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

174

CHAPTER

7.

MULTIVARIABLE

CALCULUS.

Let f(x) =

ln~x)

with f :

(0

,

00)

•

JR.

Then

f'(x)

=

弓凶

The

function

f

(x)

has one crit

iC

al point

correspondi

吨

tox

=

飞

IS

1

肌

reasi

吨

on

the

interval

(0

,

e)

and

is

decreasing on

the

interval

(e

，

∞).

We conclude

that

f

(x)

has a global maximum point

at

x = e,i.e., f

(x)

<

f

(e)

=

~

for all x > °with x

护民

and

therefore

ln(

作)

忡忡

-r<;7

which

is

eq

山

Talent

to

作

e

< e

1r

;

cf.

(7.22).

口

Problem

3:

Let

叩)

:

[0

，∞)→

[0

，∞)

be two continuous functions with

positive values. Assume

that

there exists a constant M > °such

that

u(x)

三

M

+1

x

u(t)v

(7.23)

Show

that

也

(x)

三

M

叫

1"

v(t)

dt)

,

\I

x

::::

0

(7.24)

Solution: Define

the

functionω:

[0

，∞)→

[0

，∞

)as

\111/

U

Z

/III-\

p

x

e

\1111/

,

d

U

u

z

+

M

/III-\

Z

(7.25)

丑

ecall

that

Z

u

--

\iI/

,

TL

,

?-b

Z

flo

Z

U

Z

\Ill-/

,

?-U

JU

,

Tb

4·L

WU

Z

/III-\

where

the

derivative is computed with respect

to

x.

Using

the

Product

丑

ule

，

we

find

that

州=叫

x) υ(

以

P

( -

1"

v

(t)

dt)

+

(M

+

1"

u(t)

υ(

叫(→(叫(-

1"

V(t)dt))

巾)

(u

仲

M

-

1"

u(t)v(

叫叫-

1"

V(t)dt)

(7.26)

7.3.

SOLUTIONS

TO

SUPPLEMENTAL

EXERCISES

175

In

other

words

，

ω(

x)

is a

decreasi

吨

function

on

the

i

凶

nt

怡

er

凹

va

址

I

[归

0

，∞)

an

丑

l

therefore

ω(0)

三四

(x)

for all

x

主

0.

Since

ω(0)

= M , and using (7.23), it

follows

that

M

::::

(M

+

1"

u(t)v(

叫叫-

1"

v(t) dt)

::::

u(x)exp(-

1"

V(t)dt)

\l

x::::O

,

Problem

4:

What

does

the

boundary condition

V(B

,t) = R for a down-

and-out

call with barrier B and rebate R > °correspond

to

for

the

fu

日

ction

u(x,r) defined as follows:

V(8

,

t)

=

exp(

一

α

x

- br)u(x,r),where

L

(8\(T

-

t)

σ2γ

-q

1 L

(r-q

I

1\2

2q

x =

In

I

~_

I

、

r=

、

α:

一"一一、

b=l

一"一十一

1+

一

τ.

飞

K

J 1

2σ~

2

1\σ

~

. 2

J

σ4

Sol

ω

毗

4ω

仰

ion

仰阳

7ηZ

corresponds to

°三

7

三字

Since

也

(x

，

r) =

exp(αz

十

br)V(8

，

t),

the

boundary condition corresponding

to

V(B,t) = R for all

°三

t

三

Tis

u

(In

(;)

,

7)

叫咱

)

+b

7

)

川)

(;)α

川，

V

°三

γ

三字口

Problem

5: Assume

that

the

function

V(8

,I ,t)

satis

丑

es

the

PDE

θVθ

V

1

()_()δ

2V

θV

一一十

In8~

十一

σ

'l，

8

'l，

一一十

r8

一一

-

rV

0.

θtθ

I

'

2θ8

2θS

Consider

the

following change of variables:

(7.27)

V(8

,I,t) = F(y,t), where y

cu-

a-

1Ei-

,

Tb

…

/lt\-

Show

that

F(y

,t)

satis

且

es

the

following PDE:

但十

σ

2(T

-

t)2a

2

F

十

(r

_

~2)

丘

1

笠

-

rF

= °

θ

t

I

2T2

θν2\

2)

T

θu

176

CHAPTER

7.

MULTIVARIABLE

CALCULUS.

7.3.

SOLUTIONS

TO

SUPPLEMENTAL

EXERCISES

177

Solution: Using chain rule,

it

is

easy

to

see

that

θVθ

FInS

θF

δtθ

t

T

θν7

δ

V

1θF

θ

I

T

θu?

δ

V

1

T-t

θF

δ

SST

θν?

θ

2V

1 (

(T

-

t)2

θ 2F

T-t

θF\

一一_.

一一一一.

θ

S2 S2

飞

T2

θ

y2

T

θ

y)

Then

,

the

PDE

(7.27) for

V(S

,I ,t) becomes

the

followi

鸣

PDE

for

F(y

,

t):

θVθ

V

1

')~')护

VθV

o =

一一

+

1

口

S

一一

+一

σ

L.

S

L.

一一

τ

+γS

一一

-

rV

θtθ

I

'

2-

.-

8S

2

θS

θ

FInS

θ

F

__1δF

一一一一一一一十

lnS~~

θ

t

T

θ

y

- T

θu

十

1σ2

(

(T

-

t

)Z

θ

2F

T-t

θFV+T-tθ

F

~

i

一一一一一一一一一…一一一

I

+

r

一一一一一一一一

-

2\

T2

δ

y2

T

θ

y)

, •

T

θu

θ

F

σ

2(T

-

t)2

θ

2F

,

(σ2\

T-t

θF

一一+

n

~

~十

Ir

一

-=::-1

一一一一一

-

rF.

口

θ

t'

2T2

8

γ\

2)

T

θu

which

is

what

we

wanted to

show.

己

Problem

7:

For

the

same

maturity

, options with different strikes are

traded

simultaneously.

The

goal of this problem is

to

compute

the

rate of change of

the

implied volatility

ωa

function of

the

strike of

the

options.

In

other words, assume

that

S,T ,q

and

r are given,and let

C(K)

be

the

(know

叫

value

of a call option with

maturity

T and strike

K.

Assume

that

options with all strikes K exist.

Defi

缸

fine

t

也

he

implied volatility

σ

叽

in

η1

以

K)

a

创

s

the

unique solution

to

C(K)

=

CBs(K

，

σ

imp(K))

，

where

CBs(K

，

σ

叽仰

n

η

以

1

S

阮

cl

挝

hoωoles

臼

s

value of a call

0

叩

pt

挝

io

∞

n

with strike K on an

unde

臼

rl

勾甘切

i

坊切

y

抖

rin

吨

1

鸣

ga

部

ss

附

e

创

t f

，扣

'01

如

low

叭

in

吨

ga

log

伊

nω10ωr

口

m

口

rna

丑

1

恼叫

al

model

w

呐

it

由

h

volatility

σ

叫

(K). Find an implicit differential equ

任

tion satisfied

byσ

叫

(K)

，

i.e.

，缸

d

θσ

imp(K)

θK

a

邸

saf

缸

un

旧

ct

钊

io

∞丑

of

σ

叽

imη1

以

K).

Solution: We

first

自

nd

the

partial derivative of

the

Black-Scholes value

CBs(K)

of a call option with respect

to

its strike

K.

Recall

that

CBs(S

,

K)

=

Se-

qT

N(d

1

) -

Ke-

rT

N(d

2

).

θ

d

1

俨

一币

EJd

Se

一句

'(d

1

)

;~一严

N(d

2

)

-

Ke-

r

:J.'

N'(d

2

)

坛

(7

到

Since

d

1

工

d

2

十

σ

叮

，

it

follows

that

(7.31)

(7.30)

(7.29)

θ

d

2

θK

-

e-

rT

N(d

2

)

θ

d

1

θK

We

co

肌

lude

from (7.30)

tl

川

θ

CBS

θK

Also, recall

that

Se-

qT

N'(d

1

)

-

Ke-

rT

N'(d

2

);

d.

Lemma 3.15 of

[2].

From (7.28) and (7.29),

we

find

that

到

C

'R.c:

~'T'

_

_,

• ,

~

_",

rr

~

TI

I "

(8d

1

8

ι\

#二

-

e-

rT

N(d

2

)

十

Ke

一叩'(也)

\拔一括)

Then

,

θ

CBS

θK

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 divi-

dends 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(

d

1

) -

K

e-

rT

N(

d

2

) <

Se-

qT

,

since

N(d

1

)

< 1 and

N(d

2

)

>

O.

If

Se-

qT

< S - K , which is

equiv

，由

lt

to S

>

亡

ι

T

> 0 since q > 0,

it

follows

that

CBs(

氏

K)

< S -

K.

We

conclude

that

CBS(S

,

K)

< S - K , V S

>

4K

川

178

CHAPTER

7.

MULTIVARIABLE CALCULUS.

We now differentiate with respect

to

K

the

formula

C(K)

=

CBs(K

，

σ

问

(K))

which

is

the

de

直缸缸

n

旧

l

让山

i

扰

tior

丑

1

of

σ

町仰

mη1

以

K).

Note

that

C

(K)

is

assumed

to

be known

for all

K ,as it

the

market prices. Using Chain Rule

and

(7.31)

we

find

that

r

e

+iu

p

a

h

c

θC

θK

δ

CBS

Iθ

CBS

θσ

imp(K)

一一一·

θ

K

θσθK

-e

一句

(d

2

)

+

鸣

a(CBs)

叫

ffk)?

(7.32)

Lagrange

multipliers.

N-

dimensional

Newton's

method.

Implied

volatility.

Bootstrapping.

where

θCBS

C1

_-aT

I

T_

'!J.

万

=Se-qVV=dse-PNf(do

V'T

Ke-

rT

N'(d

2

) -

K

产品生

d.

(η9).

We conclude

that

the

implied differential equation (7.32) can be

written as

vega(CBs)

8.1

Solutions

to

Chapter

8

Exercises

IT

一兰

θσ

imp(K)

VK

2

θ

-y-

=

在

+

e

一叩

(d

2

)

，

Pro

blem

1: Find

the

maximum

and

mi

山

num

ofthe

function f(Xl'

X2

，

句)=

4X2

-

2X3

subject

to

the

constraints

2X1

-

X2

-

X3

= 0 and xI +

x~

= 13.

Solution: We reformulate

the

problem as a constrained optimization problem.

Let

f :

JR3

•

JR

and 9 :

JR3

•

JR

be defined as follows:

dry

=

In

(去)十

(r

-

q)T

一切

(K)VT

h

..

(j

imp(K)VT

2'

~

削

=

4X2

-

2

叫仰)

=

(专

JZ

二

3)

with

where

x =

(X

l,

X2

,

X3).

We want

to

五

nd

the

maximum and minimum of f (x)

on

JRδsubject

to

the

constraint

g(

叫

=0.

We

自

rst

check

that

ra

出(飞

7

g(x)) = 1 for any x such

that

g(x) =

O.

Note

that

\1

g(x) =

(2;

，二

en

It

is easy

to

see

that

rank(

\7

g(

x)) = 2, unless

Xl

=

X2

= 0, in which case

g(x)

并

O.

The

Lagrangian associated to this problem

is

F(x

，

λ

)

4X2

-

2

♂

3

十

λ

1(2x1

-

X2

-

X3)

十七

(xr

+

x~

- 13), (8.1)

whereλ=(λ1

，

λ

2)t

εJR2

is

the

Lagra

口

ge

multiplier.

We

now find

the

critical points of F (

x

，

λ).

Let

x

均

0=(

阳

Z

均

0

，

1

，

X

协

Zωa

缸

an

丑

1

λ

沁

0=(λ

沁

0

，

1

，

λ

沁

O

叩

ω

，

2

公).

From (8.1) it follows

that

\7

(x

，

)..)F(xo

，

λ0)

= 0 is equivalent

179

180

CHAPTER

8.

LAGRANGE

MULTIPLIERS.

NEWTON'S

JVIETHOD. 8.1.

SOLUTIONS

TO

CHAPTER

8

EXERCISES

181

O&

nu

4t4

nu

000020·7

diU2'

ES-7

O

QU

4EA

-mhz

mm

-HM4L

fwu

'bs

。;

0;

13.

(i)

Find

the

asset allocation for a

mi

山

nal

variance portfolio with

12%

ex-

pected

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

二

1

: 4, denote by

Wi

the

weight of asset i in

the

portfolio.

Recall

that

the

expected value and

the

variance of

the

rate

of

return

of a

portfolio made of

the

four assets given above are,respectively,

and

E[R]

=

ω1μl

十

ω2μ2+ω3μ3

十

ω4μ4;

(8.4)

var(R)

=

ω?σ?

十

ω2σ~+ω;σ;

十

ωiσ~

(8.5)

十

2

(ω1ω

伊

1σ

2P1

，

2

十

ω2ω3σ

2

{J

3P2

，

3

十

ωlω3σlσ

3P1

，

3)

，

XO

,l =

-2;

XO

,2 =

3;

XO

,3 =

-7;λ

。，

1

=

-2;λ0

，

2

=

-1.

(8.3)

For

the

在

rst

solution (8.2),

we

compute

the

Hessian

D

2

月

(x)

of

凡

(x)

=

f(x)

+

λ

切

(x)

，

i.e., of

凡

(x)

-

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

xi+x~-4x1

+6x2-13

si

口

ce

Pi,4 = 0 for i = 1 :

3.

We do not require

the

weightsωi

to be positive, i.

e.

,

we

allow taking

short positions on each one of

the

assets. However,

the

following relationship

between

the

weights must hold true:

and obtain

I 2 0

0

飞

D

2

凡

(x)

工

I

0 2 0 I

\ 0 0 0 I

which

is

(semi)positive definite for

a

町

Z

εJR3.

This allows us

to

conclude

directly

that

the

point

(2

，一

3

，

7)

is

a minimum point for f(x). Note

that

f(2

,

-3

,

7)

=

-26.

Similarly, for

the

second solution (8.3),

we

自 nd

that

月

(x)

-

xi

-

x~

-

4X1

+

6X2

+

13

ω1

十

ω2

十

ω3+ω4

=

1.

(8.6)

(i) We are

looki

吨

for

a portfolio with given expected

rate

of

return

E[R] =

0.12 and minimal variance of

the

rate

of return. Using

(8

.4

-8.6),

we

obtain

that

this problem can

be

written as

the

following constrained optimization

problem:

且

ndω°

such

that

gErof(ω)

=

f(

ω0)

，

(8.7)

and

w

呐咄

t

扭

ler

陀

e

ω

=(

ω

i)

山

)i=

吐=

f(

ω

叫)

=

0.0625ωi

+

0.0625ω;+0Oω9ω:+O

04tωρi

一

0.03125w1W2

-

0.0375ω2ω3

十

0.0375ω1ω3;

/ωl

十

ω2

十

ω3

十

ω4-

1

飞

g(

ω)

= \

0

伽

1

十

O

山

2

+

0.16ω3+

0

伽

4

一

-

0.12 )

(8.8)

I

-2

0

飞

D

2

几位

)

= I

。

一

-2

0 I

飞

o

0 0 I

which

is

(semi)negative

de

主

nite

for any x

巴

JR3.

We conclude

that

the

point

(-2

，

3

，一

7)

is a maximum point for f (x). Note

that

f

(…

2,3,

-7)

=

26.

口

(8.9)

Problem

2: Assume

that

you can

trade

four assets (and

that

it

is

also

possible

to

short

the

assets).

The

expected values,

standard

deviatio

风

and

correlations of

the

rates of

return

of

the

assets are:

It

is

easy

to

see

that

rank(

\7

g(

ω))

= 2 for

缸lY

ωεJR

4

，

since

μ1

-

0.08;σ1

- 0.25;

P1

,2 = - 0.25;

μ2

-

0.12;σ2

- 0.25;

P2

,3 - - 0.25;

μ3

=

0.16;σ3

-

0.30;ρ

1.

3

工

0.25;

μ4

=

0.05;σ4

- 0.20;

PiA

- 0, V i = 1 :

3.

The

Lagrange multipliers method

can

therefore

be

used

to

丑

nd

the

minimum

variance portfolio.

8.1.

SOLUTIONS

TO

CHAPTER

8

EXERCISES

182

CHAPTER

8.

LAGRANGE

MULTIPLIERS.

NEWTON'S

METHOD.

Denote by

λ1

and

λ2

the

Lagra

且

ge

multipliers. From (8.8) and (8.9),

we

obtain

that

the

Lagrangian associated

to

this problem is

F(

ω7

人)

=

0.0625ωi

+

0.0625ω2+O

09

吟

+0.04ω2

-

0.03125ωlω2

一

0.0375ω2ω3

+

0.0375ω

1

W

3

+λ1

(ω1

十

ω2

十

ω3+ω4

-

1)

+λ2

(0.08ω1

+

0.12ω2

+

0.16ω3

+

0.05ω4

- 0.12).

The

gradient of

the

Lagra

吨

ian

is

the

following (row) vector:

0.125w1 -

0.03125ω2

+

0.0375ω3+λ1

十

0.08λ2

、

t

0.125ω2

-

0.03125ω1

-

0.0375ω3+λ1

+

0.12λ2

0.18ω3

十

0.0375ω1

-

0.0375ω2

十沁十

0.16λ2

0.08ω4+λ1

十

0.05λ2

ω1

十

ω2

十

ω3

十

ω4-

1

0.08ω1

十

0.12ω2

+

0.16ω3

十

0.05ω4

- 0.12

To

自

nd

the

critical points of

F(

叽

λ)

，

we

solve

\7

(w

，

λ)

F(

ω?λ)

= 0, which can

be written as a linear system as follows:

The

solution of

the

linear system (8.11)

is

/

0.1586\

I 0

.4

143 ,

ωo

= I

0:3295

I;λω=

0.0112;λ0.2

=一

0.3810

\0.0976

/

Let

月

(ω)

=

F(

ω7λ0

，1，

λ0.2)

，

i.e.,

凡

(ω)

=

0.0625

叫

+

0.0625w~

+

0.09

咛

+

0.04wl

-

0.03125w1

ω2

-

0.0375w2W3

+ 0.037

5w

1

W3

+ 0.0112

(ωl

十

ω2+ω3

十

ω4

- 1)

一

0.3810

(0.08ω1

+

0.12ω2

+

0.16ω3

+

0.05ω4

- 0.12),

and compute its Hessian

\7(叫)

F(

ω?λ)=

0.125

一

0.03125

0.0375

O

1 0.08

也

)1

一

0.03125

0.125

-0.0375

。

1 0.12

W2

0.0375

-0.0375

0.18

O 1 0.16

包

)3

。

O

0.08 1 0.05

W4

l

1 l

。

λ1

0.08

0.12 0.16

0.05 0

。

λ2

183

(8.10)

Note

that

the

D

2

F

o

(

w)

is

equal to twice

the

covariance

matrix

of

the

rates

of

return

of

the

four assets,i.e.,

/

σ?

σ1σ2ρ

口内

σ

3P1

，

3

0\

D

2

月

(ω

)

=

21σ1σ

2P1

，

2

作

σ2

匀

2

，

3

~

I

iσlσ3ρ1

，

3σ2σ

3P2

，

3

σ

:3

U I

飞

000σ~

J

We

co

叫

ude

that

D

2

F

o

(

ω)

is a positive definite matrix

Therefore

,

the

associated quadratic form

q(

v) = v

t

D

2

F

o

(

wo)v

is positive

definite

,and

so

will be

the

reduced quadratic form corresponding to the linear

constraints

飞

7g(

ω

。

)υ=0.

We conclude

that

the

point

ωo

= (0.1586 0

.4

143 0.3295 0.0976) is a con-

strained minimum for

f

(ω)

given

the

constraints

g(

ω)

=

O.

The

portfolio

with 12% expected rate of

return

and

minimal variance is invested 15.86%

in

the

first asset, 4

1.

43%

in

the

second asset, 32.95% in

the

third asset,and

9.76% in

the

fourth asset.

The

minimal variance portfolio has a

standard

deviation of

the

expected

rate of

return

equal to 13.13%.

(ii) Denote by

(8.11)

/σ?σ

1

0"

2P1

，

2

σ1σ

3P1

，

3

0\

M=Iσ向

P1

，

2

O"

~σ2σ

向

3

~

I

一

|σlσ

3P1

，

3

σ2σ

3P2

，

3

σ§

O~

J

\000σ

1

J

the

covariance

matrix

of

the

rates of

return

of

the

four assets.

Let

σ

p

= 0.24

be

the

required

standard

deviation of

the

rate

of

return

of

the

portfolio.

If

Wi

denotes

the

weight of

the

asset i in

the

portfolio,i = 1 : 4,

it follows from

(8

.4)

and (8.5)

that

E[R]

var(R)

μtω;

ω

tM

ω7

(8.12)

(8.13)

where

is

the

vector of

the

expected values of

the

rates of

return

of

the

four assets.

The

problem

of

五日

ding

a portfolio with

m

以

imum

expected

rate

of

return

and

standard

deviation of

the

rate

of

return

equal

to

σ

p

can be formulated

as a constrained optimization problem as follows: find

Wo

such

that

~~n_

f(

ω)

=

f(

ω

。)，

(8.14)

g(

ω)=0

185

SOLUTIONS

TO

CHAPTER

8

EXERCISES

8.1.

LAGRANGE

MULTIPLIERS.

NEWTON'S

METHOD.

184

CHAPTER

8.

However,

it

is

easy

to

see

that

山飞=

80.01

并

17.36

二

jf

σp

(8.15)

(8.16)

where

ω

=

(Wi)

仨吐

(ωtkLz)7

f(

ω)

g(

ω)

We can now proceed with finding

the

portfolio with maximum expected

return

using

the

Lagrange multipliers method. Denote by

λ1

and

λ2

the

Lagrange

ml

削

pliers.

From (8.15)

and

(8.16),

we

obtain

that

the

Lagrangian

associated

to

this problem

is

(8.21)

十

λ2(ω

tM

ω

-

(J'~)

=μtω+λ1

(It

ω-1)

The

gradient of

the

Lagrangian is

F(ω7λ)

0)

Recall

that

, if

the

function h :

JRn

•

JR

is given by h(x) =

xtAx

,where A

is

an

n

xηsymmetric

square matrix,

then

the

gradient of

h(

x)

is

where

1

M2p

2--

UM"

111

/III-11\

\l/

F

= 2(Ax)t.

(去无)

To

直缸缸

r

时

1

G(ω?λ1

，

λ2)

(8.17)

Dh(x)

Usi

吨

(8.17)

，

it

is

easy to see

that

\7

(ω

，，\)

F(ω7λ)

= 0,

'?b

rM

/III-\

\7

g(

ω)

(μ+λ11

+

2

川)

It

ω-1

,.

wt

!V

I

ω

-

(J'~

)

This is done using a six dimensional Newton's method; note

that

the

gradient

of

G(

叽

λ1

，

λ2)

is

the

followi

吨

6

x 6 matrix:

where G :

JR6

•

JR6

is

given by

G(ω?λ1

，

λ2)

In

order

to

use

the

Lagrange multipliers method for

solvi

吨

problem

(8.14),

we

first show

that

the

matrix

\7

g(

ω)

has rank 2 for any

ωsuch

that

g(

ω)

=

O.

Note

that

ra

此

(\7

g(

ω))

= 1 if and only if there exists a constant C εR

such

that

2

!V

Iw

= C

1.

Using

the

fact

that

the

covariance

matrix

!V

I of

the

assets considered here is nonsingular,

we

obtain

that

M00

-inunu

(川

It

2(M

ω

)t

飞

7(

叫)

G(ω?λ1

，

λ2)

(8.18)

From (8.16)

it

follows

that

, if

g(

ω)

= 0,

then

l

t

w

= 1

and

wtMw

工作

Usi

吨

(8.18)

，

we

find

that

;M

九

w

We find

that

the

Lagrangian (8.21) has exactly one critical point given by

(om)

0.6450

-

0.6946

J'

-0.3503

(8.19)

1=;

山一

11;

。

C

乎

C

孚

C

σ

各=一

ωι1

=

~1~ω=

r 2

~~

2 2

{=中

l

t

w = 1

wtM

ω=σ

主

λ0

，

1

=

-0.0738;λ0.2

= -0.8510.

Wo

(8.20)

From (8.19) and (8.20)

,

we

find

that

,

ifthere

exists

ωεJR4

such

that

g(

ω)

= 0

and

ra

出

(\7

g(

ω))

= 1,

then

中=丰〉

-

0.8510(ω

tM

ω

-

(J'~).

Let

月

(ω)

=

F(ω7λ0

，

1

，

λ0.2)

，

i.e.,

μtω

-

0.0738(l

t

ω

-

1)

几

(ω)

1

σ2

.

P

2

C

l

t

M-

1

186

CHAPTER

8.

LAGRANGE

1V

IULTIPLIERS.

NEWTON'S

METHOD.

8.1.

SOLUTIONS

TO

CHAPTER

8

EXERCISES

187

The

Hessian of

Fa

(

ω)

is

D

2

凡

(ω

)

= -

0.8510·

2

Jv

I = -

1.

7019M.

where B = 100

+去

and

y = 0.03340

1.

We obtain

that

the

duration of

the

bond is 4.642735 and

the

convexity of

the

bond

is

22.573118.

口

Since

the

covariance matrix M of

the

rates of

return

of

the

four assets is a

positive definite

matrix

,

it

follows

that

D

2

F

o

(

ω)

is a negative

defi

时

te

matrix

for any

ω.

Therefore,

the

associated quadratic form

q(

v)

= v

t

D

2

F

o

(

ω

o)v

is negative

definite

,

and

so

will be

the

reduced quadratic form corresponding

to

the

linear

constraints

\7

g(

ω

。

)

v =

O.

We conclude

that

the

point

ωa

= (0.0107 0.6450 0.6946 - 0.3503) is a

constrained maximum for

f

(

ω)

given

the

constraints

g(

ω)

=

O.

The

portfolio with 24%

standard

deviation of

the

rate

of

return

and max-

imal expected

return

1.07% in

the

first asset, 64.50% in

the

second asset,

69

.4

6%

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

port-

folio is

$1,000,000,

then

$350,285 of asset 4 is shorted (borrowed and sold

for cash)

, $10,715 is invested in asset 1, $644,965

is

invested in asset 2, and

$694,604 is invested in asset 3.

This portfolio has an expected rate of

return

equal

to

17.19%.

口

Problem

4: Recall

that

缸

ding

the

implied volatility from

the

given price of

a call option

is

equivalent

to

solving

the

nonlinear problem f

(x)

= 0, where

f(x)

=

Se-

qT

N(d

1

(x)) -

Ke-

rT

N(d

2

(x)) - C

一叫圣)

+

(叫+手

)T

叫到十(叫一手

)T

and d

1

(x)

-

---\1'../

xJr"':l r , d

2

(x)

=

\1'>./

xJT

(i) Show

that

limx~

∞

d

1

(x)

=

∞

and

li

皿

z

→∞

d

2

(x)

=

一∞，

and conclude

that

lim

f(x)

=

Se

一

qT

_

C.

(ii) Show

that

Problem

3:

Use Newton's method to find

the

yield of a five year semian-

nual coupon bond with

3.375% coupon

rate

and price

100

古

What

are

the

duration

and

convexity of

the

bond?

Solution: Nine

$1.

6875

coupo

丑

payments

are made every six months, and a

final payment of

$101.6875

is

made after 5 years. By writing

the

value of

the

bond in terms of its yield,

we

obtain

that

{一∞，

if Se(r-q)T

It

Ir!

d

1

(x)

=

It

Ir!

d

2

(x)

= { 0, if

Se(r-q

厅

、。

叭。

I

∞，

if

Se(r-q)T

(Recall

that

F = Se(r-q)T is

the

forward price.)

Conclude

that

(

-C.

if Se(r-q)T

lV(

叫-

1

Se-

qT

-

K

产-

C, if Se(r-q)T

<

K;

> K.

三

K;

>

K.

Uu

vhu

p

x

e

vhu

QO

nhu

nu

42·A

,

1-9"

"U

/IIiI\

p

x

e

QO

no

9ZM

1-m

+

nu

(8.22)

(iii) Show

that

f (

x)

is

a strictly

increasi

吨

function

and

-C

<

f(x)

<

Se-

qT

- C, if

Se(r-q)T

三

K;

Se-

qT

-

Ke-

rT

- C <

f(x)

<

Se-

qT

- C, if

Se(

叫

)T

>

K.

(扣)

For what range of call option values does

the

problem

f(x)

= 0 have a

positive solution? Compare your result

to

the

range

Se-

qT

-

Ke-

rT

< C <

Se-

qT

required for obtaining a positive implied volatility for a value C of

the

call

option.

We solve

the

nonlinear equation (8.22) for y using Newton's method.

With

initial guess

Xo

= 0.1, Newton's method converges in four iterations

to

the

solution y = 0.03340

1.

We conclude

that

the

yield of

the

bond is 3.3401%.

The

duration and convexity of

the

bond are given by

D

=

;(主

1ω

巾

Solution: (i) Note

that

(x) = In

(丢)

+

(γ

-

q)T , x

vIT

(8.23)

d1

‘ Z

ZZ

飞

IT

十

2

~

(飞

J、

9 2

·2

ltt/\

Z\/

飞)、

)

、‘

J

In

(圣)

+

(r

- q)T

x

vIT

C = B

Z16

吨

exp

飞

+ 10

1.

6875·

25

exp(

一切)

I,

出

(x)

(8.24)

x

vIT

2

188

CHAPTER

8.

LAGRANGE

MULTIPLIERS.

NEWTON'S

METHOD.

8.1.

SOLUTIONS

TO

CHAPTER

8

EXERCISES

189

It

is easy

to

see

that

l

豆豆

d1(x)

=

∞

and

J

豆豆

d

2

(x)

=

一∞?

(iii)

Differentiati

吨

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

飞

Tolatility

矶

which

is

equal

to

the

vega of

the

call.

In

other words,

We

conclude

that

and

therefore

lim

f(x) -

Se-

qT

- C.

出

N(ddz))=laad

J

马

N(d

2

(x))

= 0

(iv)

If

F

三

K

，

the

problem f(x) = 0 has a solution x > 0 if and only if

。<

C <

Se

一

ι(8.25)

If

F > K ,

the

problem f(x) = 0 has a solution x > 0 if and only if

Se-

qT

-

Ke-

rT

< C <

Se-qT.

(8.26)

Problem

5: A three months

at-th

e-

money call

on

an

underlying asset with

spot price

30

paying dividends continuously

at

a

2%

rate

is worth $2.5.

As-

sume

that

the

risk free interest

rate

is constant

at

6%.

(i) Compute

the

implied volatility

with

six decimal digits accuracy,using

the

bisection method on

the

i

口

terval

[0.0001,1]'

the

secant method with initial

guess

0.5, and Newton's method

with

initial guess 0.5.

f'

(x)

= vega(C)

=仇一听「

Le

一手

7

飞

/

L/

Tr

-ln(

是)

+

(r+

手

)T

where

dl=ddz)-d

Thus, f'(x) > 0,\:j x

>

队

and

f (x)

is

strictly

increasi

吨

Recall

that

limx

---t∞

f(x)

=

Se-

q

'1' - C and

r

-C.

if F <

K:

炖忡忡

1

Se-

qT

-

K;~rT

- C,

if

F

二

K7

Note

that

Se-

qT

-

Ke-

rT

=

e-rT(Se(

叫

)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-

一

qT

一

Ke-

r

气

O

吗

)

< C <

Se-

qT

Since f(

x)

is strictly

increasi

吗，

we

co

叫

ude

that

-C

< f(x) <

Se-

qT

C, if

F

三

K;

Se-

qT

-

Ke-

rT

- C < f(x) <

Se-

qT

- C, if F >

K.

xVT

In

(去)

xVT

(ii)

Let F = Se(r-q)T be

the

forward price. From (8.23)

and

(8.24)

it

follows

that

d

1

(x)

and d

2

(x)

can be written as

Ther

陀

e£

扣

ore

I

且

im♂\

oN(

何

d

出

l(

归

x))

工

1

且

im

x

八

\

oN(

伺

d

也

2(μ

x))

工

Oa

丑

:炖咆

f

仲(归

x)

尸巳一

C

In

(去)

I

xVT

d1(x)

=

一节言十一一;

也

(x)

xvT

' 2

•

If

F < K ,

then

In

(去)

< 0 and

:坦白

(z)zESd2(z)=

一∞

•

If

F = K ,

出

nd1(x)=

孚

and

也

(x)

口

一千，

a

缸侃圳

I

丑时

l

叫

d

山阳

re

伽

f

扣

b

缸

O

臼

ωr

:炖巧

d

出仰

1

叫巾(归

Z)

=

:炖坷

d

也州

2

纣

(x)

忏

=0

Therefore

li

叫\\\、、、

oN

\V(

付

d

出

l(

仪

x)

川)

=

I

且

im

♂八\

oN(

何

d

也

2(

归

x))

= 1

a

丑

:炖巧

f

仲

(μx)

忏

=S

仇

s

仇

f

旨

e

一

-qT

一 Ke

巳

f

♂一刊

T

叮

T

一

C

Thus

,1im

x\

aN(d1(x)) =

lim

八

a

N(d

2

(x)) =

~

and

一俨

r

w(Z)=;(SfqT-ke-TT)-czET(F-K)-c

=

-C.

•

If

F > K ,

then

In

(去)

> 0 and

~t~

d

1

(x)

=

l~~

d

2

(x)

=

∞-

z

\v

X

\.U

190

CHAPTER

8.

LAGRANGE

MULTIPLIER

丘

NEWTON'S

METHOD.

|σ

一

σ

zmp

，

α

pprox

vzmp

σ

imp

|σ

一

σ|

zmp

，

α

，

pprox

v zmp

I

巳

0.000948

=

0.0948%.

口

σ

zmp

191

x'

=

(-n

扣

r

Xo

工

0)

Xo

Iteration

Cou

日

t

It

eration Count

Iteration Count

Newton's Method

Approximate Newton

Approximate

时

ewton

Forward Differences

Central Differences

(

1\

{

~

}

9

40

65

43

\ 2

and

The

iteration counts are given

in

the

table

below:

We note

that

usi

鸣

central

丑时

te

differences approximates

the

gradient

DF(x)

of F(x) more accurately

than

if forward finite differences are used,

8.1.

SOLUTIONS

TO

CHAPTER

8

EXERCISES

Fi(x + hej) - Fi(x - hej)

~c

，

jFi(X)

-

~£\~

,

'--J/

2h

-

£\--

'--J/

, j = 1

:风

where

ej

is 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

obtai~ed

by

substituting

~cF(Xold)

for

~F(Xold)'

Use h =

10-

b

,

toLconsec

干

10-°

，

and

toLapprox =

10-

9

,

and

two different initial guesses:

Xo

= (1 2 3)t and

Xo

=

(2

2 2)t.

(ii) Compare these results

to

those corresponding

to

the

approxim~t~

New-

tor

白

method

with forward finite difference approximations for

~F(x).

Solution: We use Newton's

method

and

the

approximate Newton's method

both

with

forward difference approximations and with

cen~ral

difference ap-

proximations

with

toLconsec =

10-

6

and

toLapp

r<

?x

=

10-

9

.

The

parameter

h is chosen

to

be

equal

to

toLconsec,i.e.,h =

10-°.

All algorithms converged

to

the

same solutions,

/

-1.

6905507599\

f 1 \

旷工

I

1.

9831072429 I for

Xo

= I 2 I

飞

-0.8845580785

J

飞

3

J

The

approximate

gradi

怡

e

时

~cF(x

叫)

=

(~c

，

jFi(

仪

X)

)i

队

ω

i

，

j=

庐产

j

户=

put

臼

ed

us

剖

ir

卫

19

cent

忧

ra

址

1

d

也

ifference

approx

对

im

丑

lations

，

i.e.,

(8.27)

k

Secant

:N

lethod

Newton's Method

。

0.5

1 0.3969005134

0.3969152615

2

0.3970483533

0.3970481867

3

0.3970481868

Problem

6: Let F :

:lR

3

•

:lR

3

given by

(

xf

+

2XlX2

十

zi-2223

十

9\

F(x)

- (

2XI

+

2XlX~

十

zizi-ziz3

一

2

I

\

Z

问仇

+zi-23-zd-4

/

Solution: (i)

Both

the

secant

method

with

♂

-1

= 0.6

and

Xo

= 0.5

and

Newton's

method

with initial guess

Xo

= 0.5 converge in

three

iterations

to

an

implied volatility of 39.7048%.

The

approximate values obtained

at

each

iteration are given below:

(ii)

Let

σ

问

be

the

implied volatility previously computed

usi

吨

Newtor

内

method. Use

the

formula

σ

r-.JV2在

C

一与庄

s

imp

,approx

r-.J百万

1

一呼

E

to

compute an approximate value

σ

imp

，

appr

ωfor

the

implied volatility,

and

compute

the

relative error

The

bisection method on

the

interval [0.0001,

1]

co

盯

erges

in 30 iterations

to

the

same implied volatility of 39.7048%.

The

五

rst

five iterations generate

the

following intervals:

[0.250075,0.5]; [0.375063,0.5]; [0.375063,0

.4

37556];

[0.375063

,0

.4

06309]; [0.390686,0

.4

06309];

(ii)

The

approximate value for

the

implied volatility given by (8.27) is

σ

imp

，

α

pprox

= 0.3966718145 - 39.6672%.

If

σ

imp

= 0.3970481868

is

the

implied volatility obtained using Newton's

method

,

then

192

CHAPTER

8.

LAGRANGE

MULTIPLIERS.

NEWTON'S

METHOD.

resulting in algorithms with iteration counts closer

to

the

iteration counts for

Newton's

method.

口

193

十

2

.4

375

e-

1.

5r(O

,1.

5)

+

102

.4

375

e-

2r

(O

,

2)

2

.4

375

e-O.

5r

(O

,

O.5)

+ 2

.4

375 exp (

-

r(

旧的十

0

叭

A

飞1.

5

J

( .....

0.5γ(0

，

0.5)

十

z\

十

2

.4

375

exp (

-1.

5

V.VI

\

v:

V~V

J I

LV

) + 102.4375

e-2x

r-

\

_.-

1.

5 J

2

.4

375

e-O.

5r

(O

，

O.5)

十

2

.4

375

e-

r

(O

，

l)

十

2

.4

375

e-

1.

5r(O

,1.

5)

/

r(O

,

2)

+x \

+ 2

.4

375

e-

2r

(O

，

2)

十

2

侃侃

p

\

-2.5

'

\V'~d

I

LV

)

+

102

.4

375

exp

(-2x)

100

+主

32

Since

we

assumed

that

the

zero

rate

curve

is

linear between any two consecu-

tive bond maturities

,

the

zero

rate

r(O

,t) is known for any time between

the

shortest and longest bond maturities, i.

e.

, for any t

ε[0.25

，

10].

(ii) Denote by

rc(O

,t) and

r2(0

,t)

the

zero

ra~e

curves

co.rrespo~di

吨

to

identi-

cai discount factors,with

rc(O

,t)

correspondi

卫

g

to

continuously compounded

r(0,

3)

= 4.7582%.

r(0,

5)

= 4.6303%

and

r(0,

10)

= 4.6772%.

Using bootstrapping

,

we

obtain similarly

that

Thus,

we

found

the

following zero rates corresponding

to

the

maturities

of

the

four given bonds, i.e., 3 months, 6 months, 2 years, 3 years, 5 years,

and

10

years:

γ(0

，

0.25)

=

5.2341%;γ(0

，

0.5)

=

5.0636%;γ(0

，

2)

= 4.7289%;

γ(0

，

3)

=

4.7582%;γ(0

，

5)

=

4.6303%;γ(0

，

10) = 4.6772%.

Using Newton's method to solve

the

nonlinear problem above,

we

obtain

that

Z

二二

0.047582.

Therefore

We proceed by assuming

that

the

zero

rate

curve is linear between two

years

a

时

three

years.

We

note

that

r(O

,

0.5)

，

γ(0

，

1)

，

γ(0

，1.

5)

，

and

γ(0

，

2)

are

known.

If

we

let x

=

γ(0

，

3)

，

the

price of

the

three year bond can

be

written

as

Using Newton's method

to

solve

the

nonlinear equation above for x ,

we

obtain

that

x = 0.047289, and therefore

γ(0

，

2)

= 4.7289%.

8.1.

SOLUTIONS

TO

CHAPTER

8

EXERCISES

(8.28)

0.5r(0,0.5) + x

3 -

IVIonth

T-bill

6 -

IVIonth

T-bill

2 - Year T-bond

3 - Year T-bond

5 - Year T-bond

10

- Year T-bond

Assume

that

interest is continuously compounded.

(ii) How would

the

zero

rate

curves obtained by

bootstrappi

吨

from

the

bond

prices above

, one corresponding

to

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?

Sol

仰

on:

(i) For

the

Treasury bills,

the

zero rates can be computed directly:

( 100 \

γ(0

，

0.25)

=

4ln

(一~:

I = 0.052341 = 5.2341

%;

飞

98.7

)

( 100 \

r(0,0.5) =

2ln

(一一

I

=

0.050636

工

5.0636%.

飞

97.5

J

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

(2

- t)r(O,0.5) +(t - 0.5)r(0,

2)

俨

(07027V

t

ε[0.5

，

2].

1.

5

If

we

let x =

r(O

,

2)

,

we

自 nd

from (8.28)

that

r(O

,0.5) +0.5x

T(071)=;γ(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

,

100

+立=

2

.4

375

e-O.

5r

(O

，

O.5)

十

2

.4

375

e-

r

(O

,

l)

32

Problem

7: (i) Use

bootstrappi

吨

to

obtain a zero

rate

curve from

the

following prices of Treasury instruments with semiannual

coupo

口

payments:

Coupon

Rate

Price

° 98.7

° 97.5

4.875

100~

4.875

10

。在

4.625

99

玄

4.875

101

主

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

Подождите немного. Документ загружается.