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Christophe Profeta
Analíti a
k
7
Revista de Análisis Estadístico
Journal of Statistical Analysis
3.2 Proof of Theorem 5
We restrict our attention to the case
x
=
0, for which the
formulae take a simpler form. To prove Theorem 5, we first
remove the drift thanks to the Cameron-Martin formula,
which states that for every functional
F
:
E
0
[
F
(
B
s
+
ν
s
,
s
≤
t
)] =
E
0
[
e
ν
B
t
−
ν
2
t
2
F
(
B
s
,
s
≤
t
)
]
=
e
−
ν
2
t
2
E
0
[
e
ν
B
t
F
(
B
s
,
s
≤
t
)
]
.
Therefore the double Laplace transform reads:
∫
+
∞
0
e
−
(
λ
+
a
)
t
E
0
[
exp
(
−
α
∫
t
0
(
e
2
β
(
B
s
+
ν
s
)
−
1
)
+
ds
−
a
∫
t
0
(
e
2
b
(
B
s
+
ν
s
)
−
1
)
−
ds
)]
dt
=
∫
+
∞
0
e
−
(
λ
+
ν
2
2
)
t
E
0
[
e
ν
B
t
exp
(
−
α
∫
t
0
(
e
2
β
(
B
s
+
ν
s
)
−
1
)
+
ds
−
at
−
a
∫
t
0
(
e
2
b
(
B
s
+
ν
s
)
−
1
)
−
ds
)]
dt
Consider the ordinary differential equation:
1
2
ϕ
′′
(
x
) =
(
α
(
e
2
β
x
−
1
)
1
{
x
>
0
}
+
a
(
e
2
bx
−
1
)
1
{
x
<
0
}
+
a
+
ν
2
2
+
λ
)
ϕ
(
x
)
.
The function
V
defined by
V
(
x
) =
α
(
e
2
β
x
−
1
)
1
{
x
>
0
}
+
a
(
e
2
bx
−
1
)
1
{
x
<
0
}
+
a
+
ν
2
2
is positive and continuous, so we may apply the Feynman-
Kac Theorem 6. To simplify the notation, we set:
c
=
1
2
b
√
2
λ
+
ν
2
and
γ
=
1
2
β
√
2
(
λ
+
a
)
−
2
α
+
ν
2
if 2
(
λ
+
a
)
−
2
α
+
ν
2
≥
0
i
2
β
√
2
α
−
2
(
λ
+
a
)
−
ν
2
otherwise.
In this framework, the solutions
ϕ
+
and
ϕ
−
which satisfy
(9) are given by (see Section A.1):
ϕ
+
(
x
) =
K
2
γ
(
√
2
α
β
e
β
x
)
for
x
≥
0,
and
ϕ
−
(
x
) =
I
2
c
(
√
2
a
b
e
bx
)
for
x
≤
0.
Their Wronskien equals
ω
λ
=
√
2
aK
2
γ
(
√
2
α
β
)
I
′
2
c
(
√
2
a
b
)
−
√
2
α
K
′
2
γ
(
√
2
α
β
)
I
2
c
(
√
2
a
b
)
,
so the Feynman-Kac formula yields, for
x
=
0 :
∫
+
∞
0
e
−
(
λ
+
a
)
t
E
0
[
exp
(
−
α
∫
t
0
(
e
2
β
(
B
s
+
ν
s
)
−
1
)
+
ds
−
a
∫
t
0
(
e
2
b
(
B
s
+
ν
s
)
−
1
)
−
ds
)]
dt
=
2
ω
λ
(
ϕ
−
(
0
)
∫
+
∞
0
e
ν
y
ϕ
+
(
y
)
dy
+
ϕ
+
(
0
)
∫
0
−
∞
e
ν
y
ϕ
−
(
y
)
dy
)
which ends the proof.
4 Some special annuities
We now look at some particular cases of Theorem 5, ac-
cording to the values of
α
,
β
,
a
and
b
. Most of the formulae
we obtain may be found in [2].
4.1 The Black-Scholes call annuity
We first let
a
→
0 in Theorem 5. Recall the asymptotics
(see Section A.1):
I
µ
(
z
)
∼
z
→
0
z
µ
2
µ
Γ
(
µ
+
1
)
and
zI
′
µ
(
z
) =
zI
µ
+
1
(
z
) +
µ
I
µ
(
z
)
∼
z
→
0
µ
z
µ
2
µ
Γ
(
µ
+
1
)
.
We thus obtain the double Laplace transform of the Black-
Scholes call annuity:
∫
+
∞
0
e
−
λ
t
E
0
[
e
−
α
∫
t
0
(
e
2
β
(
B s
+
ν
s
)
−
1
)
+
ds
]
dt
=
2
ω
λ
∫
+
∞
0
e
ν
y
K
2
γ
(
√
2
α
β
e
β
y
)
dy
+
K
2
γ
(
√
2
α
β
)
ν
+
√
2
λ
+
ν
2
where the Wronskien
ω
λ
equals:
ω
λ
=
√
2
λ
+
ν
2
K
2
γ
(
√
2
α
β
)
−
√
2
α
K
′
2
γ
(
√
2
α
β
)
.
We may also recover the Laplace transform of the associa-
ted perpetuity as follows. Assume that
ν
<
0 and replace
λ
by
λε
to obtain:
∫
+
∞
0
e
−
λε
t
E
0
[
e
−
α
∫
t
0
(
e
2
β
(
B s
+
ν
s
)
−
1
)
+
ds
]
dt
=
1
ε
∫
+
∞
0
e
−
λ
t
E
0
[
e
−
α
∫
t
/
ε
0
(
e
2
β
(
B s
+
ν
s
)
−
1
)
+
ds
]
dt
.
Now letting
ε
→
0, we deduce that:
1
λ
E
0
[
e
−
α
∫
+
∞
0
(
e
2
β
(
B s
+
ν
s
)
−
1
)
+
ds
]
=
l´ım
ε
→
0
2
ε
w
λε
∫
+
∞
0
e
ν
y
K
2
γ
(
√
2
α
β
e
β
y
)
dy
+
K
2
γ
(
√
2
α
β
)
ν
+
√
2
λε
+
ν
2
14
Analítika,
Revista de análisis estadístico
, 4 (2014), Vol. 7(1): 7-19