Markov processes and the Kolmogorov equations
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Chapter 16
Markov processes and the Kolmogorov
equations
16.1 Stochastic Differential Equations
Consider the stochastic differential equation:
dX t= at; X t dt + t; X t dB t:
(SDE)
Here
at; x
and
t; x
are given functions, usually assumed to be continuous in
t; x
and Lips-
chitz continuous in
x
,i.e., there is a constant
L
such that
jat; x , at; y j Ljx,yj; jt; x , t; y j Ljx,yj
for all
t; x; y
.
Let
t
0
;x
be given. A solution to (SDE) with the initial condition
t
0
;x
is a process
fX tg
tt
0
satisfying
X t
0
=x;
X t= Xt
0
+
t
Z
t
0
as; X s ds +
t
Z
t
0
s; X s dB s; t t
0
The solution process
fX tg
tt
0
will be adapted to the filtration
fF tg
t0
generated by the Brow-
nian motion. If you know the path of the Brownian motion up to time
t
, then you can evaluate
X t
.
Example 16.1 (Drifted Brownian motion) Let
a
be a constant and
=1
,so
dX t=adt+dB t:
If
t
0
;x
is given and we start with the initial condition
X t
0
=x;
177
178
then
X t=x+at,t
0
+Bt,Bt
0
; t t
0
:
To compute the differential w.r.t.
t
, treat
t
0
and
B t
0
as constants:
dX t=adt+dB t:
Example 16.2 (Geometric Brownian motion) Let
r
and
be constants. Consider
dX t=rX t dt + X t dB t:
Given the initial condition
X t
0
=x;
the solution is
X t=xexp
B t , B t
0
+ r ,
1
2
2
t , t
0
:
Again, to compute the differential w.r.t.
t
, treat
t
0
and
B t
0
as constants:
dX t=r,
1
2
2
Xtdt + Xt dB t+
1
2
2
Xt dt
= rX t dt + X t dB t:
16.2 Markov Property
Let
0 t
0
t
1
be given and let
hy
be a function. Denote by
IE
t
0
;x
hX t
1
the expectation of
hX t
1
, given that
X t
0
=x
.Nowlet
2 IR
be given, and start with initial
condition
X 0 = :
We have the Markov property
IE
0;
hX t
1
F t
0
= IE
t
0
;X t
0
hX t
1
:
In other words, if you observe the path of the driving Brownian motion from time 0 to time
t
0
,and
based on this information, you want to estimate
hX t
1
, the only relevant information is the value
of
X t
0
. You imagine starting the
SDE
at time
t
0
at value
X t
0
, and compute the expected
value of
hX t
1
.
CHAPTER 16. Markov processes and the Kolmogorov equations
179
16.3 Transition density
Denote by
pt
0
;t
1
; x; y
the density (in the
y
variable) of
X t
1
, conditioned on
X t
0
=x
.Inotherwords,
IE
t
0
;x
hX t
1
=
Z
IR
hy pt
0
;t
1
; x; y dy :
The Markov property says that for
0 t
0
t
1
and for every
,
IE
0;
hX t
1
F t
0
=
Z
IR
hy pt
0
;t
1
; Xt
0
;y dy :
Example 16.3 (Drifted Brownian motion) Consider the SDE
dX t=adt+dB t:
Conditioned on
X t
0
=x
, the random variable
X t
1
is normal with mean
x + at
1
, t
0
and variance
t
1
, t
0
, i.e.,
pt
0
;t
1
; x; y=
1
p
2t
1
,t
0
exp
,
y , x + at
1
, t
0
2
2t
1
, t
0
:
Note that
p
depends on
t
0
and
t
1
only through their difference
t
1
, t
0
. This is always the case when
at; x
and
t; x
don’t depend on
t
.
Example 16.4 (Geometric Brownian motion) Recall that the solution to the SDE
dX t=rX t dt + X t dB t;
with initial condition
X t
0
=x
, is Geometric Brownian motion:
X t
1
=xexp
B t
1
, B t
0
+ r ,
1
2
2
t
1
, t
0
:
The random variable
B t
1
, B t
0
has density
IP fB t
1
, B t
0
2 dbg =
1
p
2t
1
, t
0
exp
,
b
2
2t
1
, t
0
db;
and we are making the change of variable
y = x exp
b +r,
1
2
2
t
1
, t
0
or equivalently,
b =
1
h
log
y
x
, r ,
1
2
2
t
1
, t
0
i
:
The derivative is
dy
db
= y;
or equivalently,
db =
dy
y
:
180
Therefore,
pt
0
;t
1
;x; y dy = IP fX t
1
2 dyg
=
1
y
p
2t
1
, t
0
exp
,
1
2t
1
, t
0
2
h
log
y
x
, r ,
1
2
2
t
1
, t
0
i
2
dy:
Using the transition density and a fair amount of calculus, one can compute the expected payoff from a
European call:
IE
t;x
X T , K
+
=
Z
1
0
y , K
+
pt; T ; x; y dy
= e
rT ,t
xN
1
p
T , t
h
log
x
K
+ rT , t+
1
2
2
T ,t
i
,KN
1
p
T , t
h
log
x
K
+ rT , t ,
1
2
2
T , t
i
where
N =
1
p
2
Z
,1
e
,
1
2
x
2
dx =
1
p
2
Z
1
,
e
,
1
2
x
2
dx:
Therefore,
IE
0;
e
,rT ,t
X T , K
+
F t
= e
,rT ,t
IE
t;X t
X T , K
+
= X tN
1
p
T , t
log
X t
K
+ rT , t+
1
2
2
T ,t
, e
,rT ,t
KN
1
p
T,t
log
X t
K
+ rT , t ,
1
2
2
T , t
16.4 The Kolmogorov Backward Equation
Consider
dX t= at; X t dt + t; X t dB t;
and let
pt
0
;t
1
;x; y
be the transition density. Then the Kolmogorov Backward Equation is:
,
@
@t
0
pt
0
;t
1
; x; y =at
0
;x
@
@x
pt
0
;t
1
; x; y +
1
2
2
t
0
;x
@
2
@x
2
pt
0
;t
1
; x; y :
(KBE)
The variables
t
0
and
x
in
KBE
are called the backward variables.
In the case that
a
and
are functions of
x
alone,
pt
0
;t
1
; x; y
depends on
t
0
and
t
1
only through
their difference
= t
1
, t
0
. We then write
p ; x; y
rather than
pt
0
;t
1
; x; y
,and
KBE
becomes
@
@
p ; x; y =ax
@
@x
p ; x; y +
1
2
2
x
@
2
@x
2
p ; x; y :
(KBE’)
CHAPTER 16. Markov processes and the Kolmogorov equations
181
Example 16.5 (Drifted Brownian motion)
dX t=adt+dBt
p ; x; y=
1
p
2
exp
,
y , x + a
2
2
:
@
@
p = p
=
@
@
1
p
2
exp
,
y , x , a
2
2
,
@
@
y , x , a
2
2
1
p
2
exp
,
y , x , a
2
2
=
,
1
2
+
ay , x , a
+
y , x , a
2
2
p:
@
@x
p = p
x
=
y , x , a
p:
@
2
@x
2
p = p
xx
=
@
@x
y , x , a
p +
y , x , a
p
x
= ,
1
p +
y , x , a
2
2
p:
Therefore,
ap
x
+
1
2
p
xx
=
ay , x , a
,
1
2
+
y , x , a
2
2
2
p
= p
:
This is the Kolmogorov backward equation.
Example 16.6 (Geometric Brownian motion)
dX t=rX t dt + X t dB t:
p ; x; y=
1
y
p
2
exp
,
1
2
2
h
log
y
x
, r ,
1
2
2
i
2
:
It is true but very tedious to verify that
p
satisfies the KBE
p
= rxp
x
+
1
2
2
x
2
p
xx
:
16.5 Connection between stochastic calculus and KBE
Consider
dX t=aXt dt + X t dB t:
(5.1)
Let
hy
be a function, and define
v t; x= IE
t;x
hX T ;
