2.
CONTENTS
Vl1l
5
5.1
5.2
5.3
6
7
8
咱
E
4
4
1
4
4
l
4
Taylor ￥ formula.
Taylor series.
Solutions to Chapter 5 Exercises. .
Supplemental Exercises. . . .
Solutions to Supplemental Exercises.
吨
E
4
1
i
1
土
门
，
中
。
，
"
3346
Finite Differences. BlackScholes 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 selfstudy 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
3.
Acknowledgments
"A Primer for the Mathematics of Financial Engineering" published in April
2008 , is based on material covered in a mathematics refresher course I taught
to students entering the Financial Engineering Masters Program at Baruch
College.
Using the book as the primary text in the July 2008 refresher course was a
challenging but exceptionally rewarding experience. The students from the
2008 cohort of the Baruch WIFE Program who took that course were driven to
master the material and creatively incorporate ideas at an even deeper level
than that of the WIath Primer book. WIany of the supplemental questions in
the Solutions WIanual came about as a result of the remarkable interaction
I had with 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 YuanHuang , 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
4.
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)
5.
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 2η1)(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 )(η 十 1)
勾
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)
(iii) Compute T(η ， 3 ， x) = I:~=1 k 3 x k using the recursion formula
川J)24mμ))
(iv) Note that S(η ， 3) = T(η ， 3 ， 1). U~e I'Hopital's rule to evaluate T(η ， 3 ， 1) ，
and conclude that S(n , 3) = (呼叫"
(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
杜
t iate T(η ， 2 ，
Problem 3: Compute S(η ， 4) = I:~=1 k4 using the recursion formula
c
υ
飞
i
F
J
n
t
u
一
川
MZH +.3
/Il
/
l
I

l
η
+
十
吨
3
·
A
飞
.qb
才
t
A
咱
t
4
lit/
飞
1
1
I
/
QU η
qJ
、
飞
，
，
/
(9)
飞
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
(1x)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
一
1
.
付
=
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
'
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 η1)
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 =
Problem 4: It is easy to see that the sequence (xη) 吃 1 given by X n = I::~=l k2
satisfies the recursion
X n 十 (η +1? ， V η 三 1 ，
X n十 1 
with Xl = 1.
(i) By s由stituti吨
n
+ 1 for
/Ftt
ηin
4
1
·南
X n十 1
、
、
、
2X n +l 
Xn
，
/
,
i
4
'
'
'
飞
、
V η;三
1，
3X n +2  3X n +l
+ X n 十 2 ， V η;三
亏
1
品
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 
(13)
P(z)
Solve this recursion and show that
η(η
二=
十 1)(2η+ 1)
V η>
才
'
·
A
I
I
J
+ Xn

0, V n ~三1.
(16)
= 泸 4♂ + 6z 2  4z 十 1 =
(z1)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
(111
1 2 4
1 3 9
1 4 16
rec旧slOn
0, V n 二三 1.
4xη+1 十 X n
/iEvhu
、
The characteristic polynomial associated to the recursion (16) is
、
飞
，
，
/
(12)
1，
V n 二三 1.
By subtracting (13) from (15) , we find that
Xn
X n十 3
V n ~三1.
，
，
，
、
2η+3 ，
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
+ (η 十 2?
+
3X n +l 十 X n 十 2 ，
，
with Xl 二 1 and X2 = 5.
(ii) Similarly, show that
Xn
X n +4 =
nu
Subtract (10) from (11) to find that
X n +2

(iii) We substitute η 十 1 for n in (13) and obtain that
(10) , obtain that
X n +2 
3x n十 2
1 ) C 1 )  /f 1
8 I O2
5
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
7.
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 a时导L§. There£ore， 出h re exist ∞I归t
眨，
阳 白 t 阳盯
臼 f
扣
坠e
盯
c臼∞st
on S
丑阳
n
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)
=
z24z 十 3
=
(z1)(z3) ,
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tit1巾
扰
(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 near 陀
1i丑
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
8.
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)
Assume that λis a root of multiplicity 2 of P(z). Show that the seque丑ce
(Yn)吨。 given by
Yn = cηλ飞 η;三 0 ，
where C is an arbitrary consta风 satisfies the recursion (25).
It is easy to see that
Ih(x) + h(x) I /' 1~ YY> """J h(x) I J.. lim ~;mn Ih(x) I <'
IiTftplznl: liryhγ 十 brylznl<
三
n
and therefore、 by definition , h(x) + h(x) = O(x ) as x • 0.
Let gl(X) 二可♂) and g2(X) = o(俨 )asx • 0. The凡
Then , for
x•
I
风
O
~
v
、
飞
B
t
/
1i
1
n
Z
T
汇
M
2二 α以η+ i)λ叫
Cωη2 α川+刊 十 C2二二αiλη 十
汇二
4
汇 iω 俨
叫 肿
向
i.e. , show that
Cη λn
队
沪
队
汇二 α川十 C λ
2
ηω+
礼
1
η
均
C η λ P( λ)
l
d
寸
。(的，
o(x n ) + o(x n )
= 。但n) and
lims叫 一一一 II < ∞
DI h(x)
00
aT
I
俨
2z
z
叫n
→
1. 2
lims叩 μ17;1< ∞
n一→。C
and that
+ C λη叫+礼 P'( λ )
肿 1 从川
材
10
~~_1 k 2 一手
I

as
x
•
0;
o(x n ) , as x • 0.
h(x) =
O(俨) as
and
p
limsup
x • 0. Then ,
Ih(川
1 一一~I
x→0ι 俨 i
<∞
nZ?ZM
。(的，
。(♂)，
OJ
'也，
Similarly, prove that
口
> 0. Show that
。(♂) +O(♂)
m
•
骂
问
LZn
n →∞;
as
~n
ω
汇二 ωtλ•
2 叫
H 1
t俨
沪一」
In other words , the sequence (Yn)n?:.O satisfies the rec旧sion (25).
h(x)
十
仪
?十州， asη →∞?
n 一今汉二
Sol州on: Let
lilio
/ 川
!liio iilliliQU
n3飞
n
=
/
lim sup:::='凡J.
Problem 9: Let n
o I x ,. I
Problem 10: Prove that
anyη;三 0 ，
汇 αiYn+i
• 01 x
mmm
dMt
h
刃 +L blzpk
、J n
J
臼 ρU / e
l λ
O /吨 川
• nq liilinud
9 oa1
zd/飞 zB l
u
<
L一  …J
一
m
L
m叶
一
znE
Z
一
n叫 2
让
但 Z l
k
P'(z) = 汇 ωii
°
叫四 1=0 and 叫 92(~) I =
n
Z
gW
Sol毗on: Note thatλis a root of multiplicity 2 of P(z) if and only if P(λ)=0
and P'(λ) = 0 , where
9
0.1. SOLUTIONS TO CHAPTER 0 EXERCISES
as
n →∞;
Z 十州， asη →∞
(26)
(27)
Solution: Recall that
2二 k
2
中俨十 1) 删去 k 3
一
一
n
川
口
9.
10
MATHEMATICAL PRELI]l.ifINARIES
Then ,
0.2
lim
n→∞
""""η
L:~=1
 k=l k
2
ηU
7A2
η3
im L...tk=l 气一 τ
n→∞
η"
""",, n
1认 3
limι~k=l 内
二
n→∞
ητ
lim
1日1
L:~卢 一 τ
η4
k=l
n→∞
ηu
1
~… η(η+ 1)(2η+ 1)
η→∞
唱
6η3
3
3η2 十 η
<
.
.
…
二
『
η→∞
6η2
limd(η 十 1)2
η→∞
,.
l
4
2η 十 1
1
2
..…
n→∞
4η
<
∞;
Supplemental Exercises
α>
1. Let
0 be a positive number. Compute
Vα 十卢+~
∞;
<
一
<
11
0.2. SUPPLEMENTAL EXERCISES
∞;
2. Let
α> 0
be a positive number. Compute
1
00.
α 十一一，
α 十注工
We conclude that
η
Z
T
汇
?
γ
U
M
n
Z
M
。(的?
3. (i) Find x > 0 such that
as
ZZ
η →∞;
号 + O(n2 ) ,
as
as
T 十 O(η3) ，
η →∞;
as
η 一→ 00.
=
2.
(ii) Find the largest possible value of x > 0 with such that there exists
a number b > 0 with
η →∞;
Z♂
。(的 7
Z
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ttω η
剖仇
ion
仰仰
出肮
t hat
limit by l , ther1
口
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)
10.
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uenC e (Xηρ)η o 坦 ∞ 口阴飞
配阳 乌 哈三沪 i s c on
臼 ω
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 
where 均
臼 X 0= 飞 αand
v
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 肌削丑叫 山 时
∞
剑 a d t he ref
28)，
创
8
臼盯
伽
efl
inductior1 ， we 且 卫 tllat Xη < l for all η> O.
瓦
fiIIld 扭
Xo
(ii) The sequence (x r山三 o is increasi 吨·
since
Xn
<
X叫字=} Xn
<
vaτX n 仁丰 zizn 一
< land
Xn
> 0 for all n 三 O. vVe conclude that
η→∞川
α2 十 1
Xl
a3 + 2αα4 十 3α2+1
= 一一一一，
X2
= 一一一一:
XQ =
，1， 3α3 十 2α
α2α2 十 l'
1 十飞11 十 4α
ι
2
Xn
X2
< l <
X3
<
Xl·
 α 十一一一一=
α+ 县
lin
(α2 十 l)Yn + α
αYn 十 1
?
V n>07
(33)
with Yo = α. Note that Yn = X2n for all η 主 O.
Assume that Yn < l , where l is given by (32). Recall from (31) that
l2 .al  1 = 0
(34)
and that
h
川
一
t 2 一 αt
1
α + a~...
Yn+1
< X n十 1 for all
Problem 2: Let α> 0 be a positive number. Compute
α 十一一一寸
<
To show this , let (y山
We showed that the seque口ce (xn)n~O is increasi吨 and bounded from
above. We conclude that the seque口ce is convergent. Therefore、 the limit
l=li问→00 Xn satisfies the equation l = vaτI and is given by (29) , i. e. ,
,..,
n
Based on this observation , we co时 ecture that the subsequence (X2n)吃O
made of the even terms of the sequence (xn)n~O is increasi 吨 and has limit
equal to l , and that the subseq旧lce (X2附 1) 吃o made of the odd terms of
(X n ) 吨。 is decreasing and has limit equal to l.
(xn _ 1+ V;+石) (x n  lf石)
lim X ，..，
1m x
= α，
Xo
川) (xn 十盯了一 1)
since Xn
n> O.
α+17V n 三 O.
X
few terms of the seque丑ce (xn)n三o are
α< 0 ,
> O. Note that
2:zn 一 α

We note that the terms of the sequence are alternatively larger and smaller
than the value of l given by (32) , i. e. ,
It is easy to see that
Xn
(31)
l= α+va可2.
(i) The sequence (xn )吃o is bounded from above bv l.
Note that Xo =
< l. If we assume that Xn < l ， 白en
va丰百 <
1 = 0,
(32)
2
To show that the continuous fraction (30) does have a limit , we must
prove that the sequence (xn)n主o is converge 风 where Xo = αand
seq阳
X n +1 
13
(30)
1
(t a+vt+
4
)
(t 一 αF)
(35)
11.
14
1VIATHE1VIATICAL PRELI1VIINARIES
We will show that , for allη;三 0 ， Yn十1 >νnand Yn+1 < l.
Note that , by de且nition (33) , Yn > 0 for all η 三 O. Then , from (33) , it is
easy to see that
Yn+1 > Yn 仁斗 (α2 十 I)Yn 十 α>αd+Uη 仁斗 α(ν; 一 αYn  1) < O. (36)
From (35) , and using the assumptio且 that Yn < l , it follows that
Yn+1 <
with Zo = 年It is easy to see that Zn = X2叫 for all n 三 0
As expec'ted , the sequence (zn)n三o is decreasing and h部 limit equal to l.
The proof follows by induction: assuming that Zn > 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4a
Eom(36)and(37)pwe comiude tht7if yn<L ttm un+1
O.
From (33) , we also 在nd that
>
(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 7η
阳
仇
1
by (32) , i.e. , that
一
lim X2n 二 l.
(39)
n一→ α2
lim Xn im Xn
一 (α2 + l)zn + α
αZn 十 1
' V η>0 ,
, .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l
aI
丑
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 = limnt∞ 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<αlYn 十 J 白 Yn < α217L1 = l; (指)
l α
15
n一→汉二
Y~ 一 αun1=(unJ)(r2)<O
η>
0.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
l =2l/27
which is equivalent to
ll/l = 21/ 2 .
(43)
12.
16
lVIATHElVIATICAL
PRELIMINARIES
Then
f' (t) = ~卜勺丁1 (t)
77
77?γt;J
一俨!?于
叫
Chapter 1
ex
~?te that the function f (t) is i肌reasi吨 for t < e and decreasing for t > e.
Therefore?there will be two values of t such ttlat(43)is satis五ed， i.e. , s时l
that tl/t=21/27OIIe value being equal t027and the other OIle greater tham
Since zn<2for a!1n 主 o and l = limn→∞ x n ， we co口clude that l = 2 , 缸1
a口
n
therefore that x = 飞/2 is the solution to (41).
(ii) If there exists a number b > 0 such that
xxx·

Calculus review. Plain vanilla options.
1.1
Solutions to Chapter 1 Exercises
Problem 1: Compute
J ln(x) dx.
Solution: Using integration by parts , we find that
b
for a given x > 0 , then x b = b and the时·ore x = b叭 Recall from part (i)
that the functiozlfO)=tl/t has &Il absolute maximum at t =e.We cOBelude
that
Z 二 b 1 /b 三 maxt1 / t _ e 1 /e 自 1 .4447，
JIn仪
户
f 何叫川 dx
伽

川巾刷(仪例 一 f 叭呻巾喇
x 闻阳峭 户 倒川
In叫 μZ以
叫 叫
n Z)
(In(
In
叫
n
川ln( Z
Z忡
巾叫
叫
n 巾(仪仲
t>O
and that the largest value of b such that the limit (42) exists is b = e.
口
Problem 2: Compute
J x击x) dx by using the substitution u = ln( x)
Solution: Let u = ln(x). Then du = 警 and therefore
JXl~(X)
xln(x) dx =

J~
J
du =
ln(lull
=川x)l) + C 口
Problem 3: Show that (ta川 )'=1ν/(C Oωsx )2 a丑
阳lX
缸
an 叫
巾 叫
∞
J1: 丁训dx
l+x

町肌阳I
阳叫
ctω
ta
缸
a
Solution: Using the Quotient Rule , we find that
(tan x)'
(:25)f=(sinx阳 x  sin x (cos x)'
(COSX)2
(COSX)2 + ( s i n x ) 2 l
(COSX)2
(cosx)2
r
'
'
'、
1
45i
才
t
A
、
、
，
，
，
，
13.
CHAPTER 1. CALCULUS REVIEW. PLAIN ~生NILLA OPTIONS.
18
To prove that
J 1;'x dx =
2
arctan( x) , we will show that
1.1. SOLUTIONS TO CHAPTER 1 EXERCISES
Problem 4: Use I'H6pital's rule to show that the following two Taylor approximations hold when x is close to 0:
(arctan(训 =Jτ
Jτ王旬 l+j;
i 十 x~
Let f(叫= tan x. Then arcta叫x) = fI(x). Recall that
(f I (x))1 =
♂自 1 + x 十二
J
•
In other words , show that the following limits exist and are constant:
and note that l' (x) 二 (tanx)' =讨x)2; cf. (1. 1). Therefore ,
(arcta口 (x))' = (COS(fI(X))? = (cos(arctan(x)))2.
mJ百三(1 +~)and liJ(lfz呼)
(1. 2)
X
Let α= arctan(x). Then tan(α)=ι It is easy to see that
•o
x'::;
X
•o
x
Solution: The numerator and denominator of each limit are differentiated
until a finite limit is computed. L' H6pital's rule can then be applied sequentially to obtain the value of the initialli皿it:
x 2 十 1=__1
(cos(α))2'
since (sin(α))2 十 (cos(α))2 = 1. Thus ,
(cos(arcta口 (x)))2 = (co巾)?
19
1;~ dτ王一 (1 +~)_ 1;…对古 i
l
汇;6
(1. 3)
x2 十 1
1
8
lim
m 0
•
..…一
♂巧。
x2
2x
vVe conclude that
From (1. 2) and (1. 3) , we conclude that
dτzz l+i+0(泸)，
(arcta叫 x))'
as
x
•
0
Similarly,
and therefore that
f 市 dx 二 arctan(x)
eX  (1 十 Z 十豆}
+ C
lim
•o
X
We note that tke antiderivative of a rational fuzlctiOII is often computed
using the Sl阳titution x = tan (~)
则:12Eample?to compm j 市 dx using the substitution x = 阳(~)，
dx  ; ( tan ( ~ ) ) dz x = ~ (tan (~飞dz 2)) 山
eX 一 (1 十 Z 十二)
+
1
x

(∞S(~))2
1
(sin(α))2+(COS(α))22(cos(5))2u
;
X
•o
h
十二十 O(内部户。口
吊
'
山
1
l
品
v
=
~X
lim 二
Problem 5: Use the definition e = limxtχ(1 + ~)X of e to show that
/IIiI 1z
1llI/
1e
1IZ
m
说
11+ 山~))2 2(∞J(5))2dz
I~ 命
dz
x
lim eX  1
1日1 ox
m o 一一一一
•
/ ~ '" . . . dz.
Then
f
lim eX 一 (1 + x)
z• 6
3x 2
、/
=可 n巾(归例 十
叫 叫
a 叫 Z)
C
口
Solution: Note that
1
1 一
Z
xI
Z
1
1
z
1+ 击
xI
l
6'
14.
20
CHAPTER 1. CALCULUS REVIEW. PLAIN VANILLA OPTIONS.
The口?
m
切
·
1i3
唱
E
A
町
l
hhkh
n
一
I
∞
十
Z
)
十
n
∞
口
1
Iiil/
lz
/
J
I
I
飞、
♂
才
Z
A
Sl日 ce
Solutio 衍 Let F (x) = J f (x) dx be the antiderivative of f (x). From the
Fundamental Theorem of Calculus , it follows that
Zz
IH
1
e
fα +h
F(α + h)  F( α  h)
f(x) dx =
n~_ I
2fz
2h Jαh
Using I'H6pital's rule and the fact that F'(x) = f(x) , we find that
h•
1
lim1+__. _ 1
z→∞
fα+h
n~
62h
I
J α h
F
II!
h• 6
f (x) dx =
, ,

xI
lim
h•
and
J马 (1 +击r一 =J414♂
e.
口
~左 l(X)
七
i 言 叫协刷(仰例♂叫
V)
=
since
f (x)
Note: Let F(t) =
F'( α) is
J~ f(x) dx.
F'(α)
F( α +
I
'""
、
飞
4
，
/
I
l
I
/
、
、
Therefore ,
g'(x)
f(b(x))b'(x)  f( α (x ))α '(x).
υ
V
飞12 7f T
h)
2
F( α 
.L
I~)
'""
h)
I~)
+ 0
(h
2
)
(1. 4)
,
Problem 8: Let
f : lR •lR
h
f(x) dx + O(h 2 ).
口
given by
的) = LCi e呐?
Lp 一鸣垣 . J/ ( 川一
1
h) 
I' α 十 h
= i二 I
2h J α
'
'
而

h)
as h • o (if F( 剖 (t) = fll(t) is contin∞叫. Since F'( α) = f( α) ， formula (1. 4)
can be written as
飞
Vd
UU
o
2h
1
'
o
f

2h
α + h) + f( α
f(
The central finite difference approximation of
.L
=
f(α)
Recall that
/
ftm
d
31j
ρ
I
I
'
rJ
l
'
一'
们
'
u
飞
门'
u
ω
、
凶'
F( α 
is a continuous function.
阳
wh
1
刮
切
Solut4 on:
h) 
F( α +
f( α) ，
Problem 6: Let K , T ， σand r be positive constants , and define the function
g : lR •lR as
g(x)
21
1.1. SOLUTIONS TO CHAPTER 1 EXERCISES
eXD
.L
1
凸O~
(
(b(x))2 飞
川y12在叭叭一←厂/
(一
where
xcrVT
Ci
Solution: Note that
yti
( e )'
(In (去)十 (r 十号 )T 门内
2σ 2T
I'
are
and ti , i = 1 :叽
/
e
~
l
t
飞
们
u
u
'
'
ι
、
、
E
，
，
J
positive constants. Compute
d
一
句d
一
句
J
'
a
E飞
、
、
/
，
，
‘
‘
、
、
e
U·t
t , 一
H 1
一
ρ 1
U /
八
，
NU
?b
/
一
tteuta;
AFU
一
=t?eUti
The丑 7
n
Problem 7: Let f(x) be a contin∞ us function. Show that
出去 ffz 川
f'(y) =  L Citi e一队;
i=l
n
f"(y) = LCi 中 yti
i=l
口
f' (y)
and
f" (y).
15.
22
CHAPTER 1. CALCULUS REVIEW. PLAIN VANILLA OPTIONS.
Note:The fmction f(u)represents the price ofa bond with caskHOWS Ci paid
at time tt as a function of the yield Uof tfle bond.WheIlscaled appropztate1377
the 丑rst and second derivatives of f (y) 吼叫1 respect to y give the duration
and convexity of the bond , respectively.
Problem 9: Let f : JR3
x = (Xl , X2 , X3).
•
(ii) W~ substitute the 叫ues from (1. 6) , (1. 7) and (1. 8) for f( α) ， ，Dt(α)
~nd D 2f(α) ， respectively, in the expression f (α) + Df(α) (x 一 α) +
(x 一
1
α)t D 2 f(α) (x 一 α)
and obtain that
f(α) +Df(α) (x 一 α)+1(z 一 α)t
2
JR given by f(x) = 2xr  XIX2 十 3X2均  x~ ， where
3
(i) Compute the gradient and Hessian of the fur时ion f (x) at the point α=
(1 , 1 , 0) , i.e. , compute Df(l , 1 , 0) and D2f(1， 一 1 ， 0).
(ii) Show that
+
I
X3
/
一
一
i
q
h
M
η
ο
α
o
…
0 飞/
1
Xl 
1 飞
J
3 2
飞
J
X3
3 + (5Xl  X2  3X3 6)
+ (2xi  5Xl  XIX2 + X2 十 3X2X3 + 3X3  x~ 十 3)
…
=
Z
2xi  x向 +3x仇 zi
//iz
//11111
Illi/
i1inu
ZZZ
τ
4
寸仙一 1 ， X2 十 LZ3)i103ilh+1)
Here，叽 α ， and x 一 αare 3 x 1 column vectors , i. e. ,
Z
(x 一 α)
/ Xl  1
X2 + 1 J
飞
刷工川的 +Df(仲
It/''
D 2f(α)
I
(5 , 1 , 3)
飞
/
J
I
l
飞
23
1. 1 SOLUTIONS TO CHAPTER 1 EXERCISES
ZZ
1in
一
α
吁
飞
t
牛
,," 均
·
口
I
l
1
十
一
一
f(x) ，
才
l
·
/

Problem 10: Let
Note: Formul~ (1. 5)_ is the quadratic Taylor approximation of f(x) around
the pointαSin~e_!(x) is a second order polynomial , the quadr~tic Taylor
approximation of f(x) is exact.
,
Solution: (i) Recall that
Df(x)
飞/4 时'
for t
//11111
A
I=
吐
1
i
n
u
才
i
n
U
Q
d
n
u
q
气u
11111/
n
L
Note: This exercise shows that the function u(x , t) is a solution of the heat
equation. In fact , u(x , t) is the fundamental solution of the heat equation ,
and is used in the PDE deri飞ration of the BlackScholes formula for pricing
European plain vanilla options.
Also , note that u(x , t) is the same as the density function of a normal
variable with mean 0 and variance 2t.
θu
Df(α)
D 2 f(α)
ξJR.
絮 and g主， and show that
Solution: By direct computation and using the
Then ,
f(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
_tu/~
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)
16.
24
CHAPTER 1. CALCULUS REVIE1iV. PLAIN VANILLA OPTIONS.
1
1
二
二
δx
二
币
From (1. 9) and (1. 10) , we ∞ 叫 时 e 由 创
c onc
t
比
i cll
t hat
制
θ2 U
/
t
t
‘
、
2
θ2 U
θ t'
nu
••
4t4
θu
θx 2
才
1
4
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
,,,,,,
< 30 30 < S(T) < 35 35 < S T) < 40 40 < S(T)
。
S(T)  30
。
。
O
O
mm
黯
25
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.e7plot the value of the
portfolio V (T ) at matu y ity as a funct ion of S (T )7t he price of t he m d e rl抖
臼创y
坊刷
Jrm
臼 创 at
S se t 肮 time T.
C1 (T)
1.1. SOLUTIONS TO CHAPTER 1 EXERCISES
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
17.
26
CHAPTER 1. CALCULUS REVIE1iV. PLAIN VANILLA OPTIONS.
rather aomegative)payof M maturityTtmrefore7if you are to α ss包 η
u阳
ωn
7
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)
。 
Problem 14: Call options with strikes 100 , 120 , an d 130 on the sam under丑
丑
le
凶
时 呐由 出
yin g ass们E wit ht h e same 肮刷I
盯 缸丑
机切
lemat u
吨
江哟
牛
l
(there is no bidask spread). Is there an arbitrage opportunity prese时? If
yes , how can you make a riskless pro缸?
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
X1 C 1(0)
十 X2 C2(0) 十 X3 C3(0)
At maturity T , the value of the portfolio will be
V(T)
TiT
Z1K1
X2K2
X2 K 2  X3K3
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
x2)S(T) 一 X1K1
+ 3Z22S
 x 1K 1 (Xl 十 X2 + x3)S(T)
(Xl
i2 (T)
max(S(T) 一 (X
V(O)
27
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} g asset at maturity the
也
,
value V(T) of the portfolio is as follows:
+ X2C2(0) 十 X3 C3(0)
Xl
十 x2)K3  X1 K 1  X2 K 2
(Xl
Xl + X2 十 X3
〈
〉
一
〉
一
>
一
。;
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
。;
(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
18.
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 atthemoney put and call options on this asset
are trading at $2 and $4 , respectively. The riskfree rate is constant and
equal to 5% for all times. Show that the PutCall 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 PutCall parity is not satisfied , since
P + Se
=  P(T)  S(T) + C(T) + 39.5821erT .
Recall from the proof of the PutCall 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 PutCall 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 riskfree profit made by exploiting the discrepancy from the PutCall parity, and is the same as the future value at time T
of the misprici吨 from the PutCall parity; d. (1. 15).
口
Problem 16: The bid and ask prices for a six months European call option
with strike 40 on a nondividendpaying 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 PutCall 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 riskfree 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 riskfree 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
19.
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 atthemoney 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 arbitragefree 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
20.
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
Z
/it lz
寸
土
/IIlt
+
飞
<
l
l
I
e
<
/
寸
'
·
·
·
·

5. Let
f(x)
十
z
11I/
1z+
吨
E
在
]
V Z
(
= 一古=exp(σ 〉习作
0 三 S(T) < 20;
20 三 S(T) < 40;
40 三 S(T) ，
9. Call options on the same underlying asset and with the same maturity,
with strikes K 1 < K 2 < K 3 , are trading for C1 , C2 and C3 , respectively
(no BidAsk spread) , with C1 > C2 > C3 . Find necessary and sufficie时
conditions on the prices C 1 , C 2 and C 3 such that noarbitrage exists
corresponding to a portfolio made of positions in the three options.
J
4. Show that
2S(T) , if
(
< 60  S(T) , if
l S(T)  20 , if
where S(T) is the spot price at time T of a given asset. Use plain
vanilla options with maturity T as well as cash positions and positions
in the asset itself. Assume , for simplicity, that the asset does not pay
dividends and that interest rates are zero.
J
3. Compute
33
>
才
t
4
C bid and C，αsk ， and by 1毛id and ~αsk ， respectively, the bid
and ask prices for a plain vanilla European call and for a plain vanilla
European put option , both with the same strike K and maturity T , and
0日出e same underlying asset with spot price S and paying dividends
continuously at rate q. Assume that the riskfree interest rates are constant equal to r. Find necessary and sufficient noarbitrage conditions
for Cbid , Cask , Pbid , and 凡sk·
10. Denote by
(x 一 μ)2
2σ2
i
)
Assume that g : JR • JR is a continuous function which is uniformly
bounded , i.e. , there exists a constant C such that jg( x) I 三 C for all
x E JR. Then , show that
M: 削川x
=
g(μ)
1.3
Solutions to Supplemental Exercises
Problem 1: Compute
J
n
x ln(x) dx
6. Let
内)工艺
Ci
Solution: If n 乒 1 ， we use integration by parts and find that
Compute g' (y).
f zdx = 一Mz)j zn+1·dz
x n ln(x) n + l l f
η + 1 η十 1)
7. A derivative security pays a cash amount C if the spot price of the
underlying asset at maturity is between K 1 and K 2 , where 0 < K 1 < K 2 ,
and expires worthless otherwise. How do you synthesize this derivative
security (i.e. , how do you recreate its payoff almost exactly) using plain
vanilla call options?
n十
x 11n (吟
For n
= 一1，
x
n十 1
、A
+ C.
we obtain that
rn~x) dx
一 WJ
= (ln(x)?
+
C.
口
21.
34
CHAPTER 1. CALCULUS REVIEW. PLAIN VANILLA OPTIONS.
P VAO 'D em
Problem 2: Compute
''EA
/♂♂ dx
Solution:
For every ir由ger n 主 0 ， define the function
fn(x)
h
QU σ
A
I且
/
I
i
l
民
才
U
tliI/
alZ
z
i
<
Sol仰 on:
1
儿(x) = j ♂eX dx = xneX  n j ♂1川 = xne x 一吼一 1 协 Vn;:::l
,
J
nZM
LfM
/III11
一
n
z
一

门
飞
吁
i
Vv Z >
才
i
(1. 16)
(11
< 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
fIJ
4tA
Z 十 1
By using integration by parts , it is easy to see that
G
z neZ , 2
吨
'
4
十
<
pu
Note that (1. 16) is equivalent to
一二一
PIt
Z
111/
1z+
/I111
十
as
儿 (x) 工 j xne x dx
Since fi。但)
35
1.3. SOLUTIONS TO SUPPLE1VIENTAL EXERCISES
η
'
川h
h
l
l
l
/
♂
pu
十
c
vv n >
j' (x)
414
口
g'(x)
1
一一一+.
2
x
I
X
1
一
(x + 1) 
L一十
x(x 十 1)
I
1
(x + 1)2
x2 (x + 1) < 0:
~ ,

1< 0
x(x + 1)2
We conclude that bo也 f (x) and g( x) are decreasing functions. Since
Problem 3: Compute
lim
j(1n(x))η dx
X 一→汉二
it follows that
Solution:
f(x) =
lim
X→。c
f(x) > 0 and g(x) > 0 for
For every integerη 三 0 ， let
g(x) = 0 ,
all
x > 0, and
therefore
:>咛+:)〉 ZLI7VZ>07
儿 (x) 工 j (In (x) )n dx
which is what we wanted to show; cf. (1. 17).
口
By using integration by parts , it is easy to see that , for any η;三 1 ，
Problem 5: Let
户口(x)t dx = 川)川 j (In(x))η→
仲) =忐叫一(工
f)2)
and therefore
fn(X ) = x(I n ( x))η 一 η fn1(♂)，
V η 主1.
Since fo(x) = 凯 the following general formula can be obtained by ir吐uction:
j (1卫 (x))n dx = fn(x) = x 主
L
(γk n !
.L ~nl
I f,;
(ln(x))k 十 0，
Vn 主1.口
Assume that g : R •ffi. is a continuous function which is uniformly bounded 2 ,
i.e. , there exists a constant 0 such that Ig(x)1 三 o for all x ε ffi.. Then , show
that
叫:削g(x) dx = 仲)
2The uniform boundedness condition was chosen for simplicity: and it can be l'吐axed:
e.g.: to functions which have polynomial growth at infinity.
22.
CHAPTER 1. CALCULUS REVIE 1iV. PLAIN VANILLA OPTIONS.
36
Solution: Using the change of variables y = 乎?附在nd that
汇 f(x)g(x) dx
志汇 g(x)
exp
(气;)2)
去/∞ λ dy
V
kI/I
1
→是=
dx
ο
(1. 19)
since，吗， the function 在e15is tue p时ability density function of the
standard normal variable. From (1. 18) and (1. 19) we obtain that
f∞
1
r∞
f(x)g(x) dx = 一)
(g(μ)  g(μ 十 σy)) e 号 dy.
j一∞
d订一∞

/
(1. 20)
Our goal is to show that the right hand side of (1. 20) goes to 0 as σ、 O.
Since g( x) is a continuous function , it follows that , for 创lY ε> 0 , there
exists 61 (ε) > 0 such that
Ig(μ )g(x)1 < 飞
for
V Ix 一 μ1< 61(ε) .
「LfdM)e4dy 十气: (∞片 dy < ε
J 一∞飞 !"L，1f
Since Ig( x) I 三 C for all x
r 62(£)
1
→头=
v L,1f
I
J 一∞
E
Ig(μ)  g(μ+ 句)1 e一专 dy
J 62(£)
<
2Cε
rlu
z nud z
7
α
Z
沉
沉
<
一
〈
nu
μ
'
μ
'
η
3
2ML2
十
σ
uu
e
，
α
们
u
u
Problem 6: Let
g(y)
ZJ
Compute g' (y).
'(y)
= 一安
告 (1
Citi
+ y)t i
十1
(1. 23)
I(μ+ 句)一 μ1=σIyl <61(ε) 一一::; 61 (ε) ， V y ε[62(ε) ， 62(E)].
62(ε) 一
(1. 21) and (1. 24)
we 自nd
g(μ+σy)1
(1. 24)
that
< 飞
Problem 7: A derivative security pays a cash amount C if the spot price
of the underlying asset at maturity is between K 1 and K 2 , where 0 < K 1 <
K 2 , and expires worthless otherwise. How do you synthesize this derivative
security (i.e. , how do you recreate its payoff almost exactly) usi吨 plai川anilla
call options?
Solution: The payoff of the derivative security is
Iyl
Ig(μ)
(1. 26)
JR., it follows from (1. 22) that
It is 叫 to s叫风 if () <去87then
The凡 from
∞
J 62(£)
+τ: (∞ Ig(μ)  g(μ 叫)1 汁 dy
v ζπ
μ
'
Solution:
(1. 22)
n
i
l
←」
1
一
L
陀
一
广
We conclude , by definition , that 份
仰
叫:f(z)g(Mzg(μ) 口
门
时M
(1. 21 )
Using the fact that the integral (1. 19) exists and is finite , we obtain that ,
anyε> 0 , there exists 62 (ε) > 0 such that
V L/Tr
Ig(μ)  g(μ+σν) e一号 dy < ε
From (1. 20) , (1. 23) , and (1. 26) , it follows that , for anyξ> 0 , there exist
61(E) > 0 and 62(E) > 0 such that , if σ<:237then
∞
flL
= 1,
r 62(ε)
I
v L,1f J 62(£)
JOO
g(μ)  /
37
and therefore
vi言汇 g(μ+ 仙句
Recall that
1. 3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
Vy ε[62(E)， 62(ε)]，
(1. 25)
/
B
E
E
f
飞
nucnu
ι
<ss
阿 气/
m < K
凡 〈
<
w
ι 一
m
"
Since V(T) is 出 ∞ 挝t inuo 盹 i t cannot be replicated exactly usi吨 call 址
d isc on挝
叫
让
抖 w创
岳
丑
lopt10口
whose p a;句 offs are contirlUaus.
~
丑
V T
一

ρ
·
?
i
p
?
i
p
?
·晶
·
门
，
，
"
23.
38
CHAPTER 1. CALCULUS REVIE 1iV. PLAIN
v:生NILLA
OPTIONS.
We approxi口时 e the payoff V (T) of the derivati飞肘ec盯ity by the followi吨
payoff
0,
(
I
~(T)
= <
c(S(T) 一 (K1
I
l
c,

if S(T) < K1 一 ε;
E))/ε ， if K1 一 ε 三 S(T) 三 K1 ;
if K 1 < S(T) < K 2 ;
if K2 三 S(T) 三 K2 + ε;
if K2 十 ε < S(T).
c  c(S(T)  K2 )/ε ，
0,
(1. 27)
< K1
~(T)
一 ε
。
K21
二
15εS 三十
KK1
K 2 ~二
K2 + ε < S(T)
ε
(T
;(S(T)
c
•
(fKC
(KS12
;(S(T)
+ 60 
33(T)
= 2S(T)  3(3(T)  20).
This is equivalent to a long position in two units of the underlying asset and
a short position in three calls with strike 20.
To replicate the payoff S(T)  20 of the portfolio when 40 三 S(T) ， note
that
S(T)20
=
603(T) 十 2S(T)80
= 2S(T)  3(3(T)20) + 2(S(T)40).
This is equivalent to a long position in two units of the underlying asset , a
short position in three calls with strike 20 , and a long position in two calls
with strike 40.
Summarizing , the replicating portfolio is made of
• long two units of the asset;
• short 3 call options with strike K = 20 on the asset;
• long 2 call options with strike K = 40 on the asset.
We check that the payoff of this portfolio at maturity, i.e. ,
vi (T) = 23(T)  3 max(S(T)  20 , 0) + 2 max(S(T)  40 , 0)
(1S)K(T20))() 2 1))=C
K+
)EQ+()S)f ((SK)E
(T
ε))·;;0
(T
K
39
One way is to use the underlying asset , calls with strike 20 , and calls with
strike 40.
First of all , a portfolio with a long position in two units of the underlying
asset has value 2S(T) at maturity, when S(T) < 20.
To replicate the payoff 60  S(T) of the portfolio when 20 三 3(T) < 40 ,
note that
60  3(T) = 2S(T)
Note that V(T) = ~(T) unless the value S(T) of the underlyi吨 asset at
maturity is either between K1 一 ε and K 1 , or between K 2 and K2 十 ε.
The payoff ~(T) can be realized by goi吨 long c/εbull spreads with
strikes K1 一 ε and K 1 , and shorti吨 c/εbull spreads with strikes K 2 创
an
已
K2 十 ε . In other words , the payoff V(T) of the given derivative security can
be synthesized by taking the following positions:
• long c/εcalls with strike K1 一 ε;
• short c/εcalls with strike K 1 ;
• short c/εcalls with strike K 2 ;
• 10吨 c/εcalls with strike K2 十 εIt is easy to see that the payoff ~(T) is the same as in (1. 27):
S(T)
1. 3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
(1. 29)
is the same as the payoff from (1.28):
同 (T)
23(T)
3(T) S 20
2S(T)=3(S(T)  20) = 60 3(T)
20 < S(T 三 40
否){
60  S(T) + 2(S(T)  40) = S(T)  20
40 < S(T)
Problem 8: Create a portfolio with the following payoff at time T:
V(T)
r 23(T) , if
= < 60  S(T) , if
l S(T)  20 , if
0 三 3(T) < 20;
20 三 S(T) < 40;
40 三 S(T) ，
(1. 28)
where S(T) is the spot price at time T of a given asset. Use plai丑 vanilla
options with maturity T as well as cash positions and positions in the asset
itself. Assume , for simplicity, that the asset does not pay dividends and that
interest rates are zero.
Solution: Using plain vanilla options , cash , and the underlying asset the
payoff V(T) can be replicated in different ways.
As a consequence of the PutCall parity, it follows that the payoff V(T)
from (1. 28) can also be sy毗lesized using put options. If the asset does 口ot
pay dividends and if interest rates are zero , then , from the PutCall parity,
it follows that
C = P+SK.
Denote by C20 and P20 , and by 0 40 and 凡0 ， the values of the call and put
options with strikes 20 and 40 , respectively.
Then , the replicating portfolio with payoff at maturity given by (1. 29)
can be written as
V  23  3020 + 2040 .
(1. 30)
24.
40
CHAPTER 1. CALCULUS REVIE1iV. PLAIN VANILLA OPTIONS.
To synthesize a short position in three calls with strike 20 , note that
1.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
41
which is equivalent to taking a short position in three units of the underlying
asset , taking a short position in three put options with strike 20 , and being
a long $60.
Similarly, to synthesize a long position in two calls with strike 40 , note
that
2040
2凡。十 28  80 ,
(1. 32)
• short 3 call options with strike K = 20 on the asset;
• long 2 call options with strike K = 40 on the asset.
The second replicating portfolio will be made of the following positions:
• long e qT units of the asset;
• short $20e rT cash;
• short 3 put options with strike K = 20 on the asset;
• long 2 put options with strike K = 40 on the asset.
Note that any piecewise linear payoff of a single asset can be synthesized ,
in theory, by using plai丑 vanilla options , cash and asset positions. 口
which is equivalent to a borrowing $80 , taking a long position in two units
of the underlying asset , and taking a long position in two put options with
strike 40.
Usi吨(1. 31) and (1. 32) , we obtain that the payoff at maturity given by
(1. 29) can be replicated usi吨 the followi吨 portfolio consisti吨。f put options ,
cash , and the underlying asset:
Problem 9: Call options on the same underlying asset and with the same
maturity, with strikes K 1 < K 2 < K :3' are trading for 0 1, O2 and 0 3, respecti飞吨T (∞ BidAsk spread) , with C1 > O2 > 0 3 Find necessary and
sufficient conditions on the prices 0 1 , O2 and 0 :3 such that noarbitrage exists corresponding to a portfolio made of positions in the three options.
一 3020
V

 3P20  38 + 60 ,
(1. 31)
3020 十 2040
二
28 28 
=
83马。十 2凡。一 20.
3马。一 38 +60 十 2凡。+
28  80
(1. 33)
The positions of the replicati吨 portfolio (1. 33) can be summarized as follows:
• long one unit of the asset;
• short $20 cash;
• short 3 put options with strike K = 20 on the asset;
• long 2 put options with strike K = 40 0丑 the asset.
We check that the payoff of this portfolio at maturity, i.e. ,
巧 (T)
= 8(T)  20  3 max(20  8(T) , 0) + 2 max(40  8(T) , 0)
is the same as the payoff from (1. 28):
2日
8(T ) ~ 20
页
B(T) 3(20
8(T)  20 十
20 < 8(T) 三 40
40 < 8(T)
{4S ST)T2川 S(T))~
页C到TL f芦邪o= 60  8(T)28(T)
日
8(T))
2(40
If the asset pays dividends continuously at rate q and if interest rates are
constant and equal to 飞 in order to obtain the same payoffs at maturity, the
asset positions in the two portfolios must be adjusted as follows:
The first replicating portfolio will be made of the following positions:
• long 2e qT units of the asset;
,
Solution: An arbitrage exists if and only if a nocost portfolio can be set up
with nonnegative payoff at maturity regardless of the price of the underlying
asset at maturity, and such that the probability of a strictly positive payoff
is greater than O.
Consider a portfolio made of positions in the three options with value 0
at inception , and let Xi > 0 be the size of the portfolio position in the option
with strike K i , for i = 1 : 3. Let 8 = 8(T) be the value of the underlyin
臼创 凶
坊
asset at maturity. For noarbitrage to occur , there are three possibilities:
Portfolio 1:Long the K 1option , short the K2 一 option， long the K 3option.
Arbitrage exists if we can find Xi > 0 , i = 1 : 3 , such that
X1 0 1  X2 02 十 X303 = 0;
(1. 34)
x1(8  K 1)  x2(8  K2 ) 十句 (8  K3 ) 三 0 ， V 8 主 O.
(1. 35)
We note that (1. 35) holds if and only if the following two conditions are
satisfied:
Xl  X2 十 Z32三 0;
(1. 36)
x1(K3  K 1)  x2(K3  K 2) ~ O.
(1. 37)
vVe solve (1. 34) for X3 and obtain
O2
Since we assumed that X3
01
03
.."'03
(1. 38)
> 0, the following condition must also be satisfied:
X2
01
一，一
句
O2
(1. 39)
25.
42
CHAPTER 1. CALCULUS REVIE1V. PLAIN VANILLA OPTIONS.
Recall that C1 > C2 > C 3 . Using the value of
that (1. 36) and (1. 37) hold true if and 0日ly if
C1 
X2
、
Xl
… C2 
X3
from (1. 38) , it follows
C3
(1. 40)
、
C3 '
X 2 K 3  K1
Xl

K3

(1. 41)
K2
Also , note that if (1 .40) holds true , then (1. 39) is satisfied as well , since
C1  C3
C2  C3
C1
C
二

C1
、
C3
C2 C3

(1 .4 2)
Portfolio 2:Long the K 1option , short the K 2option , short the
Arbitrage exists if we can 且口d Xi > 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 <
X2 一二 X'}. 一二
v
1
1
43
x1(8  K 1 ) + x2(8  K2 ) 一句 (8  K3 ) 三 0 ， V 8 三 O.
The inequality (1 .48) holds if and only if
+ X2
(1. 48)
 X3 主 O.
(1 .49)
It is easy to see that (1. 47) and (1. 49) cannot be simultaneously satisfied:
C,
C?
3 Xl 一二 +X2 一二 > X1 +X
C3
C3
Xl
~
~
~
since C1 > C2 > C3 .
Therefore , no arbitrage can be obtained by being long the options with
strikes K 1 and K 2 and short the optio日 with strike K 3 .
We conclude that (1. 42) , i.e. ,
2
We conclude that arbitrage happens if and only if we can find Xl > 0 and
X2 > 0 such tl时(1.40) and (1. 41) are simultaneously satis丑ed. Therefore ,
noarbitrage exists if and only if
K 3 K1
K 3 K2
1.3. SOLUTIONS TO SUPPLE1VIENTAL EXERCISES
In other words , no arbitrage can be obtained by being long the option
with strike K 1 and short the options with strikes K 2 and K 3 .
Portfolio 3:Long the K 1option , long the K2 一 option， short the K 3option.
Arbitrage exists if we can find Xi > 0 , i = 1 : 3 , such that
0;
C1 一 α
'二
K3 K2
is the only condition required for
、
C2 
C3
noarbitrage.
口
Problem 10: Denote by Cbid and Cask , and by 凡id and 凡sk ， respectively,
the bid and ask prices for a plain vanilla European call and for a plain vanilla
European put option , both with the same strike K and maturity T , and on the
same underlying asset with spot price 8 and paying dividends continuously
at rate q. Assume that the riskfree interest rates are constant equal to 7'.
Find necessary and sufficient noarbitrage conditions for Cbid , Cαsk ， Pbid , and
凡sk·
Solution: Recall the PutCall parity
C
8e qT 
P
Ke rT ,
where the right hand represents the value of a forward contract on the underlying asset with strike K.
An arbitrage would exist
• either if the purchase price of a long call short put portfolio , i.e. , Cα砾，一 Pbid
were less than the value 8e qT  K e rT of the forward contract , i.e. , if
Cαsk 一凡id<SeqTkfrT;
X')
., 十 X3·
X1 C1 十 X2C2  X3C3 
K3 K1
(1. 47)
• or if the selling price of a long call short put portfolio , i.e. , Cbid  Pask were
greater than the value 8e qT  Ke rT of the forward contract , i.e. , if
Cbid 一凡sk
We conclude that there
Call parity if and only if
>
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
26.
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 Z
ιιi;:(归户
[
((怡均牛 泸的州叫叫)川川将巾户2)讪x
2与 一 Utcc) ) d
U
U:ζ:Uω;
口
汇口口
!:
斗
L
二二
ω
斗
J;
I 泸 (5x

1 一一 (归仙川 一((仰一 1)2 一归仙川+刊 Z
叶 1
Z 川+札
呐
阳
1
1，'(σ x 一 1 一川附一仰一 1 叫十刊(川
5
忏
俨(←一泸仇川十叶 Z 川一 叫)川Z ←=←一; 口
l
川
3x 2
卜 川Z d户
白
Problem 2: Let
f : (0 ，∞)→lR denote the Gamma functio孔
的)
χ
x
=
V俨
Z
l 飞 α俨 1
俨俨←叫~一→
45
i.e. ，
le t.
27.
CHAPTER 2. NUMERICAL INTEGRATION. BONDS.
46
(i) Show that f(α) is well defined for any α> 0 , i.e. , show that both
[xα1
e x dx
[ ' x"
•
1 e x
zαlez <
_ 时12α1 e一♂ dx
dx
出 [xα一1
dx
exist and are finite.
(ii) Prove ，旧吨 i的egration by parts , that f (α) = (α 1) f( α 一 1) for any
α> 1. Show that f(l) = 1 and conclude that , for any η 三 1 positive integer ,
f(n) = (ηI)!.
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 xα1
t", 0 Jt
0，
the functio丑 Zαl 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
ex dx 
lim I
to Jt
(2.5)
We can then use
X
Jim
"~UV
I
Jl
e dx 
and
(2.6)
< ε(2.6)
to show that , for any s > η(ε) ,
J豆豆/，' X"l 川Z
X"l 川Z
<J坦/，' ~x切=出(_2e t !2 十川)
α
2e s !2 < 2e n (f. )!2 < 飞
xα1 e x
dx
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)
zαl ez <ez/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 <已
(2 .4)
/,
l
(2.3)
_x/2
1m x  e , =
2e n (f. )!2
dx 工 lim 主 zllim(ltα)
Jo
ez/27 Vx> 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
t
which is what we wanted to show; d. (2.2).
(ii) It is easy to see that
m)=fev=JI
叫tfzdz= 出 (_et + 1)
Assume that α> 1. By integration by parts , we find that
1,
JiITl (1  e ) 
f(α)
"~UV
it follows that
fzMe曰 :Etlzα1
x
e dx
(2.1)
exists and is finite.
lIIaking these intuitive arguments precise is somewhat more subtle. We
include a mathematically rigorous arguments for , e.g. , showing that the integral in (2.1) exists and is finite.
By de丑nition， we need to prove that , for a町 ε> 0 , there existsη(ε) >
such that
°
I∞川X dx < 飞 Vs> η(ε)
(2.2)
= 1
f
l
'
o
k
扫
人
w
h
χ
i
i
况
一

α
!
「U
扒
L
1
1
斗
{
旷
日
仁
川
3
川
m
叭
卜
]
α
e
f
'
'
z
一1
1
J
『
f
n
n
U吃
Ui
F
I
J
•
咄 i Zα  e ffl'o
j
d 伽
咄
0
0
/
q&
飞
Z
Z
f 7
e
飞
K
/
划
α
已
z
=
卡
飞
1
l
ι
t
n
1
·
·
A
才
t
4
ufbfqh
l
I
/
{
卜
Z
smce
lim x o: 1 e x
limxα1
z0
2J
lim t o: 1 e t
t → οc

0,
ια1
lim 二:
t• x
eL
= 0.
for α> 1;
气
Z
E
B
,m
G
曲
d
I
i
l


t
E
28.
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) = (η1)!
(η l)f(η1).
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 exdx 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 N dpoint ,
Ii
Trapezoidal , and Simpson's rules can be found in the table below:
No. Intervals NIidpoint Rule Trapezoidal Rule Simpson's Rule
4
0 .40715731
0 .4 1075744
0 .40835735
8
0 .40895737
0 .40807542
0 .40836940
16
0 .40829709
0 .40851639
0 .40837019
32
0 .40835199
0 .40840674
0.40837024
64
0 .40837937
0 .40837024
0 .40836569
128
0 .40836911
0 .4 0837253
0 .40837082
256
0 .40836996
512
0 .40837018
0 .4 0837039
0 .40837028
1024
0 .40837023
(ii) Without computi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 order 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→ 问问V
R +R gi
忖阳
(ωi) Use NIidpoint rule with tal = 10 6 to compute an approximation of
(1
(1
I
币 5/2
= Jo f (x) dx = J ~一一亨
I
I 1+
o
OJ
'I

The approximate value of the integral is 0.17917117 , and is obtained for
a partition of the interval [0 , 1] using 64 intervals.
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) / (σIT). Compl.叫 (x)
Solution: Recall that
irf叫 = f川。)
29.
CHAPTER 2. NUMERICAL INTEGRATION. BONDS.
50
Therefore ,
2.1. SOLUTIONS TO CHAPTER 2 EXERCISES
51
it follows that
g"(t) = h(t) ,
左印 ω
Vi言b (归)
七 阳
功
Z
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 6month , 12month , 18month ,
and 24month 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)
2
十 (10 0 十
∞
= h(t) .口
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
Ill/ M
M
P 叫
da PI ν叭凡 rlu
I 凡
VG
,
, α
Z
z
υ Z , 2 + rld vhu, , 'hu i'U rlu α
'
一
'
'
'
一
1 V
1 V
α
'
a
α
A similar result can be derived for improper integrals , i.e. ,
/Ill
P ∞
M
da ft ∞仙 rld YG
I 仙
rld α
i'U
z
l
Z
Z
z
υ一
一
'
α
l
1
一
α W
1 w
衍
α
Lcash且ow
4
L
ι
'
L
U
j
?U
7
飞
}
Tb
α
/
/
'
'
飞E
、
/
'
1
、
白
、
Tb 、B
J
/
、
i
ι
'
L
U
、
飞
}
/
、
Tb 、E
Tb
E
，
J
'
飞
、
、
l
t
/
ι
I
7
l
u
i
t
/
α
/
，
1
飞1
iTb
ι
'
U
、
飞
l
'
/
J
Z
a
飞E
。
，
"
v一cas且且ow
= [2.5 2.5 2.5 102.5] .
The discount factors are
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.
disc = [0.97530991 0.94885432 0.92288560 0.89583414],
and the price of the bond is B =
[' (x  t)h(x) dX)
(2.9)
Solution: Par yield is the coupon rate C that makes the value of the bond
equal to its face value. For a 2year semiannual coupon bond , the par yield
can be found by solving
100 
~
+
=  [(X) dx
T 28
U
二( 1: 川x) = 一川川7
口
Problem 8: The continuously compounded 6month , 12month , 18month,
and 24month zero rates are 5% , 5.25% , 5.35% , and 5.5% , respectively. vVhat
is the par yield for a 2year semiannual coupon bond?
From (2.72.9) , we conell由 that
= 二(
98.940623.
时
i
For our problem ,
Since
= [0.5 1 1. 5 2];
J
ι
'
U
g' (t)
ω
川川川
;• l叫 μ咐咐仰(仰队阳，2
e r叫 O矶
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 19 the bond price given the zero rate curve.
丑
Input: n = 4

、
、
l
'
/
=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
•
30.
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 6month , 12mo时h ， 18month , and 24month discount factors;
(iii) Find the price of a two year semiannual coupon bond with coupon rate
5%.
Solution:
(i) Recall that the zero rate curve r(O , t) can be obtained from the instantaneous interest rate curve r(t) as follows:
ltp''Ifo
vv > nu
nU ,
γ
俨
T
T
一
J
I
飞
飞
7
、
、
TLV
t
'
b
/
= ;
+
l丑 (ο1
)
+ t)
e r (O ,O.5)O.5
_
e r (O ,l)
0.94939392;
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100er(O?05)05 十一~V
2
+
102.5 disc(4)
Problem 10: The yield of a semiannual coupon bond with 6% coupon rate
and 30 months to maturity is 9%. What are the price , duration and convexity
of the bond?
Solution: The price , duration , and convexity of the bond can be obtained
from the yield y of the bond as follows:
B
D
= 兰 3 仪p(一到十川p(补
1 f ι3i
一 I )
: :
B 仨t 2
exp
(i 5
(5 飞
+ 103 ~exp ~y }I JI
I
飞 2
~ ~ 2
2''
I ~y I
(2.10)
,, 门
l ，
‘
中
07
呼
41·A
1
4
、
、
1
'
/
}
1 (~9i
VA"
4
(i 2 5
(51
exp (一 ~y) + 103 A~ exp (一 ~y ) I .
飞 2
4 2'" } J
07
一 ~I{
100 e r (O ,1
I)
}
'


(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， 12month , 18month , and 24month discount factors are ,
respectively,
er(O ，1. 5) 1. 5
2.5 disc(3)
口
Lcash_flow = [0.5 1 1. 5 2 2.5];
0.005(1 十 t)

+
The data below refers to the pseudocode from Table 2.7 of [2] for computing
the price , duration and convexity of a bond given the yield of the bond.
Input:η= 5; y = 0.09;
;(0 阳+ 0ω0 0 阳
∞
disc(l )
disc(2)
disc(3)
disc(4)
2.5 disc(2)
) ~
B 仨t
庐 0ω05
fh扣
才 t、 十 0ωO 0 5 l协
∞ 盯 丑叫(川 )
臼阳川叫
n(
0.045
99.267508.
=一 I
Then ,
叫OJ )
+
~({ t)飞叮
e 一 I V忡忡
ι
α
B
'
一
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 (2.102.12). 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 J + 1. 75 exp I  12'' J
I
~"y I
12''
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/
31.
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 = 112 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 po np ay 江即
∞ 叩 ∞ 归 尼I
rrr
payment.
口
Problem 14: By how much would the price of a ten year zerocoupon 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 zerocoupo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 ShhOw
叭仰飞
咄
ν
these 旬 丑c 丑 are either always positive or always 丑 肮，ive.
fUl囚
n ctiorIS
1legat 忖
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

B +!:1 B 
103.75.
口
Problem 16: Establish the following relationship between duration and
convexity:
叮
C _ D :t.
Solution: Recall that
D
一
n
lDU Bvu
σ
r
c
an以

θD
=:.一
oy
,d
C
1θ2B
B
θy2
32.
CHAPTER
56
Therefore ,
2.
NUMERICAL INTEGRATION. BONDS.
2.2
θB
一一
= DB.
θν
(2.13)
Using Product Rl由 to differentiate (2.13) with respect to y , we 直nd that
叫 :uu
切
ω B
σ B2
B
D
DB 一τ
Dhu D B
一
一
一
一
一
。

一
句θ
B
D
十
一
切
句n
y
B
一
一
ω
一
ω句
γ
一
norC
、
飞1
1
1
1
/
'
'
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 1year and 2year discount factors with six decimal digits accuracy, and compute the 3year 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 riskneutral valuation to write the value of the put as an integral
over a finite interval. Find the value of the put option with six decimal
digits accuracy using the N dpoint Rule and using Simpson's Rule.
Ii
Also , compute the 丑lackScholes 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 $9 ,
respectively. Create a butterfly spread by going long a 45call and a
55call , and shorting two 50calls. What are the payoff and the P &L
at maturity of the butterfly spread? When would the butterfly spread
be profitable? Assume , for simplicity, that interest rates are zero.
4. Dollar duration is defined as
D电=一旦
ωθu
and measures by how much the value of a bond portfolio changes for a
small parallel shift in the yield curve.
Similarly, dollar convexity is defined as
C虫二工 θ2B
eθy2·
33.
58
CHAPTER 2. NUMERICAL INTEGRATION. BONDS.
Note that , unlike classical duration and convexity, which can only be
computed for individual bonds , dollar duration and dollar convexity can
be estimated for any bond portfolio , assuming all bond yields change
by the same amount. In particular , for a bond with value B , duration
D , and convexity C , the dollar duration and the dollar convexity can
be computed as
D$ = 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 riskneutral 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 BlackScholes 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 1year, 2year , and 3year 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 riskneutral valuation , we find that the value of the put is given bγ
P 
Use Simpson's Rule to compute the 1year and 2year discount factors with
six decimal digits accuracy, and compute the 3year 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
erTER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 BlackScholes 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 approximation errors to the
吨
BlackScholes value PBS in the table below:
We 丑rst note that the approximation error does not go below 6 . 106 .
This is due to the fact that the BlackScholes value of the put , which is given
by
PBS
Ker(Tt) N( d2 )  Seq(Tt) N( d1) ,
34.
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 BlackScholes value will propagate to the
approximation errors of the numerical integration.
If we consider that convergence is achieved when the error is less than
105 , 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 45call and a 55call ,
and shorting two 50calls. 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)=i55S(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

$76 , 000 , 000.
35.
62
CHAPTER
NUMERICAL INTEGRATION. BONDS.
2.
(ii) Using dollar duration and dollar co盯exity， the approximate formula
i:::.. B
7 倡一 D i:::.. y
1
+ §C( i:::.. y?
for the change in the value of a bond ca丑 be written as
i: :. B 臼 D向十 ;C$(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. BlackScholes 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 i17else
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)
36.
64
CHAPTER 3. PROBABILITY. BLACKSCHOLES FOR l/f ULA.
kl
=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]
E[X]
十y
T(η ， 1 ，
= 去汇 X(x， y) = 与生ω (k) 十 36Tk)( 一 1)
ω (k)(13k)23k
立
(3.2)
36
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
k
k(lp) fC
E[X] =
The values of w(k) for k = 2 : 12 are as follows:
= 8;ω(6) =ω(8) = 6.2;ω(7) = 5.
品
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 ,
E [X] = p' 1 + (1  p) . (1 十 E[X]) = 1 + (1 p)E[X].
EX
1P
一

pn→ χ 仨?
η
/,,‘
Z
一
一
i口 (3 .4)， we 直nd
17P
= τ了一
十
一
(
M
1i41i肿
川 一
一 、Z
咱
i
ω
i
f
十
一
…
门
，
，
，
"
n
E[X 2 ]
一
内
，
，
却
(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(lp)日:一一 ·r= 一.
pn→χ 仨7 1  p
y
p
L 对 (1 _ p)kl p 二 1 ~?')nl~~Lk2(Ip)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)
仨71pn →∞
pl
1 P
…
(3.3)
nz +
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)?
~?') 1~ L k(1  p)k
1
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;
n
汇
M
x)
By letting x = 1  p
(x ，y) εS
We conclude that
k (1  p)kl p =
Recall that
, (k  6, 6)}.
= k exactly 13  k times. Then ,
ω(5) =ω(9)
L
= k if and only if
(x ， y) ε {(6 ， k6) ， (5 , k5) , …
In other words , x
Another way of computing E[X] is as follows: The coin will first land
heads in the kth 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)kl p .
The凡
37  k
kl
ω (k) 兰一一 oj'
If 8
65
3.1. SOLUTIONS TO CHAPTER 3 EXERCISES
p + (1  p)2
p3
2p
p2
37.
66
CHAPTER 3. PROBABILITY. BLACKSCHOLES 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]
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
Problem 4: The density function of the exponential random variable X
with parameterα> 0 is
J a e一 αzif Z >
f(x)
=
fα 川z=J叫tα 川Z
出 (rz)lZ
出(1  e一叫
Z
1
=
(ii) By integration by parts we find that
Jxe叫z
1
xe 一 α x
一一一一一+…
r
一 αx
Uw
~一
e αx
xe 
~_嘈
I e
ααl
dx
αα2
r
'
一生主+~ xe一ω dx
f 泸川Z
ααl
x2 e一αx
2xe 一 αx
2e 一 αz
αα2α3
T 'ME
n叫
W凹
一

P
I
←
α
1
1
0;
)、77if Z 二。
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
α
χ
出
k
‘
→
z
n
χ
/
rJ
l
飞
z
i
/
I
t
一
一L
z句
α
e一
α
一
J
J
产o
d
二
ω
出
ziii 一etnu
=
1
1
1
/
M
α
叫
α
/
I
l
l
h
以
优
t
叫
zdz
fl04g
pu
d
1d
e
十
d

/
l
i
l
飞
38.
68
CHAPTER 3. PROBABILITY. BLACKSCHOLES FORJYIULA.
E[X 2]
LW)dz=dueα♂ 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
垃
ii
(仙 i) 1f x < 0 , then F(x) = J~∞ f(s) ds = O.
1f x 主 0 ， the口
(iv)
1f t 三 0 ，
=
fa" αf叫
α2
= (e α s) I~
1e
一αX
l 9" 川+叫bf(Z)g(Z)dz+ff2 川三 0 ，
if and only if
,
/lilf
。
中
1 P(X < t) = 1 一
l
一l'
只
f叫 d
λρ (阳
扣 Z州)川
巳一叫 = 1
αω
一
(e ω)
VαεR
b
then
P(X 主 t)
V αεJR.
Recall that a quadratic polynomial P(x) = Ax 2 十 Bx + C is nonnegative
for all real values of x if and only if P(x) has at most one real double root ,
which happens if and only if B 2  4AC 三 O.
For our problem , it follows that
Therefore
E[X
啊[阿冈附
判]
一 (但 X
b
b
~ 0,
咐:汇川s
JR.
ρb( 叫川+忖以仰州
Uf
飞(υ α Z )川)
巾
αJ坦(一午 25ff23+ 主)
var叫州
ar伊帆(仪问
町 X
叫叫 = 州
刘 刻)
f , g : JR •
69
t
t
t
I
α
叮
，
。
rId
z
/
ny Z
'
4
飞
,Z
G
，
Ill/
"
LU
/Il
F
1
1
α
A
哇
、
、
which is equivalent to
I~
P
I
'
'
1
α
(3.6)
门
rId
，
，
"
J'i
rId
Z η
'
3
'
1飞
5
Z
,2
G
<
p
t
t
'
'
1
α
n
rld
,,"
z
、
α
、
，
，
z
/
o
/III
'D
，
Z
JU
Z
I
α
IlI门
u
u
nL
z
7
、
1
Z
α
，
〈
一
/
,/
/It
f
飞
、
、
‘
，
，
，
，
，
'O
/II
P
飞
1
1
I
/
nu
Thu
I
'
1
1
α
l
l
I
/
内
od
，h
Z
α
，
z
飞
l
i
I
/
口
Recall that the conditional probability of A given B is
=
P(AIB)
P(A n B)
P(B)
(3.7)
Let s , t 三 O. The凡 from (3.6) and (3.7) , we find that
P(X 主 t+s I X 三 t)
P((X 三 t+s)n(x 主 t))
P(X 三 t)
e 一 α (t+ s)
_,""
If
ε''0
工
e
P(X 三 t
+ s)
Problem 6: Use the BlackScholes formula to price both a put and a call
option wit且 strike 45 expiring in six months on an underlying asset with spot
price 50 and volatility 20% paying dividends continuously at 2% , if interest
rates are constant at 6%.
Solution: Input for the BlackScholes formula:
P(X 三 t)
S = 50; K = 45; T  t =
0.5;σ=
0.2; q = 0.02; r = 0.06.
一讪甲
l f(x)g(川三 (l 尸 (x) (l
dx )'
The BlackScholes price of the call is C = 6.508363 and the price of the put
is P = 0.675920.
口
g2 川)
Problem 7: What is the value of a European Put option with strike K = 07
is the value of a European Call option with strike K = 07 How do you
hedge a short position in such a call option7
毛iV hat
39.
70
CHAPTER 3. PROBABILITY. BLACKSCHOLES FORNIULA.
Solution: A put option with strike 0 will never be exercised , since it would
mean selling the underlying asset for the price K = O. The price of the put
option is O.
A call with strike 0 will always be exercised , since it gives the right to
buy one u日it of the underlying asset at zero cost. The value of the call at
maturity is V(T) 工 8(T) ， and therefore V(O) 工 e qT 8(0). This can be seen
by building a portfolio with a long position on the call option and a short
position of e qT shares , or by using riskneutral pricing:
V(O)
e rT E
RN [8(T)]
3.1. SOLUTIONS TO CHAPTER 3 EXERCISES
e.g. , for pricing volatility swaps. These two Greeks are called volga and vanna
and are de如led as follows:
θ(vega(V))θ(vega(V))
volga(V) =θσand vanna(V) =θS
It is easy to see that
θ2V
volga(V) =万万
e rT . e(r一伊 8(0) 二 e qT 8(0).
A short position i口 the call optior1 祀 hedged (static a11吵 by buying ∞
丑 1s
剖 创 伪 与y )
址 坊圳
→
7
0l1
口
share of the underlying asset.
口
=  K(T  t)er(Tt) N(d 2 ).
Solution: Recall that
θCθP
p(C) = a~
p(P) = 万;
and
and
θ2V
vanna(V) 否瓦
The name volga is the short for "volatility gamma". Also , vanna can be
interpreted as the rate of change of the Delta with respect to the volatility
of the underlying asset , i.e. ,
Problem 8: Use formula p( C) = K(T  t)er(Tt) N(也) for p( C) and the
PutCall parity to show that
p(P)
71
θ (~(V))
vanna(V)
θσ
(i) Compute the volga and vanna for a plai丑 vanilla European call option on
an asset paying dividends continuously at the rate q.
(ii) Use the PutCall parity to compute the volga and vanna for a plain
vanilla European put option.
Solution: (i) Recall that
By differentiating the PutCall parity formula
P
with respect
to 飞 we
+
8e q(Tt) 
C
vega(C)
K er(Tt)
~(C)
find that
p(P) 
p(C) =
K(Tt)er(Tt).
叶 ) + (γ q 吟) (T 
d1
p(P) = ρ (C)  K(T  t)er(Tt)
K(T  t)er(Tt) N(d 2 )  K(T  t)er(Tt)
In (去)
+ (r
…
q)(T  t)
σ VT丁 t
tσ VTτz
2
Then ,
口
volga(C)
Problem 9: The sensitivity of the vega of a portfolio with respect to volatility and to the price of the underlying asset are often important to estimate ,
t)
VT丁Z
K(T  t)er(Tt) (1  N(d 2 ))
since 1  N(d 2 ) = N(d 2 ).
飞/27了'
eq(Tt) N(d 1 ) ,
where
Therefore ,
 K(T  t)er(Tt) N( d2 ) ,
8eq (Tt)叮叮 LJ:
vanna(C)
θ(吵(C)) _  8e q (Tt哺可L
Uσ
飞/2 7r
dJf11;

dσ'
θ(~(C))_ eq(Tt) N'(d1 ) 纽  e q 叫↓ λ ad 1
θσ
〉三作
θσ
40.
72
CHAPTER 3. PROBABILITY. BLACKSCHOLES
FOR~在ULA.
In (圣)
+
(γ  q)(T  t) ，~丁I
一二
σ2VT士t
θσ
C 
2
叫) + (γ q 一手) (T  t)
Keq(Tt) N(d 1 )
Ke r(Tt)N(d
P 
σ2 飞厅可

Ker(Tt) N(出);
Ke q(Tt)N(d
2) 
1 ).
where
d2
叶子气)旧 and 叫于一~)旧
σ
We conclude that
Then
volga(C)
Se旷tL.jT丁zie千生生
(3.8)
vanna(C)
eq(Tt) 」e4d2
(3.9)
川
νd开
d石
σ7
σ
(ii) By differentiating the PutCall parity P 十 Seq(Tt)  C = K er(Tt)
with respect to 风 we find that
vega(P) 
θP
否歹
C 三 P
仁=?
eq(Tt) N(d I)  er(Tt) N(d2 )
三 er(Tt)N( d 2 ) 
中=争 e一旷一气 N(d 1 ) + N( d 1 )) 三 er(T一气N(d2 )
牛=辛> eq(Tt) > er(Tt)
牛=学
since N(d 1 )
γ 二三
eq(Tt) N( d1 )
+ N( d2 ))
q,
+ N( d1 ) =
N(d 2 )
+ N( d2 ) =
1.口
δC
歹歹工 vega(C).
Problem 11: (i) Show that the Theta of a plain vanilla E旧opean call option
on a nondividendpaying asset is always negative.
(ii) For long dated (i.e. , with T  t large) ATJVI calls on an underlying asset
paying dividends continuously at a rate equal to the constant riskfree rate ,
i.e. , with q = r , show that the Theta may be positive.
Therefore ,
Volga(P)z
73
Alternatively, the BlackScholes formulas for atthemoney options can
be written as
Note that
θd 1
3.1. SOLUTIONS TO CHAPTER 3 EXERCISES
θ(vega(P))θ(vega(C))
= z Volga(C);
θσθσ
θ(vega(P))θ(vega(C))
Solution:
vama(P)===vanm(C)?
θSθs
where volga(C) and van叫 C) are given by (3.8) and (3.9) , respectively.
口
(i) 丑ecall
that
Sσeq(Tt)
~
@ ( C ) =  e 2 十 qSeq(Tt)N(d 1 )
2 、/2π(T 

t)
r K er(Tt) N(d 2 ).
Problem 10: Show that an ATM call on an underlying asset paying dividends continuously at rate q is worth more than an ATM put with the same
maturity if and only if q 三飞 where r is the constant risk free rate. Use the
PutCall parity, and then use the BlackScholes formula to prove this result.
For a nondividendpaying asset , i.e. , for q = 0 , we find that
Solution: For atth emoney options , i.e. , with S = K , the PutCall parity
can be written as
(ii) If q = r , the Theta of an ATM call (i.e. , with S = K) is
C P =
Seq(Tt)  K er(Tt) _ K eq(Tt)  K er(Tt)
K er(Tt) (e(rq)(Tt) 
1)
Therefore , C ~三 P if and only if e(rq)(Tt) ;三 1 ， which is equivalent to r 主 q.
8(C) =
Sσe4Tker(Tt)N(d2)<0.
2 飞/2作 (T  t)
Kaer(Tt)
~
@ ( C ) =  e 2 十 r K er(Tt) N( d 1 ) 
2 飞/2作 (T
 t)
r K er(Tt) N( d 2 )
Ker(Tt) (r(N(d 1 )  N(d 2 )) 一
σJ)
2 飞/2汀 (T 
tr )
41.
CHAPTER 3. PROBABILITY. BLACKSCHOLES FORlvIULA.
74
since d 1 = d 2 十 σ VTτt and therefore
where
θd 1
d1 =
θ d2
θKθK
Note that
lim
(Tt)→∞
Then ,
li~
(Tt)→∞
d1 = ∞
N(d 1 )
and
lim
(Tt)→∞
仇=一∞.
，~li~
= 1 and
(Tt) →∞
N(d 2 )
By
(Tt)→∞
differentiati吨 (3.10)
with respect to K , we find that
空气工  er(Tt) N'(d2 ) 纽工 _ er(Tt) τLe手 252
=0
δK
8K
v 三作
θK
(3 叫
Note that
and therefore
!γ(N(d1 )  N(d2 )) 一
σe4i
2y2作 (T  t)
)
In
∞
z
d2
8(C) = Ke r 叫
(γ (N(d 1 )

飞
州十 (γ q 一手) (T  t)
In(K)
N(出))一
JTt
σJ}
2y2作 (T 
t)
Then
)
口
'
σVT丁I
θd 2
1
and , from (3.11) and (3.12) , we conclude that
θ2C
Problem 12: Show that the price of a plain vanilla European call option is
a convex function of the strike of the option , i.e. , show that
θ2C
一一==δ K2
> O.
一
Seq(Tt) N'(d 1 )

Ker(Tt) N'(d 2 ).
By differentiating the BlackScholes formula
Seq(T一忖T(d 1 ) 
Ker(Tt) N(d 2 )
Seq(T一叫鞋一 Ker(T一川也)在一 er(T一切(d2 )
_er(Tt) N(d 2 ) ,
(坐 252}
飞 θKθK
}

σK j2作 (T
 t)
er(Tt) e二Ji>O
口
Solution: The input in the BlackScholes formula for the Gamma of the call
is S = K = 50 ， σ= 0.3 , r = 0.05 , q = O. For T = 1/24 (assuming a 30 days
per month count) , T = 1/4 , and T = 1, the following values of the Gamma
of the ATM call are obtained:
with respect to K , we obtain that
Seq(Tt) N'(d 1 )
θ K2
Problem 13: Compute the Gamma of AT l!I call options with maturities of
fifteen days , three months , and one year , respectively, on a nondividendpaying underlying asset with spot price 50 and volatility 30%. Assume that
interest rates are constant at 5%. What can you infer about the hedging of
AT l!I options with different maturities?
Solution: Recall that
C 
(3.12)
θKσKJTt'
will be positive. We note that the positive value of 8( C) is nonetheless small ,
si丑ce lim(Tt)→∞ 8(C) = O.
(圣) + (γ q~)(Tt)
JTt
We conclude that , for T  t large enough ,
w
山
w
75
3.1. SOLUTIONS TO CHAPTER 3 EXERCISES
eT(T一切(出)
(3.10)
r(15days) 
0.057664; f(3months) = 0.052530; r(1year) = 0.025296.
We note that Gamma decreases as the maturity of the options increases.
This can be seen by plotting the Delta of a call option as a function of spot
price , and noticing that the slope of the Delta around the atthemoney point
is steeper for shorter maturities. The cost of Deltahedging AT l!I options
42.
76
CHAPTER
3.
PROBABILITY. BLACKSCHOLES FOR1vIULA.
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
(iii) For clarity, let
= T  t. For r = q = 0 , we obtain from (3.13) that
7
vega(C) =
M
二
豆坦豆f
而
VL1「
汀
Problem 14: (i) The vega of a plain vanilla European call or put is positive ,
smce
vega(C)  vega(P)  Seq(T一明习气Le一手
(3.13)
since , for an ATJVI option with r = q = 0,
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 nondividendpaying 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
吨a(C) 工吨a(P)
=
SVT习气LJ7
飞/三'Tr
巾
w he叫 = 巳F Cωom川附 山 仰
e t he de pe丑毗丑ce of 吨叩&叫圳州 on ti 江
叩
阴
(ρ 川 凶
C)
让I
ime
T 一 t ， i.e. ,
θ(vega( C))
θ(T  t) ,
= In (主)十
d1
VL/Tr
77
(γ q 十号)7 _aft
σJ于一
i
2
By direct computation , we find that
nuve
g β
/'1
… 一
θ T
，
l，
t
、

C
一
、
、
l
/

、
、
l
g
/

S
一旦;z
2)2 'Tr 7 ~
dS〉宇 z;工
8~石 U
忐 (1 一宁)e牛
For σ= 0.3 and for time to maturity less than one year , i.e. , for 7 三 1 ，
that
h7
we 丑nd
町
> nuQd
i
414
and therefore
巧
i
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 BlackScholes 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
nondividend 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 noarbitrage condition:
K e rT
 Se qT
S P < K e rT .
(3.15)
43.
78
CHAPTER 3. PROBABILITY. BLACKSCHOLES FORMULA.
Solution: One way to prove these bounds on the prices of European options
is by using the PutCall 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 < SeqT.
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 SeqT.
(3.19)
<
P
<
Ke rT .
A more insightful way to prove these bounds is to use arbitrage arguments
and the Law of One Price.
Consider a portfolio made of a short position in one call option with strike
K and maturity T and a long position in e qT units of the underlying asset.
The value of at time 0 of this portfolio is
V(O) = Se qT  C.
If the dividends received on the long asset position are invested continuously
in buying more units of the underlying asset , the size of the asset position at
time T will be 1 unit of the asset. Thus ,
V(T)
= S(T)  C(T) = S(T)  max(S(T)  K , 0)
since , if S(T) > K , then V(T) = S(T) 一 (S(T)  K) = K , and , if S(T) 三 K ，
then V(T) = S(T) 三 K.
From the Generalized Law of One Price we conclude that
V(O) = Se qT  C 三 KeyT?
and therefore Se qT  K e rT 三 C ， which is the left inequality from (3.14).
All the other inequalities can be proved similarly:
• To establish that C ~二 Se qT ， show that the payoff at maturity T of a
portfolio made of a long position i丑 e qT units of the underlying asset at time
o and a short position in the call option is nonnegative for any possible values
of S(T);
• To establish that K e rT  Se qT ~二 P ， show that the payoff at maturity T
of a portfolio made of a long position in e qT units of the underlying asset
at time 0 and a long position in the put option is greater than K for any
possible values of S(T);
• To establish that P ~二 K e rT , show that the payoff at maturity T of a
portfolio made of a short position in the put option and a long cash position
口
of Ke rT at time 0 is nonnegative for any possible values of S(T).
Problem 16: A portfolio containing derivative securities on only one asset
has Delta 5000 and Gamma 200. A call on the asset with 6.(C) = 0 .4
and f(C) = 0.05 , and a put on the same asset , with 6. (P)  0.5 and
f(P) = 0.07 are currently traded. How do you make the portfolio Delt aneutral and Gammaneutral?
Solution: Take positions of size Xl and X2 , respectively, in the call and put
options speci五ed above. The value II of the new portfolio is II = V + Xl C +
归 P ， where V is the value of the original portfolio. This portfolio will be
Delta and Gammaneutral , provided that Xl and X2 are chosen such that
From (3.18) and (3.19) it follows that
Ke rT _ Se qT
79
3.1. SOLUTIONS TO CHAPTER 3 EXERCISES
三 K，
6. (II)
=
6. (V)
+ xl 6.( C) + x2 6. (P)
=
5000 十 O .4Xl  0.5X2
f(II) = f(V) 十 xlf(C) 十 X2f(P) =  200 十 0. 05X l 十 0.07X2
The solution of this linear system is
250 , 000
._. ~ ~_
1= 一一一ι...::....::.. = 4716.98
53
 .     and
..
0;
o.
330 , 000
x') = 一」一== = 6226 .4 2.
 "'
53
To make the initial portfolio as close to Delt a and Gammaneutral as
possible by only trading in the given call and put options , 4717 calls must
be sold and 6226 put options must be bought. The Delta and Gamma of the
new portfolio are
6.(江) =
f(口)
=
6. (V)  4717 6.( C) 十 6226 6. (P) = 0.2;
f(V)+4717f(C)+6226f(P) = 一 0.03.
44.
80
CHAPTER
3.
PROBABILITY. BLACKSCHOLES FOR1VIULA.
To understand how well balanced the hedged portfolio II is , recall that the
initial portfolio had ~(V) = 5000 and r(V) = 200.
口
3.1. SOLUTIONS TO CHAPTER 3 EXERCISES
81
(iii) The new spot price and maturity of the option are 8 2 = 98 and T2 =
12~ /252 (there are 252 trading days in one year). The value of the call option
is $8 .4 53134 and the value of the por毛folio is
Problem 17: You are long 1000 call options with strike 90 and three months
to maturity. Assume that the underlying asset has a lognormal distribution
with drift μ= 0.08 and volatility σ= 0.2 , and that the spot price of the
asset is 92. The riskfree rate is l' = 0.05. vVhat Deltahedging position do
you need to take?
1000C2
units of the underlying asset , where
d1 = In
二一
(圣) + (1'  q 吟)T
σ 飞IT
with 8 = 92 , K = 90 , T = 1/4 ， σ= 0.2 ， γ= 0.05 , q = O.
Note that , for Deltahedging purposes , it is not necessary to know the
driftμof the underlying asset , since ~ (C) does 口ot depend on μ.
口
Problem 18: You buy 1000 six months ATM Call options on a nondividendpaying asset with spot price 100 , following a lognormal process
with volatility 30%. Assume the interest rates are constant at 5%.
(i) How much money do you pay for the options?
(ii) What Deltahedgi position do you have to take?
吨
(iii) On the next trading day, the asset opens at 98. What is the value of
your position (the option and shares positio时?
(iv) Had you not Delt ahedged , how m旧h would you have lost due to the
increωe in the price of the asset?
Solution: (i) Usi吨 the BlackScholes forml出 with input 8 1 = K = 100 ,
T = 1/2 ， σ= 0.3 , l' = 0.05 , q = 0 , we 自nd tl时 the value of one call option
is 0 1 = 9.634870. Therefore , $9 ,634.87 must be paid for 1000 calls.
(ii) The Deltahedgi position for long 1000 calls is short
吨
1000~(C)
= 1000e qT N( d 1 ) = 588.59
units of the underlying. Therefore , 589 units of the underlying must be
shorted.
=  49268.87.
(iv) If the long call position is not Deltahedged , the loss incl
decrease in the spot price of the underlying asset is
1000(C2
Solutioη:
A long call position is Deltahedged by a short position in the
underlying asset. Deltahedging the long position in 1000 calls is done by
shorting
1000~(C) = 1000e qT N( d1 ) = 653.50
58982


cd
=一 $118 1. 74.
For the Deltahedged portfolio , the loss incurred is
(100002

58982 ) 一 (1000C1  58981 ) =一 $3.74.
As expected , this loss is much smaller than the loss incurred if the options
positions is not hedged ("naked"). 口
45.
CHAPTER 3. PROBABILITY. BLACKSCHOLES FORNIULA.
82
3.2
Supplemental Exercises
1. What is the expected number of coin tosses of a fair coin in order to
get two heads in a row? What if the coin is biased and the probability
of getting heads is p?
3.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
83
8. You hold a portfolio with 6.(口) = 300 , f(II) = 100 , and vega(II)
You can trade in the underlying asset , in a call option with
6. (C) = 0.2;
f(C) = 0.1;
= 89.
vega(C) = 0.1 ,
and in a put option with
2. What is the expected number of tosses in order to get k heads in a row
for a biased coin with probability of getting heads equal to p?
6. (P) = 0.8;
vega(P) = 0.2.
What trades do you make to obtain a 6., f , and veganeutral portfolio?
3. Calculate the mean and variance of the uniform distribution on the
interval [a ,b] .
4. Let X be a normally distributed random variable with
standard deviation σ> O. Compute E[ IXI ] and E[X 2].
f(P) = 0.3;
meanμand
5. Compute the expected value and variance of the Poisson distribution ,
i.e. , of a random variable X taking only positive integer values with
probabilities
3.3
Solutions to Supplemental Exercises
Problem 1: What is the expected number of coin tosses of a fair coin in
order to get two heads in a row? What if the coin is biased and the probability
of getting heads is p?
n 一入飞 k
P(X=k) = 气卡， V k 三 0 ，
whereλ>
0 is a fixed positive number.
6. Show that the values of a plain vanilla put option and of a plai且 vanilla
call option with the same maturity and strike , and on the same underlying asset , are equal if and only if the strike is equal to the forward
pnce.
7. You hold a portfolio made of a long position in 1000 put options with
strike price 25 and maturity of six months , on a nondividendpaying
stock with lognormal distribution with volatility 30% , a long position
in 400 shares of the same stock , which has spot price $20 , and $10 ,000
如 cash. Assume that the risk自free rate is constant at 4%.
(i) How much is the portfolio worth?
(ii) How do you adjust the stock position to make the portfolio Deltaneutral?
(iii) A month later , the spot price of the underlying asset is $24. Wr时
is new value of your portfolio , and how do you adjust the stock position
to make the portfolio Deltaneutral?
Solution: If p is the probability of the coin toss resulting in heads , then the
probability of the coin toss resulting in tails is 1  p.
The outcomes of the first two tosses are as follows:
• If the first toss is tails , which happens with probability 1  p , then the
process resets and the expected number of tosses increases by 1.
• If the 趾st toss is he告ds ， and if the second toss is also heads , which happens with probability p气 then two consecutive heads were obtained after two
tosses.
·If the first toss is heads?and if the second toss is tails?which happens
with probability p(l  p) , then the process resets and the expected number
of tosses increases by 2.
If E[X] denotes the expected number of tosses in order to get two heads
in a row , we conclude that
E[X]
= (1 p)(l + E[X]) 十 2p 2 十 p(l
p)(2 + E[X]).
(3.20)
We solve (3.20) for E[X] and obtain that
1 牛竹
E[X] = 寸主.
p"
For an u~biase_d coir:, i.e. , for p = ~， we find that E[X] = 6, and therefore
the expected number of coin tosses to obtain two heads i~ a row is 6.
口
46.
CHAPTER 3. PROBABILITY. BLACKSCHOLES FORIvIULA.
84
Problem 2: What is the expected number of tosses in order to get n heads
in a row for a biased coin with probability of getting heads equal to p?
3.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
var(U)
= E [eu  E[U])2] = E [ (U 一引 j
Solution: The probability that the first n throws are all heads is pn. If the
k throws are heads and the (k + l)th throw is tails , which happen
wit h probability pk( 一 p) ， then the process resets after the k 十 1 steps; here ,
让
吐
οl
k = 0: (η1). Then , if x(η) denotes the expected number of tosses in order
to get ηheads in a row , it follows that
二 f(z132)dz= 击 ~(X_b;a)' 一
且rst
Z
η
一
一
n 二
l
飞 、
PP 气
η
十
η
η
η
4 F
k H
一
一
门
V
+
k
h
T
d
o
一
kkZK
十 忏
V
iJSSttP
UA
川
川
、
4
元
ι
1
1
「
d
A
4
八
十
P
A
Recall that
叫
Z
M
p
1 pn
lp7
八
/
'
川
。
飞
川
γ
μ
M
，
V
h
L
I
r
l
l

P
A
向
+
z n
n
吁
1

PA
飞
、
J
牛
'
K
p
2二 kpk
/
二
、
F
E
'hh
UA
η，p n 十
(1  p)2
1pη
1 一 (η 十 l)pn 十 npn+1
1p
一一十 x(η)(1
1p
and therefore
Problem 4: Let X be a normally distributed random variable with
and standard deviation σ> O. Compute E[ IXI ] and E[X 2
].
Solution: We compute E[
+
x(n)(l _ pn)
vi言汇 lμ+ 的Ie一手 dz
E[IX 门
τ~ (一μ/σ 一(μ 十的) e一句+气L
v 马丁II
V L/Jr
Jx
f
J 一 μ/σ
(μ + o z) e一手dz
tl:Vdz 一元zlJNJdz
1pη
(η) =一一一
I
+μ 一言=
e一 T
dz
ν 2i J一 μ/σ，.
ft7
+
,
一r.::= 1
~在 j一μ/σ
一 Td
山
It is easy to see that
1du
Problem 3: Calculate the mean and variance of the uniform distribution on
the interval [价].
JUJu
l
而
，
α
υ
γ
，
，
α
叩
N(~) = 1N(~);
(e一手) I二 f= 一叫一￡);
在[~" e'; dy N (;) ;
f
土
Solution: The probability density function of the uniform distribution U on
the interval [a , b] is the c∞stant functi∞ j(x) = b~a' for all x E [a , b]. Then ,
rb ~" •
1
rb
b+ α
E[U] = I xj(x) dx = 一~ I xdx = 一
j α b一 α j α 2
IXI ] in terms of the cumulative dist由ution
of the standard normal variable Z.
Note that X = μ+σ Z. Then ,
_ pn) ,
pn (1 p)'
We conclude that the expected number of tosses in order to get n headsin
a row for a biased coin with p时 ability of getti昭leads 叩al to pis 元57
If the ∞i丑 were unbia时， i.e. , for p = ~， the expected number of tosses in
order to get n heads in a row is 2η+1  2.
口
meanμ
N(t) 工 vi言 L 巾z
npn 十 (η  l)pn十1
Then , we find that
x(η)
85
(e~) J:二/σ= 叫一￡)
47.
CHAPTER 3. PROBABILITY. BLACKSCHOLES FORJVIULA.
86
87
3.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
Then , it follows that
We conclude that
一 μ (lN(;)) 十元叫一出
E[X]
二气
)
十叫;) +在叫一￡)
μ (2N (;) 1)
+岳叫一出
One way to compute E[X 2] would be to compute the
E[X 2] =
JOO
While this would provide the correct result , an easier way is to recall that
Since E[X] = μand var(X)
=σ2 ，
、
e…
(3.22)
eλ
(3.23)
From (3.21) and (3.22) , we 丑 nd that E[X] =
Similarly,
。c
E[X 2]
= 汇 P(X
e一λ 已
..
=
k)· k2 =
(k  1)λk
台
(k1)!
x
2
var(X) = E[X ] 一 (E[X]?

ZJ二:; 
followi吨 integral:
E[(μ+σ Z)2] = 仨/∞ (μ 十 M)2Jdz
V'<""
λ kl
一一一一一一一
λ.
~一 λ 飞 k
汇气产
~一
'  λ
e 兰川
k2
λk
亏
~但
..一
x
飞 k2
飞 kl
d2EtrA 罢工一
we concl 时 e that
λ2
E[X 2] = var(X) + (E[X])2 = μ2 十 σ2
十 λ7
where (3.22) and (3.23) were used for the last equality.
We conclude that
Problem 5: Compute the expected value and variance of the Poisson distribution , i.e. , of a random variable X taking only positive integer values with
probabilities
n 一入飞 k
var(X) = E[X 2]  (E[X])2 = λ
口
where λ> 0 is a fixed positive numbe r.
Problem 6: Show that the values of a plain vanilla put option and of a
plai 丑 vanilla call option with the same maturity and strike , and on the same
underlying asset , are equal if and only if the strike is equal to the forward
pnce.
Solution: We show that E[X] = λand var(X) =λ
Solution: Recall that the forward price is F = Be 一 (rq)T
P(X
= k)
=
气!'
Vk 三 0 ，
By definition ,
EX
M
∞Z
From the PutCall parity, we know that
Z∞
M
P X
川
丁
Recall that the Taylor expansion of the function eX is
et
=z;
∞
汇
M
上
…
c
P
Be qT

Ke rT .
(3.24)
If a call and a put with the same strike K have the same value , i.e. , if C = P
in (3.24) , then Be qT = Ke rT . Thu
K 
Be 一 (rq)T7
i.e. , the strike of the options is equal to the forward
price.
口
48.
88
CHAPTER 3. PROBABILITY. BLACKSCHOLES FOR1VIULA.
Problem 7: You hold a portfolio made of a long position in 1000 put options
with strike price 25 and maturity of six months , on a nondividendpaying
stock with lognormal distribution with volatility 30% , a long position in 400
shares of the same stock , which has spot price $20 , and $10 , 000 in cash.
Assume that the riskfree rate is constant at 4%.
(i) How m时1 is the portfolio worth?
(ii) How do you adjust the stock position to make the portfolio Deltaneutral?
(iii) A month later , the spot price of the underlying asset is $24. What is new
value of your portfolio , and how do you adjust the stock position to make the
portfolio Deltaneutral?
Solution: (i) The value of the portfolio is
1000P(0)
+
803 + 400 =
( 1 ( 1
I + 8038 I
飞 12 )
~_ I
飞 12
}
十 1946
= 23400.
The new Delta of the portfolio is
1000N( d l )
+ 803 = 292.
To make the portfolio Deltaneutral , you should sell 292
shares.
口
Problem 8: You hold a portfolio with i:::.. (II) = 300 , r(II) 工 100 and
vega(II) = 89. You can trade in the underlyi吨 asset， in a call option with
i:::.. (C) 工 0.2;
r(C) = 0.1;
vega(C) = 0.1 ,
89
and in a put option with
i:::..
(P) = 0.8;
r(p) = 0.3;
What trades do you make to obtain a
i:::..,
vega(P) = 0.2.
r , and veganeutral portfolio?
Soltdi07ZJ You can make the portfolio Tand Veganeutral by takiIlg positions in the call and put optiOI17respectively.By trading iII the underlying
asset7the r aad Vega of tfle portfolio would not change7and the portfolio
can be made i:::..neutral.
Formally7let z17Z27and hbe the positions i2the underlying asset7the
Cali option?and the put option7respectively.The value of the zlew portfolio
is II旧w=II 十 x 1 8 + X2 C + 句 P and therefore
i:::..(丑nω) = i:::.. (II) 十 Xl + x 2 i:::.. (C) 十 x 3 i:::.. (P);
r(II nω ) = r(口) + X2 (C) + X3 (P);
vega(口new)  vega(II) 十 x2veg a (C) 十 x3 ve g a (P).
r
r
Then , i:::..(丑nω) = r(丑ηω) = vega(II n 叫 = 0 if and only if
•
(们 0.2X2 0.8
 403.
To obtain a Deltaneutral portfolio , 403 shares must be purchased for $8 ,060.
The Deltaneutral portfolio will be made of a long position in 1000 put options a long position in 803 shares of the underlying stock , and $1 ,940 in
cash.
(iii) A month later , the spot price of the und 凹lying asset is S ( 古剖) = 24 a丑
the put options have 直ve months left urlt削1 maturity. The BlackScholes value
悦
fi飞
丘 时i
垃让
挝
of the put option is P (古) = 2.1818. The cash position has accrued interest
and its current value is 1940 exp (呼) = 1946. The portfolio is wo巾
~_
SOLUTIONS TO SUPPLEMENTAL EXERCISES
4008(0) 十 10000 = 22927 ,
where 8(0) 二 20 is the spot price of the underlyi吨 asset and the value
P(O) = 4.9273 of the put option is obtained using the BlackScholes formula.
(ii) The Delta of the put option position is 1000N( d l ) = 803. (Here
and in the rest of the problem , the values of Delta are rounded to the nearest
integer.) The Delta of the portfolio is
1000P I
3.3.
0. lx 2 十 0. 3X 3
0. lx 2 十 0. 2X 3
300;
= 一 100;

89.
The solution (rounded to the nearest integer)is Zl=254向饥= 670、
均 =1101n othr words , to make the portfolio A7Tand Vega
一丑eut叫:
one must short 254units ofthe underlying asset and sell 670call options &创1
an
口
110 put
options.
口
49.
Chapter 4
Lognormal random variables. Riskneutral
prIcIng.
4.1
Solutions to Chapter 4 Exercises
Problem 1: Let Xl = Z and X 2 = Z be two independent random variables , where Z is the standard normal variable. Show that Xl + X 2 is a
normal variable of mean 0 and variance 2, i.e. , Xl 十 X2 = V2 Z.
Solution: 丑ecall that if Xl and X 2 are independent normal random variables
with mean and variance μ1 and σ?7aMμ2 and σi ， respectively, then Xl + X 2
is a normal variable with mean μ1+μ2 and varia肌eσr+σ27and
Xl + X 2 =
f.t 1
+
f.t 2
+
For Xl = Z and X 2 = Z , it follows that
conclude that
E[X]= μ1 十问= 0
and therefore
X
and
;;;f.丐 z
μ1= 问=
0 and
σ1= 句=
1. vVe
var(X) =σ?+σ~ = 2,
= Xl +X2 = V2Z. 口
Problem 2: Assume that the normal random variables Xl , X 2 , … , X n
of meanμand variance σ2 a陀 uncorrelated， i.飞 cov(Xi ， Xj)  0 , for all
l 三 t 并 j 三 η. (This happens , e.g. , if Xl , X 2 , … , X n are independent.) If
Sn = ~ L:~=1 Xi is the average of the variables Xi , i = 1 :风 show that
E[Sn]
Solution: Recall that , for
= μand
Ci ε~ ，
叫几
町
斗 q凡斗]
E [~← ~吝辛←←X均4斗]
呐
Ci剧刷
i E耶啊啊[仄问阳
=
91
var 问
50.
CHAPTER 4. LOGNORMAL VARIABLES. RN PRICING.
92
刊主Ci
L crvar(Xi) +
4. 1. SOLUTIONS TO CHAPTER 4 EXERCISES
Problem 4: If the price S(t) of a ∞ndividend paying asset has lognor口时
distribution with drift r and volatility 风 show that
2 ~二呐∞，v(Xi ， X j )
ER 川S(t
Therefore ,
E[Sn]
var(Sn)
er8t S(t);
E
[~字]
咔字)
主 2二 var(Xi)
1
一 ·ημ
=
=μ;
η
主 zmE间十二三二nCOVM)
1
c)
σ
since COV(Xi , X j ) = 0 for all 1 三 t 泸 j 三 η.
口
Show that
ERN[S2(t 十 6t)刀
缸均 !
创
(pu + (1  p)d) S(t);
(pu 2 十 (1  p) d2) S2 (t ).
var(Y)
E[tft)j
(4.1)
(4.2)
space{U7D}ofttle possible moves ofthe price ofthe asset fromtimet to time
t .L+ 8t ~ndo~ed with the riskneutral probability function P : {U , D} • [0 , 1]
with P(U) = p and P(D) = 1  p. Then S(t + 8t) is given by
= S(t)u;
S(t 十们)(D)
S(t)d.
Then , by definition ,
ER
E RN[S2(t + 8t)]
P(U) . S(t 十们)(U) + P(D)· S(t 十 8t)(D)
(pu + (1  p)d) S(t);
P(U) . (S(t 十们)(U)? 十 P(D) . (S(t 十们)(D))2
(pu 2 十 (1  p) d2) S2 (t ) .口
优p (2μ + 0: 2 )
(i:
2

1)
盯=需在 th叫 =(γ 一手) 8t and 歹工 σv5i and 切向
vaT
(吐了))
叫 ((γ_ ~2) /it 十叩2)
仪p
e
Solution: We can regard S(t 十们) as a random variable over the probability
S(t 十们)(U)
+ σ) 8t Z.
圳叼
叫μ+ 三);
E[Y]
Problem 3: Assume we have a one period binomial model for the evolution
of the price of an underlying asset between time t and time t 十 8t:
If S(t) is the price of the asset at time t? t~en the p:~ce S.(t 十 8t) of the asset
at time t 十 8t will be either S(t)叽 with (riskne飞巾al) probability p, or S(t)d ,
with probability 1  p. Assume that u > 1 and d < 1.
ER
丑〈俨(了俨 S(tt) ) 一一 J 8t
贝例
S (川 J = I /
均
Recall tl时， if ln(Y) =μ 十歹Z is a lognormal random variable with
parameters μand 歹， the expected value and variance of Yare
η
n血
(4.4)
Solution 1: If the price S (t) of the nondividend payi吨 asset has lognormal
distrib 叫0 川ith drift 川 nd 叫at削 川
创 il r
址 lity
过
given by
2
一τ ·nσ
(4.3)
e(2r+σ2)8t S2(t).
+ c5t)]
ERN [S2(t 十 c5t)]
:土贝Xi]
93
2r8t
=eTSt;
(4.5)
(2 (γ_ ~2) 们+州)2) .(e(σ而)2 1J4 . 6 )
(产 1)
(4.7)
Note that ,
E[tft)]
mr(tf))
」一 ·E[S(仆 8t)]
S(t)
(4.8)
LVar (S(t+60)
(4.9)
S2(t)
From (4.5) and (4.8) , and from (4.7) and (4.9) , respectively, we conclude that
E[S(t 十 8t)]
var (S(t + 8t))
er8t S( t);
e叮产
1) S2(t)
(4.10)
51.
CHAPTER
94
4.
LOGNOR1VIAL VARIABLES.
RN
PRICING.
4.1. SOLUTIONS TO CHAPTER 4 EXERCISES
Note that
var (S(t 十们)) = E[S2(t
=
S2(t) 可言=
+ 8t)] 一 (E[S(t 十们)])2
E[S2(t 十 6t)]
V L7r
 e2r6ts2(t).
(4.11)
fχ(.L£ J _2£
exp 12γ们十 σ2 6t 
Jx
i
n
I
飞.<.
~
fχ
( (x  2σ V5t)2 飞
I exp {一门)
V L,7r J x
E[S2(t 十 6t)] = var (S(t 十 8t)) + e2r6t S2( t)
(X2 oV5t )2
J
I
/
S2(t)e(2r+ e> 2) 6t
From (4.10) and (4.11) it follows that
95
.<.
I
S2 (t) e (2r+e>2)6t ↓/沉 e4ds
e 2r6t十σ 26t S2(t) ,
V L,7r Jx
S2(t)e(2r+σ2)6t;
which is what we wanted to show.
Solution 2: Note that S(t 十归) can be written as a function of the standard
the substituti∞ S 可 2σv5t was used above
口
normal variable Z as follows:
I
S(t 十们)
=
I
σ2 飞
Problem 5: The results of the previous two exercises can be used to calibrate
a binomial tree model to a lognormally distributed process. This means
在nding the up and down factors u and d , and the riskneutral probability p
(of goi鸣 up) such that the values of ERN[S(t 十 6t)] and ERN[S2(t+6t)] given
by (4.1) and (4.2) coincide with the values (4.3) and (4 .4) for the lognormal
model.
In other words , we are looking for u , d , and p such that
r
S(t) 仰~ ~ r 一言) 6t 十 σ训tZ)
Then ,
E[S(t + 6t)]
刁卖 fι:S (tω)
七
i言 附
川
t均
m 左 ι 叫6←一宁 +oV5t ♂卜斗一 ν
ι
们t
ζ dx
x;)
μ1
pu 十 (1  p)d
pu2
r∞/ (x 一 σVti? 飞 J
S(t) 矿Ot 苟言儿仪p 一
+
(1 _ p)d2 =
viz 汇巾ν
(e r6t  d) (u  er6t )
1,
ud 
E[S2忻州]=去 (00 S2(均叫2 (γ_ ~2) 6t + 2σvti x) e~
I L,7r
J 一∞
I
S2(t) exp t 2r6t 
I
n
m
h
a
(4.14)
‘
/
t
t
飞
向A
U
aTb
、
寸
、
t
t
I
h
J
(4.15)
dx
/
rτ
x
嘈
o:.:I 6t + 2σ yot x  2) dx
J一
∞
/
1.
Show that the solution can be written as
SimiLr1377mobtaizltbM
I
(4.13)
(ii) Derive the Cox一RoωSS卧一
臼 εsRubinsteir1 parametri zation for a binomial tree , by
阳时 丑
忧坦 创
solving (4.144.15) with the additional condition that
since ÷eJ兰 is the density function of the standard normal variable.
去=
一
一
衍
p
where we used the substitution y = x 一 σvti and the fact that
f∞
e(2r十σ2)6t
Mlub
t
eedr
"
S(t)eT5"t?
阶旷
均
t
1
(4.12)
Since there are two constraints and three unknowns , the solution will not be
umque.
(i) Show that (4.124.13) are equivale时 to
 2')α
旷沪
仰附)汩T
)er
I L,7r
= er6t ;
p
= 主二f; 也 =
u 一 α
where
A =
A+
YA2才; d=AVA可7
~ (川十 e(r+e>2)6t)
.
(4.16)
52.
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 = u
1
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 (ue 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)
After canceling out the term _e 2r8t , we divide (4.19) by e耐 and obtain
U十d
 e r8t

仨 k2
e(r十泸州 7
are uniformly bounded , in order to conclude that the series is convergent.
This can be seen as follows:
1 L41
1
汇丰
1
4
咱
i
which can be written as
U 十 1 一 ( e r8t + e(r十σ2)叫 z
u
/
U
++j
K
气
/
立
/
l
i
l
1+ 各(kll) 工 1 十二击 i
一
〈
l
一
l
一
η
i
<
/
To show that the series 1二;二1
+12A=0;
U
2,
V n ~三 2.
t is diverge风 we will prove that
4η4
d. (4.16) for the de自nition of A.
In other words , u is a solution of the quadratic equation
u 2  2Au 十 1 = 0,
which has two solutions ,
conclude that
A + VA亡landAVA亡1.
u=A 十 J万丁I;
i丑(η) 十二<
(4.20)
Since u > 1 , we
)~二<
乞fk
In(n )
+ 1, V
n 三 1
(4.21)
The integral of the function f(x)  ~ over the interval [1 , n] can be
approximated from above and below as follows: Note that
fjdzzzf+1:dz
53.
CHAPTER 4. LOGNORNIAL VARIABLES. RN PRICING.
98
Since j (x) = ~ is a decreasing function , it is easy to see that
L<f(z)<17V
k+1 k
OJ
 /

Since
Z 巳 (k， k 十 1).
ZfzLzdz
>
which is the same as
z 击= 1 十 zi
(4.24).
口
(4.22)
Problem 7: Find the radius of convergence R of the power series
n 1z
p''''''1i
，
α
三气
Zfl:dz
<
位 k ln(k)'
Zfidz
(4.23)
and investigate what happens at the points x where Ixl = R.
Solution: It is easy to see that
The inequality (4.21) follows from (4.22) and (4.23) , since
2二 x
f:dz=1日(η
川) ln(叩门十 L <土 L
内 (η)
仨 k ln(k)'
vn>2;
立 L<ln(ln(η)) In(ln(2)) 十
…( 4.25)
1
k l B ( k ) 2 1(2)'
口
(424)
and co肌lude that the series '2二;二2 时而 is divergent
For example , (4.24) can be proved as follows:
1
~
rk+1
I~__dx)'I
儿 xln(x)ω中台儿
1η~
rk+1
一二~dx< 亨 l
xln仙
台儿
1
主~
1
一一一一7" dx = 予一一一
k ln(k) 山仨 k ln(k)'
which is equivalent to
I".'
二
' 一 … 一 ‘ '
儿土豆(幻川 I
k
2二 αJ?
(4.28)
with 句:时可， k 三 2. Note that
In a similar fashion , by considering the integral of x l~(百 over the interval
[2 , n] , we can show that
川
xk
.二
用
·
山
zi= 一;十主;
kd
(4.27)
坦(lB(n))h(ln(2))+ 」一<安 kln(k)'
1
nln(η)
仨
Z
ZfV
In (l n(2)) ,
99
we conell陇 from (4.26) and (4.27) that
flJI
1z
α
/ldz=ln(ln(η)) x ln(x)
J2
Then ,
7
4.1. SOLUTIONS TO CHAPTER 4 EXERCISES
二
nln(n) 仨 kln(k)'
(4.26)
lim
k→∞
Iα川 =lim(」YK=1(429)
k→∞ k ln(k))
Recall that , if li问→∞时11k exists , the radius of co盯erge肌e of the series
E二;二2αkXk is given by
(4.30)
一一」
 li叫→∞ αk 1/k '
From (4.29) and (4.30) the radius of co盯ergence of the series (4.28) is R = 1
We co肌lude that the series is co盯ergent if Ixl < 1 , and not co盯ergent if
Ixl > 1.
If x=工 1. the series becomes ~∞ ζ且 i:)!llce 'L ne 'L erms ζ且 have
S ~k=l kln(k)' Since the terms kln(k)
alternating signs and decrease in absolute value to 0 , the series is co丑vergent.
If x = 1, the series becomes '2二;二1 kl~(k)' which was show口 to be divergent
in Problem 6 of this chapter. 口
R
一
1
Problem 8: Consider a put option with strike 55 and maturity 4 months
on a nondividend paying asset with spot price 60 which follows a lognormal
54.
CHAPTER 4. LOGNORMAL VARIABLES. RN PRICING.
100
model with drift μ= 0.1 and volatility σ= 0.3. Assume that the riskfree
rate is constant equal to 0.05.
(i) Find the probability that the put will expire i口 the money
(ii) Find the risk一时的ral probability that the put will expire in the money.
(iii) Compute N( d 2 ).
Solution: (i) The probability that the put option will expire in the money
is equal to the probability that the spot price at maturity is lower than the
strike price , i.e. , to P(S(T) < K). Recall that
n
/111
叹 T
一
别
一
的
111/
一
一
2
σ
一2
/JIll
μ
'
q
Ill/
T
十
σ
4.1. SOLUTIONS TO CHAPTER 4 EXERCISES
The凡 d 2
101
= 0.511983 , and
N( d 2 ) = 0.304331 ,
which is the same as the riskneutral probability that the put option will
expire in the money; cf. (4.31)
To understand this result , note that
PRN(S(T)
< K)
=
P
(z
< 一叫到 +(γq4) 气
1σ VT
d z
d. (4.31) and (4.32).
N(d 2 );
=
J
口
Then ,
P(5(T)
< K)
P(;2< 忐)
=
P
(叫:2)< 叫忐))
P( (μqDT+dZ< 叫录)) )
Problem 9: (i) Consider an at拍emoney call on a nondividend paying
asset; assume the BlackScholes framewor k. Show that the Delta of the option
is always greater than 0.5.
(ii) If the underlyi吨 asset pays dividends at the continuous rate q , when is
the Delta of an atthemoney call1ess than 0.5?
Note: For most cases , the Delta of an atthemoney call option is close to 0.5.
Sol毗 on:
(i) Recall that the Delta of a call option is given by
qT 刷=严N(咄 +(γ q+ 号)T}
1σ〉于
For 5 = 60 , K = 55 , T = 1/3 ， μ= 0.1 , q = 0 ， σ= 0.3 , and 7' = 0.05 , we
obtain that the probability that the put will expire in the money is 0.271525 ,
i.e. , 27.1525%.
(ii) The riskneutral probability that the put option will expire in the money
is obtained just like the probability that the put expires i丑 the money, by
substituting the riskfree rate 7' for μ ， i.e. ,
PRN (5(T) < K)
P
(z
叫蒜)
 (γ q 一手)寸
1σVT
1
For an atth emoney call on a nondividend paying asset , i.e. , for K = S
and q = 0 , we find that
=N (
(r 十 j) ♂)三
(ii) If the underlyi吨 asset pays dividends at the ∞
c or巾
of ar1 ATM call 恒
口
iS
J
(4.31)
TN(d 1 )
=
e刊γ q+ 叮
(iii) Recall that
d2
叫到十 (7'  q 一主)T
σ叮
(4.32)
For a fixed riskfree rateγand 自xed maturity T , we conclude that ~(C) < 0.5
if and only if the dividend yield q and the volatility σof the underlying asset
55.
CHAPTER 4. LOGNORMAL VARIABLES. RN PRICING.
102
satis句T
4.1. SOLUTIONS TO CHAPTER 4 EXERCISES
When pricing a plain vanilla call using ris kneutrality, we proved that
nrT r ∞
II
_2 9
CBS(O) = 8(0) 二言=
exp ( (γ 一二) T+ σdz4:}dz
the following condition:
V 二7i
r
一 盯旷
Ke 一→
eT
This happens , for example , if r = q and T is large enough , si丑ce
lim N (
T•
00
(jV! 1 ~
103
1
and
J
lim
T•
00
I
J d 2
~
L, )
)
一
古 /ιLef勾手幻
仆;〉〉 d
叮
S (0 )N( d 1) 一 Ke T TN( 也).
μd公
出
贝创 何 均
ω2
…
0.5 严=∞口
In other words , we showed that
产去/∞ Adz=e一叩 (d2 );
(4.34)
V L,7i J d2
Problem 10: Use riskneutral pricing to price a supershare , i.e. , an option
that pays (max(8(T)  K , 0))2 at the maturity of the option. In other words ,
compute
V(O) = e rT E RN [(max(8(T)  K , 0))2 ],
where the expected value is computed with respect to the riskneutral distribution of the price 8(T) of the underlyi吨 asset at maturity T , which is
assumed to follow a lognormal process with drift r and volatility σ. Assume
that the underlying asset pays no dividends , i.e. , q = O.
Solution: Recall that
T
C
υ

cu nu exDA
/ft
/III
俨
'
qr2
T
1
1
1
飞
十
σ
d z
/
5二汇叫 (r ~2) T 十 σdz2)dz= 川 d1 )
From (4.33) , (4.34) , and (4.35) , we conclude that
V(O) = K 2e rT N(d 2)  2K8(0)N(d1 )
同一 rT
?
r∞
/
+8 2(0)
(2 …协 2σ VTx  ~ dx. (4.36)
)
~ Jd2 叫
The integral from (4.36) is computed by co 皿 pleti吨 the square as follows:
的。)
飞
1
1
1
/
~rT
e
V
/c
/
哎
r∞
/叫( (2γ  (j 2)T + 2σdzZ}dz
L,7i J d2
L, )
rT
(x
r∞
的 O)e~ / 侃p (
(一
L,7f 一出
v
飞
and note that
 2σ VT)2
z
>
叫茄)一 (γ 一手 )
一们IT
T
z 一也
V L,7i
d 2
8 2 (0) 户2)T ↓/∞
飞
工
「
2
VT? }
I
叫(一白 dy
J
L，
S 气泪栩(仰例 e
0印)诈
俨的州
创
ff7T'1
汀咽
一
与乍
的阶TTV七言
K2 e
左 l;
i 兀
衍
2
P( 伪 巾
〉仪叫叫叫(ρ σ T+2 卜 ζ ν
恬 lι: 均巾巾((←价T…一→2巧)泞 川Z 一 x;) dx
T
户(价
we used the substitution y = x  2σ IT and the fact that
巾Z
七
布仆 X
Vi言 l:〉e叫
一叫
S 0
俨
d2气协(仰
+刊呻
咐 的内
(j
L，
飞!"l，7i J 一 (d 2 +2σ IT)
dx
I
的 0) 川 )T 牛/∞叫{一 (x 一 2
J
Then ,
m
十 (2γ+σ 2)T 、1
η
L,
J
8(T) 三 K 白
(4.35)
'rorαm
Fr
￥
(4.33)
(4.36) and (4.37) , we conclude that
V(O恒 的rT N(d2) 
2K8(0)N(d1 )
+ 8 2 (0) 户 )TN(d2 十 2σ 叮) .口
56.
CHAPTER 4. LOGNORMAL VARIABLES. RN PRICING.
104
Problem 11: If the price of an 部set follows a normal process , i.e. , d8 =
μdt+ σdX ， then
8(t2)
8(tl) 十 μ (t2  t 1 ) + σd产石 Z， V 0 < tl < t2·
=
Assume that the risk free rate is constant and equal to r.
(i) Use risk neutrality to 且nd the value of a call option with strike K and
maturity T.
(ii) Use the PutCall parity to find the y~~ue of a put ~ption with s~rike K
~n'd maturity T , if the underlying asset follows a normal process as above.
Sol仰on: (i) Using riskneutral prici吗， it follows that
C(O) = e rT ER川max(8(T)  K , 0)] ,
= 8(0) +γT 十 σVTZ.
(4.38)
> K iff Z
>
K  8(0)  rT 一
〉三1r
Jd

Ke rT 去 Jd ehz
y '21r f
+
d
l一
而
挝
tllat
∞f
f
d
d
f
I
l
d
= e rT ERN [8(T)  K] = e rT (8(0)
zc
，
α
d，
z
m
曲
J!!叫
1vi言 f: e九 =
tm
一切
=
it(e 一句 I:
d2
e 2;
1 N(d) = N(d)
where N(t) is the cumulative distribution of the standard normal variable.
We conclude that
rT
rT
0(0) 工 (8(0) + rT)e N(d)  K e N(d) +
m
e rT 厅、 IT
~.L e 一 τ.
v
+ γTK)，
since ER
From (但.3ω9 and (任 .4钊)， we obtain that
4 创)
4 0
创
已rTσ vT
e
+卢J
飞 /2 汀
since 1  N( d)
严 Vi 言 [(卡归仰贝仰(仰仲
在
七
O
(问贝喇 们巾町叮 ÷ L l∞ 汁 dx
的 rTγγT「
O如)忖十忖州叫严一
叩旷)诈
Tη 巳 r
e r
Note
0(0)  P(O)
一
d2
τ
;[
 d
Then ,
0(0)
(ii) Regardless of the model used for describi吨 the evolution of the price of
the underlying asset , the PutCall parity says that a portfolio made of a long
position in a plain vanilla European call option and a short position in a plain
vanilla European put option on the same asset and with the same strike and
maturity as the call option has the same payoff at maturity as a long position
in a unit of the underlying asset and a short cash position equal to the strike
of the options. Using riskneutral pricing , this can be written as
KerTN(d) 一 (8(0)
Note that
8(T)
105
P(O) = 0(0) 一 (8(0) 十 γT)e rT 十 Ke rT
K e rT (1  N ( d)) 一 (8(0) 十 rT) ε rT(l  N( d))
where the expected value is computed with respect to 8(T) given by
8(T)
4.1. SOLUTIONS TO CHAPTER 4 EXERCISES
(4.39)
= N(d).
口
+ rT)e rT N(d)
~rT
十
b
_.
/币刀
二 V .L e一专
(4 .40)
57.
CHAPTER 4. LOGNORMAL VARIABLES. RN PRICING.
106
4.2
4.3
Supplemental Exercises
=
Solutions to Supplemental Exercises
(言。 I日(η)
Xn
is convergent to a limit between 0 and 1.
Note: The limit of this sequence is γ 臼 0.57721 ， the Euler's constant.
2. Assume that an asset with spot price 50 paying dividends continuously
at rate q = 0.02 has lognormal distribution with mean μ= 0.08 and
volatility σ= 0.3. Assume that the riskfree rates are constant and
equal to r = 0.05.
(ii) Find 95% and 99% riskneutral con在dence intervals for the spot
price of the asset in 15 days , 1 month , 2 months , 6 months , and 1 year ,
i. e. , assuming that the drift of the asset is equal to the riskfree rate.
3. If you play (American) 1
roulette 100 times , betting $100 on black each time , what is the probability of winning at least $1000 , and what is the probability of losing
at least $10007
4. Use riskneutral pricing to find the value of an option on a 口。且一
dividendpaying asset with lognormal distribution if the payoff of the
option at maturity is equal to max((S(T))α K , O). Here ， α> 0 is a
fixed constant.
5. Find a binomial tree parame~rization for a riskneutral probability (of
going up or dow时 equal to ~. In other words , find the up and down
factors u and d such that
+
(lp)d
pu2 十 (1
 p)d2
ρ
u
ρ
U
=
(剖 ln(η)
is convergent to a limit between 0 and 1.
Solution: Recall from (4.21) that
ln(η)+j<zi<ln(η) +
1,
;/
η 三 1，
which can be written as
1
(i) Find 95% and 99% confidence intervals for the spot price of the asset
in 15 days , 1 month , 2 months , 6 months , and 1 year.
pu
107
Problem 1: Show that the sequence
1. Show that the sequence
Xn
4.3. SOLUTIONS TO SUPPLENIENTAL EXERCISES
<
一
η
zn <
才
i
vv n
>
42···b
It is easy to see that
Xn+1 一句 :Lln(η+
η 十 l
Therefore , X n十 1 <
Xn
1) +
ln(η) .
if and only if
(n+ 1
tI<lnh+1)一 ln(η)= 叫丁)
T且is is 叫uivalent to 1 < (η 十 1) I口(平)， and therefore to
e
<
/η+1η+1
I 一一I
=
飞
η/
(~1 η+1
11 十一 i
飞
η/
which holds for anyη;三 1 ， from the definition of e.
We showed that the sequence (x讪=1:∞ is decreasi吨 and bounded from
below by 0 and from above by 1, The seque口ce is therefore convergent to a
limit between 0 and 1.
口
T(
"
ι
λ
ι
U
7
，A
十
门
伊
'
σ
"
，
何
) Fλ
'?L
U
ifp=i
1 American roulette has 18 red slots: 18 black slots: and two green slots (corresponding to 0 and 00). European roulette: also called Fr ench roulette: has only one green slot
corresponding to O.
Problem 2: Assume that an asset with spot price 50 paying dividends
continuously at rate q = 0.02 has lognormal distribution with mean μ= 0.08
and volatility σ= 0.3. Assume that the riskfree rates are constant and equal
to r = 0.05.
(i) Find 95% and 99% confidence intervals for the spot price of the asset in
15 days , 1 month , 2 months , 6 months , and 1 year.
58.
CHAPTER 4. LOGNORMAL VARIABLES. RN PRICING.
108
(ii) Find 95% and 99% riskneutral con且dence intervals for the spot price of
in 15 days , 1 month , 2 months , 6 months , and 1 year , i.e. , assuming
that the drift of the asset is equal to the riskfree rate.
theωset
Solution: If the asset has lognormal distribution , then
S(t) =
=
S(O) 叫( (μqf)t 十 σ0z }
)
2 }
飞飞
TXT
50 exp(O.Ol 日十 0.30z).
< Z < 1. 95996) = 0.95; P( 2.57583 < Z < 2.57583) = 0.99.
Therefore , the 95% and 99% con直dence intervals for S (t) are
[50 exp(O.Ol 日一 0.30 . 1. 95996) , 50 exp(O.Ol 日十 0.30· 1. 95996)];
[50 exp(O.Ol 日 0.30 ·2.57583) , 50 exp(0.015t + 0.30 ·2.57583) ],
respectively.
The riskneutr l confidence intervals for the spot price of the asset are
α
，
found by substituting the riskfree rate r for μin (4 .4 1) to obtain
SRN(t) = 50 exp( 0.015t + 0.30Z).
Therefor the
飞
川
r 100 ,
1.
100 ,
with probability 器;
with probability 盖
Note that
E[ Wi]
μ =
10 ,
. _ _,
= 一 (100)
19 ~/
9. __
100
+ 1~n 100 =一一'
' 19~
19
600/10
σ= 削 (Wi) = E[(Wi )2] 一 (E[Wi ]? = 二口.
Let W = L:i~~ Wi be the total value of the winr山gs after betting 100
times. Since every bet is independent of any other bet , it follows that W is
the sum of 100 independent identically distributed random variables. From
the Central Limit Theorem we 自nd that
W
自
100μ+ 10σZ
=
100006000/10
~ ~Y 一 =Z
19
19
一一一+
The probability of winning at least $1000 can be approximated as follows:
95% and 99% confidence intervals of SRN(t) are
[50 exp(O.Ol 日一 0.30 . 1. 95996) , 50 exp(O.Ol 日十 0.30 . 1. 95996)]
[50 exp( 0.01日 0.30 ·2.57583) , 50 exp( 0.01日十 0.30 . 2.57583月 7
respectively.
For t ε{ 主 ι i ，~， 1} , we obtain the following confidence intervals:
t
15 days
1 month
2 months
6 months
1 year
109
Solution: Recall that an American roulette has 18 red slots , 18 black slots ,
and two green slo~s. Therefore , every time you bet on black , you win $100
with probability 击 and lose $100 with probability 击 In other words , if V只
is the value of the winnings in the ith round of playing , then
(4 .4 1)
Recall that the 95% and 99% con直dence intervals for the standard normal
distribution Z are [1. 95996 ,1. 95996] and [2.57583 , 2.57583 ], i.e. ,
P( 1. 95996
4.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
95% CI S(t)
43.36 , 57.76
42.25 , 59.32
39 .43, 63.72
32.25 , 75.35
28.20 , 91. 38
99% CI S(t)
41. 45 , 60 .4 3
40.05 , 62.57
36.56 , 68.70
29.17 , 86.99
23 .44 ,109.90
95% CI SRN(t) 99% CI SRN(t)
43.28 , 57.66
41. 36 , 60.30
42.14 , 59.18
39.96 , 64 .4 1
39.23 , 63 .41
36.36 , 68.37
32.74 , 75.21
28.73 , 85.70
[27.36 , 88.68
22.75 ,106.65
Problem 3: If you play (American) roulette 100 times , betting $100 0丑
black each time , what is the probability of winning at least $1000 , and what
is the probability of losing at least $10007
P(W>
1000) 臼
=
[ 100006000 飞/101
PI 一一一一十
v …
Z> 1000 I
19
19
.
J
P(Z > 1. 5284) = 0.0632.
The probability of losing at least $1000 can be approximated as follows:
1000)
目
[ 100006000 飞/101
PI 一一一一十
v 一 Z < 1000 I
=
P(vV <
P(Z < 0 .4743) = 0.3176.
19
19
J
We conclude that the probability of winning at least $1000 is approximately
6% , and the probability of losing at least $1000 is approximately 32%. 口
Problem 4: Use riskneutral pricing to 且nd the value of an option on a
nondividendpaying asset with lognormal distribution if the payoff of the
option at maturity is equal to max((S(T))α K , O). Here ， α> 0 is a 在xed
constant.
59.
CHAPTER 4. LOGNORMAL VARIABLES. RN PRICING.
110
Solution: Using riskneutral pricing , we find that the value of the option is
V(O) = e rT ERN[max((S(T))  K , O) ],
α
where
盯)
=
4.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
note that the change of variables y = x 一 ασ JT was used above.
We conclude that
v nu  s nu
S(O叫 (γDT 十 σVTZ)
叫
//ll1
α
ex D
//I11
α
ι
咱
t
4
Note that (S(T))α 三 K is equivalent to S(T) 三 K /α. Using (4 .42) , we fin
牛::::::}
Z>
d
a.
The丑7
叩
V(O)
=
俨
'
where
1
S(T) 三 K 1 /α
In
α
十
川
一2
lliI/
111l/
T N
十
α
α
(吕 )+(γ4)T
σ
d
k
俨
u
ρ
'
TN
α
口
σJT
~~oblem 5: Find a binomia) tree parametrization for a riskneutral proba旧出 时
r
bility (of going up) equal to ~. In other word s ， 直缸缸1
包 aI
tL 缸ld d such that
丑
亏琴;江汇/汇川川~(川川川(牛←0创)川)
仙阳(侈问仰附附
μ创
贝
S仰俐(仰阶附附
+
Recall from (但.34) 由 创
4 刊 t hat
钊
4
δ
'
v
ρ
、
?
，
p
(1  p)d
2
pu 十 (1  p)d2
pu
5且:e句 = erTN归)
111
心
，
u
"
，
u
7
'
，
、
去
十
σ
4
η
)
U
向λ
ι
ι
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~dzZhz

u
2
+ d2
=
2e(2r 十 σ2)dt7
and therefore
KerTN(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)
卖 优叫叫 α
Vi 言 1~ex均((←(归川
七
p(
We solve (4 .4 3) and find that
优叫 ←归忻川川一斗斗
p( 机)沙俨
1)
叫叫(川川) (γ 十 子)川 左 ιι1:+Qαωσd 叫叫(一5) dν
← 1
α川一」斗
才 七 叫
T) VIi言 刊叩叮叫
而叮而)
引
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 <
u.
口
t
M
M_
= 旷♂沪叫 1 十 Je刷耐几
叮 山 02 6衍 一
气(←
工矿 M
(1 一
ν e o 2M

O
1) ,
60.
c
ap er wo
hu
4EU
Taylor's formula and Taylor series. ATM
approximation of BlackScholes formulas.
5.1
Solutions to Chapter 5 Exercises
Problem 1: Show that the cubic Taylor approxi口lation of vfτx around 0
IS
"十 Z 自 1 +王一旦+主3
2
8 ' 16
Solution: Recall that the cubic Taylor approximation of the function f(x)
around the pointα 工 o is
f(x) 臼 f(α)
+
(x 一 α) 1' (α)
1(_.
+
(x 一 α)2.£1/(_.
~
... / 'f"(α)/
2
" 
,
十
(x 一 α)3
p
fi
f(O) 十 x1'(O) 十三f响)十三 f(川0).
2"
6
,/
f(的 (α)
(5.1)
For f(x) = Vf石， we find that
1' (x) = 卢z ; f ff ( z h arld
丑
U
Z ) 3/ 2 ;
尸内 = ~
)(x)
(μ
therefore
f(O) = 1; 1' (0) 寸;ff(0)=j; 内)=;(52)
From (5.1) and (5.2) we conclude
由前
dτZ 但十三 zf+21 口
2
8 ' 16
Problem 2: Use the Taylor series expansion of the function eX to find the
value of eO. 25 with six decimal digits accuracy.
113
61.
CHAPTER 5. TAYLOR'S FORMULA. TAYLOR SERIES.
114
Solution: Recall that the Taylor series expansion of the function f(x) = ♂
around 0 converges to eX at all points ♂ ε JR.， i.e.
,
∞巾 k
♂=三示，
5.1. SOLUTIONS TO CHAPTER 5 EXERCISES
l
1 x 2
L: x
2k
= 1 + x 2 + x 4 忖 6+... ， i xE(l ， l)
Problem 4: Let
ι(_l)k+l x k
l. e.
T(x) = ):
ι (0.25)k
w
_
be the Taylor series expansion of f (x) = In (1
T(x) = f(x) for all x such that Ixl < 1.
,
eO.
25
=
lim
xn
, where
xn
= 安哗fii7h 三 0
"'.,
Let
Pn(x)
n→ 3二
k=O
Note that the sequence {xn}n=O:∞ is increasing. It is then enough to
compute Xo , Xl , x2 , …, until the first seven decimal digit~ _<!f these terms are
the same , in order to find the first six decimal digits of e ' 均. We find that
U
X3
= 1;
= 1. 28385417;
X6
= 1. 28402540;
Xo
Xl
X4
X7
= 1. 25;
= 1. 28401698;
= 1. 28402541 ,
X2
X5
= 1. 28125;
= 1. 28402507;
=
be the Taylor polynomial of degree ηcorrespondi吨 to f(x). Since T(x) =
f(x) = T(x) for all Ixl < 1 if and only if
limn→∞ Pn(x) ， it follows that
J兰在 If(x) 一凡 (x)1
= 0,
i lxl< 1.
(5.3)
(i) Show that , for any x ,
叶
w
f(x)  Pn(x)
口
fiz
zm
一
一
j
t
4
eO. 25 自 1. 284025.
十 x). Our goal is to show that
已 (_l)k+l x k
) ~
才
and conclude that
口
JR.
i xE
For x = 0.25 we find that
已0.25
115
i
6、
1
1
?
4
'
,
n
L

i
罔
刊
ν

…
, ι'
G
U
(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
1x
L: x
k
= 1 + x + x2 + x 3 十
，
i
In (1  x 2 )
= 一了 zffzz221Efzf。c 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〉 ;zz 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)
62.
CHAPTER 5. TAYLOR'S FORMULA. TAYLOR SERIES.
116
(x  t) n(1
J
I{X 一一~ (1 十)7川 n! d
/
.
Jo
n!
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<t<x<1 、
1
一'
一一
J豆豆 If(x) 一凡(x)1
f(EY T41dt 三 42Ifzdt
f(x)  Pn(x) =
巳 [0 ， 1).
0, V x
8
=
r
I
(5.7)
s ( 1)η十2(8
{s ( 1?n+2(8 十 t)n JIrJo
(1 + t)叫…。
s
(8 十 t)n
I
7
(一 l叫 1  z))I;二;
8叫n(18)
=
( x)nlln(l 十 x)l;
8 ll叫 1
n
(5.11 )
~ (川十川附
eCJVlt :
(5.12)
e 一σJ瓦
(5.13)
1
1 (T'
σ~
p  一十一 i 一一~ I v8t.
2 ' 2σ2/ . 
a
n
( 5.10)
(
d
0<Z<8< 1.'
一一
{S (8  z)η{ I:i 8 n
l
z < I 一一…
Jo (1  z)卅1 厅  Jo 1 z
8
(5.9)
u
we find that
If(x)  Pn(x)1
V x E (1 , 1) ,
Use Taylor expansions to show that , for a small time step 8t , u , d and p may
be approximated by
By taking absolute values and using the fact that
主二三 一'
<8‘
1z
where
A =
(8  z)n
{S

J
h
(1 十 t)叫…
Using the substitution t = z , we obtain that
=
 t)n
= 0,
u = A+VA亡I
d = AV万1·
er8t  d
p 一ud'
(1 + t)叫山
I
f(x)  Pn(x)
(5.8)
Problem 5: In the CoxRossRubinstein parametrization for a binomial tree ,
the up and down factors u and d , and the riskneutral probability p of the
price going up during one time step are
x. From (5.6) , it follows that
{X(_1)n+2(x  t)n J
~~
~~n+l / dt = I
(1 + t) 肿1
~V
Jo
I。
ε(1 ， 0).
and co叫ude t~at the Taylor series expansion of the function f(x) = ln(l 十 x)
converges to f (x) for a叮 x with Ixl < 1.口
We conclude that
(iii) Assume that x 巳 (1 ， 0) and let
0, V x
From (5.7) and (5.8) we obtain that
Jl忠 If(x) 一凡 (x)1
lim If (x)  Pn (x) I
0
=
,
we obtain that
lf(x) 一凡 (x)1
+ x)1
We conclude that
(ii) Let x ε[0 ， 1). By taking absolute values in (5.6) and using the fact that
+t
117
recall that , by definition 8 = x.
Note that , for a叮 Z ε(1 ， 0) ，
From (5 .4) and (5.5) it follows that
f(x)  Pn(x) =
5.1. SOLUTIONS TO CHAPTER 5 EXERCISES
 8)1
J_.
dz
内
(5.14)
In other words , write the Taylor expansion for (5.95.11) and for (5.125.14)
and show that they are identical if all the terms of order 0 (8t) and smaller
are neglected.
Solution: We will show that the Taylor expansion for (5.95.11) and for
(5.125.14) are identical if all the terms of order 0 ((8t) 3/2) and smaller are
63.
CHAPTER 5. TAYLOR'S FORMULA. TAYLOR SERIES.
118
5.1. SOLUTIONS TO CHAPTER 5 EXERCISES
neglected. In other words , we will show that
1 + a 05t +
U
eu!8i 十 0((6t)3/2);
(5.15)
d
e一σ!8i 十 0((6t)3/2).
(5.16)
1
1
a
p
2
,
1 a 05t
2σ2
(5.17)
日
ι
l+x 十二十州7 川→ 0;
1+j 十 O(此 as x • 0;
lj+0( 向 as x • 0
l 十 ?+0川)
Then ,
e
σ v&t
~
(e
r<5t
+ e(r+u
2)
+
们
叫 6t
宁 十 O仰州(仅仰阳(仔仰州
户= (1+ 平。((ot)'))' 
Therefore , (5.15) and (5.16) are established.
Finally,
1 十 (r 忖协 0(( 们 )2))
1
1
+ 0(( 们 )2;
d亡工工作26t + 0((6t)2) = σ 币、Ii十 0(6t)
0((6t)2))I
0((6t)2)
σ v6t + 0(( 们 )3/2)
and
A 十 J万τ丁
1 十字 + 0((M)2)
er<5t  d
u  d
σ yft;i + (γ ~)们十 0(( 们)附)一 σ 十 (γ
σ26t
u 
工
(1 + r6t + 0(( 圳))一 (1 一 σ yft;i 十字 + 0((6t)3/2))
(1 + a yft;i 十字+ 0((们 )3/2) )一 (1  a yft;i 十字 + 0((6t)3/升
1+
。但!l
σ v6t .(1 十一叫一 +
(
缸
6
l+a v6t 十 字 +0 ωt
叭咐州(仅
(θ
阳
仰川
1 一 d 十宁+叫川
u = eU V&t +0((M)3/2);
d = e u V&t +0((6t)3/2).
<5t)
;(1dt+0(( 们 )2)
+ a v6t + 0((M)3
0((σ 05t )3)
we conclude that
p
A =
(σ yft;i)2
而 十一门一+
eUV&t
In particular , note that
飞/i 十 O(M)
+ 字+ O( 川
Since
Recall the following Taylor approximations:
ex
A 一 J万丁I
d
2
一十一 (γ 一三 )v6t十 0((出)3/2) ，
缸
6t
宁 + 叫创仰州(付仰川
2σ yft;i + 0((6t)3/2)
1 ,
a2
一号 )yft;i + O(6t)
2σ 十 O( 们)
一十一 (γ 一巳 )v6t十 0((6t)3 j 2) ，
2
2σ2
wh
Problem 6: (i) What is the approximate value ~凡附
αpprox
丑 忖飞
丑
money put option on a nor1 dividerld一paying underlying asset with 印 创 price
s pot
S = 60 , volatility σ= 0.25 , and maturity T = 1 year , if the constant riskfree
interest rate is r = 07
(ii) Compute the BlackScholes value PB Sr严叫咆气
伫
归俨户
r
氏向仲仲如?卢
mate the relative approximate error
一
IPBS户闪=0 ~α即叫 r=O ， q=ol
PBS， r=O ， 俨 O
64.
CHAPTER 5. TAYLOR'S FORMULA. TAYLOR SERIES.
120
(iii)Assume thatT=0.06and q=0.03Compute the approximate value
Pα.pprox，r=0.06俨0.03 of an ATM put option and estimate the relative approxi日late
error
jPBS，r=0.06，俨0.03 一凡pprox，r=0.06俨0.031
PB S ,r=0.06 ,q=0.03
'
where PB S ,r=0.06 ,q=0.03 is the BlackScholes value of the put option.
u
(5.18)
Solution: (i) Using the approximate formula
Pα附町=O，q=O
_ /T
21r'
as

V
5.984134.
PBS，r=O ，俨O
Ke r
Based on the linear Taylor expansion e x 自 1 一凯 new approximation
formulas for the price of AT lVI options which satisfy the PutCall parity can
be obtained by replacing rT and qT by 1  e rT and 1  e qT , respectively.
The resulting formulas are

2
十
叮
S (e qT  e rT )
2
(5.19)
(5.20)
Solution: (i) From (5.19) and (5.20) , it is easy to see that
approximate formula
川
丁
IT (
(γ +q)T
Pappr叫附 = aSy ;1r 1 一
j
we obtain that
Pα.pprox，r=0.06，q=0.03 = 4.814848.
自nd

that
4.886985 ,
QU
7
IPBS， r=0.06， 俨0.03  ~αpprox ， r=0.06 ， q=0.031
.r BS,r=0.06 ,q=0.03
=
(1JTlq)T
2 S
rT
(寸 e ) 十 Se qT
凡，pprox_new ， r=0.06 ， 俨 0.03
_
Se rT
kfrT7

4.86103 1.
The BlackScholes value of the put option is PB S ,r=0.06 ,q=0.03 = 4.886985 , and
therefore
1.4 761 %.口
0.014761
Problem 7: It is interesting to note that the approximate formulas
γ
卜J lγγγqω
υ (r 旷
C 臼 σS占 (ν1价巳叮 )T
一 γ
丁
pc 十川=
since K = S for ATM options.
(ii) Using the new approximation formula (5.20) , we obtain that
and therefore
P 自 σ s f[;
S (e qT 一 (γ  q)T) 并 Se rT
(i) Show that the PutCall parity is satisfied by the approximations (5.19)
and (5.20).
(ii) Estimate how good the new approximatio丑 (5.20) is , for an ATM put
with S = 60 , q 二二 0.03 , σ = 0.25 , and T = 1, if r = 0.06 , by computing
the corresponding relative approximate error. Compare this error with the
relative approximate error (5.18) found in the previous exercise.
IPBS，r=O俨0 一凡pprox，r=O，q=O 1= 0.002604 = 0.26%.
PBS ,r=0.06 ,q=0.03
C
jz
and therefore
From the BlackScholes formula , we

P 自 σS1/.!.e qT+
2
PBS ,r=O ,q=O = 5.968592 ,
Usi吨 the
P 十 Se qT
JZehedSMfrT)
(ii) From the BlackScholes formula , we find that
(iii)
for ATM call and put options do not satis有T the PutCall parity:
C 用 σ Sy ;1r
we obtain that
Pα.pprox，r=O， q=O 
5.1. SOLUTIONS TO CHAPTER 5 EXERCISES
十
(r  q)TS:
IPBS ， 问 0.06 ，俨 0.03  ~α.pprox_new ， r=0.06 俨 0.031
PBS ,r=0.06 ,q=0.03
工 0.005311
= 0.5311%.
Recall from Problem 7 that the approximation error corresponding to
the original approximation formula is 1. 4761%. We conclude that , for this
particular example , the new approximation formula is more accurate. 口
2
(γ  q)T
2
σ
Problem 8: Consider an ATM put option with strike 40 on a nondividend
paying asset with volatility 30% , and assume zero interest rates.
65.
CHAPTER 5. TAYLOR'S FORMULA. TAYLOR SERIES.
122
Compute the relative approximation error of the approximation
σ币汪叫叫…
p川 1 ， 3, 5, 10 , and
P 臼
20 years
Solution: We expect the precision of the approximation formula for ATJVI
options to decrease as the maturity of the option increases. This is , indeed ,
the case:
Approximation Error
T
PBS
Pαpprox
1 4.787307 4.769417
0.38%
1. 13%
3 8.291860 8.199509
1.88%
5 10.704745 10.507368
3.76%
10 15.138795 14.589748
7.55%
20 21 .409489 19.906608
Problem 9: A 在ve year bond worth 101 has duration 1. 5 years and convexity
equal to 2.5. Use both the formula
(5.21)
which does not include any convexity adjustment , and the formula
+
~C(b:..y)2 ，
(5.22)
the price of the bond if the yield increases by ten basis points (i.e. ,
0.001) , fifty basis points , one perce风 and two perce风 respectively.
to 自nd
Solution: Denote by B n 毗 D the approximate value given by formula (5.21) for
the value of the bond corresponding to the new yield. Then , b:.. B = Bneω .DB
a叫 from (5.21) , it follows that
Bnew ,D =
B
(1  Db:.. y).
(5.23)
Similarly, let Bnew.D.C the approximate value for the value of the bond
given by formula (5.22). We obtain that
,
Bn w ,D ,C
=
B
(1  D
t>y 斗叫) .
Note that pz1017D =157and C =25i口 (5.23) and (5.24).
The following approximate values are obtained for
b:.. y ε{0.001 ， 0.005 , 0.01 , 0.02}:
Bneω ，D
，G
0.0010 100.8485 1B0η0ε.ω8，4D8 6 0.0001%
0.0050 100.2425 100.2457 0.0031%
0.01
99 .4850 99 .4976 0.0127%
0.02
97.9700 98.0205 0.0515%
b:.. y
Bne ω ，D ，G  Bne叽 D
B neω ，D
IPBS 一凡pproxl
[
PBS，r=O，q=。一
b:.. B
E一目  Db:.. y
123
The last column of the table represents the percent difference between the
approximate value miIlg d11ratioa done?and ttle approximate value usiIIg
both duration and convexity, i.e. ,
Here , the Approximation Error is the relative approximation error defined as
b:.. B
王一用一 Db:..y ，
5.1. SOLUTIONS TO CHAPTER 5 EXERCISES
(5.24)
〔
一
66.
CHAPTER
124
5.2
TAYLOR'S FORMULA. TAYLOR SERIES.
5.
Supplemental Exercises
5.2. SUPPLEMENTAL EXERCISES
6. The goal of this exercise is to compute
11In(1 x) In例
1. (i) Let g(x) be an i凶nitely differentiable function. Find the linear and
q~adrati~ Taylor approximations of eg(x) around the point O.
(ii) Use the result ~bove to compute the quadratic Taylor approximation
around 0 of e(x+I)2
(iii) Compute the quadratic Taylor approxi可lation around 0 of
by using Taylor approximations of eX and eX·.
2. Show that
e X 一 L=0(内，
as x
l 十Z
e(x+I)2
(i) Show that
:地(同  x) ln( ♂))
l'扑
PAVAOve
<
//111
已
4i
<
+
[
Z
1z十
叫
仨IηJ O
飞
VV Z
1
1
1
In(l 一叫
1
立
(iv) Use that fact that
14
>
立主 =Z
41·&
/
ha
4L
0,
(iii) Prove that
4. Recall that
z
111J/'
1z
= x/I (问  x) ln(x)) =
lim
μl1 1 一 功训巾叫
1h
〉」(
ln (μ
Z
凶丑叫
以坤口叫巾
川
around the point 0, and find its radius of convergence.
十
(5.26)
(ii) Use the Taylor series expansion of 1 叫 1 x) for jxj < 1 to show that
• o
叫;号~)
才
t
A
dx
and conclude that the integral (5.26) can be regarded as a definite
integral.
3. Compute the Taylor series expansion of
/JIl
125
4EU
to obtain that
/Iit
4ti
十
1111/
1z+ l2 b
/fiI 1z+ l2
♂
♂
<
ρ
U
<
吨
E
L
飞
十
l
i
f
/
Vv Z >
才叶到 ln(x
咱
5
4
5. (i) Find the radius of convergence of the series
~4
1 十
L
2!
、、 8
十
'
L
4!
~12
十二一+.
'
6!
(5.25)
(ii) Show that the series from (5.25) is the Taylor series expansion of
the function
X2
ez2
十 e2
。
，
"
作2
6
7. Consider an ATM put option with strike 40 on an asset with volatility
30% and paying 2% dividends continuously. Assume that the interest
rates are constant at 4.5%. Compute the relative approximation error
to the BlackScholes value of the option of the approximate value
.IT ( 1 一
Pα附 oX ， r 手 O ， q 并 o σ BV
;1T
(7'十 q)T (7' 
FJ

if the put option expires in 1 , 3, 5, 10 , and 20 years.
q)T
67.
CHAPTER 5. TAYLOR'S FORMULA. TAYLOR SERIES.
126
5.3
5.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
Problem 2: Show that
Solutions to Supplemental Exercises
Problem 1: (i) Let g(x) be an infinitely differentiable function. Find the
linear and quadratic Taylor approximations of eg(x) around the point O.
(ii) Use the result above to compute the quadratic Taylor approximation
around 0 of e(x+1?
(iii) Compute the quadratic Taylor 斗pproximation around 0 of e(x札 )2 by using
Taylor approximations of eX and eX*.
ex 一 L=0(z2)7aS
1 十 Z
= g'(x)eg(x)
1 十Z
+ (g' (x))2)e g(X).
+
O(x 2) ,
as
x
•
O.
e x
j(O)
•• _,
x 2 _"
+ x j' (O) 十五 f气。) +
。
J
O(x ) ,
x
as
•
0,
1
一l+x
as
x • 0;
x
are
•
O.
as
。但2) ，
x
•
O.
一二十 O(x3 ) ，
In
e g (川
I
x
•
1
(L
(2x)2
3
1 + 2x 十一γ + O(x ) ,
•
0,
I
口
xJ
叫;52)=l中川川
O.
=
I
0
Solution: Note that the function
quadratic Taylor approximations
(2X)2 " f ~3
J
e""x = 1 + 2x + ;一 + O(x ) ,
x2
e
_ 1 + x 2 十 O(x气 as x
t
around the point 0 , and find its radius of convergence.
as x • O.
(ii) By letting g(x) = (x 十 I? in (5.27) , we find that
as
x
(1 +纠
飞 1
(5.27)
e(叫 )2 工 e + 2ex + 3ex 2 + 0 (川)，
as
Problem 3: Compute the Taylor series expansion of
becomes
Usi吨 the
l~X
as
十。但 3) ，
l+x
The quadratic Taylor approximation
eg(x) =
0
Note that we implicitly proved that
eg(x) = eg(O) 十 xe到 O)g'(O)
(iii)
1 x + x 2
e x 一 L= 一主十 O(泸)
= j(O) 十 x f' (O) + O(x 2) , as x • 0,
can be written as
j(x)
•
Therefore ,
The linear Taylor approximation
j(x)
Ix 十二十州，
山m
ρ
υ
1
j"(x) = (g"(x)
and
Z
Solution: The qu础 atic Taylor approximations of e x and
Solution: (i) Let j(x) = eg(x). Then
f' (x)
127
as
x
•
0;
is not defined for x = lor x = 1. Therefore , the largest possible radius of
convergence of its Taylor series expansion around 0 is 1.
The Taylor series expansions of the functions In(l 十 x) and In(l  x) are
In (1
+ x)
it follows that
e(x十1)2=e dzet=e(l十 2x 十 2x 2 )(1 十 x 2 ) 十 O(x3 )
e 十 2ex + 3ex 2 + O(x 3 ) , as x → O. 口
In (1  x)
立
fz2z3z4
(OMJZZ 一一十一一一+...
k
2 '3
4
军空气 x k
x2
x3
x4
, V ♂ ε(1 ， 1];
一予:二 = x 一一一一一一一... , V x ε[1 ， 1) ，
乞fk
234
68.
CHAPTER
128
TAYLOR'S FORMULA. TAYLOR SERIES.
5.
5.3. SOLUTIONS TO SUPPLEl1ENTAL EXERCISES
Fr。因 (5.31) ， we also find that
and have radius of convergence equal to 1.
We conclude that the Taylor s~ries expansion of In (i~~) is
/IIl 1
Z 十
叫;占~)
In(1 + x) 一 l叫 1  x)
xk
=
•
, I.
,
1
2叶子+车十
土
liI/ /II11
n
EA
q
G
414
+
liI/
1z
=
1 十
至 271
，I xEε 川)
川
X+ j
Z
Recall that
Sol州on:
1
1
3(2z+1)21tF 一
1 十一)
(1+~时
(1 十 ~r l x?1
(5.29)
1+ 一」龟
12x(x 十 1)'
(]斗"
2 y 3 , 2ν5
In ( ~一立) = 2y 十一+一十... , I y E (1 , 1).
飞 1  y)
3 ' 5
(5.30)
l
l x?1
(5.33)
x> 1.
(5.34)
Using (5.34) , we 丑nd from (5.33) that
(1 十 ~r十二
Recall from (5.28) that
< e1吨
e<(l+:yl
Problem 4: Prove that
e <
1y 一」
+ 1)2k
3(2x 十 1)2 位 (2x
and therefore that
(5.28)
and has radius of ∞I1飞 ergence equal tω0 1.口
c O口v
∞
(1+D 忖 <
1十
x k λ 2x 2j + 1
t; ((一叫+讨
(
129
卢
e
< e(1+:)? VG17
and we conclude that
0'
(1 十 D 叫一百 < e ,
For any x 三 1 ， substitute y = 古 i口 (5.30) and obtain that
咔 +D =击+… 2
Jwn2
4 、民十
,
I x
> 1,
which can also be written as
(什)咔 +1)=1+ 。J 川十川一1 亏d +. .,
XI
, I x > 1.
(5 出)
From (5.31) , we find th创
1
(→)咔 +:)7VQ17
<
I x 二三 1.
The left inequality of (5.29) is therefore established.
口
Problem 5: (i) Find the radius of converge肌e of the series
t2
+泸
川 十
(5.35)
十
一
4
创
(ii) Show that the series from (5.35) is the Taylor series expansion of the
function
4t4
十
ex2
+ e x2
2
which is equivalent to
e
<
/
... x+~
(1 十二 I
, I x 二三1.
飞
X
I
The right inequality of (5.29) is therefore established.
Solution: (i) The series (5.35) can be written as a power series as follows:
(5.32)
T
Z
一

∞Z
H
α hv
z句
w n
α
句
一
一
l
回
Vv DA >
69.
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
一
门L
P
n
咱
止
川
/
自nd tl川
,,,,,,,,
n4
p
...."
Ll吨1
~
e1/2 lim
p→ x 飞/2p
11日1
，'~
,,,.

In
2
1/
I
(2p)1/
(iv) Use that fact that
limp→∞ α4p1 / 4p
立去 z
~~，
to obtain that
Sol毗 on:
;倍 ;21+ 主 i二;11)

2位
k!
1+z4z8212
37+ZI+ 石「+
which is the same as the series (5.35).
口
1
f工 x 4j
 但 (2j)!
z
才112(1 一叫) dx 工 2Z
(i) First of all , note that
?V口 (1  x) In(x))
it is easy to see that
1 王三产+ (_1)k x 2k
立
T (x) is
ez=57Vzd?
2
(iii) Prove that
[ln(1 一叫) dx
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
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
0
l l
lims叩k→况 αkl 1/k
0,
1
一…一一一τ;:;
p~叩 ((2p)! )l /2p )

=
I
2p
(
Therefore , the radius of convergence of the power series
R 一
= lim (叫 x) In(x)) =
x/1
才1ln川 h川

P叫 ((2p) !)1/ 4
p
(5.36)
=ε
,.
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
=
lim (同  x) ln(x))
(5.38)
x/1
We compute the left hand side limit of (5.38) by changing it to a limit to
infinity corresponding to y = ~ as follows:
/iBh 叽
/
1U"1
l
且 m 恼
h li →
1/fIi1/JI
i
一
l U
:与(l n(l 一仙 (x ))
俨 ∞ ，
l
/
/
内 wv 1
且
J
l
俨 m l
/ uu /飞 一 n UU
一 U
l
一
一 ∞ n
l
r
j
/
v
U
 〓 气
i
h
vhuη n
u .
l l
!
一
ο u
w l
υ
f
H俨 ∞
/
w
I
i
l
/
n
一
;
ai
，
，
，E
E、
、
、
，
U
J
J
/
t
飞t
、
、
1
1
2
J
、
飞
E
1
'
'，
/
'
'
飞
飞
、
、
l
/
70.
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
J41j)=:
and therefore
J坦咐 D")
=
(5 .40)
1
ln(ν)
1
li~ (ln (1
x) ln(x)) =
0
The integral (5.37) is equal to the de自缸l it e int egra1 betweer1 0 and 1 of the
fin 怡 丑挝
丑旧
让
怡 址 创
丑
g(l) = 0;
ln(l 一 x)l叫 x) ，
g(x) =
absolutely ∞ 盯erge旧 to
c or
1口 (1
ln(1  x) ln( x)
[ln(1X)ln(x)
V「 ZK
一
Therefore ,
汇
1.
1
x) , i. e. ,
(ii) The Taylor series expansion of ln(l
川一叫
M lι Z ) =
VO<x<
L忑
1
η(η 十 1)η+1
η(η+
1)
(η+
1)2
111
n n+ 1 (η+1 )2
∞丑
C O I1 毗
∞
g(O) =
创
1)
1
η(η 十 1)2
x/I
(5 .43)
Then , it is easy to see that
We conclude that
x) ln(x)) =
1
l
n(η+
l地 (ln(1z)lm(z))=J
马亏一= 0
炖(川 
立
(iv) Note that
From (5.39) and (5 .40) it follows that
坦
is
133
1
Vx 叫( 1 ， 均)
托叫 1υ
ε
一→
/IIlI
1
一
η
/
i
i
l
1
日
才
III/
一汇去
(5 .44)
Here , we used the fact that
立主 =Z
 x). Then ,
nl:(x)
dx =  I I) ~ ~一一 )l
n
In ....I
[信
Zifznln(Z)dz
∞
汇
叫
卜
and the telescoping series
(5 .4 1)
主 G 一击) zA 兰 G 一击)
lN

叩m
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)dzzdr
1一
川
1.
[I中叫l川
f
Then
1xnln仪
1
( x n + I l日(x) 一
η
+1
xn+ I
[
(η 十 1 )2 )
Problem 7: Consider a口 ATIVI put option with strike 40 on an asset with
volatility 30% and paying 2% dividends continuously. Assume that the interest rates are constant at 4.5%. Compute the relative approximation error
of the approximation
10
L一一 ~~.~~(x 肿Il n (x) )
1)2
(η+
dF
VGl
(5 .42)
凡，pprox ，呐
IT
(
(γ 十 q)T (γ 
q)T
71.
134
CHAPTER 5. TAYLOR'S FORMULA. TAYLOR SERIES.
if the put option expires in 1, 3, 5, 10 , and 20 years.
Solution: The approximate option values and the corresponding approximation errors are given below:
T
凡.pprox
4.1317
3 5.9834
5 6 .4652
10 5.2187
20 2.5067
l
Error
PBS
4.1491 0 .4 2%
6.1577 2.83%
6.9714 7.26%
7.3398 28.90%
6.0595 NjA
While the approximation formula is still within 3% of the BlackScholes
value when the maturity is three years or less , it deteriorates for long dated
options , and even produces a negative value for the 20years option. 口
Chapter 6
Finite Differences. BlackScholes PDE.
6.1
Solutions to Chapter 6 Exercises
Problem 1: A butterfly spread is made of a long position in a call option
with strike K 一叽 a long position in a call option with strike K 十凯 and
a short position i口 two calls with strike K. The options are on the same
underlying asset and have the same maturities.
(i) Show 也at the value of the butterfly spread is
C(K 十 x)
 2C(K)
+ C(K 
x) ,
where , e.g. , C(K + x) denotes the price of the call with strike K 十 x.
(ii) Show that , in the limiting cωe when x goes to 0 , the value of a position in
主 b耐er组y spreads as above co盯erges to the second order partial derivative
of the value of the option , C , with respect to strike K , i.e. , show that
C(K 十 x)

2C(K)
+ C(K 
x2
2、6
x)
护C
=一τ (K).
8K2
(iii) Show that , in the limiti吨 case when x • 0 , the payoff at maturity of
a po副on in ~ butte吻 spreads as above is goi吨 to approximate the payoff
of a derivative security that pays 1 if the underlying asset expires at K , and
o otherwise.
Note: A security that pays 1 in a certain state and 0 i丑 any other state is called
an ArrowDebreu security, and its price is called the ArrowDebreu price of
that state A position h
butter23rspreads as above7with Z smll7is a
synthetic way to construct a口 ArrowDebreu security for the state S(T) = K.
i
Solution: (i) The value of a butterfly spread , i. e. , of a long position in a call
option with strike K  x and value C(K  x) , a long position in a call option
with strike K 十 x and value C(K + x) , and a short position i口 two calls with
strike K and value 2C(K) is
C(K
+ x) 
2C(K)
135
+ C(K  x).
72.
136
CHAPTER 6. FINITE DIFFERENCES. BLACKSCHOLES PDE.
(ii) The value of a position in 去 butterfly spreads is
毛 (C(K 忖) ..
2C(K)
x
+ C(K 
x)).
(6.1)
The value of a call option as a function of the strike of the option is infinitely
many times differentiable (for any fixed point i丑 time except at maturity).
The;efore the expression from (6.1) represents the central finite difference
approximation of 在 (K) and we know 山t
6.1. SOLUTIONS TO CHAPTER 6 EXERCISES
Problem 2: A bull spread is made of a long position in a call option with
strike K and a short position in a call option wi出的rike K + x , both options
being on the same underlying asset and having the same maturities. Let
C(K) and C(K + x) be the values (创 time t) of the call options with strikes
KandK 十叽 respectively.
(i) The value of a position in ~ bull spreads is cmf(K刊)In the limiting
case when x goes to 0 , show that
C(K 叫一 2C (K) +C(K  x) = 些 (K) + O(泸)
z", 0
σ K2
C(K
+ x)
 2C(K)
、
、
i
J
T
=
m
U
L
X
S
l
θ2C
=一τ (K).
8K2
/
K
J
J
」
DT创
e ne
b
』
ι
d
E
f
f
3
J
阶
I
L
口
y
r
I
I
r
r
七
十
1
5
J
飞
n叫
x)
u
M U 的 d d 巴 h 忱 臼 b
刘
'DAa L eI7·1·1·1·IE 句 : sU
a ;
n
nuoo
eu 江
υ   幻 m 刘
u
Z
U7
S ?
1
 K
S
刘 c
的
S
( K
十
a/』 knu
丑
m十 、z d
i )
…n
u d
h
一
u f
f
t
t
〈
τ
h
U
+ C(K 
x2
l
z
z
l
u
r
一
b
、
、
‘
，
，/
υ
，
v
'
'
叫
乱
?
f
f
f
f
g
川
吭

川
fx(8)
K
K
K
M
矶
〈
一
<
一
十
口
K
Z
S
Z
削
一
<
一
/
一
<
哑
Z
S
T
M
C
U
L
叫
}
<
r
r
H
、
飞
『/
Q
U
M r
P J
k
υ
l
飞
i
h
J
T
L
θC
p
l
z
e
4EU
'+b
y
印
e
QU
一
= 0, V 8
并 K.
•
+ O(x) ,
O. We conclude that
2、o
x
θC
=一一 (K).
θK
(ii) The payoff at maturity of the bull spread is
(6.2)
1.
as x
C(K)  C(K 十 x)
Let 8 笋 Kbea 直xed value of the spot price of the underlyi吨 asset. Then
儿 (8) = 0 for any x such that 0 < x < IK  81 , and therefore
人j
C(K 十 x)  C(K)
θK
注
Ii闷儿(K) =
z >U
li同 fx(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
一 (K) =
Note that fx(K) = 1 for any x 并 0 ， and therefore
Z
(ii) Show that , in the limiting case when x
0 , the payoff at maturity of
a position in ; bull spreads as above is going to approximate the payoff of a
derivative security that pays 1 if the price of the underlying asset at expiry
is above K , and 0 otherwise.
Note: A position in ~ bull spreads 部 above， with x small , is a synthetic way
to construct a cashornothing call maturing at time T.
川
叫
(
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.
=
C(K)  C(K 十 x)
θC
=一 ~;/(K)
x
θK
•
Then , in the limit as x goes to 0 , we obtain that
zo
137
r 0, if 8 三 K;
max(8  K , O)  max(8 一 (K 十 x) ， O) ={ 8K , if K<8 三 K+x;
x,
if K + x < 8.
,
If gx(8) denotes the payoff at maturity of a position in ~ bull spreads , then
如何)
(6.3)
From (6.2) and (6.3) , we conclude ~hat ， in the limiti吨 case when x • 0 ,
the payoff at maturity of a position in ; butterfly spreads as above is 1 if the
underlying asset expires at K , and 0 otherwise.
口
If 8 三 K ，
r
=
0, if 8::; K;
<与互， if K < 8 三 K+x;
l 1, if K 十 x < 8.
then gx(8) = 0 for any x
Ii同 gx(8)
Z。U
=
> 0 and therefore
0, V 8 三 K.
(6 .4)
73.
CHAPTER 6. FINITE DIFFERENCES. BLACKSCHOLES PDE.
138
If S > K , then gx(S)
therefore
=
1 for a叮 x such that 0 < x < S  K , and
Ii同 gx(S)
x ,U
=
1, V S
as h
→ O.
Problem 3: Find a second order finite difference approximation for f'( α)
using f(α) ， f(α + h) , and f(α + 2h).
Note: This type of approximation is needed , e.g. , when discretizing a PDE
with boundary conditions i盯olvi吨 derivatives of the solution (also called
抖
∞ t lnuou
悦
Robin boundary conditi∞ s). For example , for Asian Op tions (cO丑时叫
computed average rate call , to be more precise) , this type of finit~ _difference
臼略
approximation is used to discretize the boundary conditio丑誓+嚣= 0 , for
R=O.
Solution: To obtain a finite difference approximation for f'(α) in terms of
f(α) ， f(α + h) , and f(α 十 2h) we use the cubic Taylor approximation of f (x)
around the point x = 风i. e. ，
口
Problem 4: Find a centralfinite difference approximation for the fourth
derivative of f at 风 i.e. ， for f(4)( α) ， usi吨 f(α  2h) , f(α h) ， f(α) ， f(α 十 h) ，
and f(α+ 2h). What is the order of this 缸ite difference approximation?
Solution: We will use the following Taylor approximation of f(x) around the
point x = α:
f(x) = f(α) 十 (x 一 α) f' (α) 十
,
(6.6)
+ 0 ((μZ 一 α) 7) ,
叫
as
f( α +2h)
h 2 _",'
h 3 _1 飞
A
f(α) + hf'(α)+Efff(α) + '~ f(3)(α) + 0 (的;
(6.7)
4h 3 川
A
f(α) + 2hf'(α)+ 约2f"(α)+ 了 f(3)(α) + 0 (的， (6.8)
as h • O.
We multiply (6.7) by 4 and subtracting the result from (6.8) to obtain
f(α + 2h)  4f(α +h) =
as h
缸lite
•
For symmetry reasons , and keeping in mind the form of the central diι
ference approximation for f" (α) ， we use (6.10) to compute
_IA.
f(α +
2h)
+ f(α 
OJ
_fA.
OJ
as h • O. We multiply (6.11) by 4 and subtract the result from (6.12).
We solve for f(4)( α) and obtain the followi口g second order 丑nite difference
approximation:
(4)(~ _ f(α + 'UJ  4f(α +h) 十 6f(α)
_ J ~ 2h) ~J ~ 'UJ ~Jh~
f~4) (α)
ash → O.
I
I
 4f(α  h) 十 f(α  2h) 十 O( 的?
I
口
2h
3f(α)  2hf'(α) 十了 f(3)(α) + 0 (的， (6.9)
difference approximation
f'( α) ， we
of f' (α):
obtain the following second order
一 f(α + 2h) + 4f(α +h)3f(α)
2h
f(α +
h4
,
h6
h) = 2f(α) + h2f"(α) 十一 f(4)(α)+f(川的
12
,/. 360
十 o (h 7 ) , as h • 0;
(6.11)
4
6
4h
,
8h
α) 十一
2h) = 2f(α) + 4h 2f"(α) + "or: f(4)(,/. 45 f(6)( α)
3
十 o (h 7 ) ,
(6.12)
f(α 十 h) 十 f(α 
3
O. By solving (6.9) for
f'( α)
(6.10)
x 一→ α
as x → α. By letti吨 Z 工 α 十 h and x = α + 2h in (6.6) , we obtain that
f( α 十 h)
(x 一 α)2 II(x 一 α)3(3)
2 J f" (α) 十
6f()(α)
+(zff(4)(α) +(
f(α) 十 (x 一的f'(α) +(气川"(α) 十(气川但) (α)
十 o ((x 一 α)4)
139
(6.5)
> K.
From (6.4) and (6.5) , we conclude th~t ， in the Iimiti吨 case when x • 0,
the payoff at maturity of a position in ~ bull spreads as above is 1 if the
underlying asset expires above K , and 0 otherwise.
口
f(x) =
6.1. SOLUTIONS TO CHAPTER 6 EXERCISES
h 2 r (3)
十'; f(3)(α) + 0 (的
2h) + 4f(α 十 h)  3f(α)2
2 h + O (的?
Problem 5: The goal of this exercise is to emphasize the importance of symmetry 坦白nite difference approximations. Recall that the central difference
approximations for the first and second order derivatives are
f'(α)
f"(α)
f(α +
h) 
2h
f(α 
f(α +h)2f(α)
~~""J
h)
+
I
十
f(α J ""
2
(h ) ;
h)
'~J
+ 0 (h2) ,
74.
140
CHAPTER 6. FINITE DIFFERENCES. BLACKSCHOLES PDE.
as h • O. In other words , f'( α) and f"(α) are approximated to second order
accuracy by using the value of f at the point αand at the pointsα hand
α + h that are symmetric with respect toα.
We investigate what happens if symmetry is not required.
(i) Find a second order finite difference approxi口lation of f' (α) usi吨 f(α) ,
f(α  h) and f(α + 2h).
(ii) Find a first order finite difference approximation of f" (α) usi吨 f(α) ,
f(α  h) and f(α + 2h). Note that , in general , a second order finite difference
approximation of f" (α) using f(α) ， f(α  h) and f(α + 2h) does 日ot exist.
Let (3 <α<γsuch thatα (3 = C(γ 一 α) ， where C is a constant.
(iii) Find a finite differe肌e approximation of f' (α) 四吨 f(α) ， f( (3), and
f(γ) which is second order in terms of b 一 αI ， i.e. , where the residual term
isO (Iγ 一 α1 2 )
(iv) Find a 直nite difference approximation of f" (α) using f(α) ， f( (3), and
f ('Y) which is first order in terms of b 一 αI. Show that , in general , a second
order finite difference approximation of f" (α) using f(α) ， f ((3) and f ('Y) is
not possible，叫esMzEFFie?mlessβandγare
sym皿etric with .resp创
to α.
Solution: (i) and (ii). vVe use the cubic Taylor approximation of f(x) around
the point x = 矶 i.e. ，
f(x) =
f(α) 十 (x 一 α) 1' (α) + (x 一 α? f"(α) 十 (x 一 α)3 f(3)(α)
+ 0 ((x 一 α)4) ，
2
as x → α(6.13)
By letting x = α  h and x = α + 2h in (6.13) , we obtain that
2
;
(6.14)
4h3
f(α + 2h) = f(α) 十 2h1'(α) + 2h f"(α) 十了 f(3)(α) +0 (的， (6.15)
2
as h • O.
We eliminate the terms ∞ t 创口旧吨 f"1 (μα) by ml 毕 趴 (仰.14) 均 4 an
cO丑川
ing 气
咄i pI抖
y
ri鸣 6 钊 by
丑
1
sub tractir the result from (仰.1 5). By solvi吨 for f'( α ), we obtain the fol由
忱
吨
忧
6 臼盯
叫
lowing second order 自nite difference approximation of l' (α) :
一 f(α + 2h) + 3f(α)
4f(α … h)h 2
1'(α)6hEf(川α) + 0 (h 3 )
f(α + 2h) + 3f(α)  4f(α  h)
2
2
6h
十 o (h ) ,
as h • O.
141
Similarly, we eliminate the terms containi吨1'(α) by multiplying (6.14)
by 2 and adding the result to (6.15). By solving for f"( α) ， we obtain the
following 负rst order finite difference approximation of f" (α) :
f"(α) = f( +叶州叫 3 豆♂扩??俨俨)忡川十叶叫附
μα 2 h川)卜一斗
孰均
圳
勾
2f
f( α
归
+ 2h) 一 3f( α) 十 2f( α 一 h)
归刨
叫
归
川
? + O (h)7
3h 2
as h • O.
(iii) a时 (iv). Denote γ 一 αby h , i.e. , let h = γ 一 α. Then ， α (3 = Ch.
We write the cubic Taylor approximation (6.13) of f(x) around the point
α for x = γ = α + h and for x 二二(3= α  Ch and obtain
f( 'Y)
h 2 _".,
h3
f(α) + h1' (α) + '; f勺) + '~ f(3)(α) + 0 (h 4 )
f( (3)
C h _".,
h _(9飞
A
f(α)  Ch1' (a) 十一f fll(α) 一 rf(3)(α) +0 (的， (6.17)
c
2 2
;
(6.16)
3 3
as h • O.
By eliminating from (6.16) and (6.17) the terms containi吨 f"(α) and
solving for 1' (α) we obtain the following finite difference approximation:
C 2 f ('Y)  (C 2

1) f (α)  f( (3)
1'(α)1C(C 十 l)h+O (的
(6.18)
Similarly, we eliminate from (6.16) and (6.17) the terms contai时日g 1' (α)
and solve for f" (α) to obtain the following finite difference approximation:
f"( α)
3
h rill
h
f(α  h) = f(α)  h1'(α)+Eff(α)lff(3)(α) + 0 (h 4 )
6.1. SOLUTIONS TO CHAPTER 6 EXERCISES
=
2
Cf(γ) 一 (C+1)f(α)
+ f( (3)
C(C 十 1)h 2
十 o
(h).
(6.19)
Note that , i口 general ， the finite difference approximation (6.18) of f'(α) is
second order , while the finite difference approximation (6.19) of f"(α) is 且rst
order. The finite difference approximation (6.19) of f"(α) would be second
order , e.g. , if C = 1 or if f(3)(α) = O.
Also , note that , for C = 1, i.e. , if (3 =α h and γ=α 十 h are symmetric
with respect to the point α ， then (6.18) becomes the central finite difference
approximation of l' (α) ， i.e. ,
f(α +
h) 
f(α 
h)
1'(α)=+O (的
门飞
The same would not be true for (6.19) , which becomes
f(α 十 h)

2f(α)
f"(α)=79
+ f(α  h)
十 o
(h) ,
(6.20)
75.
142
CHAPTER 6. FINITE DIFFERENCES. BLACKSCHOLES PDE.
6.1. SOLUTIONS TO CHAPTER 6 EXERCISES
Using the Taylor approximation In (1 +叫 :Z4 十 O(X 3 ) we find that
instead of the central finite difference approximation
f(α +h)2f(α)
f协) =ω
+ f(α  h)
十 O(的
i (l n (1
I I
+ h)  h)
y' (x) = Y (x) , V x ε[0 ， 1];
ν(0) = 1.
= 0
臼
_ 1
:叽 where
缸lite
difference
，
寸
，
什
十
P
1
[ln(l 十h)h]
十
l J
e γ←
一
Vi
=
0:η.
(ii) Note that y(x) = eX is the exact solution of the ODE. Let
ei = Yi  Y(Xi) = (1 + h)i  eih
be the approximation error of the finite difference solution at the node Xi ,
i = 0 :η. Show that this finite difference discretization is convergent , i.e. ,
that
lim J;ll ftx Iei I = O.
ih 三 1
since
。
以
2
=
=
e仇( ei (ln(l+h)
十
。
/
i
叮
vhM ,
，
"
IJ
"
飞
Since eih < e for all i
= O(h 2 ) is
0
0
'
h
M
7
几
1111/lll/
町
，
血
门
，
血
+
o
、
飞
B
，
，
，
，
、
、
4t4
//II11 以
/IlI

mb
B
，
，
，
，
oh
十
1llI/
门
4
飞
qAl/
I
才
t
l
I
/
2
4
纠
/
and therefore
一
一
。
/It1
/III
'h2
十
O
门
Ill/
飞
，4
门
vhM
，
"
、
、
t
E
J
/
I
l
l
/
一
一
。
/
l
飞
'hM
门
，
，
"
= 0 : n , we obtain that
回 iq 
=
臼眈 ei h μin
jEg野 l卡仇 (←呻巾
盟3
2窍
伊
川沪
e ii(队忡
((1
::; e ~~~ lei (ln(l+h)h) 
11
1. 2
<e i22号 Ii亏 +O(的
<已;十 O(h2 )
(1 + h)忡17
which is what we wanted to show.
(ii) Let Y(x) = eX be the exact solution of the ODE. It is easy to see tl时 the
approximation error ei can also be written as
ei =的  Y(Xi) = (1 + h)i  eih = ei1n (1+h) _ eih
十
十
仍
11llI/
门
1111/
，
nb
Solution: (i) Recall that yo = 1 and Yi+ 1 = (1 十 h)Yi ， for all i = 0 : (n  1).
It is easy to see by induction tl川的= (1 +}札 for all i = 0 :η:
Initial condition: for i = 0, we know that yo = 1 = (1 十 h)o.
Induction step: assume that Yi = (1 十 h)2. Then ,
+ h)Yi
:η ，
/JIll
,
/Il1
n一→。C 2=U:η
的+1 = (1
=0
for all i
一2
以
2
0
n
V
A
with yo = 1. Show that
+ f讥
:ηNote that the estimate iO(h 3 )
= 0
以
/lit/iil2
ei
= (1 + f仰i ， V i = 0 : (η 1) ,
(1
1. 2
sharp , since , for i =叽 the product ih is equal to ih = nh = 1.
From the Taylor approximation eX = 1 十 X + O(x 2 ) , it now follows that
Problem 6: Consider the following first order ODE:
的=
 h}
h2
s年ce ih 三 1 ， for all i
的+1
i亏+ iO(的工~亏+ O(时) ,
口
(i) Discretize the i且terval [0 , 1] using the nodes Xi = ih , i
h= 去 Use forward finite differences to obtain the following
discretization of the ODE:
1. 2
i ( ( h 一亏十 O( 的)
of f"(α). This is due to the fact that , for C = 1, the coe自cient of h from
o (h)' f~om (6.20) cancels out and the next term of order 0 (的 becomes
relevant.
143
咐= o(~)
We conclude that
lim J;llftx Iei I
n一→。o
2=u:n
 0,
and therefore that the finite difference discretization scheme of the ODE is
convergent.
76.
144
CHAPTER 6. FINITE DIFFERENCES. BLACKSCHOLES PDE.
Note that we actually showed that the finite difference discretization is
first order convergent , i. e. ,
233eii
=
°G)
6.1. SOLUTIONS TO CHAPTER 6 EXERCISES
145
From the boundary conditions (6.22) , we find that Yo = 1 a时 Un=i
(ii)For n=67tkeaIlite difemme dberetizatiOII(624)of tke ODE (621)
can be written in matrix form as
口
AY
b,
where A is a tridiagonal 5 X 5 matrix given by
Problem 7: Consider the following second order ODE:
2
3x y"(x)  xy'(x) + y(x) 
0 , I x
ε[0 ， 1];
A(i , i1)
A(i , i 十 1)
(6.21)
的 )=1;U(hj
(6.22)
I. e.
nu
Vi=l:(η 1).
(η
ν "(Xi)
y( 向一 1)
Y'(Xi)
y( 叫一 1)
2h
+
2
'h
q&
q
，
旬
一
+ UU
却
Uu十
μ
h
」
2
since Xi = ih , which can be written as
'hM
气
(3 叶)的一 1 一 (6i 2 1 如 ( 3i' 一 DYi 十 1
口
印
O
O
O
钊
」
0
0
25.5
95
77.5
引
。
illit/
n
叫
U
飞
1iqAqOAUZWO
I
t
B
E
l
t
/
Y
b
口
=
Problem 8: Show that the ODE
y"(X)
2y'(x) 十 x 2 y(X)
=
0
can be written as
Y'(X)
2
O( 的 7
f(x , Y(x)) ,
where
into (6.23) , use the approxi血 a~e values yi for the exact values y(Xi) , for
i = 0 :叽 and ignore the O( h'2) term. The followi鸣 second order finite
difference discretization of (6.21) is obtained:
q
ο
一
。
，
"
。
,
}qovhu
2
8
0
0
(6.23)
h2+0(hzh
y(X i+ 1) 
13
O
O
O
川
1) ,
We substitute the second order central difference approximations for Y"(Xi)
and y'(均)， respectively, i.e. ,
y(X i+ 1)  2Y(Xi) +
i = 1 : 5;
i = 2: 5;
i = 1 : 4,
/IEt11111
UUUUU
= 仇， i = 0: 凡 where h = ~. We look for Yo , y1 , . ..，仇
Solution: (i) Let Xi
such that yi is an approximate value of y(Xi) , f6'r all i = 0 :ηBy writing the ODE (6.21) at each interior node Xi = ih , i = 1 :
we obtain
6
4
5
A
VVVVVV
1) ,
3i2j7
,
(i) Partition the interval [0 , 1] intoηequal intervals , corresponding to ∞ des
Xi = ih , i = 0 :凡 where h = 土. Write the finite difference discretization
n
of the ODE at each node Xi , i = 1 : (η … 1) ， usi吨 central 丑nite difference
approximations for both y' (x) and y" (x).
(ii) If n = 6 , we find , from the boundary conditions , that Yo = 1 and y6 =
~. The 直丑 ite difference discretization scheme presented above will have 自 ve
equations can be written as a 5 x 5 linear system AY = b. Find A and b.
3X;Y"(Xi)  xiy'(Xi) + y(Xi)
一 (6i 2 3i 2 十乞
A( i , i)
UU
十的

0,
Y(X)
(n1) 叫
and
f(x ,Y(x))
内
，
，
"
i
Iil/
。
4
Y
/
Z
'
'
飞
、
、
飞
，
飞
，
，
，
，
‘
Solution: Note that y"(x) = 2y'(x)  x 2 y(x). Then ,
Y'(x)
= 0, Vi = 1 :
(削)
of
/IIiI
(UZj)z (勾 fdt2U(Z))
(工 2;)(38)=( 工 2
n
Y(x)
口
77.
146
CHAPTER 6. FINITE DIFFERENCES. BLACKSCHOLES PDE.
Problem 9: Consider a six months plain vanilla European call option with
strike 18 on a nondividendpaying underlying asset with spot price 20. Assume that the asset has lognormal distribution with volatility 20% and that
interest rates are constant at 5%.
(i) Compute the Greeks of the call optio凡i. e ， ~， f , P, vega , and 8.
Use finite differences to 丑nd approximate values for the Greeks. Recall that
A
θCθ2CρθC
f =一一· ρ 一一一
θ S'
θ S2'
t'
θγ7
一一一:
一
vega
…一
δC
_
τ一;。
。σ
θC
θT
Denote by C (S , K , T， σ， r) the value of the call option obtained from the
BlackScholes formula.
(ii) The forward and central difference approximations ~f and ~e for ~， and
the central difference approximation f e for fare
~f  C(S 十 dS， K， T， σ， r)  C(S， K， T， σ， r)
=
f =
dS
~e
 C(S  dS， K， T， σr);
2dS
f = C(S+dS， K， T， σ， r) 一 2C(S， K， T， σ， r) + C(S  dS , K , T， σ， r)
e
(dS)2
=
一
6.1. SOLUTIONS TO CHAPTER 6 EXERCISES
147
Solution: (i) Usi吨 the formulas for the Greeks of a plain vanilla call option
derived from the BlackScholes formula , we find that
~
= 0.839523;
f
= 0.086191;
vega = 0.01501;
P = 7.045377;
8 = 1. 394068.
(ii) The BlackScholes value of the call option is C = 2.699703. The corresponding forward and central difference approximation errors for the Delta
and Gamma of the option are:
dS
0.1
0.01
0.001
0.0001
0.00001
0.000001
6. f
6. e
0.843774
0.839952
0.839565
0.839526
0.839522
0.839522
0.839464
0.839521
0.839522
0.839522
0.839522
0.839522
fe
0.086199
0.086196
0.086195
0.086196
0.086207
0.090594
I~~fl
I~~el
If  fel
0.00425158
0.00042959
0.00004228
0.00000349
0.00000037
0.00000076
0.00005840
0.00000138
0.00000082
0.00000081
0.00000081
0.00000081
0.0000081~
0.0000042:
0.0000042l
0.0000042!
0.0000153
0.0044029
C(S+dS， K， T， 叽 γ)
一
The corresponding relative approximation errors are
σ
V e二 W σ。 a
nL
Compute the approximation errors for the following values of dS:
dS
6. f 6. e f e 6.  6. fl
1
1
6.  ~el
(iii) The 且由 e difference approximations of vega , P and 8 give川y (6.256.27)
are
vegaf
3 .448385;ρf = 7.045624;
8 f =  1. 392998.
If  fel
一
0.1
0.01
0.001
0.0001
0.00001
0.000001
b
a
一
『
可
一
乃
reJ
一
一

nunu nu
nu
18 一 8fl_
181
呼
B
E

Ip Pfl =
p
一一一一一
门
U《
n
0.000035;
nnn7C::巧内
…
Problem 10: Show that the value of a plai口 vanilla European call option
the BlackScholes PDE. In other words , show that
satis直es
(iii) Let dσ= 0.0001 ，价= 0.0001 , and dT = 在， i. e. , one day. Find the
following forward difference approximations for vega ， ρ ， and 8:
vegaf
Pf
8f
C(S， K， T， σ +dσ， r)  C(S , K , T， σ， r)
dσ
C(S , K , T， σ7 俨十 dr)  C(S , K , T， σ， r)
dr
C(S , K , T + dT， σ， r)  C(S， K， T， σ， r)
dT
(6.25)
(6.26)
(6.27)
θC
一
θt
+
'
1 叮叮 θ2C
δC
';a 2 S 2 一丁 +(γ  q)S一
2
~
8υθS

rC 
0,
where C = C (S , t) is given by the BlackScholes formula.
Solution: Although direct computation can be used to show this result , we
will use the version of the BlackScholes PDE invol飞ring the Greeks , i.e. ,
@+jd价十 (γ 一仙一 rC
0,
78.
148
CHAPTER 6. FINITE DIFFERENCES. BLACKSCHOLES PDE.
substitute for .6., rand
.6.
=
e 
e the values
qSe 一矿t)
ρ q(Tt)
rL
eq(Tt) N(d 1 );
r K er(Tt) N( d2 )

eT
O.
σ Seq(Tt)
_
d~
~ v ~~G /~
., e~ ，
2 、/2作 (T … t)
and substitute for C the value given by the BlackScholes formula , i.e. ,
C 
Se旷t) N(d 1 )  Ker(Tt) N(d 2 ).
Then ,
@+jd付 + (r …仙一 γc
qSeq(Tt) N(d 1 )
十 σ2s2
σ Seq(Tt)
r Ker(Tt) N(d 2 )

21
2 飞/2作(Tt)e2
一手
eq(Tt)
t)
2 Sσ 、/2作 (T 
十 (r 一仙一 q(Tt) N(d 1 ) 一 r(仇旷 q 阳叫)N( 出廿
川一吁呗 Tι 功
v (T t μd 1
创 归叫一→卅
队
0 .口
Problem 11: The value at time t of a forward contract struck at K and
maturing at time T , on an underlying asset with spot price S paying dividends
continuously at the rate q , is
j(S , t) = Seq(Tt) 
Ker(Tt).
Show that j(S, t) satis五es the BlackScholes PDE , i. e. , show that
θj
1 _202 θ2j
百 + ~O'~S:~a;2
I
+
δf
(γ  q)S页  rj
= O.
Solution: It is easy to see that
红 = aSeq(Tt)  r K er(Tt)
θt
Then ,
":1
.
wa
十
12
nL
σ
c
υ
坐
，
eq(Tt);
θS
u
c
υ
十
T
口
可A
θ2 j
θ S2
2 rffqa
AUnAU
oa
c
υ
O
一θ
_ rKer(Tt)
+
0
一 γ (Seq(Tt)  Ke r 叫)
d~
 t)
Sσ 、/2作 (T
qSeq(Tt) N( d1 )
6.1. SOLUTIONS TO CHAPTER 6 EXERCISES
fJS
T rlu
A

口
+
(γ 
149
q)Seq(Tt)
79.
150
6.2
CHAPTER 6. FINITE DIFFERENCES. BLACKSCHOLES PDE.
Supplemental Exercises
1. Let 1(x) = x 3 ♂ 6♂. Show that the central differe肌e approximation
for l' (x) around the point 0 is a fourth order approximatio泣
2. Let
6.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
4. Consider a six months 5% inthemoney plain vanilla European call
option with strike 30 on an underlying asset with volatility 20% , paying
dividends continuously at a 2% rate. Assume 出at the interest rates are
constant at 5%.
(i) Use central differences to compute the finite difference approximations ~c and r c for ~ and r , respectively, i.e. ,

2C(K , T) + C(K  x , T)
1(S) =
x2
where , e.g. , C(K , T) denotes the value of a plain vanilla call option
with strike K and maturity T on an underlying asset with spot price
S following a lognormal distribution. Show that , for any continuous
function g : JR. •JR.,
C(K 十队 T)
~c
rc
户。c
Ii玛 I
ZU JOC
I
可出~
2S2
r
2
(J
8
0
connecting the r and the θof plain vanilla European options is exact if
the underlying asset pays no dividends and if the riskfree interest rates
are zero. In other words , for , e.g. , call options , show that , if r = q = 0,
then
1 +σ2S2 r(C)
6.3
十 十 二 一
8(C)
2
(ii) If q = 0 but
r 并 0，
σ2S2 r(C)
1 十一一一·一一一
8(C)
1
1+
'T
N'(d2)
~ ~一~μ
2r /T习 N(d2 )
(iii) Consider a six months plain vanilla European call option on an
underlying asset with spot price 50 and volatility 30%. Assume that
the interest rates are constant at 4%. If the asset pays no dividends ,
compute
1
+
Solutions to Supplemental Exercises
Problem 1: Let 1(x) = x 3 ex  6e X • Show that the central difference approximation for 1'(x) around the point 0 is a fourth order approximation.
show that
2
 C(S  dS)
2dS
'
C(S + dS)  2C(S) + C(S  dS)
(dS)2
(ii) Compute the Delta and Gamma of the call using the BlackScholes
~c
formula , and the approximation errors 1  ~ 1and Ire  r I. Note
that these approximation errors stop improving , or even worsen , as dS
becomes too small. How do you explain this?
3. (i) Show that the approximate formula
十二一.一目
C(S 十 dS)
for dS = 102 with i  I : 12 , where , e.g. , C(S + dS) = C(S 十
dS， K， T， σ， r) denotes the BlackScholes value of the call option corresponding to a spot price S 十 dS of the underlying asset.
1(S)g(S) dS = g(K).
1
151
σ2S2 r(C)
一一一 .一一一
2
8(C)
if the options are atthemoney, 10% , 20% , 30% , and 50% inthemoney, and 10% , 20% , 30% , and 50% outofthemoney, respectively.
What happens if the asset pays dividends continuously at a 3% rate?
Solution: Recall that , in general , the central difference approximation of the
first derivative is a second order approximation , i.e. ,
f(h)
1' (0) = f(h) 2~ '''/ + 0
J '''/
2
(h 2 )
,
as h
•
O.
(6.28)
To see why, for the function f(x) = x 3 ex  6e气 the central difference
for l' (x) around the point 0 is a fourth order approximation ,
we investigate how the approxi日lation (6.28) is derived.
The Taylor approximation of f (x) around the point 0 for n = 5 is
appro均nation
f(x)
=
f(O)+x1'(O)+ 旦f"(O) 十 21f(川0) +马队)(0) +主严)(0)
2
6
24
120
6
+ 0 (x ) , as x • O.
(6.29)
J
 /
'
J
 I
'
J
 I
'
80.
152
CHAPTER 6. FINITE DIFFERENCES. BLACKSCHOLES PDE.
We let x = h and x = h in (6.29) and sum up the two resulting formulas.
After solvi鸣 for 1' (0) we obtain
1' (0) =f(h)  f( h) 一句句(0)  1~:f(5)(0) +
2h
6"
/
120
0
(的，
(6.30)
as h • O.
For f (x) = x 3eX  6e x , we find that f(3) (x) = (x 3 + 9x 2 十 18x )e X, and thus
that f(3)(0) = O. Also , f(5)(0) = 54 乒 o and (6.30) becomes
f(h)  f( h)9h4
f(h)  f( h)
f f ( 0 ) = 2 h  3 5 + O (问 :2h+O(的?
as h • O. In other words , the central difference approximation for
around the point 0 is a fourth order approximation. 口
f' (x)
where we used the s巾的itutions z = 8 一 (K x) and ω =K+x
two integrals above , respecti飞rely.
Let z = xy a丑dω= xt. Then dz = x dy , dω = x dt , and we find that
1:
f(S)g(S) dS =
l'扑
1' yg
We let x 、、 o 阳 (仰.3 2) . Since the function 9 : ~ →~ is continuous , we
in 6 臼
勾
obtain that
炖时 贝阴川 8
fι: S )归
冈川
g
=
以州(仪川
9K
which is what we wa时ed to show; ef. (6.31).
For the sake of completeness , we provide rigorous proof of the fact that
(6.32) becomes (6.33) when x ￥ O. To do so , it is eno吨h to show that
炖叫才' νω
1 川
yg
g
Problem 2: Let
f(8)
6.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
C(K+ 民 T)
 2C(K , T)
x2
+ C(K  x ,T)
where , e.g. , C(K， 1寸) denotes the value of a plain vanilla call option with
strike K and maturity T on an underlying asset with spot price 8 following
a lognormal distribution. Show that , for a口y continuous function 9 : ~ →~，
~!!J 1: f(S)川8
= g(K)
(6.31)
Let ε> 0 arbitrary. Since 9 is continuous , it follows that there exists 6 > 0
such that Ig(K)  g(7)1 < εfor all 7 such that IK  71 < 6. Let x ε(0 ， 6)
and y ε(0 ， 1). Then jK 一 (K x 十 xy) I = x(1  y) < 6 and therefore
Ig(K)  g(K  x 十 xy)1
< 飞
主 (ω(8 一 (K 一川 2 日lax(8 一 K， O) +max(8 一协叫，0))
。 … 三 Kx;
旦在纯， ifKx<8 三 K;
qiifk<S三 K+x;
l1' yg
νωg川 +忖叫叫♂均)d
ωUω
忡
<Jlydu=i
汇 f(8)g(8)d8
In other words , for any
ε>
0 there exists 6 > 0 such that
l1' yg川 + xy) dy
1
1
 g(K)
y dY ! <
 g(K)
泸州)d8 + 儿泸州)d8
主 1" zg(Kx 十忡十刊 ωg(K 十 Z 一 ω)dω7
V0 < x < 6
l' y 叫= 0
r K S 一 (K  x ) r K +x K 十 x8
儿X
~，
Then , by definition ,
划 11' yg(K 一叶 xy) dy
Then ,
x < 6, 0 < y < 1.
Therefore , it is easy to see that , for any 0 < x < 6,
Solution: From the definition of f(8) , it is easy to see that
f(8) =
:j 0 <
Problem 3: (i) Show that the approximate formula
σ28 2
r
2
e
l 十一一一·一用。
口
81.
CHAPTER 6. FINITE DIFFERENCES. BLACKSCHOLES PDE.
154
connecting the f and the 8 of plain vanilla European options is exact if the
underlying asset pays no dividends ar: d if the riskfree interest rates are zero.
In oth~r ';ords , for , e.g. , call options!,
σ28
2
f(C)
6.3. SOLUTIONS
f(C)
σ 8V27rT
σ8
8(C)
8(C)
2
155
(ii) For q = 0 , we obtain from (6.34) and (6.35) that
AU
1 十一一·一一
TO SUPPLEMENTAL EXERCISES
二百万e
_~
2
 rKe rT N(d 2 ).
Then ,
(ii) If q = 0 and r 笋 0 ， show that
σ282
2
1 十 σ28 f(C)
一.一
2
8(C)
f(C)
2
8(C)
1+ 一一一·一一一
1
l+~urn:;旦旦~.
σSfii
1 一
2V2irT V
σS ~一手
2V2irT
2rVT N(d 2 )
uS
p.一手
1
1 + ~ rVT
U
1m
8(C)
if the options are atthemoney, 10% , 20% , 30% , and 50% inthemoney,
and 10% , 20% , 30% , and 50% outofthemoney, respectively.
What happens if the asset pays dividends continuously at a 3% rate?
Solution: Recall that the f and the 8 of a plain vanilla European option are
c
eqTfi
aSV2汗
σ 8eqT
8(C)
2V27r T
where d1 = (In
(6.34)
7
e
_5.
2
r K e rT N (d 2 ) ,
十 qSe一叫V(d 1 ) 
(圣)十(叫到T) / (σvr)a叫 =
d1 一 σ ♂
(i) For r = q = 0 , we obtain from (6.34) and (6.35) that
f(C)
σS 刁万 e2;
σSfi
8(C)
2V27r T
Then ,
1+σ2S2r(C)1+σ2~2 (_~飞, = O.
2
十一.一
2θ (C)
十二
2σ28 )
lNote that~ if r = q = O~ then f( P) = f( C) and
l. e一手
S
.
Ke rT N(d 2 ) ' J2王
1
1 十一一一·一一一
pi
+γKerT N(d2 )
2 飞!27rT
f(C)
2
+γ Ke rT N(d 2 )

r K e rT N(d 2 )
(iii) Consider a sixmo nthsp lai川aani llaE盯opeaI allopμtio I 1 an underlyiIl
议 ∞
川 抖创
ω 丑川
缸li
让
让
缸
川
∞ on
ω
丑
臼创 沁
口
asset with 印 创 price 50 and volatility 30%. Assume that the interest rates
s pot
are constant at 4%. If the asset pays no dividends , compute
σ28 2
V

8(P) = 8(C).
1+ 动T
K:JX:Zii2)
(6.36)
since N'(t) = 在λfor all tεR
Recall that the "magic" of Greek computations is due to the following
result:
8 N'(d 1 ) = Ke rT N'(d 2 );
cf. Lemma 3.15 of [2] for q = O. Then , (6.36) becomes
σ282
f(C)
1
2
8(C)
l+~ulm 旦旦~.
1 十一一一·一一
(6.35)
rVT N(d 2 )
(iii) Let 8 = 50 , T = 0.5 ， σ= 0.3 , and r = 0.04. The table below records
the values of
σ28 2
f(C)
l 十一一·一一一
2
8(C)
(denoted by "Val时') both for q = 0 , and for q = 0.03 , for the following
values of the moneyness of the option:
S
K
{1 , 1. 1, 1. 2 , 1. 3 , 1. 5, 0.9 , 0.8 , 0.7 , 0.5} ,
corresponding to call options that are atthemoney, 10% , 20% , 30% , and
50% inthemoney, and 10% , 20% , 30% , and 50% outofthemoney, respectively:
82.
156
CHAPTER 6. FINITE DIFFERENCES. BLACKSCHOLES PDE.
SIK Value for q = 0 Value for q = 0.03
1
1. 1
1. 2
1. 3
1. 5
0.9
0.8
0.7
0.5
0.0238
0.0233
0.0144
0.0187
0.5503
0.0214
0.0184
0.0155
0.0107
0.1897
0.2582
0.3518
0 .4 711
0.7361
0.1412
0.1068
0.0822
0.0505
6.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
Solution: The spot price S = 3 1. 5 corresponds to a 5% ITM call with K = 30.
We find that ~ = 0.692130579727 and r = 0.077379043990.
The central 且nite difference approximations ~e and the approximation
errors I~e  ~ I are recorded in the table below:
dS
0.1
0.01
0.001
0.0001
10 b
10 0
10(
We note that the approximation
10δ
σ2S2
r
2
e
1+ 一一一·一但
10 :1
O
is better for deep outofthemoney options (corresponding to small values
of S I K) and is worse for deep inthemoney options (corresponding to large
values of S I K). Also , for this particular case , the approximation is more
accurate if the underlying asset pays dividends. 口
Problem 4: Consider a six months 5% inthemoney plain vanilla European
call option with strike 30 on an underlying asset with spot price 20 and
volatility 20% , paying dividends continuously at a 2% rate. Assume that the
interest rates are constant at 5%.
(i) Use central differences to corr职lte the finite difference approximations ~e
and f e for ~ and f , respectively, i.e. ,
~e
f
e
C(S + dS) … C(S  dS)
2dS
'
C(S + dS)  2C(S) 十 C(S  dS)
(dS)2
for dS = 102 with i = 1 : 12 , where , e.g. , C(S 十 dS) = C(S + dS , K , T， σ， r)
denotes the BlackScholes value of the call option corresponding to a spot
price S + dS of the underlying asset.
(ii) Compute the Delta and Gamma of the call using the BlackScholes formula , and the approximation errors I~e  ~I and Ife q. Note that these
approximation errors stop improving , or even worsen , as dS becomes too
small.丑ow do you explain this?
157
10 1υ
10 11
10 U
~e
I~e~I
0.692112731743
0.692131730564
0.692131920566
0.692131922513
0.692131922087
0.692131918000
0.692131916226
0.692131862934
0.692132573477
0.692104151767
0.691890988946
0.687450096848
0.000017847983
0.000001150838
0.000001340839
0.000001342786
0.000001342360
0.000001338274
0.000001336498
0.000001283207
0.000001993750
0.000026427960
0.000239590780
0.004680482879
The approximations became more precise when dS decreased , until dS =
10 8 ; the best approximation was within about 10 6 of~. However , for values
of dS smaller than 10 9 , the finite difference approximations deteriorated very
quickly.
To explain this phenomenon , denote the exact value 2 of Delta by ~exaet.
Note that the value of ~ is given by the BlackScholes formula , i. e. ,
~ =
~BS = e qT N(d 1 ).
This value is computed using a numerical approximation of N(d 1 ) that is
accurate within 7.5 . 10 7; d. [1 ], page 932. In other words , we 0日ly know
that
(6.37)
I~BS  ~exaetl < 10 6 .
When computing the 且nite difference approximation ~e ， we use a numerical estimation of the BlackScholes formula to compute C(S + dS) and
C (S  dS) which once again involves the numerical approximation of the
cumulative density of the standard normal variable. In other words ,
~e  CBS(S 十 dS)  CBS(S  dS)
=
2N ate that
~exact
2dS
is a theoretical value , and is 口at the
~
from the table above.
(6.38)
83.
158
CHAPTER 6. FINITE DIFFERENCES. BLACKSCHOLES PDE.
Deωte by Cexaet(B 十 dS) and C口αet(B 一 dB) the exact 刊alues of the opμtio丑
m
v址
叫
挝
Since the central 缸l it edi宦erence is a sec orld order approxim创 凡， 让 follows
fi川臼 出
n
让
∞丑
对 丑 tion it
1a
1
that , for exαct values of Delta and of the call options ,
~exω= Cexact(B + dB)  C口aet(B  dB) 十 o ((dB)2) ,
2dB
(6.39)
•
s dB
O.
Since CBS(S)=SfqTN(dl)keTTlV(d2)7and since lV(do and lV(d2)
are comput~d 'm~merically within 106 of their exact value , it follows that
CBs(B 十 dB)  Cexaet(S 十 dB)1 < α10j;
jCBs(B  dB)  Cexaet(B  dB)j < α10° ，
(6 .40)
(6 .4 1)
where αis a constant proportional to the values of Band K ,
Usi吨 (6.38) and(6.39) we find that
~e

(~e  ~exaet)
~BS
CBs(B +
+ (~exaet 
~BS)
dB)  Cexaet(S 十 dB)
2dB
CBs(B  dB)  C口创t(B  dB)
2dB
十 ~exaet  ~BS 十 o ((dB?) ,
(6 .4 2)
as dB • O.
The only estimate we can find using (6.37) , (6.40) , (6 .41) , and (6 .42) for
the approximation of ~BS by ~e as dB • o is
~e

~BS 三
CBS(S 十 dB)
 Cex α et(S 十 dB)1
2dS
十 ICBS(B  dB)  Cexaet(B  dB)j
2dB
+I~exαet  ~BSj + 0 ((dB)2)
α10 6
〈一一 + 10 6
dB
+
0 ((dB)2) ,
(6 .4 3)
as dB • O.
While the approximation error j~e  ~j may be better in practice , the
bound (6.43) provides the intuition behind the fact that , for dS too small , t_h~
numerical approximation error I~e  ~I = I~e  ~Bsl deteriorates as 斗主
becomes large.
The central finite difference approximations
errors Ire  r j are recorded in the table below:
re
and the approximation
6.3. SOLUTIONS TO SUPPLE1VIENTAL EXERCISES
159
dB
re
We  f/
0.1
0.077370009586 0.000009034404
0.01
0.077371495486 0.000007548504
0.001 0.077371502982 0.000007541008
0.0001 0.077371709040 0.000007334951
10b
0.077271522514 0.000107521476
100
0.074606987255 0.002772056735
10( 0.355271367880 0 .4 32650411870
101:> 71. 054273576010 70.976894532020
For dB 三 10 9 ， the values of r e increased dramatically, reaching 109 for
dB = 1012 , and were no longer recorded. The finite difference approximations of r became more precise while dB decreased to 104 , but were much
worse after that; the best approximation was within 105 of r. The reason
for this is similar to the one explained above for the finite difference approximations of ~.口
84.
c
冒n
ap 4L r
e
阿
t
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)
85.
CHAPTER 7. MULTIVARIABLE CALCULUS.
162
7.1. SOLUTIONS TO CHAPTER 7 EXERCISES
163
from (7 .4), we conclude that
Then ,
d1 (S)
1
Sσn
1' (S)
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 xaxis , the yaxis , 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( S) is 址 沁丑肌reaωaslI1
侈 川 aIsωo 1 C
盯 且丑
恒
i
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) ，
Ju
1
ι
s
mwa
s
… 
2
7
飞
(
(d1 (S))2
s
nWM/
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
SO
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)
2n
一
c
υ
1
lim 一?古 exp (一
S→∞ σV 'l.作 l2
I x 主 O， y 三 O， X 十 y 三
= 0
1 y
+ 勾 ln(y)
ds
86.
CHAPTER 7. MULTIVARIABLE CALCULUS.
164
si丑ce limyo 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 BlackScholes 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 +;)22(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
θ 2V
θVθV
1
一十 S 一十一 σ 2S2 一一 + rS 一  rV = O.
' 2θ S2
θS
θtδI
(7.10)
87.
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)，
oθVθ
V
167
it follows that
1 <)_<)θ2V
θV
=一一十 S一一十一 σ，c， 5，c， 一一一十 r5一一 θ t θI
' 2θ52θS
rV
θHθH
1 '> _，> R θ 2H
_ (θH
S一十 S一一十… σ于一一一十 γ 5IHR 一~ 1
2
Show that H (R , t) satisfies the followi吨 PDE:
θH
1 '>_'>θ2H
万J+2σ".<l R".<l 否R2
+
θ t θR
θH
θH
(1 rR) 否五口。
(7.12)
Note: An Asian call option pays the maximum between the spot price 5(T)
of the underlying asset at maturity T and the average price of the underlying
asset over the entire life of the option , i.e. ,
max
/III
c
υ
T
T
T
PI''''nu
C
liI/
T
υ
T
Thus , the value V (5, I , t) of an Asian option depends not only on the spot
price 5 of the underlying asset and on the time t , but also on the following
random variable:
I(t) =
I
5(7) d7.
JO
It can be shown that V(5 , I , t) satis丑es the PDE (7.10). Similarity sol u
沁
叫
1
of the type (7.11) are good candidates for solving the PDE (7.10). The PDE
(7.12) satisfied by H(R ,T) can be solved numerically, e.g. , by using finite
differences.
Sol仰 on: Let V(5 , I , t)
= 5 H(R , t) , with R
=~' Using Chain Rule , we 白缸l
fin
丑
that
W
一
W仿
M
W
m
θ2V
ω
δ52
S 豆豆 .
θt '
S 纽纽一 θH
一
θRθIθR'
θHθRδH (
I IθH
H+5 一一…一 =H+5 一一!一一 l=H 一一一一
θRθSθR 飞
HR
52
J
~~
δH
一一:
δR
J
2H θR
_ I θ2H
θHθRθRθHθ
一一一一一一一一一  R 一一一一…  R 一一一一一
θRθSθSθRθR2
R 2 θ 2H
S
θR2'
δ5

.L"
1 叮叮 θ2H
52 θ R2
5
θR2
飞
θ RJ
 r5H
θH
S一十 ::'(7".<l 5R".<l 一丁十 5(1 rR) 一
θt
' 2
山 LθR
By dividi吨 by 5 , we conclude that H(R , t) satis且es the PDE
θH
1 叮叮 θ2H
θH
37+rzRZ否E 十 (1  r R) 百五
which is the same as
JU
' 2 5
(7.12).
0,
口
Problem 7: The price of a nondividendpaying asset is lognormally distributed. Assume that the spot price is 40 , the volatility is 30% , and the
interest rates are constant at 5%. Find the BlackScholes values of the IT JvI
put options 0丑 the asset with strikes 45 , 48 and 51 , and maturities 3 months
and 6 months.
For which of these options is the intrinsic value max(K  5 , 0) larger
than the BlackScholes value of the option (in which case the ∞
c orrespo丑时i丑
剑
d
ArneflcaI1 put is guaranteed to be worth more than the European p 叫 ?
臼创 缸
卫
Solution: The values of the out options are summarized in the table below:
Option Type Strike Jvlaturity Value K5
6 months 5.8196
45
5
Put
5
3 months 5.3403
Put
45
8
6 months 8.0325
48
Put
3 months 7.8234
48
8
Put
6 months 10 .4862
11
51
Put
3 months 10.5476
11
51
Put
θR
In general , the values of deepinthe money European put options are
lower than the premium K  5. This feature was observed for the options
priced here: the values of the 51puts and of the three months 48put are
below their intrinsic value K  5.
Also , note th创 for the 51puts (i.e. , for deep in the money puts) , the
values of the short dated options are higher than the values of the long dated
88.
CHAPTER 7. l1ULTIVARIABLE CALCULUS.
168
options. This is to be expected; for example , if the spot price is 0 , the value
of a European put is K e r1 in which case longer dated puts are worth less
than short dated ones. 口
Problem 8: Show that the premium1 of the BlackScholes value of a European call option over its intrinsic value max( 8  K , 0) is largest at the money.
In other words , show that the maximum value of
7.1. SOLUTIONS TO CHAPTER 7 EXERCISES
plain vanilla call with strike K when B 、 O. For simplicity, assume that the
underlying asset does not pay dividends and that interest rates are zero.
Sol毗 on: Let t = 0 in formula (7.13). For r = q = 0 , we find that α=i
Therefore , the value of the downandout call is
V(8)
8 _. I β2
B
81B
B
if 8 三 K;
f(S)=i cm(S)S+K7if S >K
OBs(8) ,
d1 =
Note that j (8) is a continωus function , but it is not differentiable at 8 = K.
For 8 < K , the function j(8) is the value of a call with strike K , 缸1
a丑
n
trlerefore is irlcreasir19.
扭
丑
卫
For 8 > K , we find that
1 < N(d 1 )
…
0(8) 
1 =N(d1 )<0 ,
and therefore the function j(8) is decreasing.
We conclude that j(8) has an absolute maximum point at 8 = K.
口
γ 一/一
= 0(8 , K , t) 一(计
o
IB2
(7 叫
γ 一/一
…
σ 飞IT
Note that 0 < N(d 1 ) < 1. Then ,
且也 BN(d 1 ) = O.
(7.15)
Using I'H6pital's rule , we obtain that
8K
E~ ~;.，~
B() B
N(d 2 )
θ出
8K 问丁zSKB可LNf(d2) 一
N(d 2 )
θB
1
li囚一二eXD
B、0 飞12 1f
I
I
品
d~ 2
I
~.4
~J

.一一τ
B σ 飞IT
I口 (B) )
" }
(7.16)
As B 、 0 ， the term 一手 In(B) is on the 。阳. of 一(ln( B))2 , and 阳efore
I
3 2
』史坷吃~~←一 7?In
一峭
 n
(7.17)
From (σ7.16) a叫 (σ7.17η) we 直nd that
叫
创
6
8K
口
Problem 10: Show that the value of a downandout call with barrier B
less than the strike K of the call , i.e. , B < K , converges to the value of a
lThis premium is also called the time value of the option.
(7.14)
I川剧 _ ~2T
=
28K..
I d~
τF王三 Ulr~ exp (  ~.4
~ , K ,t ) ,
$3.398883.
d?
and
σ 〉乙 1f T BJ~
whereα= 手 ~' to find the value of a six months downandout call on a
nondividendpaying asset with price following a lognormal distribution with
30% volatility and spot price 40. The barrier is B = 35 and the strike for the
call is K = 40. The riskfree interest rate is constant at 5%.
Solution: The value of the downandout call is
I
8K
~~~ N(d 2 ) ,
B
+
BN(d 1 )
σ 飞IT
Problem 9: Use the formula
V(8 , K , t)
8
l叫到十卓T
8K
IB 2α
飞
where 0(8) is the value of the plai口 vanilla call wi出 strike K and
Solution: Let j(8) = OBs(8)  max(8  K , O). It is easy to see that

飞 8
2
0(8) 一一( ~n  ,.J. I)    _ . '  "2I J
N(d 1
KN(d ' )
is obtained for 8 = K , where OBs(8) is the BlackScholes value of the plain
vanilla European call option with strike K and spot price 8.
1' (8) = ~(CBS)  1 = e qT N(d1 )
I: J)
0(8) 一;， 0
=
OBs(8)  max(8  K , O)
J
169
问~~~ N(d 2 )
= 0,
(7.18)
and , fro皿 (7.14) ， (7.15) , and (7.18) , we conclude that
(
_. _.
U~ V(8) = h~ (0(8) O.0 飞
_
_ _, _ ,
8K
BN(d 1 ) 十一:; _. (d 2 ))
N '"'I J
B
= 0 (8) .口
89.
CHAPTER 7. MULTIVARIABLE CALCULUS.
170
Problem 11: Compute the Delta and Gamma of a downandout call with
7.2. SUPPLEMENTAL EXERCISES
7.2
171
Supplemental Exercises
B<K.
Solution: We rewrite formula (7.13) to emphasize the dependence ofthe value
V(8) of the downandout call 0日 the spot price 8 of the underlyi吨 asset
as follows:
V(8) =
B
2α
CB s(8) 一 ~')n
2α
8
(B
2
CBS (
~n
飞 8
(7.19)
).
}
1. Compute
2αB2α
(B2
B2α+2
~(CBs)(8) + 一一CBsli
2α
8
+1
.
,_.
•
(B2
飞 8 )
υj
飞
j
where ~(CBS)(X) = e qT N(d 1 (x)) , with
d1 (x)
=
E
凶
{(川
=
D
十一丁 ~(CBS) (一)
hI唱h anpe
MDTK C41ng P叹
euu
DEsi
c
牛
E
U
u
n
d
一/
(7.20)

L
臼
a naf 5] z UU n τ
υ
V
口
忻
×
州
J
'
'
S
A
1
i
QU
2
BU
,
TLU
G
吨
，
一
一
uz any
m
咽
'
QU A?U
' 、
B
A
i
e doma n D
·
唱
B
E

larger ，俨 or 7r e ?
2. Which number is
l丑(是) + (γq 十号 )T
xdX
where
By differentiati吨 (7.19) with respect to 8 , we obtain that
~(V) =
J!n
3. Let 叩) : [0 ，∞)→ [0 ，∞) be two continuous functions with positive
values. Assume that there exists a constant M > 0 such that
u(x) 三 M+fa 叫 t)v( 忖 f x~O
2α(2α 十 1)B 2 α
r(V) = r(CBs)(8) 一例。
8'2α十万
(4α +
Show that
u(x) 三 M 叫俨(t) dt) ,
2)B 2a + 2 ( B 2 B h + 4 /
~(CBS)
r(CBs) (τ) ,
~ )一百 a+4
where
r(cBS)(x) =
with d1 (x) given by
(B 2
CBS ( 一)
飞 8 J
(7.20).
e qT
1 J
_ d1 (x)2
~exp (
':
I
(
x σ 飞/三7r 1
Hint: Investigate the monotonicity of the
v'
，e，
f
/III
V
/
l
十
。
口
、
、
U
f
A'LU
υ
/,
ι
11
'
U
、
、
l
'
/
7
α
fu日 ction
/III PI'120
z
Z
J
If x ?: 0
, .、B
TLU
u
ex D&
S
E
It's//
,
?b
、
/
t
B
飞
'
,
d
ι
，
ν
'
'
,
'
Note: This is a version of Gronwall's inequality, and it is needed , e.g. ,
to prove the uniqueness of the solution of an initial value problem for
ordinary differential equations.
4. What does the boundary condition V(B , t) = R for a downandout
call with barrier B and rebate R > 0 correspond to for the function
也 (x ， T) defined as follows: V(8 , t) 二 exp( 一 αx  bT)U(X , T) , where
(8
x = In I =_ I.
飞 K J
I
T
二
(T  t) σ2γ q
1
1._
(γ  q
,
12
,
2q
α= 一一「一丁、 b=l 一"十丁 1+ 一τ·
σ4
~'σ :L
2σ :L
~J
龟
90.
CHAPTER 7. IvIULTIVARIABLE CALCULUS.
172
7.3
5. Assume that the function V(S , I , t) satisfies the PDE
θ V θVI') ~')θ2V
δV
一+ InS一十一σ乎一一 + rS一  rV
δtθ I
2θS2
θS
7.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
= O.
(7.21)
Solutions to Supplemental Exercises
Problem 1: Compute
Consider the following change of variables:
V(S , I , t)
Show that F(y , t)
I
= F(y , t) , where y = satis且es
the
+ (T  t) InS
,
2T2
8 y2
D
followi吨 PDE:
主主十 σ2(T  t)2 ~2: + (r 一到丘 t 8F
δt
where
, T
2)
f儿z 伽
= {何?们}R2 I x ~ 0 , 1 三时 2 ， 1 才三 2}
Sol仰on: The change of variables 8 = xy and t
rF
=
0
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 as a function of the strike of the options.
In other words , assume that S , T , q and r are given , and let C (K) be
the (known) value of a call option with maturity T and strike K. Assume that options with all strikes K exist. Define the implied volatility
σimp(K) as the unique solution to
~ is equivalent to
when x 主 o and y ~三 O. This change of variables maps the domain D into
the recta吨Ie n = [1 , 2] x [1 , 2]. It is easy to see that
6. One way to see that American calls on nondividendpaying assets are
never optimal to exercise is to note that the BlackScholes 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 dividendpaying assets. In
other words , prove that the BlackScholes value of the European call
is smaller than S  K for S large enough , if the underlying asset pays
dividends continuously at the rate q > 0 (and regardless of how small
q is).
=
x~andY=而?
Tθu
Note: The values of Asian options with continuously sampled geometric
average satisfy the PDE (7.21).
173
dxdv _
IJ

θzβ钊
θzβ111
I:~ ~~i1 一一一立 I
Iθsθtθtθ81
d8dt
土豆 (均丘[出dt
2VSt 20 2t0) 2ft
i2t dsdt
Then ,
J!n
XdX
[[~去 ω
H[ VS dS ) ([忐 dt)
i(;二1:)(剖)
C(K) = CBs(K， σ问 (K)) ，
where CBs(K， σimp(K)) = CBs(S , K , T， σimp(K) ， r， q) represents the
BlackScholes value of a call option with strike K on an underlying
asset following a lognormal model with volatility 向mp(K). Find an
implicit differential equation satis自ed byσ问 (K) ， i.e. , fi日d
δσimp(K)
θK
aωsafun ct 10 of σirr1以 K).
旬肌杜∞丑
町η
Problem 2: Which number is larger ，♂ or 7r e ?
Solution: We show that 7r e <♂.
By taking the natural logarithm , it is easy to see that
l叫7r) / 1 _ ln(e)
作e< ♂仁中 eln(作) <作仁学一一<一一 7·
7r
e
e
(722)
91.
174
CHAPTER 7. MULTIVARIABLE CALCULUS.
Let f(x) = ln~x) with f : (0 , 00) • JR. Then f'(x) = 弓凶 The function
f (x) has one crit iC al point correspondi吨 tox = 飞 IS 1肌reasi吨 on the interval
(0 , e) and is decreasing on the interval (e ， ∞).
We conclude that f (x) has a global maximum point at x = e, i.e. , f (x) <
f (e) = ~ for all x > with x 护民 and therefore
In other words， ω( x) is a decreasi吨 function on the 凶怡 凹址 [归 ，∞) an
i nt ervaI 0
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<;7
which is
eq山Talent to 作 e
< e1r ; cf.
(7.22).
:::: u(x)exp(
Show that
ω
/
、
、
丑ecall
+
M
一
Z
t
t
1
x
u(t)v
(7.23)
I
u
'
'
'。
、
ι
U
、
'
3
u
，
I
、
ι
B
，
/
/
I
ex p
,
d
、
ι
，
于
心
(7.24)
x :::: 0
L
( 8 ( T  t) σ2γ q
t
~_ I 、
K
J1
r=
、
'
也 (x ，
(7.25)
'
U
WU
'
'。
'U
J
F
E
E飞
、
4·L υ
飞
，
，
/
,
、
Tb、l
/
JU
,
一
?U
一
J
们U
、
Z U Z
l
'
s
'
t飞
/
、
1
2
/
Z
flo
们
U
/
1
iI/
,
?b
、
I
12
. 2
J
2q
σ4
，
α
,
TL

u
Z
r)
=
exp(αz 十 br)V(8， t) ,
the boundary condition corresponding to V (B , t) = R for all °三 t 三 Tis
u
/
I
I
I
飞
(rq
L
b=l 一"一十一 1+ 一τ.
2 1σ ~
Solω4ω 7ηZ
毗 仰阳
仰
ion
n
Ill/
1
α: 一"一一、
2σ~
u (In ( ; ) ,
'
x::::O ,
°
111/
， ι
α
丛
'
U
that
'
l
corresponds to °三 7 三字 Since
U
I
'
/III
ρ Z
V(t)dt)
Problem 4: What does the boundary condition V(B , t) = R for a downandout call with barrier B and rebate 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 ρ Z
ρ z
/III
，
°
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
7)
叫咱) +b 7 ) 川)
(;)α 川，
、
飞
，
，
/
V
°三 γ 三字口
'
where the derivative is computed with respect to x.
Using the Product 丑ule， we find that
州=叫x)υ(以P ( +
1"
Problem 5: Assume that the function V(8 , I ,t) satis丑es the PDE
v(t) dt)
(M + 1" u(t)υ(叫(→(叫( 1" V(t)dt))
巾) (u仲 M 
1" u(t)v(叫叫 1"
θ V θV
1 ()_()δ2V
θV
一一十 In8~ 十一σ'l， 8'l， 一一十 r8 一一 θtθ I ' 2θ8 2θS
rV
Consider the following change of variables:
r
i
V(8 , I , t)
V(t)dt) (7.26)
Show that F(y , t)
=
F(y , t) , where
satis且es
一
y
十
一
T
,
Tb
二
/lt
一
T
2
I
2T2
、
、1
1
'
/
acu
1Ei
…
the following PDE:
但十 σ2(T  t)2a F 十 (r _ ~2) 丘1 笠 θt
(7.27)
0.
θν2
2)
T
θu
rF =
°
92.
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
θI
T
δV
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. Defiine 也 implied volatility σin以 K) 创 the
缸 t he
f
叽 η1
as
unique solution to
C(K) = CBs(K， σimp(K)) ，
1θF
1 Tt θF
θu?
δS S T
θ2V
θν?
1 ( (T  t)2 θ2F
一一_.
θS2
S2 飞
Tt 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
θtθ
I
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 I + r 一一一一一一一一
θ2F Tt θFV+TtθF
2
T2
δy2
T
θF
σ2(T  t)2 θ2F ,
一一+
n
~ ~十
θt'
2T2
where CBs(K， σ 仰1
叽n
以
η
阮挝 臼
S cl oωoles value of a call 叩 挝∞ with strike K on an underl抖 吨 部 附 ，'01lowinga
h
s
0 pt io n
臼勾y 1 gasset f 扣如叭吨
坊切
i 甘切 鸣
rin
创
lognω10ωr口na model 呐由 volatility σ叫 (K). Find an implicit differential equ 任
伊 口丑 al
m叫
r1
恼
w it h
tion satisfied byσ叫 (K) ， i.e. ，缸d
θy)
,
•
T
θu

~
(σ2 Tt θF
Ir 一 =::1 一一一一一  rF.
8γ
2)
T
θu
θK
邸 缸旧钊 ∞丑
a safun ct io of σ imη1 K).
叽 以
口
Solution: We first 自nd the partial derivative of the BlackScholes 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 nondividendpaying
assets are never optimal to exercise is to note that the BlackScholes 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 dividendpaying assets. In
other words , prove that the BlackScholes 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 )
93.
CHAPTER 7. MULTIVARIABLE CALCULUS.
178
We now differentiate with respect to K the formula
C(K)
CBs(K， σ问 (K))
=
which is the de直缸缸i山 丑 of σ仰以 K). Note that C (K) is assumed to be known
旧 tior1
n扰
让
l
町 mη1
for all K , as it the market prices. Using Chain Rule and (7.31) we find that
θC
θK
δCBS
IθCBS
一一一·
θK
θσθK
(7.32)
where
θCBS
_aT
民
U
Lagrange multipliers. N dimensional
Newton's method. Implied volatility.
Bootstrapping.
I T_ '!J.
万 =SeqVV=dsePNf(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
一兰 θσimp(K)
2
θy = 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=( Z 0 ， 1 ， X协 Zωa丑
阳
均
缸1
an
沁
λ 0=(λ0，1 ， λO ，2). From (8.1) it follows that 7 (x，)..)F(xo， λ0) = 0 is equivalent
沁 沁公
ω
叩
179
94.
180 CHAPTER 8. LAGRANGE MULTIPLIERS. NEWTON'S JVIETHOD.
ω
t
4
ι
u
·
τ
·
·
·
4
1
w 4L S 7
QUW u
'f HσMmzs Eh
ρ b m
μ me E
η
U
b
e
U
f
Q
a
U
; 认 矶 十 2 λ m Z 000020·7
W
l
2 λ
。;
「 入川 ， 十 J J m
J diU2'Z 1
0
4 z
2
， ，
0;
l
l
1
飞
l 一
"
八
l
3
0;
h
m
O
C
ο
.
4EA
，
一
们
向
Z
Z
一
十
=
缸
，2
队
，
川
h
L
一
一
，
，
"
3
巧
t
吼
叫
一
Z
Z
2
人
(ii) Find the asset allocation for a maximum expected return portfolio with
standard deviation of the rate of return equal to 24%.
0;
13.
nu 4t4
and
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,
门
4
一
一
飞
八
nuO& 一
才
t
牛
f
'
5
1飞
。
δ
。
，
血
、
、
E
E
，
，
，
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) 
4x22x32(2x1x2x3)+xi+x~13
xi+x~4x1 +6x213
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 .48.6) , 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;
句
B
A

在
·
龟
，
8.1. SOLUTIONS TO CHAPTER 8 EXERCISES
呐 tlere
w咄 陀 ω
扭
(8.7)
=( ωi) 吐=
山
)i=
f(ω) = 0.0625ωi + 0.0625ω;+0Oω9ω:+O 04tωρi
叫
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
95.
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%.
/σ?σ1 0"2P1 ，2 σ1σ3P1 ，3
包)3
W4
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 /
,
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
σ§
000σ1
J
J
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
μtω;
E[R]
var(R)
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
where
μ
'
一
一
ω tMω7
/
/
I
I
l
l
1飞
μ
'
μ
'
μ
'
μ
'
i
q
A
n
O
A
(8.12)
(8.13)
飞
l
i
t
z
1
/
也
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)
96.
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 (Itω1) 十 λ2(ωtMω  (J'~)
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) ，
(8.17)
To
it is easy to see that
7g(ω)
。
"
!
rM
ω
/
'
4
1、
、
I
;M九
(8.18)
=1
{=中
1=;山一11;
wtMω=σ主
中=丰〉
σ各=一 ωι1
。
C
2
乎
~~
(8.19)
C 孚
C
= ~1~ω= 2
2
•
、
、
l/
一
一
、
‘A
气
，
2UM" ω
111 飞八 M2p
1
i
f
?
b
ω
飞
n
L
1
1
1
1
1
/
守
J
7 (ω，，) F(ω7λ) = 0 ,
2
1
C
σ2
P
.
JR6 is given by
G(ω?λ1 ， λ2)
(μ+λ11 + 2川), .
Itω1
wt !V
Iω  (J'~
)
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
白
It
(川
飞7(叫) G(ω?λ1， λ2)
ω
2(Mω )t
飞
I
I
I
f
/
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
l t M 1 1
A
ω
飞
八
，
/
where G : JR6
r
飞
/
1
G(ω?λ1 ， λ2)
I
I
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
ω
F
μ
'
直缸缸1
时
r
飞
'?b
飞
，
I
/
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
/III11
十
V
2(Ax)t.
=
(8.21)
The gradient of the Lagrangian is
节
(去无)
/III
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'~).
97.
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· 2Jv
I
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) +x
(叫+手)T
 1'../ Jr"':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(rq)T < K;
if Se(rq厅
K;
if Se(rq)T > K.
(Recall that F = Se(rq)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
寸
t
4
+
nu
nu
1m
一

9ZM
寸
t
4
/IIiI
QO
no 巧t
v
h
υ
,
飞
I
I
I
/
'
"U 19" 十
exp
吃
nu
$
42·A
4
QO
nhu 时 vhu
i
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 exp
9
2
B
·2
(iii) Show that
if Se(rq)T
C.
三 K;
> K.
f (x) is a strictly increasi吨 function and
C < f(x) < Se qT  C , if Se(rq)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 <
Se qT
required for obtaining a positive implied volatility for a value C of the call
option.
Solution: (i) Note that
= ;(主 1ω巾
C =
(
1Se qT  K产 C, if Se(rq)T
lV(叫
d1
‘
ltt/
Z
飞 /飞)、I ,
+ 10 1. 6875· 25 exp(一切) )、J‘
(x)
Z
出 (x)
= In (丢) + (γ  q)T , x vIT
ZZ飞IT
十
2
In (圣)
+ (r  q)T
x vIT
x vIT
2
(8.23)
(8.24)
98.
188 CHAPTER 8. LAGRANGE MULTIPLIERS. NEWTON'S METHOD.
8.1. SOLUTIONS TO CHAPTER 8 EXERCISES
It is easy to see that
(iii) Differentiati 吨 f (x) with respect to x is the same as computing the
derivative of the BlackScholes value of a European call option with respect
to the 飞Tolatility 矶 which is equal to the vega of the call. In other words ,
l豆豆 d1(x) = ∞
J豆豆 d 2 (x) = 一∞?
and
and therefore
f' (x)
We conclude that
lim f(x) 
In (去)
xVT
= 一节言十一一;
xvT
I
Se qT  C.
'
2
也 (x)
In (去)
L/Tr
ln( 是) + (r+手)T
where d l = d d z )  d
Thus , f'(x) > 0 , :j x > 队 and f (x) is strictly increasi吨
Recall that limxt∞ f(x) = Se q'1'  C and
(ii) Let F = Se(rq)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:
F 二 K7
if F
if
C,
that
Se qT 
C , if F 三 K;
C, if F > K.
(iv) If F 三 K ， the problem f(x) = 0 has a solution x > 0 if and only if
Ther e£ore 且
陀扣
I im♂oN( d l( x)) 工 1 im 八 oN( d2(μ)) 工 Oa丑
何 归
出
且 x
伺 x
也
:炖咆 f(归)
仲 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
<
Se一ι(8.25)
If F > 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Il d 山阳伽
a丑时
侃圳
叫
refOr
缸
b
扣
ω
臼
<
C
<
SeqT.
(8.26)
Note that
:炖巧 d 1 巾(归 = :炖坷 d2 (x) =0
出仰
叫 Z)
也纣 忏
州
Se qT  Ke rT
Thus ,1im x a N(d1(x)) = lim八a N(d 2 (x)) = ~ and
一俨r
w(Z)=;(SfqTkeTT)czET(FK)c
=
erT(Se(叫)T _ K)
From (8.25) and (8.26) , we conclude that the problem f(x) = 0 has a positive
solution if and only if C belongs to the following range of values:
r
max (SeqT 一 Ke气 O )
一
吗
= C.
= erT(F  K).
< C <
Se qT
• If F > K , then In (去) > 0 and
~t~ d1 (x) = l~~ d2 (x) = ∞
z v
X .U
Therefore li叫、、、oN 付 l( x) = 且
V( d 仪 川)
出
I im♂八oN( d2( x)) = 1 a丑
何 归
也
:炖巧 f (μ) =Sf旨 qT 一 Ke 一刊 一 C
仲忏 仇
x
仇
s e一
♂
f叮
巳 TT
Problem 5: A three months atthemoney call on an underlying asset with
spot price 30 paying dividends continuously at a 2% rate is worth $2.5. Assume that the risk free interest rate is constant at 6%.
(i) Compute the implied volatility with six decimal digits accuracy, using the
bisection method on the i口terval [0.0001 , 1]' the secant method with initial
guess 0.5 , and Newton's method with initial guess 0.5.
99.
190 CHAPTER 8. LAGRANGE
(ii) Let
MULTIPLIER NEWTON'S METHOD.
丘
σ问 be
the implied volatility previously computed
method. Use the formula
σ
usi吨 Newtor内
r.JV2在 C 一与庄s
r.J百万
imp ,approx
to compute an approximate value
compute the relative error
σimp， apprωfor
the implied volatility, and
σzmp， αpprox 一 σ
vzmp
σimp
Solution: (i) Both the secant method with ♂ 1 = 0.6 and Xo = 0.5 and
Newton's method with initial guess Xo = 0.5 converge in three iterations to
an implied volatility of 39.7048%. The approximate values obtained at each
iteration are given below:
k Secant :Nlethod Newton's Method
。
1
2
3
0.5
0.3969005134
0.3970483533
0.3970481868
0.5
0.3969152615
0.3970481867
0.3970481868
[0.375063 , 0.5];
[0.375063 , 0 .406309];
Fi(x + hej) 2h  £  'J/ , j = 1 :风
 Fi(x hej)
~£~
,
'J/
Solution: We use Newton's method and the approximate Newton's method
both with forward difference approximations and with cen~ral difference approximations with toLconsec = 10 6 and toLapp r<?x = 10 9 . The parameter
h is chosen to be equal to toLconsec , i.e. , h = 10°.
All algorithms converged to the same solutions ,
f 1
I
1. 9831072429 I
飞 0.8845580785 J
旷工
and
x'
[0.390686 , 0 .406309];
0.3966718145 

where ej is a vector with all entries equal to 0 with the exception of the jth
entry, which is equal to 1.
(i) Solve F(x) = 0 using the approximate Newton:s algorithm obtai~ed by
substituting ~cF(Xold) for ~F(Xold)' Use h = 10 b , toLconsec 干 10° ， and
toLapprox = 10 9 , and two different initial guesses: Xo = (1 2 3)t and Xo =
(2 2 2)t.
(ii) Compare these results to those corresponding to the approxim~t~ Newtor白 method with forward finite difference approximations for ~F(x).
[0.375063 , 0 .4 37556];
(ii) The approximate value for the implied volatility given by (8.27) is
σimp，αpprox =
~c，jFi(X)
/ 1. 6905507599
The bisection method on the interval [0.0001 , 1] co盯erges in 30 iterations
to the same implied volatility of 39.7048%. The 五rst five iterations generate
the following intervals:
[0.250075 , 0.5];
The approximate gradie时 ~cF(x) = (~c，jFi( X) )ii 庐产
怡
叫
仪 队，j=
ω=
户
j
i.e. ,
put ed us ir19 cent ra1 也
臼
剖卫
忧 址 d ifference approx im
对 丑
lations ，
(8.27)
1 一呼E
191
8.1. SOLUTIONS TO CHAPTER 8 EXERCISES
If σimp = 0.3970481868 is the implied volatility obtained using Newton's
method , then
Xo
= I
2
I
飞 3
J
(n 扣r Xo 工 0)
The iteration counts are given in the table below:
Xo
39.6672%.
=
for
Iteration Count
Iteration Count
Iteration Cou日t
Approximate Newton Approximate 时ewton
Newton's Method
Central Differences
Forward Differences
( 1
{~ }
zmp，α，pprox
一 σ
v
zmp I
巳
0.000948
=
0.0948%.
口
σzmp
(
xf + 2XlX2 十 zi2223 十 9
( 2XI + 2XlX~ 十 ziziziz3 一 2
Z问仇 +zi23zd4
9
9
65
43
2
Problem 6: Let F : :lR3 • :lR3 given by
F(x) 
9
40
σ
I
/
We note that usi鸣 central 丑时te differences approximates the gradient
DF(x) of F(x) more accurately than if forward finite differences are used ,
100.
192 CHAPTER 8. LAGRANGE MULTIPLIERS. NEWTON'S METHOD.
8.1. SOLUTIONS TO CHAPTER 8 EXERCISES
+ 102 .4375 e 2r (O ,2)
2.4375 eO. 5r (O ,O.5) + 2.4 375 exp (  r(旧的十 0 叭
十 2 .4 375
resulting in algorithms with iteration counts closer to the iteration counts for
Newton's method. 口
e 1. 5r(O ,1. 5)
A
十2 .4 375
Problem 7: (i) Use bootstrappi 吨 to obtain a zero rate curve from the
following prices of Treasury instruments with semiannual coupo口 payments:
3623510 
IVIonth Tbill
IVIonth Tbill
Year Tbond
Year Tbond
Year Tbond
Year Tbond
°
°
If we let x = r(O , 2) , we
自nd
r(O , 0.5)
+ 0.5x
t ε[0.5 ， 2].
0.5r(0 , 0.5)
1. 5
(8.28)
+x
Recall that the price of the two year bond is the discounted present value of
all the future cash flows of the bond. Then ,
100 +立= 2 .4 375
32
eO. 5r (O，O.5) 十 2.4375 e r(O ,l)
I
LV
)
e2x
= 4.7289%.
We proceed by assuming that the zero rate curve is linear between two
years a时 three years. We note that r(O , 0.5) ， γ(0 ， 1) ， γ(0 ，1. 5) ， and γ(0， 2) are
known. If we let x = γ(0 ， 3) ， the price of the three year bond can be written
as
100 +主
32
2.4 375
eO. 5r (O，O.5) 十 2 .4 375 e r (O， l) 十 2 .4 375
+ 2.4375
e 2r (O，2) 十 2 侃侃p
/
r(O , 2)
2.5 ' V'~d
e 1. 5r(O ,1. 5)
+x
I
LV
)
+ 102 .4375 exp (2x)
r(0 , 3) = 4.7582%.
from (8.28) that
T(071)=;γ(0， 1. 5)
1. 5
. ,, _._/
Z 二二 0.047582.
Bootstrapping is needed to obtain the 2year , 3year , 5year and 10year
zero rates.
For example , for the two year bond , if the zero rate curve is assumed to
be linear between six months and two years , then
+ (t  0.5)r(0 , 2)
1. 5
V.VI
Using Newton's method to solve the nonlinear problem above , we obtain that
Therefore
( 100
= 4ln (一~: I = 0.052341 = 5.2341 %;
飞 98.7 )
( 100
r(0 , 0.5) = 2ln (一一 I = 0.050636 工 5.0636%.
飞 97.5 J
γ(0 ， 0.25)
(2  t)r(O , 0.5)
( .....
exp ( 1. 5 0.5γ(0 ， 0.5) 十 z + 102.4375
V~V J
r
_.1. 5
J
γ(0 ， 2)
(i) For the Treasury bills , the zero rates can be computed directly:
俨 (07027V
v:
J
飞1. 5
Using Newton's method to solve the nonlinear equation above for x , we obtain
that x = 0.047289 , and therefore
Coupon Rate Price
98.7
97.5
100~
4.875
4.875
10。在
4.625
99 玄
4.875
101 主
Assume that interest is continuously compounded.
(ii) How would the zero rate curves obtained by bootstrappi吨 from the bond
prices above , one corresponding to semiannually compounded interest , and
the other one corresponding to continuously computed interest , compare? In
other words , will one of the two curves be higher or lower than the other one ,
and why?
Sol仰on:
193
Using bootstrapping , we obtain similarly that
r(0 , 5) = 4.6303%
and
r(0 , 10) = 4.6772%.
Thus , we found the following zero rates corresponding to the maturities
of the four given bonds , i.e. , 3 months , 6 months , 2 years , 3 years , 5 years ,
and 10 years:
γ(0 ， 0.25) = 5.2341%;γ(0 ， 0.5) = 5.0636%;γ(0 ， 2) = 4.7289%;
γ(0 ， 3) = 4.7582%;γ(0 ， 5) = 4.6303%;γ(0 ， 10) = 4.6772%.
Since we assumed that the zero rate curve is linear between any two consecutive bond maturities , the zero rate r(O ,t) is known for any time between the
shortest and longest bond maturities , i. e. , for any t ε[0.25 ， 10].
(ii) Denote by rc(O , t) and r2(0 , t) the zero ra~e curves co.rrespo~di吨 to identicai discount factors , with rc(O , t) correspondi卫g to continuously compounded
101.
195
194 CHAPTER 8. LAGRANGE MULTIPLIERS. NEWTON'S METHOD.
8.1. SOLUTIONS TO CHAPTER 8 EXERCISES
interest , and with r2(0 , t) corresponding to semiannually compounded interest. Then ,
Assume that the overnight rate is 5% and that the zero rate curve is linear
on the following time intervals:
产衍 呻
f tT叫州
川川
川 (O
川
ρ
[0 , 0.5];
By solving for rc(O , t) in (8.29) , we 在nd that
几(0， t) =咐 +23二~r)
巧(0， t)
In (
(1 + r2(~' t) )"(~"))
巧(0， t) In ( (1 十 2/句:OJ))2/T2仰
r 51
11, ~
I 5 10 I
I~' 31;
I;
[0.5 , 1];
110 I
I 言， 51
Solution: We know that r(O , O) = 0.05. The six moI1 叫
览
e
∞
丑
computed from the price of the 6ιI工丑∞时 曰 zero ∞
一囚 10I1ths
丑 s
此
c oupon bOI1d as
口
( 100
飞 97.5 J
r(0 , 0.5) = 2ln (一一 ) = 0.050636 = 5.0636%
(8.30)
Using (8.30) , we can solve for the zero rate r(O , 1) from the formula given
the price of the one year bond , i.e. ,
100 _ 2.5 e O.5r (O ,O.5) + 102.5 e r (O ,l)
and obtain that
γ(0 ， 1) = 0.049370 = 4.9370%.
for the last inequality we used the fact that
(1+ ~r <民 'v'x > 0 ,
for x = 2/r2(0 , t). In other words the semiannually compounded zero rate
curve is higher than the continuously compounded zero rate curve if both
curves have the same discount factors.
While a rigorous proof is much more technical , the same happens if the
two curves are obtained by bootstrapping from the same set of bonds , i.e. ,
the zero rates corresponding to each bond maturity are higher if interest
is compounded semiannually than if interest is compounded continuously.
This is done sequentially, beginning with the zero rates corresponding to the
shortest bond maturity and moving to the zero rates corresponding to the
longest bond maturity one bond maturity at a time.
口
The third bond pays coupons in 2, 8 , 14 , and 20 months , when it also
pays the face value of the bond. The凡
103 =
3 叫去(喝)
3 叫去(叫)
P(马(山( 20 ( ) 叫
(一))叫ZT(072))(8
12
飞，
12 ) }
}J
Since we assumed that the zero rate curve is linear on the intervals [0 , 0.5]
and [0.5 , 1], the zero rates r(O， 是) and r(O ， 击) are known and can be obtained
by linear interpolation as follows:
γ(咐
命 (070)7(0705)= 0.0502 叫
(8.32)
r(叫
4r(0 , 0.5) + 2γ(0 ， 1) = 0.050214.
(8.33)
Let x = γ(0 ，
f§).
Since r(O , t) is linear 0川he interval [1 , f§ J, we find that
(D
, 12}
Problem 8: Use bootstrapping to obtain a continuously compounded zero
rate curve given the prices of the following semiannual coupon bonds:
NIaturity Coupon Rate Price
6 months
0
97.5
1 year
5
100
20 months
6
103
40 months
5
102
4
5 years
103
+
衍 (0 ， 1) + 2x
(8.34)
8
From (8.34) , it follows that the formula (8.31) can be written as
(
103 =
1
(一(
2
(~
8
3 仪p ( ~r ( 0，专门+ 3 叫了( 0, 1~2) )
i .
/6γ(0 ， 1)
3 叫
VI V
~
,
+ 2x 103 叫 ~x 飞
( 5
~W) +
),
I
(8 蚓
102.
196 CHAPTER 8. LAGRANGE MULTIPLIERS. NEWTON'S METHOD.
where r (0 ， 元) and r (0 ， 击) are given by (8.32) and (8.33) , respecti叫y. Using
Newtor内 method to solve for x in (8.35) , we 自nd t1时 x = 0.052983 , and
therefore
( 20
(0、~= I = 5.2983%.
8.2
197
8.2. SUPPLEMENTAL EXERCISES
飞
'12
J
Bootstrapping for the fourth and fifth bonds proceed similarly. For exampIe , the fourth bond makes coupon payments in 4 , 10 , 16 , 22 , 28 , 34 , and 40
months. The zero rates corresponding to coupon dates less than 20 months ,
i.e. , to the coupon dates 4 , 10 and 16 months , can be obtained from the
part of the zero curve that was already determined. By setting x = γ(0 古)
and assuming that the zero rate curve is linear between 20 months and 40
months , the zero rates corresponding to 22 , 28 , 34 , and 40 months can be
written in terms of x. Thus , the pricing formula for the fourth bond becomes
a nonlinear equation in x which can be solved using Newton's method. The
zero rate r (0 , ~) is then determined
Using bootstrapping and Newton's method we obtain that
。一)
(40
, 12
=
4.5帆 r(0， 5)
=
3叫
Summarizing , the zero rate curve obtained by bootstrapping is given by
r(O , O) = 0.05;
(
riO ,
飞，
2/
1("
12
I
riO , 1~(" I = 0.050214;
飞， 12 }
= 0.050212;
}
民
72)=om3; r叫叶= 0.0铅326;γ(0， 5)
~D = 0.052983; (叫
(0 ,
and is linear on the intervals
nu nu vhu
nu vhu
咱
t
4
「
1
1
3
1
3
3
3
3
1
h
434
v
h
u
q
ο
气
3
3
3
1
1
1
'
i
i
d
俨
s
t
t
5
3
3
3
3
1
L
53
ω
3
叮
3
1
1
1
1
1
1
3
3
4
俨
i
t
t
E
t
l
t
t
g
L
= 0.032119 ,
『
g
a
i
i

」
1. (i) If the
c盯rent
η (0， t)
zero rate curve is
=
0.025 由叫古)十 Jv
find the yield of a four year semiannual coupon bond with coupon rate
6%. Assume that interest is compounded continuously and that the
face value of the bond is 100.
(ii) If the zero rates have a parallel shift up by 10 , 20 , 50 , 1~0 ， and ?O~
,
bs.:sis points , respectively, i.e. , ifthe zero rate curve changes from rl(O , t)
to r2(0 ,t) = rl (0 , t) +dr , with dγ= {0.001 , 0.002 , 0.005 , 0.01 , 0.02} , find
out by how much does the yield of the bond increase in each case.
Note: In general , a small parallel shift in the zero rate curves results in
a shift of similar size and direction for the yield of most bonds (possibly
with the exception of bonds with 10吨 maturity). This assumption will
be tested for the bond considered here for parallel shifts ranging from
small shifts (ten basis points) to large s1曲
2. Consider a six months atthemoney call on an underlying asset following a lognormal distribution with volatility 30% and paying dividends
continuously at rate q. Assume that the interest rates are constant
at 4%. Show that there is a unique positive value of q such that
.6.( C) = 0.5 , and find that val~e using Newton's method. How does
this value of q compare to r 十亏?
3. The following prices of the Treasury instruments are given:
m3
vhu
Supplemental Exercises
口
612 23510 
lVlonth Tbill
Month Tbill
Year Tbond
Year Tbond
Year Tbond
Year Tbond
Coupon Rate
0
0
2
4.5
3.125
4
Price
99.4565
98.6196
1011号
107~I
102 立
g
?t
103~';
The Treasury bonds pay semiannual coupons. Assume that interest is
continuously compounded.
(i) Use bootstrappi 吨 to obtain a zero rate curve from the prices of the
6months and 12monthsτ
'reasury bills , and of the 2year , 5year and
10year Treasury bonds;
103.
198 CHAPTER 8. LAGRANGE lVIULTIPLIERS. NEWTON'S METHOD.
(ii) Find the relative prici吨 error correspondi吨 to the 3year Treasury
bond if the zero rate curve obtained at part (i) is used. In other words ,
price a 3year semiannual coupon bond with 4.5 coupon rate and find
its relative error to the price 107去 of the 3year Treasury bond
8.3
8.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
(ii) If the zero rates increa毗 the value of the bond decreases , and therefore
the yield of the bond will increase. Our goal here is to investigate whether a
parallel shift of the zero curve up by dr results in an increωe of the yield of
the bond also equal to dr
When the zero rates increase from r1 (0 , t) to 俨2(0 ， t) = r1(0 ,t) + dr , the
value of the bond decreases from B 1 given by (8.36) to
B2
Solutions to Supplemental Exercises
= 0.025
 了 3 叫 ( r2 ( 0 , ~ )斗+ 103 吨(4州， 4))
仨7
飞飞， 2) 2)
The new yield of the bond , denoted by Y2 , will be larger than the initial yield
and is obtained by solving
Problem 1: (i) If the current zero rate curve is
η (0， t)
到，
+忐叫古)+τ:71
B2 =
且nd the yield of a four year semiannual coupon bond with coupon rate 6%.
Assume that interest is compounded continuously and that the face value of
the bond is 100.
(ii) If the zero rates have a parallel shift up by 10 , 20 , 50 , 100 , and 200
basis points , respecti叫y， i. e. , if the zero rate curve cha吨es from 'r1 (0 , t) to
r2(0 , t) = η(OJ)+d71with dT={0.00170.00270.00570.0170.02}7and out by
how much does the yield of the bond increase in each case.
乌
B1 工 兰 3 倪吨叫叶 η1
寸(一仆俨
p(
= 106.1995.
(8.36)
The yield of the bOIId is found by solving the formuia for the price of the
bond in terms of its yield , i.e , by solving
,
=
~ 3 exp ( y
D
+ 103
exp(一句)
(8.37)
for U7WIleEe BIis given by (836)7i.e7BI=1061995.Using NewtOI内
method , we obtain that the yield of the bond is
Y = 0.042511 = 4.2511%.
￡ 3 仪叫叫叶 Uω2
寸(一叶
p(
for Y2 , where B 2 is given by (8.38).
For parallel shifts equal to dr = {0.001 , 0.002 , 0.005 , 0.01 , 0.02} , we obtain
the following bond prices and yields:
Zero rate shift New bond price New yield Yield increase
dr
B2
的 Y
Y2
0.043511
10bp = 0.001
105.8150
0.00099979
0.044510
0.00199957
20bp = 0.002
105 .4 319
0.047510
50bp = 0.005
104.2915
0.00499893
0.052509
100bp = 0.01
102 .4 199
0.00999784
0.062506
200bp = 0.02
98.7829
0.01999562
SoJUUonJ(i)TEle bond provides coupOBpaymmts equal to3every sixmo毗1S
until 3.5 years from now , and a final cash flow of 103 i且 four years. By
disc，?u~t，i?g t~is ca~h flo;V~ t~ the. I? resent using the zero rate cur~e r1 (0 , t) ,
we 直nd that the value of the bond is
B
199
As expected , the increase of the yield of the bond is slightly smalle飞
very close to , the parallel shift of the zero rate curve , i.e. ，的 Y 自 dr.
but
口
Problem 2: Consider a six months atthemoney call on an underlying asset
following a lognormal distribution with volatility 30% and paying dividends
continuously at rate q. Assume that the interest rates are constant at 4%.
Show that there is a unique positive value of q such that .6.( C) = 0.5 , an
白口
fII1d that value usi口g Newtωon method. How does this value of q compare to
岳
旷
l'S
r+ 号?
Solution: Recall that the Delta of a plai口 vanilla call option on an underlying
asset paying dividends continuously at rate q is
.6.( C) = e qT N(d 1 ) ,
(8.39)
104.
200 CHAPTER 8. LAGRANGE MULTIPLIERS. NEWTON'S METHOD.
8.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
where
In other words , if the interest rates are flat at 4% , the Delta of a six
months atth emoney call option on a口 u日derlying asset with volatility 30%
is equal to 0.5 if the underlying asset pays 6.69% dividends continuously.
d1 =
In (去) + (rq 十号 )T
(/
‘
σ 飞IT
If q= γ + 兰 we 直缸 from (何.ι40 and (8 .4钊钊 thatdl=Oa卫
nd
8 创)
1) 创 出
For an atthemoney option , i.e. , for S = K , we find that
d1
~(C)
= 川♂ σf
一
(8 .40)
From (8.39) and (8.40) , we find that
~(C)
/ (r  q) 叮
σd 飞
)
= e qT N(d1 ) = e qT N Iσ; ~
It is easy to see that
~(C)
201
(8 .4 1)
=
f
巳 一叫V( 0) = 0.5e一q T
仰
矿
< 0.5.
Since ~(C) is a decreasing function of q , we obtain that the value of q such
that ~ (C) = 0.5 must be lower than r 十号 Indeed， the value previously
obtained for q satisfies this condition , i.e. ,
q = 0.066906
<
0 附  r 十二口
is a decreasing function of the dividend rate
q , smce
Problem 3: The following prices of the Treasury instruments are given:
θA
θq
Te 句 (d 1 ) 十 f卢 川/气(刷
俨 q叩咐川 d1)
T N' 出
ω
8d1
dr
I
‘~ jrF 飞
中
卡
J
一 T e一庐句川 d 1) 十+ 俨一吁 TτE =e一亏叮{ 一工立土
口
川 仇创
N(刷
e q
飞乒
V 'l:7r
l
σ/
< O.
…
When q = 0 , we find that
6. (0) = N
(手+手)
> N(
5
Also , since 0 < N(d 1 ) < 1, it follows that
J!!炉 (C) = J豆豆 (e句 (d 1 )) = 0
We conclude that ~(C) is a decreasing function of q and that , for q 主 0 ，
the values of ~(C) decrease from N(O) > 0.5 to O. Therefore , there exists
a unique value q > 0 such that ~(C) = 0.5. From (8 .4 1) , it follows that
this value can be obtained by solving for x the nonlinear equation f (x) = 0,
where
/ (r  x)/叮 + σnI  0.5.
f( 叫 = e X1 ' N I .
~ v ~
~.L
v
1σ
二~
612 23510 
Month Tbill
Month Tbill
Year Tbond
Year Tbond
Year Tbond
Year Tbond
Price
99 .4 565
98.6196
1011旦
107里
102 立
~?b
103~'~
The Treasury bonds pay semiannual coupons. Assume that interest is continuously compounded.
(i) Use bootstrapping to obtain a zero rate curve from the prices of the 6一
months and 12 叮
τ￥easur)γy bonds;
(ii) Find the relative pricing error corresponding to the 3year Tr easury bond
if the zero rate curve obtained at part (i) is used. In other words , price a
3year semiannual coupon bond with 4.5 coupon rate and 五日d its relative
error to the price 107~ of the 3year Treas町 bond
Solution: (i) The 6mo叫1S and 12months zero rates can be obtained directly from the prices of the τ'reasury bills , i.e. ,
r(0 , 0.5)
I
Using Newton's method , we obtain that q = x = 0.066906.
Coupon Rate
0
0
2
4.5
3.125
4
俨
'
/
I
E
飞
nu
才
t
4
、
、
，1
'
'
'
'
21n
(品5)
叫品)
= 1.09%;
=
1 究
105.
202 CHAPTER 8. LAGRANGE MULTIPLIERS. NEWTON'S IvIETHOD.
Using bootstrapping , we obtain the following 2year , 5year and 10year
zero rates:
γ(0 ， 2)
1'
1'
(0 , 5)
(0 , 10)
1. 2099%;
2.6824%;
3.7371%.
Bibliography
(ii) The zero rates computed above correspond to the followi吨 zero rate
curve which is piecewise linear between consecutive bond maturities:
1'
(0 , t)
=
r (到一 1)γ( 0, 1) + 2(1  t) l' (0 , 0.5) , if 0.5 三 t 三 1;
J (t 1) 1' (0 ， 2) 十 (2  t) 1' (0 , 1) ,
if 1 三 t 三 2;
)平 γ(0， 5) +于 γ(0， 2) ，
if 2:::;t:::;5;
t孚 γ(0 ， 10) +半响的，
if 5 < t < 10
l
[1] Milton Abramowitz a时 Irene Stegun. Handbook of Mathematical Functions. National Bureau of Standards , Gaithersburg ，扣1aryland ， 10th corrected printing edition , 1970.
[2] Dan Stefa旧ca. A lVlathematical Primer with Numerical lVlethods for Financial Engineering. FE Press , New York , 2007.
With this zero rate curve , the value of the 3year semiannual coupon bond
with 4.5 coupon rate is
t2z2
/III /It 1111/
liI/
nu
nu
nu qd
5
exDA
ex ny
2
十
一
γ
一
nu
一
μ
川
队
H
m
The price of the 3year Treasury bond was given to be 107.5625. There
B
门
，
由
v
h
u
τ
i
γ
。
，
"
。
中
vhu
/
1
、
毛J
才
i
门
。
币
'
飞
fore , the relative pricing error given by the bootstrapped zero rate curve
which does not include the 3year bond is
1107.5625  108.19301
i = 0.005862 107.5625
0.59%.
口
203
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