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FINANCIAL ENGINEERING
ADVANCED BACKGROUND SERIES
FE PRESS
New York
Financial Engineering Advanced Background Series
Published or forthcoming
1. A Primer for the Mathematics of Financial Engineering, by Dan Stefanica
2. Solutions Manual — A Primer for the Mathematics of Financial Engineering, by Dan Stefanica
3. A Probability Primer for Mathematical Finance, by Elena Kosygina
4. Numerical Linear Algebra Methods for Financial Engineering Applications, by Dan Stefanica
5. Differential Equations with Numerical Methods for Financial Engineering,
by Dan Stefanica
SOLUTIONS MANUAL
A Primer
for the Mathematics of
Financial Engineering
DAN STEFANICA
Baruch College
City University of New York
FE PRESS
New York
FE PRESS
New York
www.fepress.org
Information on this title: wwwlepress.org/mathematical_primer
@Dan Stefanica 2008
All rights reserved. No part of this publication may be
reproduced, stored in a retrieval system, or transmitted,
in any form or by any means, electronic, mechanical,
photocopying, recording, or otherwise, without the prior
written permission of the publisher.
First published 2008
Printed in the United States of America
ISBN-13 978-0-9797576-3-1
ISBN-10 0-9797576-3-0
To My Beautiful Family
Contents
Preface
ix
Acknowledgments
xi
0. Mathematical preliminaries
0.1 Solutions to Chapter 0 Exercises
0.2 Supplemental Exercises
0.3 Solutions to Supplemental Exercises
1
1
11
11
1 Calculus review. Plain vanilla options.
1.1 Solutions to Chapter 1 Exercises
1.2 Supplemental Exercises
1.3 Solutions to Supplemental Exercises
17
17
32
33
2 Numerical integration. Interest Rates. Bonds.
2.1 Solutions to Chapter 2 Exercises
2.2 Supplemental Exercises
2.3 Solutions to Supplemental Exercises
45
45
57
58
3 Probability concepts. Black—Scholes formula. Greeks and
Hedging.
63
3.1 Solutions to Chapter 3 Exercises
63
3.2 Supplemental Exercises
82
3.3 Solutions to Supplemental Exercises
83
4 Lognormal variables. Risk—neutral pricing.
4.1 Solutions to Chapter 4 Exercises
4.2 Supplemental Exercises
4.3 Solutions to Supplemental Exercises
vii
91
91
106
107
viii
CONTENTS
5 Taylor's formula. Taylor series.
5.1 Solutions to Chapter 5 Exercises
5.2 Supplemental Exercises
5.3 Solutions to Supplemental Exercises
113
113
124
126
6 Finite Differences. Black—Scholes PDE.
6.1 Solutions to Chapter 6 Exercises
6.2 Supplemental Exercises
6.3 Solutions to Supplemental Exercises
135
135
150
151
7 Multivariable calculus: chain rule, integration by substitution, and extrema.
161
7.1 Solutions to Chapter 7 Exercises
161
7.2 Supplemental Exercises
171
173
7.3 Solutions to Supplemental Exercises
8 Lagrange multipliers. Newton's method. Implied volatil179
ity. Bootstrapping.
8.1 Solutions to Chapter 8 Exercises
179
197
8.2 Supplemental Exercises
198
8.3 Solutions to Supplemental Exercises
Bibliography
203
Preface
The addition of this Solutions Manual to "A Primer for the Mathematics of
Financial Engineering" offers the reader the opportunity to undertake rigorous self—study of the mathematical topics presented in the Math 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. Many 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 Math 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 Manual 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 Programs since
its inception in 2002. Over 90 percent of the graduates of the Baruch MFE
Program are currently employed in the financial industry.
"A Primer for the Mathematics of Financial Engineering" and this Solutions
Manual are the first books to appear in the Financial Engineering Advanced
Background Series. Books on Numerical Linear Algebra, on Probability, and
on Differential Equations for financial engineering applications are forthcoming.
Dan Stefanica
New York, 2008
'Baruch MFE Program web page: http://www.baruch.cuny.edu/math/masters.html
QuantNetwork student forum web page: http://www.quantnet.org/forum/index.php
ix
Acknowledgments
"A Primer for the Mathematics of Financial Engineering" published in April
2008, is based on material covered in a mathematics refresher course I taught
to students entering the Financial Engineering Masters Program at Baruch
College.
Using the book as the primary text in the July 2008 refresher course was a
challenging but exceptionally rewarding experience. The students from the
2008 cohort of the Baruch MFE Program who took that course were driven to
master the material and creatively incorporate ideas at an even deeper level
than that of the Math Primer book. Many of the supplemental questions in
the Solutions Manual came about as a result of the remarkable interaction
I had with this very talented group. I am grateful to all of them for their
impressive efforts.
Special thanks go to those who supported the proofreading effort: Barnett
Feingold, Chuan Yuan-Huang, Aditya Chitral, Hao He, Weidong Huang, Eugene Krel, Shlomo Ben Shoshan, Mark Su, Shuwen Zhao, and Stefan Zota.
The art for the book cover was again designed by Max Rumyantsev, and
Andy Nguyen continued to lend his tremendous support to the fepress.org
website and on QuantNet.org. I am indebted to them for all their help.
This book is dedicated to my wonderful family. You give sense to my work
and make everything worthwhile.
Dan Stefanica
New York, 2008
xii
ACKNOWLEDGMENTS
Chapter 0
Mathematical preliminaries.
0.1 Solutions to Chapter 0 Exercises
Problem 1: Let f R R be an odd function.
(i) Show that x f (x) is an even function and x2 f (x) is an odd function.
(ii) Show that the function g1 R R given by gi (x) = f (x2) is an even
function and that the function g2 R R given by g2(x) = f (x3) is an odd
function.
(iii) Let h R
R be defined as h(x) -=- x' f ( ), where i and j are positive
integers. When is h(x) an odd function?
Solution: Since f (x) is an odd function, it follows that
f (—x) = — f (x), V x E R.
(1)
(i) Let f1 (x) = x f (x) and f2(x) = x 2f (x). Using (1), we find that
f1(—x) = —xf(—x) = xf(x) = f 1(x), V x E R;
f2(—x) = (—x)2 f(—x) =
x 2 f(x) =
f2(x), V x E R.
(2)
(3)
We conclude from (2) that fi (x) is an even function, and, from (3), f2(x) is
an odd function.
(ii) From (1), it follows that
(4)
91( — x) = f ((—x)2) = f (x2) = g1(x), V x E IR;
g2(—x) = f ((—x)3 ) = f (—x3) = — f (x 3) = — 92(x), V x E R. (5)
We conclude from (4) that g1(x) is an even function, and, from (5), that g2(x)
is an odd function.
(iii) If j is a positive integer, it follows from (1) that
f ((—x)3) = (-1)3 f (x), V x E R.
1
(6)
2
MATHEMATICAL PRELIMINARIES
Then, using (6), we find that
h(—x) =- (—x)i f ((—x)3) = (-1)i ((-1)3f (x3 )) = (-1)i+3 f (x3)
= (-1)t±3 h(x), V x E R.
Therefore, if i + j is an even integer, the function h(x) is an even function,
and, if i + j is an odd integer, the function h(x) is an odd function. ❑
Problem 2: Let S(n, 2) = ELI k2 and S(n, 3) = ELI k3.
(i) Let T(n, 2, x) = >k=1k2 Xk ' Use the recursion formula
T(n, 2, x) = x —1 (T (n, 1, x)),
and the fact that
T (n, 1, x) =
x
—
(n i)
xn+i
nxn+2
(1 — x ) 2
to show that
T(n, 2, x)
x + x2 — (n 1)2xn+1 + (2n2+ 2n — i ) xn+2 — n2xn+3
(1— x)3
(ii) Note that S(n, 2) = T(n, 2,1). Use l'HOpital's rule to evaluate T(n, 2,1),
n(n+1)(2n+1)
and conclude that S(n, 2) =
(iii) Compute T (n, 3, x) =
k3 5k using
T(n,3,x) = x
the recursion formula
(T(n,2,x)).
(iv) Note that S(n, 3) = T(n, 3, 1). Use l'HOpital's rule to evaluate T(n, 3, 1),
and conclude that S(n, 3) -=
(n(n2+1) )
Solution: (i) The result follows from (7) and (8) by using Quotient Rule to
differentiate T (n, 1, x).
(ii) It is easy to see that T(n, 2, 1) =
k2 = S(n, 2). By using l'HOpital's
rule we find that T (n, 2, x) is equal to
x x
2
(n +1)2 xn+1
(2n2 + 2n — 1)sn+2 n2 xn-1-3
(1 — x)3
1 + 2x — (n + 1)3xn + (2n2+ 2n — 1)(n + 2)xn+1— n2(n + 3)xn+2
lim
x-4
—3(1 — x)2
lim
3
0.1. SOLUTIONS TO CHAPTER 0 EXERCISES
( 2 - (n + 1)3nxn-1+ (2n2 + 2n - 1)(n + 2)(n +1)xn
-n2(n + 3)(n + 2)xn+1
lim
x.-41
6(1 - x)
( -(n + 1)3n(n - 1)xn-2 + (2n2+ 2n - 1)(n + 2)(n + 1)nxn-1
-n2(n + 3)(n + 2)(n + 1)xn
lim
x-4
-6
-(n + 1)3n(n - 1) + (2n2 + 2n - 1)(n + 2)(n + 1)n
-n2 (n + 3)(n + 2)(n + 1)
6
n(n +1) (-(n + 1)2(n - 1) + (2n2 + 2n - 1)(n + 2) - n(n + 3)(n+ 2))
6
n(n + 1)(2n + 1)
6
Therefore,
n(n + 1)(2n + 1)
S(n, 2) =
6
(iii) Finding the value of T(n, 3, x) requires using Quotient Rule to differentiate T(n, 2, x).
(iv) The solution follows similarly to that from part (ii), albeit with more
complicated computations. ❑
Problem 3: Compute S(n, 4) = Erkl=i 1c4using the recursion formula
S(n,i) =
1
((n +1)i+1 - 1 i +1
( t1
S(n, j))
i=0
for i = 4, given that
S(n, 0) = n; S(n,1) =
n(n
1) .
2
S(n,3) =
S(n, 2) =
(n(n2+ 1))2 .
Solution: For i = 4, the recursion formula (9) becomes
S(n,
=
1
3
5
((n + 1)5 - 1 -
(
o
) S(n, j))
n(n+ 1)(2n + 1) .
6
(9)
4
MATHEMATICAL PRELIMINARIES
1
= — ((n + 1)5— 1 — S(n, 0) — 5S(n, 1) — 10S(n, 2) — 10S(n, 3))
5
n(n + 1)(6n3 +9n2+ n — 1).
❑
30
Problem 4: It is easy to see that the sequence (xn)n>1given by xn = Erk1=1 k2
satisfies the recursion
xn+i = xn + (n + 1)2 , V n > 1,
(10)
with xi = 1.
(i) By substituting n + 1 for n in (10), obtain that
xn+2 = xn+i. + (n + 2)2.
Subtract (10) from (11) to find that
Xn+2 = 2Xn-F1
Xn+ 2n
+ 3, V n > 1,
(12)
with x1= 1 and x2 = 5.
(ii) Similarly, show that
xn+3 = 3xn+2 — 3xn+1 + xn + 2, V n > 1,
(13)
with x1 = 1, x2 = 5, and x3 = 14.
(iii) Prove that the sequence (xn)n>1 satisfies the linear recursion
Xn+4
—
4Xn+3
6 X71,4-2 —
4xn+1 + xn = 0, V n > 1.
Solve this recursion and show that
n(n+ 1)(2n + 1)
, V n > 1.
6
Conclude that
n(n + 1)(2n + 1)
, V n > 1.
S(n 2) = E k2 _
6
k=1
Solution: From (11), we obtain that the first terms of the sequence (xn)n>i
are x1 = 1, x2 = 5, x3 = 14, X4 = 30.
(i) The recursion (12) follows immediately by subtracting (10) from (11).
5
0.1. SOLUTIONS TO CHAPTER 0 EXERCISES
(ii) We substitute n + 1 for n in (12) and obtain that
xn+3 = 2x,+2 — xn+i + 2(n + 1) + 3.
(14)
By subtracting (12) from (14), we find that
Xn+3 = 3Xn+2 3Xn+1 + x n +2, V n > 1.
(iii) We substitute n + 1 for n in (13) and obtain that
Xn+4 = 3Xrz+1 3Xn+2 + Xn+1 + 2, V n > 1.
(15)
By subtracting (13) from (15), we find that
Xn+4
4xn+3 + 6 xn+2 4Xn+1 + xn = 0, V n > 1.
(16)
The characteristic polynomial associated to the recursion (16) is
P(z) = z4— 4z3 + 6z2— 4z +1 = (z — 1)4.
The polynomial P(z) has root A = 1 with multiplicity 4. We conclude that
the there exist constants CZ, i = 1 : 4, such that
xn =
C2n + C3n2 + C4n3, V n > 1.
Since x1 = 1, x2 = 5, x3 = 14, x4 = 30, it follows that C1, C2, C3 and C4
satisfy the following linear system
(1
1
1
1
1
2
3
4
1
4
9
16
1 )
8
27
64
We obtain that C1 = 0, C2 = , C3 =
Ci
( 1
5
14
30
C2
C3
C4
a
n n2 n
3
n(n+
xn = 6 + — +
=
— 2 3
and C4 =
1)(2n + 1)
6
and therefore
, V n > 1. ❑
Problem 5: Find the general form of the sequence (xn)n>0 satisfying the
linear recursion
xn+3 = 2xn±1 + xn, V n > 0,
with xo= 1, x1 = —1, and x2 = 1.
First Solution: By direct computation, we obtain x3= —1, x4 = 1, x5= —1,
x6 = 1. It is natural to conjecture that xn = (-1)n for any positive integer
n. This can be easily checked by induction.
6
MATHEMATICAL PRELIMINARIES
Second Solution: Alternatively, we note that the sequence (xn )n>0 satisfies
the linear recursion x n+3 — 2xn+i — xn= 0, with characteristic polynomial
P(z) = z3— 2z +1 = (z +1)(z2— z — 1).
The roots of P(z) are 1, 1+1
5, and 1 -1̀52 . Therefore, there exist constants
C1, C2, and C3 such that
xn = C1(-1)n + C2
(
1 ±1
2
n + C3 (1
n , V n > O.
1
2
By solving the 3 x 3 linear system for C1, C2, and C3 obtained by requiring
that xo = 1, xi = —1, and x2 = 1 we find that Ci = 1, C2 = 0, and C3 = O.
We conclude that
xn = (-1)n, V n > O. ❑
Problem 6: The sequence (xn)n>0 satisfies the recursion
Xn+1 = 3xn + 2, V n > 0,
(17)
with xo = 1.
(i) Show that the sequence (xn)n>0 satisfies the linear recursion
Xn+2 = 4xn+1 — 3xn, V n > 0,
with xo = 1 and xi = 5.
(ii) Find the general formula for xn, n > 0.
Solution: (i) Let n = 0 in (17) to find that x1 = 5. By substituting n +1 for
n in (17), it follows that
Xn+2 = 3Xn+i + 2.
(18)
We subtract (17) from (18) and obtain that
xn+2 — 4xn-ki + 3xn = 0, V n > O.
(ii) The characteristic polynomial of the linear recursion (19) is
P(z) = z2— 4z + 3 = (z — 1)(z — 3),
which has roots 1 and 3. Thus,
xn = Ci + C2371, V n > O.
(19)
0.1. SOLUTIONS TO CHAPTER 0 EXERCISES
7
Since xo = 1 and x1= 5, we obtain that C1= —1 and C2 = 2 and therefore
xr, = 2 . 3' — 1, V n > O.
❑
Problem 7: The sequence (xn)n>0 satisfies the recursion
xn+1 = 3x n + n + 2, V n > 0,
(20)
with xo =1.
(i) Show that the sequence (xn)n>0 satisfies the linear recursion
Xn+3 = 5xn+2
3xn) V n > 0,
7Xn+1
with xo= 1, x1 = 5, and x2 = 18.
(ii) Find the general formula for xr„ n > 0.
Solution: (i) The first three terms of the sequence can be computed from
(20) and are xo = 1, x1 = 5, and x2 = 18.
By substituting n + 1 for n in (20) we obtain that
xn+2 = 3xn+1 + n + 3, V n > 0.
(21)
Subtract (20) from (21) to find that
Xr1+2 = 4xn+1 3xn 1, V n > 0.
(22)
Substitute n + 1 for n in (22) to obtain that
xr,±3 = 4xn+2 — 3xn+1 + 1, V n > 0.
(23)
Subtract (22) from (23) to find that
xn+3 = 5Xn+2
7xn+1
3xn) V n > 0.
(24)
(ii) The characteristic polynomial of the linear recursion (24) is
P(x) = z3— 5z2 +7z — 3 = (z — 1) 2(z — 3).
Therefore, there exist constants C1, C2, C3 such that
xn = Ci3n + C2n + C3, V n > O.
Since xo = 1, x1 = 5, and x2 = 18, we find that= )C9- —
We conclude that
xn —
3n+2 — 2n — 5
4
, V n > O.
❑
5
4 ) C3 =
MATHEMATICAL PRELIMINARIES
8
Problem 8: Let P(z) = Ek a.zi
o 2 be the characteristic polynomial corresponding to the linear recursion
aixri±i = 0, V n > O.
E
i=o
(25)
Assume that A is a root of multiplicity 2 of P(z). Show that the sequence
(yn)„>0 given by
yri = CnAn, n > 0,
where C is an arbitrary constant, satisfies the recursion (25).
Solution: Note that A is a root of multiplicity 2 of P(z) if and only if P(A) = 0
and Pi(A) = 0, where
P'(z) =
E
i=i
Then, for any n > 0,
Eaiyn±i = E aiC(n + i)An±i
i=o
i=0
= Cn
E
i=0
= CnAn
+ C
E icti An+i
i=0
E aiAi + CAn+1 E
k
i=i
i=o
CnAnP(A) + CAn+1P1( A)
=0.
In other words, the sequence (yn)n>osatisfies the recursion (25).
❑
Problem 9: Let n > 0. Show that
O(xn) + 0(xn) = 0(xn), as x —> 0;
o(xn) + o(xn) = o(xn), as x —› 0.
Solution: Let fi (x) = 0(xn) and f2(x) = 0(xn) as x --> 0. Then,
lim sup
fi(x)
xn
< oo and lim sup
x,o
f2(x)
< oo.
xn
(26)
(2 7)
0.1. SOLUTIONS TO CHAPTER 0 EXERCISES
9
It is easy to see that
< lirn sup fl (x) + lim sup f2(x)
Xn
x-03
x--f0
fi(x) + f2(x)
lirn sup
xn
< 00,
and therefore, by definition, fi(x) + f2(x) = 0(e) as x -+ 0.
Let gl (x) = o(xn) and 92(x) = o(xn) as x
0. Then,
. 92 (x)
91(x)0 and lim
lira=
= O.
xn
We note that
lira
x—>0
Mx) + g2(x)
91 x
(
< lim
xn
)
xn
x•—∎0
+ lim
92 ( x )
= 0,
xn
and therefore, by definition, gi(x) + g2(x) = o(xn) as x
O.
❑
Problem 10: Prove that
k2 = 0(n3),
co;
as n
k=1
3
k2 = —
11
3
k=1
0(n2),
as n
co,
i.e., show that
lim sup Ek=l k2 <
n3
n—∎ x
00
and that
lim sup
EZk-1 2 n2
< Do.
Similarly, prove that
n
k3 = 0(n4), as n -+ 00;
k=1
4
n
k3 = —
4 +
k=1
0(n3),
as n co.
Solution: Recall that
E
n
k=1
k2
n(n + 1)(2n + 1)
6
71
and
Ek3
k=1
=
n2 (n + 1 ) 2
4
MATHEMATICAL PRELIMINARIES
10
Then,
lim
n---co
n3
k 2_=
=lira
n—o
K>
Ern ELI k2
n—>cra
n2
v-,rt
iimL-4=1. r‘'
n-“:,0
n4
n(n+ 1)(2n + 1)
6n3
3n2 + n
n—*Do 6n2
Ern Ek=1 k3 1
n3
n—oo
1
- 3
1
2
= — < 00
1
2 (n + 1)2
lirn n
n-,00
47/4
4<
2n + 1
1
= - < oo.
n->oo 4n
2
CXD;
We conclude that
n
k2 = 0(n3), as n -> CO;
k=1
n3
k2 = - + 0(n2),
3
k=1
as n oo;
n
k3 = 0(n4), as n --> oo;
k=1
n4
E k3 = —4 + 0(n3),
k=1
as n -> oo.
❑
oo;
11
0.2. SUPPLEMENTAL EXERCISES
0.2
Supplemental Exercises
1. Let a > 0 be a positive number. Compute
\/a+Va+Va+...
2. Let a > 0 be a positive number. Compute
1
3. (i) Find x > 0 such that
= 2.
(ii) Find the largest possible value of x > 0 with such that there exists
a number b > 0 with
xxr
= b.
Also, what is the largest possible value of b?
0.3
Solutions to Supplemental Exercises
Problem 1: Let a > 0 be a positive number. Compute
\/a+Va+Va+...
Solution: If we know that the limit of Va + -Va +
that limit by 1, then it follows that / must satisfy
/ = 1/a+1a+Va+... =
+ . .. exists, and denote
+1,
(28)
which can be solved for 1 to obtain that
/
1 +-V1 + 4a
2
•
(29)
MATHEMATICAL PRELIMINARIES
12
We now show that, for any a > 0, the limit of \I a+ Va + Va+ ... does
exist, which is equivalent to proving that the sequence (xn)n>ois convergent,
where xo-= -VT2 and
xn+i = -Va + xn, V n > O.
We will how that the sequence (xn) n,0 is bounded from above and is
increasing.
Let 1 be given by (29), i.e., let 1 —
(i) The sequence (x„)„>0 is bounded from above by 1.
Note that xo
< 1. If we assume that xi, <1, then
= Va + xn < Va +1 = 1,
Xn+1
since 1 is the positive solution of (28), and therefore 1 = \/a + 1. Thus, by
induction, we find that xn <1 for all n > 0.
(ii) The sequence (xn)n>0 is increasing.
It is easy to see that
xn< xn+1 < > xn <
2
Xn — X n —
+ xn <
a < 0,
since xn > 0. Note that
2
Xn — X n —
a =
(x7, 1
+ -V1+ 4a) (
xn 1
—
+ 4a)
2
2
= (xn — 1) (xn
V1 + 4a — 1)
2
< 0,
since xn < 1 and xn > 0 for all n > 0. We conclude that xn < xn± i for all
n > 0.
We showed that the sequence (xn)n>0is increasing and bounded from
above. We conclude that the sequence is convergent. Therefore, the limit
/ = limn, xn satisfies the equation 1 = Va +1 and is given by (29), i.e.,
lim xn
n-+co
= 1 + V1+ 4a
2
Problem 2: Let a > 0 be a positive number. Compute
a+
1
a + a+1•.. •
(30)
0.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
13
Solution: If we know that the continuous fraction (30) has a limit 1, then 1
must satisfy
1
(31)
1 = a + < > /2— al — 1 = 0,
and therefore
a + Va2 ±4
(32)
2
To show that the continuous fraction (30) does have a limit, we must
prove that the sequence (x„),>0 is convergent, where xo = a and
/=
Xn+1 = a+ 1 V n O.
xn
,
The first few terms of the sequence (x,)n>o are
xo = a; x1 =
a4 3a2 + 1
a2 + 1
a3 + 2a
; x2 = ,
; x3 =
.
a
a3 + 2a
a' + 1
We note that the terms of the sequence are alternatively larger and smaller
than the value of 1 given by (32), i.e.,
x0 < x2 < 1 < x3 < xi.
Based on this observation, we conjecture that the subsequence (x2n)n>o
made of the even terms of the sequence (x„)„>0is increasing and has limit
equal to 1, and that the subsequence (x2„±1)„>0 made of the odd terms of
(x,),>0 is decreasing and has limit equal to 1.
To show this, let (y„),>0 be the sequence given by the recursion
Yn+i = a +
1
(a2 +
+a
=
, V n 0,
ay,
+
1
a+1
—
(33)
with yo = a. Note that y, = x2, for all n > 0.
Assume that y„ < I, where l is given by (32). Recall from (31) that
/2— a/ — 1 = 0
(34)
and that
t2— at — 1
(t a +Va2 + 4) (
t a — \/a2 +4)
Va2 +4— a )
(35)
14
MATHEMATICAL PRELIMINARIES
We will show that, for all n > 0, yn±i > yr, and yr,±1 < 1.
Note that, by definition (33), yr, > 0 for all n > 0. Then, from (33), it is
easy to see that
Yn+1 > Yn < > (a2+ 1)yn + a > aYn2 +Yn < > a(Yn2— ayn — 1) < 0. (36)
From (35), and using the assumption that yr, < 1, it follows that
Y72.1 —ayn — 1 = (yn — 1 (yn +
\ia2 + 4 — a
)
2
< O.
(37)
From (36) and (37), we conclude that, if yn < 1, then y7,±1 > yn, for any
n > 0.
From (33), we also find that
Yn-Fi < /
/—a
(a2+ 1)Yn + a < a/yn + / < > Yn < 2
a — al +1
1; (38)
the last equality can be derived as follows:
1— a =
1 -< > 1
a2— al +1
a = a21
a/2 + / t > a(12— al — 1) = 0,
where the last equality is the same as (34).
We conclude from (38) that, if yr, < 1, then yn±i < I for all n > 0.
In other words, we showed by induction that the sequence (yn)n>ogiven
by the recursion (33) with yo= a is increasing and bounded from above by
1. Therefore, the sequence (yn)n>ois convergent. Denote by /1 =
y71
the limit of the sequence (yn)n>o. From (33) and using (35), we obtain that
/1 =
(a2 + 1)/ +a
all +11
> a(1? — all— 1) = 0
< > a(/1
1) (11+
A/a2+ 4 — a
2
= 0.
Since /1> 0, it follows that /1 =1, i.e., that limn yn = I.
Recall that yn = x27, for all n > 0. We showed that the subsequence made
of the even terms of (x7,),>0is increasing and converges to the limit 1 given
by (32), i.e., that
lim X2n = 1.
(39)
n—"
Similarly, we define the sequence (zr,),>0by the recursion
Zn+1 =
(a2 + 1)zr, + a
azn + 1
n > 0,
0.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
15
with zo = a2a+1. It is easy to see that zn= x271-1-1 for all n > 0.
As expected, the sequence (zn),>0 is decreasing and has limit equal to 1.
The proof follows by induction: assuming that zn > 1, we show that zn+i < zn
and zn±i > 1. This proof is very similar to that given above for the sequence
(yn)n>o and is left to the reader as an exercise. We conclude that
lim
1.
X2n+1 =
(40)
From (39) and (40), we find that
lim xn = / =
+4
a+
2
n—■
Dc
. CI
Problem 3: (i) Find x > 0 such that
= 2.
(41)
(ii) Find the largest possible value of x > 0 with such that there exists a
number b > 0 with
(42)
xxx = b.
Also, what is the largest possible value of b?
Solution: (i) If there exists x such that (41) holds true, then x2 = 2 and
therefore x = VI We are left with proving that
= 2.
Consider the sequence (xn )n>0 with xo
recursion:
xn±i = \/Zn =
and satisfying the following
V n > 0.
It is easy to see by induction that the sequence is increasing and bounded
from above by 2, since
Xn+1 >
Xn+i <
xn
2< >
> 2x r, /2 > 2x,_1/2 < >
2xn/2 < 2
xn > Xn-1;
< > xn /2 < 1 < > xn < 2.
We conclude that the sequence (xn)n>o is convergent. If 1 =
xn, then
l=21/2
which is equivalent to
11/1
= 21/2.
(43)
MATHEMATICAL PRELIMINARIES
16
Let f : (0, oo) -- (0, oo) be given by
f (t) = tilt = exp (1n(tt))
Then
f'(t) =
1 — ln(t)
t2
exp
(111 (0
t )
Note that the function f(t) is increasing for t < e and decreasing for t > e.
Therefore, there will be two values of t such that (43) is satisfied, i.e., such
that tilt = 21/2, one value being equal to 2, and the other one greater than
e. Since xr, < 2 for all n > 0 and 1 = limn,
xn, we conclude that 1 = 2, and
therefore that x = -V2 is the solution to (41).
(ii) If there exists a number b > 0 such that
xx' = b
for a given x > 0, then xb = b and therefore x = bl/b. Recall from part (i)
that the function f(t) = tilt has an absolute maximum at t = e. We conclude
that
x = bi/b < max t i/t = eve P---1 1.4447,
t>o
and that the largest value of b such that the limit (42) exists is b = e.
❑
Chapter 1
Calculus review. Plain vanilla options.
1.1 Solutions to Chapter 1 Exercises
Problem 1: Compute
f ln(x) dx.
Solution: Using integration by parts, we find that
f
dx = I (x)' ln(x) dx = xln(x) —
(x)
ln
= xln(x) —
Problem 2: Compute
f xln1(s)
f
x(ln(x))' dx
i 1 dx = xln(x) — x + C.
❑
dx by using the substitution u = ln(x).
Solution: Let u = ln(x). Then du = ''s and therefore
f
1
j x ln(x)
1
dx = f u du = lnaul) = Ina ln(x)1) + C.
Problem 3: Show that (tan x)' = 11 (cos x)2 and
f 1 +1 s2 dx =
arctan(x) + C.
Solution: Using the Quotient Rule, we find that
(tan x)' =
=
( sin x '
(sinx)/ cos x — sin x(cos x)/
cos x)
(cos x)2
(cos x)2 + (sin x)2
1
_=
(cos X)2
(
17
cos X)2
El
18
CHAPTER 1. CALCULUS REVIEW. PLAIN VANILLA OPTIONS.
To prove that
f1
dx = arctan(x), we will show that
(arctan(x))' =
1
1 + x2 '
Let f(x) = tan x. Then arctan(x) = f-1(x). Recall that
(f_1(x \\
j)
1
.r(f 1(x))
and note that f(x) = (tan x)' — (cosx)2; cf. (1.1). Therefore,
(arctan(x))' = (cos(f-1(x)))2 = (cos(arctan(x)))2.
(1.2)
Let a = arctan(x). Then tan(a) = x. It is easy to see that
x2
1
+ 1 = (cos(a))2'
since (sin(a))2 (cos(a))2 = 1. Thus,
(cos(arctan(x)))2 = (cos(a))2 =
1
x2 + 1•
(1.3)
From (1.2) and (1.3), we conclude that
(arctan(x))' =
1
x2 +1
and therefore that
f
1
j 1+x2
dx = arctan(x) + C.
We note that the antiderivative of a rational function is often computed
using the substitution x = tan (0 .
For example, to compute f +
i xldx using the substitution x = tan (0,
note that
dz.
dx = d (tan (z)) dz =
2)
dz
2(cos(0)2
Then
1+1x2
I
dx
f
1+ (tan (0)2 2(cos1(0)2
f
J
dz
(cos (0) 2
1
dz
(sin(a))2 + (cos(a))2 2(cos (0)2
f1
z
- dz = - = arctan(x) + C. ❑
2
2
19
1.1. SOLUTIONS TO CHAPTER 1 EXERCISES
Problem 4: Use l'Hopital's rule to show that the following two Taylor approximations hold when x is close to 0:
+x
1 + 2;
X2
1+ x +
ex
2
In other words, show that the following limits exist and are constant:
(1 + x + 4)
+ x - (1 + i)
and lim ex
x->0
x2
lim
X3
Solution: The numerator and denominator of each limit are differentiated
until a finite limit is computed. L'El6pital's rule can then be applied sequentially to obtain the value of the initial limit:
lim
\/1 + X -
(1 + 0
X2
x-40
= lim
x-∎0
1
2
1
2V1d-s
2x
1
= lim
x---,0
4(1+x)3 / 2
2
-
1
8
We conclude that
x
+x =1+ 0(x2),
2+
as
x -> 0.
Similarly,
lim ex - (1+x + 4) =
lim
s-,c)
x3
ex - (1+ x)
3x2
ex
= lim
x-A3 6x
lim— =
=m
x-,13 6
6'
and therefore
ex
-(1+x+
X2)
X2
= 1 + x + — + 0(x 3)
2
e
lim (1 - 1-) x .
x
Solution: Note that
1
x
0. ❑
(1 + is of e to show that
Problem 5: Use the definition e
1
as
1 _ x-1
x
x
1
1
x = 1_,_ 1 •
x-1
' x-i
20
CHAPTER 1. CALCULUS REVIEW. PLAIN VANILLA OPTIONS.
Then,
lira
firm
s-,00
1
lim
x->co (1 j_ 1 •\ x-1 1
1
'-1
11
since
1
1
lim 1 +
=1
x-400
x-1
and
hill (1 +
,
x-,00
x-
= e.
= lira
x-,00
0
Problem 6: Let K, T, a-and r be positive constants, and define the function
g
b( x)
1
2
g(x) =-dy
T7r
where b(x)
+ (r + 1)
e
,
/ (o--M Compute g' (x).
Solution: Recall that
d
dx
fa(x
b(x)
f(Y) dy
)
= f (b(x))bl (x) f (a(x))d (x).
Therefore,
1
_oc.»2
e 2
bf (X) —
exp
(x) =
47r
-/27r
1
exp
xo- V27T
-=
-
(b(x))2
1
2 ) xa--VT
(k) + (r + 1) T) 2
❑
2o-2T
Problem 7: Let f (x) be a continuous function. Show that
1
h -,o 2h
f
a+h
f (x) dx = f (a),
—h
V a E R.
1.1. SOLUTIONS TO CHAPTER 1 EXERCISES
21
Solution: Let F(x) = f f (x) dx be the antiderivative of f (x). From the
Fundamental Theorem of Calculus, it follows that
f a+h
F(a + h) - F(a - h)
f(x) dx 2h ja_h
21/
Using l'Hopital's rule and the fact that P(x) = f (x), we find that
1
urn —
h—+0 2h
a+h
F(a + h) F(a - h)
2h
f (a + h) + f (a - h)
= lim
h—■
0
2
= f (a),
f (x) dx
lim
a—h
since f (x) is a continuous function.
Note: Let F(t) = fctlf (x) dx. The central finite difference approximation of
F'(a) is
F'(a) = F(a + h) - F(a
+ 0 (h2) ,
(1.4)
2h
as h
0 (if F(3) (t) = f"(t) is continuous). Since F'(a) = f (a), formula (1.4)
can be written as
1a+h
f (a) =
2h fa-h
f (x) dx + 0(h2). ❑
Problem 8: Let f : R R given by
n
=E
f(y)
Cie
—yti
,
i=1
where ci and ti, i = 1 : n, are positive constants. Compute f(y) and f"(y).
Solution: Note that
(e-ytt)'
d (e-yt\
)
dy
- tie-Yt';
(e-yt,\"
) = — (-tie-Yt9 =
dy
Then,
1(y) = _E citieYti;
t=i
f"(y ) =
ciqe-Yt'.
❑
22
CHAPTER 1. CALCULUS REVIEW. PLAIN VANILLA OPTIONS.
Note: The function f (y) represents the price of a bond with cash flows ci paid
at time tias a function of the yield y of the bond. When scaled appropriately,
the first and second derivatives of f (y) with respect to y give the duration
and convexity of the bond, respectively.
Problem 9: Let f : R3 R given by f (x) = 2xi — x1x2 + 3x2x3 — x3, where
x = (xi, X2> X3) .
(i) Compute the gradient and Hessian of the function f (x) at the point a =
(1, —1, 0), i.e., compute D f (1, —1, 0) and D2f (1, —1, 0).
(ii) Show that
f(x) = f (a) + D f (a) (x — a) ± —
1
2 (x
—
a)t D2f (a) (x — a).
(1.5)
Here, x, a, and x — a are 3 x 1 column vectors, i.e.,
(
X
=
(
Xi
X2
;
a=
x3
1
—1
0
;
x— a =
( x1— 1
x2 + 1
x3
Note: Formula (1.5) is the quadratic Taylor approximation of f (x) around
the point a. Since f(x) is a second order polynomial, the quadratic Taylor
approximation of f (x) is exact.
Solution: (i) Recall that
D f (x) = ( 1-(x)
Pl.(x))
ax J
\axi
Ox2
(4x1 — x2, — x1 + 3x3, 3x2 — 2x3) ;
-ax?f(x)
D2f (x)
=
821 (X)
axiax2
\ axiax3
82 f (x)
82f (x)
ax2ax,
ax38x,
1(x)
ax2
5
f ()
ax2ax3
ux3,X2
84 x )
(
)
/
4 —1
0
—1
0
3
0
3 —2
Then,
f (a) = f (1, —1, 0) = 3
Df(a) = Df(1, —1,0) = (5, — 1, — 3);
( 4 —1 0
—1 0 3 ) .
D2 f (a) = D2f (1, —1, 0) =
0
3 —2
(1.6)
(1.7)
(1.8)
23
1.1. SOLUTIONS TO CHAPTER 1 EXERCISES
(ii) We substitute the values from (1.6), (1.7) and (1.8) for f (a), D f (a)
and D2f (a), respectively, in the expression f (a) + D f (a) (x - a) + ;12 (x a)t D2f (a) (x - a) and obtain that
f (a) + D f (a) (x - a) +
1
(x - a)t D2f (a) (x - a)
-1
= 3 + (5, -1, -3) x2 + 1
x3
4 -1 0 )
1
+ - (xi - I, x2 + I, x3) -1 0 3
2
0 3 -2
-1
x2 + 1
x3
-= 3 + (5x1 - x2 - 3x3 - 6)
+ (2xT - 5x1 - xix2 + x2 + 3x2x3 + 3x3 -4 + 3)
_=. -x,
9 — x1X2 3x2x3 — x:23
= f (x). ❑
Problem 10: Let
1
u(x, t)
Compute c-'; and
x2
e 4t ,
7rt
for t > 0, x E R.
0, and show that
a2u
au
at
ax2.
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 derivation of the Black-Scholes 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.
Solution: By direct computation and using the Product Rule, we find that
au
1
at
1
au
ax
_
1
3 /2
x2
V47r
r2
- 2tV4irt
1
e
e4t
2t \/4lrt
.r2
e-4t •
1
47rt
2
X
4
A/47a
X2
e-1T + —
4t 2
.
2 (
4
t
e
x2
4t •
.
1 ))
t2
(
(1.9)
24
CHAPTER 1. CALCULUS REVIEW. PLAIN VANILLA OPTIONS.
a2u
axe
=
1
x2
e--47
2Wr
4t
1
4t2 .V47rt 6
4t
(1.10)
From (1.9) and (1.10), we conclude that
au
at
(92u
axe
11
Problem 11: Consider a portfolio with the following positions:
• long one call option with strike K1= 30;
• short two call options with strike K2 = 35;
• long one call option with strike K3 = 40.
All options are on the same underlying asset and have maturity T. Draw
the payoff diagram at maturity of the portfolio, i.e., plot the value of the
portfolio V(T) at maturity as a function of S(T), the price of the underlying
asset at time T.
Note: This is a butterfly spread. A trader takes a long position in a butterfly
spread if the price of the underlying asset at maturity is expected to be in
the K1 < S(T) < K3 range.
Solution: A butterfly spread is an options portfolio made of a long position
in one call option with strike K1 a long position in a call option with strike
K3, and a short position in two calls with strike equal to the average of the
strikes K1and K3, i.e., with strike K2 = K1+21(3; all options have the same
maturity and have the same underlying asset.
The payoff at maturity of a butterfly spread is always nonnegative, and
it is positive if the price of the underlying asset at maturity is between the
strikes K1 and K3, i.e., if K1 < S(T) < K3.
For our particular example, the values of the three call options at maturity
are, respectively,
,
C1(T) = max(S(T) — Kl 0) = max(S(T) — 30, 0);
C2(T) = max(S(T) — K2, 0) = max(S(T) — 35, 0);
C3(T) = max(S(T) — K3, 0) = max(S(T) — 40, 0)
,
and the value of the portfolio at maturity is
V(T) = C1(T) — 2C2(T) + C3(T).
Depending on the values of the spot S(T) of the underlying asset at
maturity, the value V(T) of the portfolio at time T is given below:
1.1. SOLUTIONS TO CHAPTER 1 EXERCISES
CI. (T)
C2 (T)
C3 (T)
V (T)
S(T) < 30 30 < S(T) < 35 35 < S(T) < 40
0
S(T) — 30
S(T) 30
0
0
S(T) — 35
0
0
0
S(T) — 30
0
40 — S(T)
—
25
40 < S(T)
S(T) — 30
S(T) — 35
S(T) — 40
0
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
V1(T) = max(S(T) — 30,0) — max(S(T) — 35,0).
Depending on the value of the spot price S(T), the value of the bull spread
at maturity T is
Vi (T)
S(T) < 30 30 < S(T) < 35 35 < S(T)
0
S(T) — 30
5
The value of the bear spread at maturity T is
V2(T) = max(20 — S(T), 0) — max(15 — S(T), 0),
which can be written in terms of the value of S(T) as
5
15 < S T <20 20 < S T
20 — S T
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. ❑
Problem 13: Which of the following two portfolios would you rather hold:
• Portfolio 1: Long one call option with strike K = X — 5 and long one call
option with strike K = X + 5;
• Portfolio 2: Long two call options with strike K = X?
(All options are on the same asset and have the same maturity.)
Solution: Note that being long Portfolio 1 and short Portfolio 2 is equivalent
to being long a butterfly spread, and therefore will always have positive (or
26
CHAPTER 1. CALCULUS REVIEW. PLAIN VANILLA OPTIONS.
rather nonnegative) payoff at maturity. Therefore, if you are to assume a
position in either one of the portfolios (not to purchase the portfolios), you
are better off owning Portfolio 1, since its payoff at maturity will always be
at least as big as the payoff of Portfolio 2.
More precisely, note that
V(T) = V(T)—V2(T)
= max(S(T) — (X — 5), 0) + max(S(T) — (X + 5), 0)
— 2 max(S(T) — X, 0).
The value of the portfolio at time T is detailed below:
V(T)
S(T) < X — 5
0
X — 5 < S(T) < X S(T) — (X — 5)
X < S(T) < X + 5 (X + 5) — S(T)
0
X + 5 < S(T)
Problem 14: Call options with strikes 100, 120, and 130 on the same underlying asset and with the same maturity are trading for 8, 5, and 3, respectively
(there is no bid—ask spread). Is there an arbitrage opportunity present? If
yes, how can you make a riskless profit?
Solution: For an arbitrage opportunity to be present, there must be a portfolio made of the three options with nonnegative payoff at maturity and with
a negative cost of setting up.
Let K1 = 100 < K2 = 120 < K3 = 130 be the strikes of the options.
Denote by x1, x2, x3the options positions (which can be either negative or
positive) at time 0. Then, at time 0, the portfolio is worth
V(0) = xiC1(0) + x2C2(0) + x3C3(0)
At maturity T, the value of the portfolio will be
V(T) = xiCi.(T) + x2C2(T) + x3C3 (T)
= x1max(S(T) — K1, 0) + x2max(S(T) — K2, 0)
+ X3 max(S(T) — K3, 0),
respectively.
Depending on the value S(T) of the underlying asset at maturity, the
value V(T) of the portfolio is as follows:
1.1. SOLUTIONS TO CHAPTER 1 EXERCISES
S(T) < Ki
K1 < S(T) < K2
27
V(T)
0
Xi S(T) - X1 K1
(xi + x2)S(T) - x1K1 - x2K2
K2 < 8(T) < K3
K3 < S(T)
(x1 + x2 + x3)S(T) - X1 Ki - x2K2 - x3K3
Note that V(T) is nonnegative when S(T) < K2 only if a long position is
taken in the option with strike K1, i.e., if xi> 0. The payoff V(T) decreases
when K2 < S(T) < K3, accounting for the short position in the two call
options with strike K2, and then increases when S(T) > K3.
We conclude that V(T) > 0 for any value of S(T) if and only if x1 > 0,
if the value of the portfolio when S(T) = K3 is nonnegative, i.e., if (xi +
x2)K3 — xiKi — x2K2 > 0, and if xi + x2 + x3 > 0.
Thus, an arbitrage exists if and only if the values C1(0), C2(0), C3(0) are
such that we can find xi, x2, and x3 with the following properties:
xi Ci (0) + x2C2(0) + x3C3(0) < 0;
xi > 0;
(x1 + X2)K3 - X1 Ki - x2K2 > 0;
Xi ± X2 ± X3 > 0.
For C1(0) = 8, C2(0) = 5, C3(0) = 3 and K1 = 100, K2 = 120, K3 = 130,
the problem becomes finding x1 > 0, and x2 and x3such that
8x1 +5x2 + 3x 3 < 0;
30x1 + 10x2 > 0;
xi + x2 + x3 > 0.
(1.11)
(1.12)
(1.13)
(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 xi, x2, and x3satisfying 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 position in the
option with middle strike (note the similarity to butterfly spreads). Then,
choosing x3= —x1— x2 would be optimal; cf. (1.13). The constraints (1.11)
and (1.12) become
5x1 + 2x2 < 0;
3x1 + x2 > 0.
These constraints are satisfied, e.g., for x1
corresponds to x3 = 2.
1 and x2 = —3, which
28
CHAPTER 1. CALCULUS REVIEW. PLAIN VANILLA OPTIONS.
Buying one option with strike 100, selling three options with strike 120,
and buying two options with strike 130 will generate a positive cash flow of
$1, and will result in a portfolio that will not lose money, regardless of the
value of the underlying asset at the maturity of the options. ❑
Problem 15: A stock with spot price 40 pays dividends continuously at a
rate of 3%. The four months at-the-money put and call options on this asset
are trading at $2 and $4, respectively. The risk-free rate is constant and
equal to 5% for all times. Show that the Put-Call parity is not satisfied and
explain how would you take advantage of this arbitrage opportunity.
Solution: The following values are given: S = 40; K = 40; T = 1/3; r = 0.05;
q = 0.03; P = 2; C = 4.
The Put-Call parity is not satisfied, since
P + Se-qT - C = 39.5821 > 39.3389 = Ke'T .
(1.14)
Therefore, a riskless profit can be obtained by "buying low and selling
high", i.e., by selling the portfolio on the left hand side of (1.14) and buying
the portfolio on the right hand side of (1.14) (which is cash only). The riskless
profit at maturity will be the future value at time T of the mispricing from
the Put-Call parity, i.e.,
(39.5821 - 39.3389)erT = 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:
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 buying shares on
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: +$39.5821.
The initial value of the portfolio is zero, since no money were invested:
V(0) = - P(0) - 3(0)e-qT + C(0) + 39.5821 = 0.
Note that by shorting the shares you are responsible for paying the accrued
dividends. Assume that the dividend payments are financed by shorting
1.1. SOLUTIONS TO CHAPTER 1 EXERCISES
29
more shares of the underlying asset and using the cash proceeds to make the
dividend payments. Then, the short position in e-qT shares at time 0 will
become a short position in one sharer at time T.
The value of the portfolio at maturity is
V(T) = - P(T) - S(T) + C(T) + 39.5821erT.
Recall from the proof of the Put-Call parity that
P(T)+S(T)-C(T) = max(K -S(T),0)+ S(T)-max(S(T)-K, 0) = K,
regardless of the value S(T) of the underlying asset at maturity.
Therefore,
V(T)
-(P(T) + S(T) - C(T)) + 39.5821erT
= -K + 39.5821erT = - 40 + 40.2473 = 0.2473.
This value represents the risk-free profit made by exploiting the discrepancy from the Put-Call parity, and is the same as the future value at time T
of the mispricing from the Put-Call parity; cf. (1.15). ❑
Problem 16: The bid and ask prices for a six months European call option
with strike 40 on a non-dividend-paying stock with spot price 42 are $5
and $5.5, respectively. The bid and ask prices for a six months European
put option with strike 40 on the same underlying asset are $2.75 and $3.25,
respectively. Assume that the risk free rate is equal to 0. Is there an arbitrage
opportunity present?
Solution: For r = 0, the Put-Call parity becomes P + S - C = K, which in
this case can be written as C - P = 2.
Thus, an arbitrage occurs if C - P can be "bought" for less than $2 (i.e.,
if a call option is bought and a put option is sold for less than $2), or if C - P
can be "sold" for more than $2 (i.e., if a call option can be sold and a put
option can be bought for more than $2).
From the bid and ask prices, we find 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 $5.5$2.75=$2.75, which is more than $2. Therefore, no risk-free profit can be
achieved this way.
Also, a call can be sold for $5 and a put can be bought for $3.25. Therefore, C - P can be "sold" for $5-$3.25=$1.75;which is less than $2. Again,
no risk-free profit can be achieved.
❑
1This is similar to converting a long position in e 9T shares at time 0 into a long position
in one share at time T. through continuous purchases of (fractions of) shares using the
dividend payments, which is a more intuitive process.
30
CHAPTER 1. CALCULUS REVIEW. PLAIN VANILLA OPTIONS.
Problem 17: You expect that an asset with spot price $35 will trade in
the $40-$45 range in one year. One year at-the-money calls on the asset
can be bought for $4. To act on the expected stock price appreciation, you
decide to either buy the asset, or to buy ATM calls. Which strategy is better,
depending on where the asset price will be in a year?
Solution: For every $1000 invested, the payoff in one year of the first strategy,
i.e., of buying the asset, is
VI (T) = 1000
—
35 S(T),
where 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
V2(T) =
1000
1°°° (S(T) - 35), if S(T) > 35;
4 max(S(T) - 35, 0) = { 4
0,
if S(T) < 35.
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 profitable if the asset appreciates in value, i.e., is S(T) is significantly
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
100
1 000
S(T) =
(S(T) - 35) -‹ > S(T) = 39.5161.
35
4
°
If the price of the asset will, indeed, be in the $40-$45 range in one year,
then buying the call options is the more profitable strategy. ❑
Problem 18: The risk free rate is 8% compounded continuously and the
dividend yield of a stock index is 3%. The index is at 12,000 and the futures
price of a contract deliverable in three months is 12,100. Is there an arbitrage
opportunity, and how do you take advantage of it?
Solution: The arbitrage-free futures price of the futures contract is
12000e'q)T
= 12000e(0.08-0.03)/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.,
e0.08/4 (e-0.03/4 12000) — 12100 = 150.94.
❑
32
CHAPTER 1. CALCULUS REVIEW. PLAIN VANILLA OPTIONS.
1.2
Supplemental Exercises
1. Compute
J xn ln(x) dx.
2. Compute
I
xnex dx.
3. Compute
f
(1n(x))71 dx,
4. Show that
1
(1 + )x < e < (1 + —
1) x+1
5. Let
f (x) =
1
exp
(x
V x > 1.
it)2 )
crA/27
(
20-2 )
Assume that g : R —› R is a continuous function which is uniformly
bounded, i.e., there exists a constant C such that Ig(x)1 < C for all
x E R. Then, show that
00
lim 1 f (x)g(x) dx = g(p).
0\o _co
6. Let
n
Ci
9(y) =
77: (1 +y)
1
Compute g' (y),
7. A derivative security pays a cash amount c if the spot price of the
underlying asset at maturity is between K1and K2, where 0 < K1 < K2)
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?
1.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
33
8. Create a portfolio with the following payoff at time T:
{28(T), if 0 < S(T) < 20;
60 — S(T), if 20 < S(T) < 40;
V(T) =
S(T) — 20, if 40 < S(T),
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.
9. Call options on the same underlying asset and with the same maturity,
with strikes K1 < K2 < K3, are trading for C1, C2 and C3, respectively
(no Bid—Ask spread), with C1 > C2 > C3. Find necessary and sufficient
conditions on the prices C1, C2 and C3 such that no—arbitrage exists
corresponding to a portfolio made of positions in the three options.
10. Denote by Cbid and Cask, and by Pbid and Pask,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 S and paying dividends
continuously at rate q. Assume that the risk—free interest rates are constant equal to r. Find necessary and sufficient no—arbitrage conditions
for Cbid) Cask, -Rid) and Pask.
1.3
Solutions to Supplemental Exercises
Problem 1: Compute
J
eln(x) dx.
Solution: If n # —1, we use integration by parts and find that
f xn ln(x) dx =
xn+i
1ln(x)
n+
xn+1ln(x)
n+1
1 1
. xn+1 , _1
x dx
n+1
xn+1
C.
(n + 1)2 +
For n = —1, we obtain that
ln(x)
J
X
dx = (ln(x))2 + C.
❑
34
CHAPTER 1, CALCULUS REVIEW. PLAIN VANILLA OPTIONS.
Problem 2: Compute
f xnex dx.
Solution: For every integer n > 0, define the function fn(x) as
fn(x) = f xnex dx.
By using integration by parts, it is easy to see that
fn ( x) f xnex dx = xnex - n x72-1ex dx = xnex n fri_i
(x), V n > 1.
Since fo(x) = es, the following general formula can be obtained by induction:
f xnex dx = f n(x) = n! (
171-(-1)"" ex + C, V n > 1.
❑
k=
Problem 3: Compute
f
(1n(x))n dx.
Solution: For every integer n > 0, let
fn(x) = f (1n(x))71 dx.
By using integration by parts, it is easy to see that, for any n > 1,
f
(1n(x))n dx = x(In(x))n - n f (1n(x))n-1dx,
and therefore
fn(x) = x(ln(x))n - nfr,_1(x), V n > 1.
Since fo(x) = x, the following general formula can be obtained by induction:
f (1n(X))n dx = f n(x) =
xE(-1)72-kn!
,
k=0
ICI
n x))k + C, V n > 1.
(
❑
35
1.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
Problem 4: Show that
1 +1
(1± _ly
, V x > 1.
< e < (1 + —
x
x
(1.16)
Solution: Note that (1.16) is equivalent to
1)
1
1
< — , Vx> 1.
< 141+x
x
x+1
Let
1
x
1) ;
f(x) -= — — ln (1 + —
g(x) = ln (1 +
1
x
(1.17)
1
x + 1.
Then,
fl(x)
=
g' (x) =
1
1
=
+
x2 x(x + 1)
1
1
x(x + 1) + (x + 1)2
1
< 0;
x2(x + 1)
1
0
x(x + 1)2 <
We conclude that both f (x) and g(x) are decreasing functions. Since
lim f (x) = lim g(x) = 0,
X -. OC
x—, DC
it follows that f(x) > 0 and g(x) > 0 for all x > 0, and therefore
1
1
1
— > 141+— ) >
V x > 0,
x
x+1'
x
which is what we wanted to show; cf. (1.17).
❑
Problem 5: Let
f (x) =
exp
(
2a-2
)
Assume that g : R R is a continuous function which is uniformly bounded2,
i.e., there exists a constant C such that I g(x)1 < C for all x E R. Then, show
that
lim f f (x)g(x) dx = g(p)•
2 The uniform boundedness condition was chosen for simplicity, and it can be relaxed;
e.g., to functions which have polynomial growth at infinity.
36
CHAPTER 1. CALCULUS REVIEW. PLAIN VANILLA OPTIONS.
Solution: Using the change of variables y = Q , we find that
co
f oa f(x)g(x) dx = af27
1- r 1: g(x) exp
ay)dy.
11
2,7r f
Recall that
1
142 ) dx
20-2
(1.18)
f°
r j_00 - 1. dy = 1,
(1.19)
e
since, e.g., the function Zee- is the probability density function of the
standard normal variable. From (1.18) and (1.19) we obtain that
'
1
g(A) - f f(x)g(x) dx = —
T7r f (g(p,) g(iL + o-y))
.V,
-co
dy.
(1.20)
Our goal is to show that the right hand side of (1.20) goes to 0 as a \ 0.
Since g(x) is a continuous function, it follows that, for any E > 0, there
exists (51(€) > 0 such that
I9(A) - g(x)I < e, V x — µI < .51(E).
(1.21)
Using the fact that the integral (1.19) exists and is finite, we obtain that,
for any € > 0, there exists (52(€) > 0 such that
1f -62(0
f2Tr J_.
1
dy
co
”2
(1.22)
\71271-
dY <
Since Ig(x)1 < C for ralls2x: R, it follows from (1.22) that
c 19(i-t) - XII + aY)i e-'Y2 dy
4,7 L
fc°
+ 11 f Ig(0) - g(A+ ay)1 e4 dy < 2C€.
v 27 82(6)
(1.23)
It is easy to see that, if a < s51(e) , then
IYI
(1/ + ay) - PI = crly1 < (51(6) j2()
81(6), V
[-82(6), 82(6)].
(1.24)
Then, from (1.21) and (1.24) we find that
I9(u) - g(p+ 0'01 < 6, V y E [---62(6), (52(6)17
(1.25)
37
1.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
and therefore
1. [ 62(')
Ig(1.1) — g(it
N/T7r
- i-52(6)
ay)1
dy < 6.
(1.26)
From (1.20), (1.23), and (1.26), it follows that, for any E > 0, there exist
then
(Si( €) > 0 and S2(6) > 0 such that, if a <
1
g(p) — f:f(x)g(x) dx
1.9(0)
goi ay)(
e 2 ay
< (2C + 1)6.
We conclude, by definition, that
lim I f(x)g(x) dx = g(p).
❑
crNO
Problem 6: Let
n
Ci
g(Y) =
(1 +Y)t
Compute g'(y).
Solution:
n
g'(y) =
4
Citi
(1 + y)ti+1'
Problem 7: A derivative security pays a cash amount c if the spot price
of the underlying asset at maturity is between Ki. and K2, where 0 < Ki <
K2, 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?
Solution: The payoff of the derivative security is
V(T) =
{0, if S(T) < Kl ;
c, if Ki < S(T) < K2;
0, if K2 < S(T).
Since V(T) is discontinuous, it cannot be replicated exactly using call options,
whose payoffs are continuous.
38
CHAPTER 1. CALCULUS REVIEW. PLAIN VANILLA OPTIONS.
We approximate the payoff V(T) of the derivative security by the following
payoff
V,(T) =
0,
if S(T) < Ki — c;
c(S(T) — (K1 — e))/e, if Ki — E < S(T) < Ki ;
c,
if K1 < S(T) < K2;
c — c(S(T) — K2)/E, if K2 < S(T) < K2 ± E;
0,
if K2 + e < S(T).
(1.27)
Note that V(T) = VE(T) unless the value S(T) of the underlying asset at
maturity is either between Ki. — c and K1, or between K2 and K2 + E.
The payoff VE(T) can be realized by going long c/c bull spreads with
strikes Ki. — c and K1, and shorting c/c bull spreads with strikes K2 and
K2 + E. In other words, the payoff V(T) of the given derivative security can
be synthesized by taking the following positions:
• long c/c calls with strike Ki. — 6;
• short c/c calls with strike K1;
• short c/c calls with strike K2;
• long c/c calls with strike K2 + E.
It is easy to see that the payoff VE(T) is the same as in (1.27):
Ve(T)
S(T) < Ki — E
0
(S(T) — (Ki.— €))
Ki. - E < S(T) < Ki
Ki < S(T) < K2
- (S(T) - (K1 - 6)) - (S(T) - KO) = C
c— (S(T) — K2)
K2 < S(T) < K2 + c
c — (S(T) — K2) + (S(T)— (K2 + 6)) = 0
K2 ± E < S(T)
Problem 8: Create a portfolio with the following payoff at time T:
V(T) =
{28(T), if 0 < S(T) < 20;
60 — S(T), if 20 < S(T) < 40;
S(T) — 20, if 40 < S(T),
(1.28)
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.
Solution: Using plain vanilla options, cash, and the underlying asset the
payoff V(T) can be replicated in different ways.
1.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
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 < S(T) < 40,
note that
60 — S(T) = 2S(T) + 60 — 3S(T) = 2S(T) — 3(S(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 < S(T), note
that
S(T)-20 = 60—S(T) + 2S(T)-80 = 2S(T) — 3(S(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.,
Vl (T) = 28(T) — 3 max(S(T) — 20, 0) + 2 max(S(T) — 40, 0)
(1.29)
is the same as the payoff from (1.28):
Vi (T)
S(T) < 20
28(T)
20 < S(T) < 40
28(T) — 3(S(T) — 20) = 60 — S(T)
40 < S(T)
60 — S(T) + 2(S(T) — 40) = S(T) — 20
As a consequence of the Put—Call parity, it follows that the payoff V(T)
from (1.28) can also be synthesized using put options. If the asset does not
pay dividends and if interest rates are zero, then, from the Put—Call parity,
it follows that
C = P + S — K.
Denote by C20 and P20 i and by C40 and P40, 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 = 2S — 3C20 + 2C40.
(1.30)
40
CHAPTER 1. CALCULUS REVIEW. PLAIN VANILLA OPTIONS.
To synthesize a short position in three calls with strike 20, note that
—3C20 = — 3P20 — 3S + 60,
(1.31)
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
2C40 = 2P40 + 2S — 80,
(1.32)
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.
Using (1.31) and (1.32), we obtain that the payoff at maturity given by
(1.29) can be replicated using the following portfolio consisting of put options,
cash, and the underlying asset:
V = 2S — 3C20 + 2C40
= 2S — 3P2o — 3S + 60 + 2P40 + 2S — 80
= S — 3P20 + 2P40 — 20.
(1.33)
The positions of the replicating 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 on the asset.
We check that the payoff of this portfolio at maturity, i.e.,
V2(T) = S(T) — 20 — 3 max(20 — S(T), 0) + 2 max(40 — S(T),0)
is the same as the payoff from (1.28):
VI (T)
S(T) — 20 — 3(20 — 8(T)) + 2(40 — S(T)) = 28(T)
S(T) < 20
S(T) — 20 + 2(40 — S(T)) = 60 — S(T)
20 < S(T) < 40
S(T) — 20
40 < S(T)
If the asset pays dividends continuously at rate q and if interest rates are
constant and equal to r, 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 26-0' units of the asset;
1.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
41
• 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-4T 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 plain vanilla options, cash and asset positions. ❑
Problem 9: Call options on the same underlying asset and with the same
maturity, with strikes Ki < K2 < K3, are trading for C1, C2 and C3, respectively (no Bid-Ask spread), with Ci > C2 > C3. Find necessary and
sufficient conditions on the prices C1, C2 and C3 such that no-arbitrage exists corresponding to a portfolio made of positions in the three options.
Solution: An arbitrage exists if and only if a no-cost portfolio can be set up
with non-negative 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 0.
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 K3, for i = 1 : 3. Let S = S(T) be the value of the underlying
asset at maturity. For no-arbitrage to occur, there are three possibilities:
Portfolio 1: Long the K1-option, short the K2-option, long the K3-option.
Arbitrage exists if we can find xi >0, i = 1 : 3, such that
- x2C2 + x3C3 = 0;
(1.34)
xi (S - Ki) - x2(S - K2) + x3(S K3) > 0, V S > O.
(1.35)
We note that (1.35) holds if and only if the following two conditions are
satisfied:
xl - X2 ± X3 > 0;
(1.36)
xi(K3 - - x2(K3 - K2) > 0.
(1.37)
We solve (1.34) for x3and obtain
C2
X3 = X27,- - xl, .
C3
Cs
(1.38)
Since we assumed that x3> 0, the following condition must also be satisfied:
X2
xi
> -.
C2
(1.39)
42
CHAPTER 1. CALCULUS REVIEW. PLAIN VANILLA OPTIONS.
Recall that C1 > C2 > C3. Using the value of x3 from (1.38), it follows
that (1.36) and (1.37) hold true if and only if
X2
- C3
>
X1
(1.40)
C2 - C3
X2
K3 -
Xi
K3 - K2
(1.41)
Also, note that if (1.40) holds true, then (1.39) is satisfied as well, since
C1 -C3
Cl
C2 - C3 > C2
We conclude that arbitrage happens if and only if we can find x1> 0 and
x2 > 0 such that (1.40) and (1.41) are simultaneously satisfied. Therefore,
no—arbitrage exists if and only if
K3 - K1
Cl - C3
(1.42)
K3 - -2 < -2 - -3 '
Portfolio 2:Long the Ki—option, short the K2—option, short the K3—option.
Arbitrage exists if we can find xi > 0, i = 1 : 3, such that
X1C1 - X2C2
xi (S
—
X3 C3 =
0;
K1) — x2(.5 — K2) — x3(S — K3) > 0, V S > 0.
(1.43)
(1.44)
The inequality (1.44) holds if and only if the following two conditions are
satisfied:
Xi - X2 - X3 > 0;
(1.45)
xi (K3— K1) — x2(K3 — K2) > 0. (1.46)
However, (1.43) and (1.45) cannot be simultaneously satisfied. Since C1 >
C2 > C3, it is easy to see that
C
2
C3
Xi = X20- X3- < X2 ± X3.
In other words, no arbitrage can be obtained by being long the option
with strike K1 and short the options with strikes K2 and K3.
Portfolio 3:Long the Ki—option, long the K2—option, short the K3—option.
Arbitrage exists if we can find xi > 0, i = 1 : 3, such that
X1C1
X2C2 - X3 C3 =
0;
(1.47)
43
1.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
(1.48)
xi (S + x 2( S - K2 - x3(S - K3) > 0, V S > O.
The inequality (1.48) holds if and only if
(1.49)
xi + x2 - x3 > 0.
It is easy to see that (1.47) and (1.49) cannot be simultaneously satisfied:
)
X3 = Xi —r.
C3
C2
X2 Z7 > X1 + x2,
3
since C1 > C2 > C3.
Therefore, no arbitrage can be obtained by being long the options with
strikes K1 and K2 and short the option with strike K3.
We conclude that (1.42), i.e.,
K3 — KI
CI — C3
K3 — n2 < C2 — C3
is the only condition required for no-arbitrage.
Problem 10: Denote by Cbid and Cask, and by Pbid and Pass, 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 S and paying dividends continuously
at rate q. Assume that the risk-free interest rates are constant equal to r.
Find necessary and sufficient no-arbitrage conditions for Cbid, Cask, Pbid, and
Pask•
Solution: Recall the Put-Call parity
C - P = Se —qT — 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., Cask — Pbid
were less than the value Se-gT -Ke-rT of the forward contract, i.e., if
Cask — Pbid
< se —qT
Ke —rT
• or if the selling price of a long call short put portfolio, i.e., Cbid — Pask were
greater than the value Se-qT - Ke-rT of the forward contract, i.e., if
> se qT
Ke-rT .
Cbid Pask
We conclude that there is no-arbitrage directly following from the PutCall parity if and only if
Cas k — Pbid <
Se-qT - K e'T < Cbid Pask.
❑
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) = x2— 2y on the
region bounded by the parabola y = (x + 1)2and the line y = 5x — 1.
Solution: We first identify the integration domain D. Note that (x + 1)2 =
5x —1 if and only if x = 1 and x = 2, and that (x + 1)2< 5x — 1 if 1 < x < 2.
Therefore,
D = {(x,y) 1 < x < 2 and (x +1)2< y < 5x — 11.
Then,
I
f2
f 5s-1
f(x y)dxdy
V(s+1)2(x2 —
,
2y)dy) dx
2
=
=f
(x 2 y — Y 2)15
(:+11)2 ) dx
2
x 2(5x —1— (x + 1)2) - ((5x — 1)2— (x +1)4)dx
2
(5x — 1 — (x + 1)2)(x2— (5x — 1
=
= f1
(x + 1)2 )) dx
2
(-X2 ±3x — 2)(-7x) dx = — 4
7. 0
Problem 2: Let f : (0, co) —p R denote the Gamma function, i.e., let
f (a) =
fo
e' dx.
c
45
46
CHAPTER 2. NUMERICAL INTEGRATION. BONDS.
(i) Show that f (a) is well defined for any a > 0, i.e., show that both
foi
xa-1
and
0.0
dx = Hrn f x"-1 e'dx
t\o
e-x dx
lirn
t-,00
f x"-1e-x dx
exist and are finite.
(ii) Prove, using integration by parts, that f (a) = (a - 1) f (a - 1) for any
a > 1. Show that f (1) =1 and conclude that, for any n > 1 positive integer,
f (n) = (n - 1)!.
Solution:
(i) Let a > 0. Intuitively, note that, as x
0, the function x'-le-x is on
the order of xa-1, since limx \o e' = 1. Since
1
1
1
lim (1 - La) =
a t\o
Jim f xa-1dx = lim
t\o t
t\o a
it follows that
i
fo lx'-1 e-x dx = two f
t\o t
e' dx
exists and is finite.
In a similar intuitive way, note that, as x
oo, the function xa-le'
is on the order of e-x, since the exponential function dominates any power
function at infinity. Since
lim f e-x dx = lim(1 - e-t) = 1,
t,c0 1
t-,00
it follows that
xa-1 e-x dx = lim
t-.co
xa-1 e' dx
(2.1)
exists and is finite.
Making 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 definition, we need to prove that, for any e > 0, there exists n(e) > 0
such that
co
xa-1 e-x dx < E, V s > n(e).
(2.2)
1.
2.1. SOLUTIONS TO CHAPTER 2 EXERCISES
Note that there exists N > 0 such that
xa-1 e-x < e-x/2, V x > N,
47
(2.3)
since
lim xa-1 e-x/2 = 0.
x-vx
Also, since lim,oc e-x/2 = 0, it follows that, for any E > 0, there exists
m(E) > 0 such that
2e-m(E)/2 < E.
(2.4)
Choose n(e) = max(m(E, N)). From (2.3) and (2.4) we obtain that
xa-1 e-x < e-x/2, v x> n(e);
(2.5)
2e-n(E)/2 < E.
We can then use (2.5) and (2.6) to show that, for any s > n(E),
t
e-1 e-x dx = limf xa-1 e' dx
r
t-,DC
s
t
< lim f e-x/2 dx = lim (-2e-t/2 + 2e-5/2)
t-+x s
t-+x
= 2e-s/2 < 2e-n(')/2 < E,
which is what we wanted to show; cf. (2.2).
(ii) It is easy to see that
t
/x
f(1) =
e'
dx = lim f e-x dx
lim (-e-t + 1) = 1.
t-,x 0
t-x
o
Assume that a > 1. By integration by parts, we find that
pc
t
e-1 e'dx = lim x'-1 e-x dx
f(a) =
t-,Dc o
ft
lim [( -xa-1 e-x x=ot + (a -1)
xa-2 e-x dx]
t-,pc \
t
= (a -1) lim f xa-2 e-x dx
t-,x 0
= (a - 1) f(a - 1) ,
fo
)
since
lim x a-i e-x = lim x'-1
= 0, for a > 1;
xv
x\„o
lim to-1 e-t = lim
t-,
t*
DC
to-1
4
e,
= 0.
(2.6)
48
CHAPTER 2. NUMERICAL INTEGRATION. BONDS.
For any positive integer n > 1, we find that f (n) = (n— 1)f(n— 1). Since
f (1) = 1, it follows by induction that f (n) = (n — 1)! ❑
Problem 3: Compute an approximate value of fi.3 fi e-Tdx using the Midpoint rule, the Trapezoidal rule, and Simpson's rule. Start with n = 4 intervals, and double the number of intervals until two consecutive approximations
are within 10-6of each other.
Solution: The approximate values of the integral found using the Midpoint,
Trapezoidal, and Simpson's rules can be found in the table below:
No. Intervals Midpoint Rule Trapezoidal Rule Simpson's Rule
4
0.40715731
0.41075744
0.40835735
8
0.40807542
0.40895737
0.40836940
16
0.40829709
0.40851639
0.40837019
32
0.40840674
0.40835199
0.40837024
64
0.40836569
0.40837937
0.40837024
128
0.40836911
0.40837253
256
0.40836996
0.40837082
512
0.40837039
0.40837018
1024
0.40837023
0.40837028
The approximate value of the integral is 0.408370, and is obtained for a
256 intervals partition using the Midpoint rule, for a 512 intervals partition
using the Trapezoidal rule, and for a 16 intervals partition using Simpson's
rule. ❑
Problem 4: Let f : lik. -- R given by gx) = ix+51:2.
(i) Use Midpoint rule with to/ = 10-6to compute an approximation of
1
I
= f
0
l X5/2
f(x) dx = fo
1 + x2 .
(ii) Show that f (4) (x) is not bounded on the interval (0,1).
(iii) Apply Simpson's rule with n = 2k, 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:
2.1. SOLUTIONS TO CHAPTER 2 EXERCISES
49
No. Intervals Midpoint Rule
4
0.17715737
8
0.17867774
16
0.17904866
32
0.17914062
64
0.17916354
128
0.17916926
256
0.17917070
512
0.17917105
(ii) Without computing f(4)(x), note that the denominator 1 + x2 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-312, which is not defined at 0, and is unbounded in
the limit as x \ 0.
(iii) Using Simpson's rule, the following approximate values of the integral
are obtained:
No. Intervals
4
8
16
32
64
128
256
Simpson's Rule
0.179155099725056
0.179169815603871
0.179171055067087
0.179171162051226
0.179171171372681
0.179171172188741
0.179171172260393
The approximate value of the integral is 0.17917117, and is obtained for
a partition of the interval [0, 1] using 64 intervals. ❑
Problem 5: Let K, T, a and r be positive constants. Define the function
g :118—÷R as
1 f b(s) _2.2_
g(x) =
e 2 dy,
-V27 -Dc
where b(x) = (ln (i) + (r + 5) T) / (ail). Compute g'(x).
Solution: Recall that
d
Clt (
f
b(t)
„
f (x) dx) = f (b(t))bi(t).
CHAPTER 2. NUMERICAL INTEGRATION. BONDS.
50
Therefore,
g/(x) =
1,
cbc.»2
b (x) e 2
V271-
(ln (k) +
1
+4
( ) T)2)
.
2o-2T
x27T
o- V
exP
Problem 6: Let h(x) be a continuous function such that
exists. Define g(t) by
❑
f7,3 1xh(x)1dx
g(t) = ft °°(x — t)h(x) dx,
and show that
g"(t) = h(t). ❑
Solution: Recall that, if a(t) and b(t) are differentiable functions and if f (x, t)
is a continuous function such that 4(x, t) exists and is continuous, then
b (t)
d
dt
f a(t)
f
(x, t) dx
f
N
=f
(t)O
at
t) dx
f (b(t), t)I1 (t) — f (a(t),
(t).
A similar result can be derived for improper integrals, i.e.,
co
amf (x, t) dx
dt ( l
= f
af
(x, t) dx — f (a(t),
a (t)
ut
(t) .
(2.7)
For our problem,
a(t) = t and f (x, t) = (x — t)h(x),
(2.8)
where h(x) is continuous. Then, 4(x, t) = -h(x) is continuous. Note that
f(a(t),t) = f(t,t) = (t — t)h(t) = 0.
From (2.7-2.9), we conclude that
g/(t) =
ft c°(x — t)h(x) dx I =
jith(x) dx.
Since
1(
f (x) dx)
— f (a(t)) (t),
(2.9)
51
2.1. SOLUTIONS TO CHAPTER 2 EXERCISES
it follows that
g"(t) = h(t),
which is what we wanted to show.
Problem 7: The continuously compounded 6-month, 12-month, 18-month,
and 24-month zero rates are 5%, 5.25%, 5.35%, and 5.5%, respectively. Find
the price of a two year semiannual coupon bond with coupon rate 5%.
Solution: The value B of the semiannual coupon bond is
• 2
100 e-r(C14)
B= — 100 e-r(0.0.5)0.5 +
2
C
+ (100 + — 100)
2
+ - 100
2
e-r(0.1.5)1.5
e-r(0,2)2,
where C = 0.05, and r(0,0.5) = 0.05, r(0,1) = 0.0525, r(0,1.5) = 0.0535,
r(0, 2) = 0.055.
The data below refers to the pseudocode from Table 2.5 of [2] for computing the bond price given the zero rate curve.
Input: n = 4
t_cash_flow = [0.5 1 1.5 2] ; v_cash_flow = [2.5 2.5 2.5 102.5] .
The discount factors are
disc = [0.97530991 0.94885432 0.92288560 0.89583414],
and the price of the bond is B = 98.940623.
❑
Problem 8: The continuously compounded 6-month, 12-month, 18-month,
and 24-month zero rates are 5%, 5.25%, 5.35%, and 5.5%, respectively. What
is the par yield for a 2-year semiannual coupon bond?
Solution: Par yield is the coupon rate C that makes the value of the bond
equal to its face value. For a 2-year semiannual coupon bond, the par yield
can be found by solving
C
C
100 = — 100 e-r(c).")" + — 100 e-r(°:1) + — 100 e-r0.1:5)1:5
2
2
2
C
+ 1100 + — 100 e-r(°:2)2.
2
Thus,
C = e-r(0.0.5)0.5
2(1 — e-r(°,2)2)
e -r(0.1) e-r(04.5)1.5
e-r(0.2)2'
52
CHAPTER 2. NUMERICAL 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%. ❑
Problem 9: Assume that the continuously compounded instantaneous interest rate curve has the form
r(t) = 0.05 + 0.0051n(1 + t), V t > 0.
(i) Find the corresponding zero rate curve;
(ii) Compute the 6-month, 12-month, 18-month, and 24-month discount factors;
(iii) Find the price of a two year semiannual coupon bond with coupon rate
5%,
Solution:
(i) Recall that the zero rate curve r(0, t) can be obtained from the instantaneous interest rate curve r(t) as follows:
ft
r(0,t) = t
r(r) dr, V t > 0.
Then,
r(0,t) =
f t 0.05 + 0.0051n(1 +r) dr
0
1
- (0.05t + 0.005 ( (1 + t) ln(1 + t) - t )
t
ln(l + t)
= 0.045 + 0.005(1 + t)
t
(ii) The 6-month, 12-month, 18-month, and 24-month discount factors are,
respectively,
disc(1) = e-r(0,0.5)0.5 = 0.97478242;
disc(2) = e-r(0'1) = 0.94939392;
disc(3) = e-r(°'1.5)1.5 = 0.92408277;
disc(4) = e-r(o,2)2 = 0.89899376.
(iii) The price of the two year semiannual coupon bond with 5% coupon rate
is
B -=
0.05
100 e-r(0,0.5)0.5
2
0.05
0.05
100 e-r(0,1) +
100 e-r(C)'1' 5)1'5
2
2
53
2.1. SOLUTIONS TO CHAPTER 2 EXERCISES
+0.05100) e-r(0.2)2
+ (100
2
= 2.5 disc(1) + 2.5 disc(2) + 2.5 disc(3) + 102.5 disc(4)
= 99.267508. ❑
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:
4
5 y) ;
B = E 3 exp (--y) + 103 exp(—2
2
D=
'i 3i
1 (x.
2_,
2)
- y
exp
+
103 exp Hy)) ;
(
9i
1
i=1
i \
exP
+ 103 4exp (- y))
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 = 0.09;
t_cash_flow = [0.5 1 1.5 2 2.5] ; v_cash_flow = [3 3 3 3 103] .
Output: bond price B = 92.983915, bond duration D = 2.352418, and bond
❑
convexity C = 5.736739.
Problem 11: The yield of a 14 months quarterly coupon 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 101.75 at maturity in 14 months. The formulas for the
price, duration, and convexity of the bond in terms of the yield y of the bond
are similar to those from (2.10-2.12). For example, the price of the bond can
be computed as follows:
2
12
5
12
B = 1.75 exp (--y) + 1.75 exp (--y) + 1.75 exp (-8 y)
l
14
+1.75 exp (—L u + 101.75 exp ky) .
12-)
12
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;
t_cash_flow = 2 5 8 11 14
v_cash_flow = [2 2 2 2 102] .
12 12 12 12 12 '
Output: bond price B = 101.704888, bond duration D = 1.118911, and bond
convexity C = 1.285705. ❑
Problem 12: Compute the price, duration and convexity of a two year
semiannual coupon bond with face value 100 and coupon rate 8%, if the zero
rate curve is given by r(0, t) = 0.05 + 0.011n (1 + 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: n = 4; zero rate r(0, t) = 0.05 + 0.011n (1 + 1);
t_cash_flow = [0.5 1 1.5 2] ; v_cash_flow =- [4 4 4 104] .
Discount factors:
disc = [0.97422235 0.94738033 0.91998838 0.89238025].
Output: Bond price B = 104.173911.
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.
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 computing PZ,- and F,-, and showing that
these functions are either always positive or always negative.
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 coupon rate increases,
the duration of the bond decreases. This is due to the fact that the earlier
cash flows equal to the coupon payments become a higher fraction of the
payment made at maturity, which is equal to the face value of the bond plus
2.1. SOLUTIONS TO CHAPTER 2 EXERCISES
55
increases as c increases, where c is one coupon
a coupon payment, i.e.,
❑
payment.
Problem 14: By how much would the price of a ten year zero-coupon bond
change if the yield increases by ten basis points? (One percentage point is
equal to 100 basis points. Thus, 10 basis points is equal to 0.001.)
Solution: The duration of a zero—coupon bond is equal to the maturity of the
bond, i.e., D = T = 10. For small changes Ay in the yield, the percentage
change in the value of a bond can be estimated as follows:
AB
— Ay D = — 0.001 • 10 = — 0.01.
B
We conclude that the price of the bond decreases by 1%.
❑
Problem 15: A five year bond with duration 32 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.,
AB
— Ay D.
For B = 102, D = 3.5 and Ay = —0.005 (since 1% = 100 bp), we find that
— Ay D B = 1.785.
AB
The new value of the bond is
B,„ = B AB = 103.75. I=1
Problem 16: Establish the following relationship between duration and
convexity:
c
D2
ap
ay
Solution: Recall that
D=
1 aB
1 82B
and C =
B ay
B aye
CHAPTER 2. NUMERICAL INTEGRATION. BONDS.
56
Therefore,
8B
— — DB.
(2.13)
ay
Using Product Rule to differentiate (2.13) with respect to y, we find that
82BaD
0B
aD
= ,- B — D 77— = — w i B
1 3 — D(—DB)
ay
aye
0Y
aD
ap
= — B ± B D2 = B (D2— w)
ay
We conclude that
1 32B
8D
.
C = Bat = D2—
ay
❑
57
2.2. SUPPLEMENTAL EXERCISES
2.2
Supplemental Exercises
1. Assume that the continuously compounded instantaneous rate curve
r(t) is given by
0.05
r(t) =
1 + exp(—(1 + t)2) •
(i) Use Simpson's Rule to compute the 1—year and 2—year discount factors with six decimal digits accuracy, and compute the 3—year discount
factor with eight decimal digits accuracy.
(ii) Find the value of a three year yearly coupon bond with coupon rate
5% (and face value 100).
2. Consider a six months plain vanilla European put option with strike 50
on a lognormally distributed underlying asset paying dividends continuously at 2%. Assume that interest rates are constant at 4%.
Use risk—neutral valuation to write the value of the put as an integral
over a finite interval. Find the value of the put option with six decimal
digits accuracy using the Midpoint Rule and using Simpson's Rule.
Also, compute the Black—Scholes value PBS of the put and report the
approximation errors of the numerical integration approximations at
each step.
3. The prices of three call options with strikes 45, 50, and 55, on the
same underlying asset and with the same maturity, are $4, $6, and $9,
respectively. Create a butterfly spread by going long a 45—call and a
55—call, and shorting two 50—calls What are the payoff and the P&L
at maturity of the butterfly spread? When would the butterfly spread
be profitable? Assume, for simplicity, that interest rates are zero.
4. Dollar duration is defined as
D5 =
aB
ay
and measures by how much the value of a bond portfolio changes for a
small parallel shift in the yield curve.
Similarly, dollar convexity is defined as
azi3
C$= ay2 .
CHAPTER 2. NUMERICAL INTEGRATION. BONDS.
58
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 Cs = BC.
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, find 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 immunize your portfolio (i.e.,
to obtain a portfolio with zero dollar duration and dollar convexity)?
2.3
Solutions to Supplemental Exercises
Problem 1: Assume that the continuously compounded instantaneous rate
curve r(t) is given by
0.05
r(t) = 1 + exp(—(1 + t)2)
Use Simpson's Rule to compute the 1—year and 2—year discount factors with
six decimal digits accuracy, and compute the 3—year discount factor with
eight decimal digits accuracy.
(ii) Find the value of a three year yearly coupon bond with coupon rate 5%
(and face value 100).
Solution: (i) Recall that the discount factor corresponding to time t is
t
exp (— f r(r) dr) .
o
Using Simpson's Rule, we obtain that the 1—year, 2—year, and 3—year discount
factors are
disc(1) = 0.956595; disc(2) = 0.910128; disc(3) = 0.86574100.
2.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
59
(ii) The value of the three year yearly coupon bond is
B = 5 disc(1) + 5 disc(2) + 105 disc(3) = 100.236424.
❑
Problem 2: Consider a six months plain vanilla European put option with
strike 50 on a lognormally distributed underlying asset paying dividends continuously at 2%. Assume that interest rates are constant at 4%.
Use risk-neutral valuation to write the value of the put as an integral
over a finite interval. Find the value of the put option with six decimal digits
accuracy using the Midpoint Rule and using Simpson's Rule. Also, compute
the Black-Scholes value PBS of the put and report the approximation errors
of the numerical integration approximations at each step.
Solution: If the underlying asset follows a lognormal distribution, the value
S(T) of the underlying asset at maturity is a lognormal variable given by
0.2
ln(S(T)) = ln(S(0)) + (r - q -—
2)T +
where o is the volatility of the underlying asset. Then, the probability density
function h(y) of S(T) is
2
h(y) =
1
yo- N/27rT
(ln y - ln(S(0)) - - q - 2.0
exp
2o-2T
, (2.14)
if y > 0, and h(y) = 0 if y < O.
Using risk-neutral valuation, we find that the value of the put is given by
e-rT
P
[max(K - S(T), 0)]
= e-rT f (K YAW ClY
(2.15)
where h(y) is given by (2.14).
The Black-Scholes value of the put is PBS = 4.863603. To compute a
numerical approximation 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
approximations to (2.15) and the corresponding approximation errors to the
Black-Scholes value PBS in the table below:
We first note that the approximation error does not go below 6 • 10-6.
This is due to the fact that the Black-Scholes value of the put, which is given
by
111( d2) — Se-q(T-t)N(
PBS = Ke-r(T-t)Ar
60
CHAPTER 2. NUMERICAL INTEGRATION. BONDS.
No. Intervals Midpoint Rule
4
5.075832
8
4.922961
4.878220
16
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 using numerical approximations to estimate the terms N(-dl)
and N(-d2). The approximation error of these approximations is on the
order of 10-7. Using numerical integration, the real value of the put option
is computed, but the error of the Black-Scholes value will propagate to the
approximation errors of the numerical integration.
If we consider that convergence is achieved when the error is less than
10-5, then convergence is achieved for 512 intervals for the Midpoint Rule
and for 32 intervals for Simpson's Rule. This was to be expected given the
quadratic convergence of the Midpoint Rule and the fourth order convergence
of Simpson's Rule. ❑
Problem 3: The prices of three call options with strikes 45, 50, and 55, on
the same underlying asset and with the same maturity, are $4, $6, and $9,
respectively. Create a butterfly spread by going long a 45-call and a 55-call,
and shorting two 50-calls. What are the payoff and the P&L at maturity
of the butterfly spread? When would the butterfly spread be profitable?
Assume, for simplicity, that interest rates are zero.
Solution: The payoff V(T) of the butterfly spread at maturity is
V(T) .=
0,
if S(T) < 45;
S(T) - 45, if 45 < S(T) < 50;
55 - S(T), if 50 < S(T) < 55;
if 55 < S(T).
0,
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) < 45;
S(T) — 46, if 45 < S(T) < 50;
P&L(T) =
54 — S(T), if 50 < S(T) < 55;
—1,
if 55 < S(T).
The butterfly spread will be profitable 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 0, it follows similarly that the butterfly spread is profitable if
45 ± erT < S(T) < 55 — erT
0
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, find 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 immunize your portfolio (i.e., to obtain a portfolio
with zero dollar duration and dollar convexity)?
Solution: Recall that the dollar duration and the dollar convexity of a position
of size B in a bond with duration D and convexity C are
Ds = BD and Cs = BC.
(i) The value, duration and convexity of the two bond positions are
B1= 1, 000, 000; D1 = 3.2; C1 = 16;
B2 = 2,500,000; D2 = 4; C1= 24.
Denote by B = B1 +B2 the value of the bond portfolio. The dollar
duration and dollar convexity of the portfolio are
_aB _
8B1
aB2
ay
ay
ay
= B1 D1+ B2D2 = $13,200,000.
8 2B _ a2 B .
a2 B
Ds =
Cs = aye
=
Bic,
± ay2
+ B2C2 = $76,000,000.
62
CHAPTER 2. NUMERICAL INTEGRATION. BONDS.
(ii) Using dollar duration and dollar convexity, the approximate formula
AB
DAy
1
— C(Ay)2.
2
for the change in the value of a bond can be written as
AB
1
DsAy + 2 Cs(Ay)2.
(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
durations and of the dollar convexities of the bonds making up the portfolio,
respectively.
Using (2.16), we find that the new value of the bond portfolio is
Bnew = B ABP.--1 $3,500,000 — $13, 200 + $38 -= $3,486,838.
(iii) Let B3 and B4 be the value of the positions taken in the bond with
duration D3 = 1.6 and convexity C3 = 12 and in the bond with duration
D4 = 3.2 and convexity C4 = 20, respectively.
If II = B B3 + B4 denotes the value of the new portfolio, then
DWI) = D$(B) + Ds(B3) + Ds(B4)
= $13.2mil + D3B3 + D4-134;
Cs(II) = Ds(B) + Ds(B3) + Ds(B4)
= $76mil + C3B3 + C4B4.
Then, D$(II) = 0 and Cs(II) = 0 if and only if
$13.2mil + 1.6B3 + 3.2B4 = 0;
$76mi1 + 12B3 + 20B4 = 0,
(2.17)
The system (2.17) has solution B3 = $3.25mi1 and B4 = —5.75mi1.
We conclude that, to immunize 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 and convexity 20. ❑
Chapter 3
Probability concepts. Black—Scholes formula.
Greeks and Hedging.
3.1
Solutions to Chapter 3 Exercises
Problem 1: Let k be a positive integer with 2 < k < 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 w(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) x = 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) has probability
of occurring.
Formally, the discrete probability function P : S
[0, 1] is
A
P(x,y) = — V (x, y) E S.
36'
Let k be a fixed positive integer with 2 < k < 12. The value X of your
winning (or losses) is the random variable X : S —> JR given by
X(x,y) =
w(k), if x + y = k;
—1, else.
If 2 < k < 7, then x + y = k if and only if
(x,y) E {(1, k — 1), (2,k — 2),..., (k —1,1)1.
In other words, x + y = k for exactly k — 1 of the total of 36 outcomes from
S. Then,
E[X] =
E
p(x,y)x(x,y)
=
(x.y)ES
1
>
(x,y)ES
63
X(x,y)
64
CHAPTER 3. PROBABILITY. BLACK—SCHOLES FORMULA.
k— 1
36 — (k — 1)
w(k)(k — 1) — 37 + k
( 1) =
w(k) +
(3.1)
36
36
36
The game is worth playing if E[X] > 0. From (3.1), it follows that the
game should be played if
=
w(k) >
37 —
k — 1k '
for 2 < k < 7.
If 8 < k < 12, then x + y = k if and only if
(x,y) e {(6,k — 6), (5, k — 5), ... , (k — 6,6)1.
In other words, x + y = k exactly 13 — k times. Then,
1
E[X] = —
36
E
X (x, y) =
(x,y)ES
13 k
36 — (13 — k)
w(k) +
( 1)
3-6
36
w(k)(13 — k) — 23 — k
=
36
(3.2)
From (3.2), it follows that the game is worth playing if E[X] > 0, i.e., if
w(k) >
23 + k
13 — k'
for 8 < k < 12.
The values of w(k) for k = 2 : 12 are as follows:
w(2) = w(12) = 35; w(3) = w(11) = 17; w(4) = w(10) = 11;
w(5) = w(9) = 8; w(6) = w(8) = 6.2; w(7) = 5. ❑
Problem 2: A coin lands heads with probability p and tails with probability
1 — p. Let X be the number of times you must flip the coin until it lands
heads. What are E[X] and var(X)?
Solution: If the first coin toss is heads (which happens with probability p),
then X = 1. If the first 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 + E[X]) = 1 + (1 — p)E[X].
We conclude that
E[X] = 1—.
P
3.1. SOLUTIONS TO CHAPTER 3 EXERCISES
65
Another way of computing E[X] is as follows: The coin will first land
heads in the k-th toss, which corresponds to X = k, for a coin toss sequence
of T T
T. ..T H, i.e., the first k - 1 tosses are tails, followed by heads once.
This coin toss sequence occurs with probability P(T)k-1P(H) = (1 -Ak-rp.
Then,
oc
E k (1 - p)k-1 p -
E[X]
k=1
E k(1 - p)k.
P
1—p
(3.3)
k=1
Recall that
Ekxk
T(n,l,x) =
X - (n + 1)xn+1+ nxn+2
(3.4)
(1 - x)2
k=1
By letting x = 1- p in (3.4), we find that
+1 n
A n+2
- p (n + 1)(1 - prn(1
ko_ p)k
p2
p2
(3.5)
•
P2
k=1
From (3.3) and (3.5) and since 0 < 1 - p < 1, we conclude that
E[X] =
1
p
lim
slim
k=i
k(1.
p)k
1
P
-p
1
p p2
=
1
Similarly,
E[X 2] =
x
E k2
n
(1 _ p)k-1p = PH. Ek2 (1 - p)k.
1 - p n--∎ x
k=1
k=1
Since
T(n, 2, x) =
E k2 Xk
k=1
(n 1)2xn+1 + (2n2 + 2n - 1)xn+2
_= x + x 2 (1 - x)3
n2xn+3
we find that
n
E[X2] = P lim E k2(1
1-7)11-' x k=1
p)k =
P
lira T(n 2 1 - p)
1 — p n—>oc
p 1 - p + (1 - p) 2 _ 2 - p
p2 '
p3
1-p
"
66
CHAPTER 3. PROBABILITY. BLACK-SCHOLES FORMULA.
Therefore,
1
var(X) = E[X 2] - (EIXD2 = 2pe p p2
1pe p . ❑
Problem 3: Over each of three consecutive time intervals of length 'r = 1/12,
the price of a stock with spot price So = 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[ST] and var(ST).
Solution: The probability space S is the set of all different paths that the
stock could follow three consecutive time intervals, i.e.,
S = {UUU, UU D, U DU, UDD, DUU, DUD, DDU, DDD},
where U represents an "up" move and D represents a "down" move.
The value ST of the stock at time T is a random variable defined on S,
and is given by
ST(DDD) = Sod3;
ST(UUU) = Sou3;
ST (UUD) = ST (U DU) = ST (DUU) = Sou2d;
ST(UDD) = ST(DUD) = ST(DDU) = Soud2.
Note that
P(DDD) =- (1 - p)3;
P(UUU) = p3;
P(UUD) = P(UDU) = P(DUU) = p2(1 - p);
P(UDD) = P(DUD) = P(DDU) = p(1 - p)2.
We conclude that
E[ST] = Sou3.133 + 3Sou2d • p2 (1 — p) + 3Soud2 • p(1 p)2
+ Sod' (1 — 13) 3
= 41.7036;
E[(ST)2] = (Sou3)2 • p3 + 3(Sou2d) 2 • p2 (1 — p) + 3(Soud2 )2 • p(1 — p)2
+ (Sod3 )2 • (1 — p)3 = 1749.0762;
var(ST) = E[(ST)2] (E[ST])2 = 9.8835. ❑
Problem 4: The density function of the exponential random variable X
with parameter a > 0 is
a e-a , if x > 0;
0, if x < 0.
f (x) =
{
3.1. SOLUTIONS TO CHAPTER 3 EXERCISES
67
(i) Show that the function f(x) is indeed a density function. It is clear
that f (x) > 0, for any x E R. Prove that
fx
L
f (x) dx = 1.
.
(ii) Show that the expected value and the variance of the exponential
random variable X are E[X] = a and var(X) -=
(iii) Show that the cumulative density of X is
F(x) -=
{ 1 - e-", if x > 0
0,
otherwise
(iv) Show that
P(X > t) =
ft
f (x) dx = e-at, V t > 0.
Note: this result is used to show that the exponential variable is memoryless,
i.e., P(X >t+s X> t) = P(X > s).
Solution: (i) It is easy to see that
fx
f (x) dx =
e' dx = lim f e -as dx
L.
0
0
x=t
= lim -e-ax I x=0
)
lim (1 - e't) = 1.
t-,0c
(ii) By integration by parts we find that
xe' 1 f
e-'x dx =
+
a
a
x 2 e-ax 2
+
xe-' dx
f xe-ax dx =
x 2e' dx =
x2e-ax 2xe's
a2
a
xe-ax
e-ax
a
a2
2e-'s
a3
Then,
f.
x
f (x) dx = a / xe-"x dx =
E[X] =
= lim
a
xe-ax
a
eack:
lim
0 =
lim f xe-ax dx
t-,x 0
t e -at C
at
1)
a2
CHAPTER 3. PROBABILITY. BLACK-SCHOLES FORMULA.
68
E[X 2]
x2 f(x) dx = a f
x2e-ax dx = a lim f x 2e-ax dx
t_>00
0
x2e-ax
= a lira
li
t-“x)
= a lirn
t->D0
2
2xe-ax
a2
a
t2 e-at
2te-'t
a
a2
2e-ax
a3
2e «t
a3
0
t
0
2
cd)
a2 .
Therefore,
2
var(X) = E[X 2] - (E[X])2 = (7 (iii) If x < 0, then F(x) =
If x > 0, then
a2
f x0,3 f(s) ds = 0.
fo
x a e' ds = (-e-as)Io = 1 - e'.
F(x) = f x f (s) ds =
(iv) If t > 0, then
P(X >t) = 1- P(X < t) = 1 - f f(x) dx
= 1- f a e'rx dx = 1- (-e-as) to
-at.
= e
(3.6)
Recall that the conditional probability of A given B is
P(A1.13) =
P(An B)
p(B) .
(3.7)
Let 8, t > 0. Then, from (3.6) and (3.7), we find that
P(X>t+sIX>t) =
P((X > t + s) n (X > t))
P(X > t)
e-a(t+s)
e -at
P(X > t + .9)
P(X > t)
= e' = P(X > s).
Problem 5: Show that
b
f(x)g(x) dx
fab
(fa f2 (X) dx 1 2 (f g2(x) dx) ,
a
69
3.1. SOLUTIONS TO CHAPTER 3 EXERCISES
for any two continuous functions f, g : R R.
Solution: Let a E R be an arbitrary real number. Note that
b
f
(f 2(x)+ 2a f (x)g(x) + a2g2(x)) dx
b (f (x) + ag(x))2dx
I
=
b
b
b
a2 f g2(x)dx + 2a I f (x)g(x)dx + f f 2(x)dx
a
a
a
> 0, V a E IR.
Recall that a quadratic polynomial P(x) = Axe + 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 B2— 4AC < 0.
For our problem, it follows that
b
b
b
g2(x) dx + 2a f f(x)g(x) dx + I f 2(x) dx
a2 Ia
>_
0, VaER
a
a
if and only if
2
b_
)
b
b
(2 f f ( x ) g ( x ) dx ) — 4 (f f 2(x) dx) (fa g2(x) dx <
a
0,
a
which is equivalent to
fib
b
1
b
1
f(x)g(x) dx < (fa f2(X) dx) (f g2(x) dx)
0
a
Problem 6: Use the Black-Scholes formula to price both a put and a call
option with 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 Black-Scholes formula:
S = 50; K = 45; T — t = 0.5; a = 0.2; q = 0.02; r = 0.06.
The Black-Scholes price of the call is C = 6.508363 and the price of the put
is P = 0.675920. 0
Problem 7: What is the value of a European Put option with strike K = 0?
What is the value of a European Call option with strike K = 0? How do you
hedge a short position in such a call option?
70
CHAPTER 3. PROBABILITY. BLACK-SCHOLES FORMULA.
Solution: A put option with strike 0 will never be exercised, since it would
mean selling the underlying asset for the price K = 0. The price of the put
option is 0.
A call with strike 0 will always be exercised, since it gives the right to
buy one unit of the underlying asset at zero cost. The value of the call at
maturity is V(T) = S(T), and therefore V(0) = e-qTS(0). This can be seen
by building a portfolio with a long position on the call option and a short
position of e-qTshares, or by using risk-neutral pricing:
V(0) = e-rTERiv[s(T)]
= e-rT
e(r-e)Ts(0)
e-4Ts(0).
A short position in the call option is hedged (statically) by buying one
share of the underlying asset.
❑
Problem 8: Use formula p(C) = K(T - t)e-r T-t
Put-Call parity to show that
72
)\ for p(C) and the
p(P) = - K(T - t)e-r(T-t) N(-d2).
Solution: Recall that
p(C) =
ac and p(P) = ap
—
ar
ar
By differentiating the Put-Call parity formula
P
se-q(T-t)
C = Ke-r(T-t)
with respect to r, we find that
p(P) - p(C) = - K(T _ o e-r(T-t).
Therefore,
p(P) = p(C) - K(T - t)e-r(T-t)
= K(T - t)e-r(T-t) N(d2)
K(T - t)e-r(T-t)
it
t
)
N
(1
H
dN
(2_,
2)(:/
= -K(T-t)e-:(T
2 ))
-K(T - t)esince 1 - N(d2) = N(-d2).
❑
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,
71
3.1. SOLUTIONS TO CHAPTER 3 EXERCISES
e.g., for pricing volatility swaps. These two Greeks are called volga and vanna
and are defined as follows:
a(vega(V))
volga(V) =
aa-
a( ega(V))
and vanna(V) =
v
as
.
It is easy to see that
a2 v
volga(V) = 8
2v
&sac.
0
.2 and vanna(V) =
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.,
vanna(V) =
a(ACV))
ao-
.
(i) Compute the volga and vanna for a plain vanilla European call option on
an asset paying dividends continuously at the rate q.
(ii) Use the Put-Call parity to compute the volga and vanna for a plain
vanilla European put option.
Solution: (i) Recall that
vega(C) = Se-q(T-tYT
1 A
e 2
21r
-
t
;
A(C) = e-q(T-t) N(di ),
where
ln (i) + (r - q + I) (T - t)
d1 =
.--
o-VT - t
ln (t
sc') + (r - q)(T - t)
o-VT - t
+
a--VT - t
2
Then,
volga(C) =
vanna(C) =
a(vega(C)) _
ao.
Se-q(T-t) VT
a(A(0)— e_q(T_t) N,(di)ad,
ao-
aa-
t
1 , 4 ad,
al e 2
aoV2 Tr
= e-g(T -t)
-
1
. 0 Tr 6
;
_ <121 ad
awl
.
72
CHAPTER 3. PROBABILITY. BLACK-SCHOLES FORMULA.
Note that
ad,
au
ln (1
4) + (r q)(T t)
+
o-2-VT - t
-t
2
In(x) + (r-q- 2) (T - t)
o-2 VT - t
d2
0
We conclude that
1 A d id2 •
ec
- VTrr
1 e _2
4 d2vanna(C) -= - e-q(T-t)
4 Tr
a
volga(C) = Se-q(T-t)-VT
t
(3.8)
(3.9)
(ii) By differentiating the Put-Call parity P + Se-q(T-t) - C = Ke-r(T-t)
with respect to a, we find that
vega(P) =
oPac
=
= vega(C).
Therefore,
volga(P) =
vanna(P) =
a(vega(P))
a(vega(C))
au
au
a(vega(P))
a(vega(C))
as
as
volga(C);
vanna(C),
where volga(C) and vanna(C) are given by (3.8) and (3.9), respectively.
❑
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 < r, where r is the constant risk free rate. Use the
Put-Call parity, and then use the Black-Scholes formula to prove this result.
Solution: For at-the-money options, i.e., with S = K, the Put-Call parity
can be written as
C - P = Se-q(T-t) - Ke-r(T-t) = Ke-q(T-t) Ke-r(T-t)
= K e-r(T-t) (e(r-q)(T-t)
1)
Therefore, C > P if and only if e(r-q)(T-t)> 1, which is equivalent to r > q.
73
3.1. SOLUTIONS TO CHAPTER 3 EXERCISES
Alternatively, the Black-Scholes formulas for at-the-money options can
be written as
C = Ke-q(T-t) N(di ) - Ke-r(T-t) N(d2);
P = Ke-r(T-t) N(-d2) - Ke-q(T-t)
where
di = Cr q+ -VT
2
t and d2 = (r q a ) VT - t.
2
Then
C > P < > e-q(T-ON(di) - e -r(T-0 N (d2) > e-r(T-oN(_d2) - e-qp--0N(_d1)
e-q(T-t) (N(di) N(-d1)) > e-r(T-t)(N ' '2‘ + N(-d2))
-q(T-t) > e-r(T-t)
e
< > r> q,
since N(d1) + N(-d1) = N(d2) + N(-d2) = 1.
❑
Problem 11: (i) Show that the Theta of a plain vanilla European call option
on a non-dividend-paying asset is always negative.
(ii) For long dated (i.e., with T - t large) ATM calls on an underlying asset
paying dividends continuously at a rate equal to the constant risk-free rate,
i.e., with q = r, show that the Theta may be positive.
Solution: (i) Recall that
SCIe-q(T-t)
O(C) =
gse-q(T-o N (di )
rKe-r(T-o N(d2).
2 N/271-(T - t)
For a non-dividend-paying asset, i.e., for q = 0, we find that
(C) =
Sa
2 -V27(T - t)
- rKe-r(T-t) N(d2) < 0.
(ii) If q = r, the Theta of an ATM call (i.e., with S = K) is
Ko.e-r(T-t) e a?
O(C)
=
2V27(T -
+ rKe-r(T-t) N (di) — rKe-r(T-t) N(d 2)
K e -r(T-i) (r (N(d0
- N(d2))
a
2,07 (T - t)
e
d2
2)
CHAPTER 3. PROBABILITY. BLACK-SCHOLES FORMULA.
74
where
d =
T-t
o-\/T
2
and d2 =
o-VT - t
2
Note that
lim dl = oo and
lim d2 = —0°.
(T—t)—,Do
(T—t)—+co
Then,
lim
(T—t)--K:o
N(di) = 1 and
lim
(T—t)—)oo
N(d2) = 0
and therefore
a
e 2
2 V27r(T - t)
lim (r(N(di) - N(d2))
(T-t)--,00
= 00.
We conclude that, for T - t large enough,
(
- N(d2))
O (C) = Ke-r(T-t) r(N(di)
4)
2- /27(T - t) e
will be positive. We note that the positive value of e(C) is nonetheless small,
since lim(T_t),,, e(C) = 0. ❑
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
a2 c
> 0.
aK2 —
Solution: Recall that
Se-q(T-t)
= Ke-r(T-t) N'(d2).
By differentiating the Black-Scholes formula
C = Se-q(T-t) N(d .j ) - Ke-r(T-t) N(d2)
with respect to K, we obtain that
, ,ad 2
ad,
ac =
e-r(T-t) N(d2)
- Ke-r(T -t)N (a
Se-q(T-t)N/(d1)
2)0K
arc
ad,
e—r(T—t)N(d2)
_ se_q(T_t)
aK
(8K ad2
= -e-r(T-t) N(d2),
(3.10)
75
3.1. SOLUTIONS TO CHAPTER 3 EXERCISES
since d1 =- d2 + aN/T - t and therefore
ad1
aK
ad,
aK.
By differentiating (3.10) with respect to K, we find that
a2c
e-r(T-t)
aK
NI(d2 )
e-r(T-t)
aK
1
ad,
aK.
V271- e 2
(3.11)
Note that
-q -
ln(K) +
d2
(T - t)
o-VT - t
ln(S) + (r - q - 1)(T - t)
ln(K)
a-VT - t
Then
o- VT - t
ad,1
aK
d.x-vT —t'
(3.12)
and, from (3.11) and (3.12), we conclude that
a2c
8K2
1
o-K V27(T - t)
e-r(T-t) e2> 0. 0
Problem 13: Compute the Gamma of ATM call options with maturities of
fifteen days, three months, and one year, respectively, on a non-dividendpaying 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
ATM options with different maturities?
Solution: The input in the Black-Scholes formula for the Gamma of the call
is S = K = 50, o = 0.3, r = 0.05, q = 0. 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:
F(15days) = 0.057664; F(3months) = 0.052530; F(lyear) = 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 at-the-money point
is steeper for shorter maturities. The cost of Delta-hedging ATM options
76
CHAPTER 3. PROBABILITY. BLACK-SCHOLES FORMULA.
may be higher for short dated options, since small changes in the price of
the underlying asset lead to higher changes in the Delta of the option, and
therefore may require more often hedge rebalancing. ❑
Problem 14: (i) The vega of a plain vanilla European call or put is positive,
since
1 _A
e 2
vega(C) = vega(P) = Se-q(T-t) T t
(3.13)
V-27
Can you give a financial explanation for this?
(ii) Compute the vega of ATM Call options with maturities of fifteen days,
three months, and one year, respectively, on a non-dividend-paying underlying asset with spot price 50 and volatility 30%. For simplicity, assume zero
interest rates, i.e., r = 0.
(iii) If r = q = 0, the vega of ATM call and put options is
vega(C) = vega(P) = S VT
where d1 =
T - t, i.e.,
2
t
1
e
_5I.1
2 ,
2
Compute the dependence of vega(C) on time to maturity
a(vega(C))
8(T
— t)
and explain the results from part (ii) of the problem.
Solution: (i) The fact the vega of a plain 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, depending on
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 on 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
"long volatility" .
(ii) The input in the Black-Scholes formula for the Gamma of the call is
S = K = 50, a = 0.3, r = q =- 0. For T = 1/24, T 1/4, and T = 1, the
following values of the vega of the ATM call are obtained:
vega(15days) = 4.069779;
vega(3months) = 9.945546;
vega(lyear) = 19.723967.
77
3.1. SOLUTIONS TO CHAPTER 3 EXERCISES
(iii) For clarity, let T = T - t. For r = q = 0, we obtain from (3.13) that
vega(C) -=
SVFSvIF
e2 =
27r
\/27r
e 8
,
since, for an ATM option with r = q = 0,
0.,\F-r
ln (i) + (7- - q + 1-2
,_ ) T
d1 =
=
a,\FT
2
By direct computation, we find that
a (vega(C)) =
ar
_,2,
S
2.\/271-7-
=
-
e 8
a2S0-- _ a 2,
8\/27r
e 8
S
( a2 T
1 - - - ) e- e
4
2-V2irr
02 T
For a = 0.3 and for time to maturity less than one year, i.e., for T < 1,
we find that
2
1and therefore
T>
4
0.9775,
a(vega(C))
> O.
—
aT
We conclude that, for options with moderately large time to maturity, the
vega is increasing as time to maturity increases. Therefore we expect that
vega(lyear) > vega(3months) > vega(15days),
which is what we previously obtained by direct computation.
❑
Problem 15: Assume that interest rates are constant and equal to r. Show
that, unless the price C of a call option with strike K and maturity T on a
non-dividend paying asset with spot price S satisfies the inequality
S -qT - Ke-rT < C < Se-qT ,
(3.14)
arbitrage opportunities arise.
Show that the value P of the corresponding put option must satisfy the
following no-arbitrage condition:
K e-rT Se-qT < P < Ke-rT
(3.15)
78
CHAPTER 3. PROBABILITY. BLACK-SCHOLES FORMULA.
Solution: One way to prove these bounds on the prices of European options
is by using the Put-Call parity, i.e, P + Se-qT - C = Ke-rT.
To establish the bounds (3.14) on the price of the call, note that
C = Se-qT - Ke-rT + P.
(3.16)
The payoff of the put at time T is max(K - S(T), 0) which is less than the
strike K. The value P of the put at time 0 cannot be more than Ke-rT, the
present value at time 0 of K at time T. Also, the value P of the put option
must be greater than 0. Thus,
0 < P < Ke-rT ,
(3.17)
and, from (3.16) and (3.17), we obtain that
se-qT
Ke-rT < se-qT Ke-rT
p
c < se-qT
To establish the bounds (3.15) on the price of the put, note that
P = Ke'T - Se-qT + C.
(3.18)
The payoff of the call at time T is max(S(T) - K, 0) which is less than S(T).
The value C of the call at time 0 cannot be more than Se-qT , 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 0. Thus,
0 < C < Se-qT .
(3.19)
From (3.18) and (3.19) it follows that
Ke-rT se-qT <
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(0) = 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) < K,
79
3.1. SOLUTIONS TO CHAPTER 3 EXERCISES
since, if S(T) > K, then V(T) = S(T)- (S(T)- K) = K, and, if S(T) < K,
then V(T) = S(T) < K.
From the Generalized Law of One Price we conclude that
V(0) = Se-qT — C < Ke-''T
KerT < C, which is the left inequality from (3.14).
,
and therefore Se-qT
All the other inequalities can be proved similarly:
• To establish that C < Se-qT , 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 short position in the call option is nonnegative for any possible values
of S(T);
• To establish that Ke-''T - Se-qT < 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 < Ke-''T , show that the payoff at maturity T of a
portfolio made of a short position in the put option and a long cash position
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 A(C) = 0.4
and F(C) = 0.05, and a put on the same asset, with A(P) = -0.5 and
F(P) = 0.07 are currently traded. How do you make the portfolio Deltaneutral and Gamma-neutral?
Solution: Take positions of size x1and x2, respectively, in the call and put
options specified above. The value H of the new portfolio is H = V + x1C +
x2P, where V is the value of the original portfolio. This portfolio will be
Delta- and Gamma-neutral, provided that xi and x2 are chosen such that
A(1-1) = A(V) + x10(C) + x2A(P) = 5000 + 0.4x1 - 0.5x2 = 0;
r(n) = F(V) + xiF(C) + x2P(P) = - 200 + 0.05x1+ 0.07x2 = 0.
The solution of this linear system is
250, 000
000
= -4716.98 and x2 = 330'
= 6226.42.
xi =
53
53
To make the initial portfolio as close to Delta- and Gamma-neutral 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
0(1-1) = A(V)
4717A(C) + 6226A(P) = 0.2;
r(n) = F(V) + 4717F(C) + 6226P(P) = - 0.03.
80
CHAPTER 3. PROBABILITY. BLACK-SCHOLES FORMULA.
To understand how well balanced the hedged portfolio H is, recall that the
initial portfolio had A(V) = 5000 and F(V) = -200.
❑
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 cr = 0.2, and that the spot price of the
asset is 92. The risk-free rate is r = 0.05. What Delta-hedging position do
you need to take?
Solution: A long call position is Delta-hedged by a short position in the
underlying asset. Delta-hedging the long position in 1000 calls is done by
shorting
1000A(C) = 1000e-qTN(d1) = 653.50
units of the underlying asset, where
ln
dl =
+ (r - q +1) T
o-N IT
with S = 92, K -= 90, T = 1/4, -= 0.2, r = 0.05, q = O.
Note that, for Delta-hedging purposes, it is not necessary to know the
driftµ of the underlying asset, since A(C) does not depend on pt. ❑
Problem 18: You buy 1000 six months ATM Call options on a nondividend-paying 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 Delta-hedging position do you have to take?
(iii) On the next trading day, the asset opens at 98. What is the value of
your position (the option and shares position)?
(iv) Had you not Delta-hedged, how much would you have lost due to the
increase in the price of the asset?
Solution: (i) Using the Black-Scholes formula with input S1 = K = 100,
T = 1/2, a- = 0.3, r = 0.05, q = 0, we find that the value of one call option
is C1 = 9.634870. Therefore, $9,634.87 must be paid for 1000 calls.
(ii) The Delta-hedging position for long 1000 calls is short
10000(C) = 1000e-4TN(d1) = 588.59
units of the underlying. Therefore, 589 units of the underlying must be
shorted.
3.1. SOLUTIONS TO CHAPTER 3 EXERCISES
81
(iii) The new spot price and maturity of the option are S2 = 98 and T2 =
125/252 (there are 252 trading days in one year). The value of the call option
is $8.453134 and the value of the portfolio is
1000C2 — 589S2 = — 49268.87.
(iv) If the long call position is not Delta-hedged, the loss incurred due to the
decrease in the spot price of the underlying asset is
1000(C2 — C1) = — $1181.74.
For the Delta-hedged portfolio, the loss incurred is
(1000C2 — 589S2) — (1000C1 — 589S1) = — $3.74.
As expected, this loss is much smaller than the loss incurred if the options
positions is not hedged ("naked").
0
CHAPTER 3. PROBABILITY. BLACK-SCHOLES FORMULA.
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?
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?
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 meanµ and
standard deviation a > 0. Compute E[ IX( ] and E[X2].
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
e-')k
, V k > 0,
k!
where A > 0 is a fixed positive number.
P(X -= k) —
6. Show that the values of a plain vanilla put option and of a plain 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
price.
7. You hold a portfolio made of a long position in 1000 put options with
strike price 25 and maturity of six months, on a non-dividend-paying
stock with lognormal distribution with volatility 30%, a long position
in 400 shares of the same stock, which has spot price $20, and $10,000
in cash. Assume that the risk-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. What
is new value of your portfolio, and how do you adjust the stock position
to make the portfolio Delta-neutral?
3.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
83
8. You hold a portfolio with O(II) = 300, F(II) = 100, and vega(II) = 89.
You can trade in the underlying asset, in a call option with
A(C) = 0.2; F(C) = 0.1; vega(C) = 0.1,
and in a put option with
A(P) = —0.8;
F(P) = 0.3; vega(P) = 0.2.
What trades do you make to obtain a A—,
folio?
3.3
r-, and vega—neutral port-
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?
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 first toss is heads, and if the second toss is also heads, which happens with probability p2, 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(1 — 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)(1 + E[X]) + 2p2+ p(1 — p)(2 + E[X]).
(3.20)
We solve (3.20) for E[X) and obtain that
E[X] =
1+p
p2
For an unbiased coin, i.e., for p = .-, we find that E[X] -= 6, and therefore
the expected number of coin tosses to obtain two heads in a row is 6. ❑
84
CHAPTER 3. PROBABILITY. BLACK-SCHOLES FORMULA.
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?
Solution: The probability that the first n throws are all heads is pm. If the
first k throws are heads and the (k + 1)-th throw is tails, which happens
with probability pk(1 -p), then the process resets after the k + 1 steps; here,
k = 0 : (n - 1). Then, if x(n) denotes the expected number of tosses in order
to get n heads in a row, it follows that
n-1
x(n) = npn + Epk(1-p)(k+ 1 + x(n))
k=0
n-1
= 71pn + (1 —
n-1
n-1
p) (Epk +Ekpk) + x(n)(1 - p) Epk.
k=1
k=0
k=0
Recall that
n-1
n-1
pn . Ekpk = p _ npn ± (n _ 472+1
(1 _ 0
1- p '
k=1—
Ep
k=0
k=1
Then, we find that
n
1 - (n + 1)pn ± Thpn+1
+ x(n)(1 - pn)
x(n) = np +
1-p
1 - pn
+ x(n)(1 - In),
=1 and therefore
x(n) =
1 - pn
IP (1- p).
We conclude that 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 is 1-1P
P"(1-P)'
If the coin were unbiased, i.e., for p = -., the expected number of tosses in
order to get n heads in a row is 2n+1— 2. ❑
Problem 3: Calculate the mean and variance of the uniform distribution on
the interval [a,b].
Solution: The probability density function of the uniform distribution U on
t -7,,, for all x E [a, b]. Then,
the interval [a, b] is the constant function f (x) = 4
b
E[U] = i x f(x) dx =
a
1
b
- f x dx =
b-a a
b+a
•
2 '
-
3.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
2
b+
2 a)1
var(U) = E [(U - E[U])2] = E [(U
fb
1
2
b
b - a Ja
(b -a)2
12 •
1
1
b-a 3
dx
2 )
85
x=b
b+aV
2 )
x=a
❑
Problem 4: Let X be a normally distributed random variable with mean p,
and standard deviation a > 0. Compute E[ IXI] and E[X2].
Solution: We compute E[ IXI ] in terms of the cumulative distribution
2
N(t) =
1 ft
dx
of the standard normal variable Z.
Note that X = p + aZ. Then,
E[ IX 1] =
f
1
.47
7r _„
± azi e-4 dz
foc.
1-121'
12
(,u + az)
2 dz +
- .\ /7r
1
f -to, ,
— IVcr
1
.N/Tir
e- 2 dz -
2
1
e 2 dz
+ 277 _Iv,
a
27r
J_Dc
a
.477
2
(ft + az) e-2dz
2
ze-T dz
oc
w
o. ze 2 dz.
It is easy to see that
1
-A/a
e-T dz = N
27r
i
2
ze-7 dz
1
CC
2
e-T dz
57r
f-A/Q
_2. ,
ze 2 aZ
a
= 1 - N (11 ;
o-
z=-A/0z—oc
- exp
;
1 f A/a _Z
e 2 dy = N (1 ,
a
-V27
Z DC
= exp
2a2) •
86
CHAPTER 3. PROBABILITY. BLACK-SCHOLES FORMULA.
We conclude that
E[ IX! ]= - ii (1 - N (L)) +
+ µN
(Q)
oexp ( 2:
2)
vr27
21
/ 2)
+ va27r exp (- 4
2
ii (2N (L-1) — 1) ± \/±
2 exp (
a
27r
A )
20.2 /
One way to compute E[X 2] would be to compute the following integral:
E[X2] =
E[(µ + 0-Z)2]
=
1
\/27r
fCC (it + az)2 e-2T dz.
_cc,
While this would provide the correct result, an easier way is to recall that
var(X) = E[X2] - (E[X])2 .
Since E[X] = u and var(X) = cr2, we conclude that
E[X2] = var(X) + (E[X])2 = /12 + 0.2 .
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
CA Ak
P(X = k) = kA , V k > 0,
where A > 0 is a fixed positive number.
Solution: We show that E[X] = A and var(X) -= A.
By definition,
cc e-AAk
C°Ak -1
E P(X = k) • k = E k! k = e-AAE (k - 1)r
co
E[X] =
k=0
k=1
k=1
Recall that the Taylor expansion of the function ex is
cc k
et = V" t
L-•
k! .
k=0
(3.21)
87
3.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
Then, it follows that
Ak-1
(k - 1)!
k=1
DC
Ak-2
E
(k - 2)!
=e•
(3.22)
= eA .
(3.23)
From (3.21) and (3.22), we find that E[X] = A.
Similarly,
E[X 2] =
E P(X = k) k 2 = E k! k2 =
k=1
k=0
e
k=1
(k -1)Ak
+
(k - 1)!
= e —AA2
k=2
A k-2
(k - 2)!
pc
CA
E (kkAk
- 1)!
k=1
xle
(k - 1)!
k=1
x
eAA
k=1
k-1
(k - 1)!
A2 ±
where (3.22) and (3.23) were used for the last equality.
We conclude that
var(X) = E[X 2] - (E[X])2 = A. ❑
Problem 6: Show that the values of a plain vanilla put option and of a
plain 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
price.
Solution: Recall that the forward price is F = Se-(r-q)T .
From the Put-Call parity, we know that
C - P = Se-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 Se-qT = Ke-rT. Thus,
K = Se-(r-q)T,
i.e., the strike of the options is equal to the forward price.
❑
88
CHAPTER 3. PROBABILITY. BLACK—SCHOLES FORMULA.
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 non—dividend—paying
stock with lognormal distribution with volatility 30%, a long position in 400
shares of the same stock, which has spot price $20, and $10,000 in cash.
Assume that the risk-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 Delta—neutral?
(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 Delta—neutral?
Solution: (i) The value of the portfolio is
1000P(0) + 4008(0) + 10000 = 22927,
where S(0) = 20 is the spot price of the underlying asset and the value
P(0) = 4.9273 of the put option is obtained using the Black—Scholes formula.
(ii) The Delta of the put option position is —1000N(—d1) = —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
—803 + 400 =- — 403.
To obtain a Delta—neutral portfolio, 403 shares must be purchased for $8,060.
The Delta—neutral 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 underlying asset is S (12) = 24 and
the put options have five months left until maturity. The Black—Scholes value
of the put option is P (T.1 ) = 2.1818. The cash position has accrued interest
and its current value is 1940 expoio2(4 ) = 1946. The portfolio is worth
1000P
1
2
+ 8033 (
1\
) + 1946 = 23400.
12
The new Delta of the portfolio is
—1000N(—d1) + 803 = 292.
To make the portfolio Delta—neutral, you should sell 292 shares.
❑
Problem 8: You hold a portfolio with A(1-1) .= 300, r(11) = 100 and
vega(H) = 89. You can trade in the underlying asset, in a call option with
A(C) = 0.2;
r(c) = 0.1; vega(C) = 0.1,
3.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
89
and in a put option with
A(P) -= -0.8;
F(P) = 0.3; vega(P) = 0.2.
What trades do you make to obtain a A-, F-, and vega-neutral portfolio?
Solution: You can make the portfolio F— and vega- neutral by taking positions in the call and put option, respectively. By trading in the underlying
asset, the F and vega of the portfolio would not change, and the portfolio
can be made A-neutral.
Formally, let x1, x2, and x3 be the positions in the underlying asset, the
call option, and the put option, respectively. The value of the new portfolio
is nnew H + xi S+ x2C + x 3P and therefore
A(IInetv) = A(H) + x1 + x2A(0) + x3A(P);
r(IInew) = r(11) + x2F(C) + x3F(P);
vega(IIne„) = vega(11) + x2vega(C) + x3vega(P).
Then, A(I-Inew) = F(II„,) = vega(Ffne„) = 0 if and only if
xi+ 0.2x2 - 0.8x3 = -300;
0.1x2+0.3x3
= -100;
0.1x2 + 0.2x3
= -89.
{
The solution (rounded to the nearest integer) is x1 = -254, x2 = -670,
x3= -110. In other words, to make the portfolio A-, F— and vega- neutral,
one must short 254 units of the underlying asset and sell 670 call options and
110 put options.
❑
Chapter 4
Lognormal random variables. Risk—neutral
pricing.
4.1
Solutions to Chapter 4 Exercises
Problem 1: Let X1 = Z and X2 = — Z be two independent random variables, where Z is the standard normal variable. Show that X1 + X2 is a
normal variable of mean 0 and variance 2, i.e., X1+ X2 = N/2Z
Solution: Recall that if X1 and X2 are independent normal random variables
with mean and variance pi and a?, and t2 and 4, respectively, then X1+ X2
is a normal variable with mean pi + p2 and variance a +
and
4,
X1 + X2 = 01 + ,u2 +
\/a? +
Z.
For X1 = Z and X2 = — Z , it follows that pi = 112 = 0 and al = a2 = 1. We
conclude that
E[X] = pi + p2 = 0 and var(X) = + =2,
and therefore
X = X1 + X2 = N/2Z.
❑
Problem 2: Assume that the normal random variables X1, X2, • • • , Xn
of mean it and variance a2 are uncorrelated, i.e, cov(Xi, = 0, for all
1 1 Xiis the average of the variables Xi, i = 1 : n, show that
,
E[Sn] = i
and var(Sn) =
Solution: Recall that, for ci E R,
11
E
E cixi] = E ci E[X j];
i=i
[
i=1
91
a2
92
CHAPTER 4. LOGNORMAL VARIABLES. RN PRICING.
E
E cicicov(xi, Xi).
var E ciXi) =
c,2var(Xi) + 2
(
i=1 i=1
1 1 and d < 1.
Show that
ERN [S(t + St)] = (pu + (1 — p)d) S(t);
ERN [S2(t St)] = (pu2 + (1 — p)d2) S2(t).
(4.1)
(4.2)
Solution: We can regard S(t + St) as a random variable over the probability
space {U, D} of the possible moves of the price of the asset from time t to time
t + St endowed with the risk—neutral probability function P : {U, D}
[0, 1]
with P(U) = p and P(D) = 1 — p. Then S(t St) is given by
S(t + St)(U) = S(t)u; S(t + St)(D) = S(t)d.
Then, by definition,
ERN[S(t + St)] = P(U) • S(t + St)(U) + P(D) • S(t + St)(D)
(pu + (1 — p)d) S(t);
ERN[S2(t + St)] =. P(U) • (S(t + St)(U))2+ P(D) • (S(t + St)(D))2
(pu2+ (1 — Ad2) S2 (t).
❑
4.1. SOLUTIONS TO CHAPTER 4 EXERCISES
93
Problem 4: If the price S(t) of a non—dividend paying asset has lognormal
distribution with drift r and volatility o, show that
ERN[S(t + St)] = er8tS(t);
E RN[S2 (t Si)] = e(2rd-o2)(St s2 (t)
(4.3)
(4.4)
Solution 1: If the price S(t) of the non—dividend paying asset has lognormal
distribution with drift r and volatility o, then s(strtts)t)is a lognormal variable
given by
(r
(S(t + St)\
In
a 2 ) St + affiZ.
S(t) )
—
Recall that, if ln(Y) = ft + az is a lognormal random variable with
parameters it and a, the expected value and variance of Y are
E[Y] = exp + -T) ;
var(Y) = exp (2/..t + 6:2) (e(12— 1) .
If Y = s(Vt) , then µ = —
E
var
IS(t + St)]
a- = o-Agi and therefore
exp (r
L s(t)
(S(t + (5t)
S(t)
St and
=
a2 ) St + (O-1(5)2
= er6t ;
(4.5)
2
2
0.2
exp (2 (r — -y) St + (o-18i)2) (e('18i)2— 1)(4.6)
)
e2rot (ea2f5t
1)
(4.7)
S t) E[S(t + St)]
i
(4.8)
1
var (S(t + St)) .
S2(t)
(4.9)
Note that,
rS(t + St)]
E
S(t) j
(S(t + St)
var
S(t) )
From (4.5) and (4.8), and from (4.7) and (4.9), respectively, we conclude that
E[S(t + St)] = er5t S(t);
var (8(1 + St)) = ezrst (eu2st — 1) S 2(t).
(4.10)
CHAPTER 4. LOGNORMAL VARIABLES. RN PRICING.
94
Note that
var (S(t + (5t)) = E[S2(t + St)] - (E[S(t + 8t)])2
= E[S2(t + St)] -
e2r6ts2(t).
(4.11)
From (4.10) and (4.11) it follows that
E[S2(t + St)] = var (S(t + St))
e 2r6ts2(t)
= e2rbt+o-26ts2(t),
which is what we wanted to show.
Solution 2: Note that S(t + St) can be written as a function of the standard
normal variable Z as follows:
St) = S(t) exp ((r
- 2- St + o--VKZ) .
S(t
0-2)
Then,
E[S(t + (5t)] =
1
fw S(t) exp
-
= S(t) f27
1. r r
oo exp (rSt
rot 1
S(t)e
p.
x)
dx
a2 St
2 + (nick x - X;) dx
(x - 0-0i)2)
dx
2
J_.exP
S(t)er6t 1 f
27r J-00
St +
dy
S(t)er6t ,
where we used the substitution y = x - affi and the fact that
W
1
e _4 dy =
71-1_,„
[27
since ,-/-1
rr e-2Y1is the density function of the standard normal variable.
Similarly, we obtain that
E[S2(t + (St)] =
=
°- ) St +2cr■
fift x) e-4 dx
1 fcc S2(t) exp (2 (r - -y
r -00
1
s2 (t) exp (2rbt - o-28t +2o- f6t x - x2)
r
dx
x
95
4.1. SOLUTIONS TO CHAPTER 4 EXERCISES
= S2 (t)
(x - 2a--ai)2)
dx
2
f
1
xexp2rbt
( + o-28t
..VTr
2 _.
= s2 (t)e(2r+a2)St
= S2(t)e(2r+a2)"
1 fp. exp
g-J-Dc
( (x
- 2o-Vrt)2 )
2
dx
f
1
pc e-4 ds
-■i.r -x ,
= S2(t)e(2r+a2)";
❑
the substitution s = x - 2o-N/Tt was used above.
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
finding the up and down factors u and d, and the risk-neutral probability p
(of going up) such that the values of E RN[S (t + 80] and ERN[S2(t ± St)] 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
pu + (1 - p)d = erst;
+ (1 p)d2 = e(2r+o-2)(5t
pu2
(4.12)
(4.13)
Since there are two constraints and three unknowns, the solution will not be
unique.
(i) Show that (4.12-4.13) are equivalent to
er5t - d
P
(erst - d) (u - erst)
=
u—d'
e2rdt (ecT26t — 1).
(4.14)
(4.15)
(ii) Derive the Cox-Ross-Rubinstein parametrization for a binomial tree, by
solving (4.14-4.15) with the additional condition that
ud = 1.
Show that the solution can be written as
P=
er8t - d
u—d ;
u = A + V A2 - 1; d = A - V A2 - 1,
where
A = 1- ( e— rat + e(r+0 2 )st •
2
j
(4.16)
96
CHAPTER 4. LOGNORMAL VARIABLES. RN PRICING.
Solution: (i) Formula (4.14) can be obtained by solving the linear equation
(4.12) for p.
To obtain (4.15), we first square formula (4.12) to obtain
p2u2
2p(1 — p)ud + (1 — 23)2d2 =e2rot
and subtract this from (4.13). We find that
p(1 — p)u2— 2p(1 — p)ud + p(1 — p)d2
e (2r+a2)5t
e2r6t ,
which can be written as
p(1— p)(u d)2 =
e2r6t (eo26t
1).
(4.17)
Using formula (4.14) for p, it is easy to see that
P(1— 1))
= (er5t —
(u — er6t)
(u — d)2
(4.18)
From (4.17) and (4.18), we conclude that
(er"— d) (u — est) = e2r6t (eo 25t
1).
(ii) By multiplying out (4.15) and using the fact that ud -= 1, we obtain that
uer5t —
1_
e2rSt
derdt = e(2r+o-2 )St
e2r8t
(4.19)
After canceling out the term —521*, we divide (4.19) by er6t and obtain
u d — e —r6t =
e(r+c2 )8t,
which can be written as
u + — (e—rst
1
e(r+o-2)6t) = U + —
—
2A = 0;
cf. (4.16) for the definition of A.
In other words, u is a solution of the quadratic equation
u2— 2Au + 1 = 0,
(4.20)
which has two solutions, A +./A2— 1 and A — ./A2 — 1. Since u > 1, we
conclude that
u A + VA2— 1;
4.1. SOLUTIONS TO CHAPTER 4 EXERCISES
97
the other solution of the quadratic equation (4.20) corresponds to the value
of d, since
1
1
= A — ✓A2— 1. ❑
d=—=
u A + VA2— 1
tie
Problem 6: Show that the series EkD.e 1 is convergent, while the series
1
Ekpc=i and
2 kln(k) are divergent, i.e., equal to oo.
E°c
Note: It is known that
k=1
and
( E
n
liM
n—*x
7/.2
1
=
6
k2
1
— — ln(n)
=7,
k=1
where -y 0.57721 is called Euler's constant.
1 are positive, it is enough
Solution: Since all the terms of the series \--, Dc k2
to show that the partial sums
n
\--. i
Z—• k2
k=1
are uniformly bounded, in order to conclude that the series is convergent.
This can be seen as follows:
n
1
Z--1 k2 — 1 +
k2
k=1
k=2
n
1
1 ± Ek(k
k=2
— 1)
=1+
1
1
,--, k — 1 k
k=2
1
= 1 + (1 — — ) < 2, V n > 2.
To show that the series EL -1kt is divergent, we will prove that
1
ln(n) + — <
E_ <
n
k=1
ln(n) + 1,
V n > 1.
(4.21)
k
The integral of the function f(x) =
over the interval [1, n] can be
approximated from above and below as follows: Note that
n-1 f k+i 1
71 1
- dx = E
x
k-1
— dx.
k
X
CHAPTER 4. LOGNORMAL VARIABLES. RN PRICING.
98
Since f (x) =1 is a decreasing function, it is easy to see that
1
1
< f (x) <
k +1
,
V x E (k, k +1).
Then,
n-1 f k+1 1
n1
— dx =
1 x
k=1
E
k =1
A
k+ 1
k=1
,
1
n-1
k=1
(4.22)
k=-1
n-1 k+1 1
— dx <
k=1 J k
i
1+ Z-d k
k+1 1
=E
dx
n ,
v-.
k+1
n-1
In —ldx
dx
1
k X
n-1
=
n-1 f k+1
— dx >
X
1
Ef
k=1
1
k = n +
k— dx
k
1
k=1
(4.23)
k
The inequality (4.21) follows from (4.22) and (4.23), since
n1
i —xdx =
ln(n).
In a similar fashion, by considering the integral of sin(x)over the interval
[2, n], we can show that
1
1
+ nln(n) < k=2 k ln(k)'
V n > 2;
(4.24)
1
< ln(ln(n)) ln(ln(2)) +
21n(2)'
k ln (k)
V n > 2,
(4.25)
ln(ln(n)) — ln(ln(2))
n
E
k=2
E
and conclude that the series Eke k in1(k) is divergent.
For example, (4.24) can be proved as follows:
n-1
n-1 f k+1
n-1f k+1 1
1
1
in 1
dx =
dx <
1
E kln(k)'
kln(k)
E jk
A xln(x) dx = E k x ln(x)
which is equivalent to
fn
12
1
x ln(x) dx + n ln(n) <
1
n
K
1
kln(k) •
(4.26)
4.1. SOLUTIONS TO CHAPTER 4 EXERCISES
99
Since
fn 1
ln(ln(n)) — ln(ln(2)),
2 xln(x) dx =
we conclude from (4.26) and (4.27) that
ln(ln(n)) — ln(ln(2)) +
1
nln(n) <
k=2
(4.27)
1
kln(k)'
❑
which is the same as (4.24).
Problem 7: Find the radius of convergence R of the power series
c>o
k=2
X
kln(k)'
and investigate what happens at the points x where Ix' = R.
Solution: It is easy to see that
oo
k=2
with ak =
k ln(k) '
Xk
k ln(k)
Eakxk,
(4.28)
k=2
k > 2. Note that
liM I ak
k---,00
1/k
li.fic = lim ( 1 )
= 1.
k-40.0 V chi(k)
(4.29)
Recall that, if limk_,colakillkexists, the radius of convergence of the series
k-2 akx k is given by
1
R=
(4.30)
From (4.29) and (4.30) the radius of convergence of the series (4.28) is R =1
We conclude that the series is convergent if Ixl < 1, and not convergent if
ixi > 1.
1,11)
(:
) • Since the terms (-1)k
If x = —1, the series becomes EaD
k=1 ln()have
alternating signs and decrease in absolute value to 0, the series is convergent.
If x = 1, the series becomes Ekt1 k lni(k) which was shown to be divergent
in Problem 6 of this chapter. ❑
Problem 8: Consider a put option with strike 55 and maturity 4 months
on a non-dividend paying asset with spot price 60 which follows a lognormal
CHAPTER 4. LOGNORMAL VARIABLES. RN PRICING.
100
model with drift pt = 0.1 and volatility o = 0.3. Assume that the risk-free
rate is constant equal to 0.05.
(i) Find the probability that the put will expire in the money.
(ii) Find the risk-neutral probability that the put will expire in the money.
(iii) Compute N(-d2).
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
< K). Recall that
P(S(T)
2
(p,- q uT)T + a TZ.
ln (S(T)
✓
S(0) )
Then,
P(S(T) < K) = P
(S(T) <
S(0)
S(0) )
P (in (S((°)) < in ( ZOO)
rr
P ((tt, - q - ‘- ) T + o-VT Z < In
(40)
= p z
< In
(
(4))
(it - q
- c)T
off'
=N
S
For
= 60, K = 55, T = 1/3, A = 0.1, q = 0, o- = 0.3, and r = 0.05, we
obtain that the probability that the put will expire in the money is 0.271525,
i.e., 27.1525%.
(ii) The risk-neutral probability that the put option will expire in the money
is obtained just like the probability that the put expires in the money, by
substituting the risk-free rate r for
i.e.,
it,
PR N( S(T) < K) = P
(z < ln
(*) - (r - q - 2-0T
= 0.304331 = 30.43%.
a-V7'
(4.31)
(iii) Recall that
(4.32)
4.1. SOLUTIONS TO CHAPTER 4 EXERCISES
101
Then, d2= 0.511983, and
N(-d2) = 0.304331,
which is the same as the risk-neutral probability that the put option will
expire in the money; cf. (4.31).
To understand this result, note that
ln (54) + (r - q - 4)T)
PRN(S(T) < K) = P Z <
N(-d2);
(
cf. (4.31) and (4.32).
❑
Problem 9: (i) Consider an at-the-money call on a non-dividend paying
asset; assume the Black-Scholes framework. Show that the Delta of the option
is always greater than 0.5.
(ii) If the underlying asset pays dividends at the continuous rate q, when is
the Delta of an at-the-money call less than 0.5?
Note: For most cases, the Delta of an at-the-money call option is close to 0.5.
Solution: (i) Recall that the Delta of a call option is given by
ln (12-s?')+ (r - q + 2-;) T)
A(C) = e-qT N(d1) = e-gT N
a-VT
(
For an at-the-money call on a non-dividend paying asset, i.e., for K = S
and q = 0, we find that
A(C) = N(d1) = N (
(r +
"c19
> N(0) = 0.5.
(ii) If the underlying asset pays dividends at the continuous rate q, the Delta
of an ATM call is
-q +
VT)
A(C) = e-gT N(di) = e-gT N
For a fixed risk-free rate r and fixed maturity T, we conclude that A(C) < 0.5
if and only if the dividend yield q and the volatility a of the underlying asset
CHAPTER 4. LOGNORMAL VARIABLES. RN PRICING.
102
satisfy the following condition:
(r - q + c) -VT)
N
< 0.5 OT .
a
This happens, for example, if r = q and T is large enough, since
lim N
T—+oc
(7\11' =1 and Tim .5 egg' = oo. ❑
2
T-,00
Problem 10: Use risk-neutral pricing to price a supershare, i.e., an option
that pays (max(S(T) - K, 0))2at the maturity of the option. In other words,
compute
V(0) = e—rT ERN [(max(S(T) - K, 0)) 2],
where the expected value is computed with respect to the risk-neutral distribution of the price S(T) of the underlying asset at maturity T, which is
assumed to follow a lognormal process with drift r and volatility a. Assume
that the underlying asset pays no dividends, i.e., q = 0.
Solution: Recall that
0.2
S(T) = S(0) expr((- — ) T + aN/77)
2
and note that
In
- (r -
S(T) > K < > Z >
o-VT
- - d2.
Then,
V(0) =
1
2771.
f d2 (S(0)exp-((r
2 T + a-V71x) - K) 2 e-T dx
2
K2e —rT 1 f 3c e- 2 dx
—d2
e-rT J pc
— 2K S(0)
exp ((r
-
Tix
+a - )VT
2-) dx
2
V 27 —d2
e—rT f oc
x
xpe ((2r - a2)T + 2o-VT'x - — ) dx
+ 82(0)
2
—d2
(4.33)
4.1. SOLUTIONS TO CHAPTER 4 EXERCISES
103
When pricing a plain vanilla call using risk-neutrality, we proved that
p rT
co
a2
Cgs(0) = S(0)
f exp (7. - L9 T + aV-77x - '12-) dx
21- -d2
- Ke-rT
=
S(0)N(dl)
f c°
dx
27 -d 2 e
- K e-rT N(d2).
1
In other words, we showed that
l
c
2 dx = e'T N(d2);
2-Ir -d2 e
e rT 1
e-rT ff o e
(
4.34)
x2
I7r d2 exp
(r - -2-) T +
x - -2-)
= N (di)
(4.35)
From (4.33), (4.34), and (4.35), we conclude that
V(0) = K2 e'T N (d2) - 2K S (0)N (di)
e rT Do
x2
I exp ((2r - a2 )T + 2a-V7x - —2 ) dx. (4.36)
+82(0)
27r -d2
The integral from (4.36) is computed by completing the square as follows:
x2
e -rT oo
S2(0)
f exp ((2r - a2 )T + 2afi"x - — ) dx
2
v27 -(12
co
p -rT f
(x — 2o-.V7)2
exp (
= S2 (0) -/-+ (2r + o-2)T dx
2
27r _d2
(x — 20-4) 2)
= s2(0)e(r+,72)T 1 /cc' exp
-411- - d2
2
Y
dy
2
4T -(d2+20-0") exp (-2)
= S2(0)e(r+a 2)TN(d2 + 20- V1 );
=
s2 (0) e(r + a-2)T
dx
1 l
c.°
(4.37)
we used the substitution y = x - 2a-V7 and the fact that
47r
I i:exp
2
y2 ) dy
V271-
f
a
( y2
— y dy = N (a).
From (4.36) and (4.37), we conclude that
V(0) = K2CrT N (d2) - 2K S (0)N (di) + S2(0)e(r 4-'2)TN (d2 + 2a-V-7).
❑
CHAPTER 4. LOGNORMAL VARIABLES. RN PRICING.
104
Problem 11: If the price of an asset follows a normal process, i.e., dS
pdt + cdX, then
S(t2) = S(ti) + ii(t2 - ti) + 0'02 - ti Z, V 0 < ti < t2.
Assume that the risk free rate is constant and equal to r.
(i) Use risk neutrality to find the value of a call option with strike K and
maturity T.
(ii) Use the Put-Call parity to find the value of a put option with strike K
and maturity T, if the underlying asset follows a normal process as above.
Solution: (i) Using risk-neutral pricing, it follows that
C(0) = e-rT ERN [max(S (T) - K, 0)],
where the expected value is computed with respect to S(T) given by
S(T) = S(0) + rT + o--VT Z.
(4.38)
Note that
S(T) > K if Z >
K - S(0) - rT
crVT
= d.
Then,
C(0) = e—rT
1
x2
27r f:
(S(0) + rT + cIN/T x - K)
(S(0)
f DC
rT)e-rT 1
IT
d
x
2
e-T dx
6— T
1 fpC
,7,
Ke'
dx
2
fgr d
dx
.2
e-rTo-VT x
xe-Tdx.
fgr
Note that
f
X xe - dx
d
1
2
'2
fx x2
fg- d
lim
t---■
ac
e-Tdx = 1
ft
.2
.2 )
xe-Tdx = lim (-CT
t— ■oc
d
1
t
d
= e 452-;
id
-47r L o,
e-Tdx = 1- N(d) = N(-d),
where N(t) is the cumulative distribution of the standard normal variable.
We conclude that
C(0) = (S(0)+rT)e-rT N(-d) - Ke-rT N(-d) + e
-rT a-VT _d2
e T. (4.39)
-47'r
4.1. SOLUTIONS TO CHAPTER 4 EXERCISES
105
(ii) Regardless of the model used for describing the evolution of the price of
the underlying asset, the Put-Call parity says that a portfolio made of a long
position in a plain vanilla European call option and a short position in a plain
vanilla European put option on the same asset and with the same strike and
maturity as the call option has the same payoff at maturity as a long position
in a unit of the underlying asset and a short cash position equal to the strike
of the options. Using risk-neutral pricing, this can be written as
C(0) - P(0) = e-rT ERN[S(T) - K] = e'T (S(0) + rT - K),
since ERN[S(T)] = S(0) + rT; cf. (4.38).
From (4.39) and (4.40), we obtain that
P(0) = C(0) - (8(0) + rT)e-rT
Ke-TT
K e-rT (1 - N(-d)) - (S(0) + rT)e-rT (1 - N(-d))
e-rT a ff , d2
e_
= Ke'T N(d) - (S(0) + rT)e-rT N(d) +
since 1 - N(-d) = N(d).
❑
e-rTo-ff
-r e
(4.40)
CHAPTER 4. LOGNORMAL VARIABLES. RN PRICING.
106
4.2
Supplemental Exercises
1. Show that the sequence
n
xn
=
k=1
k
— ln(n)
is convergent to a limit between 0 and 1.
Note: The limit of this sequence is -yR.,- 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 p, = 0.08 and
volatility a = 0.3. Assume that the risk—free 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.
(ii) Find 95% and 99% risk—neutral confidence intervals for the spot
price of the asset in 15 days, 1 month, 2 months, 6 months, and 1 year,
i.e., assuming that the drift of the asset is equal to the risk—free rate.
3. If you play (American)1
roulette 100 times, betting $100 on black each time, what is the probability of winning at least $1000, and what is the probability of losing
at least $1000?
4. Use risk—neutral pricing to find the value of an option on a nondividend—paying asset with lognormal distribution if the payoff of the
option at maturity is equal to max((S(T))a — K, 0). Here, a > 0 is a
fixed constant.
5. Find a binomial tree parametrization for a risk—neutral probability (of
going up or down) equal to 1. In other words, find the up and down
factors u and d such that
pu + (1 — p)d = erbt.,
put + (1 p)d2 = e(2r+cr2 )5t
if p = 2 .
'American roulette has 18 red slots. 18 black slots, and two green slots (corresponding to 0 and 00). European roulette. also called French roulette;has only one green slot
corresponding to 0.
4.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
4.3
107
Solutions to Supplemental Exercises
Problem 1: Show that the sequence
1
xn = (n — ) —
ln(n)
k=1
is convergent to a limit between 0 and 1.
Solution: Recall from (4.21) that
1
—k < ln(n) + 1, V n > 1,
ln(n) + —
n<
k=1
which can be written as
1
— < xn < 1, V n > 1.
rt
It is easy to see that
1
Xrt+1
Xn =
n+1
ln(n + 1) + ln(n).
Therefore, xn+1 < xr, if and only if
1
(n + 11
n + 1 < ln(n + 1) ln(n) = ln
n )
This is equivalent to 1 < (n + 1) In (V), and therefore to
e<
(n + 1)n+1
n )
(
1 n+1
1 + —)
which holds for any n > 1, from the definition of e.
We showed that the sequence (xn)n=1;,„ is decreasing and bounded from
below by 0 and from above by 1, The sequence is therefore convergent to a
limit between 0 and 1. ❑
Problem 2: Assume that an asset with spot price 50 paying dividends
continuously at rate q = 0.02 has lognormal distribution with mean p, = 0.08
and volatility a-= 0.3. Assume that the risk—free 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.
108
CHAPTER 4. LOGNORMAL VARIABLES. RN PRICING.
(ii) Find 95% and 99% risk-neutral confidence intervals for the spot price of
the asset in 15 days, 1 month, 2 months, 6 months, and 1 year, i.e., assuming
that the drift of the asset is equal to the risk-free rate.
Solution: If the asset has lognormal distribution, then
0.2
S(t) = S(0) exp (/2 - q - --2-) t + o-ViZ)
(4.41)
= 50 exp(0.015t + 0.3ViZ).
Recall that the 95% and 99% confidence intervals for the standard normal
distribution Z are [-1.95996, 1.95996] and [-2.57583, 2.57583], i.e.,
P(-1.95996 < Z < 1.95996) = 0.95; P(-2.57583 < Z < 2.57583) = 0.99.
Therefore, the 95% and 99% confidence intervals for S(t) are
[50 exp(0.015t - 03-Vi • 1.95996), 50 exp(0.015t 0.3-Vi • 1.95996)];
[50 exp(0.015t - 0.3-Vi • 2.57583), 50 exp(0.015t + 0.3fi • 2.57583)],
respectively.
The risk-neutral confidence intervals for the spot price of the asset are
found by substituting the risk-free rate r for /2 in (4.41) to obtain
SRN(t) = 50 exp(-0.015t +
Therefore, the 95% and 99% confidence intervals of SRN(t) are
[50 exp(-0.015t - 0.3 Nfi • 1.95996), 50 exp(-0.015t 0.3-Vt • 1.95996)]
[50 exp(-0.015t - 0.3
respectively.
For t E {A,
t
95% CI S(t)
15 days
43.36, 57.76
1 month 42.25, 59.32
2 months 39.43, 63.72
6 months 32.25, 75.35
1 year
28.20, 91.38
• 2.57583), 50 exp(-0.015t + 0.3\fi • 2.57583)],
we obtain the following confidence intervals:
99% CI S(t) 95% CI SRN(t) 99% CI SRN(t)
[41.45, 60.43.
43.28, 57.66
[41.36, 60.30
[40.05, 62.57
42.14, 59.18
39.96, 64.41
36.56, 68.70
39.23, 63.41
36.36, 68.37
29.17, 86.99.
32.74, 75.21
[28.73, 85.70
[23.44,109.90]
27.36, 88.68
[22.75,106.65]
Problem 3: If you play (American) 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 $1000?
4.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
109
Solution: Recall that an American roulette has 18 red slots, 18 black slots,
and two green slots. Therefore, every time you bet on black, you win $100
with probability A and lose $100 with probability E. In other words, if Wi
is the value of the winnings in the i—th round of playing, then
Wz
E;
f —100, with probability
100, with probability
Note that
10
19
9
19
µE[W]
=
= — (-100) + —100 =
—
o = std(Wi) = E[(Wi)2] — (E[WiD2 =
100.
19
6000Th
19
Let W = Eroi Wibe the total value of the winnings 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 find that
W 100p + 10aZ = —
10000 6000 10
19
19
Z.
The probability of winning at least $1000 can be approximated as follows:
P(W > 1000)
P
10000 6000 0T
19
19
Z > 1000)
= P(Z > 1.5284) = 0.0632.
The probability of losing at least $1000 can be approximated as follows:
P(W < —1000)
P
1N
0000 6000 /TO
19 + 19
Z < —1000)
= P(Z < —0.4743) = 0.3176.
We conclude that the probability of winning at least $1000 is approximately
6%, and the probability of losing at least $1000 is approximately 32%. ❑
Problem 4: Use risk—neutral pricing to find the value of an option on a
non—dividend—paying asset with lognormal distribution if the payoff of the
option at maturity is equal to max((S(T))a — K, 0). Here, a > 0 is a fixed
constant.
CHAPTER 4. LOGNORMAL VARIABLES. RN PRICING.
110
Solution: Using risk-neutral pricing, we find that the value of the option is
V(0) =
CrT ERN[MaX((5
where
(T))a — K,0)],
2
(S(T))a > K
Note that
that
-
S(0)exp
S(T)
is equivalent to
+ o- VT
2
.
(4.42)
S(T) > K11'. Using (4.42), we find
In (V) - (r - '4) T
Z>
S(T) > K11a
- a.
o-,5"
Then,
e-rT
0.2
fc
((S(0))aexp (a - — ) T+ cto- A/Tx) - K)
2
f2-Tr -a
V(0) =
2
dx.
Recall from (4.34) that
x
e-rT r
27r
2
dx
= e-rTN(a) .
Therefore,
ev0.2
V(0) = (S(0))
11g-T -a
exp)rT
exp
((a 1
2
T + ao- tx -
x2
✓
dx
- Ke-rT N (a).
By completing the square for the argument of the exponential function
under the integral sign we obtain that
1
r
(10.2
exp ((a 1)rT
7I -a
2
T + ao- ftx -
a 2T+ aa
22
2 T) /,r
= exp( (a-1)rT- —
2
= exp ((a - 1) (r + a(7-2 °
) T)
2
a0.2
= exp ((a - 1) (7. + —) T
)
2
C
(10.2
X2
f:exp
dx
(
x -ac -P2
2
Y
exp (--) dd
y
2
Nig _(,±a,,m
1 fa+a0VT
2
exp (--0 dy
1 fG
-‘/Yr j,
exp ((a - 1) r + 2 )
T) N(a + ao-ft);
)
dx
4.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
111
note that the change of variables y = x — ao- ff" was used above.
We conclude that
(10. 2
V(0) = (S(0))aeXp
((a — 1)
+ --2—)T) N(a + ao-VT) — Ke'T N(a),
where
a=
In Klio
s(°) ) + (r — 2
)
o /T
Problem 5: Find a binomial tree parametrization for a risk—neutral probability (of going up) equal to Z. In other words, find the up and down factors
u and d such that
Pu + (1 — p)(1 = en5t
pu2 + (1 —Ad2 = e(2r+cr2 )(5t
if p= 2.
Solution: It is easy to see that, if p = 1, then
u + d = 2er5t;
u2 d2 = 2e(2r+cr2)bt ,
and therefore
ud =
(u + d)2— (u2+ d2) = 2e2r5t
2
e(2r+o 2)(5t
Note that u and d are the solutions of z2 — + d)z + ud = 0, which is
the same as
(4.43)
z2— 2er5t z + 2e2r8t— e(2'+`72)6t = 0.
We solve (4.43) and find that
u --= er8t (1 + ✓ea26t — 1) ;
d = erbt (1 — V eu26t — 1) ,
since d < u.
❑
Chapter 5
Taylor's formula and Taylor series. ATM
approximation of Black—Scholes formulas.
5.1
Solutions to Chapter 5 Exercises
Problem 1: Show that the cubic Taylor approximation of \/1 + x around 0
is
V-1+x
-
1+
X
2
X
—
2
+
8
X
3
16'
Solution: Recall that the cubic Taylor approximation of the function f (x)
around the point a = 0 is
f(a) + (x
f(x)
a)l(a)
(x
2
f(x) =
6
f(3) (0).
= 1(0) + xf(0) + 1-/"(0) +
For f (x) = 11
f(3)(a)
(x
a)2 1"(a)
(5.1)
x, we find that
11
; f" (x) = 4(1 x)32 '
2-V1 + x
f (3)(x) =
3
81
(
+ x 5/2'
)
and therefore
1(0) = 1; 1/(0) = 2
. f(3)(0
)
4'
f „(0)
;
From (5.1) and (5.2) we conclude that
A/1+X
1+
X
2
—
X
2
X
3.
8
(5.2)
3
8 + 16'
L
Problem 2: Use the Taylor series expansion of the function e2' to find the
value of e0.25with six decimal digits accuracy.
113
114
CHAPTER 5. TAYLOR'S FORMULA. TAYLOR SERIES.
Solution: Recall that the Taylor series expansion of the function f(x) = ex
around 0 converges to ex at all points x E R, i.e.,
Xk
v■
ex =
V x E R.
k!'
k=0
For x = 0.25 we find that
e0.25 =
E
E(0.25)k
k=0
e0.25 =
n-• Qc
Xn,
k!
E
n (0.25)k
where xn =
k!
k=0
v n > O.
Note that the sequence {x,}n=0,,cis increasing. It is then enough to
compute xo, xi, x2, ... , until the first seven decimal digits of these terms are
the same, in order to find the first six decimal digits of 60.25. We find that
x2= 1.28125;
xi = 1.25;
xo = 1;
x3 = 1.28385417; x4 = 1.28401698; x5 = 1.28402507;
x6= 1.28402540; x7 =1.28402541,
and conclude that
e0.25 Re, 1.284025.
❑
Problem 3: Find the Taylor series expansion of the functions
ln(1 — x2) and
1
1 — x2
around the point 0, using the Taylor series expansions of ln(1 — x) and 1. 1 .
Solution: Recall that
\--", X k
X2
X3
X4
,VXE1-1, 1);
k=1
DO
=E
xk =1±X±X2 ±X 3 +...,e X E(-11).
k=0
By substituting x2 for x in the Taylor expansions above, where Ix' < 1, we
find that
x T 2k
2 X 4 X 6 X8
1.11 — 2X) = —
= x
—
—— —
, V x E (-1, 1);
3
4
k=1
-
-2
5.1. SOLUTIONS TO CHAPTER 5 EXERCISES
1
1 — x2
E x2k = ± x2 ± x4 ± x6 ±
115
,dx E
1).
❑
k=0
Problem 4: Let
co (_i)k+1 x k
T(x)
k=1
be the Taylor series expansion of f (x) = In (1 + x). Our goal is to show that
T(x) = f (x) for all x such that Ix! < 1.
Let
Pn(X)
=E
(_i)k+1 xk
k
k=1
be the Taylor polynomial of degree n corresponding to f (x). Since T(x) =
limn,,, Pi,(x), it follows that f (x) = T(x) for all Ix! < 1 if and only if
I
lim f (x) — Pn(x)I = 0, V Ix! < 1.
n-+ oo
(5.3)
(i) Show that, for any x,
f (x) — Pr,(x) =
A
(-1)n±2(x —
(1 + on+1
dt
dt.
(ii) Show that, for any 0 < x < 1,
I f (x) — P9,(x)I < xn ln(1 + x)
and prove that (5.3) holds for all x such that 0 < x < 1.
(iii) Assume that —1 < x < 0. Show that
f (x) — Pn(x)I < (—
ln(1 + x)!
and conclude that (5.3) holds true for all x such that —1 < x < 0.
Solution: (i) From the integral formula for the Taylor approximation error
we know that
f(x) — Pn(x) =
Jo
(x —
n!
f
(n+i)
(t) dt.
(5.4)
Since f(x) = In (1 + x), we obtain by induction that the derivatives of
f(x) are
fuo (x)
(-1)k+1(k — 1)!
(5.5)
V k > 1.
(1 + x)k
CHAPTER 5. TAYLOR'S FORMULA. TAYLOR SERIES.
116
From (5.4) and (5.5) it follows that
f (x) — Pri(x)
x (x
jo
—
77,1
on
(
s (-1)71+2(x
=-
fo
dt
(1
)
(1 + tr+1-
dt
(5.6)
.
(ii) Let x E [0, 1). By taking absolute values in (5.6) and using the fact that
x—t
1+t
< x, V0 0.
The quadratic Taylor approximation
f (x) = f (0) + x (0) + 2- f"(0) + 0(x 3),
as
x
0,
becomes
e9( x)
+ xe9Mg' (0) + x2e9(°) 9ll(0) + (9i(13))2 + 0(x3),
2
as x
0.
(ii) By letting g(x)
(5.27)
(x + 1)2in (5.27), we find that
e(x+1)2 = e + 2ex + 3ex2 + 0(x3),
as
x
0.
(iii) Using the quadratic Taylor approximations
)2 0(x3),
+ (2x
1 + 2x + (2x)2 + 0(x3),
2
2
ex2 = 1 + x2 + 0(x4), as x
0,
e2x
1 + 2x
as
it follows that
e(x+1)2 = e e2x ex2
e(1 + 2x + 2x2)(1 + x2) + 0(x 3)
= e + 2ex + 3ex2 + 0(x3), as x
0.
❑
x —> 0;
5.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
127
Problem 2: Show that
1
= 0(x2),
1+x
e-x
as
x
0.
Solution: The quadratic Taylor approximations of e-x and 1. _11 are
x2
e-x = 1 - x +
1
=1-x
1+x
+ 0(x3),
as
x 0;
X2 + 0(x3),
as
x
0.
Therefore,
x2
e—x
1+x
0(x3) = 0(x2),
2
as
x -> 0.
Note that we implicitly proved that
1
x2
= -— + 0(x 3),
1+x
2
e-x
as x 30. 1=1
-
Problem 3: Compute the Taylor series expansion of
In
(1 + x)
1 -x
around the point 0, and find its radius of convergence.
Solution: Note that the function
ln
(1 + x)
1-x
= ln(1 + x) - ln(1 - x)
is not defined for x = -1 or 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 ln(1 + x) and ln(1 - x) are
X2
co
i)k+1"'
7-- X —
k=1
-E-k
cc) Xk
k=1
X
2
2
+
X3
X3
3
X4
X4
4
+
,V x E
, VxE[-1, 1),
11;
CHAPTER 5. TAYLOR'S FORMULA. TAYLOR SERIES.
128
and have radius of convergence equal to 1.
We conclude that the Taylor series expansion of In (N) is
In
(1 + x)
1— x
ln(1 + x) — ln(1 — x)
=
'
x'
2x2,1+1
—.., 2j + 1
Xk )
t ((_1)1c.+1Xk
k + Tc
22 22
= 2x + T+ —5— + ... , V x E (-1,1),
and has radius of convergence equal to 1.
Problem 4: Prove that
x
1
'2
1
12.r
(5.28)
❑
11
X+
< e < (1 + -
2
, VX
1.
(5.29)
Solution: Recall from (5.28) that
2y5
2y
In (1+Y ) = 2y+- +- +... , V y E
—y
3
5
(5.30)
For any x > 1, substitute y = 22+1in (5.30) and obtain that
ln (1 +
1
=
2
2
+ 2 +
+... , V x > 1,
2x + 1 3(2x + 1)3 5(2x + 1)5
which can also be written as
(x + 1
— ) ln (1 + —) = 1+
+... , V x > 1. (5.31)
2
3(2x+1)2+ 5(2x + 1)4
From (5.31), we find that
1 < (x+
ln (1 + I") , V x > 1,
which is equivalent to
1)x+i
e < (1+ —
, V x > 1.
x
The right inequality of (5.29) is therefore established.
(5.32)
5.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
129
From (5.31), we also find that
1
1
+ 1)141
+ 1)
- < 1+
2
x
3(2x + 1)2 E (2x + 1)2k
k=0
1
1
1
-= 1 +
= 1+
3(2x + 1)2 1 (2x+
12x(x + 1)'
1 1)2
and therefore that
(
1 x+2
x±i
+ -1
< e 375-T' 1), V x > 1.
x
(5.33)
Recall that
ly+1
e < (1 + x
V x > 1.
(5.34)
Using (5.34), we find from (5.33) that
x-F l
(1+ =-)
2
< e (1 +
<
X
1
1) 12'
, V x > 1,
and we conclude that
71+ 1 x+1-11x
< e, V x > 1.
The left inequality of (5.29) is therefore established.
❑
Problem 5: (i) Find the radius of convergence of the series
X4
x8
X 12
1+2i + z +7,- +...
!
(5.35)
(ii) Show that the series from (5.35) is the Taylor series expansion of the
function
-x2
ex2 e
2
Solution: (i) The series (5.35) can be written as a power series as follows:
03
T(x) =
a4P X4 '
p=1
with a4p =
1
(2p)!' V P
CHAPTER 5. TAYLOR'S FORMULA. TAYLOR SERIES.
130
From Stirling's formula we know that
k!
Fun
= 1.
11\k
k—' Dc e VI271-k
It is then easy to see that
( 011k
lim
k
k—>x
and therefore that
1
=—
e
2p
lim ( (20)1/2p = e
(5.36)
Using (5.36), we find that
2p
1
p—■
DC
e1/2
VVP
/1 2
)
lim (((20)1/2p
p_,,e
lim
p--40c ((2p)!)1/4P
lm la4p1114P
(20/2
= O.
Therefore, the radius of convergence of the power series T(x) is
R—
1
1
lim supk„ I aklific
limp ocI a4p 11/4P
-= co,
which means that the series (5.35) is convergent for all x E R.
(ii) Using the Taylor series expansion
Xk
e"
V x E R,
Ls k!
k=0
it is easy to see that
ex2
e
2
_x 2
=
=
1
N(• (X2
1
k=0
k=0
x2k ± (_i)k x 2k
2 (L--1 k! + E
2
)k
E
k=o
=1+
k!
X4
DG
X2
—
( )k
k!
—
x8
x12
+ — + — + ... ,
2!
4!
6!
which is the same as the series (5.35).
❑
,
x . x4j
v
6'
3=o(2j)!
5.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
131
Problem 6: The goal of this exercise is to compute
fo1
ln(1 — x) ln(x) dx.
(5.37)
(i) Show that
lira (ln(1 — x) ln(x)) = lim (ln(1 — x) ln(x)) =- 0,
x\o
x/1
and conclude that the integral (5.37) can be regarded as a definite integral.
(ii) Use the Taylor series expansion of ln(1 — x) for Ix( < 1 to show that
1
ln(1 — x) ln(x) dx = —
En=ec—'a1 of l
xn ln(x) dx.
1
(iii) Prove that
Li ln(1 — x) ln(x) dx =
00
1
E n(n + 1)2.
k=1
(iv) Use that fact that
oo
1
2 =
k=1 n
to obtain that
1
io
71'
2
6
72
ln(1 — x)ln(x) dx = 2 — w.
Solution: (i) First of all, note that
lim (ln(1 — x) ln(x)) -= lim (ln(1 — x) ln(x)) .
x/1
x\o
(5.38)
We compute the left hand side limit of (5.38) by changing it to a limit to
infinity corresponding to y = 1 as follows:
lim (ln(1 — x) ln(x))
x\o
11)
lu (lim In (1 — —
Y—,0
0
Y
Y)
Y )v ) • Y ) (— ln(y))
lim (ln ( (1 — —
Y—'00
)v) ln(y) .
1
= — lim In (( l. — —
y—>co
Y
Y) )
(5.39)
CHAPTER 5. TAYLOR'S FORMULA. TAYLOR SERIES.
132
Recall that
1) =.
1
-- —
e
y
lim (1 — —
Y—Qc
and therefore
lim In ( (1 — -I--Y) = — 1.
y
y—,x
From (5.39) and (5.40) it follows that
urn (ln(1 — x)
x \O
= lim
y—'
ln(y)
y
(5.40)
O.
We conclude that
Ern (ln(1 — x) ln(x)) = him (ln(1 — x) ln(x)) = 0.
x\o
x/1
The integral (5.37) is equal to the definite integral between 0 and 1 of the
continuous function g : [0, 1] -- IR given by
g(0) -= g(1) -= 0;
g(x) = ln(1 — x)ln(x), V 0 < x < 1.
(ii) The Taylor series expansion of ln(1 — x), i.e.,
x xk
ln(1 — x) = —E T V x E (-1,1),
k=1
is absolutely convergent to ln(1 — x). Then,
1
ln(x)
o ln(1 — x)ln(x) dx =
xn
t
dx
J0
n=1
(
x1
= — E -Iii x" ln(x) dx.
n=1 n 0
f
1:1
(5.41)
(iii) Using integration by parts, it is easy to see that
J
xnln(x) dx =
xn+11n(x)
n+ 1
X
n+1
(n + 1)2 + C.
Then,
/1x" ln(x) dx
1
(xn+1 1n(x)
xn+1
n+ 1
(n+ 1)2 ) 0
1
1 1;
=
.c, (xn+1 1n(x))
(n + 1)2
n + 1 ;ri
1
=
V n > 1.
(n+ 1)2
—
(5.42)
5.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
From (5.41) and (5.42), it follows that
fo i
ln(1 - x) ln(x) dx -=
CO
133
1
E
n=in(n + 1)2 '
(5.43)
(iv) Note that
11 1
n(n+ 1) = n n+1.
Then, it is easy to see that
1
n(n + 1)2
1
1
1
=
n(n + 1) n + 1
n(n + 1)
1
1
1
n n+1 (n+ 1)2.
1
(n + 1)2
Therefore,
00
r.,=1n(n + 1)2
n=1 (n
n2
1 ) - 2-•
n=2
72 — 1) = 2 - '762
= 1 — (6
(5.44)
.
Here, we used the fact that
00 1
k=1 n
and the telescoping series
00
1
2=
72
T
N
) = lim x--, (1
1 )
=- lim 1
N-400
Z-i n n + 1
N-,00 Z
-, n n
n+1
n=1 (1
n=1
From (5.43) and (5.44), we conclude that
i
72
ln(1 - x)ln(x) dx = 2 w. ❑
\---,
N +1 = 1,
fo
Problem 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
of the approximation
, T (
1
Papprox,r0,00 = Cr,D \I ,Tr
(r + q)n
2 )
(r -OT
2
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:
Error
Papprox
PBS
1 4.1317 4.1491 0.42%
3 5.9834 6.1577 2.83%
5 6.4652 6.9714 7.26%
10 5.2187 7.3398 28.90%
20 -2.5067 6.0595 N/A
T
While the approximation formula is still within 3% of the Black—Scholes
value when the maturity is three years or less, it deteriorates for long dated
options, and even produces a negative value for the 20—years option. U
Chapter 6
Finite Differences. Black—Scholes 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 — x, a long position in a call option with strike K + x, and
a short position in two calls with strike K. The options are on the same
underlying asset and have the same maturities.
(i) Show that the value of the butterfly spread is
C(K + x) — 2C(K) + C(K — x),
where, e.g., C(K + x) denotes the price of the call with strike K + x.
(ii) Show that, in the limiting case when x goes to 0, the value of a position in
butterfly spreads as above converges to the second order partial derivative
Of the value of the option, C, with respect to strike K, i.e., show that
lim
x \o
C (K + x) — 2C (K) + C(K — x) - a2c
(K ).
aK2
x2
(iii) Show that, in the limiting case when x —> 0, the payoff at maturity of
a position in i butterfly spreads as above is going to approximate the payoff
of a derivative security that pays 1 if the underlying asset expires at K, and
0 otherwise.
Note: A security that pays 1 in a certain state and 0 in any other state is called
an Arrow-Debreu security, and its price is called the Arrow-Debreu price of
that state. A position in 1 butterfly spreads as above, with x small, is a
synthetic way to construct an Arrow-Debreu security for the state S(T) = K.
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 + x and value C(K + x), and a short position in two calls with
strike K and value —2C(K) is
C(K + x) — 2C(K) + C(K — x).
135
136
CHAPTER 6. FINITE DIFFERENCES. BLACK-SCHOLES PDE.
(ii) The value of a position in
butterfly spreads is
1
-i
x (C(K + x) - 2C(K) + 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 in time except at maturity).
Therefore the expression from (6.1) represents the central finite difference
approximation of g2,4 (K) and we know that
C(K + x) -2C(K) + C(K - x)
8 2c
= 0K2 (K) + 0(x2).
x2
Then, in the limit as x goes to 0, we obtain that
lim
x\o
C(K + x) - 2C(K) +C(K - x)02C ( re.\
x2
ar-c2IL ).
(iii) The payoff at maturity of the butterfly spread is
max(S - (K - x), 0) - 2 max(S - K, 0) + max(S - (K + x),0)
if S < K - x;
0,
if K - x < S < K;
S
(K
x),
=_
K + x - S, if K < S < K + x;
if K + x < S.
0,
Denote by fx (S) the payoff at maturity of a position in 1 butterfly spreads.
Then,
{0,
if S < K - x;
S-(K-x)
if K - x < S < K;
fx(S) K-Fxs-s if K 0,
the payoff at maturity of a position in I- butterfly spreads as above is 1 if the
underlying asset expires at K, and 0 otherwise. ❑
6.1. SOLUTIONS TO CHAPTER 6 EXERCISES
137
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 with strike 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 (at time t) of the call options with strikes
K and K + x, respectively.
(i) The value of a position in bull spreads is
case when x goes to 0, show that
lim
C( K)-C(K+x)
In the limiting
ac(K).
C(K) — C(K + x)
8K
(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 as above, with x small, is a synthetic way
to construct a cash—or—nothing call maturing at time T.
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
finite difference approximation of E(K) is
ac
ax
as x
(K) =
C(K + x) — C(K)
+ o(x),
0. We conclude that
lim
x\o
ac(K).
C(K)— C(K + x)
ax
(ii) The payoff at maturity of the bull spread is
if S < K;
S — K, if K < S < K + x;
x,
if K + x < S.
{0,
max(S — K, 0) — max(S — (K + x), 0)
=
If gx (S) denotes the payoff at maturity of a position in bull spreads, then
{
gx(S) =
0, if S < K;
if K < S < K + x;
1, if K + x < S.
If S < K, then gx(S) = 0 for any x > 0 and therefore
lim gx(S) = 0, V S
x\o
< K.
(6.4)
138
CHAPTER 6. FINITE DIFFERENCES. BLACK—SCHOLES PDE.
If S > K, then gm (S) = 1 for any x such that 0 < x < S – K, and
therefore
lim gx (S) = 1, V S > K
(6.5)
x\o
From (6.4) and (6.5), we conclude that, in the limiting 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. ❑
Problem 3: Find a second order finite difference approximation for f'(a)
using f (a), f (a + h), and f (a + 2h).
Note: This type of approximation is needed, e.g., when discretizing a PDE
with boundary conditions involving derivatives of the solution (also called
Robin boundary conditions). For example, for Asian Options (continuous
computed average rate call, to be more precise), this type of finite difference
approximation is used to discretize the boundary condition -5§ +
= 0, for
R = 0.
Solution: To obtain a finite difference approximation for f'(a) in terms of
f (a), f (a + h), and f (a + 2h) we use the cubic Taylor approximation of f (x)
around the point x = a, i.e.,
f(x) = f (a) + (x – a)[ (a) +
(x – a)2
(x – a)3(3)
2 f u(a) +
6
f (a)
+ O ((x – a)4) ,
(6.6)
as x –+ a. By letting x = a + la and x = a + 2h in (6.6), we obtain that
h2
(a) + — f n(a) +
(3) (a) + 0 (h4) ;
(6.7)
2
6f
4h3
f (a + 2h) = f (a) + 2h t (a) + 2h2 f ll (a) +
f (3) (a) + 0 (h4) , (6.8)
f (a + h) = f (a) +
as h –+ 0.
We multiply (6.7) by 4 and subtracting the result from (6.8) to obtain
2h3
f (a + 2h) – 4f (a + h) = – 3f (a) – 2h f'(a) + — f (3) (a) + 0 (h4) , (6.9)
3
as h
0. By solving (6.9) for [ (a), we obtain the following second order
finite difference approximation of (a):
[ (a) =
– f (a+ 2h) + 4f (a + h) – 3,f(a)
h2 f(3) ) + 0 (h3)
2h
3
–f (a + 2h) + 4f (a + h) – 3f (a)
+ 0 (h2) ,
2h
6.1. SOLUTIONS TO CHAPTER 6 EXERCISES
as h
0.
139
❑
Problem 4: Find a central finite difference approximation for the fourth
derivative of f at a, i.e., for f (4)(a), using f (a — 2h), f (a — h), 1 (a), f (a+ h),
and f (a + 2h). What is the order of this finite difference approximation?
Solution: We will use the following Taylor approximation of f (x) around the
point x = a:
f (x) = f (a) + (x — a) f' (a) +
(x a)2
—
(x a)3(3)
—
2 f "(a) +
6
f (a)
f (4) (a) + (x — ar f(5)(a) + (x — ar f(6) (a)
—
24a)4
(a
) ' 120
\'
720
\'
+ 0 ((x — a)7) ,
(6.10)
+ (x
as x —› a.
For symmetry reasons, and keeping in mind the form of the central difference approximation for f " (a), we use (6.10) to compute
h4
h6
f (a + h) + f (a — h) = 2f (a) + h2f" (a) + —
12 f (4) (a) + W f (6) (a)
6
(6.11)
+ 0 (hi) , as h —› 0;4
+43_
h f(4)(a) +84h5 f(6)(a)
f(a + 2h) + f(a — 2h) = 2f (a) + 4h2 f u(a)
+ 0 (hi) ,
(6.12)
as h —> 0. We multiply (6.11) by 4 and subtract the result from (6.12).
We solve for f(4) (a) and obtain the following second order finite difference
approximation:
f(4) (a) = f (a + 2h) — 4f (a + h) + 6f (a) — 4f (a — h) + f (a — 2h)+0 (h2),
h4
as h —> 0. ❑
Problem 5: The goal of this exercise is to emphasize the importance of symmetry in finite difference approximations. Recall that the central difference
approximations for the first and second order derivatives are
2
f(a + h) — f(a — h)
+ 0 (h ) ;
2h
f(a + h) — 2f (a) + f(a — h)
+ 0 (h2) ,
f'' (a) =
f' (a) =
h2
140
CHAPTER 6. FINITE DIFFERENCES. BLACK—SCHOLES PDE.
as h
0. In other words, f'(a) and f " (a) are approximated to second order
accuracy by using the value of f at the point a and at the points a — h and
a + h that are symmetric with respect to a.
We investigate what happens if symmetry is not required.
(i) Find a second order finite difference approximation of fl(a) using f (a),
f (a — h) and f (a + 2h).
(ii) Find a first order finite difference approximation of f " (a) using f (a),
f (a — h) and f (a + 2h). Note that, in general, a second order finite difference
approximation of f (a) using f (a), f (a — h) and f (a + 2h) does not exist.
Let /3 < a < 7 such that a —
C(-y — a), where C is a constant.
(iii) Find a finite difference approximation of f'(a) using f (a), AO), and
f (7) which is second order in terms of 17 — al, i.e., where the residual term
is 0 (1-y — a12).
(iv) Find a finite difference approximation of no) using f (a), AO), and
f (7) which is first order in terms of 17 — al. Show that, in general, a second
order finite difference approximation of f" (a) using f (a), f (0) and f (y) is
not possible, unless a = P+27, i.e., unless /3 and -y are symmetric with respect
to a.
"
Solution: (i) and (ii). We use the cubic Taylor approximation of f (x) around
the point x = a, i.e.,
f (x) = f (a) + (x — a) f'(a) + (x 2a)2f" (a) + (x
+ 0 ((x
a)3
6
f (3)(a)
a)4) , as x te a.
(6.13)
By letting x = a — h and x = a + 2h in (6.13), we obtain that
h
h3
f (a — h) = f (a) — hf'(a) + —
2f " (a) — —
6 f (3)(a) + 0 (h4) ; (6.14)
f (a + 2h) = f (a) + 2hf'(a) + 2h2 f u(a) +
433
f `31(a) + 0 (h4) , (6.15)
as h —> 0.
We eliminate the terms containing f" (a) by multiplying (6.14) by 4 and
subtracting the result from (6.15). By solving for f'(a), we obtain the following second order finite difference approximation of f (a):
f(a) =
-=
as h —> 0.
f (a + 2h) + 3f (a) — 4f (a — h)
h2
6h
3
f (3) (a) + 0 (h3)
f (a + 2h) + 3f (a) — 4f (a — h)
+ O(h2)
6h
6.1. SOLUTIONS TO CHAPTER 6 EXERCISES
141
Similarly, we eliminate the terms containing f'(a) by multiplying (6.14)
by 2 and adding the result to (6.15). By solving for 7(a), we obtain the
following first order finite difference approximation of f "(a):
f (a + 2h) — 3f (a) + 2 f (a — h)
h /3\
f"(a) + 0 (h2)
3 h2
3
= f (a + 2h)— 3f (a) + 2f (a — h) + O(h)
3h2
f" (a) —
as h —> 0.
(iii) and (iv). Denote 1/ — a by h, i.e., let h = 7 — a. Then, a — 0 = Ch.
We write the cubic Taylor approximation (6.13) of 1(x) around the point
a for x = 7 = a + h and for x = (3 = a — Ch and obtain
h2
h3
f (-y) = f (a) + h f' (a) + -T f" (a) + T f (3)(a) + 0 (h4) ;
2
f (0) = f (a) — Ch f' (a) + C 2h2 f " (a)
(6.16)
C6 3
f(3) (a) + 0 (h4) , (6.17)
as h --- 0.
By eliminating from (6.16) and (6.17) the terms containing f" (a) and
solving for f'(a) we obtain the following finite difference approximation:
f r(a) = C2 f (7) - (c2 -1)f (a) - f (0) + 0 (h 2) .
c(c + 1)h
(6.18)
Similarly, we eliminate from (6.16) and (6.17) the terms containing f'(a)
and solve for fll(a) to obtain the following finite difference approximation:
f n (a) — 2 Cf (7) — (C + 1)f (a) + f (0) + 0 (h) .
C (C + 1)h2
(6.19)
Note that, in general, the finite difference approximation (6.18) of f'(a) is
second order, while the finite difference approximation (6.19) of f"(a) is first
order. The finite difference approximation (6.19) of r(a) would be second
order, e.g., if C = 1 or if f (3)(a) = 0.
Also, note that, for C = 1, i.e., if p = a — h and 7 = a + h are symmetric
with respect to the point a, then (6.18) becomes the central finite difference
approximation of f'(a), i.e.,
2
f (a + h) — f (a — h)
+ 0 (h )
2h
The same would not be true for (6.19), which becomes
f'(a) =
f "(a) =
.
f (a + h) — 2f(a) + f (a— h)
+ 0 (h) ,
h2
(6.20)
142
CHAPTER 6. FINITE DIFFERENCES. BLACK—SCHOLES PDE.
instead of the central finite difference approximation
2
„ (a) = f (a + h) — 2f (a) + f (a— h)
0 (h )
h2
of f"(a). This is due to the fact that, for C = 1, the coefficient of h from
O (h) from (6.20) cancels out and the next term of order 0 (h2) becomes
relevant. ❑
f
Problem 6: Consider the following first order ODE:
y/(x) = y(x), V x E [0,1];
y(0) = 1.
(i) Discretize the interval [0,1] using the nodes xi = ih, i = 0 : n, where
Use forward finite differences to obtain the following finite difference
h = 1.
rz
discretization of the ODE:
yi±i = (1 + h)yi, V i = 0 : (n — 1),
with yo = 1. Show that
yi = (1 + h)i, V = 0 : n.
(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 : n. Show that this finite difference discretization is convergent, i.e.,
that
lim max leiI = 0.
i=0:n
Solution: (i) Recall that yo = 1 and yi±i = (1 + h)yi, for all i = 0 : (n — 1).
It is easy to see by induction that yi= (1 + h)i, for all i = 0 : n:
Initial condition: for i = 0, we know that yo = 1 = (1 + h)°.
Induction step: assume that yi = (1 + h)i. Then,
Yi+i = (1 + h)yi = (1 + h)i+1 ,
which is what we wanted to show.
(ii) Let y(x) = ex be the exact solution of the ODE. It is easy to see that the
approximation error ei can also be written as
ei = yi — y(xi) =
(1 + h)i
= eih (ei(ln(l+h)—h)
1)
eih
= ei ln(l+h) — eih
•
6.1. SOLUTIONS TO CHAPTER 6 EXERCISES
143
Using the Taylor approximation In (1 + x) = x - z + 0(x3) we find that
i (1n(1 + h) - h) = ((h - T
h2+ 0(h3))
h2
2
+ i0(h3) = -
h)
h2
+ 0(h2),
2
since ih < 1, for all i = 0 : n. Note that the estimate i0(h3) = 0(h2) is
sharp, since, for i = n, the product ih is equal to ih = nh = 1.
From the Taylor approximation ex = 1 + x + 0(x2), it now follows that
ei
[1n(1+h)-hl
h2
1 = exp (-i —
+ 0(h2)) - 1
2
2
h2
= 1 + (-i-y 0(h2)) + 0 ((-i 2 +0(h2))
2
=
-1
h2
+ 0(h2),
2
since ih < 1 for all i = 0 : n, and therefore
0
h2
-i- + 0(h2))
2
+ 0(h2))
= 0
= 0(h2).
((
Since
eh < e for all i = 0 : n, we obtain that
max lei(
i=0:n
max eih (eiono.+N-h) _ 1)
i=0:n
< e max ei (In(1-Fh)-h) - 1
i=0:n
h2
< e max -i + 0(h2)
i=0:n
z
< e - + 0(h2)
2
= 0(h) = 0 (1)
We conclude that
lim max led =- 0,
n->co i=0:n
and therefore that the finite difference discretization scheme of the ODE is
convergent.
144
CHAPTER 6. FINITE DIFFERENCES. BLACK-SCHOLES PDE.
Note that we actually showed that the finite difference discretization is
first order convergent, i.e.,
11 • 11
= 0 (+
n
max
i=0:n
Problem 7: Consider the following second order ODE:
3x2y"(x) — xy' (x) + y(x) = 0, V x E [0,1];
(6.21)
y(0) = 1;
(6.22)
y(1) = 1.
(i) Partition the interval [0,1] into n equal intervals, corresponding to nodes
Write the finite difference discretization
xi = ih, i = 0 : n, where h
of the ODE at each node xi, i = 1 : (n — 1), using central finite difference
approximations for both yi (x) and y"(x).
(ii) If n = 6, we find, from the boundary conditions, that yo = 1 and y6 =
The finite difference discretization scheme presented above will have five
equations can be written as a 5 x 5 linear system AY = b. Find A and b.
We look for yo, yi, • • • , Yn
Solution: (i) Let xi =ih, i = 0 : n, where h =
such that yiis an approximate value of y(xi), for all i = 0 : n.
By writing the ODE (6.21) at each interior node xi =ih, i = 1 : (n — 1),
we obtain
3x2y"(xi) — xiy/(xi) + y(xi) = 0, V i = 1 : (n — 1).
(6.23)
We substitute the second order central difference approximations for y"(xi)
and y/(xi), respectively, i.e.,
y"(xi ) =
y'(x,) =
y(xj+i) — 2y(xi) + y(xi-i)
0(h2);
h2
y(xi+i) —
2h
0(h 2),
into (6.23), use the approximate values yi for the exact values y(xi), for
i = 0 n, and ignore the 0(h2) term. The following second order finite
difference discretization of (6.21) is obtained:
3i 2h2 Yi+i — 2yi + Yi-1
h2
since xi =ih, which can be written as
(3i2 + 2) yi_i — (67:2-1)yi
(3i2 —
ih
— yi-1
2h
+ yi = 0,
yi±i = 0, V i = 1 : (n-1). (6.24)
6.1. SOLUTIONS TO CHAPTER 6 EXERCISES
145
From the boundary conditions (6.22), we find that yo = 1 and yr, = Z.
(ii) For n = 6, the finite difference discretization (6.24) of the ODE (6.21)
can be written in matrix form as
A Y = b,
where A is a tridiagonal 5 x 5 matrix given by
A(i, i)
= —(6i2 — 1), V i = 1 : 5;
A(i, — 1) = 3i2 +2,
V i = 2 : 5;
V i = 1 : 4,
A(i, i + 1) = 3i2— 2,
i.e.,
0 \
0
0
—5 2.5
0
0
13 —23 11
0
0 28.5 —53 25.5
A=
0
0
50 —95 46
0 77.5 —149
\0
0
and Y and b are the following column vectors:
Y=
( Y1 \
Y2
y3 ; b =
Y4
\y5 /
/
—3.5 \
0
0
.
0
\ —36.25 /
❑
Problem 8: Show that the ODE
y"(x) — 2y'(x) + x 2y(x) = 0
can be written as
Y'(x) = f(x,Y(x)),
where
Y(x) = (y(x)
y/( x))
and
f(x,Y(x)) =
22
Y(x)
Solution: Note that y"(x) = 2y'(x) — x2y(x). Then,
Y'(x) =
yY/11((x )
02
2)
2y/(x '—(xx)2y(x)
02
_
(x)
)
2
) Y(x).
❑
146
CHAPTER 6. FINITE DIFFERENCES. BLACK—SCHOLES PDE.
Problem 9: Consider a six months plain vanilla European call option with
strike 18 on a non—dividend—paying 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 option, i.e, A, F, p, vega, and O.
Use finite differences to find approximate values for the Greeks. Recall that
A—
ac • F
as'
a2c
=
3,52'
p =
ac •
ar '
vega =
ac •
ao-'
8=
ac
aT.
Denote by C(S, K, T, o, r) the value of the call option obtained from the
Black—Scholes formula.
(ii) The forward and central difference approximations Af and A, for 0, and
the central difference approximation F, for F are
Of =
C(S + dS, K,T, a, r) — C (S, K , T, a, r) .
dS
C(S+ dS, K , T, a, r) — C(S — dS,K,T,a,r)
,
2dS
C(S + dS, K,T, a, r) — 2C (S , K,T, c, r)+C(S — dS, K , T, a, r)
F, —
(dS)2
Ac =
Compute the approximation errors for the following values of dS:
Of Ac F, IA
dS
0.1
0.01
0.001
0.0001
0.00001
0.000001
— Af l
IA
— Acl Ir — rcl
(iii) Let do- = 0.0001, dr = 0.0001, and dT = 5
2 2 , i.e., one day. Find the
following forward difference approximations for vega, p, and 8:
vegaf
Pt
Of
C(S, K,T, a + do-, r) — C(S, K,T, a-, r)
doC (S, K , T, a, r + dr) — C (S, K,T, c,r)
,
dr
C(S, K, T + dT, a, r) — C(S, K, T, a, r)
dT
(6.25)
(6.26)
(6.27)
6.1. SOLUTIONS TO CHAPTER 6 EXERCISES
147
Solution: (i) Using the formulas for the Greeks of a plain vanilla call option
derived from the Black-Scholes formula, we find that
A = 0.839523; P = 0.086191;
p = 7.045377;
vega = 0.01501; 8 = -1.394068.
(ii) The Black-Scholes 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
Of
0.843774
0.839952
0.839565
0.839526
0.839522
0.839522
Ac
0.839464
0.839521
0.839522
0.839522
0.839522
0.839522
rc
0.086199
0.086196
0.086195
0.086196
0.086207
0.090594
IA Afl
0.00425158
0.00042959
0.00004228
0.00000349
0.00000037
0.00000076
-
IA - AEI
0.00005840
0.00000138
0.00000082
0.00000081
0.00000081
0.00000081
Ir - rcl
0.0000081!
0.0000042:
0.00000421
0.0000042'
0.0000153
0.0044029
(iii) The finite difference approximations of vega, p and 8 given by (6.25-6.27)
are
vegaf = 3.448385; pf = 7.045624; 81 = - 1.392998.
The corresponding relative approximation errors are
Ivega - vegaf
vega
P.fl
0.000213;
le - ef I - 0.000767.
= 0.000035;
❑
Iel
Problem 10: Show that the value of a plain vanilla European call option
satisfies the Black-Scholes PDE. In other words, show that
ac
at +
52c
2
352
+ (r
os
ac
as
rC = 0,
where C = C(S, t) is given by the Black-Scholes formula.
Solution: Although direct computation can be used to show this result, we
will use the version of the Black-Scholes PDE involving the Greeks, i.e.,
s2r +
e + 1a2
2
(r - q)SA - rC = 0,
CHAPTER 6. FINITE DIFFERENCES. BLACK-SCHOLES PDE.
148
substitute for A,
F and 8 the values
,—q(T—t)
0 = e-q(T-t)N(di); r=
a2
_L
e2
•
SoV271-(T - t)
S e—q(T—t)
8 = qSe-97-0 N(di ) - rKe-''(T-t)N(d2)
e
2 ,
2 V27 (T - t)
and substitute for C the value given by the Black-Scholes formula, i.e.,
C = Se-q(T-t) N(di ) - Ke-r(T-t) N(d2).
Then,
e+
2 cr2S2 r + (r - q)SA - rC
e-q(T-t)
0.252
o-Se-t)
-t) e
2 V271- (T -
rKe-t)
-t) N(d 2)
= qS q(T -t)N (di )
e
2
A
2
2 S0.V27(T - t)
+ (r - q)Se-q(T-t) N(di ) - r (Se-q(T-t) N(di) - Ke-r(T-t) N(d2))
= O. ❑
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
f (S, t) = Se-q(T-t)
Ke-r(T-t).
Show that f (S, t) satisfies the Black-Scholes PDE, i.e., show that
af
at
+ 12
0"
2
f + (r - q)Sas
,52 1
as
- rf = O.
Solution: It is easy to see that
of
at
qSe-q(T-t) -rKe-r(T-0.
o
asf = e—q(T—t),
Then,
a—f
at
+ 21-0- 2 s2a2f + (r - q)S a—f as2
as
rf
—0
82! =
882
6.1. SOLUTIONS TO CHAPTER 6 EXERCISES
= qSe-q(T-t) -rKe-r(T-t) + 0 + (r - q)Se-q(T-t )
r (se- q(T-t) -K e-r (T-0 )
= 0. ❑
149
150
6.2
CHAPTER 6. FINITE DIFFERENCES. BLACK—SCHOLES PDE.
Supplemental Exercises
1. Let f (x) = X 3ex — 65x. Show that the central difference approximation
for f'(x) around the point 0 is a fourth order approximation.
2. Let
f( 5) =
C(K + x,T) — 2C(K,T) +C(K — x,T)
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 : R —4 R,
pc
lim I f(S)g(S) dS = g(K).
xNo _oc
3. (i) Show that the approximate formula
0.282
r
1+
,---,-,, 0
2 8
connecting the F and the 8 of plain vanilla European options is exact if
the underlying asset pays no dividends and if the risk—free interest rates
are zero. In other words, for, e.g., call options, show that, if r = q = 0,
then
1 + 0.252 F(C)
= 0.
2 8(C)
(ii) If q = 0 but r # 0, show that
1 + v.252 F(C)
2 8(C)
1+
1
a
Ni(d2) •
27.07- 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
a252 F(C)
1+
2 O(C)
if the options are at—the—money, 10%, 20%, 30%, and 50% in—themoney, and 10%, 20%, 30%, and 50% out—of—the—money, respectively.
What happens if the asset pays dividends continuously at a 3% rate?
6.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
151
4. Consider a six months 5% in-the-money plain vanilla European call
option with strike 30 on an underlying asset with volatility 20%, paying
dividends continuously at a 2% rate. Assume that the interest rates are
constant at 5%.
(i) Use central differences to compute the finite difference approximations A, and F, for A and F, respectively, i.e.,
A, =
C(S + dS) -C(S-dS) .
2dS
C(S + dS) - 2C(S) + C(S - dS)
(dS)2
for dS = 10-i with i = 1 : 12, where, e.g., C(S + dS) = C(S +
r) denotes the Black-Scholes 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 Black-Scholes
formula, and the approximation errors IA, - AI and Ire — rl. Note
that these approximation errors stop improving, or even worsen, as dS
becomes too small. How do you explain this?
6.3
Solutions to Supplemental Exercises
Problem 1: Let f(x) = x3ex - 6ex. Show that the central difference approximation for f'(x) around the point 0 is a fourth order approximation.
Solution: Recall that, in general, the central difference approximation of the
first derivative is a second order approximation, i.e.,
(h) )
f (-h
+ 0 (h2) , as h
0.
(6.28)
To see why, for the function f (x) = x3ex - 6ex, the central difference
approximation for f'(x) around the point 0 is a fourth order approximation,
we investigate how the approximation (6.28) is derived.
The Taylor approximation of f(x) around the point 0 for n = 5 is
x3
x4
x2
(5)(0)
f"(0) + — f (3)(0) + — f(4)(0) +
f (x) = f (0) + xf(0)+ —
24
120 f
6
2
(6.29)
+ 0 (x6) , as x 0.
CHAPTER 6. FINITE DIFFERENCES. BLACK—SCHOLES PDE.
152
We let x = h and x = —h in (6.29) and sum up the two resulting formulas.
After solving for /(0) we obtain
f (h) — f (—h)
t(0) =
h2
h4
f (3)(0) —
f (5) (0) + 0 (h5) ,
6
120
2h
(6.30)
as h —› 0.
For f (x) = x3ex — 6ex, we find that f (3) (x) = (x3+ 9x2 +18x)ex, and thus
that f (3) (0) = 0. Also, f (5)(0) = 54 0 and (6.30) becomes
f(0)
f(h)
—
f(—h)
9h4
f (h) — f (—h)
0 (h5) =
+ 0 (h4) ,
20 +
2h
2h
as h —> 0. In other words, the central difference approximation for f'(x)
around the point 0 is a fourth order approximation.
❑
Problem 2: Let
f (S) =
C(K + x,T)-2C(K,T)+C(K — x,T)
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 : R —> R,
lim f f(S)g(S) dS = g(K).
s\,,o _pc
(6.31)
Solution: From the definition of f (S), it is easy to see that
f (S) =
1
(max(S — (K — x), 0) — 2 max(S — K, 0) + max(S — (K + x), 0))
0,
S-(K-x)
if 0 < S < K — x;
if K — x < S < K;
ifK 0 arbitrary. Since g is continuous, it follows that there exists S > 0
such that Ig(K) — g(r)1 < E for all 7 such that 1K — 71 < 6. Let x E (0,8)
and y E (0, 1). Then (K — (K — x + xy)1= x(1 — y) < 6 and therefore
1g(K) — g(K — x + xy)I < 6, V <
X
<
0 < y < 1.
Therefore, it is easy to see that, for any 0 < x <
1
31yg(K — x + xy)dy — g(K) I ydy <
ylg(K)— g(K — x + xy)Idy
1
<
y dy = — .
2
e f
In other words, for any € > 0 there exists d > 0 such that
foi
i
yg(K — x + xy) dy — g(K) f y dy < 2' V 0 < x < 6.
o
Then, by definition,
11
urn f yg(K — x + xy) dy — g(K) f y dy = O. 1=1
x\,0
Problem 3: (i) Show that the approximate formula
0.282
1+
2
r
0
154
CHAPTER 6. FINITE DIFFERENCES. BLACK-SCHOLES PDE.
connecting the F and the 8 of plain vanilla European options is exact if the
underlying asset pays no dividends and if the risk-free interest rates are zero.
In other words, for, e.g., call options',
a 2s2
r(c)
1 +
= O.
2 8(C)
(ii) If q = 0 and r 0, show that
1+
o-2S2 F(C)
8(C)
2
1
cr
1+
N'(d2) •
2rfT 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
0.2s2 F(C)
1+
2 8(C)
if the options are at-the-money, 10%, 20%, 30%, and 50% in-the-money,
and 10%, 20%, 30%, and 50% out-of-the-money, 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
e-gT
d2
e2 •
F(C) =
o- Sf27rT
S e-qT _
8(C) =
e 2+
2,/21-T
where d1= (ln
+ (r +1)
(6.34)
17Se—qTN(d1)
rKe-TTN(d2),
/ (oVT) and d2 = d1 - °VT.
(i) For r = q = 0, we obtain from (6.34) and (6.35) that
F(C)
=
1
USV27rT
e
_,11
;
8(C) =
v5 e _‘11
2
2-/27T
Then,
0-282 F(C)
2 ) = 0.
2 ( o- 2S2
2
e(C) = 1 + v.252
'Note that; if r = q = 0, then F(P) = r(c) and e(P) = e(C).
1+
(6.35)
6.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
155
(ii) For q = 0, we obtain from (6.34) and (6.35) that
1
_
e 2
aSA/21-T
crS e _ 2
2-V27T
r(c) =
e(C)
rKe-rTN(d2).
-
Then,
1+
o-282 r(c)
2 e(C)
0.s
=
p-
2V277
-T -
1
d2
2
k
rT e--21-
rKe-rTN(d2)
rKe-rT N(d2)
crs
64
rKe—rTN(d2)
1
1 + 27-VT
S
1 0-1221
Ke-rT N(d2)
1
1 ± 2rff
SI\P(di )
Ke-'2'N(d2)
(6.36)
t2
since N'(t) = -1
-e-T for all t E R.
V-27r
Recall that the "magic" of Greek computations is due to the following
result:
-= Ke-rT N'(d2);
S
cf. Lemma 3.15 of [2] for q = 0. Then, (6.36) becomes
1+
v.2 52 F(C)
2
e(c)
1
1+
2r-
a
(iii) Let S = 50 , T = 0.5, a = 0.3, and r = 0.04. The table below records
the values of
r(c)
,252
1+
2 O(C)
(denoted by "Value") both for q = 0, and for q = 0.03, for the following
values of the moneyness of the option:
s
= {1,1.1,1.2,1.3,1.5, 0.9, 0.8, 0.7, 0.5},
corresponding to call options that are at-the-money, 10%, 20%, 30%, and
50% in-the-money, and 10%, 20%, 30%, and 50% out-of-the-money, respectively:
156
CHAPTER 6. FINITE DIFFERENCES. BLACK-SCHOLES PDE.
S/K 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
0.4711
0.7361
0.1412
0.1068
0.0822
0.0505
We note that the approximation
0.2 s2
1+
r
2 8
:,---,
0
is better for deep out-of-the-money options (corresponding to small values
of S/K) and is worse for deep in-the-money options (corresponding to large
values of S/K). Also, for this particular case, the approximation is more
accurate if the underlying asset pays dividends.
❑
Problem 4: Consider a six months 5% in-the-money 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 compute the finite difference approximations A,
and r, for A and F, respectively, i.e.,
C(S + dS) - C(S - dS).
2dS
C(S + dS) - 2C(S) + C(S - dS)
F, =
(dS)2
A, =
for dS = 10-' with i = 1 : 12, where, e.g., C(S + dS) = C(S + dS, K, T, o, r)
denotes the Black-Scholes 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 Black-Scholes formula, and the approximation errors IA, - AI and I Fc - F. Note that these
approximation errors stop improving, or even worsen, as dS becomes too
small. How do you explain this?
6.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
157
Solution: The spot price S = 31.5 corresponds to a 5% ITM call with K = 30.
We find that A = 0.692130579727 and r = 0.077379043990.
The central finite difference approximations A, and the approximation
errors IA, - AI are recorded in the table below:
dS
0.1
0.01
0.001
0.0001
10-'
10-6
10-7
10-8
10-9
10-10
10-11
10-12
Ac
0.692112731743
0.692131730564
0.692131920566
0.692131922513
0.692131922087
0.692131918000
0.692131916226
0.692131862934
0.692132573477
0.692104151767
0.691890988946
0.687450096848
iAc
-
AI
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 10-6of A. However, for values
of dS smaller than 10-9, the finite difference approximations deteriorated very
quickly.
To explain this phenomenon, denote the exact value2 of Delta by -exact
A •
Note that the value of A is given by the Black-Scholes formula, i.e.,
A = ABS -= e-gT N (di).
This value is computed using a numerical approximation of N(di) that is
accurate within 7.5 • 10-7; cf. [1], page 932. In other words, we only know
that
(6.37)
IBS - exact' < 10-6 .
When computing the finite difference approximation Ac, we use a numerical estimation of the Black-Scholes 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,
A, =
2 Note
CBs(S + dS)
- CBs(S - dS)
2dS
that Aexact is a theoretical value, and is not the 0 from the table above.
(6.38)
158
CHAPTER 6. FINITE DIFFERENCES. BLACK-SCHOLES PDE.
Denote by Cexact(S +dS) and Cexact(S - dS) the exact values of the options.
Since the central finite difference is a second order approximation, it follows
that, for exact values of Delta and of the call options,
Aexact
Cexact(S
-
+ dSexact(S
2dS
- dS) + 0 ((dS)
2) ,
(6.39)
as dS -± 0.
Since CBS(S) = Se-gT N(di) - Ke-rT N(d2), and since N(di ) and N(d2)
are computed numerically within 10-6of their exact value, it follows that
IC
B
Cexact(S +
a10-6;
+ dS) - (7
CBs(S
ICBs(S - dS) - (7
Cexact(S
(S - dS)1
dS)1 < a10-6,
(6.40)
(6.41)
where a is a constant proportional to the values of S and K,
Using (6.38) and(6.39) we find that
Ac - ABS = (Ac — Aexact) + (Aexact - ABS)
+ dS)
= CBS(S + dS) - Cexact(S
2dS
CBs(S - dS) - Cexact(S - dS)
2dS
+ Aexact - ABS + 0 ((dS) 2) ,
(6.42)
as dS -4 0.
The only estimate we can find using (6.37), (6.40), (6.41), and (6.42) for
the approximation of ABS by Ac as dS —> 0 is
CBS(S + dS) - Cexact(S
exact(S + dS)I
2dS
ICBs(S - dS) - Cexact(S
0
- dS)I
+
2dS
+ IAexact — ABS' + 0 ((dS)2)
iAc - ABSI
<
-
a10-6
+ 10-6 + 0 ((dS)2) ,
dS
(6.43)
as dS -- 0.
While the approximation error IA, - Almay be better in practice, the
bound (6.43) provides the intuition behind the fact that, for dS too small, the
numerical approximation error IA, - Al = lAc - ABSI deteriorates as alN6
becomes large.
The central finite difference approximations Pcand the approximation
errors lc, - 11 are recorded in the table below:
6.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
159
dS
r,
IF, - ri
0.1
0.077370009586 0.000009034404
0.01
0.077371495486 0.000007548504
0.001 0.077371502982 0.000007541008
0.0001 0.077371709040 0.000007334951
10-5
0.077271522514 0.000107521476
10' 0.074606987255 0.002772056735
10-7 -0.355271367880 0.432650411870
10' 71.054273576010 70.976894532020
For dS < 10-9, the values of 1 increased dramatically, reaching 109 for
dS = 10-12, and were no longer recorded. The finite difference approximations of F became more precise while dS decreased to 10-4, but were much
worse after that; the best approximation was within 10-5of F. The reason
for this is similar to the one explained above for the finite difference approximations of A. ❑
Chapter 7
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=
Scr-V27T
exp
(d1(2S))2 )
(7.1)
where
(S) =
(7.2)
crA/T
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
lim r(S) = 0 and lim F(S) = 0.
(7.3)
s\o
s-.00
Solution: From (7.1) we find that F can be written as
F(S) —
oV27T
exp
(i(2
c/ S))2
ln(S)) ,
(7.4)
where di (S) is given by (7.2).
Since F(S) > 0, it follows that the functions F(S) and ln(F(S)) have the
l given by
same monotonicity intervals. Let f : (0, oo)
f(S) = ln(F(S)) -=
(di(S ))2
ln(S) ln(o- 271-T).
2
✓
161
CHAPTER 7. MULTIVARIABLE CALCULUS.
162
Then,
f'(S) = —cli(S)
—
-
a(di(s))
1dl(S)
as
S
f cii (s)
S
1± a, ) •
so--vT
1
S
(7.5)
Recall from (7.2) that
ln (i) + (r + '20 T
dl (S) =-
o-V7'
It is easy to see that dl (S) is an increasing function of S and that
(7.6)
lim di (S) = —co; Ern di(S) = co.
S-*x
S \O
From (7.5) we find that f(S) has one critical point, denoted by S*, with
di (S*) = —afft From (7.5) and (7.6) it follows that f'(S) > 0 if 0 < S < S*
and f'(S) < 0 if S* < S.
In other words, the function f(S) = ln(F(S)) is increasing when 0 < S <
S* and is decreasing when 5* < S. We conclude that F(S) is also increasing
when 0 < S < S* and decreasing when S* < S.
We now compute lims\o F(S) and lims,,GF(S).
Note that lims,c/i(S) = oo. Therefore,
lim F(S)
s--40c
=
lim
S--,x
1
exp ( (d1(S))2
o -V27rT
-
2
ln(S)) = 0.
From (7.2), and using the fact that lims\o ln(S) = —co, it follows that
m
S O
(d,(s)2
2
ln(S)
= m
s o
—(1n(S))2
(ln(S) — ln(K) + (r + 1 ) T) 2
2o 2T(ln(S))2
1
= 2o-2T •
Since lims\o (exp (—(1n(S))2)) = 0, we obtain that
lim exp (
s\n
—((S))2
c/i
2
ln(S)) = 0.
1
+
ln(S)
7.1. SOLUTIONS TO CHAPTER 7 EXERCISES
163
from (7.4), we conclude that
lim c(S) = lsirn
1
exp (
s\o
\o 0-V27T
(di(S))2
2
ln(S)) = O. 11
Problem 2: Let D be the domain bounded by the x—axis, the y—axis, and
the line x + y = 1. Compute
f fD xxy
y dxdy.
(7.7)
Solution: Note that
x > 0,y > 0,x +y < 1}.
We use the change of variables s = x + y and t = x — y, which is equivalent
D = {(x,y)
to
s— t
;
Y
=
2
2
It is easy to see that (x, y) E D if and only if (s, t) E S2, where
s+t
= {(s, t) 0 < 8 < 1, -8 < t < 8}.
The Jacobian of the change of variable (x, y) E D --+ (s, t) E C2 is
dxdy =
ax ay
as at
ax ay
at as
dsdt =
1
— dsdt,
2
and therefore
I fpxx Yy
t 1dsdt = f 1 (f —t dt) ds
dxdy = f f _
2
=
1
—
2 0 s
o
2s
t dt) ds = 0.
The integral (7.7) can also be estimated directly as follows:
_ y dx
(
x
—y
dy
fp x+y dxdy =
Jo Vo x +
)
= fo ((x — 2y ln(x + Y))i s11Y dy)
= f 1— y + 2y ln(y) dy
y=1
Y2 ± Y2in(Y)) ly=0
= 0,
CHAPTER 7. MULTIVARIABLE CALCULUS.
164
since limy \ o y2 ln(y) = 0.
❑
Problem 3: Use the change of variables to polar coordinates to show that
the area of a circle of radius R is 7rR2, i.e., prove that
I
1 dxdy = 7R2.
ID(O,R)
Solution: We use the polar coordinates change of variables
(x, y) =- (r cos 0, r sin 0) with (r, 0) E S2 = [0, R] x [0, 2r).
Recall that dxdy = rdOdr. Then,
I
R
1 dxdy =
ID(O,R)
f 27
o
r &kir =
which is equal to the area of a circle of radius R.
f rdr = irR2,
❑
Problem 4: Let V(S, t) = exp(—ax — bT)u(x, 7), where
x = In ( S
K
T
(T —t)a2
2
2
r —q 1
(r — q
2q
= a2
b
=
a=
+2
2,
u2_
2+ a 2.
This is the change of variables that reduces the Black—Scholes PDE for V(S, t)
to the heat equation for u(x,r).
(i) Show that the boundary condition V(S,T) = max(S — K, 0) for the European call option becomes the following boundary condition for u(x, r) at
time T = 0:
u(x,0) = K exp(ax) max(ex — 1,0).
(ii) Show that the boundary condition V(S,T) = max(K — S, 0) for the
European put option becomes
u(x, 0) = K exp(ax) max(1 — ex, 0).
Solution: Note that t = T if and only if r = 0. Then,
V(S,T) = exp(—ax)u(x, 0).
Here, x = In (i), which can also be written as S = Kex.
(7.8)
7.1. SOLUTIONS TO CHAPTER 7 EXERCISES
165
(i) For a call option, V(S,T) = max(S — K, 0). From (7.8), we find that
u(x, 0) = exp(ax)V(S, T) = exp(ax) max(S — K, 0)
= exp(ax) max(Kex — K, 0) = Kexp(ax) max(ex — 1, 0).
(ii) For a put option, V(S,T) = max(K — S, 0). From (7.8), we find that
u(x, 0) = exp(ax)V(S, T) = exp(ax) max(K — S, 0)
= exp(ax) max(K — ice, 0) = Kexp(ax) max(1 — ex, 0). ❑
Problem 5: Solve for a and b the following system of equations:
{2a + 1 2(r--2.q)
b+a2+a(1 2(r-2q)) (2r
72
,
=
-
0;
0.
Solution: From the first equation, it is easy to see that
r— q
a = 0-2
1
2
(7.9)
Using (7.9), we note that the second equation can be written as
b = —a2
a
2( r — q)) 2r
+7
( -2
0-2
1
(r — q 1)2
2
a2
2r
(r — q 1) 2 (1 r — q)
+ u2
2
2
o-2
2
a
1
\2
2r
(r — q 1)2
+
2
(
=
+
0
-2
a2
2
r a2 q
2)
2
(r — q 1) Zr
=
0-2
2
+02
=
(r — q 1\2(r — q)
a2
0-2 + 2 )2
(r — q 1)2 2q
=
+ -c;-2-- ' ❑
a2 + 2
=
2r
± u2
Problem 6: Assume that the function V(S, I, t) satisfies the following PDE:
av
av
av 1 2 2 a2v
+ swi + -2-0- 5 ' 85,2 + rSy-s- — rV =
w
.-
0.
(7.10)
CHAPTER 7. MULTIVARIABLE CALCULUS.
166
Consider the following change of variables:
V(S,I,t) = S H(R,t),
where
R = —.
s
(7.11)
Show that H(R, t) satisfies the following PDE:
8H
1
at + 2 Q
a2H
aR2
8R2
+
(1 — rR)
ax
aR
= 0.
(7.12)
Note: An Asian call option pays the maximum between the spot price S(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 (S(T) —
7,
1
I
T Ser) dr) .
Thus, the value V(S, I, t) of an Asian option depends not only on the spot
price S of the underlying asset and on the time t, but also on the following
random variable:
I(t)
f S(T)
dr.
It can be shown that V(S, I, t) satisfies the PDE (7.10). Similarity solutions
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.
Solution: Let V(S, I, t) = S H(R, t), with R = S. Using Chain Rule, we find
that
av
aH .
at = S at'
at '
avax
aR an
=
S
alan alan'
av = H+S ax an
an I
= H+s
as
aR as
aRs2)
Can=H—R
aR"
a2vax
aR aR an
=
as2
aR as
R2 a2n
S aR2'
as aR
R
82H aR
aR2 as
= R
u I an
11 s an
182H
s2 aR2
7.1. SOLUTIONS TO CHAPTER 7 EXERCISES
167
By substituting into (7.10), it follows that
0 =
ay
_ay + i a,s2a2v
ay
+ rS— — rV
at + '5 al
2
as2
as
pH
1
R2 52.11
+ 2s2
II ) — rSH
S— + s
+ rS (H — R P--at
at
aR
2
s DR2
aR
S pH
1
521-1
ail
+ -2sR2
0+ S(1 — rR)
=
at
2
aR2
aR•
By dividing by S, we conclude that H(R, t) satisfies the PDE
pH
1
a2H
ail
(1-rR) — = 0,
at ± ia2R2DR2 +
aR
which is the same as (7.12).
❑
Problem 7: The price of a non-dividend-paying 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 Black—Scholes values of the ITM
put options on 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 — S, 0) larger
than the Black—Scholes value of the option (in which case the corresponding
American put is guaranteed to be worth more than the European put)?
Solution: The values of the out options are summarized in the table below:
Option Type Strike Maturity Value K — S
5
45
6 months 5.8196
Put
5
3 months 5.3403
Put
45
6
months
8.0325
8
48
Put
3 months 7.8234
8
48
Put
10.4862
11
6
months
Put
51
11
51
3 months 10.5476
Put
In general, the values of deep—in—the money European put options are
lower than the premium K — S. This feature was observed for the options
priced here: the values of the 51—puts and of the three months 48—put are
below their intrinsic value K — S.
Also, note that for the 51—puts (i.e., for deep in the money puts), the
values of the short dated options are higher than the values of the long dated
CHAPTER 7. MULTIVARIABLE CALCULUS.
168
options. This is to be expected; for example, if the spot price is 0, the value
of a European put is Ke-rT, in which case longer dated puts are worth less
than short dated ones.
❑
Problem 8: Show that the premiumsof the Black-Scholes value of a European call option over its intrinsic value max(S - K, 0) is largest at the money.
In other words, show that the maximum value of
CBS(S) — max(S - K, 0)
is obtained for S = K, where CBS(S) is the Black-Scholes value of the plain
vanilla European call option with strike K and spot price S.
Solution: Let f (S) = CBS(S) - max(S - K, 0). It is easy to see that
f (S)
CBs(S),
if S < K;
CBS(S) - S + K, if S > K.
Note that f(S) is a continuous function, but it is not differentiable at S = K.
For S < K, the function f(S) is the value of a call with strike K, and
therefore is increasing.
For S > K, we find that
f'(S) = .(CBs ) - 1 = e-9T N(d1) - 1 < N(di) - 1 = -N(-d1) < 0,
and therefore the function f (S) is decreasing.
We conclude that f (S) has an absolute maximum point at S = K.
❑
Problem 9: Use the formula
B 2a c
V(S,K,t) = C(S,K,t) - (—
S
K ,t) ,
(7.13)
where a = a2a -1, to find the value of a six months down-and-out call on a
non-dividend-paying 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 risk-free interest rate is constant at 5%.
Solution: The value of the down-and-out call is $3.398883.
❑
Problem 10: Show that the value of a down-and-out call with barrier B
less than the strike K of the call, i.e., B < K, converges to the value of a
'This premium is also called the time value of the option.
7.1. SOLUTIONS TO CHAPTER 7 EXERCISES
169
plain vanilla call with strike K when B
0. For simplicity, assume that the
underlying asset does not pay dividends and that interest rates are zero.
Solution: Let t = 0 in formula (7.13). For r = q = 0, we find that a =
Therefore, the value of the down—and—out call is
"Be
V(S) = C(S) — S C
B
S)
S B2N(d
1) KN(d2))
= C(S) — —
B S
SK
N(d2),
= C(S) — BN(di) +
(7.14)
where C(S) is the value of the plain vanilla call with strike K and
In ('4
s + 4T
(7\5
.,
in (sec) _
and d2 =
a--\5"
•
Note that 0 < N(di) < 1. Then,
urn BN(di) = 0.
.B\o
(7.15)
Using l'HOpital's rule, we obtain that
ad 2
SK
N(d2)
N(d2) = SK lim B = SK Bli\MoN1(d2)
lim
aB
B\c) B
1
2
mo
= S K Bli\
2/r exp ( 2 ) Baff
d2
2SK
exp
— ln(B))
2
0- /-27T B
4■
As B
0, the term
(7.16)
— ln(B) is on the order of —(ln(B))2, and therefore
lim
B\O
d2
— ln(B)) = — oo.
2
(7.17)
From (7.16) and (7.17) we find that
lim
B\o
SK
N(d2) = 0,
(7.18)
and, from (7.14), (7.15), and (7.18), we conclude that
SK
Br V(S) = lim (C(S) — BN(cli) + — N(d2)) = C(S). ❑
B
/3\0
B\O
CHAPTER 7. MULTIVARIABLE CALCULUS.
170
Problem 11: Compute the Delta and Gamma of a down-and-out call with
B < K.
Solution: We rewrite formula (7.13) to emphasize the dependence of the value
V(S) of the down-and-out call on the spot price S of the underlying asset
as follows:
B 2a
( B2
(7.19)
V(S) = CBS(S)
CBS -,75T •
By differentiating (7.19) with respect to S, we obtain that
A(V)
2aB2a
(B2 \
s2ad-i CBS S
(CBs)(S)
2a+2
B
S2a+2
p(CBs) (Bs2)
where A(CBs)(x) = e-qT N(di(x)), with
ln (k) + (7.
dl (x) =
q + 2) T
(7.20)
o- VT
Similarly,
2a(2a + 1)B2a
r(v) = r(cBs)(s)
(4a + 2)B2a+2
S2a+3
where
r(CBs)
CBS
S2a+2
(B2 )
A(CBS)
x)
with di (x) given by (7.20).
xcr-/2T-T
B2a+4
S2a+4
e —gT
(
(B2)
exp
r(CBS)
cli (2x)2 )
7.2. SUPPLEMENTAL EXERCISES
7.2
171
Supplemental Exercises
1. Compute
I
where
fox dxdy,
D = {(x,y) ER2I x > 0, 1 < xy < 2, 1 < y < 2}.
x —
Hint: The change of variables s = xy and t = ! maps the domain D
into the rectangle [1,2] x [1,2].
2. Which number is larger, elr or Ire?
3. Let u, v : [0, oo) —4 [0, oo) be two continuous functions with positive
values. Assume that there exists a constant M > 0 such that
x
u(x) < M + f u(t)v(t) dt, V x > 0.
o
Show that
u(x) < M exp (f v(t) dt) , V x > O.
o
Hint: Investigate the monotonicity of the function
x
x
(M + f u(t)v(t) dt) exp (— f v(t) dt) .
o
o
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 down—and—out
call with barrier B and rebate R > 0 correspond to for the function
u(x,r) defined as follows: V(S,t) = exp(—ax — br)u(x,T), where
/S
x = ln
K)
C—
(T — t)a2
7=
2
2
r— q 1
r — q 1) 2q
a = 0-2
a2
2 , b = ( 0.2 +2 +—.
CHAPTER 7. MULTIVARIABLE CALCULUS.
172
5. Assume that the function V(S, I, t) satisfies the PDE
1 2 2 a2v
av
av
In
+
+n
+ s
at
av
— rV = 0.
(7.21)
Consider the following change of variables:
= I + (T — t)inS
V(S, I, t) = F(y,t), where y
T
Show that F(y, t) satisfies the following PDE:
aF
0.2(T
at + '2T2
,92F
ay2
C
rr
2
T — t aF
T ay
rF = 0.
Note: The values of Asian options with continuously sampled geometric
average satisfy the PDE (7.21).
6. One way to see that American calls on non—dividend—paying assets are
never optimal to exercise is to note that the Black—Scholes value of the
European call is always greater than the intrinsic premium S — K, for
S > K.
Show that this argument does not work for dividend—paying assets. In
other words, prove that the Black—Scholes value of the European call
is smaller than S — K for S large enough, if the underlying asset pays
dividends continuously at the rate q > 0 (and regardless of how small
q is).
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
az,p(K) as the unique solution to
C(K) = CBs(K,o-imp(K)),
where CBS(K, uimp(K)) = CBS(S, K, T, crimp(K), r, q) represents the
Black—Scholes value of a call option with strike K on an underlying
asset following a lognormal model with volatility 72,,,p (K). Find an
implicit differential equation satisfied by uznip(K), i.e., find
acrimp(K)
ax
as a function of aimp(K).
7.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
7.3
173
Solutions to Supplemental Exercises
Problem 1: Compute
I fox dxdy
where
{(x,E
x—
Solution: The change of variables s = xy and t = is equivalent to
X =-
and y =
t
when x > 0 and y > 0. This change of variables maps the domain D into
the rectangle S2 = [1,2] x [1, 2]. It is easy to see that
dxdy
ax ay
ax ay
as at at as dsdt
1
(
2t-vi) 2.\/75
dsdt
1
dsdt.
= 2t
Then,
II
f 2 f2e
dxdy, =
t
A.
2t
d dt
8
(1
2 N/79 d8) (f2
1 (23 2
2 3s2 1
—6
3
dt)
t1it-
( 22
. o
Problem 2: Which number is larger, e or re?
Solution: We show that 776 < e.
By taking the natural logarithm, it is easy to see that
Ire <
eln(7r) < 7r < >-
1n(r) < 1 = ln(e)
e
e
7r
(7.22)
.
CHAPTER 7. MULTIVARIABLE CALCULUS.
174
Let f(x) =-- 11x 1with f : (0, oo) —> R. Then 1(x) = 1-x2(x) . The function
f (x) has one critical point corresponding to x = e, is increasing on the interval
(0, e) and is decreasing on the interval (e, oo).
We conclude that f (x) has a global maximum point at x = e, i.e., f (x) <
f(e) = e for all x > 0 with x e, and therefore
f(7) =
ln(7r) < 1
7r
e,
❑
which is equivalent to Ire < er; cf. (7.22).
Problem 3: Let u, v : [0, oo) —› [0, oo) be two continuous functions with
positive values. Assume that there exists a constant M > 0 such that
u(x) < M + f u(t)v(t) dt, V x > 0.
(7.23)
Show that
u(x) 5 M exp (f v(t) dt) , V x
Solution: Define the function w : [0, oo)
0.
(7.24)
[0, oo) as
w(x) = (M + fo x u(t)v(t) dt) exp (— I v(t) dt) .
(7.25)
Recall that
CJ
fo
x
u(t )v(t) dt)
u(x)v(x );
(f
v(t)
dt )/ = v(x),
0
where the derivative is computed with respect to x.
Using the Product Rule, we find that
w'(x)
u(x)v(x) exp (— fo x v(t) dt)
+ (M + fo xu(t)v(t) dt) (—v(x) exp (— of x v(t) dt))
v(x) (u(x) — M— fo x u(t)v(t) dt) exp (— fo x v(t)dt) (7.26)
Using (7.23) and the fact that v(x) > 0 for all x > 0, we conclude from (7.26)
that
w'(x) < 0, V x > O.
7.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
175
In other words, w(x) is a decreasing function on the interval [0, oo) and
therefore w(0) > w(x) for all x > 0. Since w(0) = M, and using (7.23), it
follows that
M > (M + f u(t)v(t) dt) exp (— f x v(t) dt)
> u(x) exp (— f v(t) dt)
which is equivalent to (7.24).
x > 0,
❑
Problem 4: What does the boundary condition V(B,t) = R for a downand—out call with barrier B and rebate R> 0 correspond to for the function
u(x,r) defined as follows: V(S, t) = exp(— ax — br)u(x, r), where
r — q 11 2 2q
r— q 1
b
=
x = ln S
T
(T t)a2 a = u 2
2>
u 2 +2 + 2
2
Solution: Note that S = B corresponds to x = In
corresponds to 0 < T < 74' Since
(f)
and 0 < t < T
u(x, r) = exp(ax + br)V(S, t),
the boundary condition corresponding to V(B, t) = R for all 0 < t < T is
B +br) V (B ,t)
u (ln ( B ) ,r) = exp (a In (-k
= (—KB Y ebr R, V 0 < < T2
2
Problem 5: Assume that the function V(S, I, t) satisfies the PDE
ay
, ay
2
+ mS ai + 2 a
_2(3592v + _ay
,2
— rV = 0.
Consider the following change of variables:
V(S, /,t) = F(y,t), where y = /±(T—t)lnS
T
Show that F(y, t) satisfies the following PDE:
(
0-2(T — t)2 a2F
OF
2T2 a +
at +
0-2 T— t OF
) T ay
rF = 0.
(7.27)
CHAPTER 7. MULTIVARIABLE CALCULUS.
176
Solution: Using chain rule, it is easy to see that
av _— aF lns aF .
at
T ay'
at
av _ i. aF
Tay '
al1 T - t aF
av
S T ay ;
as
a2v1 RT — t)2 a2F
T2 ay2
352
52
T - tan
T ay ) •
Then, the PDE (7.27) for V(S,/,t) becomes the following PDE for F(y,t):
av 1 , a2v
av
+ r.5...s, - rV
+ 2 a2'-'2 aS2
1 aF
OF
In 5
=
+
T ay
T ay
at
1 2 ((T -t)2 82F T - t aF)
T - t aF
rF
±r
+ (7T2 ay2
T ay
T ay
aF
02(T — t)2 02 F
( 6,2 T - t OF
rF. ❑
=
2T
2 ay2 +
2 ) T ay —
at ±
av
0 = .-,
+ 1nS-,
id/
In s aF
Problem 6: One way to see that American calls on non-dividend-paying
assets are never optimal to exercise is to note that the Black-Scholes value
of the European call is always greater than the intrinsic premium S - K, for
S > K.
Show that this argument does not work for dividend-paying assets. In
other words, prove that the Black-Scholes value of the European call is
smaller than S - K for S large enough, if the underlying asset pays dividends continuously at the rate q > 0 (and regardless of how small q is).
Solution: We want to show that, if the dividend rate of the underlying asset
is q > 0, then CBS(S, K) < S - K for S large enough.
Note that
CBs(S,K) = Se-qT N(di) - Ke'T N(d2) < Se-4T ,
since N(d1) < 1 and N(d2) > 0.
If Se-qT < S - K, which is equivalent to S > 1 1,47-7, > 0 since q > 0, it
follows that CBS(S, K) < S - K. We conclude that
CBS(S, K) < S - K, V S >
K
1- e-qT '
7.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
177
❑
which is what we wanted to show.
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 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 crimp(K) as the
unique solution to
C(K) = CBs(K, aimp(K)),
where cgs (K , tiimp(K)) = CBs(S, K, T, o-imp(K), r, q) represents the BlackScholes value of a call option with strike K on an underlying asset following a
lognormal model with volatility aimp(K).Find an implicit differential equation satisfied by climp(K), i.e., find
ao-imp (K)
ax
as a function of crimp(K).
Solution: We first find the partial derivative of the Black-Scholes value
CBS(K) of a call option with respect to its strike K. Recall that
CBS(S, K) = Se-qT N (di ) - Ke-rT N(d2).
Then,
8CBS
OK
= Se-qT N' (di )
ad,
aK
e-rT N (d2)
ad, (7.28)
K CrT Ni(d2) aK .
(
Also, recall that
Se-gT Ni (di ) = K e'T (d2);
cf. Lemma 3.15 of [2]. From (7.28) and (7.29), we find that
acBs
ax
e _rT N (ot
,
-2 )
Ke'T (d2)
(ad,
ax- ax ).
(7.29)
(7.30)
Since di. = d2 + c•-VT, it follows that
a(d2+ °VT) _ ad,
ad,
ax
ax
aK .
We conclude from (7.30) that
aCss = - e-rT N(d2).
OK
(7.31)
CHAPTER 7. MULTIVARIABLE CALCULUS.
178
We now differentiate with respect to K the formula
C(K) = CBs(K,a,mp(K))
which is the definition of crimp(K). Note that C(K) is assumed to be known
for all K, as it the market prices. Using Chain Rule and (7.31) we find that
acacBs acBs ao-imp(K)
ax
axaoax
( K)
= —e'TN(d2) + vega(CBs) acri,
aKP
(7.32)
where
vega(CBs) =
=
aCBs
a°. =Se-4T
Te
27r
2
= .VT'Se-qT N/(di)
Ke-rT (d2) = K
cf. (7.29). We conclude that the implied differential equation (7.32) can be
written as
Ke'
T e _4
aorimp(K)
ac
2
=
27r
ax
aK
_T
e r N(d2),
with
d2
In (Tsc-) + (r -q)T
crimp (K).VT
a-imp(K)VT
2
❑
Chapter 8
Lagrange multipliers. N— dimensional
Newton's method. Implied volatility.
Bootstrapping.
8.1
Solutions to Chapter 8 Exercises
Problem 1: Find the maximum and minimum of the function f (xi, x2, x3) =
4x2— 2x3 subject to the constraints 2xi — x2 — x3 = 0 and 4 + 4 = 13.
Solution: We reformulate the problem as a constrained optimization problem.
Let f : R3—± R and g : R3—* 1(8 be defined as follows:
f (x) = 4x2— 2x3;
g(x) =
2x1 - X2 - X3
4 + 4 — 13 ) '
where x -= (x1, x2, x3). We want to find the maximum and minimum of f (x)
on 1183subject to the constraint g(x) = 0.
We first check that rank(Vg(x)) = 1 for any x such that g(x) = 0. Note
that
—1 —1
V g(x) =
)
22x, 2x2
It is easy to see that rank(Vg(x)) = 2, unless x1 = x2 = 0, in which case
g(x) O.
The Lagrangian associated to this problem is
F(x,
= 4x2— 2x3 +
(2x1 — x2 — x3) + A2(x1 +
4
- 13),
(8.1)
where A =
A2)t E R2is the Lagrange multiplier.
We now find the critical points of F(x, A). Let xo= (x0,1, x0,2, x0,3) and
= (A0,1, Ao 2) • From (8.1) it follows that V(x,),)F(xo, A0) = 0 is equivalent
179
180 CHAPTER 8. LAGRANGE MULTIPLIERS. NEWTON'S METHOD.
to the following system:
24,1 + 2 4,2xo,1
4 — A0,1 + 24,24,2
—2 — Ao,i
2x0,1 — X0,2 - X0,3
.A.,04
= 0;
= 0;
= 0;
= 0;
= 13.
This system has two solutions:
xo,1 = 2; x0,2 = —3;
X0,3 =
7; A0,1 = —2;
A0,2 = 1
(8.2)
and
(8.3)
—2; x0.2 = 3; X0,3 = —7; Ao.1 = —2; A0.2 = -1.
Fo(x)
=
D2F0(x)
of
For the first solution (8.2), we compute the Hessian
f (x) + Ato g(x), i.e., of
X0,1 =
Fo (x) = 4x2 — 2x3 — 2 (2xi — x2 — x3) + 4 + 4 — 13 = + 4 — 4xi + 6x2 — 13
and obtain
D2F0(x)
= (2 0 0
020
000
,
which is (semi)positive definite for any x E R3. This allows us to conclude
directly that the point (2, —3, 7) is a minimum point for f(x). Note that
f (2, —3, 7) = —26.
Similarly, for the second solution (8.3), we find that
Fo(x) =
— x2 — 4x1+ 6x2 + 13
and
D2F0(x) =
( —2 0 0
0 —2 0 ,
0 0 0
which is (semi)negative definite for any x E R3. We conclude that the point
(-2,3, —7) is a maximum point for f (x). Note that f (-2, 3, —7) = 26. ❑
Problem 2: Assume that you can trade four assets (and that it is also
possible to short the assets). The expected values, standard deviations, and
correlations of the rates of return of the assets are:
0.25;
= 0.08; 01 = 0.25; P1,2 =
0.25;
it2 = 0.12; 02 = 0.25; P2,3 =
it3 = 0.16; 03 = 0.30; p1 3 = 0.25;
/24 = 0.05; 04 = 0.20; pi,4 = 0, V i = 1 : 3.
8.1. SOLUTIONS TO CHAPTER 8 EXERCISES
181
(i) Find the asset allocation for a minimal variance portfolio with 12% expected rate of return;
(ii) Find the asset allocation for a maximum expected return portfolio with
standard deviation of the rate of return equal to 24%.
Solution: For i = 1 : 4, denote by withe weight of asset i in the portfolio.
Recall that the expected value and the variance of the rate of return of a
portfolio made of the four assets given above are, respectively,
E[R] = w1µ1 + W2A2 + W3/13 + W4114;
var(R) = w?a? + tv3c4 + w3o-3 + w4a4
+2 (wiw2cia2P1,2 + W2W30-20-3P2,3
(8.4)
(8.5)
wiw3rig3P1,3),
since pi,4 = 0 for i = 1 : 3.
We do not require the weights wi 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:
wl
+ w2 + w3 + w4 = 1.
(8.6)
(i) We are looking for a portfolio with given expected rate of return E[R]
0.12 and minimal variance of the rate of return. Using (8.4-8.6), we obtain
that this problem can be written as the following constrained optimization
problem: find wo such that
mM f(w)
g(w)=
no
= f(wo),
(8.7)
where w = (wi)j=1:4, and f : liti -4 R and g :R4 -> R2 are defined as
f (w) = 0.0625w? + 0.062574 + 0.09w3 + 0.04w4
- 0.03125w1w2 - 0.0375w2w3+ 0.0375w1w3;
wi+w2+w3+w4 -1
0.08w1+ 0.12w2 + 0.16w3+ 0.05W4 - 0.12 ) •
g(w)
(8.8)
(8.9)
It is easy to see that rank(Vg(w)) = 2 for any w E R4, since
Vg(w)
1
1
1
1
( 0.08 0.12 0.16 0.05 ) •
The Lagrange multipliers method can therefore be used to find the minimum
variance portfolio.
182 CHAPTER 8. LAGRANGE MULTIPLIERS. NEWTON'S METHOD.
Denote by Al and A2 the Lagrange multipliers. From (8.8) and (8.9), we
obtain that the Lagrangian associated to this problem is
F(w, A) = 0.0625w? + 0.0625/4 + 0.09/4 + 0.04/4
(8.10)
- 0.03125w1w2 - 0.0375w2w3+ 0.0375w1w3
+ Al (w/ + w2 + W3 ± W4 - 1)
+ A2 (0.08w1 0.12w2 + 0.16w3 + 0.05w4 - 0.12).
The gradient of the Lagrangian is the following (row) vector:
/ 0.125w1 - 0.03125w2 + 0.0375w3 + Al + 0.08A2
0.125w2 - 0.03125w1 - 0.0375w3 + Al + 0.12A2
0.18w3 + 0.0375w1 - 0.0375w2 + Al + 0.16A2
V(w.A) F(w, A) =
0.08w4 + Al+ 0.05A2
wl + w2 + W3 + W4 - 1
0.08w1 + 0.12w2+ 0.16W3 + 0.05w4 - 0.12 /
t
To find the critical points of F(w, A), we solve V(w,A) F(w, A) = 0, which can
be written as a linear system as follows:
/
0.125
-0.03125 0.0375
0 1
-0.03125
0.125
-0.0375 0 1
0.0375
-0.0375
0.18
0 1
0
0
0
0.08 1
1
1
1
1 0
\
0.08
0.12
0.16
0.05 0
0.08
0.12
0.16
0.05
0
0 /
/ w1
/ 0
w2
0
w3
w4
0
0
Al
\ A2 /
1
\ 0.12 /
(8.11)
The solution of the linear system (8.11) is
WO =
(
0.1586
0.4143
0.3295
0.0976
= 0.0112;
A0.2 = -0.3810.
Let Fo(w) = F(w, A0.1, A0.2), i.e.,
Fo(w) = 0.0625w? + 0.0625/4 + 0.09/4 + 0.04w4
- 0.03125w1w2 - 0.0375w2w3 + 0.0375w1w3
+ 0.0112 (■
w/ + w2 + w3 + w4 - 1)
- 0.3810 (0.08w1+ 0.12w2+ 0.16w3 + 0.05w4 - 0.12),
and compute its Hessian
D2F0(w)
0.125
-0.03125 0.0375
0
-0.03125
0.125
-0.0375 0
0.0375
-0.0375
0.18
0
0
0
0
0.08
•
8.1. SOLUTIONS TO CHAPTER 8 EXERCISES
183
Note that the D2F0(w) is equal to twice the covariance matrix of the rates
of return of the four assets, i.e.,
/
D2F0 (w) = 2
2
0-1 Cr2P1,2 U10'3191,3
al
„.2
al 0-2P1,2
0'20'3P2,3
'2
„.2
,. , 3
0-10-3P1,3 (720 3P2,3
0
0
0
0
\
0
0
cri i
We conclude that D2 F0(w) is a positive definite matrix.
Therefore, the associated quadratic form q(v) = vtD2Fo(wo)v is positive
definite, and so will be the reduced quadratic form corresponding to the linear
constraints Vg(wo) v = 0.
We conclude that the point w0= (0.1586 0.4143 0.3295 0.0976) is a constrained minimum for f (w) given the constraints g(w) = 0. The portfolio
with 12% expected rate of return and minimal variance is invested 15.86%
in the first asset, 41.43% 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%.
(ii) Denote by
(
M =
CT?
a10-2P1,2 Cr103P1,3 0 \
,.,.2
0-2g3P2,3 0
0-10-2P1,2
'2
0'32
0
al a3P1,3 0-20-3P2,3
02
0
0
a4 j
the covariance matrix of the rates of return of the four assets.
Let o-p = 0.24 be the required standard deviation of the rate of return of
the portfolio. If widenotes the weight of the asset i in the portfolio, i = 1 : 4,
it follows from (8.4) and (8.5) that
E[R] = ittw;
var(R) = wt Mw,
(8.12)
(8.13)
where
Pl
1-12
=
I-t3
Pt4
is the vector of the expected values of the rates of return of the four assets.
The problem of finding a portfolio with maximum expected rate of return
and standard deviation of the rate of return equal to ap can be formulated
as a constrained optimization problem as follows: find w0 such that
min f(w) = f(w0),
g(w)=0
(8.14)
184 CHAPTER 8. LAGRANGE MULTIPLIERS. NEWTON'S METHOD.
where w = (wi),=1,4, and f : R4
R and g : 1184
it tw
f (w)
g(w)
R2 are defined as
_
(8.15)
itw- 1
wtMw ,72p
(8.16)
where
1=
I
)
Recall that, if the function h : R71
R is given by h(x) = xtAx, where A
is an rt x n symmetric square matrix, then the gradient of h(x) is
/
Dh(x) =
2(Ax)t.
aaxhri)
(8.17)
Using (8.17), it is easy to see that
i
Vg(w)
= 2(m
ittot
In order to use the Lagrange multipliers method for solving problem (8.14),
we first show that the matrix Vg(w) has rank 2 for any w such that g(w) = 0.
Note that rank(Vg(w)) = 1 if and only if there exists a constant C e R
such that 2Mw = Cl. Using the fact that the covariance matrix M of the
assets considered here is nonsingular, we obtain that
W=
c
(8.18)
2
From (8.16) it follows that, if g(w) = 0, then ltw = 1 and wtMw = cr2p .
Using (8.18), we find that
ltw = 1 <
wt Mw = cr2p <
c
> 1 = — ltM-11;
2
2 Ct
C t
a P=
=
(8.19)
C
(8.20)
From (8.19) and (8.20), we find that, if there exists w E R4such that g(w) = 0
and rank(Vg(w)) = 1, then
itm-i
= 2 = 1
.7.f •
C
8.1. SOLUTIONS TO CHAPTER 8 EXERCISES
185
However, it is easy to see that
lt M-11 = 80.01
17.36 =
1
o-r,
We can now proceed with finding the portfolio with maximum expected
return using the Lagrange multipliers method. Denote by Al and A2 the
Lagrange multipliers. From (8.15) and (8.16), we obtain that the Lagrangian
associated to this problem is
F(w, A) = litw + Ai(itw — 1) + A2(wtMw — 4).
(8.21)
The gradient of the Lagrangian is
V (w,A) F(w, A) =
(it + all + 2A2Mw
ltw — 1
wt Mw — a2p
t
To find the critical points of F(w, A), we must solve
G(w, A1, A2) = V(,,,,,A) F(w, A) = 0,
where G :R6—> R6 is given by
(tz + Ail + 2A2Mw
ltw — 1
G(w, Ai, A2) =
wt Mw — cr2p
.
This is done using a six dimensional Newton's method; note that the gradient
of G(w, A1, A2) is the following 6 x 6 matrix:
V(w,),) G(w, Ai , A2) =
(2A2M 1 2Mw
0
0
1t
0
2(Mw)t 0
.
We find that the Lagrangian (8.21) has exactly one critical point given by
wo
(
0.0107
0.6450 ) .
'
0.6946
—0.3503
A0,1 = —0.0738; A0,2 = —0.8510.
Let Fo(w) = F(w, A0,1, A0.2), i.e.,
Fo(w) = [ttw — 0.0738(1t w — 1) — 0.8510(wt Mw — 4).
186 CHAPTER 8. LAGRANGE MULTIPLIERS. NEWTON'S METHOD.
The Hessian of Fo(w) is
D2F0(w)
— 0.8510 • 2M = — 1.7019M.
Since the covariance matrix M of the rates of return of the four assets is a
positive definite matrix, it follows that D2F0(w) is a negative definite matrix
for any w.
Therefore, the associated quadratic form q(v) = vtD2Fo(wo)v is negative
definite, and so will be the reduced quadratic form corresponding to the linear
constraints Vg(wo) v = 0.
We conclude that the point wo = (0.0107 0.6450 0.6946 — 0.3503) is a
constrained maximum for f (w) given the constraints g(w) = 0.
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%. ❑
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
What are the
duration and convexity of the bond?
h.
Solution: Nine $1.6875 coupon 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
1
100 + - =
32
9
E 1.6875 exp ( —y 2
+ 101.6875 exp(-5y).
(8.22)
i=1
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 = 0.033401. We conclude that the yield of the bond is 3.3401%.
The duration and convexity of the bond are given by
D=
C
=
(E 1.6875—iexp
x (—
2
Y
E 1.687571exp I —y —
2)
( i=i
+ 101.6875 . 5 exp(-5y))
101.6875 • 25 exp(-5y))
8.1. SOLUTIONS TO CHAPTER 8 EXERCISES
187
th
where B = 100 +
and y = 0.033401. We obtain that the duration of the
bond is 4.642735 and the convexity of the bond is 22.573118. ❑
Problem 4: Recall that finding 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(di(x)) — Ke-rTN(d2(x)) — C
ln(K) + (r—q+)T
and di (x)
xl/T
(i) Show that lim,„
10) + (r—q—)T
, d2(x) =
x./T
(x) = oo and lima.
d2(x) = —oo, and conclude that
lim f(x) = Se —qT — C.
x—■co
(ii) Show that
ess,,o
—oo, if Se(r-OT < K;
0, if Se(r-q)T = K;
T
oo, if Se(r-q) > K.
= limd2(x) =
x\„o
T
(Recall that F = Se(r-q) is the forward price.)
Conclude that
lim f (x) =
x\O
—C,
— K e —rT
Se
if Se(r-q)/' < K;
if se(r—q)T > K.
(iii) Show that f (x) is a strictly increasing function and
—C < f(x) < Se-qT — C, if Se(r-q)T < K;
Se-qT — Ke-rT — C < f(x) < Se-qT — C, if Se(r-q)T > K.
(iv) 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
di(x) =
In (*) +
In
d2(x) =
(*) +
(r — q)T
(r — q)T
x-VT
+
x N/T
2
xV
2
(8.23)
(8.24)
188 CHAPTER 8. LAGRANGE MULTIPLIERS. NEWTON'S METHOD.
It is easy to see that
lim di(x) = oo and lim d2(x) = -oo,
x.,0c
and therefore
lim N(di(x)) = 1 and lira N(d2(x)) = 0.
x-,x
We conclude that
lim f (x) = Se-qT - C.
X--∎ DC
(ii) Let F = Se(r-q)T be the forward price. From (8.23) and (8.24) it follows
that di(x) and d2(x) can be written as
di(x) =
ln
xvt +
x \if"
d2(x)
2
'
ln
x N-
x-VT
2•
• If F < K, then ln (ifc) < 0 and
lim
x\o
(x) = lim d2(x) = -co.
x\o
Therefore limx N(di(x)) = limx
N(d2(x)) = 0 and
lim f(x) = -C.
x\o
• If F = K, then di(x) = x4" and d2(x) = -x477 , and therefore
lim dl (x) = lim d2(x) = 0.
x\o
x\o
Thus, lims\o N(di(x)) = limx
N(d2(x)) = a and
e-rT
1
lim f(x) = - (Se-qT - Ke-rT ) - C = 2 (F - K) - C
x\O
2
= -C.
• If F > K, then ln
> 0 and
lim dl (x) = lim d2(x) = oo.
x\o
x\o
Therefore limx \ N(di(x)) = limx
N(d2(x)) = 1 and
o f (x) = Se-qT - K e-rT - C.
s\
8.1. SOLUTIONS TO CHAPTER 8 EXERCISES
189
(iii) Differentiating f(x) with respect to x is the same as computing the
derivative of the Black-Scholes value of a European call option with respect
to the volatility o-, which is equal to the vega of the call. In other words,
f'(x) = vega(C) = Se-qT /7" 1 -`41
Te
1n04) + (r+)T
where d1 = di(x) =
Thus, f'(x) > 0, V x > 0, and f(x) is strictly increasing.
Recall that lima„, f (x) = Se - C and
lim f(x) =
x\o
if F < K;
if F > K.
—C,
se-qT Ke-rT
Since f(x) is strictly increasing, we conclude that
-C < f(x) < Se-gT - C, if F < K;
Se-9T — Ke-rT — C < f (X) < Se-qT — C, if F > K.
(iv) If F < K, the problem f (x) = 0 has a solution x > 0 if and only if
0 < C < Se
-9T
(8.25)
If F > K, the problem f (x) = 0 has a solution x > 0 if and only if
se-qT - K e-rT < C < se-qT
(8.26)
Note that
se-qT K e-rT = e-rT (se(r-q)T K)
e-rT (F 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:
max (Se-qT -Ke-rT , 0) < C <
Problem 5: A three months at-the-money 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 interval [0.0001, 1], the secant method with initial
guess 0.5, and Newton's method with initial guess 0.5.
190 CHAPTER 8. LAGRANGE MULTIPLIERS. NEWTON'S METHOD.
(ii) Let crimpbe the implied volatility previously computed using Newton's
method. Use the formula
.N/ 77r
2q)T S
C
(8.27)
S15" 1 2
to compute an approximate value azinp.app„x for the implied volatility, and
compute the relative error
0-imp: approx
(r+g)T •
1°-imP,aPProx — wimp
Crimp
Solution: (i) Both the secant method with x_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 Method Newton's Method
0.5
0.5
0.3969005134
0.3969152615
0.3970483533
0.3970481867
0.3970481868
0.3970481868
0
1
2
3
The bisection method on the interval [0.0001, 1] converges in 30 iterations
to the same implied volatility of 39.7048%. The first five iterations generate
the following intervals:
[0.250075, 0.5];
[0.375063, 0.5];
[0.375063, 0.406309];
[0.375063, 0.437556];
[0.390686, 0.406309];
(ii) The approximate value for the implied volatility given by (8.27) is
Crirrip.approx
= 0.3966718145 = 39.6672%.
If crimp= 0.3970481868 is the implied volatility obtained using Newton's
method, then
lain1P' aPPr°x aimPi
Crimp
= 0.000948 = 0.0948%. ❑
Problem 6: Let F : R3 R
3given by
F(x) =
2xix2
— x2x3 ± 9
2xi + 2x14 + 4x3 — 4x3— 2
x1x2x3+ xi — x3 — xlx2 — 4
8.1. SOLUTIONS TO CHAPTER 8 EXERCISES
191
The approximate gradient AcF(x) =- (A, jFi(x))ij=1,7., of F(x) is computed using central difference approximations, i.e.,
= Fi(x + hey) - Fi(x - hey)
Ac
=1:
2h
where eiis a vector with all entries equal to 0 with the exception of the j-th
entry, which is equal to 1.
(i) Solve F(x) = 0 using the approximate Newton's algorithm obtained by
substituting AcF(xo/d) for AF(xo/d). Use h = 10-6, tol_consec = 10-6, and
tol_approx = 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 approximate Newton's method with forward finite difference approximations for AF(x).
Solution: We use Newton's method and the approximate Newton's method
both with forward difference approximations and with central difference approximations with tol_consec = 10-6and tol_approx = 10-9. The parameter
h is chosen to be equal to tol_consec, i.e., h = 10-6.
All algorithms converged to the same solutions,
x*
( -1.6905507599 )
1.9831072429
-0.8845580785
=
for xo =
1
2
3
and
-1
3 )
x* = (
for
x0
=
(2
2).
2
The iteration counts are given in the table below:
x0
1
3
2
Iteration Count
Iteration Count
Iteration Count
Newton's Method Approximate Newton Approximate Newton
Forward Differences
Central Differences
9
9
9
40
65
43
2
2
2
We note that using central finite differences approximates the gradient
DF(x) of F(x) more accurately than if forward finite differences are used,
192 CHAPTER 8. LAGRANGE MULTIPLIERS. NEWTON'S METHOD.
resulting in algorithms with iteration counts closer to the iteration counts for
Newton's method. ❑
Problem 7: (i) Use bootstrapping to obtain a zero rate curve from the
following prices of Treasury instruments with semiannual coupon payments:
3 - Month T-bill
6 - Month T-bill
2 - Year T-bond
3 - Year T-bond
5 - Year T-bond
10 Year T-bond
-
Coupon Rate Price
0
98.7
0
97.5
4.875
100a
4.875
100-7,
4.625
992232
4.875
101k
Assume that interest is continuously compounded.
(ii) How would the zero rate curves obtained by bootstrapping from the bond
prices above, one corresponding to semi-annually 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?
Solution: (i) For the Treasury bills, the zero rates can be computed directly:
100
r(0, 0.25) = 41n V ) = 0.052341 = 5.2341%;
8.7
100
r(0, 0.5) = 21n V ) = 0.050636 = 5.0636%.
7.5
Bootstrapping is needed to obtain the 2-year, 3-year, 5-year and 10-year
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
r(0,t) =
(2 - t)r(0, 0.5) + (t -0.5)r(0,2)
Vt E [0.5, 2].
.5
(8.28)
If we let x = r(0, 2), we find from (8.28) that
r(0 1) = r(0, 0.5) + 0.5x
0.5r(0, 0.5) + x
,
. r(0 1.5) =
1.5
1.5
Recall that the price of the two year bond is the discounted present value of
all the future cash flows of the bond. Then,
5
100 + — = 2.4375 e-o.5r(o.o.5) + 2.4375 e-r(°'1)
32
8.1. SOLUTIONS TO CHAPTER 8 EXERCISES
193
+ 2.4375 e-1.5r(°'1.5)+ 102.4375 e-2r(o,2)
r(0, 0.5) +0.5x1
= .4375 e -0.5r(0,0.5) + 2.4375 exp
2
1.5
0.5r(0, 0.5) + x)
+ + 102.4375 e-2x.
2.4375 exp (-1.5
1.5
Using Newton's method to solve the nonlinear equation above for x, we obtain
that x = 0.047289, and therefore
r(0, 2) = 4.7289%.
We proceed by assuming that the zero rate curve is linear between two
years and three years. We note that r(0, 0.5), r(0, 1), r(0, 1.5), and r(0, 2) are
known. If we let x = r(0, 3), the price of the three year bond can be written
as
100 +
= 2.4375 e-ur(".5) + 2.4375 e-r(o°1) + 2.4375 e-1'5r(13'1.5)
0, 2)
+ 2.4375 e-2r(°'2) + 2.4375 exp (-2.5 r(
+ xl
2
+ 102.4375 exp (-2x)
Using Newton's method to solve the nonlinear problem above, we obtain that
x = 0.047582. Therefore
r(0, 3) = 4.7582%.
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:
r(0, 0.25) = 5.2341%;
r(0, 3) = 4.7582%;
r(0, 0.5) = 5.0636%; r(0, 2) = 4.7289%;
r(0, 5) = 4.6303%; r (0, 10) = 4.6772%.
Since we assumed that the zero rate curve is linear between any two consecutive bond maturities, the zero rate r(0, t) is known for any time between the
shortest and longest bond maturities, i.e., for any t E [0.25, 10].
(ii) Denote by re(0, t) and r2(0, t) the zero rate curves corresponding to identical discount factors, with r,(0, t) corresponding to continuously compounded
194 CHAPTER 8. LAGRANGE MULTIPLIERS. NEWTON'S METHOD.
interest, and with r2 (0, t) corresponding to semi-annually compounded interest. Then,
2t
(1Jr_ r2(0,L.))
e-tr,(04)
(8.29)
, V t > O.
2
By solving for r,(0, t) in (8.29), we find that
2)
r,(0, t) = In I 1 + r20,
2)
= r2(0, t) In ( (1 + r2(° t) ) '2(21
2/r2(0.t)
= r2 (0, t) In ( (1 +
2/r2 (0,t))
< r2(0, t);
for the last inequality we used the fact that
(1 + —
1)x < e, Vx > 0,
for x = 2/r2(0, t). In other words the semi-annually 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 semi-annually 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.
❑
Problem 8: Use bootstrapping to obtain a continuously compounded zero
rate curve given the prices of the following semiannual coupon bonds:
Maturity Coupon Rate Price
6 months
0
97.5
1 year
5
100
20 months
6
103
40 months
5
102
5 years
4
103
8.1. SOLUTIONS TO CHAPTER 8 EXERCISES
195
Assume that the overnight rate is 5% and that the zero rate curve is linear
on the following time intervals:
[0, 0.5] ;
[0 .5, 1] ;
[1,
;
[35 1301 ;
r1O
51 •
Solution: We know that r(0, 0) = 0.05. The six months zero rate can be
computed from the price of the 6-months zero coupon bond as
00
r(0, 0.5) = 21n (1 ) = 0.050636 = 5.0636%.
97.5
(8.30)
Using (8.30), we can solve for the zero rate r(0, 1) from the formula given
the price of the one year bond, i.e.,
100 = 2.5 e-a5r".5) + 102.5 e-r(")
and obtain that
r(0, 1) = 0.049370 = 4.9370%.
The third bond pays coupons in 2, 8, 14, and 20 months, when it also
pays the face value of the bond. Then,
103 = 3 exp (-i2y (0) i.2-))
2
+ 3 exp (-u
8 r (0, l))
+ 3 exp (_,7
1427, (0, i_
142))
103 exp
12 02))
12 r (
0,
(8.31)
Since we assumed that the zero rate curve is linear on the intervals [0, 0.5]
are known and can be obtained
and [0.5, 1], the zero rates r(0, and r(0,
by linear interpolation as follows:
A)
= 4r(0, 0) + 2r(0, 0.5)
0.050212;
6
4r(0, 0.5) + 2r(0, 1)
=
=- 0.050214.
r 0, 8
6
12/
r (0 12
Let x = r (0, ;A). Since r(0, t) is linear on the interval [1,
(8.32)
(8.33)
N] , we find that
6r(0, 1) + 2x
14)
r (0, —
=8
12
(8.34)
From (8.34), it follows that the formula (8.31) can be written as
103 = 3 exp
+ 3 exp
2 ( 8 ))
2
1 /0, T2-)) + 3 exp (--dr 0, u
6•
5
6r(0, 1) + 2x)
+ 103 exp (-- x) , (8.35)
3
8
196 CHAPTER 8. LAGRANGE MULTIPLIERS. NEWTON'S METHOD.
A)
are given by (8.32) and (8.33), respectively. Using
where r (0, i?) and r (0,
Newton's method to solve for x in (8.35), we find that x = 0.052983, and
therefore
0, 2
102 )
T(
= 5.2983%.
Bootstrapping for the fourth and fifth bonds proceed similarly. For example, 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 = r (0,
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
11)
il)
/,, 40
r v, u) =4.5326%; r(0, 5) = 3.2119%.
Summarizing, the zero rate curve obtained by bootstrapping is given by
2
r(0, 0) = 0.05; r (0, i .) = 0.050212; r (0, 8 ) = 0.050214;
20
40)
r (0 — ) = 0.052983; r (0 —
= 0.045326; r(0, 5) = 0.032119,
' 12
' 12
and is linear on the intervals
[0,0.5] ;
[0.5,1] ;
[1, 31 ;
[5 1°1 •
3' 3 '
[10 5
Ti
3'1• 1
8.2. SUPPLEMENTAL EXERCISES
8.2
197
Supplemental Exercises
1. (i) If the current zero rate curve is
1
ri(0, t) = 0.025 + —exp (100
100)
100(t + 1)'
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, 100, and 200
basis points, respectively, i.e., if the zero rate curve changes from ri (0, t)
to r2(0, t) = ri (0, t)+dr, with dr = {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 long maturity). This assumption will
be tested for the bond considered here for parallel shifts ranging from
small shifts (ten basis points) to large shifts (two percent).
2. Consider a six months at-the-money 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
A(C) = 0.5, and find that value using Newton's method. How does
this value of q compare to r + 2
?
3. The following prices of the Treasury instruments are given:
6 - Month T-bill
12 - Month T-bill
2 - Year T-bond
3 - Year T-bond
5 - Year T-bond
10 - Year T-bond
Coupon Rate Price
0
99.4565
0
98.6196
10117.5
2
4.5
10711
1021
3.125
4
1032
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-months and 12-months Treasury bills, and of the 2-year, 5-year and
10-year Treasury bonds;
198 CHAPTER 8. LAGRANGE MULTIPLIERS. NEWTON'S METHOD.
(ii) Find the relative pricing error corresponding to the 3—year Treasury
bond if the zero rate curve obtained at part (i) is used. In other words,
price a 3—year semiannual coupon bond with 4.5 coupon rate and find
its relative error to the price 1074 of the 3—year Treasury bond.
8.3
Solutions to Supplemental Exercises
Problem 1: (i) If the current zero rate curve is
1
t
(—
-ri(0, t) = 0.025 +100exp
100) 1+100(t + 1)'
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, 100, and 200
basis points, respectively, i.e., if the zero rate curve changes from ri(0, t) to
r2(0, t) = ri (0, t) + dr, with dr = {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.
Solution: (i) The bond provides coupon payments equal to 3 every six months
until 3.5 years from now, and a final cash flow of 103 in four years. By
discounting this cash flows to the present using the zero rate curve r i (0, t),
we find that the value of the bond is
7
131 =
E 3 exp (-7.1CO,
i:=1
= 106.1995.
) —i) + 103 exp(-4r1(0, 4))
2) 2
(8.36)
The yield of the bond is found by solving the formula for the price of the
bond in terms of its yield, i.e, by solving
7
B1 =
E 3 exp (—y 2i ) + 103 exp(-4y)
(8.37)
i=1
for y, where B1is given by (8.36), i.e., B1= 106.1995. Using Newton's
method, we obtain that the yield of the bond is
y = 0.042511 = 4.2511%.
8.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
199
(ii) If the zero rates increase, 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 increase of the yield of
the bond also equal to dr.
When the zero rates increase from ri(0, t) to r2(0, t) = ri (0, t) + dr, the
value of the bond decreases from B1 given by (8.36) to
7
B2 =
E 3 exp (-r2 CO, 0 0 + 103 exp(-4r2(0, 4)).
i=i
The new yield of the bond, denoted by y2, will be larger than the initial yield
y, and is obtained by solving
7
B2 =
E 3 exp (-y2 i2
+ 103 exp( -4y2)
(8.38)
i=i
for y2, where B2 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
P2
Y2 Y
10bp -= 0.001
105.8150
0.043511
0.00099979
20bp = 0.002
105.4319
0.044510
0.00199957
50bp = 0.005
104.2915
0.047510
0.00499893
100bp = 0.01
102.4199
0.052509
0.00999784
0.062506
200bp = 0.02
98.7829
0.01999562
-
As expected, the increase of the yield of the bond is slightly smaller, but
very close to, the parallel shift of the zero rate curve, i.e., y2 -y dr. ❑
Problem 2: Consider a six months at-the-money 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 A(C) = 0.5, and
find that value using Newton's method. How does this value of q compare to
r+ c?
Solution: Recall that the Delta of a plain vanilla call option on an underlying
asset paying dividends continuously at rate q is
A(C) = e-qT N(di),
(8.39)
200 CHAPTER 8. LAGRANGE MULTIPLIERS. NEWTON'S METHOD.
where
ln (K) + (7- - q + 2) T
d1 -
o- NiT
For an at-the-money option, i.e., for S = K, we find that
d1
= (r - q).\ff + a-VT
a
(8.40)
2
From (8.39) and (8.40), we find that
(r
A(C) = e-qTN(d1) = e-qT N (
- q)fT-- ±a -VT
a
2 ).
(8.41)
It is easy to see that 6,(C) is a decreasing function of the dividend rate
q, since
OA
aq
= —Te-qT N(do + e-qT ir (di )
=
—Te-qT N(di )
+
e—qT
ad1
aq
1 A.
VT
2 ——
e
(
)
-
A/T7i
a
< 0.
When q = 0, we find that
A(C)
N (r-VT' +a-f77)
Q
2
> N(0) = 0.5
Also, since 0 < N(d1) < 1, it follows that
lim A(C) = lim (e-qT N(di)) = 0.
q—,oc
q--oc
We conclude that A(C) is a decreasing function of q and that, for q > 0,
the values of A(C) decrease from N(0) > 0.5 to 0. Therefore, there exists
a unique value q > 0 such that A(C) = 0.5. From (8.41), it follows that
this value can be obtained by solving for x the nonlinear equation f (x) = 0,
where
f (x) = e-xT N ((r - x)07 + aft)
- 0.5.
Q
2
Using Newton's method, we obtain that q = x = 0.066906.
8.3. SOLUTIONS TO SUPPLEMENTAL EXERCISES
201
In other words, if the interest rates are flat at 4%, the Delta of a six
months at-the-money call option on an underlying asset with volatility 30%
is equal to 0.5 if the underlying asset pays 6.69% dividends continuously.
If q = r + -c, we find from (8.40) and (8.41) that di. = 0 and
A(C) = e-9TN(0) = 0.5e-qT < 0.5.
Since A(C) is a decreasing function of q, we obtain that the value of q such
that A(C) = 0.5 must be lower than r + 1. Indeed, the value previously
obtained for q satisfies this condition, i.e.,
0.2
q = 0.066906 < 0.085 = r +
2 '
❑
Problem 3: The following prices of the Treasury instruments are given:
6 - Month T-bill
12 - Month T-bill
2 - Year T-bond
3 - Year T-bond
5 - Year T-bond
10 - Year T-bond
Coupon Rate Price
0
99.4565
0
98.6196
2
1011,4,2
4.5
107
3.125
1028
4
103-325
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 6months and 12-months Treasury bills, and of the 2-year, 5-year and 10-year
Treasury bonds;
(ii) Find the relative pricing error corresponding to the 3-year Treasury bond
if the zero rate curve obtained at part (i) is used. In other words, price a
3-year semiannual coupon bond with 4.5 coupon rate and find its relative
error to the price 1074 of the 3-year Treasury bond.
Solution: (i) The 6-months and 12-months zero rates can be obtained directly from the prices of the Treasury bills, i.e.,
100 )
= 1.09%;
99.4565
0)
= 1.39%.
r(0 , 1) = In
(98.16196
r(0,0.5) = 214
202 CHAPTER 8. LAGRANGE MULTIPLIERS. NEWTON'S METHOD.
Using bootstrapping, we obtain the following 2-year, 5-year and 10-year
zero rates:
r(0, 2) = 1.2099%;
r(0, 5) -= 2.6824%;
r(0, 10) = 3.7371%.
(ii) The zero rates computed above correspond to the following zero rate
curve which is piecewise linear between consecutive bond maturities:
{ (2t - 1) r(0,1) + 2(1 - t) r(0, 0.5), if 0.5 < t < 1;
if 1 < t < 2;
(t - 1) r(0, 2) + (2 - t) r(O, 1),
y r(0, 5) + 1
=1r(0, 2)
if 2 < t < 5;
io-tr(0, 5),
'
r(0, 10) +0
if 5 < t < 10.
r (0, t)
y-
With this zero rate curve, the value of the 3-year semiannual coupon bond
with 4.5 coupon rate is
5
B=
E 2.25 exp (-r (0, )
+ 102.25 exp(-3r(0, 3))
i=i
= 108.1930.
The price of the 3-year Treasury bond was given to be 107.5625. Therefore, the relative pricing error given by the bootstrapped zero rate curve
which does not include the 3-year bond is
1107.5625 - 108.19301
= 0.005862 = 0.59%. ❑
107.5625
Bibliography
[1] Milton Abramowitz and Irene Stegun. Handbook of Mathematical Functions. National Bureau of Standards, Gaithersburg, Maryland, 10th corrected printing edition, 1970.
[2] Dan Stefanica. A Mathematical Primer with Numerical Methods for Financial Engineering. FE Press, New York, 2007.
204
BIBLIOGRAPHY
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