Versión de HTML Básico
Christophe Profeta
Analíti a
k
7
Revista de Análisis Estadístico
Journal of Statistical Analysis
Using this transform and the change of variable
u
=
A
s
,
we obtain :
∫
+
∞
0
E
2
s
(
E
2
s
+
2
a
E
s
+
b
)
2
ds
=
∫
+
∞
0
R
2
A
s
(
R
2
A
s
+
2
aR
A
s
+
b
)
2
ds
=
∫
+
∞
0
1
(
R
2
u
+
2
aR
u
+
b
)
2
du
.
(2)
To compute the law of this perpetuity, we shall construct
the appropriate martingale, thanks to the following lem-
ma:
L
EMMA
3
([13])
.
Let
∆
=
a
2
−
b and
2
λ
+
∆
>
0
. Set for
x
≥
0
:
η
(
x
) =
∫
x
0
dz
z
2
+
2
az
+
b
.
Then, the functions
ϕ
±
(
x
) =
√
x
2
+
2
ax
+
b
exp
(
±
η
(
x
)
√
2
λ
+
∆
)
are two independent solutions of the equation:
f
′′
−
2
λ
(
x
2
+
2
ax
+
b
)
2
f
=
0.
Let
(
W
t
,
t
≥
0
)
be a Brownian motion. Applying Itô’s for-
mula, we deduce that the process
M
t
=
√
W
2
t
+
2
aW
t
+
b
sinh
(
η
(
W
t
)
√
2
λ
+
∆
)
exp
(
−
λ
∫
t
0
du
(
W
2
u
+
2
aW
u
+
b
)
2
)
is a continuous martingale, and with
T
0
=
´ınf
{
t
≥
0,
W
t
=
0
}
, Doob’s optional stopping theorem implies:
√
x
2
+
2
ax
+
b
sinh
(
√
2
λ
+
∆
η
(
x
)
)
=
E
x
[ √
W
2
t
∧
T
0
+
2
aW
t
∧
T
0
+
b
sinh
(
η
(
W
t
∧
T
0
)
√
2
λ
+
∆
)
exp
(
−
λ
∫
t
∧
T
0
0
du
(
W
2
u
+
2
aW
u
+
b
)
2
)]
=
E
x
[√
W
2
t
+
2
aW
t
+
b
sinh
(
η
(
W
t
)
√
2
λ
+
∆
)
exp
(
−
λ
∫
t
0
du
(
W
2
u
+
2
aW
u
+
b
)
2
)
1
{
t
<
T
0
}
]
.
Now, from the absolute continuity formula between Brow-
nian motion and the three-dimensional Bessel process (14),
we deduce that:
√
x
2
+
2
ax
+
b
sinh
(
η
(
x
)
√
2
λ
+
∆
)
=
x
E
(
3
)
x
[
1
R
t
√
R
2
t
+
2
aR
t
+
b
sinh
(
η
(
R
t
)
√
2
λ
+
∆
)
exp
(
−
λ
∫
t
0
du
(
R
2
u
+
2
aR
u
+
b
)
2
)]
and, letting
t
go towards
+
∞
, we obtain from the domina-
ted convergence theorem :
E
(
3
)
x
[
exp
(
−
λ
∫
+
∞
0
du
(
R
2
u
+
2
aR
u
+
b
)
2
)]
=
1
x
√
x
2
+
2
ax
+
b
sinh
(
η
(
x
)
√
2
λ
+
∆
)
sinh
(
η
(+
∞
)
√
2
λ
+
∆
)
,
(3)
which, thanks to (2), gives the first part of Theorem 2.
Next, the analogous formula for
M
follows by symmetry.
Indeed, since when
B
0
=
0,
M
t
=
exp
(
B
t
−
t
2
)
(law)
=
exp
(
−
B
t
−
t
2
)
=
1
E
t
,
(4)
we obtain :
∫
+
∞
0
x
2
M
2
s
(
x
2
M
2
s
+
2
ax
M
s
+
b
)
2
ds
=
∫
+
∞
0
x
2
E
2
s
(
x
2
+
2
ax
E
s
+
b
E
2
s
)
2
ds
=
x
2
∫
+
∞
0
b
2
E
2
s
(
bx
2
+
2
axb
E
s
+
b
2
E
2
s
)
2
ds
and
E
x
[
exp
(
−
λ
∫
+
∞
0
M
2
s
(
M
2
s
+
2
a
M
s
+
b
)
2
ds
)]
=
√
b
+
ax
+
x
2
√
b
sinh
(
x
√
2
λ
+
∆
∫
b
0
dz
z
2
+
2
axz
+
bx
2
)
sinh
(
x
√
2
λ
+
∆
∫
+
∞
0
dz
z
2
+
2
axz
+
bx
2
)
=
√
b
+
2
ax
+
x
2
√
b
sinh
(
√
2
λ
+
∆
∫
b
/
x
0
dy
y
2
+
2
ay
+
b
)
sinh
(
√
2
λ
+
∆
∫
+
∞
0
dy
y
2
+
2
ay
+
b
)
.
after the change of variable
y
=
xz
.
Remark 4.
Letting
x
tend towards 0 in (3), we obtain the
simple formula:
E
(
3
)
0
[
exp
(
−
λ
∫
+
∞
0
du
(
R
2
u
+
2
aR
u
+
b
)
2
)]
=
√
2
λ
+
∆
√
b
sinh
(
η
(+
∞
)
√
2
λ
+
∆
)
.
10
Analítika,
Revista de análisis estadístico
, 4 (2014), Vol. 7(1): 7-19