All Solutions for Yang

8/10/2019 All Solutions for Yang

1/48

Solutions to Stochastic Calculus for Finance II (Steven Shreve)

Dr. Guowei Zhao

Dept. of Mathematics and Statistics

McMaster University

Hamilton,ON L8S 4K1

October 18, 2013

Contents

1 Chapter 1 21.1 Exercise 1.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Exercise 1.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Exercise 1.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.4 Exercise 1.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.5 Exercise 1.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.6 Exercise 1.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.7 Exercise 1.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.8 Exercise 1.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Chapter 2 7

2.1 Exercise 2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Exercise 2.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.3 Exercise 2.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.4 Exercise 2.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.5 Exercise 2.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.6 Exercise 2.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

3 Chapter 3 12

3.1 Exercise 3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.2 Exercise 3.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.3 Exercise 3.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.4 Exercise 3.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133.5 Exercise 3.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.6 Exercise 3.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.7 Exercise 3.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

4 Chapter 4 19

4.1 Exercise 4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194.2 Exercise 4.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204.3 Exercise 4.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

Email: [email protected]

1
http://localhost/var/www/apps/conversion/tmp/scratch_3/[email protected]://localhost/var/www/apps/conversion/tmp/scratch_3/[email protected]


2/48

4.4 Exercise 4.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224.5 Exercise 4.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224.6 Exercise 4.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234.7 Exercise 4.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244.8 Exercise 4.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264.9 Exercise 4.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.10 Exercise 4.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.11 Exercise 4.18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284.12 Exercise 4.19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

5 Chapter 5 30

5.1 Exercise 5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305.2 Exercise 5.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315.3 Exercise 5.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335.4 Exercise 5.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345.5 Exercise 5.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365.6 Exercise 5.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

6 Chapter 6 40

6.1 Exercise 6.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 406.2 Exercise 6.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416.3 Exercise 6.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416.4 Exercise 6.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436.5 Exercise 6.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

7 Chapter 7 47

8 Chapter 8 47

9 Chapter 9 47

10 Chapter 10 47

11 Chapter 11 4711.1 Exercise 11.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

Introduction

This solution manual will be updated anytime, and is NOT intended for any business use. The authorsuggests this manual as a reference book to the metioned book by Steven Shreve. Also anyone involvedin any mathematical finance courses shall not copy the solutions in this book directly. This is ideallyfor self-check purpose.

If anyone find any mistakes or misprints in this book, please inform me. Thanks.

1 Chapter 11.1 Exercise 1.4

(i) On the infinite coin-toss space , for n = 1, 2, . . . we define

Yn() =

1, ifn= H

0, ifn= T

2


3/48

We setX=

n=1Yn2n Then by Example 1.2.5, Xis uniformly distributed on [0, 1]. Furthermore,

let

N(z) =

z

()d where () = 1

2e

2/2

Then for Z= N1(X), by Example 1.2.6,

Z[a, b] =P(: a Z() b) = P(: a N1(X()) b) = P(: N(a) X N(b)) = N(b)N(a)for any < a b < . So we conculde thatZ= N1(X) is a standard normal distributiononon (, F, P).

(ii) Let

Zn() = N1(Xn()) where Xn() =

nk=1

Yk()

2k

It is clear that

limnZn() = limnN

1(Xn()) = N1( limnXn()) = N

1(X()) = Z() for every

SinceYk() only depends on the kth coin toss, Xn will only depend on the first n coin tosses,which meansZn will also only depend on the first n coin tosses.

1.2 Exercise 1.5

First, since

I(0,X())(x) =

0, ifx / (0, X())1, ifx (0, X()) (1.1)

we can compute the double integral as:

0

I(0,X())(x)dxdP() =

[X() 0]dP() = E[X] (1.2)

In the other hand, we may also compute the same double integral by changing the order f the integrationas:

0

I(0,X())(x)dxdP() =

0

I(0,X())(x)dP()dx

Now let us consider the innner integral with respect to P(), in which we shall consider the functionI(0,X())(x) as a function of instead ofx. Now note the conditionx (0, X()) in (1.1) is equivalenttoX() x. So

I(0,X())(x)dP() =

I:X()x()dP() = P(: X() x) = P(X x) = 1P(X x) = 1F(x)

Then

0

I(0,X())(x)dxdP() =

0 I(0,X())(x)dP()dx=

0

[1

F(x)]dx (1.3)

By equating (1.1) and (1.3), we conculde that

E[X] =

0

[1 F(x)]dx

3


4/48

1.3 Exercise 1.6

1.

E[euX ] =

euxf(x)dx

=

1

2 expux (x )2

22 dx=

12

exp

ux x

2 2x + 222

dx

= 1

2

exp

1

22

x2 (2 + 2u2)x 222

dx

= 1

2

exp

1

22

x ( + u2)2 + 122

( + u2)2 2

22

dx

= 1

2

exp

1

22

x ( + u2)2 dx exp 122

( + u2)2 2

22

= 1

2 exp 1

22y2 dy exp

2u2

22 +

24

22 = 1 exp

u +

1

2u22

2.E[(X)] = E[euX ] =eu+

12u

22 eu =() = (EX)

1.4 Exercise 1.7

1. f(x) = 0 for any x;

2. limn

fn(x)dx= 1 = 0 =

f(x)dx;

3. The sequencefi is not monotone. Refer to the following diagram for the graphs of the functions:

Figure 1: Graphs of Exercise 1.7 with n=1,2,3,4 from the top respectively

4


5/48

1.5 Exercise 1.8

(i) We consider the sequence

Yn() =etX() esnX()

t snFirst note that

limsnt Yn() = ddt [etX()] =X()etX()

Also by the Mean value Theorem, there exists n [sn, t] (or [t, sn]) such that

Yn() =etX() esnX()

t sn =X()en()X()

Note that for n sufficiently large, n max{t, sn} 2|t| since sn t, so

|Yn()| =X()en()X() X()e2|t|X()

Also note that for any n and E[XetX ]< for any t, so by Dominated Convergence Theorem,we have

limnEYn = limn

Yn()dP() =

limnYn()dP() =

X()etX()dP()

=E[XetX ]

i.e.,lim

nEYn= E[ lim

nYn] =E[Xe

tX ]

(ii) If X can take both positive and negative values, then write X = X+ X, where X+ =max{X, 0} and X = max{X, 0}. Then by the Mean value Theorem, there exists n [sn, t](or [t, sn]) such that

Yn() = X()en()X()

Then

|Yn()| = X()en()X() |X| en()X() |X| e2|t||X| (1.4)Now we need to show thatE[|X| et|X|]< for anytto apply Dominated Convergence Theorem.For any t,

E[|X| et|X|] =

|X| et|X|dP()

=

1X()0 |X| et|X|dP() +

1X()0 |X| et|X|dP()

=

1X()0 X+etX+

dP() +

1X()0 XetX

dP()

=E[1X()0

X+etX+

] + E[1X()0

XetX

]

SinceE[|X| etX ]< for any t, so

E[|X| etX ] =

|X| etXdP()

=

1X()0 |X| etXdP() +

1X()0 |X| etXdP()

5


6/48

=

1X()0 X+etX+

dP() +

1X()0 XetX

dP()

=E[1X()0 X+etX+

] + E[1X()0 XetX

]<

which implies for any t

E[1X()0 X+etX+

]< and E[1X()0 XetX

]< Or equivalently for any t, we have

E[1X()0 X+etX+

]< and E[1X()0 XetX

]<

HenceE[|X| et|X|] = E[1X()0 X+etX

+

] + E[1X()0 XetX

]< for any t. Then E[|X| e|t||X|] 0, P(A) = 1/20

Z()d=1/20

0d= 0.

1.7 Exercise 1.11

By (1.6.4) of Text, we have for any u R

EeuY

=EeuYZ= Ee

u(X+)eX12

2

=E

e(u)X

eu 12 2 =e 12 (u)2 eu 12 2=e

12u

2u+ 12 2+u 12 2

=eu2/2

thereforeYis standard normal under P.

6


7/48

1.8 Exercise 1.14

Since fora 0P(X a) = 1 ea =

a

f(x)dx

where

f(x) = ex, x 00, x


8/48

3. Here we only investigate the non-trivial cases:

A (X) B (S1) A B P(A)P(B) P(A B){H T , T H } {H T , H H } {HT} (14 + 14) (14 + 14) = 14 14{H T , T H } {T H , T T } {TH} (14 + 14) (14 + 14) = 14 14

{H H , T T

} {H T , H H

} {HH

} (14 +

14)

(14 +

14) =

14

14

{H H , T T } {T H , T T } {TT} (14 + 14) (14 + 14) = 14 14

therefore for anyA (X) andB (S1), we have P(AB) = P(A)P(B). So(X) and(S1)are independent under P.

4. Again we only investigate the non-trivial cases:

A (X) B (S1) A B P(A)P(B) P(A B){H T , T H } {H T , H H } {HT} ( 29 + 29) (29 + 49) = 827 29{H T , T H } {T H , T T } {TH} ( 29 + 29) (29 + 19) = 427 29{H H , T T } {H T , H H } {HH} ( 49 + 19) (29 + 49) = 1027 49{H H , T T } {T H , T T } {TT} ( 49 + 19) (29 + 19) = 527 19

therefore for some A (X) and B (S1) (indeed for any non-trivial elements as shown), wehave P(A B) =P(A)P(B). So (X) and (S1) are not independent under P.

5. If we are told thatX= 1, then we have S2 = 4, i.e., the event must be H T or T H. But sinceP(HT) = P(T H) = 2/9, we canot use the un-conditioned probability forS1 any longer. In otherwords, based on the current information, we shall estimate P(S1 = 8|X= 1) = 1/2 = P(S1 =2|X= 1).

2.2 Exercise 2.3

SinceV is normal random variable and

E[V] =E[Xcos + Y sin ] = cos E[X] + sin E[Y] = cos

0 + sin

0 = 0

V ar[V] =V ar[Xcos + Ysin ] = cos2 V ar[X] + sin2 V ar[Y] = cos2 1 + sin2 1 = 1soVis standard normal. Similarly Wis also standard normal.

To show the independence, we quote the fact that ifZis a standard normal random variable, thenfor anyu, E[euZ] = exp(u2/2) (Exercise 1.6). Then for any real numbers a and b,

E[eaV+bW] = E[ea(Xcos +Y sin )+b(Xsin +Y cos )]

=E[eX(a cos b sin )+Y(a cos +b sin )]

=E[eX(a cos b sin )] E[eY(a cos +b sin )] by the independence ofXand Y

= exp

(a cos b sin )2

2 +

(a cos + b sin )2

2

= exp a2 cos2 + b2 sin2 2ab cos sin 2 + a2 cos2 + b2 sin2 + 2ab cos sin

2

= exp

a2

2 +

b2

2

=E[eaV]E[ebW]

SoV andW are independent.

8


9/48

2.3 Exercise 2.4

1. By the definition,

E

euX+vY

=

eux+vxzdZ(z)dX(x)

= 12 eux+vx1 +12 eux+vx(1) dX(x)=

1

2

e(u+v)xdX(x) +1

2

e(uv)xdX(x)

=E[e(u+v)X ] + E[e(uv)X ]

=1

2e(u+v)

2/2 +1

2e(uv)

2/2

=e(u2+v2)/2 e

uv + euv

2

for any u and v .

2. Letu = 0 in the last result, then

E[evY ] =e(02+v2)/2 e

0v + e0v

2 =ev

2/2

So Y must be a standard normal random variable.

3. IfXand Yare independent, then by Theorem 2.2.7 of text, the joint moment generating functionmust factor as:

EeuX+vY =EeuXEevY

But apparently

EeuX+vY =e(u2+v2)/2 e

uv + euv

2 =eu2/2 ev2/2 =EeuXEevY

So they are not independent.

2.4 Exercise 2.5

fX(x) =

fX,Y(x, y)dy

=

|x|

2|x| + y2

exp

(2|x| + y)

2

2

dy

=

|x|

z2

exp

z

2

2

dz (z = y + 2|x|)

= 12

exp[z2

2]

z=

z=|x|

= 12

exp[x22

]

fY(y) =

fX,Y(x, y)dx

=

|x|y

2|x| + y2

exp

(2|x| + y)

2

2

dx

9


10/48

ify 0, then

fY(y) =

2|x| + y2

exp

(2|x| + y)

2

2

dx

= 2

0

2x + y2

exp

(2x + y)

2

2 dx

= 2y

z2

expz

2

2

dz2

(z= 2x + y)

= 1

2exp[y

2

2]

ify 0, then

fY(y) =

y

2|x| + y2

exp

(2|x| + y)

2

2

dx +

y

2|x| + y2

exp

(2|x| + y)

2

2

dx

=

y

2x + y2

exp

(2x + y)

2

2

dx +

y

2x + y2

exp

(2x + y)

2

2

dx

= 2y

2x + y2

exp (2x + y)22 dx= 2

y

z2

exp

z

2

2

dz

2 (z= 2x + y)

= 1

2exp[y

2

2]

Therefore bothX andY are standard random variable.It is clear that fX,Y(x, y) =fX(x) fY(y), so they are not independent.And it is also clear that E[X] = E[Y] = 0, and

E[XY]

=

xyfX,Y(x, y)dydx

=

|x|

xy 2|x| + y2

exp

(2|x| + y)

2

2

dydx

=

0

x

xy 2x + y2

exp

(2x + y)

2

2

dydx +

0

x

xy 2x + y2

exp

(2x + y)

2

2

dydx

=

0

x

xy 2x + y2

exp

(2x + y)

2

2

dydx +

0

t

(t)y 2t + y2

exp

(2t + y)

2

2

dy(dt)

(t= x)

=

0

x

xy 2x + y2

exp

(2x + y)

2

2 dydx

0

x

xy 2x + y2

exp

(2x + y)

2

2 dydx

=0

soE[XY ] =E[X]E[Y], which implies X andY are uncoorelated.

10


11/48

2.5 Exercise 2.7

Let = E[Y X], then we consider

V ar(Y X) =E(Y X )2=E((Y E[Y|G]) + (E[Y|G] X ))

2

=E

(Y E[Y|G])2 + (E[Y|G] X )2 + 2 (Y E[Y|G]) (E[Y|G] X )

E(Y E[Y|G])2 + 0 + 2E[(Y E[Y|G]) (E[Y|G] X )]=V ar(Err ) + 2E[(Y E[Y|G]) (E[Y|G] X )]

(2.1)

The last step is since E(Err ) = 0, so V ar(Err) = E

(Y E[Y|G])2.Now note that both X andE[Y|G] areGmeasurable random variables,

E[(Y E[Y|G]) (E[Y|G] X )]=E[Y E[Y|G] XY Y E[Y|G]E[Y|G] + XE[Y|G] + E[Y|G]]=E[Y E[Y|G] XY Y E[Y E[Y|G]|G] + E[XY|G] + E[Y|G]] byGmeasurable=E[Y E[Y

|G]]

E[XY]

E[y]

E[E[Y E[Y

|G]

|G]] + E[E[XY

|G]] + E[E[Y

|G]]

=E[Y E[Y|G]] E[XY] E[y] E[Y E[Y|G]] + E[XY ] + E[Y] by iterated conditioningE[E[Z|G]] =E[Z=0

By substituting this result into (2.1), we will have

V ar(Y X) V ar(Err)

2.6 Exercise 2.10

By the notations, for any A (X),

A g(X)dP={:X()B}

g(X())dP()

=

g(X()) (X B)dP()

=

g(x)fX(x) (x B)dx

=

yfX,Y(x, y)

fX(x) dy

fX(x) (x B)dx

=

y (x B)fX,Y(x, y)dxdy

=E[Y

(X

B)] by the notation of (2.6.3)

=E[Y (A)] by the notation of (2.6.2) and A = {X B}=

A

Y dP

whic verifies the partial averaging property, hence proves E[Y|X] =g(X).

11


12/48

3 Chapter 3

3.1 Exercise 3.2

For 0 s t,E[W2(t)

t

|F(s)]

=E[(W(t) W(s))2 + 2W(t)W(s) W2(s) t|F(s)]=E[(W(t) W(s))2 t|F(s)] + E[(2W(t)W(s)|F(s)] + E[W2(s)|F(s)]=[(t s) t] + 2W(s)E[W(t)|F(s)] W2(s)= s + 2W2(s) W2(s)=W2(s) s

so martingale.

3.2 Exercise 3.3

(u) = Eeu(X)

= e12u

22

(u) = E

(X )eu(X)

= u2e12u

22

(u) = E

(X )2eu(X)

= (2 + u24)e12u

22

(3)(u) = E

(X )3eu(X)

=

2u4 + (2 + u24) u2 e 12u22=

3u4 + u36

e12u

22

(4)(u) = E

(X )4eu(X)

=

(34 + 3u26) + (3u4 + u36) u2 e 12u22=

34 + 6u26 + u48

e12u

22

In particular, (3)(0) = 0 and (4)(0) = 34. The latter implies the kurtosis of a normal randomvariable is(4)(0)/(2)2 = 3.

3.3 Exercise 3.4

1. Since

n1j=0

(W(tj+1) W(tj))2 max0kn1

|W(tk+1 W(tk)| n1j=0

|W(tj+1) W(tj)|

By re-writing it, we have

n1

j=0|W(tj+1) W(tj)|

n1j=0 (W(tj+1) W(tj))2

max0kn1 |W(tk+1) W(tk)|T

0 =

for almost every path of the Brownina motion W as n and|| 0.

12


13/48

2.

0 n1j=0

|W(tj+1) W(tj)|3

= max0kn1 |

W(tk+1)

W(tk)

|

n1

j=0 |W(tj+1) W(tj)|2

0 T = 0

for almost every path of the Brownina motion W as n and|| 0.

3.4 Exercise 3.5

Note that W(T) N(0, T), with probability density function f(x) = 12T

exp(x2/2T). Then bythe definition of the expectation

E

erT(S(T) K)+=

erT S(0)e(r 12

2)T+x

K

+

f(x)dx

=

A

erT

S(0)e(r12

2)T+x K

f(x)dx

where A is the constant such that S(T) K for any x A and S(T) K for any x A. Moreprecisely,

A= min{x: (r 12

2)T+ x K}

=min{x: x log

KS(0)

(r 122)T

}

=

log KS(0) (r

12

2)T

=

T d(T, S(0))

Therefore,

E

erT(S(T) K)+=

Td

erTS(0)e(r12

2)T+xf(x)dx Td

erTKf(x)dx

=erTS(0)e(r12

2)T

Td

exf(x)dx

I

erTKTd

f(x)dx

II

13


14/48

Now we compute the two integrals separately:

I= 1

2T

Td

ex e x2

2T dx

= 1

2

d

exp

T t t2

2 dt (t=

xT

)

= 1

2e

2T/2d

exp1

2(t T)2

dt (completing the square)

= 1

2e

2T/2

dT

ex2

2 dx (x= t

T)

= 1

2e

2T/2

d+

ex2

2 dx (d+= d+

T)

=e2T/2N(d+)

and

II= 1

2T

Tde

x2

2T dx

= 1

2

d

exp

t

2

2

dt (t=

xT

)

=N(d)

By substituting IandI I,we have the expected payoff of the option

E

erT(S(T) K)+=erTS(0)e(r

12

2)Te2T/2N(d+) erTKN(d)

=S(0)N(d+) erTKN(d)

3.5 Exercise 3.6

1. WriteE[f(X(t))|F(s)] = E[f(X(t) X(s) + X(s))|F(s)]

We defineg(x) = E[f(X(t)X(s)+x)], then we haveE[f(X(t)X(s)+X(s))|F(s)] = g(X(s)).Now since

X(t) X(s) = (t s) + W(t) W(s)is normally distibuted with mean(t s) and variance t s,

g(x) = 12(t s)

f(+ x)exp

( (t s))

2

2(t s)

d

= 1

2(t s)

f(y)exp(y x (t s))2

2(t

s) dyby changing variable y = + x.

ThereforeE[f(X(t))|F(s)] =g(X(s)), i.e. Xhas Markov property.2. Write

E[f(S(t))|F(s)] = E

f

S(t)

S(s) S(s)

|F(s)

14


15/48

We defineg(x) = E

f

S(t)S(s) x

|F(s)

, then since

logS(t)

S(s)=(t s) + (W(t) W(s))

is normally distibuted with mean(t

s) and variance 2(t

s),

g(x) = 12

t s

f(ex)exp

( (t s))

2

22(t s)

d

= 12

t s

0

1

yf(y)exp

(log

yx (t s))222(t s)

dy

by changing variable y = ex ( d = dy/y and x implies 0 y ).Therefore

g(x) =

0

f(y)p(,x,y)dy with p( , x , y) = 12y

exp

(log

yx )2

22

satisfiesE[f(S(t))|F

(s)] = g(S(s)) for any f. HenceShas Markov property.

3.6 Exercise 3.7

1.

E

Z(t)F(s)

=E

Z(t)

Z(s)Z(s)

F(s)

=Z(s)E

Z(t)

Z(s)

F(s)

=Z(s)Eexp(t s) + (W(t) W(s)) ( +1

2

2)(t

s) F(s)

=Z(s)e(ts)E

e(W(t)W(s))

e(+ 122)(ts) (by independence)=Z(s)e(ts) e 122(ts) e(+ 122)(ts) (sinceW(t) W(s) N(0, s t))=Z(s)

2. since m is a stopping time, and a martingale which is stoped at a stopping time is still amartingale,

E[Z(t m)] =E[Z(0 m)] =E[Z(0)] =Z(0) = 1which is exactly

E[exp{X(t m)

+12

2

(t m)}] = 1 (3.1)

3. Form< ,limt

Z(t m) = Z(m) = exp

m ( +12

2)m

(3.2)

Form= ,limt

Z(t m) = limt exp

X(t) ( +1

22)t

(3.3)

15


16/48

since

0 eX(t) em limt exp

( +1

22)t

= 0 (3.4)

limtZ(t m) = limt exp[X(t)] limt exp

( +1

22)t

= lim

t eX(t) 0 = 0 (3.5)

therefore

limt exp

X(t m)

+

1

22

(t m)

= Im


17/48

4. By differentiating

d

dE

em

= d

d

emm

2+2

E

m em

=

m22 +

2 2 emm

2+2

lim0+ E

m em= lim0+ m2 + 2 emm2+2

E[m] = m0 + 2

emm

0+2

E[m] =m

since >0

And E[m] =m < follows immediately.

5. For 2 > 0, we still have (3.2) and (3.3) respectively. The slight difference tothe argument in (3.4) will be,

0

limt

exp( +1

2

2)t limt exp (+ 2)

2

t= 0as both >0 and + 2 > 0.

Then (3.5)-(3.7) follow directly as part (iii), which is

E

Im


18/48

=

eu/

n

rn+ 1 e/

n

e/n e/n

eu/

n

rn+ 1 e/

n

e/n e/n

n

(ii) First by the substitution

n= 1/x, we have

log 1/x2(u) =

t

x2 log eux rx2 + 1

ex

ex ex eux rx

2 + 1

ex

ex ex =

t

x2log

(rx2 + 1)

eux euxex ex +

e(u)x e(u)xex ex

= t

x2log

(rx2 + 1) sinh ux + sinh( u)x

sinh x

Then by sinh(A B) = sinh A cosh B cosh A sinh B, we have

log 1/x2(u) = t

x2log

(rx2 + 1) sinh ux + sinh( u)x

sinh x

= t

x

2log

(rx2 + 1) sinh ux + sinh x cosh ux cosh x sinh ux

sinh x =

t

x2log

cosh ux +

(rx2 + 1 cosh x)sinh uxsinh x

(iii) Since sinh z = z + O(z3),

sinh ux

sinh x=

ux + O(x3)x + O(x3)=

u

+ O(x3)

This is because of the identity

(x + O(x3)) ( u

+ O(x3)) = ux + O(x3) + O(x6) = ux + O(x3)

And since cosh z= 1 + 12z2 + O(z4),

1 cosh z = 12

z2 + O(z4)

Therefore

cosh ux +(rx2 + 1 cosh x)sinh ux

sinh x

=1 +u2x2

2 + O(x4) + (rx2

2x2

2 + O(x4)) ( u

+ O(x3))

=1 +u2x2

2 +

rux2

1

2ux2+ O(x4) + O(x5) + O(x7)

=1 + u2

x2

2 + rux

2

1

2ux2+ O(x4)

(iv) Since

log 1/x2(u) = t

x2log

1 +

u2x2

2 +

rux2

1

2ux2+ O(x4)

18


19/48

= t

x2log

1 +

u2

2 +

ru

u

2

x2 + O(x4)

= t

x2

u2

2 +

ru

u

2

x2 + O(x4) + O(x6) + O(x8)

= t

x2

u2

2

+ru

u

2 x2 +

O(x4)

=

u2

2 +

ru

u

2

t + O(x2)

we have the moment generating function

limx0

log 1/x2(u) =

u2

2 +

ru

u

2

t

=u

rt

t

2

+

u2

2 t

This implies that1

nMnt,n

Nrt

t

2 , t as n Now by the following property of the normal random variables

Z N(a, b) = cZ N(ca,c2b)

we have

nMnt,n N

rt

t

2

, 2t

= N

r

2

2

t, 2t

asn , i.e., the limiting distribution of

nMnt,nis normal with mean (r2/2)tand variance

2t.

4 Chapter 4Theorem 1. Ifdx = a(x, t)dt + b(x, t)dW, wheredWis a Wiener process/Brownian motion then forG(x, t), we have

dG=

a

G

x +

G

t +

1

2b2

2G

x2

dt + b

G

xdW (4.1)

4.1 Exercise 4.1

For any 0 s t T, without loss of generality, we assumes = tl and t = tk 1. Now

E[I(t)|F(s)]

=El1

j=0 tj [M(tj+1 M(tj)) +

l

j=k1 tj [M(tj+1 M(tj))]F(tl)

=l1j=0

E

tj [M(tj+1 M(tj))F(tl) + k1

j=l

E

tj [M(tj+1 M(tj))F(tl)

1since we can always re-arrange the indices to have a new partition over 0 to T

19


20/48

=l1j=0

[tj [M(tj+1 M(tj))] +k1j=l

E

E

tj [M(tj+1 M(tj))F(tj) F(tl)

=I(s) +k1j=l

E

E

tj [M(tj+1 M(tj))F(tj)

F(tl)

=I(s) +k1j=l

E

[tj [M(tj M(tj))]F(tl)

=I(s) +

k1j=l

E

0F(tl)

=I(s) + 0

=I(s)

soI(t) is a martingale for 0 t T.

4.2 Exercise 4.2

(i) Without loss of generality, we assumet = tk and s = tl where tl < tk. In other words, to showthatI(t) I(s) in independent ofF(s) if 0 s < t T, it is sufficient to show thatI(tk) I(tl)is independent ofF(tl) for 0 tl< tk T. Hence

I(t) I(s) =k

j=0

(tj)[W(tj+1) W(tj)] l

j=0

(tj)[W(tj+1) W(tj)]

=k

j=l+1

(tj)[W(tj+1) W(tj)]

Now since (t) is a nonrandom simple process, each (tj) will be a constant; also since W(t) isa brownian motion, each W(tj+1) W(tj) will be independent ofF(tj) for j > l. Futhermore,sinceF(tl) F(tj) for all j < l, each W(tj+1) W(tj) will be independent ofF(tl), whichimplies each (t

j)[W(t

j+1)

W(tj

) is independent ofF

(tl). BY taking the summation, we have

I(t) I(s) =kj=l+1(tj)[W(tj+1) W(tj)] is independent ofF(tl).(ii) Again, we consider

I(t) I(s) =k

j=l+1

(tj)[W(tj+1) W(tj)]

Now since each W(tj+1) W(tj) is a normal distribution random variable with mean 0 andvariance tj+1 tj , (tj)[W(tj+1) W(tj)] will have a normal distribution with mean 0 andvariance 2(tj)(tj+1 tj). Since the summation of any normal random variables is still a normalrandom variable, then

I(t) I(s) N0,

k

j=l+12(tj)(tj+1 tj)

As (t) is a nonrandom simple process, N

0,k

j=l+12(tj)(tj+1 tj)

is exactly the defi-

nition of the Riemann sumtktl

2(u)du. Hence

I(t) I(s) N

0,

ts

2(u)du

20


21/48

(iii) Since

E[I(t)|F(s)] =E[I(t) I(s) + I(s)|F(s)]=E[I(t) I(s)|F(s)] + E[I(s)|F(s)]=E[I(t) I(s)] + I(s) sinceI(t) I(s) is independent ofF(s)=0 + I(s) since I(t)

I(s) is normally distributed with mean 0

=I(s)

I(t) is a martingale.

(iv) First we write

I2(t) t0

2(u)du

=(I(t) I(s))2 + 2I(t)I(s) I2(s) t0

2(u)du

=(I(t) I(s))2 + 2(I(t) I(s))I(s) + 2I2(s) I2(s) t0

2(u)du

then for 0 s t T,E

I2(t)

t0

2(u)du|F(s)

=E

(I(t) I(s))2 + 2(I(t) I(s))I(s) + I2(s)

t0

2(u)du|F(s)

=E

(I(t) I(s))2 + 2(I(t) I(s))I(s)|F(s) + I2(s) t0

2(u)du

=E

(I(t) I(s))2 + 2E[2(I(t) I(s))|F(s)] I(s) + I2(s) t0

2(u)du

=

t

s

2(u)du + 2 0 I(s) + I2(s)

t

0

2(u)du

=I2(s) s0

2(u)du

which implies thatI2(t) t02(u)du is a martingale.4.3 Exercise 4.5

1. In Theorem1, G(x, t) = log x, then

d log S(t) =

(t)S(t)

1

S(t)+ 0 +

1

22S2(t)

1

S2(t)

dt + S(t)

1

S(t)dW(t)

=

(t) 1

22

dt + (t)dW(t)

2.

log S(t) log S(0) = t0

(s) 1

22

ds +

t0

(s)dW(s)

so

S(t) = S(0)exp

t0

(s) 1

22

ds +

t0

(s)dW(s)

21


22/48

4.4 Exercise 4.6

Define Itos process

X(t) =

t0

((s) 12

2)ds +

t0

(s)dW(s)

Then

dX(t) = ( 12

2)dt + dW(t)

dX(t)dX(t) = 2dt

Now we use two different notations to compute d(Sp(t)):

1. Under the notations in Theorem1, G(x, t) = Sp(0)epx, a = 122 andb = . Then

d(Sp(t)) =

( 1

22) pSp(0)epX(t) + 0 + 1

22p2 Sp(0)epX(t)

dt + Sp(0)pepX(t)dW(t)

=

p 1

2p2 +

1

2p22

Sp(t)dt + pSp(t)dW(t)

2. Under the notations in Theorem 4.4.6 of Text,Sp(t) = f(X(t)), thenf(x) = Sp(0)epx , f(x) =pSp(0)epx andf(x) = p2Sp(0)epx. Now

d(Sp(t)) = df(X(t))

=pSp(0)epX(t)dX(t) +1

2p2Sp(0)epX(t)dX(t)dX(t)

=pSp(t)dX(t) +1

2p2Sp(t)dX(t)dX(t)

=pSp(t)

( 12

2)dt + dW(t)

+

1

2p2Sp(t) 2dt

=

p 1

2p2 +

1

2p22

Sp(t)dt + pSp(t)dW(t)

4.5 Exercise 4.7

Here we use the following version (4.4.1)of Itos lemma:

df(W(t)) = f(W(t))dW(t) +1

2f(W(t))dt (4.2)

1. Setf(x) = x4 in (4.2), then f(x) = 4x3 andf(x) = 12x2. Then (4.2) becomes

dW4(t) = 4W3(t)dW(t) +1

2 12W2(t)dt= 4W3(t)dW(t) + 6W2(t)dt

We may also write it in the integral form as:

W4(T) = W4(0)+4 T0

W3(t)dW(t)+6 T0

W2(t)dt= 4 T0

W3(t)dW(t)+6 T0

W2(t)dt (4.3)

2. First we note thatW(t) N(0, t), then by the symmetry of the integrand

E[W3(t)] =

x3 1

2te

x2

2t dx= 0

22


23/48

Now we take the expectation in (4.3),

E

W4(T)

=E

4

T0

W3(t)dW(t) + 6

T0

W2(t)dt

= T

0

4EW3(t) dW(t) + 6 T

0

EW2(t) dt=

T0

40dW(t) + 6

T0

t2dt

=3T2

3. Setf(x) = x6 in (4.2), then f(x) = 6x5 andf(x) = 30x4. Then (4.2) becomes

dW6(t) = 6W5(t)dW(t) +1

2 30W4(t)dt= 6W5(t)dW(t) + 15W4(t)dt

We may also write it in the integral form as:

W6(T) = W6(0) + 6 T

0

W5(t)dW(t) + 15 T

0

W4(t)dt= 6 T

0

W5(t)dW(t) + 15 T

0

W4(t)dt

It is easy to see thatE[W5(t)] = 0. Then by taking expectation in the above equation, we have

E

W6(T)

=E

6

T0

W5(t)dW(t) + 15

T0

W4(t)dt

=

T0

6E

W5(t)

dW(t) + 15

T0

E

W4(t)

dt

=6

T0

0dW(t) + 15

T0

3t2dt

=15T3

4.6 Exercise 4.8

1. In (4.4.23) of text, let f(t, x) = etx, then ft = etx, fx = e

t and fxx = 0. Then (4.4.23)becomes

d

etR(t)

=etR(t)dt + etdX(t)

=etR(t)dt + et [( R(t))dt + dW(t)]=etdt + etdW(t)

2. Integrating the both sides of the above equation, we have

etR(t) =e0R(0) + t

0

esds + t

0

esdW(s)

=R(0) +

et 1 + t

0

esdW(s)

which implies

R(t) = etR(0) +

1 et + et t

0

esdW(s)

23


24/48

4.7 Exercise 4.9

1. It is clear that

N(y) = 1

2e

y2

2

and

d = d(T t, x) = 1

T t log xK + (r 2

2)(T t)Then

N(d+)N(d)

= exp

d

2+

2 +

d22

= exp 1

22(T t)

log

x

K + (r

2

2)(T t)

2

log x

K+ (r+

2

2)(T t)

2

= exp 1

22(T t)4

log

x

K + r(T t)

2

2(T t)

= exp

log x

K + r(T t)

= K

xer(Tt)

2. Firstdx

= 1

T txBy the definition

c(t, x) = xN(d+(T t, x)) Ker(Tt)N(d(T t, x))

cx= c

x

=N(d+(T

t, x)) + xN(d+)

d+

x Ker(Tt)N(d

)

d

x=N(d+) +

xN(d+) Ker(Tt)N(d)

1

T tx=N(d+)

3.

ct =c

t

=xN(d+)d+

t Ker(Tt)rN(d) Ker(Tt)N(d) d

t

= Ker(Tt)rN(d) + N(d+)

xd+

t Ker(Tt) d+

t xKer(Tt)

= Ker(Tt)rN(d) + xN(d+)d+t

dt

Now note that

d+ d=

T t(d+ d)

t =

2

T t

24


25/48

soct = Ker(Tt)rN(d) x

2

T t N(d+)

4.

ct+ rxcx+1

2

2x2cxx

= rKer(Tt)N(d) x2

T t N(d+) + rxN(d+) +

1

22x2

d

dx[N(d+(x))]

= rKer(Tt)N(d) x2

T t N(d+) + rxN(d+) +

1

22x2N(d+)

d+x

= rKer(Tt)N(d) x2

T t N(d+) + rxN(d+) +

1

22x2N(d+) 1

T tx=rxN(d+) rKer(Tt)N(d)=rc

5. As t T, T t 0. And ln xK >0 ifx > K, and ln xK K, if 0< x < k

Then

limtT

N(d)

N() = 1, ifx > KN() = 0, if 0< x < k

So

limtT

c(t, x) =

x 1 Ker(TT) 1 = x K, ifx > Kx 0 Ker(TT) 0 = 0, if 0< x < K

=(x K)+

6. For 0 t < T, ln xK as x 0+.

limx0+

d = limx0+

1T t

log

x

K+ (r 1

22(T t))

=

which implies N(d)

N(

) = 0 as x

0+. Thereforec(t, x)

x

0

Ker(Tt)

0 = 0

as x 0+.7. For 0 t < T, ln xK as x .

limx

d= limx

1T t

log

x

K+ (r 1

22(T t))

=

25


26/48

which implies N(d) N() = 1 as x .limx

c(t, x)

x er(Tt)K

= lim

x

xN(d+) Ker(Tt)

x er(Tt)K

= lim

x

x(N(d+)

1)

= limx

N(d+) 1x1

= limx

ddx [N(d+) 1]

x2

= limx

N(d+) xTtx2

= 1

T t limxx3N(d+)

= 1

T t limxx3 1

2ed

2+/2

= 1

2T t limxK3

exp3T td+ 3(T t)(r+ 2

2) ed2+/2=

1

2

T t exp3(T t)(r+

2

2)

limx exp

d

2+

2 + 3

T td+

= 1

2

T t exp3(T t)(r+

2

2)

0

=0

4.8 Exercise 4.11

Without confusion, we shorten the notations when it is proper through this problem:

S= S(t), c= c(t, S(t)), cx= cx(t, S(t)), ct = ct(t, S(t)), cxx= cxx(t, S(t))

By dS(t) = Sdt + 2SdW(t)

we havedS(t)dS(t) = 2S2dtdt + 2S2SdtdW(t) +

22SdW(t)dW(t) =

22Sdt

and

dc(t, S(t)) = ctdt + cxdS+1

2cxxdSdS

Now we compute

d(ertX(t)) = rertX(t)dt + ertdX(t)

= rertX(t)dt + ert

dc cxdS+ r(X c + Scx)dt 12

(22 21)S2cxxdt

=ert ctdt + cxdS+ 12 cxxdSdS cxdS+ r(X c + Scx)dt 12 (22 21)S2cxxdt

=ert

ct+1

221S

2cxx rc + rScx 12

(22 21)S2cxx

dt + ert (cx cx) dS

=ert(ct rc + rScx+12

21S2cxx)dt

= 0

26


27/48

which implies ertX(t) is a non-random constant and ertX(t) = er0X(0) = X(0) = 0 for any t.Sincer >0, so X(t) = 0 for any t.

4.9 Exercise 4.13

Since B1(t) and B2(t) are Brownian motions, dB1(t)dB1(t) = dt and dB2(t)dB2(t) = dt. By substi-

tuting the definition

dB1(t) =dW1(t)

dB2(t) =(t)dW1(t) +

1 2(t)dW2(t)

Note that these equations imply thatW1(t) is Brownian motion andW2(t) is a continuous martingalewith W2(0) = 0 (since we can write dW2(t) = dB2(t)/

1 2(t)(t)dW1(t)/

1 2(t) with

1<


28/48

Then we compute the quadratic variation, or equivalently,

dBi(t)dBi(t) =

d

j=1

ij(t)

i(t)dWj(t)

d

k=1

ik(t)

i(t)dWk(t)

=

1

2i (t)

dj=1

ij(t)dWj(t) dk=1

ik(t)dWk(t)=

1

2i (t)

dj=1

ij(t)dWj(t)ij(t)dWj(t) by (4.7)

= 1

2i (t)

dj=1

2ij(t)dt

=1

2.

dBi(t)dBk(t) = d

j=1

ij(t)

i(t)dWj(t)

dl=1

kl(t)

k(t)dWl(t)

= 1

i(t)k(t)

dj=1

ij(t)dWj(t)

dl=1

kl(t)dWl(t)

= 1

i(t)k(t)

dj=1

ij(t)dWj(t)kj(t)dWj(t) by (4.7)

= 1

i(t)k(t)

dj=1

ij(t)kj(t)dt

=ik(t)

4.11 Exercise 4.18

1. In (4.4.13), let

f(t, x) = ex(r+2/2)t

then

ft(t, x) = (r+ 2/2)ex(r+2/2)t

fx(t, x) = ex(r+2/2)t

fxx(t, x) = 2ex(r+

2/2)t

Then Itoe Formula of (4.4.13) becomes

d(t) = (r+ 2/2)eW(t)(r+2/2)tdt eW(t)(r+2/2)tdW(t)+

1

22eW(t)(r+

2/2)tdW(t)dW(t)

= (r+ 2/2)(t)dt (t)dW(t) +12

2(t)dt

= (t)dW(t) r(t)dt

28


29/48

2. By Itos Product rule,

d((t)Z(t)) =(t)d(Z(t)) + Z(t)d((t)) + d((t))d(Z(t))

= [rXdt + ( r)Sdt + SdW(t)] + X [dW(t) rdt]+ [rXdt + ( r)Sdt + SdW(t)] [dW(t) rdt]

= [rX+ (

r)S

rX

S] dt

+ [S X] dW(t)=S[( r) ] dt + [S X]dW(t)=[S X ]dW(t) (since = ( r)/)

i.e.,d((t)Z(t)) has no dt term. Therefore (t)Z(t) is a martingale.

3. If an investor begin with initial capitalX(0) amd has portfolio value V(T) at time T, which isX(T) = V(T), then

(T)X(T) = (T)V(T)

SinceX(T) andV(T) are random, we take the expectation in the above formula

E[(T)X(T)] =E[(T)V(T)]

Now since (t)X(t) is a martingale,

E[(T)X(T)] =(0)X(0) =X(0)

By equating the above two equations, we have X(0) =E[(T)V(T)].

4.12 Exercise 4.19

1. By the definition, dB(t) = sign(t)dW(t), B(t) is a martingale with continuous path. Further-more, since dB (t)dB(t) = sign2(t)dW(t)dW(t) = dt, B (t) is a Brownian motion.

2. By Itos product formula,

d[B(t)W(t)] =d[B(t)]W(t) + d[W(t)]B(t) + d[B(t)]d[W(t)]

=sign(W(t))W(t)dW(t) + B(t)dW(t) + sign(W(t))dW(t)dW(t)=[sign(W(t))W(t) + B(t)]dW(t) + sign(W(t))dt

Integrate both sides, we will have

B(t)W(t) =B(0)W(0) +

t0

sign(W(s))ds +

t0

[sign(W(s))W(s) + B(s)]dW(s)

=

t0

sign(W(s))ds +

t0

[sign(W(s))W(s) + B(s)]dW(s)

Note that the expectation of an Itos integral is 0 and W(s) N(0, s), which implies

E[sign(W(s))] =

sign(x)

1

2ex

2/2sdx=

0

1

2ex

2/2sdx

0

1

2ex

2/2sdx= 0

Then by taking expectation on both sides,

E[B(t)W(t)] =

t0

E[sign(W(s))]ds= 0

Note thatE[W(t)] = 0, the above result also implies that B (t) and W(t) are uncorrelated.

29


30/48


d[W(t)W(t)] =d[W(t)]W(t) + d[W(t)]W(t) + d[W(t)]d[W(t)]

=W(t)dW(t) + W(t)dW(t) + dW(t)dW(t)

=2W(t)dW(t) + dt


d[B(t)W2(t)] =d[B(t)]W2(t) + d[W2(t)]B(t) + d[B(t)]d[W2(t)]

=sign(W(t))W2(t)dW(t) + [2W(t)dW(t) + dt] B(t)

+ sign(W(t))dW(t) [2W(t)dW(t) + dt]

=[sign(W(t))W2(t) + 2B(t)W(t)]dW(t) + [B(t) + 2sign(W(t))W(t)]dt

By taking the integral on both sides,

B(t)W2(t) =

t0

[sign(W(s))W2(s) + 2B(s)W(s)]dW(s) +

t0

[B(s) + 2sign(W(s))W(s)]ds

Note that again the expectation of an Itos integral is 0 and W(s) N(0, s). Then by takingthe expectation, we have

E[B(t)W2(t)] =

t0

E[[B(s) + 2sign(W(s))W(s)] ds

=2

t0

E[sign(W(s))W(s)] ds

Now

E[sign(W(s))] =

sign(x) 1

2ex

2/2sdx

=

0

x

2ex

2/2sdx

0

x

2ex

2/2sdx= 0

=2

0

x2

ex2/2sdx

>0

ThererforeE[B(t)W2(t)]> 0 = 0 = E[B(t)]E[W2(t)].IfB (t) andW(t) are independent, then B(t) andW2(t) will also be independent, which impliesE[B(t)W2(t)] = E[B(t)]E[W2(t)]. Contradiction. Therefore B(t) and W(t) are uncorrelatedbut not independent.

5 Chapter 5

5.1 Exercise 5.1

Recall thatdtdt= 0, dtdW= 0 and dW(t)dW(t) = dt.

1. First note that

dX(t) = (t)dW(t) + ((t) R(t) 12

2(t))dt

30


31/48

Forf(x) = S(0)ex, we havef(x) = f(x) = f(x) andf(X(t)) = S(0)eX(t) =D(t)S(t), thereforeby Itos Lemma,

d(D(t)S(t)) =df(X(t))

=f(X(t))dX(t) +1

2f(X(t))dX(t)dX(t)

=S(0)eX(t)dX(t) +12

S(0)eX(t)dX(t)dX(t)

=D(t)S(t)

(t)dW(t) + ((t) R(t) 1

22(t))dt +

1

2

2dt + 0 + 0

=D(t)S(t) [(t)dW(t) + ((t) R(t))dt]=D(t)S(t)(t) [dW(t) + (t)dt]

with (t) = ((t) R(t))/(t).2. The other way is by Itos Product Rule,

d(D(t)S(t)) =S(t)dD(t) + D(t)dS(t) + dD(t)dS(t)

=S(t) [R(t)D(t)dt] + D(t) [(t)S(t)dt + (t)S(t)dW(t)]+ [R(t)D(t)dt] [(t)S(t)dt + (t)S(t)dW(t)]

= [(t)S(t)D(t) S(t)R(t)D(t)] dt + D(t)(t)S(t)dW(t)=D(t)S(t) [(t)dW(t) + ((t) R(t))dt]=D(t)S(t)(t) [dW(t) + (t)dt]

5.2 Exercise 5.4

1. Let us considerd(ln S(t)), by Itos Lemma

d(ln S(t)) = 1

S(t)S(t) +

1

2

1S2(t)

dS(t)dS(t)

=r(t)dt + (t)dW(t)

2

2dW(t)dW(t)

=

r(t)

2(t)

2

dt + (t)dW(t)

By writing it in the integral form, we have

ln S(T) =ln S(0) +

T0

r(t)

2(t)

2

dt +

T0

(t)dW(t)

S(T) =S(0) exp

T0

r(t)

2(t)

2

dt +

T0

(t)dW(t)

Under the form S(T) = S(0)eX ,

X= T0

r(t) 2(t)

2

dt +

T0

(t)dW(t)

By Theorem 4.4.9 of Text,T0

(t)dW(t) is a normal random variable with mean 0 and varianceT0

2(t)dt, which implies Xis normal distributed with meanT0

r(t) 2(t)2

dt and varianceT

0 2(t)dt.

31


32/48

2. To simplify the notations, denote

R=1

T

T0

r(t)dt

=1

T T

0

2(t)dt

ThenXis normal with mean RT T2/2 and variance T2.

c(0, S(0))

=E

eRT(S(T) K)+=eRT

x:S(T)K

12

T

(S(0)ex K)exp

x RT+ T2

2

2/2T2

dx

=S(0)

x:S(T)K

eRT2

T

ex exp

x RT+ T2

2

2/2T2

dx

KeRT

x:S(T)K

1

2T expx RT+ T2

2 2

/2T2 dx

(5.1)

Now x:S(T)K

12

T

exp

x RT+ T2

2

2/2T2

dx

=P(S(T) K)

=P

X lnS(0)

K

=P

X (RT 12T2)

T

ln KS(0) (RT 12T2)

T

=P

X (RT 12T2)

T ln

S(0)K + (RT 12T2)

T

=N

1

T

ln

S(0)

K + (RT 1

2T2)

(5.2)

32


33/48

and x:S(T)K

eRT2

T

ex exp

x RT+ T2

2

2/2T2

dx

=x:S(T)K1

2expx RT

T2

2 2

/2T2 dx=

12

ln K

S(0)

exp

x RT T2

2

2/2T2

dx

=P

X (RT+ 12T2)

T

ln KS(0) (RT+ 12T2)

T

=P

X (RT+ 12T2)

T ln

S(0)K + (RT+

12T

2)

T

=N

1

T

ln

S(0)

K + (RT+

1

2T2)

(5.3)

By substituting (5.2) and (5.3) into (5.1), we have

c(0, S(0)) =BSM(T, S(0); K,R, )

5.3 Exercise 5.5

1. By the definition1

Z(t)= exp

t0

(u)dW(u) +1

2

t0

2(u)du

Let

X(t) =

t0

(u)dW(u) +1

2

t0

2(u)du

then

d

1

Z(t)

=

d

dt

eX(t)

=eX(t)dX(t) +

1

2eX(t)dX(t)dX(t)

= 1

Z(t)

(t)dW(t) +

1

22(t)dt

+

1

2Z(t)2(t)dt

= 1

Z(t)

(t)dW(t) + 2(t)dt

2. By Lemma 5.2.2 of text, since M(t) is a martingale, for 0 s t,

E

Z(t)M(t)F(s)= Z(s)E 1

Z(t)Z(t)M(t)

F(s)= Z(s)E M(t)F(s)= Z(s)M(s)which means M(t)Z(t) is a martingale.

33


34/48

3. Note thatdM(t) = (t)dW(t), then

d

M(t)

=d

M(t) 1

Z(t)

=M(t)d 1Z(t)

+ 1Z(t)

d [M(t)] + d [M(t)] d 1Z(t)

=M(t)

(t)

Z(t)dW(t) +

2(t)

Z(t)dt

+

1

Z(t) (t)dW(t) +

(t)

Z(t)dW(t) +

2(t)

Z(t)dt

(t)dW(t)

= 1

Z(t)[(t) + M(t)(t)] dW(t) +

1

Z(t)

(t)(t) + M(t)2(t)M(t)

dt

4. Now since d W(t) = dW(t) + (t)dt, if we define

(t) =(t) + M(t)(t)

Z(t)

then

d

M(t)

= 1

Z(t)[(t) + M(t)(t)] dW(t) +

(t)

Z(t)[(t) + M(t)(t)M(t)] dt

=(t)dW(t) +(t)(t)dt

=(t)dW(t)

Integrate both sides, we will have

M(t) = M(0) +

t0

(u)dW(u)

5.4 Exercise 5.12

1. First each Bi(t) is a continuous martingale. Secondly, since dBi(t) =

jij(t)/i(t)dWt(t),dBi(t)dt= 0 and furthermore we have

dBi(t)dBi(t) = [dBi(t) + i(t)] [dBi(t) + i(t)] =dBi(t)dBi(t) = dt

Bi(t) must be a brownian motion for every i.

2. BydBi(t) = dBi(t) + i(t)

we have

dSi(t) =i(t)Si(t)dt + i(t)Si(t)dBi(t)

=i(t)Si(t)dt + i(t)Si(t)[dBi(t) i(t)]= [i(t)Si(t) i(t)i(t)Si(t)] dt + i(t)Si(t)dBi(t)=R(t)Si(t)dt + i(t)Si(t)dBi(t)

34


35/48

3. SincedBi(t)dBk(t) = ik(t)dt,

dBi(t)dBk(t)

= [dBi(t) + i(t)dt] [dBk(t) + k(t)dt]=dBi(t)dBk(t) + i(t)dtdBk(t) + k(t)dtdBi(t) + i(t)k(t)dtdt

=dBi(t)dBk(t)

=ik(t)dt

4. By Itos product rule,

d[Bi(t)Bk(t)] = Bi(t)dBk(t) + Bk(t)dBi(t) + dBi(t)dBk(t)

then since dBi(t)dBk(t) = ik(t)dt, we have

Bi(t)Bk(t) =

t0

Bi(u)dBk(u) +

t0

Bk(u)dBi(u) +

t0

ik(u)du

Now note the fact that the expectation of an Itos integral is 0, if we take the expectation on theabove formula, then

E[Bi(t)Bk(t)] = E t

0

ik(u)du

= t0

ik(u)du

Similarly we haveE[Bi(t)Bk(t)] =t0

ik(u)du.

5. For E[B1(t)B2(t)], since dB1(t) = dW2(t) and dB2(t) = sign(W1(t))dW2(t) by Itos productrule,

d[B1(t)B2(t)] =B1(t)dB2(t) + B2(t)dB1(t) + dB1(t)dB2(t)

=W2(t)dB2(t) + B2(t)dW2(t) + sign(W1(t))dW2(t)dW2(t)

By dW2(t)dW2(t) = dt, we may write it in the integral version as:

B1(t)B2(t) = B1(0)B2(0) + t

0

W2(u)dB2(u) + t

0

B2(u)dW2(u) + t

0

sign(W1(u))du

Note that the expectation of an Itos integral is 0 and W1(u) N(0, u), which implies

E[sign(W1(u))] =

sign(x) 1

2uex

2/2udx=

0

12u

ex2/2udx

0

12

ex2/2udx= 0

Then we haveE[B1(t)B2(t)] = 0 (5.4)

ForE[B1(t)B2(t)], sincedB1(t) = dB1(t) = dW2(t) = d W2(t) (5.5)

and

dB2(t) = dB2(t) = sign(W1(t))dW2(t) = sign(W1(t) t)dW2(t) (5.6)by Itos product rule,

d[B1(t)B2(t)] =B1(t)dB2(t) + B2(t)dB1(t) + dB1(t)dB2(t)

=W2(t)sign(W1(t) t)dW2(t) + B2(t)dW2(t) + sign(W1(t))dt=W2(t)sign(W1(t) t)dW2(t) + B2(t)dW2(t) + sign(W1(t) t)dt

35


36/48

we may write it in the integral version as:

B1(t)B2(t) = B1(0)B2(0)+

t0

[W2(u)sign(W1(u) u) + B2(u)]dW2(u) + t0

sign(W1(u) u)du

Note that the expectation of an Itos integral is 0 and W1(u) N(0, u), which implies W1(u)u N(

u, u) and

E[sign(W1(u) u)] =

sign(x) 1

2ue(xu)

2/2udx

=

0

12u

e(xu)2/2udx

0

12u

e(xu)2/2udx

=

0

12u

e(xu)2/2udx

0

12u

e(x+u)2/2udx

= 1

2u

0

e(xu)

2/2u e(x+u)2/2u

du

>0 since u >0

Then we have

E[B1(t)B2(t)] = t0

E[sign(W1(u) u)du]> 0 (5.7)

(5.4) and(5.7) are the exmaples that E[B1(t)B2(t)]> 0 = 0 = E[B1(t)B2(t)].

5.5 Exercise 5.13

1.

E[W1(t)] =E[ W1(t)] = 0

E[W2(t)] =E[ W2(t) t0

W1(u)du] = E[ W2(t)] t0

E[W1(u)]du= 0

2. Note that the covariance

Cov [W1(T), W2(T)] =E[W1(T)W2(T)] E[W1(T)]E[W2(T)]=E[W1(T)W2(T)]

because ofE[W1(T)] = 0 = E[W2(T)].

Now by Itos product rule, we have

W1(T)W2(T) = W1(0)W2(0) +

T0

W1(t)dW2(t) +

T0

W2(t)dW1(t)

Here since

d W1(t) =dW1(t)

d W2(t) =dW2(t) + W1(t)dt

therefore,

dW1(t) =d W1(t)

dW2(t) =d W2 W1(t)dt

36


37/48

then

W1(T)W2(T) = W1(0) W2(0) +

T0

W1(t)d W2(t) T0

W1(t) W1(t)dt +

T0

W2(t)d W1(t)

Now since the expectation of any Itos integration is 0, and W1(0) = 0 =W2(0),

E[W1(T)W2(T)] =

T

0

E[

W1

2

(t)]dt= T

0 tdt= T2

2

here we used the fact that E[ W12

(t)] = Cov[ W1] =t.

5.6 Exercise 5.14

1. Since

d

ertX(t)

= rertX(t)dt + ertdX(t)= rertX(t)dt + ert(t)dS(t) erta(t)dt + rert (X(t) (t)S(t)) dt=ert

(t)

rS(t)dt + S(t)dW(t) + adt

a(t)dt r(t)S(t)dt

=S(t)ert(t)dW(t)(5.8)

contains no dt term, ertX(t) is a martingale.

2. (a) Let Z(t) = W(t) + (r 122)t, thendY(t) =d

eZ(t)

=eZ(t)dZ(t) +

1

2eZ(t)dZ(t)dZ(t)

=Y(t)

d W(t) + (r 1

22)dt

+

1

2Y(t)2(t)

=rY(t)dt + d W(t)

(b) Sinced

ertY(t)

= rertY(t)dt + ertdY(t)= rertY(t)dt + ertY(t)dt + ertY(t)dW(t)=Y(t)ertdW(t)

contains nodt term, ertY(t) is a martingale.(c) By Itos product rule,

dS(t) =S(0)dY(t) +

t0

1

Y(s)dsdY(t) + Y(t) a

Y(t)dt + dY(t) a

Y(t)dt

=S(0) + t

0

1

Y(s)ds dY(t) + adt since d Y(t)dt= 0

=S(t)

Y(t)dY(t) + adt

=S(t)

Y(t)

rY(t)dt + Y(t)dW(t)

+ adt

=rS(t)dt + S(t)dW(t) + adt

37


38/48

which is (5.9.7) of Text.

3. First

E[S(T)|F(t)]

=E[S(0)Y(T) + Y(T) T

0

a

Y(s)

ds

|F(t)]

=S(0)E[Y(T)|F(t)] + E

Y(T)

t0

a

Y(s)ds|F(t)

+ E

Y(T)

Tt

a

Y(s)ds|F(t)

=S(0)E[Y(T)|F(t)] + E[Y(T)|F(t)] t0

a

Y(s)ds + a

Tt

E

Y(T)

Y(s)|F(t)

ds

(5.9)

by taking out the what is known.

Secondly, since ertY(t) is a martingale under P, we have

E[Y(T)|F(t)] = E[Y(T)erT|F(t)]erT =Y(t)erterT =Y(t)er(Tt)

And by the definition ofY(t), we have

Y(T)

Y(s) =exp

W(T) W(s)

+ (r 12

2)(T s)

=exp

W(T) W(s)

e(r12

2)(Ts)(5.10)

Since W(T) W(s) is normally distributed with mean 0 and variance T s, by taking theexpectation on both sides of the above equation, we have

E

Y(T)

Y(s)

=E[e(

W(T)W(s))]e(r12

2)(Ts)

=e12

2(Ts) e(r 122)(Ts)=er(T

s)

(5.11)

Now by substituting (5.10) and (5.11) into (5.9), we have

E[S(T)|F(t)]

=S(0)er(Tt)Y(t) + er(Tt)Y(t) t0

a

Y(s)ds + a

Tt

er(Ts)ds

=er(Tt)S(t) + a Tt

er(Ts)ds

=er(Tt)S(t) +a

r

er(Tt) 1

(5.12)

4. Now we differentiate (5.12)

d(E[S(T)|F(t)])=d

er(Tt)S(t)

+

a

rd

er(Tt) 1

=erTd

ertS(t) aer(Tt)dt

(5.13)

38


39/48

Since

d

ertS(t)

= rertS(t)dt + ertdS(t)= rertS(t)dt + ertrS(t)dt + ertS(t)dW(t) + aertdt

=ert

S(t)dW(t) + ae

rt

dt

(5.14)

by substituting (5.14) into (5.13), we have

d(E[S(T)|F(t)])=erTd

ert)S(t)

+ aer(Tt)dt

=er(Tt)S(t)dW(t) + aer(Tt)dt aer(Tt)dt=er(Tt)S(t)dW(t)

There is no dt term ind(E[S(T)|F(t)]). Therefore E[S(T)|F(t)] is a martingale under P.5. Since

E

er(Tt)(S(T) K)|F(t)

=E

er(Tt)S(T)|F(t)

er(Tt)K by linearility=er(Tt)E[S(T)|F(t)] er(Tt)K by taking out what is known

E

er(Tt)(S(T) K)|F(t)= 0 implies that K= E[S(T)|F(t)], i.e.,F orS(t, T) = K= E[S(T)|F(t)] = F utS(t, T)

6. According to the hedgeing strategy, the value of the portfolio at timet is governed by (5.9.7) ofText with (t) = 1, i.e.,

dX(t) = dS(t) adt + r(X(t) S(t))dt

Then (5.8) implies thatd

ertX(t)

= S(t)ertdW(t)

By integrating both sides from 0 to T, we have

erTX(T) X(0) = T0

S(t)ertdW(t)dt

=

T0

ert (dS(t) rS(t)dt adt) by (5.9.7) of Text

= T

0

d ertS(t) T

0

aertdt

=erTS(T) S(0) ar

1 erT

Associated with the fact that X(0) = 0, we can rewrite it as

X(T) = S(T) S(0)erT ar

erT 1

39


40/48

By (5.12),

S(0)erT +a

r

erT 1= F utS(0, T) = F orS(0, T)

SoX(T) = S(T) F orS(0, T)

Therefore eventually, at time T, she delivers the asset at the price F orS(0, T), which is exactly

the amount that she needs to cover the debt in the money market X(T) S(T) = S(0)erT +ar

erT 1.

6 Chapter 6

6.1 Exercise 6.1

1. It is easy to seeZ(t) = e0 = 1 sincett (v)dW(v) = 0 =

tt(b(v) 2(v)/2)dv. Let

S(u) =

ut

(v)dW(v) +

ut

(b(v) 2(v)/2)dv

. ThenZ(u) = eS(u) and

dS(u) = (u)dW(u) + b(u) 2(u)/2 dvNow setf(t, x) =ex in Itos Formula (4.4.23) of text, then fx = fxx= ex and ft = 0. Then wehave

dZ(u) =df(t, S(u))

=eS(u)dS(u) +1

2eS(u)dS(u)dS(u)

=Z(u) [(u)dW(u) + b(u) 2(u)/2 dv]+ Z(u)

1

2 [(u)dW(u) + b(u) 2(u)/2 dv] [(u)dW(u) + b(u) 2(u)/2 dv]

=Z(u)

b(u) 2(u)/2

+1

22(u)

du (sincedW(u)dW(u) = du)

+ Z(u) [(u) + 0] du (since dW(u)du= 0 =dudu)=b(u)Z(u)du + (u)Z(u)dW(u)

2. It is clear that ifX(u) = Y(u)Z(u), then X(t) = Y(t)Z(t) = x 1 = x. Furthermore, by Itosproduct rule, dW(u)dW(u) = du and dW(u)du= 0 = dudu, we have

dX(u) =d[Y(u)Z(u)]

=Y(u)dZ(u) + Z(u)dY(u) + dZ(u)dY(u)

= [bzdu + zdW(u)] Y(u) +

a Z

+

ZdW(u)

Z(u)

+ [bzdu + zdW(u)] a

Z +

ZdW(u)

= [(a )du + dW(u)] + [bZY du + ZY dW(u)] +

Z

Z du + 0

= [a + bZY + ] du + [+ Z Y] dW(u)= (a(u) + b(u)X(u)) du + ((u) + (u)X(u)) dW(u) (since X= Z Y)

i.e. X(u) = Z(u)Y(u) solve the stochastic differential equation (6.2.4).

40


41/48

6.2 Exercise 6.3

1.d

ds

e

s

0 b(v)dvC(s, T)

= b(s)e

s

0 b(v)dvC(s, T) + e

s

0 b(v)dvC(s, T) = e

s

0 b(v)dv

2. Tt

d

ds

e

s

0 b(v)dvC(s, T)

ds=

Tt

es

0 b(v)dvds

eT

0 b(v)dvC(T, T) e

t

0 b(v)dvC(t, T) =

Tt

es

0 b(v)dvds

By C(T, T) = 0,

C(t, T) =

Tt

es

0 b(v)dvds

et

0b(v)dv

=

Tt

es

0 b(v)dv

et

0b(v)dv

ds=

Tt

est b(v)dvds

3.

A(T, T)

A(t, T) = T

t a(s)C(s, T) +2(s)

2 C2(s, T) ds

The result follows fromA(T, T) = 0.

6.3 Exercise 6.4

1.

(t) = (t) ddt

2

2

Tt

C(u, T)dt

=

2

2(t)C(t, T)

(t) = 2

2(t)C(t, T)

2

2(t)C(t, T)

therefore,

C(t, T) =(t) 22 (t)C(t, T)

2

2(t)

= 2(t)

2(t)

(t)(t)

C(t, T) = 2(t)

2(t)+

2

2C2(t, T)

2. Rearrange (6.5.14) into

C 2

2C2 =b C 1

we have

2(t)

2(t)= b 2

(t)2(t)

1

which implies (6.9.10)

3. By the characteristic equation2

b

1

2

2 = 0, we have

= b b2 + 22

2 =

b

2

Then(t) = a1e

( b2+)t + a2e( b2)t

41


42/48

Indeed, by the facts (T) = 1 and (T) = 22 (T)C(T, T) = 0, we have a set of equations:a1e

( b2+)T + a2e( b2)T = 1

a1(b2 + )e

( b2+)T + a2(b2 )e(

b2)T = 0

which imples a1 =

b2

2 e( b2+)T = 2

4(r+ b2 )e( b2+)T

a2 = + b22

e(b2)T =

2

4(r b2 )e(

b2)T

so c1 = c2 = 2/4.

4. (6.9.12) is straightforward from (6.9.11) by differentiatin w.r.t t. Again, since (T) = 22 (T)C(T, T) =0, we have

c1 e0 c2 e0 = 0i.e. c1 = c2.

5. By =

b2 + 22/2, we have

2 b2

4 = (+ b

2)( b

2) =

2

2

then

(t) = c1e b2 (Tt)

1b2 +

e(Tt) 1b2

e(Tt)

=c1e b2 (Tt)

b2

2 + b24e(Tt)

b2 +

2 + b24e(Tt)

=c1e b

2 (Tt) 22

( b

2 )e(Tt) + ( b

2+ )e(Tt)

=c1eb2 (T

t) 2

2 b e(Tt)

e(Tt)

2 + 2 e(Tt) + e(Tt)

2 =c1e

b2 (Tt) 22

[b sinh((T t)) + 2cosh((T t))]

and

(t) = c1eb2 (Tt)

e(Tt) e(Tt)

= 2c1e b2 (Tt) sinh((T t))

Therefore

C(t, T) = 2(t)

2(t)

= 4c1e b2 (Tt) sinh((T t))

2c1eb2 (Tt) 2

2 [b sinh((T t)) + 2cosh((T t))]

= sinh((T t))

b2sinh((T t)) + cosh((T t))

42


43/48

6. By the fact(T) = 1, we have

c1e b2 (TT) 2

2[b sinh((T T)) + 2cosh((T T))] = 1

c12

2 2= 1

c1 =2

4

Eventually,

A(t, T) = A(T, T) Tt

A(s, T)ds

= 0 Tt

2a(s)2(s)

ds

=

Tt

a C(s, T)ds

=a log((t))

2/2

= 2a

2log

2c12

b sinh((T t)) + 2cosh((T t))eb2 (Tt)

= 2a2

log

e

b2 (Tt)

b sinh((T t)) + 2cosh((T t))

6.4 Exercise 6.5

1. Sinceg(t, x1, x2) = E

t,x1,x2h(x1(T), x2(T))

we have for any 0 s t T similarly as Theorem 6.3.1 of Text,

E

h(X1(T), X2(T))F(t)= g(t, X1(t), X2(t))

E

h(X1(T), X2(T))F(s)= g(s, X1(s), X2(s))

Then

E

g(t, X1(t), X2(t))F(s)= EE[h(X1(T), X2(T))|F(t)] F(s)

=E

h(X1(T), X2(T))F(t)

=g(s, X1(s), X2(s))

Similarly, Since

ertf(t, x1, x2) = Et,x1,x2

erTh(x1(T), x2(T))

= erTEt,x1,x2 [h(x1(T), x2(T))]

ertf(t, X1(t), X2(t)) is also a martingale.

43


44/48

2. IfW1 and W2 are independent, dW1(t)dW1(t) = dt = dW1(t)dW1(t) anddW1(t)dW2(t) = 0. ByItos formula,

dg(t, X1(t), X2(t))

=gtdt + gx1dX1(t) + gx2dX2(t) +1

2gx1x1dX1(t)dX1(t) + gx1x2dX1(t)dX2(t) +

1

2gx2x2dX2(t)dX2(t)

=gtdt + gx1 [1dt + 11dW1(t) + 12dW2(t)]+ gx2 [2dt + 21dW1(t) + 22dW2(t)]+

1

2gx1x1[1dt + 11dW1(t) + 12dW2(t)] [1dt + 11dW1(t) + 12dW2(t)]

+ gx1x2 [1dt + 11dW1(t) + 12dW2(t)] [2dt + 21dW1(t) + 22dW2(t)]+

1

2gx2x2 [2dt + 21dW1(t) + 22dW2(t)]

=

gt+ 1gx1+ 2gx2+

1

2gx1x1

211+

212

+ gx1x2(1121+ 1222) +

1

2gx2x2

221+

222

dt

+ [11gx1+ 21gx2 ] dW1(t) + [12gx1 + 22gx2 ] dW2(t)

Since g(t, X1(t), X2(t)) is a martingale, there is no dt term in dg(t, X1(t), X2(t)). Therefore by

setting thedtterm to zero in the above differential form, we will have (6 .6.3) of Text immediately.

Smilarly by Itos formula,

d

ertf(t, X1(t), X2(t))

=rertf+ ertft dt + ertfx1dX1(t) + ertfx2dX2(t)+ ert

1

2fx1x1dX1(t)dX1(t) + e

rtfx1x2dX1(t)dX2(t) +1

2ertfx2x2dX2(t)dX2(t)

=ertrf+ ft+ 1fx1 + 2fx2+

1

2fx1x1

211+

212

+ fx1x2(1121+ 1222) +

1

2fx2x2

221+

222

dt

+ ert [11fx1+ 21fx2 ] dW1(t) + ert [12fx1 + 22fx2 ] dW2(t)

Since ertf(t, X1

(t), X2

(t)) is a martingale, there is no dt term in d [ertf(t, X1

(t), X2

(t))].(6.6.4) of Text follows immediately by setting the dt term to zero in the above differential form.

3. IfdW1(t)dW2(t) = dt, dW1(t)dW1(t) = dt = dW1(t)dW1(t), then by Itos formula,

dg(t, X1(t), X2(t))

=gtdt + gx1dX1(t) + gx2dX2(t) +1

2gx1x1dX1(t)dX1(t) + gx1x2dX1(t)dX2(t) +

1

2gx2x2dX2(t)dX2(t)

=gtdt + gx1 [1dt + 11dW1(t) + 12dW2(t)]+ gx2 [2dt + 21dW1(t) + 22dW2(t)]+

1

2gx1x1[1dt + 11dW1(t) + 12dW2(t)] [1dt + 11dW1(t) + 12dW2(t)]

+ gx1x2

[1dt + 11dW1(t) + 12dW2(t)]

[2dt + 21dW1(t) + 22dW2(t)]

+12

gx2x2 [2dt + 21dW1(t) + 22dW2(t)]

=

gt+ 1gx1+ 2gx2+

1

2gx1x1

211+

212

+ gx1x2(1121+ 1222) +

1

2gx2x2

221+

222

dt

+ [gx1x11112+ gx1x2(1122+ 1221) + gx2x22122] dt

+ [11gx1+ 21gx2 ] dW1(t) + [12gx1 + 22gx2 ] dW2(t)

44


45/48

Since g(t, X1(t), X2(t)) is a martingale, there is no dt term in dg(t, X1(t), X2(t)). Thereforeby setting the dt term to zero in the above differential form, we will have (6.9.13) of Textimmediately.

Smilarly by Itos formula,

d ertf(t, X1(t), X2(t))= rertf+ ertft dt + ertfx1dX1(t) + ertfx2dX2(t)+ ert

1

2fx1x1dX1(t)dX1(t) + e

rtfx1x2dX1(t)dX2(t) +1

2ertfx2x2dX2(t)dX2(t)

=ertrf+ ft+ 1fx1 + 2fx2+

1

2fx1x1

211+

212

+ fx1x2(1121+ 1222) +

1

2fx2x2

221+

222

dt

+ ert [gx1x11112+ gx1x2(1122+ 1221) + gx2x22122] dt

+ ert [11fx1+ 21fx2 ] dW1(t) + ert [12fx1 + 22fx2 ] dW2(t)

Since ertf(t, X1(t), X2(t)) is a martingale, there is no dt term in d [ertf(t, X1(t), X2(t))].(6.9.14) of Text follows immediately by setting the dt term to zero in the above differential form.

6.5 Exercise 6.71. By the iterated conditioning, for 0 s t T, we have

E

ertc(t, S(t), V(t))F(s)=EertEer(Tt)(S(T) K)+F(t) F(s)

=E

erter(Tt)(S(T) K)+F(s)

=E

erT(S(T) K)+F(s)

=ersE

er(Ts)(S(T) K)+F(s)

=ersc(s, S(s), V(s))

Then since it is a martingale, the dt term in the differential d[ert

c(t, S(t), V(t))] should be 0,d[ertc(t, S(t), V(t))]

=ertrcdt + ctdt + csdS(t) + cvdV(t) +1

2cssdS(t)dS(t) + csvdS(t)dV(t) +

1

2cvvdV(t)dV(t)

=ertrc + ct+ csrS+ cv(a bV) +1

2cssV S

2 + csvV S +1

2cvv

2V

dt

+ ert

cs

V S

d W1(t) + ert

cv

V

d W2(t)

which is

rc + ct+ csrs + cv(a bv) +12

cssvs2 + csvvs +

1

2cvv

2v= 0

This is the same as (6.9.26) of text.

2. To simplify the notations, we denotef=f(t, log s, v),g = g(t, log s, v) and their paritial deriva-

45


46/48

tives as well. Then

c(t,s,v) =sf(t, log s, v) er(Tt)Kg (t, log s, v)ct =sft er(Tt)Kgt rer(Tt)Kgcs=f+ sfx 1

s er(Tt)Kgx 1

s=f+ fx er(Tt)Kgx 1

s

cv =sfv er(Tt)Kgvcss=

1

sfx+

1

sfxx er(Tt)Kgxx 1

s2

csv=fv+ fxv er(Tt)Kgxv 1s

cvv =sfvv er(Tt)KgvvThen the left handside of (6.9.26) of Text becomes

ct+ rscs+ (a bv)cv+ 12

s2vcss+ sv+1

22vcvv

=s ft+ rf+ rfx+ (a bv)fv+v

2

fx+v

2

fxx+ vfv+ vfxv+2v

2

fvv+ er(Tt)K

gt+ rg rgx+ (a bv)gv+ v

2gxx+ vgxv+

2v

2 gvv

=rsf rer(Tt)Kg=rc

which verifies (6.9.26) of Text.

3. One method is to observe first thatf(t, X(t), V(t)) is a martingale, and take differentiation andthen setdt = 0 to get the equation.

Here we apply the Two Dimensional Feynman-Kac Theorem (Exercise 6.5 of Text)directly 2 Nowby the definition ofdX(t) and dV(t),

1 = r +12

v , 11 =v, 12 = 0

2 = a bv+ v , 21 = 0, 12 =

v

Then (6.9.13) yields

ft+ (r+1

2v)fx+ (a bv+ v)fv+ v

2fxx+ vfxv+

1

22vfvv = 0

Also for the boundary condition, in (6.6.1)of Text h(x, v) = IxlogK, then

f(T , x , v) = IxlogK

4. Here we still apply the Two Dimensional Feynman-Kac Theorem directly, by the definition ofdX(t) anddV(t),

1 = r 12

v , 11 =

v, 12 = 0

2 = a bv , 21 = 0, 12 =

v

2Here we cannot apply (6.6.3) or (6.6.4) of Text since they are for independent Brownian motions.

46


47/48

Then (6.9.13) yields

gt+ (r 12

v)gx+ (a bv)gv+ v2

gxx+ vgxv+1

22vgvv = 0

Also for the boundary condition, in (6.6.1)of Text h(x, v) = IxlogK, then

gT,x,v) = IxlogK

5. Apparently the functionc(t,x,v) defined by (6.9.34) satisfies the equation (6.9.26). Now by theprecise definition

c(T , s , v)

=sf(T, log s, v) er(TT)Kg(T, log s, v)

=

s K, if log s log K

s 0 K 0, otherwise=(s K)+

which implies that such c satisfies the boundary condition.

7 Chapter 7

8 Chapter 8

9 Chapter 9

10 Chapter 10

11 Chapter 11

11.1 Exercise 11.1

Since

M2(t) M2(s)=

N2(t) N2(s) + 2(t2 s2) 2tN(t) + 2sN(s)=

(N(t) N(s))2 + 2N(t)N(s) 2N2(s) + 2(t2 s2) 2t[N(t) N(s)] + 2N(s)(s t)=[N(t) N(s)]2 + 2N(s)[N(t) N(s)] + 2(t2 s2) 2t[N(t) N(s)] + 2N(s)(s t)

we have

E[M2(t) M2(s)F(s)]

=E[[N(t) N(s)]2 + 2N(s)[N(t) N(s)] + 2(t2 s2)

2t[N(t)

N(s)] + 2N(s)(s

t)F(

s)]

=2(t s)2 + (t s) + 2N(s) (t s) + 2(t2 s2) 2t (t s) + 2N(s)(s t)

=2(t2 + s2 2ts + t2 s2 2t2 + 2ts) + (t s)=(t s)

Therefore, for any 0 s t,

47


48/48

1.

E[M2(t)F(s)]

=E[M2(t) M2(s) + M2(s)F(s)]

=E[M2(t)

M2(s)F(s)] + M2(s)=(t s) + M2(s)M2(s)

i.e., M2(t) is submartingale.

2.

E[M2(t) tF(s)]

=E[M2(t) M2(s) (t s) + M2(s) sF(s)]

=E[M2(t) M2(s)F(s)] (t s) + M2(s) s

=(t s) (t s) + M2(s) s=M2(s) s

i.e., M2(t) t is martingale.

Documents

All Solutions for Yang