Lecture #28: Calculations with Itoˆ’s Formula - University of Regina
[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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