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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 +
2
γ 1 )t , where the standard Brownian motion
Σ ε W t +
W is independent of Y 1 .Weset
x = x + 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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