271I xE ) X+ jZRecall thatSolon:113(2z+1)21-tF 1 ) (1+~(1 ~r l x?1(5.29)1+ 12x(x 1)'(]" 2 y 3 , 25 In ( ~) = 2y +... , I y E (-1 , 1). 1 - y) 3 ' 5(5.30)ll x?1(5.33)x> 1.(5.34)Using (5.34) , we nd from (5.33) that(1 ~rRecall from (5.28) that< e1e 0 , the Gamma of the call option is first increasing until it reaches a maximum point and then decreases. Also , show that (7.3) 31lr(S)=O and liIII Y( =SSolon:From (7.1)we thatr can be written as/ (d 1 (S)? r(s) = exp ( 2ln(S) J , ---'- / )(7.4)where d1 (S) is given by (7.2). Since r(S) > 0, it follows that the functions r(S) and ln(r(S)) have the same monotonicity intervals. Let 1 : (0 )1R given by 1(S) = ln(r(S))=2 -ln(S) -1) 85. CHAPTER 7. MULTIVARIABLE CALCULUS.1627.1. SOLUTIONS TO CHAPTER 7 EXERCISES163from (7 .4), we conclude thatThen ,d1 (S)1Sn1' (S)S1 ( (d 1 (S))2li~ exp ( -ln(S))li~ r(S) -3 v-~ (1+ )a uVL:7r 1 /= O.(7.5) Problem 2: Let D be the domain bounded by the x-axis , the y-axis , and the line x = 1. ComputeRecall from (7.2) that(7.7)fLEjdzdud1 (S)Solution: Note thatD = {(x , y)It is easy to see that d1 (S) is an increasing function of S and that gsdl(S)= ;lim )=(7.6)SFrom (7.5) we find that I(S) has one critical poi denoted by S* , with d1 (S*) = n. From (7.5) and (7.6) it follows that 1'(S) > 0 if 0 < S < S* and 1' (S) < 0 if S* < S. In other words , the function I(S) = ln(r(S)) is increasi when 0 < S < S* and is decreasing when S* < S. We conclude that r( S) is reaaslI1 aIso 1 C i when 0 < S < S* and decreasing when S* < S. We now compute liro r(S) and lims r(s). Note that lims d 1 (S) = . Therefore , Fm r(S) =SG"Fro (7.2) Ju1 smwas- -2 7 ((d1 (S))2sn-WM/I=22T(1 (S))22 2T/ (d 1 (S)?( ln(S)I/It is easy to see that (x , y)n D= {(8 , t)if and only if (8 , t)I0 S n where1 -8 t 8 }.The Jacobian of the change of variable (x , y) D (8 t) E dxdy=lzuzyl I ~V I lstt81j = O.d8dt -n is~d8dt 2I'ij1:(l:tdt)ds=01 ) 1 (S) Jf :jM=f(/:)fL7fjdzdul' U'-Y~ dyL((x - 2yln()) I~tu )l'Since Ii (exp ( (ln(S)?)) = 0, we obtain thatSOWe use the change of variables 8 = X y and t = x - y , which is equivalent to S t 8 - t x - -2-; y - -2-it follows that1: {(ln(S) - 1(K) +(+ ~2) T) s~o1}and therefore -ln(S)I)and using the fact that lims o ln(S)2-n c 1lim ? exp (S V 'l. l2I x O y O X y = 01- y+ ln(y)ds 86. CHAPTER 7. MULTIVARIABLE CALCULUS.164 sice limyo y2 In (y) = O.7. 1. SOLUTIONS TO CHAPTER 7 EXERCISES,,,(i) For a call option V(S T) = max(S - K O). From,u(x O) = Problem 3: Use the change of variables to polar coordinates to show that the area of a circle of radius R is 1f R2 , i.e. , prove that,= exp(x) max(S - K O)max(Ke X- K , O) -,(7.8) , we find that,Kexp(x) max(e X 1 0).,(7.8) , we that(ii) For a put option V(S T) = max(K - S O). From u(x O)fL)ldzdu=d2exp(x)V(S T) exp(x)-165 exp(x)V(S T)-exp(x)max(K,= exp(x) max(K - S O) Kexp(x) max(1 - eX , 0).- KeX , O) -Solution: We use the polar coordinates change of variables,,(x y) = (r cos e r sin e)with(r,e) it = [0,R] x [0 2) .Problem 5: Solve for(Recall that dxdy = rdedr. Then , dxdy flo:R =R27 which is equal to the area of a circle of radius,R.Solution: From1~ 2'Usi (7.9) 1~b=( -qI n 22 i '2 J This is the change ofvariables that reduces the Black-Scholes PDE for V(S , t) to the heat equation for u(x ,T). (i) Show that the boundary condition V(S , T) = max(S - K , O) for the European call option becomes the following boundary condition for u(x , T) at time T = 0: u(x , O) = K exp(x) max(eX - 1, 0). (ii) Show that the boundary condition V(S , T) = max(K - S , O) for the European put option becomesu(x , O) = Kexp(x)- 0; b+ (1-~) ~ z the rstequation , it is easy to see that(7.9), ,,b the following system of equations:Problem 4: Let V(S t) = exp( x - bT)U(X T) where L (S(T - t) 2 -q x = In I -:;:-; I T =' ~ = n - K J" 22andmax(l - eX , 0).b=we note that the second equation can bewritten ~a2 - a (1 2( -D ( -D'2G )+Z ( ~r 2( -D ( -32+ ( +;)2-2() ( )24 2Solution: Note that t = T if and only if T = O. Then ,V(S , T) =exp( x)u(x O).Here , x = In () which can also be written as S = K eX(7.8)Problem 6: Assume that the function V(S , I , t)satis esthe following PDE:V 2V VV 1 S 2S2 + rS - rV = O. ' 2 S2 S tI(7.10) 87. CHAPTER 7. MULTIVARIABLE CALCULUS.1667.1. SOLUTIONS TO CHAPTER 7 EXERCISES By substitutingConsider the following change of variables:V()=5H 7where=iR(7.11)in(7.10)oV V167it follows that1 R 2H _ (H S- S 5IH-R ~ 1 2Show that H (R , t) satisfies the followi PDE: H1 '>_'>2H J+2".K OBs(8) ,d1 =Note that j (8) is a continus function , but it is not differentiable at 8 = K. For 8 < K , the function j(8) is the value of a call with strike K , 1 a n trlerefore is irlcreasir19. For 8 > K , we find that 1 < N(d 1 )0(8) -1 =-N(-d1 )K. Show that this argument does not work for dividend-paying assets. In other words , prove that the Black-Scholes value of the European call is smaller than S - K for S large enough , if the underlying asset pays dividends continuously at the rate q > 0 (and regardless of how small q is).=x-~andY=?TuNote: The values of Asian options with continuously sampled geometric average satisfy the PDE (7.21).173dxdv _ IJ-|zz111I:~ ~~i1 I Istt81d8dt|- (-[dt 2VSt 202t0) 2fti2t dsdt Then ,J!nXdX[[~ H[ VS dS ) ([ dt)i(;1:)()C(K) = CBs(K (K)) where CBs(K imp(K)) = CBs(S , K , T imp(K) r q) represents the Black-Scholes value of a call option with strike K on an underlying asset following a lognormal model with volatility mp(K). Find an implicit differential equation satised by (K) i.e. , fid imp(K) Kasafun ct 10 of irr1 K). Problem 2: Which number is larger or 7r e ? Solution: We show that 7r e correspond to for the fuction u(x , r) defined as follows: V(8 , t) = exp( x - br)u(x , r) , wherex = In I function: [0 ) [0 )as 1111/ /III- Z z/III- (x) M 1" v(t) dt) ,Solution: Define the1"Problem 3: Let ) : [0 ) [0 ) be two continuous functions with positive values. Assume that there exists a constant M > such thatu(x) M +1757.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES7)) +b 7 ) ) (;) /V 'where the derivative is computed with respect to x. Using the Product ule we find that=x)(P ( +1"Problem 5: Assume that the function V(8 , I ,t) satises the PDEv(t) dt)(M + 1" u(t)((((- 1" V(t)dt))) (u M -1" u(t)v(- 1" V V 1 ()_()2V V In8~ 'l 8'l r8 t I ' 28 2SrVConsider the following change of variables: r iV(8 , I , t)V(t)dt) (7.26)Show that F(y , t)=F(y , t) , wheresatisesy T, Tb/lt-T2I2T2 1 1 ' / a-cu-1Ei-the following PDE: 2(T - t)2a F (r _ ~2) 1 t(7.27)0.22)TurF = 92. CHAPTER 7. MULTIVARIABLE CALCULUS.176Solution: Using chain rule , it is easy to see that7.3. SOLUTIONS TO SUPPLEMENTAL EXERCISESwhich is what we wanted toshow.177VFInSF ttT7VITVProblem 7: For the same maturity, options with different strikes are traded simultaneously. The goal of this problem is to compute the rate of change of the implied volatility a function of the strike of the options. In other words , assume that S , T , q and r are given , and let C(K) be the (know value of a call option with maturity T and strike K. Assume that options with all strikes K exist. Defiine implied volatility in K) the t he f 1 as unique solution to C(K) = CBs(K imp(K)) 1F1 T-t Fu?S S T2V?1 ( (T - t)2 2F_.S2S2 T-t F.T2y2Ty)Then , the PDE (7.27) for V(S , I , t) becomes the followi PDE for F(y , t):o=V V+1 ')~') VV1 S + L. SL. +S ' 2- .- 8S 2 SrVt I FInS F tTy__1FlnS~~-Timp(K)u 12 i( (T - t)