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d
)
.
Then
,
there is C
1
>
0
E
(g(x
+
Z
1
,T
))
− E
(g(x
+
Y
1
,T
))
≤
C
1
C
2
(
∈
R
Proposition 14.7.4
Assume g
ε
d
z
j
2
ν
j
(
d
z
j
),
d
.
∀
x
∈ R
−
ε
j
=
1
(14.37)
d
) and
R
C
4
(
Furthermore
,
assume g
∈
R
|
z
|
ν(
d
z) <
∞
.
Then
,
for some C
2
>
0
d
ε
d
E
Y
2
,T
))
≤
z
j
3
ν
j
(
d
z
j
),
Z
2
,T
))
d
.
(g(x
+
− E
(g(x
+
C
2
∀
x
∈ R
−
ε
j
=
1
(14.38)
Proof
Consider the Taylor series expansion of
g(x)
at
x
0
1
2
(x
−
x
0
)
·
D
2
g(x
0
)(x
−
x
0
)
+
O
(
|
x
−
x
0
|
3
),
g(x)
=
g(x
0
)
+∇
g(x
0
)
·
(x
−
x
0
)
+
where
D
2
g
is the Hessian matrix of
g
. Define
R
ε
Y
1
. The Lévy process
R
ε
has Lévy measure
ν
ε
and is independent of
Y
1
. Since
Z
ε,j
Z
1
−
=
and
Y
ε,j
1
,T
,
j
=
1
,...,d
,
1
,T
E
(R
ε,j
T
have the same expected value, we have
)
=
0,
j
=
1
,...,d
. Thus, we obtain
E
(g(x
+
Z
1
,T
))
− E
(g(x
+
Y
1
,T
))
d
∂
j
g
x
Y
1
,T
d
d
R
ε,
T
E
R
ε,
T
+
E
R
ε,j
T
=
1
E
+
1
O
j
=
j
=
1
k
=
R
ε,j
T
2
.
d
≤
C
1
1
E
j
=
Equation (
14.37
) follows from
ε
R
ε,j
T
2
z
j
2
ν
j
(
d
z
j
),
E
=
j
=
1
,...,d.
−
ε
Furthermore,
Y
2
,t
=
Y
1
,t
+
(γ
2
−
γ
1
)t
, where the standard Brownian motion
Σ
ε
W
t
+
W
is independent of
Y
1
.Weset
x
=
x
+
(γ
2
−
γ
1
)T
and obtain
E
Y
2
,T
))
Z
2
,T
))
(g(x
+
− E
(g(x
+
E
Y
2
,T
))
Z
1
,T
))
Y
1
,T
))
Y
1
,T
))
=
(g(
x
+
− E
(g(
x
+
+ E
(g(
x
+
− E
(g(x
+
∂
j
∂
k
g
1
,T
R
ε,j
T
3
d
d
d
E
|
1
2
E
Y
ε,j
R
ε,j
T
R
ε,k
T
=
x
+
E
+
1
O
|
j
=
1
k
=
1
j
=
∂
j
∂
k
g
1
,T
4
d
d
E
|
1
2
E
Y
ε,j
−
x
+
Q
ε,jk
+
O
Σ
ε
W
T
|
j
=
k
=
1
1
ε
2
ν
j
(
d
z
j
)
2
.
ε
d
z
j
z
j
3
ν
j
(
d
z
j
)
≤
C
2
+
−
ε
−
ε
j
=
1
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