Solutions Manual - A Primer for the Mathematics of Financial Engineering

Contents Preface

IX

Acknowledgments

Xl

o.

1 1 11 11

Mathematical 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 83

4 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

CONTENTS

Vl1l

7

8

Taylor series. Solutions to Chapter 5 Exercises. . Supplemental Exercises. . . . Solutions to Supplemental Exercises.

吨E - 4

5.1 5.2 5.3 6

咱E 4 4 1 4 4 l 4

Taylor ￥ formula.

1i1 土门，中。，"

5

3346

Finite 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 孔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. . . .

Bibliography

by substitu161 . . . . . . . . . 161 . . . . . . . . . 171 . . . . . . . . . 173

Implied volatil179 . . . . . . . . . . . 179 . . . . . . . . . . . 197 . . . . . . . . . . . 198 203

Preface 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. 如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口 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 Backgrou日d 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 山tml QuantNetwork student forum web page: httr旷/ www.quantnet.org/forum/index 抖lp IX

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 也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 , 2008

Xl

Xll

ACKN01iVLEDGJVIENTS

Chapter 0 Mathematical preliminaries. 0.1

Solutions to Chapter 0 Exercises

Problem 1: Let f : IR • IR be an odd function. (i) Show that xf(x) is a丑 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 如何) = f(x :J) is an odd function. (iii) Let h : IR • IR be de主ned as h(x) 工 xtf(x J ) ， where i and j are positive integers. When is h(x) an odd function? Solutio衍 Since

f(x) is an odd function , it follows that f( -x) = - f(x) ,

(i) Let h (叫 = xf(x) and

f2 (x) = x 2f(x).

\j

x

εIR.

(1)

Using (1) , we 自nd that

h(-x) = -xf(-x) = xf(x) = h(x) ,

\j

x

εIR;

(2)

f2 (-x) = (… X)2 f( -x) = - x 2f(x) = - f2 (x) , \j x εIR. (3) We co卫clude from (2) that h (x) is an even function , and , from (3) , f2 (x) is an odd function. (ii) From (1) , it follows that

gl(-X) = f ((_X)2) = f(x 2) = gl(X) , \j x εIR; g2(-X) = f ((_x)3) = f(-x 3) = - f(x 3) = - g2(X) ,

(4) \j

x εIR. (5)

We conclude from (4) that gl(X) is an even function , and , from (5) , that 如何) is an odd function. (iii) If j is a positive integer , it follows from (1) that

f ((- x )J)

=

(-1 )J f (x) , \j x εIR.

(6)

2

JVIATHE1VIATICAL PRELIMINARIES

Then , usi吨 (6) ， we 在nd that

h(-x) = (-x)if((-x)j) = (-1)ν ((-l)jf(x j )) 工( - 1)i+ jxi f (x j ) = (-1)忡jh(x) ， V x εJR. TTIerefore7if t +j is aIIeven iMeger7the fmctionfz(z)is an eveIl function , and , if i 十 j is an odd integer , the function h(x) is a~ ~dd function. 口?

- 吐 (T(n ， 1, x)) ,

3

( 2 一 (η 十 1)3ηx n 一 1 十 (2η2 十 2η-1)(η 十 2)(η 十 l)x n \ 2 n十 ) l\ -n (η+ 3)(η + 2)x 1 6(1 - x) 1 日1

、，

/一 (η 十 1)3η(η - 1)x n - 2 十 (2η2 十 2η -1)(n+2)( η 十 1)η x n 一 1\

Ii\ -n 2 (η 十 3)(η 十 2)(η 十 1)zn6/ 1日1

、 ，

(一加川一川仇 2η 一阳仙 O 一η2(η 十 3)(η 十 2 勾)(η 十 1)

Problem 2: Let S(n , 2) = I:~=1 k 2 and S(η ， 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 EXERCISES

6 η(η+ 1) (一 (η 十 1)2(η- 1) + (2η2 十 2η- 1)(η 十 2) … η(η+ 3)(η+ 2))

(7)

6 η (n+1)(2η+

and the fact that

1)

6 T(η ， 1 ，

x)

Z 一 (η 十 1)x n + 1 十 nx n + 2

(1 - X)2

(8)

Therefore , S(η ， 2)

=

η(η+ 1)(2η+

1)

to show that T(η ， 2 ，

x)

2 X +x 一 (η +

1)2 x n+1 十 (2η2 十 2η - 1)x n + 2 _ n 2x n + 3 (1 - x)3

(ii) Note that S(η， 2) = T(η ， 2" l);"Ys~ yHopital's rule to evaluate T(η ， 2 ， 1) ， and conclude that S (川)= 7加1i(2叫)

\lit-/ -qJ 、飞，，/

QU

η

咱t4

\飞 11I/

MZH /Il--\+.3 .qb

才tA

吨3·-A

η

十

/Ill--飞飞

+

(9)

for i = 4 , given that

=η(ηf1);

~.aiution: (i) ~he res~lt follows from (7) and (8) by usi鸣 Quotient Rule to differentiate T (η ， 1 ， x).

(ii) It is easy to see that T(η ， 2 ， 1) = I:~=1 炉工 S(η ， 2). By using I'Hopital's rule we 盘时 that T(η ， 2 ， x) is equal to lim Z 十 x 2 一 (η 十 1)2 x 饥十 1 十 (2η2 十 2η - 1)x n + 2 一 η 2 X n十3 z• 1 (1-x)3 lim 1 +2x 一 (η 十 1)3 xn 十 (2η2 十 2η- 1)(η 十 2)x n十 1 _ n2(η +3)x n + 2 -3(1 - x)2

.付 tu

n

一一

and conclude that S(n , 3) = (呼叫"

JFi 飞

(iv) Note that S(η ， 3) = T(η ， 3 ， 1). U~e I'Hopital's rule to evaluate T(η ， 3 ， 1) ，

cυ

川J)24mμ))

Problem 3: Compute S(η ， 4) = I:~=1 k4 using the recursion formula

1川 -

(iii) Compute T(η ， 3 ， x) = I:~=1 k 3 x k using the recursion formula

(iii) Finding the value of T(凡 3 ， x) requires usi吨 Quotient Rule to differen x). (iv) The solution follows similarly to that from part (ii) , albeit with more complicated computations. 口

t杜iate T(η ， 2 ，

=

S(川)

(叮 γ

Solution: For i = 4 , the recursion formula (9) becomes

仰， 4)

=

H价十户一言 0)

s(n ,

j))

n(η+ 1)(2η 十 1)

6

'

4

l\!fATHENIATICAL PRELIMINARIES

~(川)5 _ 1 -

S(n , 0)

一臼川一则川)一叫川))

η(η 十 1)(6η3 十 9η2 十 η-1)

5

0.1. SOLUTIONS TO CHAPTER 0 EXERCISES (ii) We substitute η 十 1 for n in (12) and obtain that

口

X n +3 -

2x n 十 2 - Xn+l 十 2(η+ 1) + 3.

(14)

By subtracti吨 (12) from (14) , we find that X n +3 =

ηin

,i 4

2X n +l -

Xn

+

2η+3 ，

V η;三

1，

(12)

with Xl 二 1 and X2 = 5. (ii) Similarly, show that 3X n +2 - 3X n +l

+ X n 十 2 ， V η;三

with Xl = 1, X2 = 5, and 句= 14. (iii) Prove that the sequence (x n ) 吃 1 satisfies the linear X n +4 - 4X n +3

+ 6X n +2 -

1，

(13)

Solve this recursion and show that η(η Xn

P(z)

二=

十 1)(2η+ 1)

V η>

+ Xn

-

0, V n ~三1.

(16)

= 泸 -4♂ + 6z 2 - 4z 十 1 =

(z-1)4.

The polynomial P (z) has root λ= 1 with multiplicity 4. We conclude that the there exist constants Ci , i = 1 : 4 , such that = 01 十 O2η 十 03η2 十 04n3 ， V η;三1.

Since Xl = 1, X2 = 5, X3 = 14 , X4 = 30 , it follows that 0 1, O2 , 0 3 and 04 satisfy the following linear system (1 11 2 14 1 3 9 1 4 16

rec旧slOn

0, V n 二三 1.

4xη+1 十 X n

/iE\vhu

The characteristic polynomial associated to the recursion (16) is

Xn

X n十 3

V n 二三 1.

By subtracting (13) from (15) , we find that

、飞，，/

/'''飞、、

+ (η 十 2?

亏1品

X n十 1

3x n十 1 一 3X n +2 十 Xn+l 十 2 ，

X n +4 - 4X n +3 十 6X n +2 - 4X n +l

Subtract (10) from (11) to find that X n +2

X n +4 =

(10) , obtain that

X n +2 -

V n ~三1.

、I I J

+ 1 for

3X n +l 十 X n 十 2 ，

(iii) We substitute η 十 1 for n in (13) and obtain that

nu 、、、，，，，，

n

41· 南

X n 十 (η +1? ， V η 三 1 ，

X n十 1 -

with Xl = 1. (i) By s由stituti吨

/Ftt\

-

才'·-A

Problem 4: It is easy to see that the sequence (xη) 吃 1 given by X n = I::~=l k2 satisfies the recursion

3x n十 2

1 8 ) \ C I O1 5 2 \ ) - /f 1 27 I I 0 3 I - I 14 6 4 \ 04 \ 30

We obtain that 0 1 = 0 , O2 = ~， α=i aBdq=i and therefore 2η3η(η + 1)(2n 十 1) n - …十一十一= ,

1.

V n;三1.口

Conclude that S(η ， 2)

三二 k

2

η(η+ 1)(2η+

6

1)

Vη

> 1.

Problem 5: Find the general form of the sequence (x n)n 2: 0 satisfyi吨 the linear recursion X n +3 = 2X n +l 十 X n , V n 二三 0, with

Solution: From (11) , we obtain that the first terms of the sequence (xn)n三 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 , and

X2

= 1.

First Solution: By direct computation , we obtain X3 = -1 , X4 = 1 ，圳工 -1 ， = 1. It is nat 盯al to conjecture that X n = (-l)n for a町 positive integer η. This can be easily checked by induction. X6

1VIATHE1VIATICAL PRELIMINARIES

6

Second

Sol州on: Alternati 叫y，

the linear recursion X n 十 3 P(z)

-

=

we note that the seque丑ce (x n ) 吃o satis且es 0 , with characteristic polynomial

+1 =

(z 十 1)(z2 -

Z -

The roots of P(z) are 一 -1 ，与E 眨， a时导L§. Th 阳 e臼re£ 白 f扣ore， 出t h 阳 坠e盯re 盯 exist ∞ c臼∞ on I丑n归st S阳 t 严叩\

/

=

01 (-1) η 十 O2

I \

严-\

~

11 - vi月 1 03 ←」二 I U\ 2 I

+

I

2

/

η

1 斗、 15 \ 卜」二 I

=

(-1)η ，

Vn 三 o.

X n +1

, Vn

主 O.

口

3X n + 2, V η;三 0 ，

4x n十 1 -

with Xo = 1 and Xl = 5. (ii) Find the general forml由 ηi丑 (ο17η) ，

-

(18)

=

z2-4z 十 3

=

(z-1)(z-3) ,

which has roots 1 and 3. Thus , X n 二 01

4X n+1 - 3x n 十 1 ，

4xη+2 - 3xη十 1

(19)

5X n+2 - 7X n十 1

=

z3 - 5z2

+ 7z -

V η;三 o.

V n 二三

o.

(22)

+ 1,

V η;三 o.

(23)

+ 3x n, V n

~三 o.

(24)

3 = (z - 1) 2 (z - 3).

Therefore , there exist constants 0 1 , O2 , 0 3 such that Xn

=

01 3η 十 O2η

+ 0 3,

V η 二三

o.

Since Xo = 1 ，问= 5, and X2 = 18 , we find that 0 1 = ~， O2 = -~， α=-i We conclude that 3η+2 -

+ 023n ,

(21)

(ii) The characteristic polynomial of the linear recursion (24) is P (x)

~三 o.

3X n +1 十 η+3 ， V η 三 o.

Subtract (22) from (23) to find that

Sl郎ti扰t1巾

(ii) The characteristic polynomial of the linear rec旧sion (19) is P(z)

Xn+2 -

X n十 3

0, V n

V η 二兰三 0,

Substitute η+ 1 for n in (22) to obtain that

(17) from (18) and obtain that

+ 3x n

+ 3x 阳 r1 ，

with Xo = 1, Xl = 5, and X2 = 18. (ii) Find the general formula for X n , n 主 o.

Xn+3 -

Xn+2 - 4xη十1

(20)

+2 ， V n 二三 0,

5xη+2 一 7X n+1

-

Xn+2 -

it follows that

Sl削ract

η

Subtract (20) from (21) to find that

3x n , V η 二三 0,

3x n十 1 十 2.

3x n 十

By substituting n + 1 for ηi口 (20) we obtain that

for X n ， η 主 o.

Zη+2 -

We

Zη+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 ) 吨。 satisfies the linear recursion X n十 2 -

口

with Xo = 1. (i) Show that the sequence (Xn)n:::::O satisfies the 且 1i丑 near 陀 r ec盯S

Problem 6: The sequence (x n )吃o satisfies the recursion X n十 1

1, V η 二三 O.

-

Problem 7: The sequence (x n ) 吨。 satisfies the recursion

η

By 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· 3n

Xn

1).

C1 , O2 , and 0 3 such that Xn

Since Xo = 1 and Xl = 5, we obtain that C 1 = -1 and C 2 = 2 and therefore

2xη刊 - X n =

z3 - 2z

7

0.1. SOLUTIONS TO CHAPTER 0 EXERCISES

Xn -

2η

4

~ Vη> o. 口

JVIATHEJVIATICAL PRELIMINARIES

8

Problem 8: Let P(z) = 2:立。 αiZi be the characteristic polynomial corresponding to the linear recursion

2二 α向十i

°

0, V n ~

=

(25)

Ih(x) + h(x) I /' 1~ YY> """J h(x) I J.. lim ~;mn Ih(x) I 0 with

η →∞;

Z

。(的 7

Z

♂

ηZT 汇 ?γUMnZM

We conclude that

z

b.

Also , what is the largest possible value of b? 口

0.3

Solutions to Supplemental Exercises

Problem 1: Let

α>

0 be a positive number. Compute

户十卢十日τ 灿t剖tω ion 仇 仰仰 η

t出 ha 肮t

limit by l , ther口1

挝 1t

follows that l must satisfy

jα +jα+ 币二=

d丰 l ，

(28)

which can be solved for l to obtain that 1+ VI+石 2

(29)

12

!\/fATHEMATICAL

PRELIMINARIES

.-.

We now show that , for anyα> 0, the limit of 飞/α+ 、/α +vaτ-. does e于风 whick i?quimieM to provi丑g that the 叫uen 配阳 C臼e (X ωηρ)η 乌 哈三沪o 坦 is ∞ c on口阴飞

0.3. SOLUTIONS TO SUPPLE1VIENTAL EXERCISES

Solution: If we know that the continuous fraction (30) has a limit l, then l must satisfy

l= α+; 白 l2 - al -

whe 臼re 均 X 0= 飞v αand

va百二，

X n +1 -

We will how that the

\I η 主 O.

and therefore

l丑 creasm 丑g.

Let l be given by (羽)， i.e. , let l = 与严

va

X n +1

va:+l = l ,

The 自rst

since l is the positive 叫叫O∞ n of (闵 剑 28 创)， 肌a削丑叫 8 d 山 t he 臼时 re 盯 伽 eflf inductior 瓦 1 ， we 且 fiII卫ld tl扭lat Xη < l for all η> O.

Xo

(ii) The sequence (x r山三 o is increasi 吨·

since

Xn


l , we show that Zn+1 < Zn and Zn礼 > l. This proof is very similar to that given above for the seque丑ce (Yr山三o and is left to the reader as an exercise. We conclude that lim X2n十1 = l.

/ d可4-a \

Eom(36)and(37)pwe comiude tht7if yn

(37) Um for any

the last equality can be derived as follows: 2

α2 - al 十 1 = l 仁中 l 一 α=α2l - al 十 l 仁斗 α (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 η 主 O. III other words7we showed by induction that the seque口ce (Yn)吨。 giv by the recursion (33) with Yo = αis increasi 吨 and bounded from above by l. Therefore , the seque口ce (Yn)吃o IS co盯ergent. Denote by h = limn→∞ Yn the limit of the sequence (Yn)n '2 o. From (33)-and using (35)v, ;;e obtai~ th~t

J 一 (α2 十 l)h 十 α 《 1 一 αh 十 l 仁斗 α (lt 一 αII - 1) = 0

白叫It - I) (It +咛 -a) =0 Since h > 0 , it follows that h = l , i.e. , that limn→∞ Yn = l. Recall that Yn 二 X2n for allη;三 O. We showed that the subsequence made of the even terms of (x 阳 仇 7η 1 by (32) , i.e. , that 一 lim X2n 二 l. (39) n一→ α2

lim im Xn Xn -

一 (α2 + l)zn + α αZn 十 1

η>0 ' V , .v:::-v ,

l -

α+ v'歹τz

一一一一一

口

Problem 3: (i) Find x > 0 such that

XX"

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 缸 aI丑l 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: 队+1 - -/2x n _ γrz/27V 71 主 O. It is easy to see by induction that the sequence is increasing and bounded from above by 2 , since X n 十 1 > Xn 牛二中 2Xn/2 > 2 xn - I/ 2 牛工:> Xn > X n 一 1 ;

Xn+1 < 2 伫二? 2xn / 2 < 2 宇::} x n /2 < 1 仁::} Xn < 2. We conclude that the seque口ce (x n)n'2 0 is convergent. If l = limn--t∞ X n ， then

Similarly, we define the sequence (zn)n '2 o by the recursion 十

(40)

From (39) and (40) , we find that

l 白 (α2 + I)Yn 十 α 0, x

θtθz2·

- Xl + 3X3 , 3X2 - 2X3) ;

在 (x) 忐 (x) 忐 (x) 品 (x) ~(x) 蔬 (x)

Compute

2

~-已一古，

u(x , t)

θuθ2也

(θfθfθf) 瓦 (x) 瓦 (x) 瓦(叮 (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)

的(1， -1 ←(一;-i e-;)

(1. 8)

一一

at

=

一

1 2

'J问

_t-u/~ V

1

一-=

y4王

e

.r 2

"4τ+

x2

.r 2

1

一一=

d召 1

Product 丑ule ， 2

.r e- 4i

.r 2

=一一一=e 右十一·一 =e 一石; 4t 2 y4言: 2t y4巧 θu

一一一

x

ω

ax

-

-…-叩】

1

.--…-一

2t v41r t

ρ ~

.r 2 4t·

,

(

x2

(

we find that 1\\

I 一一. I 一 ~II 2

\4\

t ))

(1. 9)

CHAPTER 1. CALCULUS REVIE1iV. PLAIN VANILLA OPTIONS. nu••

4t4

From (1. 9) and (1. 10) , we ∞ c onc 叫 tcll时 e 由 i比 t ha创t θ2 U

θu

θx 2

θ t'

A

Note: This is a but臼rgy pead tMer mks a 10吨 Pωi阳1 in a b飞阳rfly pread if the price of the underWIngmset at maturity is expected to be h the K1 三 S(T) 三 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 二且寿县 ; 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) 皿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

mm 黯

25

< 30 30 < S(T) < 35 35 < S T) < 40 40 < S(T)

。

S(T) - 30

。

。

O O

O S(T) - 30

S(T - 30 S(T - 35

S(T) - 30

。

sjTj35 SrT) - 40

40 - S(T)

。

L....J

Problem 11: Co斗sider 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吨 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臼r创l坊刷 抖 yrm J 臼se S 创t 肮 at time T.

C1 (T)

1.1. SOLUTIONS TO CHAPTER 1 EXERCISES

、、、

才1 4

/ t t‘ 、

δx 2

1二 币

θ2 U

1二 二 制

24

2C2 (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 只 (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 Tis

Depe口ding

I 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 巧 (T)

= max(20 - S(T) , 0) - max(15 - S(T) , 0) ,

which can be written

i口 terms

of the value of S(T) as

A 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. 口 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 十 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吨 long a butte

26

CHAPTER 1. CALCULUS REVIE1iV. PLAIN VANILLA OPTIONS.

rather aomegative)payof M maturity-Ttmrefore7if you are to α ss包 uη阳 ω n7 POSItion ln either om of th portfolios (口 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 that

V(T)

问 (T)

-

1.1. SOLUTIONS TO CHAPTER 1 EXERCISES

当〈几

VeT) O

K 1 < SeT) < K 2 K 2 < S(T ) < K 3 K 3 < SeT)

页

- 5) ， 0) 十 max(S(T) 一 (X + 5) , 0) - 2max(S(T) - X , O).

The value of the portfolio at time

T

is detailed below:

V(T) S(Tf<X - 5 O X - 5 < S(T) < X S(T) 一{~- 5) X < S(T) < X + 5 (X + 5) S(T X_+ 5 < S(T) 。 --

Z1K1

X1 C 1(0)

十 X2 C2(0) 十 X3 C3(0)

At maturity T , the value of the portfolio will be

X1 C1(T) + X2 C2(T) 十 X3 C3(T) Xl max(S(T) - K 1, 0) + X2 max(S(T) - K 2, 0) + X3 四 ax(S(T) - K 3, 0) ,

respectively. Dependi鸣 on the value S(T) of the underl}也 g asset at maturity the , value V(T) of the portfolio is as follows:

Xl K 十 K x2)K (Xl 3 X1 1 X2 2 Xl + X2 十 X3

X2K2 X2 K 2 - X3K3

。;

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 主 0 ， and X2 and X3 such that

8X1

+ 5X2 + 3X3 30X1 + 10x2 Xl 十 X2 + X3

〈〉一>一

SoJTLUOTZJFor marbitrage opPOEtunity to be preseIlt7there must be a portm folio made of the three options with nonnegative payoff at maturity and with a negative cost of setting up. Let K 1 = 100 < K 2 = 120 < K 3 = 130 be the strikes of the options. Denote ,by x~ ， X2 ,!3 ..!~e options positions (which can be either negative or positive) at time 0. Then , at time 0 , the portfolio is worth

+ X2C2(0) 十 X3 C3(0)

〈〉一〉一>一

Problem 14: Call options with strikes 100 , 120 , an 丑 d 130 on the sam 丑le undern 呐it由 yi凶 g ass们E时 w 出 盯 ht 缸 肮 机 刷 切 h e sa 丑 me lema t I u 吨 江哟 l牛 (there is no bid-ask spread). Is there an arbitrage opportunity prese时? If yes , how can you make a riskless pro缸?

V(T)

TiT-

Note that VeT) is no 日 negative whe 丑 SeT) 三 K2 only if a long position is taken in the option with strike K 1, i.e. , if Xl 三 0. The payoff VeT) decreases when K 2 < SeT) < K 3, accounting for the short position in the two call options with strike K 2, and then increases when SeT) 三 K3 · We conclude that VeT) 三° for any value of SeT) if and ∞ly if Xl 主 0 ， if the value of the portfolio when S (T) = K3 is nonnegative , i.e. , if (Xl 十 x2)K3 - X1 K 1 - X2 K 2 ~三 0 ， and if Xl 十 b 十均主 0. Thus , an arbitrage exists if and only if the values C 1(0) , C2(0) , C3(0) are such that we can find Xl , X2 , and X3 with the following properties:

Xl C1 (0)

=

1δS

--Z X1K1 x2)S(T) 一 + 3Z22S x 1K 1 (Xl 十 X2 + x3)S(T) (Xl

\i2 (T)

max(S(T) 一 (X

V(O)

27

。;

(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日出 the option with middle strike (note the similarity to butterfly spreads). Then , choosi 吨 X3 = -Xl - X2 would be optimal; ef. (1. 13). The constraints (1. 11) and (1. 12) become

5X1 + 2X2 < 3X1 + X2 >

nunu

These constraints are satisfied , e.g. , for Xl corresponds to X3 = 2.

1 and X2 -

-3 , which

28

CHAPTER 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. 口

more shares of the underlying asset and usi吨 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 is

V(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 丑at satis丑ed and explain how would you take advantage of this arbitrage opportunity.

Solution: The followi吨 values are give口 : 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) = 一 (P(T) 十 S(T) - C(T)) + 39.5821erT - - K + 39.5821e rT - - 40 + 40.2473 -

The Put-Call parity is not satisfied , since

P + Se-

= - P(T) - S(T) + C(T) + 39.5821erT .

Recall from the proof of the Put-Call parity that

q = 0.03; P = 2; C = 4. qT

rT

.

(1. 14)

Therefore , a riskless profit can be obtained by "buying low and selling by selling the portfolio a丑 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. ,

且igh飞 i.e. ，

(39.5821 - 39.3389) 产= 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:

29

1.1. SOLUTIONS TO CHAPTER 1 EXERCISES

0.2473.

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吨 from the Put-Call parity; d. (1. 15). 口 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) = 一 P(O) - S(O)e- qT 十 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 直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.ι $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. 口

Note that by shorting the shares you are responsible for paying the accrued dividends. Assume that the dividend p句rments are financed by shorting

lThis is similar to converting a long pm;ition in c qT shares at time 0 into a long position share at time T: thro吨h continωus purchases of (fractions of) shares usi吨 the dividend payments: which is a more intuitive process.

P + Se- qT

-

C

39.5821 ,

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 b可i吨 shares a丑 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丑 one

30

CHAPTER 1. CALCULUS REVIEVV. PLAIN VANILLA OPTIONS.

Problem 17: You expect that an asset with spot price $35 will trade in the $40一$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口 one year of the first strategy, i.e. , of buying the asset , is

叫T) = 1~~0 S(T) , w且ere

S(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 , is

1I2 (T)

= 半 max(S(T) - 35， 0) 工(平叫)-35)iijzjZZJ

It 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 pro自table if the asset appreciates in value , i.e. , is S(T) is signi五cantly 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 _ ._, 35

1000

一:~ S(T) = 一一 (S(T)

- 35) 仁斗 S(T)

=

39.516 1.

If the price of the asset will , indeed , be i口 the $40一$45 range in one year , then buying the call options is the more profitable strategy. 口 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(0.08一0.03)/4

12150.94 > 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 EXERCISES

31

At 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

32

CHAPTER 1. CALCULUS REVIEW. PLAIN VANILLA OPTIONS.

1.2

Supplemental Exercises

1. 3. SOLUTIONS TO SUPPLENIENTAL EXERCISES 8. Create a portfolio with the following payoff at time T:

1. Compute

户口(x) dx 2. Compute

V(T) =

xne x dx

(I n ( x) )n dx

5. Let

f(x)

z+ \11I/ 1-z

(

]

V Z

= 一古=exp(σ 〉习作

>-

才t4


-

414

口

♂

pu

十

'h 川h

吁i

η

门

一-

z

\飞 lll-/

nZM Lf-M

/III-11\

n

(1. 16)

(1\1 < In I 1 + -=. I < 一、 Vx> 1.

Let

x)

:一咛 +D;

f(x)

x

一

(1. 17)

咔+~)一击

g(x)

The丑 7

= eX , the followi吨 general forml 一­

,J

Vv Z >-

才i

1

Z 十 1\

儿(x) = j ♂eX dx = xneX - n j ♂-1川 = xne x 一吼一 1 协 Vn;:::l fIJ

4t-A

Note that (1. 16) is equivalent to

By using integration by parts , it is easy to see that

,2 z neZ G

吨'-4

Sol仰 on:

一二一-

PIt-


0 , i = 1 : 3 , such that 的 C1 - X2C2 - X3C3 Xl (8

0;

- K 1 ) - x2(8 - K2 ) 一句 (8 - K3 ) 三 0 ， V 8 三 O.

K3 一option.

(1. 43) (1. 44)

The ineql叫ity (1. 44) holds if and only if the following two conditions are satisfied: Xl - X2 - X3 主 0; (1. 45) Xl (K3 -

K 1) -

x2(K3 -

K2 ) 三 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? C

C'}. C

Se-

qT

- K e- rT

is 丑。-arbitrage directly

Cαsk …凡id 三 8e- qT

-

K

following from the Put-

e- r1、三 Cbid 一凡 sk·

口

Chapter 2 Improper integrals. Numerical integration. Interest rates. Bonds. 2.1

Solutions to Chapter 2 Exercises

Problem 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 三 5x - 1 if 1 < x < 2.

Therefore , D = {(x , y) 11 三 Z 三 2 and (x+ 1)2 三 ν 三 5x - I}.

Then ,

fin 川白

[(汇 ιιi;:(归户 Z [

((怡均牛 2与U 一 泸的州叫叫)川川将巾 U|tcc U汇口口 口 :ζ:Uω; !: 斗 L 二二 J;斗ω户)2)讪 ) dx

I 泸 (5x

-

1 一一- (归仙川 Z叶川+札1呐一((仰一 1)2 一归仙川+刊1阳Z

1，'(σ5x 一 1 一川附一仰一 1忏叫十刊(川 俨(←一泸仇川十叶川 l 3xZ卜川一 -2川Z叫)川d白Z户←=←一; 口 Problem 2: Let

f : (0 ，∞)→lR denote the Gamma functio孔

的)

χ x

=

飞 俨1 V俨Zα←叫~一→ l 俨俨 45

i.e. ，

le t.

CHAPTER 2. NUMERICAL INTEGRATION. BONDS.

46

(i) Show that f(α) is well defined for any α> 0 , i.e. , show that both

[xα-1

e- x dx

[ ' x"

•

1 e- x

zα-le-z
1. Show that f(l) = 1 and conclude that , for any η 三 1 positive integer , f(n) = (η-I)!.

Solution: (i) Let α> 0. Intuitively, note that , as x \、 the order of x o:- 1 , since limx\", 0 e- x = 1. Since

~tIl}

(1 xα-1 t\", 0 Jt

0，

the functio丑 Zα-l e -x is 0丑

°

Choose n(ε) = max(m(ε ， N)). From (2.3) and (2 .4) we obtain that

t\、oαIt

xα一1

e-x dx -

lim I t\o Jt

X

dx

to show that , for any s > η(ε) ,

Z

smce lim x o:- 1 e- x

limxα-1

z\0

2\J

lim t o:- 1 e- t t → οc

-

0,

ια-1

lim 二--:-

t• x

eL

= 0.

for α> 1;

e

Z

,G

m 曲

q&

气 BEEt--liId

αz

才t 4

ffl'o

飞/

划已

e

1··A

伽

FI--00/ 飞

JZα

1lιtn

(2.2)

卜

I∞川-X dx < 飞 Vs> η(ε)

K

°

『f n n U 吃i U

exists and is finite. l\IIaking 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 de丑nition， we need to prove that , for a町 ε> 0 , there existsη(ε) > such that

•ijd

Z Z7 ufbfqh 飞 lI/{

(2.1)

m叭 卜

e- x dx

一 -11J

it follows that

咄咄-f

f(α)

=卡

1,

"~UV

fzMe曰 :Etlzα-1

= 1

Assume that α> 1. By integration by parts , we find that ]αef''

JiITl (1 - e- ) -

m)=fev=JI叫tfzdz= 出 (_e-t + 1) -z

t

which is what we wanted to show; d. (2.2). (ii) It is easy to see that

斗{旷日仁川

e- dx -

(2.6)

2e- s !2 < 2e- n (f. )!2 < 飞 xα-1 e- x

- 3川

I

Jl

and

<J坦/，' ~-x切=出(_2e- t !2 十川)

α ! 「U扒 L 1 1 χii 况一-

"~UV

< ε(2.6)

J豆豆/，' X"-l 川Z

X"-l 川Z

fl'ok 扫人 wh

Jim

(2.5)

We can then use

α

exists and is finite. In a similar intuitive way, note that , as x →∞， the function x O:一 l e -x is 0日出e order of e- x , since the exponential function dominates any power function at infi丑ity. Since x

(2.5)

zα-l e-z <e-z/27V Z 〉 η(ε);

it follows that

I

x

Also , since limx →。c e- x !2 = 0 , it follows that , for any ε> 0 , there exists m(E) > such that 2e- m (f. )!2 N ,

Iiα -1 X-•

e- x

°such that

Note that there exists N >

Sl日ce

and

47

2.1. SOLUTIONS TO CHAPTER 2 EXERCISES

CHAPTER

48

2.

NUMERICAL INTEGRATION. BONDS.

For any positive integerη> 1, we 五nd that f(n) = f(l) = 1, it follows by induction that f(η) = (η-1)!

(η -l)f(η-1). 口

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 512

Since

3

Problem 3: Compute an approximate value of J1 -IX e-xdx using the NIidpoint rule , the Trapezoidal rule , and Simpson's rule. Start with η= 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 NIidpoint , 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吨 f(4)(X)， note tl时 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吨 Simpsor内 rule ， the following approximate values of the integral are obtained: No. Intervals 4 8 16 32 64 128 256

The 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 τ'rapezoidal rule , and for a 16 intervals partition using Simpson's rule.

49

Simpson's Rule 0.179155099725056 0.179169815603871 0.179171055067087 0.179171162051226 0.179171171372681 0.179171172188741 0.179171172260393

口

Problem 4: Let f : 卜 R→ --+R 问问 gi忖阳 V (ωi) Use NIidpoint rule with tal = 10- 6 to compute an approximation of (1

(1

I

币 5/2

= JoI f (x) dx = JI ~一一亨 o 1+ 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. 口 Problem 5: Let K , T ， g :R • R as

σand r

x~

(ii) Show that f(4) (x) is not bounded on the interval (0 , 1). (iii) Apply Si皿psor内 rule with n = 2飞 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) =

(In

be positive constants. Define the function

= 在 rhu?

(圣)十(叫去) T) / (σ-IT). Compl.叫 (x)

Solution: Recall that

irf叫 = f川。)

CHAPTER 2. NUMERICAL INTEGRATION. BONDS.

50 Therefore ,

2.1. SOLUTIONS TO CHAPTER 2 EXERCISES

51

it follows that

g"(t) = h(t) ,

V左 i七言b印(归ω 阳 Z功)

g'(x)

which is what we wanted to show.

…(

l

xO"-J2百…y

(In

(圣)十 (γ 十号) T)2 \ 2σ2T

\

n

Problem 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巾) be a continuous fun创0口 such that J之o Ixh(x)ldx exists. Define g(t) by

以t)

=

[>O(川)h(x) = h(t) .口

,Tb

α

,Tb

、 、E E，J '

、飞}/

ι'LU

rlu

i'U

/ ' ' E飞 、

'hu

α

、、BJ/

,Tb

JZaE 飞

。，"

For our problem , α (t)

= t and f(x , t) = (x - t)h(x) ,

(2.8)

where h(x) is conti丑∞us. Then , ~{ (x , t) = -h(x) is continuous. Note that f(α (t) ， t)

= f(t , t) = (t - t)h(t) = O.

Since

= 二(

[' (x - t)h(x) dX)

口

(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 28U

二( 1: 川x) = 一川川7

98.940623.

Problem 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由 that

g' (t)

= [2.5 2.5 2.5 102.5] .

时i

ι'U

、飞 l'/

iTb

v一cas且且ow

disc = [0.97530991 0.94885432 0.92288560 0.89583414], and the price of the bond is B =

i'U

α

rld

α

z

/ ，1 1 飞

ιtu

Z

7α

W

Mυ 一衍

PIllα

∞仙

一一

YG Z

\飞 liI/

、、 lt/

z

ι'U

rld

w

ft'11α

∞仙

/Ill-\

= [0.5 1 1. 5 2];

The discount factors are

A similar result can be derived for improper integrals , i.e. ,

d-a

r叫咐咐仰(仰队阳 O矶ω 川 ;• l叫 μ e-川川 ，2

where C 工 0ω.β05 ， and r(0 , 0.5) = 0.05 ， γ(0 ， 1) = 0.0525 , r(0 ,1. 5) = 0.0535 , γ(0 ， 2) = 0.055. The data below refers to the pseudocode from Table 2.5 of [2] for ∞ c om putir丑19 the bond price given the zero rate curve. Input: n = 4 Lcash且ow

vhu ,Tb

/'1 、白

+ rld

2

飞}/

Z ,?-U

7α

、iJ

Mυ-a

-4Lj

M叫凡 V PI''1α

一一

ι'LU

VG Z

、、 l'/

z

V

PI''1'α

ν叭 凡

\Ill-/ rlu

2

十 (10∞0 十

Solution: Recall that , ifα (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\ d-a

=至 100 e- r (O，O.5)O.5 十空 100 e- r (即)十三 100 e- r (阳)1.5

d

and show that gil ( t)

B

100 e- r (O ,O.5)O.5 (100

+

+ ~ 1叫

C

~

100 e- r (O ,1)

2

户

+ 二 100 e- r (O ，1. 肌5

2

e- r (O，2)

QU

C

2(1 - e- r (O ,2)2) e- r (O ， O.5)O.5 十 e- r (O ， l) 十 e- r (O ，1. 5)L5 十 e- r (O ， 2)2

•

CHAPTER

52

2.

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%. 口

2.1. SOLUTIONS TO CHAPTER 2 EXERCISES 十

(

~ ~

I 100

2.5 disc(l) Problem 9: Assume that the continuously compounded instantaneous interest rate curve has the form r(t) = 0.05 十 0.0051泣。 + t) , V t 三 O.

(i) Find the correspondi吨 zero rate curve; (ii) Compute the 6-month , 12-mo时h ， 18-month , and 24-month discount factors; (iii) Find the price of a two year semiannual coupon bond with coupon rate 5%.

+

99.267508.

Solution: The price , duration , and convexity of the bond can be obtained from the yield y of the bond as follows:

B D

= 兰 3 仪p(一到十川p(补 1 f ι3i

一 I )

: :-

B\ 仨t 2

1 (~9i

l丑 (ο1

+ t))

e- r (O ,O.5)O.5

_

e- r (O ,l)

0.94939392;

-

e-r(O ，1. 5) 1. 5 已 -r(O， 2)2 -

0.97478242;

二 0.92408277 ;

0.89899376.

(iii) The price of the two year semiannual coupon bond with 5% coupon rate IS

β:

0.05 ~({\ (\ ,,\{\ " 0.05 刁-100e-r(O?05)05 十一~V 2

4

(i \ 5 (5\ 飞 + 103 ~exp I -~y I I 飞 2 ~ ~ 2 \ 2'-' } J

I -~y I

1\ 100 e-一 r~I{\ (O ,I)

(2.10) ,, l

‘\

41·A

07

}

(i \ 2 5 (5\1 exp (一 ~y) + 103 -A~ exp (一 ~y ) I . 飞 2 4 \ 2'" } J 07

}

'

-

-

(2.12)

-

v一cash_自ow

= [3 3 3 3 103] .

Output: bond price B = 92.983915 , bond duration D = 2.352418 , and bond convexity C = 5.736739. 口

(ii) The 6咄onth， 12-month , 18-month , and 24-month discount factors are , respectively, disc(l ) disc(2) disc(3) disc(4)

VA"

exp

Lcash_flow = [0.5 1 1. 5 2 2.5];

0.005(1 十 t)

102.5 disc(4)

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:η= 5; y = 0.09;

;(0 阳+ 0ω0∞0 阳 +

+

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?

) ~ B\ 仨t

庐t、 f 才 h扣 0ω05 十 0ωO∞0臼5盯阳川叫 l协丑叫(川 n( )

0.045

2.5 disc(3)

、、 1'/

= ;

+

呼14

叫OJ )

2.5 disc(2)

口

=一 I

Then ,

~({\ t)飞叮

e 一 I \V忡忡

门，中

ι'b

、、 Bt/

7α

俨'

TLV

一一

γ

JI 飞飞

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

0.05. __\ I

十丁一 IUU

53

0.05

+ 丁一 100 e-咐1. 5) 1. 5

Problem 11: The yield of a 14 months quarterly coupo且 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 \ / 民\ + 1. 75 exp I飞 - ~-"y I + 1. 75 exp I - ~-"y I 12'-' J \ 12'-' J

I - ~-"y I 飞

+1. 75 exp

12'-' J

( 1 1 \ ( 14\ + 10 1. 75 exp I - -,,- y I

I 一 ~~y I 飞

12'" J

飞/

CHAPTER 2. NUMERICAL INTEGRATION. BONDS.

54

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: n = 5; y = 0.07; I 2 5 8 11 141 t_c础且ow = 11-2 1~2 1~2 ~; ~;

I;

v_cash_flow = [2 2 2 2 叫

Output: bond price B = 10 1. 704888 , bond duration D = 1. 118911 , and bond convexity C = 1. 285705. 口 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:η= 4; zero rateγ (0 ， t) = 0.05 + O.Olln (1 十 i); t_cash一丑ow

= [0.5 1 1. 5 2] ; v_cash且ow = [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. 口

2.1. SOLUTIONS TO CHAPTER 2 EXERCISES

55

ac ∞ ou 叩 po ∞ np 归 ay尼 r江即 rr I

payment.

口

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时 age poi时 IS equal to 100 basis points. Thus , 10 basis points is equal to 0.00 1.) Solution: The duration of a zero-coupo丑 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

王一目

-!:1y

D =

…

= - 0.0 1.

0.001 . 10

We conclude that the price of the bond decreases by 1%.

口

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

万一句

- !:1 y

D

For B = 102 , D = 3.5 and !:1 y = -0.005 (since 1% = 100 bp) , we 五nd that !:1B 用

-

!: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 cor叩灿gggand gg7and ShhO 叭仰飞 咄 w ν these 旬 fUl丑 n囚 c ctior丑IS are either always positive or always 丑 1lega 肮，ti忖 ve.

-

B +!:1 B -

103.75.

口

Problem 16: Establish the following relationship between duration and convexity: 叮

C _ D :t.

Solution: Recall that

l-DU B-vu

a

n 以

nσ-rc

D

一­

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日 rate increases , the duration of the bond decreases. This is due to the fact that the earlier cash flows equal to the coupo丑 payments become a higher fraction of the payment made at maturity, which is equal to the face value of the bond plus

Bnew

-

θD

-=:.一

oy

,d

C

1θ2B

B

θy2

CHAPTER

56 Therefore ,

2.

NUMERICAL INTEGRATION. BONDS.

2.2

θB 一一

= -DB.

θν

(2.13)

B

\、飞 1111/''

no-rC

一

一一

ω一 句ω γ

一 -B

一切

切一句 ny

一十

BD

一一

ω一 句θ B

一-

叫σ τ 。

Using Product Rl由 to differentiate (2.13) with respect to y , we 直nd that :-uu B-2 D DB DB -D-hu

We conclude that

C

1θ2B

B

2.2. SUPPLEMENTAL EXERCISES

θν2

D2_ ~D. θu

0

57

Supplemental Exercises

1. Assume that the continuously compounded instantaneous rate curve γ (t) is given by γ (t) = 0.05 exp( 一 (1

+ t)2) .

(i) Use Simpsor内 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 NIidpoint Rule and using Simpson's Rule. Also , compute the 丑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 as

D电=一旦 ωθu 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 as

C虫二工 θ2B

eθy2·

58

CHAPTER 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 dollar

d旧ation

and dollar convexity of your portfolio?

(ii) If the yield goes up by ten basis points，五叫 new approximate values for each of the bonds. What is the new value of the portfolio? (iii) You can buy or sell two other bo时s ， one with d旧时ion 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 co盯exity)?

2.3. SOLUTIONS TO SUPPLEJYIENTAL EXERCISES

59

(ii) The value of the three year yearly coupon bond is B

= 5 disc(l) + 5 disc(2) + 105 disc(3) =

100.236424.

口

Problem 2: Consider a six months plai丑 vanilla European put option with strike 50 on a lognormally distributed underlying asset paying dividends co坠 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 且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)) +

/σ2\ 广(γ - q - v ) T + σ 飞ITZ，

2

where σis the volatility of the underlying asset. Then , the probability density function h(ν) of S(T) is

2.3

/

Solutions to Supplemental Exercises

h(y)

Problem 1: Assume that the continuously compounded instantaneous rate curve r(t) is given by

γ (t) 二

0.05

xp(一 (1

+ t)2) .

Solution: (i) Recall that the discount factor corresponding to time t is

叫-fa'州市 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)

~呵!一\时

= 0.86574100.

LL "

(2.14)

if y > 0 , and h(y) = 0 if y 三 O. Using risk-neutral valuation , we find that the value of the put is given bγ 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叼一附(0))一 (r - q 一手) T)2 i

e-rTER川 max(K - S(T) , 0)] I' K

= e- rT / (K - y)h(ν) dy ,

(2.15)

JQ

where 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 appro均nations to (2.15) and the correspondi吨 approximation errors to the Black-Scholes value PBS in the table below: We 丑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) ,

60

CHAPTER 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 4.863610

Error 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 4.863609

Error 0.007696 0.000351 0.000027 0.000007 0.000006 0.000006 0.000006 0.000006 0.000006 0.000006 0.000006 0.000006

is computed usi吨 numerical approximations to estimate the terms N( -d1 ) and N( -d2 )工 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. 口 Problem 3: The prices of three call options with strikes 45 , 50 , and 55 , 0日 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) 三 45; 45 , if 45 < S(T) 三 50; V(T)=i55-S(TL if50<S(T)< 吮 l 0, if 55 三 S(T). (

0,

J S(T) -

The cost to set up the butterfly spread is $4 - $12 + $9 = $1.

2.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES

61

The P&L at maturity is equal to the payoff V(T) minus the future value of $1 , the setup cost. Since interest rates are zero , the future value of $1 is $1 , and we conclude that ( -1 , if S(T) 三 45;

PU(T)={:jTJfii;:35; 二::; l

-1 ,

if 55 三 S(T)

The b时terfly spread will be pro主table if 46 < S(T) < 54 , i.e. , if the spot price at maturity of the underlying asset will be between $46 and $54. If r 并 0 ， it follows similarly that the butterfly spread is pr。在table if 45

+ erT < S(T) < 55 -

erT

口

Problem 4: 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 dollar duration and dollar convexity of your portfolio? (ii) If the yield goes up by ten basis points ，五nd new approximate values for each of the bonds. What is the new value of the portfolio? (iii) You can buy or sell two other bonds , one with duration 1. 6 and convexity 12 and another one with duration 3.2 and convexity 20. What positions could you take in these bonds to immur山e your portfolio (i.e. , to obtain a portfolio with zero dollar duration and dollar co盯exity)?

Solution: Recall that the dollar duration and the dollar convexity of a position of size B in a bond with duration D and convexity Care D$ = BD (i) The value , duration and

and

co盯exity

C$ = BC.

of the two bond positions are

B 1 = 1, 000 , 000; D 1 = 3.2; C 1 = 16; B 2 = 2, 500 , 000; D 2 = 4; C1 = 24. Denote by B 工 B 1 + B 2 the value of the bond portfolio. The dollar duration and dollar convexity of the portfolio are

D虫=一坐立一旦一 θB2

ψθuθuθU

B 1D 1 C虫 =θ2B

+

B 2 D 2 - $13 , 200 , 000. θ2B+θ2β

川一一一­

eθν2θν2θν2

B 1C 1

+

B 2C2

-

$76 , 000 , 000.

62

CHAPTER

NUMERICAL INTEGRATION. BONDS.

2.

(ii) Using dollar duration and dollar co盯exity， the approximate formula i:::.. B 7 倡一- D i:::.. y

1

+ §C( i:::.. y?

for the change in the value of a bond ca丑 be written as

i: :. B 臼 -D向十 ;C$(AU)2

Chapter 3 (2.16)

Formula (2.16) also holds for bond portfolios , since the dollar duration and the dollar convexity of a bond portfolio are equal to the sum of the dollar duratioIIs azld of th dollar convexities of the bOIlds making up the portfolio7 respectively. Using (2.16) , we 五nd that the new value of the bond portfolio is Bne ω = B 十 i:::.. B 臼 $3 ， 500 ， 000 一 $13 ， 200 十 $38 二 $3 ， 486 ， 838.

(iii) Let B 3 and B 4 be the value of the positions taken i口 the bond with duration D 3 = 1. 6 and convexity C3 - 12 and in the bond with duration D 4 = 3.2 and convexity C4 = 20 , respectively. If II = B + B3 十 B4 denotes the value of the new portfolio , the丑

D$(II) = D$(B) + D$(B3 ) + D$(B4 ) $13.2mil + D 3 B 3 十 D 4 B4 ; C$(口) = D$(B) + D$(B3 ) + D$(B4 ) Then , D$(口)

=

°

=

3.1

Solutions to Chapter 3 Exercises

Problem 1: Let k be a positive integer with 2 三 k 三 12. You throw two fair dice. If the sum of the dice is k , you win w (k ), or lose 1 otherwise. Find the smallest value of ω (k) thats makes the game worth playing.

Solution: Consider the probability space S of all possible outcomes of throwing of the two dice , i.e. , S =

{(x , y) I x

°

if and only if

(mml 十1. 6川 2B4 $76mil + 12B3

-

0;

+ 20马- 0,

二 1

: 6, y

=

1 : 6}.

Here , x and y denote the outcomes of the first and second die , respectively. Since the dice are assumed to be fair and the tosses are assumed to be independent of each other , every outcome (x , y) 且as probability 去 of occurring Formally, the discrete probability function P : S • [0 , 1] is

$76mil + C3 B 3 十 C4 B4 .

and C$ (II)

Probability concepts. Black-Scholes formula. Greeks and Hedging.

P(x , y) (2.17)

The system (2.17) has solution B 3 = $3.25mil and B 4 = -5.75mil. We conclude that , to imm飞lnize your portfolio , one should buy $3.25 million worth of the bond with duration 1. 6 and convexity 12 and sell $5.75 million worth of the bond with duration 3.2 创ld convexity 20. 口

= 二 ， V(x ， y) εS 36

Let k be a fixed positive integer with 2 ~二 k ~二 12. The value X of your winning (or losses) is the random variable X : S →~ given by

J w(k) , if x 十 y = k; X(川)巳 i-17else If 2 三 k ~二 7 ，

then x

+ y = k if and only if

(x ， y) ε{(1 ， k - 1) , (2 , k - 2) ,…

In other words , x S. Then ,

E[X]

十y

, (k - 1, 1)}.

= k for exactly k - 1 of the total of 36 outcomes from

= 汇 P(x ， y)X(x , y) = 去 L (X ,Y)ES

(x ， y) εS

63

X(x , y)

64

CHAPTER 3. PROBABILITY. BLACK-SCHOLES FOR l\/f ULA. k-l

=丁百一 ω (k) 十

36 一 (k ~~

- 1) ~/

(-1) 一_

ω (k)(k ~\'U/\'U

- 1) - 37 + k ;~

VI

I

1~(3.1)

The game is worth playing if E[X] 三 O. From (3.1) , it follows 出at the game should be played if

~二 k 三 12 ，

then

x十y

for 2 ::; k ::; 7.

E[X] Recall that

一-

EX

1-P

内，，却­

E [X] = p' 1 + (1 - p) . (1 十 E[X]) = 1 + (1- p)E[X].

n 一

Sol毗on: If the 自rst coin toss is heads (which happens with probability p) , then X = 1. If the 自rst coin toss is tails (which happens with probability 1 - p) , then the coin tossing process resets and the number of steps before the coin lands heads will be 1 plus the expected number of coin tosses until the coin lands heads. In other words ,

(3.4)

that

(η 十川 _ p)n+l

I

n(1 _ p)n+2

(3.5)

"ι

公

p

1- p

1

~ lim) ~k(l-p)日:一一 ·-r= 一. -pn→χ 仨7 1 - p y p

Similarly,

口

Problem 2: A coin lands heads with probability p and tails with probability 1 - p. Let X be the number of times you lllUSt flip the coin until it lands heads. What are E[X] and var(X)?

…门，，，"

The values of w(k) for k = 2 : 12 are as follows:

= 8;ω(6) =ω(8) = 6.2;ω(7) = 5.

咱i ω i f

E[X] =

E[X 2 ]

L 对 (1 _ p)k-l p 二 1 ~?')nl~~Lk2(I-p)k 1 - pn→ χ 仨?

Since T(η ， 2 ，

x)

三二 k

2

X +x 2

x

k

一

(n 十 1)2 x n+l

+ (2η2 十 2η (1

l)x n + 2

一

η 2 X n十3

x)3

we find that n

E[X 2]

.. p _ lim ~ k 2(1- p)k = ~ lim T(n , 2, 1 - p) 仨71-pn →∞

pl 1- P

…

(3.3)

+z-n-

From (3.3) and (3.5) and since 0 < 1 - p < 1, we conclude that

for 8 三 k < 12.

ω(2) =ω(12) = 35;ω(3) =ω(11)=17;ω(4) =ω(10) = 11;

17P

= τ了一

十一

k

k(l-p) fC

i口 (3 .4)， we 直nd

1i-41i

肿 一 、Z

From (3.2) , it follows that the game is worth playi吨 if E[X] 三 0 ， i.e. , if

23+ k ω (k) 三一一" 13 - k

立

(3.2)

36

/,,‘\-

M川 一 ­

By letting x = 1 - p

(x ，y) εS

ω (k)(13-k)-23-k

x)

pn→ χ 仨?

η 一

= 去汇 X(x， y) = 与生ω (k) 十 36-T-k)( 一 1)

We conclude that

品

Z 一

T(η ， 1 ，

~?') 1~ L k(1 - p)k

1

十一(

, (k - 6, 6)}.

= k exactly 13 - k times. Then ,

ω(5) =ω(9)

k (1 - p)k-l p =

n汇 M

E[X]

十y

L

= k if and only if

(x ， y) ε {(6 ， k-6) ， (5 , k-5) , …

In other words , x

Another way of computing E[X] is as follows: The coin will first land heads in the k-th toss , which corresponds to X = k , for a coin toss sequence of T T. . .T H , i. e. , the first k - 1 tosses are tails , followed by heads once. This c。如 toss sequence occurs with probability P(T)川 P(H) = (1 - p)k-l p . The凡

37 - k k-l

ω (k) 兰一一 oj'

If 8

65

3.1. SOLUTIONS TO CHAPTER 3 EXERCISES

p + (1 - p)2 p3

2-p p2

66

CHAPTER 3. PROBABILITY. BLACK-SCHOLES FORNIULA.

Therefore , var(X) = E[X 2 ] 一 (E[X])2

2- p

1

p2

p2

匕主口

(i) Show that the function f(x) is indeed a density function. It is clear that f(x) 三 0 ， for any x εJR. Prove that

汇 f川=

P'"

Problem 3: Over each of three consecutive time i时 ervals ofle丑gth 7 = 1/12 , the price of a stock with spot price 8 0 = 40 at time t = 0 will either go up by a factor u = 1. 05 with probability p = 0.6 , or down by a factor d = 0.96 with probability 1 - p = 0 .4. Compute the expected value and the variance of the stock price at time T = 37 , i.e. , compute E[8T ] and var(8T ).

Solution: The probability space 8 is the set of all different paths that the stock could follow three consecutive time intervals , i.e. ,

i

r 1- e 一αx if x > 0 F(x) = γ0770阳飞Nis (iv) Show that

M 三 t)

Sol础。即 (i)

E[(8T)2]

Jxe叫z

1

=

一 αx

e~一 α x

xe --

~_嘈

Uw

dx

αα2

r

'

一生主+~ xe一ω dx ααl

x2 e一αx

2xe 一 αx

2e 一 αz

αα2α3

fl04-g

z-d pu

z

1-d \飞 lil/

十

叫 de-d

α/Ill\

α

h以 优 - t

z--iii etnu \111/

一

一

一 -M 叫

出

ω=

二

α

-e 一α

/It--\

z-

一 -L

句

zi

znχ

/ l飞

α11

出 k ‘→

α

rJ

J J产 o -d

T 'ME

0;

)、77if Z 二。

r

I e-

ααl

f 泸川Z

PI-←

=

1

xe 一 α x

一一一一一+…

χ

J a e一 αzif Z >

f(x)

出(1 - e一叫

Z

(ii) By integration by parts we find that

一-

Problem 4: The density function of the exponential random variable X with parameterα> 0 is

fα 川z=J叫tα 川Z 出 (-rz)lZ

n 叫 W 凹

var(8T )

80u3 . p3 十 380 饥 2d.p2(1_p)+ 380 叫 2 . p(l _ p)2 十 80 d3 . (1 _ p)3 4 1. 7036; 但OU 3 )2 . p3 + 3(80u 2d)2. p2(1 - p) + 3(80ud2)2. p(l _ p)2 十 (80 d 3 )2 . (1 - p)3 = 1749.0762; E[(8T )2] 一 (E[8T])2 工 9.8835. 口

f 川x = 川 Vt 主 O

It is easy to see that

汇 f川

We conclude that

E[8T]

=

Note: this result is used to show that the exponential variable is memoryless , i.e. , P(X 三 t 十 s I X 三 t) = P(X 三 s ).

Note that

P(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时 the expected value and the variance of the expo阳 rando'm variable X are E[X] = and var(X) =去 (iii) Show that the cumulative de丑sity of X is

8 = {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 de五日ed on 8 , and is given by 8T(UUU) 二 S旷 ; 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.

67

3.1. SOLUTIONS TO CHAPTER 3 EXERCISES

68

CHAPTER 3. PROBABILITY. BLACK-SCHOLES FORJYIULA.

E[X 2]

LW)dz=due-α♂ dx 工 αEtjtpzdz

3.1. SOLUTIONS TO CHAPTER 3 EXERCISES for any two continuous functions

Solution: Let

叫一午一平 _ 2~3ax) I:

αεJR

be an arbitrary real number. Note that

叫ω

l

α21 g2(x)dx + 叫 仲州川 )g 川

1 α2

拄 i i垃i) 1f x < 0 , then F(x) = J~ f(s) ds = O. (仙 ∞ 1f x 主 0 ， the口 1-e

一αX

'O /II--\

\IlI-

〈一

z

、1 ，/

nL

7α

门u u

z

PI--α

、、，，/

Z

，α

门，，"

rId

\飞 11I/

F--11α

A哇

,GZ

Z

nu

,/

JU

Z

/It--\

Thu

od

Z

z

口

z

，α

rldn,,"

\飞 liI/

- nuQd

巧i 町i

巾 w he叫 = 巳F Cωom川附e 山 t he 仰 de叩pe丑毗丑ce of 吨叩&叫圳州 阴 (ρ C) on川 ti让ime 江凶I T 一 t ， i.e. , θ(vega( C)) θ(T - t) ,

/'1\-

C 一

飞/三'Tr

nu-v-e

g 一θ

SVT习气LJ7

2

By direct computation , we find that ，， lt 、-

=

(γ -q 十号)7 _aft σJ于一

i

β-T

吨a(C) 工吨a(P)

= In (主)十

d1

VL/Tr

Can you give a financial explanation for this? (ii) Compute the vega of ATM Call options with maturities of 自fteen days , three months , and one year , respectively, on a non-dividend-paying underlying asset with spot price 50 and volatility 30%. For simplicity, assume zero interest rates , i.e. ， γ=0. (iii) If r = q = 0 , the vega of ATJVI call and put options is

77

M二而

may be higher for short dated options , since small changes i口 the price of the underlying asset lead to higher changes in the Delta of the option , and therefore may require more often hedge rebalancing. 口

3.1. SOLUTIONS TO CHAPTER 3 EXERCISES

414

and therefore

vhu

θ(vega(C))

θ;三 0. T

and explain the results from part (ii) of the problem.

Solution: (i) The fact the vega of a plai口 vanilla European call or put is positive means that , all other things being equal , options on underlying assets with higher volatility are more valuable (or more expensive , dependi吨 0日 whether you have a long or short options position). This could be understood as follows: the higher the volatility of the underlying asset , the higher the risk associated with writing options 0日 the asset. Therefore , the premium charged for selling the option will be higher. If you have a long position in either put or call options you are essentially "10日g volatility" . (ii) The i即ut in the Black-Scholes formula for the Gamma of the call is S = K = 50 ， σ= 0.3 ， γ = q = 0. For T = 1/24 , T = 1/4, and T = 1, the following values of the vega of the ATJVI call are obtained: vega(15days) vega(3months) vega(lyear)

4.069779; 9.945546; 19.723967.

We conclude that , for options with moderately large time to maturity, the vega IS mcreasmg as time to maturity increωes. Therefore we expect that vega( 1year) > vega(3months) > vega( 15days), which is what we previously obtained by direct computation.

口

Problem 15: Assume that interest rates are constant and equal toγ. Show tflat7unless tfle price C of a ca11optiOIlwith strike k amd mat11rity T OIl a non-dividend paying asset with spot price S satisfies the inequality Se- qT -

K e- rT 三 C 三 SfqT7

(3.14)

arbitrage opportunities arise. Show that the value P of the corresponding put option must satisfy the following no-arbitrage condition:

K e- rT

- Se- qT

S P < K e- rT .

(3.15)

78

CHAPTER 3. PROBABILITY. BLACK-SCHOLES FORMULA.

Solution: One way to prove these bounds on the prices of European options is by using the Put-Call parity, i.e , P 十 Se- qT - C 二 Ke- rT . To establish the bounds (3.14) on the price of the call , no

C - Se- qT -

Ke- rT 十 P.

(3.16)

The payoff of the put at time T is 即以 (K - S(T) , 0) which is less th叩 the strike K. The value P of the put at time 0 cannot be more than K e- r1 , the present value at time 0 of K at time T. Also , the value P of the put option lllust be greater than O. Thus , O 三 P 三 K e- rT ,

(3.17)

and , from (3.16) and (3.17) , we obtain that Se- qT - Ke- rT < Se- qT - Ke- rT

+

P -

C < Se-qT.

To establish the bounds (3.15) on the price of the put , note that P

Ke- rT - Se- qT 十 C.

(3.18)

The payoff of the call at time T is max(S(T) - K , 0) which is l~s than S(T). The value C of the call at time 0 cannot be more than Se- q1 , the present value at time 0 of one unit of the underlying asset at time T , if the dividends paid by the asset at rate q are continuously reinvested in the asset. Also , the value C of the call option must be greater than O. Thus , 。三 C 三 Se-qT.

(3.19)


2)6t)

.

(4.16)

CHAPTER 4. LOGNORIvIAL VARIABLES. RN PRICING.

96

Solution: (i) Formula (4.14) can be obtained by solving the linear equation (4.12) for p. To obtain (4.15) , we 在rst square formula (4.12) to obtain

4.1. SOLUTIONS TO CHAPTER 4 EXERCISES

the other solution of the quadratic equation (4.20) corresponds to the value of d, since

L二 =A 一州 -1 A2 _ 1

d = u1

e2r8t

p2 U 2 + 2p(1 - p)ud + (1- p)2d2 -

97

A 十飞!

v 哥…

-

口

and subtract this from (4.13). We 缸d that

p(l - p)u 2 - 2p(1 - p)ud 十 p(l - p)d2

Prabl 巳ill 6: Show that the series 1二;二l 击 is conve带风 while the series 2:%二1i and E二;二2 世而 are diverge风 i.e. ， equal to ∞

已 (2r十σ2)8t _ e2r 8t

Note: It is known that

which can be written as

p(l - p)(u - d)2 = e 2r 们 (eo-

2

8t

1).

_

Using formula (4.14) for p , it is easy to see that

p(l - p)

=

and

( e r 8t 一份 (u-e r 8t ) (u - d)2

(e r8t

d) (u - e

-

2 ) = e2r 8t (e o- 8t

-

1r

2

6

战(兰 ;-h(η))

(4.18) where

From (4.17) and (4.18) , we conclude that r们

汇丰

(4.17)

γ 自 0.57721

=

~

is called Euler's constant.

Solution: Since all the terms of the series 2:%二l 击 are positive , it is eno吨h to show that the partial sums ~ 1

1)

·一

(ii) By

mt山iplyi吨 out

ue r8t

(4.15) and

usi吨 the

fact that ud = 1, we obtain that

1 _ e 2r8t 十 de r 8t = e(2r十σ2) 8t

-

_ e 2叫

( 4.19)

\

/

U

\li/

u

〈一

U 十 1 一 ( e- r8t + e(r十σ2)叫 z

一η 一

咱i

which can be written as

jl 1-K

e(r十泸州 7

++ /lil\

-

汇丰

1 气/立

- e- r8t

are uniformly bounded , in order to conclude that the series is convergent. This can be seen as follows: L-41 14

After canceling out the term _e 2r8t , we divide (4.19) by e耐 and obtain U十d

仨 k2

1+ 各(kll) 工 1 十二击 -i


2;

立 -L

α

牛::::::}

T N

十

S(T) 三 K 1 /α

In α

\lliI/\111l/ α

Note that (S(T))α 三 K is equivalent to S(T) 三 K /α. Using (4 .42) , we fin

俨'

where

1

十

叫

咱t 4

S(O叫 (γ-DT 十 σVTZ)

//I11\

α

=

//ll1\

ex

Dι

盯)

v nu - s nu

α

where

111

note that the change of variables y = x 一 ασ JT was used above. We conclude that

川 一2

V(O) = e- rT ERN[max((S(T))α - K , O) ],

4.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES

ifp=i Sol州附It is

easy to see that , if p = ~， then U 十 d =2eTdt;

Therefore ,

叩) =嘿~1~ exp ((α- 1)ι 子~dz-Zhz -

u

2

+ d2

=

2e(2r 十 σ2)dt7

and therefore

Ke-rTN(α)

ud

( 叶 d)27(u2+d2)

- 2e2rM 一俨σ2)M

Note that u and d are the solutions of Z2 一(也十 d)z 十 ud = 0 , which is the same as Z2 _ 2erbt z 十 2e 2rM _ e(2r 十 σ 2)M _ 0. (4 .43)

V卖 i七言 1~ex优叫叫 均((←(归川 p( α

We solve (4 .4 3) and find that

优叫p(←归忻川川一斗斗机)沙俨 1)

叫叫(川川 α←川一」斗1) (γ 十 子)川 T才) 左 VIi七言 ιι1:+Q刊叩叮叫 αω叫 而叮而) 叫叫(-一5引) dν σd 叫(α 一刊+子) T) 在 1: 叫引 dy OUVT

叫川 (r 十字)T) N(a+QσVT);

u

d since d
i1

,

4'Lν-

f\iz z-m 一-刊

f(x) - Pn(x)

口

罔

eO. 25 自 1. 284025.

\i lxl< 1.

(i) Show that , for any x , 才 t4

and conclude that

十 x). Our goal is to show that

叶一 i? wj-6 、1

X3

= 1; = 1. 28385417;

Xo

口

JR.

\i xE

For x = 0.25 we find that 已0.25

115

(ii) Show that , for any 0 三 x < 1, Problem 3: Find the Taylor series expansion of the functions

If(x) - Pn(x)1

叶泸) and 击 around the point 0 , using the Taylor series expansions of In(l - x) and 占

If(x) 。c

.;:::.. . . x k

z,.

x2

x3

x4

1

1-x

L: x

k

= 1 + x + x2 + x 3 十

，

\i

In (1 - x 2 )

= 一了 。c zffz-z2-21-Ef-zf一仨t k

234

7V Z

ε(-1 ， 1);

Pn(x)1 三( -x)nlln(l 十 x)1

and conclude that (5.3) holds true for all x such that -1 < x < O.

Solution: (i) From the integral formula for the Taylor approximation error we know that

xE(-l , l)

By substituting x 2 for x in the Taylor expansions above , where Ixl < 1, we find that

n

and prove that (5.3) holds for all x such that 0 三 x< 1. (iii) Assume that -1 < x < O. Show tl时

Solution: Recall that In (1 - x) = 一〉 ;-z-z 一一一一一一一…， V x E [-1 , 1); 乞t k 2 3 4

:三 x In(l 十 x)

f(x) - Pn(x) Since f(x) = In (1

=

+ x) , we

I{X

(x - t)n An+l)

一石「 ftmj(t)dt.

(5 .4)

obtain by inductio旦出at the de巾atives of

f(x) are (k)

fwj(z)=

( -l)k十 l(k

- 1)! 』 (1 + x)k

V

k >1

(5.5)

CHAPTER 5. TAYLOR'S FORMULA. TAYLOR SERIES.

116

(x - t) n(-1 )7川 n! J I{X 一一~ \ -/ . d Jo n! (1 十 t) 叫山 {X ( -1)η十2(X - t)n A Jo (1 + t)叫一

卫鬼 (-x)nlln(l

(5.6)

主二!< ι \7 0 1,

which can also be written as

From (5.31) , we find th创 1

(5.35)

(ii) Show that the series from (5.35) is the Taylor series expansion of the function

(→)咔 +:)7VQ17


1.

The left inequality of (5.29) is therefore established.

十

(什)咔 +1)=1+ 。J 川十川一1 亏\d +. ., XI

\I x 二三 1.

ex2

+ e- x2 2

which is equivalent to

hv

z

w -n

一一

Z

α

The right inequality of (5.29) is therefore established.

T

句

I

α

X

一-

飞

Solution: (i) The series (5.35) can be written as a power series as follows:

(5.32)

l回

/ ...\ x+~ (1 十二 I , \ I x 二三1.

句


-

130

CHAPTER

TAYLOR'S FORMULA. TAYLOR SERIES.

5.

__k!

1.

一一----，一

k→χ(~)κd 在

[

们 (k!)l / k ..…-

:地(问- x) In (x))

1 e

k

k 二二x

咱止

一门 L

P-n 川/

...."

Ll吨1

,,,,,,,, n4p

~

e1/2 lim

p→ x 飞/2p

(5.36)

=ε

,.

(

11日1

I

2p ，'~

\

,,,.

-

I

(2p)1/

(iv) Use that fact that

limp→∞ |α4p\1 / 4p

立去 z

~~，

ez=57Vzd?

to obtain that

Sol毗 on:

(i) First of all , note that

(5.38)

x/1

/lil\wv

叽→ U \ / 气l /

h 内 Ur-ww

1/fIi1\/JI--\

一

、飞 E1'' ，

UU

、 112J

n

-ai

/tt 飞、

、 、l /

/''飞飞

vhu

nuυ ηο

一;

Uu-f

一

ll

， ， ，E E 、 、 - 、 ，U J J

〓

一\jv.

/飞

1-uul

\Iil/n

一一

/ll\/i!\

口

∞

which is the same as the series (5.35).

m∞ h H俨

37+ZI+ 石「+

Jl\n

1+z4z8212

:与(l n(l 一仙 (x ))

恼，

k!

lim (同 - x) ln(x))

We compute the left hand side limit of (5.38) by changing it to a limit to infinity corresponding to y = ~ as follows: /i\Bh 1-U"1 且俨且俨一

2位

f工 x 4j - --但 (2j)!

=

m∞

-

z

\lil/

;倍 ;21+ 主 i二;11) 1 王三产+ (_1)k x 2k

1

才112(1 一叫) dx 工 2-Z ?V口 (1 - x) In(x))

it is easy to see that

2

立

T (x) is

which means that the series (5.35) is co盯ergent for all x E JR (ii) Using the Taylor series expansion

ez2 十 e- X2

(iii) Prove that

[ln(1 一叫) dx

l l lims叩k→况 |αkl 1/k

1 一…一一一τ;:;-

0

Therefore , the radius of convergence of the power series R 一

In

2

1/

p~叩\ ((2p)! )l /2p )

-

=

0,

才1ln川 h川

-

P叫 ((2p) !)1/ 4p

= lim (叫- x) In( x)) = x/1

that the integral (5.37) can be regarded as a de直nite integral. (ii) Use the Taylor series expansio口 of ln(l - x) for Ixl < 1 to show that

自nd tl川

J坦 lα年 111年

(5.37)

a时 co肌lude

25

lp

m明

Using (5.36) , we

In(1 一叫)

(i) Show that

It is then easy to see that

and therefore that

131

Problem 6: The goal of this exercise is to compute

From Stirling's formula we know that lim 1日1

5.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES

CHAPTER 5. TAYLOR'S FORMULA. TAYLOR SERIES.

132 丑ecall

5.3. SOLUTIONS TO SUPPLE1VIENTAL EXERCISES From (5 .4 1) and (5 .4 2) , it follows that

that

[忡

J41-j)=: and therefore

J坦咐 -D")

=

(5 .40)

-1 ln(ν)

1

li~ (ln (1-

x) ln(x)) =

0

1口 (1

1)

(η+

1)2

Vx 托叫 ε 叫( -11υ ， 均1)

一→

\III-/

一汇去 (5 .44)

Here , we used the fact that

立主 =Z

- x). Then ,

nl:(x) dx = - I I) ~ ~一一 )Il n In \....-

[信

and the telescoping series

(5 .4 1)

主 G 一击) zA 兰 G 一击)

-叩 m

-Zifznln(Z)dz

/IIlI\

lN

From (5 .4 3) and (5 .44) , we conclude that

(iii) Using integration by parts , it is easy to see that xn+I ln(x) xn+I znln(z)dzzd-r

1 一川

ln(1 - x) ln( x) [ln(1-X)ln(x)

L忑

η(η+

1日 才

absolutely ∞ c or盯erge旧 to

V「 ZK

一

汇

1.

x) , i. e. ,

(ii) The Taylor series expansion of ln(l 川lι M 一 Z叫) =

VO<x
U

C(K 十 x) - C(K)

θK

( 0, if 8 < K - x; j 仨竖立1 ， if K - x ::; 8 ::; K; {刑， -- ~-1 互年-8 ， if K::; 8 ::; K + x; I 0, if K + x < 8.

=

Solution: (i) Since the value C(K) of a call option as a function of the strike K of the option is infinitely many times differentiable , the first order forward 且nite difference approxim~tion ~f ~去 (K) is

川叫

ZSTMCUL 叫

S} 71. 054273576010 70.976894532020

For dB 三 10- 9 ， the values of r e increased dramatically, reaching 109 for dB = 10-12 , and were no longer recorded. The finite difference approximations of r became more precise while dB decreased to 10-4 , but were much worse after that; the best approximation was within 10-5 of r. The reason for this is similar to the one explained above for the finite difference approximations of ~.口

ap 4Ler

阿t

冒 n

c

Multivariable calculus: chain rule , integration by sUbstitution , extremum points. Barrier options. Optimality of early exercise. 7.1

Solutions to Chapter 7 Exercises

Problem 1: For q = 0 , the formula for the Gamma of a plain vanilla European call option reduces to

r

=s中叫一叮?〉

(7.1)

where

d1 (S)

=

In (丢)十 (γ 十号 )T 识 飞IT

(7.2)

I

Show that , as a function of S > 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( 句=

°

S→∞

Sol仰on:

From (7.1)

we 自时 that

r can be written as

/ (d 1 (S)? r(s) = 一古二 exp (一 飞

2

\

ln(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))

=一叩2 -ln(S) -1巾而)

CHAPTER 7. MULTIVARIABLE CALCULUS.

162

7.1. SOLUTIONS TO CHAPTER 7 EXERCISES

163

from (7 .4), we conclude that

Then ,

1' (S)

d1 (S)

1

Sσn

S

1 ( (d 1 (S))2 \ li~ 一万亏 exp (一 门 -ln(S))

li~ r(S) -

3 、v

-~ (1+ 扭)

a 、uσV

L:7r 1

\

中/

= O.

口

(7.5) Problem 2: Let D be the domain bounded by the x-axis , the y-axis , and the line x 十 ν= 1. Compute

Recall from (7.2) that

(7.7)

fLEjdzdu

d1 (S)

Solution: Note that

D = {(x , y)

It is easy to see that d1 (S) is an increasing function of S and that gsdl(S)= 一∞;

lim 州)=∞

(7.6)

S→∞

From (7.5) we find that I(S) has one critical poi风 denoted by S* , with d1 (S*) = 一 σn. 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 吨 when 0 < S < S* and is decreasing when S* < S. We conclude that r(侈 S)川is 址 aIsωo 沁1 丑肌 C盯reaωasl 且i恒 I丑1 when 0 < S < S* and decreasing when S* < S. We now compute lir町、o r(S) and lims→∞ r(s). Note that lims→∞ d 1 (S) = ∞. Therefore , Fm r(S) =

S→。G

"

Fro皿 (7.2) ，

(

(d1 (S))2

s

n-WM/

\

I

=一∞，

\

2σ2T(1日 (S))2

2σ 2T

/一 (d 1 (S)?

( 门 一 ln(S)

\

I

乙/

It is easy to see that (x , y)

n

ε D

= {(8 , t)

if and only if (8 , t)

I0 三 S 三

εn ，

where

1 ， -8 三 t 三 8 }.

The Jacobian of the change of variable (x , y) εD → (8 ， t) E dxdy

=

lθzθuδzθyl I 一一一一 ~V I lθsθtθtθ81

j 儿可叫

= O.

d8dt -

n is

~d8dt ，

2I

'

ij1:(l:tdt)ds=0

1 ) 1口 (S) J

f 儿 :jM=f(/:二战)

fL7fjdzdu

l' U'-Y~号刊 dy

L((x - 2yln(川)) I~二tu 句)

l'

Since Ii町、。 (exp (一 (ln(S)?)) = 0, we obtain that

S\O\

We use the change of variables 8 = X 十 y and t = x - y , which is equivalent to S 十 t 8 - t x - -2-; y - -2-

it follows that

1: … {(ln(S) - 1日(K) +(γ+ ~2) T) s~o

1}

and therefore -ln(S) \ I)

and using the fact that lims \o ln(S)

2-n 一 -cυ

1ιs

mwa

…-

s - 2 7飞

Ju

1

lim 一?古 exp (一

S→∞ σV 'l.作 l\2

I x 主 O， y 三 O， X 十 y 三

= 0

1- y

+ 勾 ln(y)

ds

CHAPTER 7. MULTIVARIABLE CALCULUS.

164 si丑ce limy\o y2 In (y) = O.

7. 1. SOLUTIONS TO CHAPTER 7 EXERCISES

,

口

,

,

(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(αx) max(S - K O)

max(Ke X- K , O) -

,

(7.8) , we find that

,

Kexp(αx) max(e X 一 1 ， 0).

,

(7.8) , we 且时 that

(ii) For a put option V(S T) = max(K - S O). From u(x ， O)

fL)ldzdu=d2

exp(αx)V(S， T) exp(αx)

-

165

工 exp(αx)V(S， T)

-

exp(αx)max(K

,

= exp(αx) max(K - S O) Kexp(αx) max(1 - eX , 0).

- KeX , O) -

口

Solution: We use the polar coordinates change of variables

,

,

(x y) = (r cos e r sin e)

with

(r

,e) εit = [0,R] x [0 ， 2作) .

Problem 5: Solve for

(

Recall that dxdy = rdedr. Then ,

川旷 dxdy 工 flo川价:叫R 巾=作R27 which is equal to the area of a circle of radius

,

R.

Solution: From

1

一~， 2'

Usi吨 (7.9) ，

1~

b=

(γ -q

I 一 n 飞 σ2

\2

十一 i '2 J

十一τ 俨

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(αx) max(eX - 1, 0). (ii) Show that the boundary condition V(S , T) = max(K - S , O) for the European put option becomes

u(x , O) = K

exp(αx)

川一半- 0; b+ ια(1-~) ~去 z 。

the 且rst

equation , it is easy to see that

(7.9)

, ,

,

b the following system of equations:

口

Problem 4: Let V(S t) = exp( 一 αx - bT)U(X T) where L (S \ (T - t) σ2γ -q x = In I -:;:-; I T =' ~ ， α= … n -飞 K J" 2σ2

αand

max(l - eX , 0).

b

=

we note that the second equation can be

written 创

~a2 - a (1 一勺牛二 2

一(子 -D 一(子 -D'2G 一子)+Z (子 ~r 十 2(子 -D 十二 (子 -32+ 主 (子 +;)2-2(γ) 十三 (子寸 )24 口 2

Solution: Note that t = T if and only if T = O. Then ,

V(S , T) =

exp( 一 αx)u(x ， O).

Here , x = In (丢)， which can also be written as S = K eX

(7.8)

Problem 6: Assume that the function V(S , I , t)

satis 自 es

the following PDE:

θV θVθV 1 2S2 θ 2V rS 一 - rV = O. σ 一十 S 一十一 一一 ' 2θ S2 + θS θtδI

(7.10)

CHAPTER 7. MULTIVARIABLE CALCULUS.

166

7.1. SOLUTIONS TO CHAPTER 7 EXERCISES By substituting

Consider the following change of variables:

V(川)=5H川 7where

=i

R

(7.11)

inω(7.10)，

o θ V θV

167

it follows that

1 _，> R θ 2H _ (θH\ S一-十 S一一十… σ于一一一十 γ 5IH-R 一~ 1 2

Show that H (R , t) satisfies the followi吨 PDE: θH

1 '>_'>θ2H 万J+2σ". with x 护民 and therefore

In other words， ω( x) is a decreasi吨 function on the 凶 i nt怡 er凹 va 址I [归 0 ，∞) an 丑l therefore ω(0) 三四 (x) for all x 主 0. Since ω(0) = M , and using (7.23) , it follows that

M:: : (M + 1" u(t)v(叫叫- 1" v(t) dt)

°

ln(作)

忡忡 -r correspond to for the fu日ction u(x , r) defined as follows: V(8 , t) = exp( 一 αx - br)u(x , r) , where

x = In I飞

functionω: [0 ，∞)→ [0 ，∞)as \1111/ /III-\ Z z

/III-\

丑ecall

°

也(x) 三 M 叫1" v(t) dt) ,

Solution: Define the

1"

口

Problem 3: Let 叩) : [0 ，∞)→ [0 ，∞) be two continuous functions with positive values. Assume that there exists a constant M > such that

u(x) 三 M +

175

7.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES

θν2\

2)

T

θu

rF =

°

CHAPTER 7. MULTIVARIABLE CALCULUS.

176

Solution: Using chain rule , it is easy to see that

7.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES

which is what we wanted to

show.

177

己

θVθFInSθF δtθt

T

θν7

δV

1θF

θI

T

δV

1 T-t θF

θu?

δS S T

θ2V

θν?

1 ( (T - t)2 θ2F

一一_.

θS2

Problem 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 ωa 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叫 value of a call option with maturity T and strike K. Assume that options with all strikes K exist. Defi 缸 fine 也 t he implied volatility σ 叽inη1以 K) 创 as the unique solution to C(K) = CBs(K， σimp(K)) ，

S2 飞

T-t θF\

一一一一.

T2

θy2

T

θy)

Then , the PDE (7.27) for V(S , I , t) becomes the followi鸣 PDE for F(y , t):

o=

θVθ V

一一

+

1 ')~')护 VθV

1口 S 一一 +一 σL. SL. 一一τ +γS一一 ' 2- .- 8S 2 θS

rV

θ t θI

θ FInS θF 一一一一一一一十 θt

T

θy

__1δF

lnS~~

-T

θσimp(K)

θu

十 1σ2 i( (T - t)Z 一一一一一一一一一…一一一 θ2F T-t θFV+T-tθF I + r 一一一一一一一一 2\

T2

δy2

T

θF σ2(T - t)2 θ2F , 一一+ n ~ ~十 θt'

2T2

where CBs(K， σ 叽仰nη以 1 阮cl挝 S hoωoles 臼s value of a call 叩 0 pt挝io ∞ n with strike K on an unde 臼rl勾i坊抖 y 甘切 切 rin 吨 1鸣 ga 部ss附 创t ，f'01 e 扣如 low 叭in 吨 ga log 伊 nω10ωr m 口 口 r丑na 1恼al 叫 model 呐 w it由 h volatility σ叫 (K). Find an implicit differential equ 任 tion satisfied byσ叫 (K) ， i.e. ，缸d

θy)

,

•

T

θu

-

~

(σ2\ T-t θF Ir 一 -=::-1 一一一一一 - rF.

8γ\

2)

T

θu

θK

邸saf缸 a un 旧ct钊io ∞丑 of σ 叽imη1以 K). 口

Solution: We first 自nd the partial derivative of the Black-Scholes value CBs(K) of a call option with respect to its strike K. Recall that CBs(S , K) = Se- qT N(d 1 )

Problem 6: One way to see that American calls on non-dividend-paying assets are never optimal to exercise is to note that the Black-Scholes value of the European call is always greater than the intrinsic premium S - K , for S>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). Solution: We want to show that , if the dividend rate of the underlying asset is q > 0, then CBS(S , K) < S - K for S large enough. Note that CBS(S , K) = Se- qT N( d1 )

-

K e- rT N( d2)

< Se- qT ,

since N(d 1 ) < 1 and N(d 2) > O. If Se- qT < S - K , which is equiv，由lt to S > 亡ιT > 0 since q > 0, it follows that CBs(氏 K) < S - K. We conclude that CBS(S , K) < S - K , V S

> 4K 川

Ke- rT N(d 2).

-

Then , θd 1

θCBS

一币

俨

EJd

Se一句'(d 1 ) ;~一严 N(d 2 ) - Ke- r :J.'N'(d 2 ) 坛

θK

Also , recall that

Se- qT N'(d 1 )

Ke- rT N'(d 2);

-

(7 到 (7.29)

d. Lemma 3.15 of [2]. From (7.28) and (7.29) , we find that 到C 'R.c:

#二 Since d 1

~'T' _ _,

• ,

~~

rr ~ TI I

(8d 1

8ι\

- e- rT N(d2 ) 十 Ke一叩'(也) \拔一括)

工 d 2 十 σ叮，

_",

"

(7.30)

it follows that θd 1

θd 2

θK

θK

We co肌lude from (7.30) tl川 θCBS θK

- e- rT N(d 2 )

(7.31 )

CHAPTER 7. MULTIVARIABLE CALCULUS.

178

We now differentiate with respect to K the formula

C(K)

CBs(K， σ问 (K))

=

θC θK

δCBS

IθCBS

一一一·

θK

θσθK

(7.32)

where θCBS

_-aT

Lagrange multipliers. N- dimensional Newton's method. Implied volatility. Bootstrapping.

I T_ '!J.

万 =Se-qVV=dse-PNf(do C1

V'T Ke- rT N'(d 2 )

-

K产品生

d. (η9). We conclude that the implied differential equation (7.32) can be written as

IT VK with

e

+iu r

θσimp(K)

-e一句(d2 ) + 鸣a(CBs) 叫ffk)?

vega(CBs)

ch ap

民U

which is the de直缸缸 旧 nl让i扰 山tior丑1 of σ 町仰mη1以 K). Note that C (K) is assumed to be known for all K , as it the market prices. Using Chain Rule and (7.31) we find that

一兰 θσimp(K) 2

θ-y- = 在 + e一叩 (d 2 ) ，

dry = In (去)十 (r - q)T 一切 (K)VT h .. (j imp(K)VT 2' ~

8.1

Solutions to Chapter 8 Exercises

Problem 1: Find the maximum and mi山num ofthe function f(Xl' X2 ， 句)= 4X2 - 2X3 subject to the constraints 2X1 - X2 - X3 = 0 and xI + x~ = 13.

Solution: We reformulate the problem as a constrained optimization problem. Let f : JR3 • JR and 9 : JR3 • JR be defined as follows:

削 =

4X2 -

2叫仰)

=

(专JZ二3)

where x = (X l, X2 , X3). We want to 五nd the maximum and minimum of f (x) on JRδsubject to the constraint g(叫 =0. We 自rst check that ra出(飞7 g(x)) = 1 for any x such that g(x) = O. Note that \1 g(x) =

(2;，二 en

It is easy to see that rank( \7 g( x)) = 2, unless Xl = X2 = 0, in which case g(x) 并 O.

The Lagrangian associated to this problem is F(x ， λ)

4X2 - 2♂3 十 λ1(2x1 - X2 - X3) 十七 (xr + x~ - 13) ,

whereλ=(λ1 ， λ2)t εJR2

(8.1)

is the Lagra口ge multiplier. We now find the critical points of F ( x ， λ). Let 均 x 0=(阳 均0 ， 1 ， X协 Zωa Z 缸 a丑 n1 沁0=(λ λ 沁0，1 ， λ 沁O叩 ω ，2公). From (8.1) it follows that \7 (x，)..)F(xo， λ0) = 0 is equivalent 179

180 CHAPTER 8. LAGRANGE MULTIPLIERS. NEWTON'S JVIETHOD. eη

飞八 ZZ

缸， 2 队，川

-3

"ZZ

在-·龟，

0 ，

hm

12132

ZZ

mm 一一十=

λλJzll

十十

22J4

句B A - -

矶，

认入川，

;「

(ii) Find the asset allocation for a maximum expected return portfolio with standard deviation of the rate of return equal to 24%.

0; 0; 0; 13.

Solution: For i 二 1 : 4 , denote by Wi the weight of asset i in the portfolio. Recall that the expected value and the variance of the rate of return of a portfolio made of the four assets given above are , respectively, 、 、E E， ， ，

f'51 飞

。，血

。δ

才t牛

nuO&

一­

飞八

门4

一一

巧t

人

吼叫

一

一一

nu 4t4

and

XO ,l = -2; XO ,2 = 3; XO ,3 = -7;λ。， 1 = -2;λ0， 2 = -1. (8.3) For the 在rst solution (8.2) , we compute the Hessian D 2 月 (x) of 凡 (x) = f(x) + λ切 (x) ， i.e. , of 凡 (x) -

4x2-2x3-2(2x1-x2-x3)+xi+x~-13

xi+x~-4x1 +6x2-13

and obtain I 2 0 0 飞 I 0 2 0 \ 0 0 0 I

which is (semi)positive definite for a町 Z εJR3. This allows us to conclude directly that the point (2 ，一 3 ， 7) is a minimum point for f(x). Note that f(2 , -3 , 7) = -26. Similarly, for the second solution (8.3) , we 自nd that 月 (x)

- xi - x~ - 4X1

=

I -2 0 0 飞 。 一-2 0 I 飞 o 0 0 I

I

which is (semi)negative de主nite for any x 巴 JR3. We conclude that the point (-2 ， 3 ，一 7) is a maximum point for f (x ). Note that f (… 2 , 3, -7) = 26. 口 Problem 2: Assume that you can trade four assets (and that it is also possible to short the assets). The expected values , standard deviatio风 and correlations of the rates of return of the assets are: μ1 μ2 μ3 = μ4 =

0.08;σ1 0.12;σ2 0.16;σ3 0.05;σ4

- 0.25; P1 ,2 = - 0.25; - 0.25; P2 ,3 - - 0.25; - 0.30;ρ1. 3 工 0.25; - 0.20; PiA - 0, V i = 1 : 3.

(8.4) (8.5)

十2 (ω1ω伊1σ2P1 ，2 十 ω2ω3σ2{J3P2 ，3 十 ωlω3σlσ3P1 ，3) ，

= 0 for i = 1 : 3. We do not require the weightsωi to be positive , i. e. , we allow taking short positions on each one of the assets. However , the following relationship between the weights must hold true: si口ce Pi ,4

ω1 十 ω2 十 ω3+ω4

= 1.

(8.6)

(i) We are looki吨 for a portfolio with given expected rate of return E[R] = 0.12 and minimal variance of the rate of return. Using (8 .4-8.6) , we obtain that this problem can be written as the following constrained optimization problem: 且ndω° such that

gErof(ω) = f(ω0) ，

+ 6X2 + 13

and 2 D 几位 )

E[R] = ω1μl 十 ω2μ2+ω3μ3 十 ω4μ4; var(R) = ω?σ? 十 ω2σ~+ω;σ; 十 ωiσ~

I

D 2 凡 (x) 工

181

(i) Find the asset allocation for a mi山nal variance portfolio with 12% expected rate of return;

000020·7 。;

hL

，，"

，

一­

4EA

diU2'

们向

Cο.

O

EflJll

σba

WUW

1μe

4ιu·τ···4

ρUQU

ωt

f w u H M 4L m h z-7 'b s QU m m E S

8.1. SOLUTIONS TO CHAPTER 8 EXERCISES

呐咄 w t扭ler陀e ω

(8.7)

=( ωi) 山 )i= 吐= f(ω 叫) = 0.0625ωi + 0.0625ω;+0Oω9ω:+O 04tωρi 一 0.03125w1W2 - 0.0375ω2ω3 十 0.0375ω1ω3;

(8.8) 飞

/ωl 十 ω2 十 ω3 十 ω4- 1

g(ω) = \ 0 伽1 十 O 山2 + 0.16ω3+ 0 伽4 一- 0.12 )

(8.9)

It is easy to see that rank( \7 g(ω)) = 2 for 缸lY ωεJR 4 ， since

The Lagrange multipliers method can therefore be used variance portfolio.

to 丑nd

the minimum

182 CHAPTER 8. LAGRANGE MULTIPLIERS. NEWTON'S METHOD.

8.1. SOLUTIONS TO CHAPTER 8 EXERCISES

Denote by λ1 and λ2 the Lagra且ge multipliers. From (8.8) and (8.9) , we obtain that the Lagrangian associated to this problem is

Note that the D 2 Fo(w) is equal to twice the covariance matrix of the rates of return of the four assets , i.e. ,

F(ω7 人) = 0.0625ωi + 0.0625ω2+O 09吟 +0.04ω2

-

0.03125ωlω2 一 0.0375ω2ω3

+λ1 (ω1 十 ω2 十 ω3+ω4 +λ2 (0.08ω1

The gradient of the

Lagra吨ian

\7(叫) F(ω?λ)=

+

(8.10)

D2

0.0375ω1 W3

- 1)

0.08ω4+λ1 十 0.05λ2 ω1 十 ω2 十 ω3 十 ω4- 1 0.08ω1 十 0.12ω2 + 0.16ω3 十 0.05ω4

- 0.12

To 自nd the critical points of F(叽 λ) ， we solve \7 (w ，λ) F(ω?λ) = 0 , which can

be written as a linear system as follows: 0.125

一 0.03125

0.0375

0.125 -0.0375

。

。

l

1 0.12

0.08

0.0375 O -0.0375 。 0.18 O O 0.08 l

0.16

1 1 1 1

0.08 0.12 0.16 0.05

。

。

0.05 0

。

也)1

000σ~

J

We co叫ude that D 2Fo (ω) is a positive definite matrix Therefore , the associated quadratic form q( v) = v t D 2 Fo(wo)v is positive definite , and so will be the reduced quadratic form corresponding to the linear constraints 飞7g(ω。 )υ=0.

We conclude that the point ωo = (0.1586 0 .4143 0.3295 0.0976) is a constrained minimum for f (ω) given the constraints g(ω) = O. The portfolio with 12% expected rate of return and minimal variance is invested 15.86% in the first asset , 4 1. 43% in the second asset , 32.95% in the third asset , and 9.76% in the fourth asset. The minimal variance portfolio has a standard deviation of the expected rate of return equal to 13.13%.

包)3

W4

/σ?σ1 0"2P1 ，2 σ1σ3P1 ，3

0\

|σlσ3P1 ，3 σ2σ3P2，3

O~

M=Iσ向P1，2

λ1 λ2

一

(8.11 ) / 0.1586\ I 0.4 143 I 0:3295 I;λω= 0.0112;λ0.2 =一 0.3810 \0.0976 /

,

σ§

\000σ1

μtω;

E[R] var(R)

ω tMω7

J

J

(8.12) (8.13)

-iqAnOA

\飞 -litz-1/

也

μ'μ'μ'μ'

μ'

//IIll--1 飞

where 一一

凡 (ω) = 0.0625叫 + 0.0625w~ + 0.09咛 + 0.04wl - 0.03125w1 ω2 - 0.0375w2W3 + 0.0375w1W3 + 0.0112 (ωl 十 ω2+ω3 十 ω4 - 1) 一 0.3810 (0.08ω1 + 0.12ω2 + 0.16ω3 + 0.05ω4 - 0.12) ,

O"~σ2σ向3 ~ I

the covariance matrix of the rates of return of the four assets. Let σp = 0.24 be the required standard deviation of the rate of return of the portfolio. If Wi denotes the weight of the asset i in the portfolio , i = 1 : 4 , it follows from (8 .4) and (8.5) that

Let 月 (ω) = F(ω7λ0，1， λ0.2) ， i.e. ,

and compute its Hessian

σ:3

0\ ~ I U I

(ii) Denote by

W2

The solution of the linear system (8.11) is ωo =

iσlσ3ρ1 ， 3σ2σ3P2 ， 3

+ 0.12ω2 + 0.16ω3 + 0.05ω4 - 0.12).

0.125w1 - 0.03125ω2 + 0.0375ω3+λ1 十 0.08λ2 、 t 0.125ω2 - 0.03125ω1 - 0.0375ω3+λ1 + 0.12λ2 0.18ω3 十 0.0375ω1 - 0.0375ω2 十沁十 0.16λ2

一 0.03125

/ σ? σ1σ2ρ口内 σ3P1 ，3 月 (ω ) = 21σ1σ2P1， 2 作 σ2匀2，3 飞

is the following (row) vector:

183

is the vector of the expected values of the rates of return of the four assets. The problem of 五日ding a portfolio with m以imum expected rate of return and standard deviation of the rate of return equal to σp can be formulated as a constrained optimization problem as follows: find Wo such that ~~n_ f(ω) = f(ω。)，

g(ω)=0

(8.14)

184 CHAPTER 8. LAGRANGE MULTIPLIERS. NEWTON'S METHOD.

8.1.

where ω = (Wi)仨吐

However , it is easy to see that f(ω)

(ωtkLz)7

σp

(8.16) We can now proceed with finding the portfolio with maximum expected return using the Lagrange multipliers method. Denote by λ1 and λ2 the Lagrange ml削pliers. From (8.15) and (8.16) , we obtain that the Lagrangian associated to this problem is

where

0)

1

F(ω7λ) =μtω+λ1 (Itω-1) 十 λ2(ωtMω - (J'~)

G(ω?λ1 ， λ2)

1

C

σ2

P

.

)

It (川

飞7(叫) G(ω?λ1， λ2)

2(Mω )t

ω

2

wt !VIω - (J'~

\飞 III-f/

l t M- 1 1

(μ+λ11 + 2川) Itω-1 ,.

This is done using a six dimensional Newton's method; note that the gradient of G(叽 λ1 ， λ2) is the followi吨 6 x 6 matrix: M00 -inunu

We find that the Lagrangian (8.21) has exactly one critical point given by

(8.20)

From (8.19) and (8.20) , we find that , ifthere exists ωεJR4 such that g(ω) = 0 and ra出 (\7g(ω)) = 1, then

11111/

C

\飞

C 孚

= ~1~ω= 2 2

JR6 is given by

。白

乎

~~

ω

2

守J

C

nL

。

r

飞八 1if

σ各=一 ωι1

ω

中=丰〉

?b

wtMω=σ主

(8.19)

，

1=;山一11;

、‘ A气 ，

{=中

μ'

=1

一一

(8.18)

•

From (8.16) it follows that , if g(ω) = 0 , then l tw = 1 and wtMw 工作­ Usi吨 (8.18)， we find that l tw

飞八

where G : JR6

;M九

2-UM" M2p 111

\7 (ω，，\) F(ω7λ) = 0 ,

G(ω?λ1 ， λ2)

In order to use the Lagrange multipliers method for solvi吨 problem (8.14) , we first show that the matrix \7 g(ω) has rank 2 for any ωsuch that g(ω) = O. Note that ra此 (\7g(ω)) = 1 if and only if there exists a constant C εR such that 2!VIw = C 1. Using the fact that the covariance matrix !VI of the assets considered here is nonsingular , we obtain that w

/1 、、

飞， I/

/'41 、、

。"

'?b

\l/

直缸缸时 r1

\飞 III/

!ω

rM

F

ω

To /III-\

ω

(8.17)

飞A

2(Ax)t.

=

it is easy to see that \7g(ω)

/III-11\

节V

(去无)

(8.21)

The gradient of the Lagrangian is 十

Recall that , if the function h : JRn • JR is given by h(x) = xtAx , where A is an n xηsymmetric square matrix , then the gradient of h( x) is

Usi吨 (8.17) ，

185

山飞= 80.01 并 17.36 二 jf

(8.15)

g(ω)

Dh(x)

SOLUTIONS TO CHAPTER 8 EXERCISES

Wo

(om)

0.6450 0.6946 J' -0.3503

λ0 ， 1

=

-0.0738;λ0.2

= -0.8510.

Let 月 (ω) = F(ω7λ0 ， 1 ， λ0.2) ， i.e. , 几 (ω)

μtω - 0.0738(l t ω - 1) -

0.8510(ωtMω - (J'~).

186 CHAPTER 8. LAGRANGE 1VIULTIPLIERS. NEWTON'S METHOD.

8.1. SOLUTIONS TO CHAPTER 8 EXERCISES

The Hessian of Fa (ω) is

where B = 100 +去 and y = 0.03340 1. We obtain that the duration of the bond is 4.642735 and the convexity of the bond is 22.573118. 口

D 2 凡 (ω )

= -

0.8510· 2JvI

187

= - 1. 7019M.

Since the covariance matrix M of the rates of return of the four assets is a positive definite matrix , it follows that D 2 Fo (ω) is a negative defi时te matrix for any ω. Therefore , the associated quadratic form q( v) = v t D 2 Fo (ωo)v is negative definite , and so will be the reduced quadratic form corresponding to the linear constraints \7 g(ω。 ) v = O. We conclude that the point ωa = (0.0107 0.6450 0.6946 - 0.3503) is a constrained maximum for f (ω) given the constraints g(ω) = O. The portfolio with 24% standard deviation of the rate of return and maximal expected return 1.07% in the first asset , 64.50% in the second asset , 69 .46% in the third asset , while shorting an amount of asset four equal to 35.03% of the value of the portfolio. For example , if the value of the portfolio is $1 ,000 ,000 , then $350 ,285 of asset 4 is shorted (borrowed and sold for cash) , $10 ,715 is invested in asset 1 , $644 ,965 is invested in asset 2 , and $694 ,604 is invested in asset 3. This portfolio has an expected rate of return equal to 17.19%. 口

Problem 4: Recall that 缸ding the implied volatility from the given price of a call option is equivalent to solving the nonlinear problem f (x) = 0 , where

f(x) = Se- qT N(d 1 (x)) - Ke- rT N(d 2 (x)) - C and d1 (x)

一叫圣) (叫+手)T - ---\1'../ +xJr"':l r ,d

2

(x)

=

叫到十(叫一手)T xJT \1'>./

(i) Show that limx~∞ d 1 (x) = ∞ and li皿z→∞ d 2 (x) = 一∞， and conclude that lim f(x) = Se 一qT _ C. (ii) Show that {一∞，

ItIr! d 1 (x) = ItIr! d 2 (x) = { 、。

叭。

I

0,

∞，

if Se(r-q)T < K; if Se(r-q厅 K; if Se(r-q)T > K.

(Recall that F = Se(r-q)T is the forward price.) Conclude that Problem 3: Use Newton's method to find the yield of a five year semiannual coupon bond with 3.375% coupon rate and price 100 古 What are the duration and convexity of the bond? Solution: Nine $1. 6875 coupo丑 payments are made every six months , and a final payment of $101.6875 is made after 5 years. By writing the value of the bond in terms of its yield , we obtain that /IIiI\

,

时i

nu

42·A

吃$ 4

"U 1-9"

exp

十

寸t4

noQO

\飞 III/'

9ZM

vhυ 巧t

1-m 一-

寸t4

+

nunu

nhuQO vhu

exp

Uu

vhu

(8.22)

We solve the nonlinear equation (8.22) for y using Newton's method. With initial guess Xo = 0.1 , Newton's method converges in four iterations to the solution y = 0.03340 1. We conclude that the yield of the bond is 3.3401 %. The duration and convexity of the bond are given by

D

~ (飞J、 Z16吨 9 2 exp

B

·2

(iii) Show that

if Se(r-q)T

-C.

三 K;

> K.

f (x) is a strictly increasi吨 function and

-C < f(x) < Se- qT - C , if Se(r-q)T 三 K; Se- qT - Ke- rT - C < f(x) < Se- qT - C, if Se(叫)T > K. (扣) For what range of call option values does the problem f(x) = 0 have a positive solution? Compare your result to the range

Se- qT - Ke- rT

< C
0 , \:j x > 队 and f (x) is strictly increasi吨­ Recall that limx---t∞ f(x) = Se- q'1' - C and

(ii) Let F = Se(r-q)T be the forward price. From (8.23) and (8.24) it follows that d1 (x) and d2 (x) can be written as d1(x)

=仇一听「Le一手 7 飞/

J马 N(d 2 (x)) = 0

出 N(ddz))=laad

= vega(C)

189

xVT

xVT

• If F < K , then In (去) < 0 and

r

炖忡忡 1 Since

f (x) is strictly increasi吗，

-

we co叫ude

C < f(x)
K.

(iv) If F 三 K ， the problem f(x) = 0 has a solution x > 0 if and only if

Ther陀 e£扣 ore 且 I im♂\oN(何 出l(归 d x)) 工 1且im 八 也2(μ d x)) 工 Oa丑 x \ oN(伺

:炖咆 f 仲(归x)尸巳一 C

-C.

K;~rT

-C < f(x) < Se- qT

Se- qT - Ke- rT -

:坦白(z)zESd2(z)= 一∞

Se- qT -

。< C


K , the problem f(x) = 0 has a solution x > 0 if and only if Se- qT - Ke- rT

• If F = K , 出nd1(x)= 孚 and 也 (x) 口 一千， 缸aI丑侃圳 l叫d时 山阳re伽 缸 b f扣 ωr O臼