Lecture #28: Calculations with Itoˆ’s Formula

[Pages:7]Statistics 441 (Fall 2014) Prof. Michael Kozdron

November 10, 2014

Lecture #28: Calculations with Ito^'s Formula

Example 17.1 (Assignment #4, problem #10). Suppose that { Bt, t

Brownian motion with B0 = 0. Determine an expression for

Z

t

sin( ) d Bs Bs

0

that does not involve Ito^ integrals.

0} is a standard

Solution. Since Version I of It^o's formula tells us that

() f Bt

Z

Z

f (B0) =

t

0( ) d

+1

f Bs Bs

0

2

t

00( ) d f Bs s,

0

if we choose 0( ) = sin( ) so that ( ) = cos( ) and 00( ) = cos( ), then

fx

x

fx

x fx

x

Z

t

1Z t

cos( ) Bt

+

cos(B0)

=

sin( ) d +

0

Bs Bs 2

cos( ) d Bs s.

0

The fact that B0 = 0 implies

Z

t

sin( ) d = 1 Bs Bs

0

cos( ) Bt

1Z t

cos( ) d

20

Bs s.

Example 17.2 (Assignment #4, problem #1). Suppose that {

0} is a Brownian

Bt, t

motion starting at 0. If the process {

0} is defined by setting

Xt, t

= exp{ }

Xt

Bt ,

use It^o's formula to compute d . Xt

Solution. Version I of It^o's formula tells us that

d ( ) = 0( ) d + 1 00( ) d f Bt f Bt Bt 2 f Bt t

so that if ( ) = x, then fx e

1 d exp{ } = exp{ } d + exp{ } d

Bt

Bt Bt 2

Bt t.

Equivalently, if = exp{ }, then

Xt

Bt

d = d + Xt d Xt Xt Bt 2 t.

17?1

Example 17.3 (Assignment #4, problem #8). Suppose that {

0} is a standard

Bt, t

Brownian

motion

with

B0

= 0.

Consider

the

process

{ Yt, t

0} defined by setting = k

Yt

B t

where is a positive integer. Use It^o's formula to show that satisfies the SDE

k

Yt

d=

1 1/k d

+

( kk

1) 1 2/k d

Yt

kY t

Bt

2

Y t

t.

Solution. Version I of It^o's formula tells us that

d ( ) = 0( ) d + 1 00( ) d f Bt f Bt Bt 2 f Bt t

so that if ( ) = k, then 0( ) = k 1 and 00( ) = ( 1) k 2 so that

fx x

f x kx

f x kk x

d k = k 1 d + k(k 1) k 2 d

B kB

t

t

Bt

2

B t

t.

Writing = k gives

Yt

B t

d= 1

Yt

kY t

1 /k

d Bt

+

k(k 2

1) 1 Y

t

2 /k

d

t.

Example 17.4 (Assignment #4, problem #5). Consider the It^o process {

0} de-

Yt, t

scribed by the stochastic dierential equation

d = 0 4d +0 1d Yt . Bt . t.

If the process {

0} is defined by = 0.5Yt, determine d .

Xt, t

Xt e

Xt

Solution. Version III of Ito^'s formula tells us that

d ( ) = 0( ) d

1 +

00(

) dh

i

f Yt f Yt Yt 2 f Yt Y t

so that if ( ) = 0.5y, then fy e

(0 5)2

d exp{0 5 } = (0 5) exp{0 5 } d + . exp{0 5 } dh i

. Yt

.

. Yt Yt

2

. Yt Y t.

Since d = 0 4 d + 0 1 d , we conclude that dh i = (0 4)2 d = 0 16 d and so

Yt . Bt . t

Yt . t . t

d exp{0 5 } = (0 5) exp{0 5 }(0 4 d

+

0

1

d

)

+

(0 5)2 .

exp{0

5

}(0 16 d )

. Yt

.

. Yt . Bt . t

2

. Yt . t .

Writing

=

05 . Yt

and

collecting

like

terms

gives

Xt e

d = 0 2 d + 0 07 d Xt . Xt Bt . Xt t.

17?2

Example 17.5 (Assignment #4, problem #11). Suppose that {

0} is a standard

Bt, t

Brownian

motion

with

B0

=

0,

and

suppose

further

that

the

process

{ Xt,

t

0}, X0 = a > 0,

satisfies the stochastic dierential equation

d = d +1d

Xt Xt Bt

t.

Xt

(a) If ( ) = 2, determine d ( ).

fx x

f Xt

(b) If ( ) = 2 2, determine d ( ).

f t, x t x

f t, Xt

Solution. Version III of Ito^'s formula tells us that

d ( ) = 0( ) d + 1 00( ) dh i f Xt f Xt Xt 2 f Xt X t

so that

d( 2) = 2 d + dh i

X t

Xt Xt

X t.

Version IV of It^o's formula tells us that

d( f t,

) Xt

=

( f t,

) Xt

d t

+

0( f t,

) Xt

d Xt

+

1 00( 2f t,

) Xt

dh i Xt

so that Since we conclude that Thus,

d( 2 2) = 2 2 d + 2 2 d + 2 dh i

tX t

tX t t

t Xt Xt t X t.

1

d = d+ d

Xt Xt Bt

t,

Xt

dh i = 2 d

Xt

X t. t

(a) d( 2) = 2 2 d + (2 + 2) d , and

X t

X t

Bt

Xt t

(b) d( 2 2) = 2 2 2 d + (2 2 + 2 2 + 2 2) d .

tX t

tX t

Bt

tX t t X t

t

t

Example 17.6 (Assignment #4, problem #7). Suppose that g : R ! [0, 1) is a bounded,

piecewise continuous, deterministic function. Assume further that 2 2([0 1)) so that gL,

the Wiener integral

Z

t

= ( )d

It

g s Bs

0

is well defined for all 0. Define the continuous-time stochastic process {

t

Mt, t

setting

Z

Z

2 Z

t

t

t

=2

Mt

I t

2( ) d = gs s

( )d g s Bs

2( ) d g s s.

0

0

0

0} by

Use It^o's formula to prove that {

0} is a continuous-time martingale.

Mt, t

17?3

Solution. If

Z

t

= ( )d

It

g s Bs,

0

then d = ( ) d so that dh i = 2( ) d . If

It g t Bt

It g t t

Z

=2

Mt

I t

t

2( ) d g s s,

0

then written in dierential form we have

d = d( 2) 2( ) d

Mt

I g t t. t

Version III of It^o's formula implies

d( 2) = 2 d + dh i

I t

It It

I t.

Substituting back therefore gives

d = d( 2) 2( ) d = 2 d + dh i 2( ) d = 2 ( ) d + 2( ) d 2( ) d

Mt

I t

gt t

It It

I t g t t g t It Bt g t t g t t

=2 () d g t It Bt.

Since Ito^ integrals are martingales, we conclude that {

0} is a continuous-time mar-

Mt, t

tingale.

Example 17.7 (Assignment #4, problem #9). Suppose that {

0} is a time-inhomogeneous

Xt, t

Ornstein-Uhlenbeck-type process defined by the SDE

d = ( )d

(

( )) d

Xt t Bt a Xt g t t

where and are (su ciently regular) deterministic functions of time. If = exp{ + },

g

Yt

Xt ct

use It^o's formula to compute d .

Yt

Solution. If d = ( ) d

(

( )) d and = exp{ + }, then Version IV of

Xt t Bt a Xt g t t Yt

Xt ct

Ito^'s formula implies that

d = d + d + Yt dh i Yt cYt t Yt Xt 2 X t.

Since

dh i = 2( )

Xt

t dt,

we conclude that

d

2( )

Yt = ( ) d +

(

( )) + t d

t Bt c a Xt g t

2 t.

Yt

Since we want a stochastic dierential equation for , we should really substitute back for

Yt

in terms of . Solving = exp{ + } for gives = log( ) so that

Xt

Yt

Yt

Xt ct Xt

Xt

Yt ct

d

2( )

Yt = ( ) d +

(log( )

( )) + t d

t Bt c a Yt ct g t

2t

Yt

2( )

= ( ) d + (1 + ) log( ) + ( ) + t d

t Bt c at a Yt ag t

2

t.

17?4

Example 17.8 (Assignment #4, problem #2). Suppose that the price of a stock {

0}

Xt, t

follows geometric Brownian motion with drift 0 05 and volatility 0 3 so that it satisfies the

.

.

stochastic dierential equation

d = 0 3 d + 0 05 d Xt . Xt Bt . Xt t.

If the price of the stock at time 2 is 30, determine the probability that the price of the stock at time 2.5 is between 30 and 33.

Solution. Since the price of the stock is given by geometric Brownian motion

d = 0 3 d + 0 05 d Xt . Xt Bt . Xt t,

we can read o the solution, namely

Xt = X0 exp

03 + . Bt

0 05 .

0 32 . 2t

=

X0

exp{0 30 . Bt

+

0 005 } . t.

Therefore,

P{30

8X2.5

33|X2

=

30}

9

< log 30 0 0125

log 33 0 0125

log 30 0 01 =

= P:

X0

.

0 30

B2.5

X0

.

0 30

B2 =

X0

.

0 30

;

.

.

.

8

9

< log 30 0 0125

=P

X0

.

:

0 30

.

log 30 0 01

log 33 0 0125

X0

0 30

.

B0.5

X0

.

0 30

.

.

log 30 X0

0 01= .

0.30

;

(

)

=P

0 0025

log

. 0 30

B0.5

33 30

0 0025 .

0 30

.

.

p using the stationarity of Brownian increments. If Z N (0, 1) so that B0.5 0.5 Z, then

P { 0.00833 B0.5 0.3094} = P{ 0.0118 Z 0.4375} = 0.1587.

Remark. The solution to the previous exercise can be generalized as follows. Suppose that

{

0} is geometric Brownian motion given by

Xt, t

d = d+ d Xt Xt Bt ?Xt t

so that If s 0, t > 0, then

Xt = X0 exp

+ Bt ?

2 2 t.

log Xt+s = (Bt+s

Xs

)+ Bs ?

2 2 t.

17?5

Using the facts that (i) Bt+s

Bs

is

independent

of

, Bs

and

(ii)

Bt+s

implies

that

(i)

log

(Xt+s/Xs)

is

independent

of

log

, Xs

and

(ii)

log Xt+s Xt N

?

Xs

X0

2

2

2 t, t .

N (0 )

Bs Bt

,t

Therefore, we can conclude that if 0

and 0 are constants, then

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