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

Подождите немного. Документ загружается.

CHAPTER

NUMERICAL

INTEGRATION.

BONDS.

2.1.

SOLUTIONS

CHAPTER

EXERCISES

The

data

below refers

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 =

y = 0.07;

I 2 5 8

141

t_c

础且

11-2

1~2

v_cash_flow =

2 2

叫

∞

叩

∞

归

尼

rrr

江即

payment.

口

Output:

bond

price B =

704888,

bond

duration

D =

118911,

and

bond

convexity C =

285705.

口

Problem

14:

By how much would

the

price of a

ten

year zero-coupon

bond

change if

the

yield increases by

ten

basis points? (One

perce

时

age

poi

时

equal

100 basis points.

Thus

, 10 basis points is equal

0.00

1.)

Solution:

The

duration

of a

zero-coupo

丑

bond

is equal

the

maturity

the

bond

, i.e., D = T = 10. For small changes

!:1

y in

the

yield,

the

percentage

change in

the

value of a

bond

can

estimated as follows:

!:1

王一目

-!:1

D = … 0.001 . 10 = - 0.0

We conclude

that

the

price of

the

bond

decreases by

1%.

口

Problem

12: Compute

the

price,

duration

and

convexity of a two year

semiannual coupon

bond

with

face value 100

and

coupon

rate

,if

the

zero

rate

curve is given by

r(O

,t) = 0.05 + O.Olln

Solution:

The

data

below refers

the

pseudocode from Table 2.5 of

[2]

for

computing

the

price of a

bond

given

the

zero

rate

curve.

Input:η=

zero

rateγ

，

t) = 0.05 + O.Olln

十

i);

t_cash

一丑

[0.5

;

v_cash

且

4 4

104]

disc = [0.97422235 0.94738033 0.91998838 0.89238025].

Output:

Bond price B = 104.17391

Note: To compute

the

duration

and

convexity of

the

bond

the

yield would

have

known.

The

yield

can

computed

, e.g., by using Newton's

method

, which is discussed

Chapter

obtain

that

the

yield of

the

bond

is 0.056792, i.e., 5.6792%,

and

the

duration

and

convexity of

the

bond

are D =

8901

and

C = 3.6895,

respectively.

口

Problem

15:

A five year

bond

with

duration

years is worth 102.

Find

approximate price of

the

bond

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

approx-

imated

the

duration

the

bond

multiplied by

the

parallel shift in

the

yield curve,

with

opposite sign, i.e.,

!:1

万一句

!:1

y D

For B = 102, D = 3.5

and

!:1

y =

-0.005

(since

= 100 bp),

we 五

that

!:1

用

!:1

y D B =

785.

The

new value

the

bond

Discount factors:

Problem

13:

the

coupon

rate

of a

bond

goes up,

what

can

said

about

the

value of

the

bond

and

its duration? Give a financial argument. Check

your answer mathematically

,i.e.,by

cor

叩灿

gggand

gg7and

ShhO

叭仰飞

咄

these

旬

丑

囚

ctior

丑

are either always positive or always

丑

lega

肮，

忖

ve.

Solution:

the

coupon

rate

goes

the

coupon payments increase

and

therefore

the

value of

the

bond

increases.

The

duration

the

bond

the

time

weighted average of

the

cash flows,

discounted

with

respect

the

yield of

the

bond.

the

coupo

日

rate

increases,

the

duration

the

bond

decreases.

This

is due

the

fact

that

the

earlier

cash flows equal

the

coupo

丑

payments

become a higher fraction of

the

payment made

maturity

,which is equal

the

face value of

the

bond

plus

new

B +!:1B -

103.75.

口

Problem

16:

Establish

the

following relationship between

duration

and

convexity:

叮

θD

C _ D

:t.

-=:.一

Solution: Recall

that

B-vu

l-DU

1θ

CHAPTER

NUMERICAL

INTEGRATION.

BONDS.

2.2.

SUPPLEMENTAL

EXERCISES

Therefore,

θB

一一

-DB.

θν

Using

Product

由

differentiate (2.13)

with

respect

直

that

2.2

Supplemental

Exercises

(2.13)

Assume

that

the

continuously compounded instantaneous

rate

curve

(t)

is given by

(t)

= 0.05

exp(

一

t)2)

-D-hu

no-rC

B-2

:-uu

1θ

θν2

D2_

~D.

θu

(i) Use

Simpsor

内

Rule

compute

the

1-year

and

2-year discount fac-

tors

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

(and face value 100).

We conclude

that

Consider a six months plain vanilla European

put

option with strike

on a lognormally distributed underlying asset paying dividends contin-

uously

2%.

Assume

that

interest rates are constant

4%.

Use

risk-neutral

valuation

write

the

value of

the

put

integral

over a finite interva

Find

the

value of

the

put

option with six decimal

digits accuracy using

the

dpoint Rule

and

using Simpson's Rule.

Also

, compute

the

丑

lack-Scholes

value

PBS

the

put

and report

the

approximation errors of

the

numerical integration approximations

each step.

The

prices of three call options with strikes

, and

, on

the

same underlying asset

and

with

the

same

maturity

, are

, and

respectively. Create a butterfly spread by going long a 45-call

and

55-call

and

shorting two 50-calls.

What

are

the

payoff and

the

maturity

the

butterfly spread?

When

would

the

butterfly spread

be profitable? Assume

,for simplicity,

that

interest rates are zero.

Dollar

duration

is defined as

电=一旦

ωθu

and measures by how much

the

value of a bond portfolio changes for a

small parallel shift in

the

yield curve.

Similarly

, dollar convexity is defined as

虫二工

eθy2·

CHAPTER

NUMERICAL

INTEGRATION.

BONDS.

2.3.

SOLUTIONS

SUPPLEJYIENTAL

EXERCISES

Note

that

, unlike classical

duration

and

convexity, which can only

computed for individual bonds,dollar duration

and

dollar convexity can

estimated for any bond portfolio, assuming all bond yields change

the

same amount. In particular, for a bond

with

value B,

duration

D ,

and

convexity C,

the

dollar

duration

and

the

dollar convexity can

computed as

and

BG.

You invest

million in a

bond

with

duration 3.2

and

convexity

and

$2.5 million

a bond with

duration

and

convexity

24.

(i)

What

are

the

dollar

旧

ation

and

dollar convexity of your portfolio?

(ii)

the

yield goes up

ten

basis

points

，五叫

new

approximate values

for each of

the

bonds.

What

the

new value of

the

portfolio?

(iii) You can buy or sell two

other

时

，

one

with

旧时

ion

and

convexity

and

another one

with

duration

3.2

and

convexity

20.

What

positions could you take in these bonds

immunize your portfolio

(i.

obtain

a portfolio

with

zero dollar

duration

and

dollar

盯

exity)?

2.3

Solutions

Supplemental

Exercises

Problem

1: Assume

that

the

continuously compounded instantaneous

rate

curve r(t) is given by

(t)

二

0.05

xp(

一

t)2)

Use Simpson's Rule

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

(and face value 100).

Solution: (i) Recall

that

the

discount factor corresponding

time

t is

叫-fa'州市

Using Simpson's Rule,

obtain

that

the

1-year

, 2-year,

and

3-year

discount

factors are

disc(l)

= 0.956595; disc(2) = 0.910128; disc(3) = 0.86574100.

(ii)

The

value of

the

three year yearly coupon

bond

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

100.236424.

口

Problem

2: Consider a six months

plai

丑

vanilla

European

put

option with

strike

on a lognormally distributed underlying asset paying dividends

坠

tinuouslyat

2%.

Assume

that

interest rates are constant

4%.

Use

risk-neutral

valuation

write

the

value of

the

put

integral

over

且

nite

interval.

Find

the

value of

the

put

option

with

six decimal digits

accuracy using

the

Midpoint Rule

and

using Simpson's Rule. Also,compute

the

Black-Scholes value PBS of

the

put

and

report

the

approximation errors

the

numerical integration approximations

each step.

Solution:

the

underlying asset follows a lognormal distribution,

the

value

S(T)

the

underlying asset

maturity

is a lognormal variable given by

/σ2\

广-

In(S(T))

In(S(O))

(γ

q - v

) T

飞

ITZ

，

where

σis

the

volatility of

the

underlying asset.

Then

the

probability density

function

ν)

S(T)

(l叼一附

(0)

)一

一手)

T)2

h(y)

~呵!一\时

" (2.14)

y > 0,

and

h(y)

= 0 if

三

Using

risk-neutral

valuation,

find

that

the

value of

the

put

is given

bγ

P -

e-rTE

川

max(K

S(T)

0)]

y)h(

)

dy, (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 appr()ximation of

the

integral (2.15),

start

with a partition

the

interval

,K] into 4 intervals,

and

double

the

numbers of -intervals

8192 intervals. We report

the

Midpoint Rule and Simpson's Rule

appro

均

nations

(2.15) and

the

correspondi

吨

approximation

errors

the

Black-Scholes value PBS in

the

table below:

丑

rst

note

that

the

approximation error does

not

go below 6 .

10-

This is due

the

fact

that

the

Black-Scholes value of

the

put

,which is given

PBS

Ke-r(T-t)

-d

) - Se-q(T-t)

-d

)

The

P&L

maturity

is equal

the

payoff

V(T)

minus

the

future value of

the

setup cost. Since interest rates are zero,

the

future value of

and

conclude

that

(

-1

, if

S(T)

三

45;

PU(T)={:jTJfii;:35;

二::;

-1

, if

三

S(T)

The

时

terfly

spread will

pro

主

table

S(T)

, i.e., if

the

spot

price

maturity

the

underlying asset will be between

$46

and

$54.

并

，

follows similarly

that

the

butterfly spread is

。在

table

erT

S(T)

erT

口

CHAPTER

NUIVIERICAL

INTEGRATION.

BONDS.

No. Intervals Midpoint Rule Error

Simpson's Rule

Error

4 5.075832 0.212228

4.855908

0.007696

4.922961

0.059357 4.863955

0.000351

16 4.878220 0.014616 4.863631

0.000027

32 4.867248 0.003644 4.863611

0.000007

64 4.864518 0.000914 4.863610

0.000006

128 4.863837 0.000233

4.863610

0.000006

256 4.863666 0.000020 4.863609

0.000006

512 4.863624 0.000009 4.863609

0.000006

1024 4.863613 0.000006 4.863609

0.000006

2048 4.863610 0.000006 4.863609

0.000006

4096 4.863610 0.000006 4.863609

0.000006

8192 4.863610 0.000006 4.863609

0.000006

2.3.

SOLUTIONS

SUPPLEMENTAL

EXERCISES

is computed

usi

吨

numerical

approximations

estimate

the

terms

-d

)

and

-d

)

工

The

approximation error of these approximations is

the

order of

10-

•

Using numerical integration,

the

real value of

the

put

option

is computed

but

the

error of

the

Black-Scholes value will propagate

the

approximation errors of

the

numerical integration.

consider

that

convergence

achieved when

the

error is less

than

10-

then

convergence is achieved for 512 intervals for

the

Midpoint Rule

and

for

intervals for Simpson's Rule. This was

expected given

the

quadratic convergence of

the

Midpoint Rule

and

the

fourth order convergence

of Simpson's

Rule.

口

Problem

The

prices of

three

call options with strikes

, and

日

the

same underlying asset

and

with

the

same

maturity

, are

and

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

maturity

the

butterfly spread?

When

would

the

butterfly spread

profitable?

Assume

,for simplicity,

that

interest rates are zero.

Solution:

The

payoff

V(T)

the

butterfly spread

maturity

( 0

, if

S(T)

三

45;

S(T)

, if

S(T)

三

50;

V(T)=i55-S(TL

if50<S(T)<

吮

l 0, if

三

S(T).

The

cost

set up

the

butterfly spread is

$4 - $12

$1.

Problem

4: You invest

million in a bond with duration 3.2 and convexity

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)

the

yield goes up by

ten

basis

points

，五nd

new approximate values for

each of

the

bonds.

What

the

new value of

the

portfolio?

(iii) You can buy or sell two

other

bonds,one with

duration

and

convexity

and another one with duration 3.2

and

convexity

20.

What

positions could

you take in these bonds

immur

山

your portfolio (i.e.,

obtain

a portfolio

with zero dollar

duration

and dollar

盯

exity)?

Solution: Recall

that

the

dollar duration and

the

dollar convexity ofa position

of size

B in a bond with duration D and convexity

Care

and

BC.

(i)

The

value, duration

and

盯

exity

the

two bond positions are

= 1,000,000; D

= 3.2; C

16;

= 2,

500

,000; D

24.

Denote by

工

+ B

the

value of

the

bond portfolio.

The

dollar

duration

and

dollar convexity of

the

portfolio are

虫=一坐立一旦一

ψθuθuθU

+ B

- $13,200,000.

虫

=θ

θ2B+θ2β

川一一一

eθν2θν2θν2

+ B

- $76,000,000.

The

system (2.17) has solution B

= $3.25mil

and

-5.75mi

We conclude

that

imm

飞

lnize

your portfolio, one should buy $3.25 mil-

lion worth of

the

bond with duration

and

convexity

and

sell $5.75

million worth of

the

bond with duration

3.2

创

convexity

20.

口

Formula (2.16) also holds for bond portfolios, since

the

dollar duration

and

the

dollar convexity of a bond portfolio are equal

the

sum of

the

dollar

duratioIIs azld of

dollar convexities of

the

bOIlds making

the

portfolio7

respectively.

Using (2.16)

五 nd

that

the

new value of

the

bond portfolio is

= B

十

i:::..

B 臼

，

500

，

000

一

$13

，

200

十

$38

二

，

486

，

838.

E[X]

汇

P(x

，

y)X(x

去

X(x

,Y)ES

，

εS

三

~二

，

then

x + y = k if

and

only if

，

ε{(1

，

k -

,k -

,… ,

- 1,1)}.

In other words

十

= k for exactly k - 1 of

the

total

outcomes from

Then

Chapter

Probability

concepts.

Black-Scholes

formula.

Greeks

and

Hedging.

3.1

Solutions

Chapter

Exercises

Problem

Let k be a positive integer with

三

12.

You throw two

fair dice.

the

sum

the

dice

k,you win w(k

or lose 1 otherwise.

Find

the

smallest value of

(k)

thats

makes

the

game worth playing.

Solution: Consider

the

probability space S of all possible outcomes of throw-

ing of

the

two dice, i.e.,

S = {(x,y) I

二

: 6, y = 1 : 6}.

Here

, x

and

y denote

the

outcomes of

the

first

and

second die, respectively.

Since

the

dice are assumed

fair and

the

tosses are assumed

be inde-

pendent of each

other

, every outcome

且

probability

去

occurring

Formally

the

discrete probability function P : S •

P(x,

二

，

V(x

，

εS

Let

k be a fixed positive integer with 2

~二

12.

The

value X of your

winning (or losses) is

the

random variable X : S

→~

given by

J w(k), if

十

川)巳

i-17else

(2.16)

(2.17)

CHAPTER

NUMERICAL

INTEGRATION.

BONDS.

(ii) Using dollar duration

and

dollar

盯

exity

，

the

approximate formula

i:::..

B 1

倡一

i:::..

y + §C(

i:::..

for

the

change in

the

value of a

bond

丑

written

i:::..

B 臼

-D

向十

;C$(AU)2

(iii) Let B3 and B

the

value of

the

positions taken

口

the

bond

with

duration D

6 and convexity C

and

the

bond

with

duration

= 3.2

and

convexity C

, respectively.

II = B +

十

denotes

the

value of

the

new portfolio,

the

丑

D$(II)

D$(B)

D$(B

)

D$(B

)

$13.2mil +

十

;

C$(

口)

D$(B)

D$(B

)

D$(B

)

$76mil +

十

Then

D$(

口)

= °

and

(II) = °if

and

only if

(mml

十1.

川

$76mil + 12B

马-

CHAPTER

PROBABILITY.

BLACK-SCHOLES

FORl\/fULA.

k-l

一

(k)(k

- 37 + k

=丁百一

(k)

十

(-1) _

~\'U/\'U

1~(3.1)

The

game is worth playing if

E[X]

三

From (3.1),

follows

出

the

game should be played if

37 -

(k)

兰一一

oj'

for 2

::;

k-l

~二

三

，

then

十

= k if

and

only if

，

{(6

，

k-6)

，

k-5)

,… , (k - 6,6)}.

other

words, x

十

= k exactly

- k times.

Then

E[X]

去汇

X(x

，

与生

(k)

十

36-T-k)(

一

，

εS

(k)(13-k)-23-k

From (3.2),

follows

that

the

game is worth

playi

吨

E[X]

三

，

i.e.,if

23+

(k)

三一一"

for

三

< 12.

- k

(3.2)

The

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

ω(2)

=ω(12)

35;ω(3)

=ω(11)=17;ω(4)

=ω(10)

11;

ω(5)

=ω(9)

8;ω(6)

=ω(8)

6.2;ω(7)

口

Problem

2: A coin lands heads

with

probability p

and

tails with probability

1 -

p. Let X

the

number of times you

lllUSt

flip

the

coin until it lands

heads.

What

are E[X] and

var(X)?

Sol

毗

on:

the

自

rst

coin toss is heads (which happens

with

probability

then

X =

the

自

rst

coin toss is tails (which happens with probability

1 -

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,

[X]

1 +

十

E[X])

= 1 +

(1-

p)E[X].

We conclude

that

1-P

3.1.

SOLUTIONS

CHAPTER

EXERCISES

Another way of computing E[X] is as follows:

The

coin will first land

heads in

the

k-th toss,which corresponds

X = k,for a coin toss sequence

..T H , i.

the

first k - 1 tosses are tails, followed by heads once.

This

。如

toss

sequence occurs with probability

P(T)

川

P(H)

- p)k-l

The

凡

E[X] L k

- p)k-l

= 1

~?')

L k(1 - p)k

品

→

仨?

(3.3)

Recall

that

，

.4)

1i-41i

/,,‘\-

By letting x = 1 - p

口

.4)，

直

that

立

17P

(η

十川

p)n+l

I n(1 _

p)n+2

k(l-p)

了一

(3.5)

From (3.3) and (3.5)

and

since 0 < 1 - p < 1,

conclude

that

"ι

公

p 1

E[X] =

lim)

~k(l-p)

日:一一

·-r=

一.

-pn

→

仨

71-p

y p

Similarly,

E[X

]

对

p)k-l

二

~?')nl~~Lk2(I-p)k

→

仨?

Since

，

三二

一

十

1)2

n+l

(2η2

十

2η

l)x

一

十

x)3

find

that

E[X

]

p _ lim

(1-

p)k =

lim

T(n

,2,1 -

仨

71-pn

→∞

… p +

p)2

2-p

CHAPTER

PROBABILITY.

BLACK-SCHOLES

FORNIULA.

Therefore,

var(X)

E[X

]

一

(E[X])2

2 - p 1

匕主口

P'"

Problem

Over each of

three

consecutive

time

时

ervals

ofle

丑

gth

7 =

1/12

the

price of a stock

with

spot

price 8

= 40

time

t = 0 will either go

by a factor u =

with

probability p = 0.6,

down by a factor d = 0.96

with

probability 1 - p = 0

.4.

Compute

the

expected value

and

the

variance

the

stock price

time

T = 37,i.e.,

compute

E[8

]

and

var(8

Solution:

The

probability space 8 is

the

set of all different

paths

that

the

stock could follow

three

consecutive

time

intervals,i.e. ,

8 = {UUU,UUD,UDU,

UDD

DUU

DUD

DDU

DDD}

where U represents

"up" move

and

D represents a "down" move.

The

value 8

the

stock

time

T is a

random

variable

五日

8 ,

and

is given by

8T(UUU)

二

旷

;

(DDD)

= 8

;

(UUD)

= 8

(UDU) = 8T(DUU) = 8

(UDD) = 8

(DUD) = 8

(DDU) = 8

Note

that

P(UUU) =

p3;

P(DDD)

p)3;

P(UUD) = P(UDU) =

P(DUU)

p2(1_

p);

P(UDD)

P(DUD)

P(DDU)

p(l-

We conclude

that

E[8T]

80u3

十

饥

2d.p2(1_p)+

叫

p(l

p)2

十

p)3

7036;

但

3(8

d)2.

p2(1

3(8

)2.

p(l

p)2

十

. (1 -

p)3

= 1749.0762;

E[(8T

)2]

一

(E[8T])2

工

9.8835.

口

E[(8

)2]

var(8

)

Problem

The

density function of

the

exponential

random

variable X

with

parameterα>

0 is

J a

一

αzif

Z > 0;

f(x)

)、

77if

二。

3.1. SOLUTIONS

CHAPTER

EXERCISES

(i) Show

that

the

function

f(x)

is indeed a density function.

is clear

that

f(x)

三

，

for any x

εJR.

Prove

that

汇

川=

(ii) Show

时

the

expected value

and

the

variance of

the

expo

阳

rando

variable X are E[X] = i

and

var(X)

=去

(iii) Show

that

the

cumulative

丑

sity

of X is

一

if x > 0

F(x)

γ0770

阳飞

Nis

(iv) Show

that

三

川

主

Note: this result is used

show

that

the

exponential variable is memoryless,

i.e.,

P(X

三

十

三

P(X

三

Sol

础。即

(i)

is easy

see

that

汇

川

fα

川

z=J

叫

tα

川

出

(-rz)lZ

出

一叫

= 1

(ii) By integration by

parts

find

that

叫

泸川

一

1 r

~_嘈

一一一一一+…

ααl

一生主+~

一

ααl

一

2xe

一

αz

αα2α3

一

~一

αα2

1-d

fl04-g

etnu

z--iii

'ME

CHAPTER

PROBABILITY.

BLACK-SCHOLES

FORJYIULA.

3.1.

SOLUTIONS

CHAPTER

EXERCISES

E[X

]

LW)dz=due-α

♂

工

αEtjtpzdz

叫一午一平

2~3ax)

αJ

坦(一午

-25ff-23+

主)

1-e

一

for

any

two

continuous

functions

g :

•

JR.

Solution:

Let

αεJR

arbitrary

real number. Note

that

ρb

飞

U(υ

叫川+忖

以仰州

叫

巾)川)

g2(x)dx

叫

仲州川

川

αεJR.

Recall

that

quadratic

polynomial

十

C is nonnegative

for all real values

and

only if P(x)

has

most

one real

double

root

which

happens

and

only

4AC

三

For

our

problem

follows

that

α2

川+叫

bf(Z)g(Z)dz+ff2

川三

，

VαεR

and

only

Therefore

var

叫叫

町叫州

刘伊帆(仪问

刻)

州

一

(但

E[X

啊[阿冈附

判]

(仙

拄

垃

then

F(x)

∞

f(s)

主

，

the

口

咐:汇川

fa"

αf

叫

(-e-

(iv)

三

，

then

α2

P(X

主

P(X

一

λρ

扣

只(阳

叫州)川

一l'

αω

巳一叫

一

(-e

ω)

\IlI-

/II--\

rId

/Il--\

\Ill-/

rId

/lil\

which is

equivalent

(3.6)

Thu

/It--\

,,"

rld

-o

/III-\

J'i\

rId

P(X

主

t+s

三

Recall

that

the

conditional

probability

of A

given

B is

P(A

P(AIB)

P(B)

Let

三

The

凡

from

(3.6)

and

(3.7)

we find

that

P((X

三

t+s)n(x

主

t))

P(X

三

P(X

三

P(X

三

一

(

_,""

工

一讪-甲-

ε'-'0

(3.7)

Problem

Use

the

Black-Scholes

formula

price

both

put

and

a call

option

wit

且

strike

45 expiring

six

months

underlying

asset

with

spot

price

and

volatility 20%

paying

dividends

continuously

interest

rates

are

constant

6%.

Solution:

Input

for

the

Black-Scholes

formula:

50;

45;

0.5;σ=

0.2;

0.02;

0.06.

The

Black-Scholes

price

the

call is

6.508363

and

the

price of

the

put

is P

0.675920.

口

f(x)g(

川三

尸

(x)

川)\

Problem

What

the

value

European

Put

option

with

strike

毛iV

hat

the

value of

European

Call

option

with

strike

How

do you

hedge

short

position

in such a call

option

CHAPTER

PROBABILITY.

BLACK-SCHOLES

FORNIULA.

3.1.

SOLUTIONS

CHAPTER

EXERCISES

Solution: A

put

option with strike 0 will never

exercised, since

would

mean selling

the

underlying asset for

the

price K =

The

price of

the

put

option is

A call

with

strike 0 will always be exercised, since

gives

the

right

buy

one

日

the

underlying asset

zero cost.

The

value of

the

call

maturity

V(T)

工

8(T)

，

and

therefore

V(O)

工

8(0). This can

seen

by building a portfolio with a long position on

the

call option and a short

position of

shares,or by using

risk-neutral

pricing:

V(O)

[8(T)]

e(r

一伊

8(0)

二

8(0).

A short position i

口

the

call

optior

丑

祀

hedged

剖

创

tic

伪

址

与坊圳

吵

→)

by buying

∞

口

share of

the

underlying

asset.

口

e.g.,for pricing volatility swaps. These two Greeks are called volga

and

vanna

and are

如

led

as follows:

θ(vega(V))θ(vega(V))

volga(V)

=θσand

vanna(V)

=θS

is easy

see

that

volga(V)

=万万

and

vanna(V)

-否瓦

The

name volga is

the

short for "volatility gamma". Also, vanna can

interpreted as

the

rate

of change of

the

Delta

with

respect

the

volatility

the

underlying asset,i.e.,

(i) Compute

the

volga

and

vanna for a

plai

丑

vanilla

European call option on

asset paying dividends continuously

the

rate

(ii) Use

the

Put-Call

parity

compute

the

volga

and

vanna for a plain

vanilla European

put

option.

Solution: (i) Recall

that

vanna(V)

Problem

8: Use formula p(

K(T

t)e-r(T-t)

也)

for p(C)

and

the

Put-Call

parity

show

that

p(P)

= -

K(T

t)e-r(T-t)

N(-d

Solution: Recall

that

θCθP

p(C)

and

p(P)

万;

By differentiating

the

Put-Call

parity

formula

P +

8e-

(T-t)

- C K e-r(T-t)

with

respect

飞

find

that

p(P)

p(C)

-K(T-t)e-r(T-t).

where

vega(C)

~(C)

(~(V))

θσ

8e-

(T-t)

叮叮

-LJ:

飞

/27

了'

e-q(T-t)

N(d

)

θ(

吵

(C))

8e-

(T-t

哺可

-L

dJf11;

Uσ

飞

σ'

(~(C))_

e-q(T-t)

N'(d

)

纽

叫↓

θσ

〉三作

θσ

Therefore,

p(P)

(C)

K(T

t)e-r(T-t)

K(T

t)e-r(T-t)

N(d

)

K(T

t)e-r(T-t)

-K(T

t)e-r(T-t)

N(d

))

K(T

t)e-r(T-t)

-d

)

since 1 -

N(d

) =

N(-d

口

Problem

The

sensitivity of

the

vega of a portfolio

with

respect

volatil-

ity

and

the

price of

the

underlying asset are often

important

estimate,

Then

volga(C)

vanna(C)

叶

)

(γ

-q

吟)

- t)

丁

(去)

… q)(T -

tσ

VTτz

丁

CHAPTER

PROBABILITY.

BLACK-SCHOLES

FOR~

在

ULA.

3.1.

SOLUTIONS

CHAPTER

EXERCISES

(ii) By differentiating

the

Put-Call

parity

十

Se-q(T-t)

- C = K e-r(T-t)

with

respect

风

find

that

Note

that

θσ

We conclude

that

volga(C)

vanna(C)

(圣)

(γ

q)(T

，~丁

一二

σ2VT

士

叫)

(γ

-q

一手)

σ2

飞厅可

Se-

旷

-tL

丁

zi-e

千生生

川

νd

开

σ7

-e-q(T-t)

」

-e-4d2

石

(3.8)

(3.9)

Alternatively

the

Black-Scholes formulas for

at-th

money options can

be written as

C -

Ke-q(T-t)

N(d

) -

Ke-r(T-t)

出);

P -

Ke-

(T-t)N(-d

) -

Ke-

(T-t)N(-d

where

叶子气)旧

and

叫于一~)旧

Then

三

仁=?

e-q(T-t)

N(d

- e-r(T-t)

N(d

)

三

e-r(T-t)

-d

) -

e-q(T-t)

-d

)

中=争

一旷一气

N(d

)

-d

))

三

e-r(T

一气

N(d

)

-d

))

牛=辛

e-q(T-t) > e-r(T-t)

牛=学

二三

since

N(d

) +

-d

) =

N(d

) +

-d

) =

1.口

θ(vega(P))θ(vega(C))

Volga(P)z

Volga(C);

θσθσ

θ(vega(P))θ(vega(C))

vama(P)===vanm(C)?

θSθs

where volga(C)

and

van

叫

are given by (3.8)

and

(3.9),respectively.

Therefore

θP

vega(P) -

否歹

δC

歹歹工

vega(C).

口

Pro

blem

11:

(i) Show

that

the

Theta

of a plain vanilla

旧

opean

call option

on a non-dividend-paying asset is always negative.

(ii) For long

dated

(i.e., with T - t large)

ATJVI

calls on an underlying asset

paying dividends continuously

rate

equal

the

constant risk-free

rate

i.e., with q = r , show

that

the

Theta

may

positive.

Solution:

(i)

丑

ecall

that

Sσ

e-q(T-t)

@(C)=-e

十

qSe-q(T-t)

N(d

) -

rK e-r(T-t)

N(d

2 、

/2π(T

- t)

Problem

10: Show

that

ATM call on

underlying asset paying divi-

dends continuously

rate

q is worth more

than

ATM

put

with

the

same

maturity

if and only if

三飞

where

r is

the

constant risk free rate. Use

the

Put-Call

parity,

and

then

use

the

Black-Scholes formula

prove this result.

Solution: For

at-th

money options, i.e.,

with

S = K ,

the

Put-Call

parity

can

written

C - P = Se-q(T-t) - K e-r(T-t) _ K e-q(T-t) - K e-r(T-t)

K e-r(T-t) (e(r-q)(T-t) -

Therefore, C

~三

and

only if e(r-q)(T-t)

;

三

，

which is equivalent

主

For a non-dividend-paying asset,i.e., for q = 0,

find

that

8(C)

-Sσe-4-Tke-r(T-t)N(d2)<0.

2 飞

作

(ii)

q = r ,

the

Theta

of an ATM call (i.e.,

with

S =

Kae-r(T-t)

@(C)=-e2

十

K e-r(T-t)

) -

rK e-r(T-t)

)

2 飞

作

Ke-r(T-t)

(r(N(d

) -

N(d

))

一

σJ)

飞

汀

)