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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 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 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 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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