A two-dimensional market model
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Chapter 19
A two-dimensional market model
Let
B t=B
1
t;B
2
t; 0 t T;
be a two-dimensional Brownian motion on
; F ; P
.Let
F t; 0 t T;
be the filtration generated by
B
.
In what follows, all processes can depend on
t
and
!
, but are adapted to
F t; 0 t T
.To
simplify notation, we omit the arguments whenever there is no ambiguity.
Stocks:
dS
1
= S
1
1
dt +
1
dB
1
;
dS
2
= S
2
2
dt +
2
dB
1
+
q
1 ,
2
2
dB
2
:
We assume
1
0;
2
0;,11:
Note that
dS
1
dS
2
= S
2
1
2
1
dB
1
dB
1
=
2
1
S
2
1
dt;
dS
2
dS
2
= S
2
2
2
2
2
dB
1
dB
1
+ S
2
2
1 ,
2
2
2
dB
2
dB
2
=
2
2
S
2
2
dt;
dS
1
dS
2
= S
1
1
S
2
2
dB
1
dB
1
=
1
2
S
1
S
2
dt:
In other words,
dS
1
S
1
has instantaneous variance
2
1
,
dS
2
S
2
has instantaneous variance
2
2
,
dS
1
S
1
and
dS
2
S
2
have instantaneous covariance
1
2
.
Accumulation factor:
t = exp
Z
t
0
rdu
:
The market price of risk equations are
1
1
=
1
, r
2
1
+
q
1 ,
2
2
2
=
2
, r
(MPR)
203
204
The solution to these equations is
1
=
1
, r
1
;
2
=
1
2
, r ,
2
1
, r
1
2
p
1 ,
2
;
provided
,1 1
.
Suppose
,1 1
. Then (MPR) has a unique solution
1
;
2
;wedefine
Z t = exp
,
Z
t
0
1
dB
1
,
Z
t
0
2
dB
2
,
1
2
Z
t
0
2
1
+
2
2
du
;
f
IP A=
Z
A
ZTdIP; 8A 2F:
f
IP
is the unique risk-neutral measure. Define
e
B
1
t=
Z
t
0
1
du + B
1
t;
e
B
2
t=
Z
t
0
2
du + B
2
t:
Then
dS
1
= S
1
h
rdt+
1
d
e
B
1
i
;
dS
2
= S
2
rdt+
2
d
e
B
1
+
q
1 ,
2
2
d
e
B
2
:
We have changed the mean rates of return of the stock prices, but not the variances and covariances.
19.1 Hedging when
,1 1
dX =
1
dS
1
+
2
dS
2
+ rX ,
1
S
1
,
2
S
2
dt
d
X
=
1
dX , rX dt
=
1
1
dS
1
, rS
1
dt+
1
2
dS
2
, rS
2
dt
=
1
1
S
1
1
d
e
B
1
+
1
2
S
2
2
d
e
B
1
+
q
1 ,
2
2
d
e
B
2
:
Let
V
be
F T
-measurable. Define the
f
IP
-martingale
Y t=
f
IE
V
T
Ft
; 0tT:
CHAPTER 19. A two-dimensional market model
205
The Martingale Representation Corollary implies
Y t=Y0 +
Z
t
0
1
d
e
B
1
+
Z
t
0
2
d
e
B
2
:
We have
d
X
=
1
1
S
1
1
+
1
2
S
2
2
d
e
B
1
+
1
2
S
2
q
1 ,
2
2
d
e
B
2
;
dY =
1
d
e
B
1
+
2
d
e
B
2
:
We solve the equations
1
1
S
1
1
+
1
2
S
2
2
=
1
1
2
S
2
q
1 ,
2
2
=
2
for the hedging portfolio
1
;
2
. With this choice of
1
;
2
and setting
X 0 = Y 0 =
f
IE
V
T
;
we have
X t=Yt; 0t T;
and in particular,
X T = V:
Every
F T
-measurable random variable can be hedged; the market is complete.
19.2 Hedging when
=1
The case
= ,1
is analogous. Assume that
=1
.Then
dS
1
= S
1
1
dt +
1
dB
1
dS
2
= S
2
2
dt +
2
dB
1
The stocks are perfectly correlated.
The market price of risk equations are
1
1
=
1
, r
2
1
=
2
, r
(MPR)
The process
2
is free. There are two cases:
206
Case I:
1
,r
1
6=
2
,r
2
:
There is no solution to (MPR), and consequently, there is no risk-neutral
measure. This market admits arbitrage. Indeed
d
X
=
1
1
dS
1
, rS
1
dt+
1
2
dS
2
, rS
2
dt
=
1
1
S
1
1
, r dt +
1
dB
1
+
1
2
S
2
2
, r dt +
2
dB
1
Suppose
1
,r
1
2
,r
2
:
Set
1
=
1
1
S
1
;
2
= ,
1
2
S
2
:
Then
d
X
=
1
1
, r
1
dt + dB
1
,
1
2
, r
2
dt + dB
1
=
1
1
, r
1
,
2
, r
2
| z
Positive
dt
Case II:
1
,r
1
=
2
,r
2
:
The market price of risk equations
1
1
=
1
, r
2
1
=
2
, r
have the solution
1
=
1
, r
1
=
2
, r
2
;
2
is free; there are infinitely many risk-neutral measures. Let
f
IP
be one of them.
Hedging:
d
X
=
1
1
S
1
1
, r dt +
1
dB
1
+
1
2
S
2
2
, r dt +
2
dB
1
=
1
1
S
1
1
1
dt + dB
1
+
1
2
S
2
2
1
dt + dB
1
=
1
1
S
1
1
+
1
2
S
2
2
d
e
B
1
:
Notice that
e
B
2
does not appear.
Let
V
be an
F T
-measurable random variable. If
V
depends on
B
2
, then it can probably not
be hedged. For example, if
V = hS
1
T ;S
2
T;
and
1
or
2
depend on
B
2
, then there is trouble.
CHAPTER 19. A two-dimensional market model
207
More precisely, we define the
f
IP
-martingale
Y t=
f
IE
V
T
Ft
; 0tT:
We can write
Y t=Y0 +
Z
t
0
1
d
e
B
1
+
Z
t
0
2
d
e
B
2
;
so
dY =
1
d
e
B
1
+
2
d
e
B
2
:
To get
d
X
to match
dY
,wemusthave
2
=0:
