Stefanica D. Solutions Manual: A Primer For The Mathematics Of Financial Engineering

34

CHAPTER

1.

CALCULUS

REVIEW.

PLAIN

VANILLA

OPTIONS.

Problem

2:

Compute

/♂♂

dx

Solution:

For every

ir

由

ger

n

主

0

，

define

the

function

fn(

x)

as

儿

(x)

工

j

xne

x

dx

By

using

integration

by

parts

,

it

is

easy

to

see

that

儿

(x)

=

j

♂

eX

dx

= x

n

eX

- n

j

♂

-1

川

=

xne

x

一吼一

1

协

Vn;:::l

Since

fi

。但)

=

eX

,

the

followi

吨

general

forml

414

>-

n

vv

c

pu

Lf-M

nZM

/III-11\

z

n

fIJ

2

,

G

Z

e

n

z

PIt-

,

J

Problem

3:

Compute

j

(1

n(x))

η

dx

Solution:

For every

integerη

三

0

，

let

儿

(x)

工

j

(I

n

(x)

)n

dx

By

using

integration

by

parts

,

it

is

easy

to

see

that

,for

any

η;

三 1

，

户口

(x)t

dx

=

川)川

j

(I

n(x))η

→

and

therefore

fn(X ) = x

(I

n(x)

)η

一

η

fn-1(

♂)，

V

η

主1.

Since fo(x)

=

凯

the

following general formula

can

be

obtained

by

ir

吐

uction:

j

主

(γ

k

n

!

(1

卫

(x))n

dx

=

fn(x)

= x L \

.L

~nl

I

f,;

(ln(x))k

十

0

，

Vn

主1.口

1.3.

SOLUTIONS

TO

SUPPLE

1V

IENTAL

EXERCISES

35

>-

Z

Vv

+

Z

\111/

1-z

4t-A

/I111\

<

pu

<

z

t\liI/

al-Z

h

QU

m

e

''EA

'D

O

VA

P

(1.

16)

Sol

仰

on:

Note

that

(1.

16) is equivalent

to

1

(1\1

一二一-

< In I 1 +

-=.

I <

一、

Vx>

1.

Z

十

1\

x)

x

一

(1.

17)

Let

:一咛

+D;

g(x)

咔+~)一击

f(x)

The

丑

7

j'

(x)

1 1

一一一+.

-

一

<

0:

x

2

I X

(x

+

1)

x

2

(x

+

1)

~

-,

L

一十

1 -

1<

0

x(x

十

1)

I

(x

+

1)2

x(x +

1)2

g'(x)

We

conclude

that

bo

也

f

(x)

and

g(

x)

are decreasing functions. Since

lim f(x) = lim

g(x)

= 0,

X

一→汉二

X-→。

c

it

follows

that

f(x) > 0

and

g(x)

> 0 for all x > 0,

and

therefore

:>咛+:)〉

ZLI7VZ>07

which is

what

we

wanted

to

show;

cf.

(1.

17).

口

Problem

5:

Let

仲)

=忐叫一(工

f)2)

Assume

that

g : R

•ffi.

is a continuous function which is uniformly

bounded

2

,

i.e.,

there

exists a

constant

0 such

that

Ig(x)1

三

o

for all x

ε

ffi..

Then

,show

that

叫:削

g(x)

dx

=

仲)

2The uniform boundedness condition was chosen for simplicity:

and

it

can be

l'吐

axed:

e.g.:

to

functions which have polynomial growth

at

infinity.

36

CHAPTER

1.

CALCULUS

REVIE

1i

V.

PLAIN

VANILLA

OPTIONS.

Solution:

Using

the

change of variables y

=

乎?附在

nd

that

汇

f(x)g(x)

dx

志汇

g(x)

exp

(气

;)2)

dx

vi

言汇

g(

μ+

仙句

ο

Recall

that

去/∞

λ

dy

= 1,

(1.

19)

V

kI/I

J-OO

since

，吗，

the

function

在

e-15is

tue

p

时

ability

density function of

the

standard

normal variable. From

(1.

18)

and

(1.

19)

we

obtain

that

f

∞

1

r

∞

g(

μ)

- / f(x)g(x)

dx

=

一)

- /

(g(

μ)

-

g(

μ

十

σ

y))

e-

号

dy.

(1.

20)

j

一∞

d

订一∞

Our

goal is

to

show

that

the

right

hand

side of

(1.

20) goes

to

0 as

σ\

、

O.

Since

g(

x)

is

a continuous function,

it

follows

that

, for

创lY

ε>

0, there

exists

61

(ε)

> 0 such

that

Ig(

μ

)-g(x)1

<

飞 V

Ix

一

μ1<

61(ε)

.

(1.

21

)

Using

the

fact

that

the

integral

(1.

19) exists

and

is finite,

we

obtain

that

,

for

anyε>

0,

there

exists

62

(ε)

> 0 such

that

「

LfdM)e4dy

十气:

(∞片

dy

<

ε

V L/

Tr

J

一∞飞

!"L，1f

J

62(£)

Since

Ig(

x)

I

三

C

for all x E

JR.,

it follows from

(1.

22)

that

1

r-

6

2(£)

→头

=

I

Ig(

μ)

-

g(

μ+

句

)1

e

一专

dy

v

L,1f

J

一∞

+τ:

(∞

Ig(

μ)

-

g(

μ

叫

)1

汁

dy

<

2Cε

v

ζπ

J

62(£)

(1.

22)

(1.

23)

It

is

叫

to

s

叫风

if

()

<去

87then

Iyl

I

(μ+

句)一

μ1=σIyl

<61(ε)

一一::;

61

(ε)

，

V y

ε[-62(ε)

，

62(E)].

(1.

24)

62(ε)

一

The

凡

from

(1.

21) and

(1.

24)

we

自

nd

that

Ig(

μ)

g(

μ+σ

y)1

<

飞

Vy

ε[-62(E)

，

62(ε)

]，

(1.

25)

1.

3.

SOLUTIONS

TO

SUPPLEMENTAL

EXERCISES

37

and therefore

1

r

6

2(

ε)

→是

=

I

Ig(

μ)

-

g(

μ+σν)\

e

一号

dy

<

ε-

v

L,1f

J -62(£)

From

(1.

20)

,

(1.

23), and

(1.

26), it follows

that

, for

anyξ>

0, there exist

61(E)

> 0

and

62(E)

> 0 such

that

,if

σ<:237then

(1.

26)

2ML2

e

uu

nu

Z

z

nud

z

rlu

flL

We

conclude,

by

definition,

that

叫

:f(z)g(Mzg(μ)

口

Problem

6: Let

g(y)

ZJ

Compute

g'

(y).

Solution:

'(y)

=

一安

Citi

告

(1

+

y)t

i

十

1

Problem

7: A derivative security pays a cash amount C if

the

spot price

of

the

underlying asset

at

maturity

is between K

1

and

K

2

, where 0 < K

1

<

K

2

, and expires worthless otherwise. How do you synthesize this derivative

security (i.e.

,how do you recreate its payoff almost exactly)

usi

吨

plai

川

anilla

call options?

Solution:

The

payoff of

the

derivative security is

K

<

<-ss

nucnu

T

V

Since

V(T)

is

d

出

isc

∞

on

挝叫

t

挝

inuo

盹

i

让

t

cannot be

rep

抖

lica

创

ted

exactly

us

岳

i

吨

call

丑

lopt

址

10

口

whose

p

a;句

~offs

are

contir

丑lU

aus.

38

CHAPTER

1.

CALCULUS

REVIE

1i

V.

PLAIN

v:生

NILLA

OPTIONS.

We

approxi

口时

e

the

payoffV (T) of

the

derivati

飞肘

ec

盯

ity

by

the

followi

吨

payoff

Solution: Using plain vanilla options, cash, and

the

underlying asset

the

payoff

V(T)

can

be

replicated in different ways.

39

(1.

30)V - 23 -

30

20

+

20

40

.

同

(T)

3(T)

S

20

23(T)

20

<

S(T

否){

三

40

2S(

T)=3(S(T)

- 20) =

60

3(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

sy

毗

lesized

using

put

options.

If

the

asset

does

口

ot

pay dividends

and

if interest rates are zero,

then

, from

the

Put-Call

parity,

it follows

that

C =

P+S-K.

Denote by C

20

and P

20

,

and

by 0

40

and

凡

0

，

the

values of

the

call and

put

options with strikes

20

and

40

, respectively.

Then

,

the

replicating portfolio with payoff

at

maturity

given by

(1.

29)

can

be

written

as

One way is

to

use

the

underlying asset,calls

with

strike

20

,and calls with

strike 40.

First of all

, a portfolio with a long position in two units of

the

underlying

asset has value

2S(T)

at

maturity

, when

S(T)

<

20.

To replicate

the

payoff

60

-

S(T)

of

the

portfolio when

20

三

3(T)

<

40

,

note

that

60

-

3(T)

=

2S(T)

+

60

-

33(T)

=

2S(T)

-

3(3(T)

- 20).

This is equivalent

to

a long position in two units of

the

underlying asset and

a short position in three calls

with

strike

20.

To replicate

the

payoff

S(T)

-

20

of

the

portfolio when

40

三

S(T)

，

note

that

S(T)-20

=

60-3(T)

十

2S(T)-80

=

2S(T)

-

3(3(T)-20)

+

2(S(T)-40).

This is equivalent

to

a long position in two units of

the

underlying asset, a

short position in three calls with strike

20

,

and

a long position in two calls

with strike 40.

Summarizing

,

the

replicating portfolio is made of

• long two units of

the

asset;

• short

3 call options

with

strike K =

20

on

the

asset;

• long 2 call options

with

strike K =

40

on

the

asset.

We

check

that

the

payoff of this portfolio

at

maturity

, i.e.,

vi

(T)

=

23(T)

- 3

max(S(T)

-

20

,0) + 2

max(S(T)

-

40

,0)

(1.

29)

is

the

same as

the

payoff from (1.28):

1.

3.

SOLUTIONS

TO

SUPPLEMENTAL

EXERCISES

(1.

28)

(1.

27)

~(T)

S(T)

<

K

1

一

ε

。

KK1

15εS

二三十

K2

1

ε

;(S(T)

•

(fKC

(KS12

(T

)EQ+()S)f

(T

((SK)E

(T

(1S)K(T20))

()

K

2

K+

1))=C

K

2

~二

K

2

+

ε

<

S(T)

c -

;(S(T)

ε))·;;0

Problem

8: Create a portfolio

with

the

following payoff

at

time T:

r

23(T)

, if

0

三

3(T)

<

20;

V(T)

= <

60

-

S(T)

, if

20

三

S(T)

<

40;

l

S(T)

-

20

, if

40

三

S(T)

，

where

S(T)

is

the

spot price

at

time

T of a given asset. Use

plai

丑

vanilla

options

with

maturity

T as well as cash positions

and

positions in

the

asset

itsel

f.

Assume,for simplicity,

that

the

asset does not pay dividends

and

that

interest rates are zero.

( 0

, if

S(T)

<

K

1

一

ε;

I

c(S(T)

一

(K

1

-

E))/ε

，

if

K

1

一

ε

三

S(T)

三

K

1

;

~(T)

= < c, if K

1

<

S(T)

< K

2

;

I c -

c(S(T)

- K

2

)/

ε

，

if

K

2

三

S(T)

三

K

2

+

ε;

l 0, if

K

2

十

ε

<

S(T).

Note

that

V(T)

=

~(T)

unless

the

value

S(T)

of

the

underlyi

吨

asset

at

maturity

is either between

K

1

一

ε

and

K

1

, or between K

2

and

K

2

十

ε.

The

payoff

~(T)

can

be

realized by

goi

吨

long

c/εbull

spreads

with

strikes

K

1

一

ε

and

K

1

,

and

shorti

吨

c/εbull

spreads with strikes K

2

a

创

an

K

2

十

ε

已

.

In

other words,

the

payoff

V(T)

of

the

given derivative security can

be

synthesized by taking

the

following positions:

• long

c/εcalls

with strike

K

1

一

ε;

• short

c/εcalls

with strike K

1

;

• short

c/εcalls

with strike K

2

;

•

10

吨

c/εcalls

with strike

K

2

十

ε-

It

is easy

to

see

that

the

payoff

~(T)

is

the

same as in

(1.

27):

40

CHAPTER

1.

CALCULUS

REVIE1

iV.

PLAIN

VANILLA

OPTIONS.

To synthesize a short position

in

three

calls

with

strike

20

, note

that

巧

(T)

=

8(T)

-

20

- 3 max(20 -

8(T)

,

0)

+ 2max(40 -

8(T)

,

0)

is

the

same as

the

payoff from

(1.

28):

41

(1.

39)

(1.

36)

(1.

37)

X2

0

1

一，一

句

O

2

Xl

-

X2

十

Z32

三 0;

x1(K

3

- K

1

) -

x2(K

3

- K

2

)

~

O.

vVe

solve

(1.

34) for

X3

and

obtain

O

2

0

1

(1.

38)

0

3

.."'0

3

Since

we

assumed

that

X3

> 0,

the

following condition must also

be

satisfied:

Problem

9: Call options on

the

same underlying asset

and

with

the

same

maturity

,

with

strikes K

1

< K

2

< K

:3'

are

trading

for 0

1

, O

2

and

0

3

, re-

specti

飞吨

T

(∞

Bid-Ask

spread),

with

C

1

> O

2

> 0

3

,

Find

necessary

and

sufficient conditions on

the

prices 0

1

, O

2

and

0

:3

such

that

no-arbitrage

ex-

ists 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

O.

Consider a portfolio

made

of positions in

the

three

options

with

value 0

at

inception,

and

let

Xi

> 0

be

the

size of

the

portfolio position in

the

option

with

strike K

i

, for i = 1 :

3.

Let 8 =

8(T)

be

the

value of

the

unde

臼创

rl

坊

yi

凶

n

asset

at

maturity. For

no-arbitrage

to

occur,

there

are three possibilities:

Portfolio

1:Long

the

K

1

-option

,

short

the

K

2

一

option

，

long

the

K

3

-option.

Arbitrage exists if we can find

Xi

> 0, i = 1 : 3, such

that

X1

0

1-

X2

0

2

十

X303

=

0;

(1.

34)

x1(8

- K

1

) -

x2(8

-

K

2

)

十句

(8

-

K

3

)

三

0

，

V

8

主

O.

(1.

35)

We note

that

(1.

35) holds if

and

only if

the

following two conditions are

satisfied:

1.3.

SOLUTIONS

TO

SUPPLEMENTAL

EXERCISES

• short 3 call options

with

strike K =

20

on

the

asset;

• long 2 call options

with

strike K =

40

on

the

asset.

The

second replicating portfolio will

be

made of

the

following positions:

• long

e-

qT

units of

the

asset;

• short

$20e-

rT

cash;

• short 3

put

options

with

strike K =

20

on

the

asset;

• long 2

put

options

with

strike K =

40

on

the

asset.

Note

that

any piecewise linear payoff of a single asset can be synthesized,

in theory, by using

plai

丑

vanilla

options,cash

and

asset

positions.

口

(1.

31

)

一

30

20

- -

3P20

-

38

+

60

,

8(T

页

)

~

20

B(T) -

2

日

3(20

页

{4S

C

到

L-Tf

芦

S

T-)

邪

T2

川

日

o

--S(T))~

28(T)

20

<

8(T)

三

40

8(T)

-

20

十

2(40

-

8(T))

=

60

-

8(T)

40

<

8(T)

If

the

asset pays dividends continuously

at

rate

q

and

if interest

rates

are

constant

and

equal

to

飞

in

order

to

obtain

the

same payoffs

at

maturity

,

the

asset positions

in

the

two portfolios

must

be

adjusted

as follows:

The

first replicating portfolio will

be

made

of

the

following positions:

• long

2e-

qT

units of

the

asset;

V

28

-

30

20

十

20

40

二

28

-

3

马。一

38

+60

十

2

凡。+

28

-

80

=

8-3

马。十

2

凡。一

20.

(1.

33)

The

positions of

the

replicati

吨

portfolio

(1.

33) can

be

summarized as follows:

• long one

unit

of

the

asset;

•

short

$20 cash;

•

short

3

put

options with strike K =

20

on

the

asset;

• long 2

put

options

with

strike K = 40

0

丑

the

asset.

We check

that

the

payoff of

this

portfolio

at

maturity

, i.e.,

which

is

equivalent

to

taking a

short

position in

three

units of

the

underlying

asset

,

taking

a

short

position in

three

put

options

with

strike

20

,

and

being

a long $60.

Similarly

,

to

synthesize a long position in two calls

with

strike

40

, note

that

2040

2

凡。十

28

-

80

,

(1.

32)

which is equivalent

to

a borrowing $80,

taking

a long position

in

two

units

of

the

underlying asset,

and

taking a long position in two

put

options

with

strike 40.

Usi

吨(1.

31)

and

(1.

32),

we

obtain

that

the

payoff

at

maturity

given

by

(1.

29) can

be

replicated

usi

吨

the

followi

吨

portfolio

consisti

吨。

f

put

options,

cash,

and

the

underlying asset:

42

CHAPTER

1.

CALCULUS

REVIE1V.

PLAIN

VANILLA

OPTIONS.

1.3.

SOLUTIONS

TO

SUPPLE

1V

IENTAL

EXERCISES

43

Recall

that

C

1

> C

2

> C

3

. Using

the

value of X3 from

(1.

38), it follows

that

(1.

36)

and

(1.

37) hold

true

if

and

0

日

ly

if

C

1

- C

3

C

1

二



C

2

-

C

3

- C

2

We conclude

that

arbitrage happens if

and

only if

we

can find

Xl

> 0 and

X2 > 0 such

tl

时(1.

40)

and

(1.

41) are simultaneously

satis

丑

ed.

Therefore,

no-arbitrage

exists

if

and

only if

X2K

3

-K

1

(1.

41)

Xl

- K

3

-

K

2

Also,note

that

if

(1

.4

0)

holds

true

,

then

(1.

39) is satisfied as well,since

since

C

1

> C

2

> C

3

.

Therefore, no arbitrage can be obtained by being long

the

options with

strikes

K

1

and K

2

and short

the

optio

日

with

strike K

3

.

We

conclude

that

(1.

42), i.e.,

K

3

-K

1

C

1

一

α

'二

K

3

-K

2

、

C

2

-

C

3

is

the

only condition required for

no-arbitrage.

口

x1(8

- K

1

) +

x2(8

-

K

2

)

一句

(8

-

K

3

)

三

0

，

V

8

三

O.

(1.

48)

The

inequality

(1

.4

8)

holds if

and

only if

Xl

+X2 -

X3

主

O.

(1

.4

9)

It

is easy

to

see

that

(1.

47) and

(1.

49) cannot

be

simultaneously satisfied:

C,

C?

3 -

Xl

一二

+X2

一二

>

X1

+X

~

C

3

~

C

3

~

(1.

40)

X2

、

C

1

-

C

3

、-

Xl

…C

2

-

C

3

'

Portfolio

2:Long

the

K

1

-option

, short

the

K

2

-option

,

short

the

K

3

一

option.

Arbitrage exists if

we

can

且口

d

Xi

> 0, i = 1 : 3, such

that

Xl

- X2 -

X3

主

0;

(1.

45)

Xl

(K

3

- K

1

) -

x2(K

3

-

K

2

)

三

O.

(1.

46)

However

,

(1.

43)

and

(1.

45) cannot

be

simultaneously satisfied. Since C

1

>

C

2

> C

3

,

it

is easy

to

see

that

的

C

1

-

X2C2

-

X3C3

-

0;

(1.

43)

Xl

(8

- K

1

) -

x2(8

-

K

2

)

一句

(8

-

K

3

)

三

0

，

V

8

三

O.

(1.

44)

The

ineql

叫

ity

(1.

44) holds

if

and only if

the

following two conditions are

satisfied:

C?

C'}.

1 =

X2

一二

-X'}.

一二

<

X')

十

X3·

C

1

--v

C

1

-

--.,

In

other

words, no arbitrage can

be

obtained by being long

the

option

with

strike K

1

and

short

the

options

with

strikes K

2

and K

3

.

Portfolio

3:Long

the

K

1

-option

,long

the

K

2

一

option

，

short

the

K

3

-option.

Arbitrage exists if we can find

Xi

> 0, i = 1 : 3, such

that

We conclude

that

there

is

丑。

-arbitrage

directly following from

the

Put-

Call parity if

and

only if

Cα

sk

…凡

id

三

8e-

qT

- K

e-

r1

、三

C

bid

一凡

sk·

口

Problem

10:

Denote by

Cbid

and

Cask

,

and

by

凡

id

and

凡

sk

，

respectively,

the

bid and ask prices for a plain vanilla European call and for a plain vanilla

European

put

option,

both

with

the

same strike K and

maturity

T,and on

the

same underlying asset

with

spot price 8 and paying dividends continuously

at

rate

q.

Assume

that

the

risk-free interest rates are constant equal

to

7'.

Find

necessary

and

sufficient

no-arbitrage

conditions for

Cbid

,

Cα

sk

，

Pbid, and

凡

sk·

Solution: Recall

the

Put-Call

parity

C - P

8e-

qT

-

Ke-

rT

,

where

the

right

hand

represents

the

value of a forward contract on

the

un-

derlying asset

with

strike

K.

An arbitrage would exist

• either if

the

purchase price of a long call short

put

portfolio,i.e.,

Cα

砾，一

Pbid

were less

than

the

value

8e-

qT

- K

e-

rT

of

the

forward contract,i.

e.

,if

Cα

sk

一凡

id<Se-qT-kfrT;

• or if

the

selling price of a long call short

put

portfolio,i.

e.

,

Cbid

-

Pask

were

greater

than

the

value

8e-

qT

-

Ke-

rT

of

the

forward contract, i.

e.

,

if

C

bid

一凡

sk

>

Se-

qT

- K

e-

rT

(1

.4

2)

(1.

47)

K

3

-K

1

C

1

-

C

3

K

3

-K

2

、

C

2

-C

3

X1

C

1

十

X2C2

-

X3C3

-

0;

Chapter

2

Improper

integrals.

Numerical

integration.

Interest

rates.

Bonds.

2.1

Solutions

to

Chapter

2

Exercises

Problem

1: Compute

the

integral of

the

function

f(x

,y) = x

2

- 2y on

the

region bounded by

the

parabola y =

(x

+

I?

and

the

line y =

5x

-

1.

Solution: We first identify

the

integration domain

D.

Note

that

(x

+

I?

=

5x

-1

if and only if x = 1 and x

=2

,and

that

(x

+

1)2

三

5x

- 1 if 1 < x <

2.

Therefore,

D =

{(x

,

y)

11

三

Z

三

2

and

(x+

1)2

三

ν

三

5x

- I}.

Then

,

fin

川白

[(

汇

ιιi;:

(归户

Z

[

((怡均牛

2

与

U

一

泸的州叫叫)川川将巾

U|tcc

口

U!:

汇口口

:ζ:Uω;

二二

L

斗

J;

斗

ω)

户

)2

)讪

dx

I

泸

(5x

-

1

一一-

(归仙川

Z

叶川+札

1

呐一((仰一

1)2

一归仙川+刊

1

阳

Z

1，'

(σ

5x

一

1

一川附一仰一

1

忏叫十刊(川

l

俨(←一泸仇川十叶

3x

川

Z

卜川一

-2

川

Z

叫)川

d

白

Z

户←=←一;

口

Problem

2:

Let f :

(0

，∞)→lR

denote

the

Gamma

functio

孔

i.e.

，

le

t.

的

) =

l

x

俨俨

χ

飞

VZ

俨俨

α

←叫~一→

1

45

46

CHAPTER

2.

NUMERICAL

INTEGRATION.

BONDS.

(i) Show

that

f(

α)

is

well defined for any

α>

0, i.e. ,show

that

both

[x

α

-1

e-

x

dx

_

时

12α-1

e

一♂

dx

and

['

x"• 1

e-

x

dx

出

[x

α

一

1

e-

x

dx

exist

and

are finite.

(ii)

Prove

，旧吨

i

的

egration

by

parts

,

that

f

(α)

=

(α-

1)

f(

α

一

1)

for any

α>

1.

Show

that

f(l)

= 1

and

conclude

that

,for any

η

三

1

positive integer,

f(n)

=

(η-I)!.

Solution:

(i) Let

α>

0.

Intuitively,

note

that

, as x

\、

0

，

the

functio

丑

Zα

-l

e

-x

is

0

丑

the

order of x

o:-

1

, since lim

x\

",

0

e-

x

=

1.

Since

~t

Il}

(1

x

α

-1

dx

工

lim

主

|zllim(l-tα)

t\

",

0 Jt

t

\

、

oαIt

it

follows

that

l

α

I

x

α

一

1

e

-x

dx

- lim I

x

α

-1

e-

x

dx

Jo

t

\o

Jt

exists

and

is finite.

In

a similar intuitive way,

note

that

, as x

→∞，

the

function

x

O:一

l

e

-x

is

0

日出

e

order of

e-

x

, since

the

exponential function dominates any power

function

at

infi

丑

ity.

Since

Jim I

e-

x

dx

- Ji

ITl

(1

-

e-

t

)

- 1,

"~UV

Jl

"~UV

it

follows

that

fzMe

曰

:Etlzα

-1

e-

x

dx

(2.1)

exists

and

is finite.

l\IIaking these intuitive arguments precise is somewhat more subtle. We

include a mathematically rigorous arguments for

, e.g., showing

that

the

in-

tegral in

(2.1)

exists

and

is

finite.

By

de

丑

nition

，

we

need

to

prove

that

,for

a

町

ε>

0,

there

existsη(ε)

> °

such

that

I

∞川

-X

dx

<

飞

Vs>

η(ε)

(2.2)

2.1.

SOLUTIONS

TO

CHAPTER

2

EXERCISES

47

Note

that

there

exists N > °such

that

zα-le-z

< e-z/27

Vx>

N,

(2.3)

Sl

日

ce

Iiα

-1

_-x/2

1m

x-

- e

-,-

=

X-

• x

Also, since

lim

x

→。

c

e-

x

!2

= 0,

it

follows

that

, for

any

ε>

0,

there

exists

m(E)

> °such

that

2e-

m

(

f.

)!2

<已

(2

.4)

Choose

n(

ε)

=

max(m(ε

，

N)).

From

(2.3)

and

(2

.4)

we

obtain

that

zα-l

e-z

<e-z/27V

Z

〉

η(ε);

(2.5)

2e-

n

(

f.

)!2

<

ε(2.6)

We

can

then

use

(2.5)

and

(2.6)

to

show

that

,for any s

>

η(ε)

,

J

豆豆/，'

X"-l

川

Z

<J

坦/，'

~-x

切=出(

_2e-

t

!2

十川)

2e-

s

!2

<

2e-

n

(

f.

)!2

<

飞

which is

what

we

wanted

to

show;

d.

(2.2).

(ii)

It

is easy

to

see

that

m)=fev=JI

叫

tfzdz=

出

(_e-

t

+1) = 1

Assume

that

α>

1.

By integration by

parts

,

we

find

that

/,

X

X"-l

川

Z

f(

α)

,

G

Z

e

q&

ffl'o

•-

ij-d

ufbfqh

ZZ7

smce

lim

x

o:-

1

e-

x

z\0

limx

α-1

- 0, for

α>

1;

2\J

ια-1

lim

二--:-

=

0.

t• x e

L

lim t

o:-

1

e-

t

→

οc

48

CHAPTER

2.

NUMERICAL

INTEGRATION.

BONDS.

2.1.

SOLUTIONS

TO

CHAPTER

2

EXERCISES

49

For any positive

integerη>

1,

we

五 nd

that

f(n)

=

(η

-l)f(

η-1).

Since

f(l)

= 1, it follows by induction

that

f(

η)

=

(η-1)!

口

Problem

3: Compute

an

approximate value of

J1

3

-IX

e-xdx

using

the

N

Ii

d-

point rule,

the

Trapezoidal rule,

and

Simpson's rule.

Start

with

η=

4 inter-

vals

,

and

double

the

number of intervals until two consecutive approximations

are within

10-

6

of each other.

Solution:

The

approximate values of

the

integral found using

the

N

Ii

dpoint,

Trapezoidal,

and

Simpson's rules

can

be

found in

the

table

below:

No.

Intervals

NIidpoint Rule

Trapezoidal Rule Simpson's Rule

4 0

.4

0715731 0

.4

1075744

0

.4

0835735

8

0

.4

0807542 0

.4

0895737

0

.4

0836940

16 0

.4

0829709 0

.4

0851639 0

.4

0837019

32

0

.4

0835199 0

.4

0840674 0.40837024

64

0

.4

0836569 0

.4

0837937 0

.4

0837024

128 0

.4

0836911 0

.4

0837253

256 0

.4

0836996 0

.4

0837082

512 0

.4

0837018 0

.4

0837039

1024 0

.4

0837023 0

.4

0837028

The

approximate value of

the

integral is 0

.4

08370,

and

is obtained for a

256 intervals

partition

using

the

N

Ii

dpoint rule, for a

512

intervals

partition

using

the

τ

'r

apezoidal

rule,

and

for a

16

intervals

partition

using Simpson's

rule.

口

Problem

4: Let f :

R

卜→

--+R

问问

gi

忖阳

V

(ωi)

Use NIidpoint rule with tal =

10-

6

to

compute

an

approximation of

(1

币

5/2

I = I f(x) dx = I

~一一亨

Jo

OJ

'--I

----

J

o

1 +

x~

(ii) Show

that

f(4)

(x) is

not

bounded on

the

interval

(0

,1).

(iii) Apply

Si

皿

psor

内

rule

with

n =

2

飞

k

= 2 : 8, intervals

to

compute

the

integral

I.

Conclude

that

Simpson's rule converges.

Solution:

(i)

The

approximate value of

the

integral

is

0.179171,

and

is obtained for a

partition

of

the

interval

[0

,

1]

using

512

intervals:

No. Intervals

NIidpoint 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

computi

吨

f(4)(X)

，

note

tl

时

the

denominator 1 + x

2

of f

(x)

is bounded away from 0, and

that

the

fourth derivative of

the

numerator of

f(x)

is on

the

o-rder of

X-

3

/

2

,which is not defined

at

0,

and

is unbounded in

the

limit as x

、

o.

(iii)

Usi

吨

Simpsor

内

rule

，

the

following approximate values of

the

integral

are obtained:

No. Intervals

Simpson's Rule

4 0.179155099725056

8

0.179169815603871

16

0.179171055067087

32

0.179171162051226

64

0.179171171372681

128

0.179171172188741

256

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.

口

Problem

5: Let K ,

T

，

σand

r

be

positive constants. Define

the

function

g

:R

• R as

g(x)

=

在

rhu?

where

b(x)

=

(In

(圣)十(叫去

)

T)

/

(σ

-IT).

Comp

l.叫

(x)

Solution: Recall

that

irf

叫

=

f

川。)

50

CHAPTER

2.

NUMERICAL

INTEGRATION.

BONDS.

2.1.

SOLUTIONS

TO

CHAPTER

2

EXERCISES

51

Therefore,

it follows

that

g'(x)

V

左

i

七言

b

印(归

ω

阳

Z

功)

l

…(

(In

(圣)十

(γ

十号)

T)2

\ n

x

O"-J2

百…

y

\

2σ

2T

J'

~

g"(t) = h(t),

which is

what

we

wanted

to

show.

gil(t) = h(t)

.口

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

B

=至

100

e-

r

(O

，

O.5)O.5

十空

100

e-

r

(

即)十三

100 e-

r

(

阳)1.

5

2

十

(10

∞

0

十

;•

l

叫

e-

r

μ

川川川

r

叫咐咐仰(仰队阳

O

矶

ω

，

2

where

C 工

0ω.β05

，

and

r(0,0.5) =

0.05

，

γ(0

，

1)

= 0.0525, r(0

,1.

5)

= 0.0535,

γ(0

，

2)

= 0.055.

The

data

below refers

to

the

pseudocode from Table 2.5 of

[2]

for

c

∞

om

putir

丑

19

the

bond price given

the

zero

rate

curve.

Input:

n = 4

Lcash

且

ow

=

[0.5

1

1.

5 2];

v

一

cas

且且

ow

=

[2.5

2.5 2.5 102.5] .

The

discount factors are

Problem

6: Let

1

巾)

be

a continuous

fun

创

0

口

such

that

J

之

o

Ixh(x)ldx

exists. Define g(t) by

以

t)

=

[>O(川

)h(x)

d

and

show

that

Solution: Recall

that

,

ifα

(t)

and

b(

t) are differentiable functions

and

if f

(x

,t)

is a continuous function such

that

?It

(x

,t) exists

and

is continuous,

then

,

Tb

,

Tb

rlu

i'U

'hu

,

Tb

,

Tb

vhu

rld

+

2

,

?-U

Z

\Ill-/

Z

VG

z

rlu

/-11iI\

d-a

iTb

i'U

rld

z

Z

YG

z

rld

/Ill-\

d-a

disc = [0.97530991 0.94885432 0.92288560 0.89583414

],

and

the

price of

the

bond

is

B =

98.940623.

口

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

For our problem,

Problem

8:

The

continuously compounded 6-month, 12-month, 18-month,

and 24-month zero rates are

5%

,5.25%, 5.35%,

and

5.5%,respectively. vVhat

is

the

par

yield for a 2-year semiannual coupon bond?

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

户

100 -

~

100

e-

r

(O

,

O.5)O.5

+

~

100

e-

r

(O

,1)

+

二

100

e-

r

(O

，1.

肌

5

2 2

+

(100

+

~

1

叫

e-

r

(O

，

2)

α

(t)

= t and

f(x

,t) =

(x

- t)h(x),

(2.8)

where

h(

x) is

conti

丑∞

us.

Then

,

~{

(x

,t) =

-h(x)

is continuous. Note

that

f(

α

(t)

，

t) =

f(t

,t) =

(t

- t)h(t) =

O.

(2.9)

From (2.7-2.9)

,

we

conell

由

that

g'

(t)

=

二(

['

(x

- t)h(x)

dX)

= -

[\(X)

dx

Since

二(

1:

川

x)

=

一川川

7

QU

U

28

T

C

2(1 -

e-

r

(O

,

2)2)

e-

r

(O

，

O.5)O.5

十

e-

r

(O

，

l)

十

e-

r

(O

，1.

5)L5

十

e-

r

(O

，

2)2

•

52

CHAPTER

2.

NUNIERICAL

INTEGRATION.

BONDS.

2.1.

SOLUTIONS

TO

CHAPTER

2

EXERCISES

53

Problem

9: Assume

that

the

continuously compounded instantaneous in-

terest

rate

curve has

the

form

(

~

0.05.

__\

~({\

t)飞叮

十

I

100

十丁一

IUU

I

e

一

I

\V

忡忡

2.5 disc(l) + 2.5 disc(2) + 2.5 disc(3) + 102.5 disc(4)

99.267508.

口

For

the

zero rates given in this problem,

the

corresponding value of

the

par

yield is C = 0.05566075,i.e. ,

5.566075%.

口

r(t)

=

0.05

十

0.0051

泣。

+

t)

, V

t

三

O.

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?

(i)

Find

the

correspondi

吨

zero

rate

curve;

(ii)

Compute

the

6-month,

12-mo

时

h

，

18-month, and 24-month discount fac-

tors;

(iii)

Find

the

price of a two year semiannual coupon bond with coupon

rate

5%.

nu

>

vv

TT

p''Ifo

l--t

,

TLV

nU

Solution:

The

price, duration,

and

convexity of

the

bond

can be obtained

from

the

yield y of

the

bond as follows:

B

=

兰

3

仪

p(

一到十川

p(

补

1

f

ι3i

(i

\5

(5\

飞

一

I

) :

:-

exp I

-~y

I + 103

~exp

I

-~y

I I

B\

仨

t

2

飞

2

07

}

~

2\

2'-'

} J

1

(~9i

(i

\25

(5\1

=一

I

)

~

VA"

exp

(一

~y)

+ 103

-A~

exp

(一

~y

) I. (2.12)

B\

仨

t

4

飞

2

07

} ' - - -

4\

2'"

} J

The

data

below refers

to

the

pseudocode from Table 2.7 of

[2]

for computing

the

price, duration and convexity of a bond given

the

yield of

the

bond.

Input:η=

5;

y = 0.09;

Lcash_flow =

[0.5

1

1.

5 2 2.5];

v

一

cash_

自

ow

=

[3

3 3 3

103]

.

Output:

bond

price B = 92.983915, bond duration D = 2.352418, and bond

convexity

C =

5.736739.

口

(2.10)

Solution:

(i) Recall

that

the

zero

rate

curve

r(O

,

t)

can

be

obtained from

the

instanta-

neous interest

rate

curve

r(t)

as follows:

D

41·A

,,

l

‘\

Then

,

叫

OJ ) = ;

才

f

庐扣

t

、

h0ω05

十

0ωO

∞

0

臼阳川叫

5

盯协

ln(

丑叫(川

)

;(0

阳+

0ω0

∞阳

0

l

丑

(ο1

+

t)

)

0.045 +

0.005(1

十

t)

(ii)

The

6

咄

onth

，

12-month, 18-month,

and

24-month discount factors are,

respectively,

disc(l)

disc(2)

disc(3)

disc(4)

e-

r

(O

,

O.5)O.5

_ 0.97478242;

e-

r

(O

,

l)

- 0.94939392;

e

-r(O

，1.

5)

1.

5

二

0.92408277

;

已

-r(O

，

2)2

- 0.89899376.

Problem

11:

The

yield of a

14

months quarterly

coupo

且

bond

with

8%

coupon

rate

is

7%.

Compute

the

price, duration,

and

convexity of

the

bond.

Solution:

The

quarterly bond will pay a cash flow of

1.

75

in

2, 5, 8, and

11

months, and will pay

10

1.

75

at

maturity

in 14 months.

The

formulas for

the

price,

duration

,

and

convexity of

the

bond in

terms

of

the

yield y of

the

bond

are similar

to

those from (2.10-2.12). For example,

the

price of

the

bond can

be

computed as follows:

(

2\(

5\/

民\

B =

1.

75

exp I -

~-"y

I +

1.

75

exp I -

~-"y

I +

1.

75

exp I -

~-"y

I

飞

12'-'

J

飞

12'-'

J \

12'-'

J

(

11\(

14\

+1.

75

exp

I

一

~~y

I +

10

1.

75

exp I -

-,,-

y I

飞

12'"

J

飞/

(iii)

The

price of

the

two year semiannual coupon

bond

with

5%

coupon

rate

IS

0.05

~({\

(\

,,\{\

"

0.05

一

~I{\

1\

0.05

β:

刁

-100e-r(O?05)05

十一

~V

100

e-

r

(O

,I)

+

丁一

100

e-

咐1.

5)

1.

5

2

Stefanica D. Solutions Manual: A Primer For The Mathematics Of Financial Engineering

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