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Applying Theorem 4.1.4, we can obtain the value
V
1
of the underlying option by
solving the partial differential equation
∂
t
V
1
+
A
V
1
−
rV
1
=
0in
(
0
,T
1
)
× R
+
,V
1
(T
1
,s)
=
g
1
(s)
in
R
+
,
where
denotes the Black-Scholes generator (4.5). In a second step, we get the
price of the compound option by solving
A
∂
t
V
+
A
V
−
rV
=
0in
(
0
,T)
× R
+
,V T, )
=
g(V
1
(T , s))
in
R
+
.
As in the European plain vanilla case, we change to log-price
x
and localize the
problem to a bounded domain
G
R,R)
. The corresponding value of the barrier
option for the underlying option is denoted with
v
1
,R
,i.e.
=
(
−
e
−
r(T
1
−
t)
g
1
(e
X
T
1
)
1
x
,
v
1
,R
(t, x)
= E
T
1
<τ
G
}
|
X
t
=
(6.7)
{
and the value of the barrier option for the compound option by
v
R
,i.e.
e
−
r(T
−
t)
g(v
1
,R
(T , X
T
))
1
x
.
v
R
(t, x)
= E
}
|
X
t
=
(6.8)
{
T<τ
G
Again, the compound barrier option price
v
R
converges to the compound option
price
v
exponentially fast as
R
→∞
in a Black-Scholes model.
Theorem 6.3.1
Suppose the payoff functions g,g
1
: R → R
of a compound contract
in a Black-Scholes market satisfy
(4.10)
and that g is Lipschitz continuous
.
Then
,
there exist C(T,T
1
,σ),γ
1
,γ
2
>
0,
such that
C(T,T
1
,σ)e
−
γ
1
R
+
γ
2
|
x
|
.
|
v(t,x)
−
v
R
(t, x)
| ≤
Proof
Applying Theorem 4.3.1 to the underlying option value
v
1
, we obtain
v
1
(t, x)
v
1
,R
(t, x)
≤
C
1
(T
1
,σ)e
−
γ
3
R
+
γ
4
|
x
|
,
−
for some
C
1
(T
1
,σ),γ
3
,γ
4
>
0. If we denote by
v
R
the barrier compound option
on
v
1
,
e
−
r(T
−
t)
g(v
1
(T , X
T
))
1
{
T<τ
G
}
|
x
,
v
R
(t, x)
= E
X
t
=
we have with Lipschitz constant
C
2
C
2
E
v
1
(T , X
T
)
v
1
,R
(T , X
T
)
1
{
T
≥
τ
G
}
|
x
|
v
R
(t, x)
−
v
R
(t, x)
| ≤
−
X
t
=
e
γ
4
|
X
T
|
1
{
T<τ
G
}
|
x
C
1
C
2
(T
1
,σ)e
−
γ
3
R
≤
E
X
t
=
C
3
(T , T
1
,σ)e
−
γ
3
R
+
γ
4
|
x
|
.
≤
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