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On Dufresne’s Translated Perpetuity and Some Black-Scholes Annuities
Analíti a
k
7
Revista de Análisis Estadístico
Journal of Statistical Analysis
2.2 A probabilistic approach to Dufresne’s
translated perpetuity
We now give another proof of Theorem 2, with a slightly
more probabilistic approach. Let
x
>
0 and assume that
M
0
=
x
. From the Dambis, Dubins-Schwarz theorem (see
[15, Theorem 1.6, p.181]), there exists a Brownian motion
(
B
t
,
t
≥
0
)
started from
x
such that:
M
t
=
B
⟨M⟩
t
with
⟨M⟩
t
=
∫
t
0
(
M
s
)
2
ds
.
Therefore:
∫
+
∞
0
M
2
s
(
M
2
s
+
2
a
M
s
+
b
)
2
ds
=
∫
+
∞
0
B
2
⟨M⟩
s
(
B
2
⟨M⟩
s
+
2
aB
⟨M⟩
s
+
b
)
2
ds
=
∫
T
0
0
1
(
B
2
u
+
2
aB
u
+
b
)
2
du
,
(5)
where
T
0
=
´ınf
{
u
≥
0,
B
u
=
0
}
. By the classic time-
reversal result of Brownian motion (see Section A), this last
expression is seen to be identical in law with:
∫
T
0
0
1
(
B
2
u
+
2
aB
u
+
b
)
2
du
=
∫
T
0
0
1
(
B
2
T
0
−
s
+
2
aB
T
0
−
s
+
b
)
2
ds
(law)
=
∫
G
x
0
1
(
R
2
s
+
2
aR
s
+
b
)
2
ds
(6)
where
(
R
s
,
s
≥
0
)
is a three-dimensional Bessel process
starting from 0 and
G
x
=
sup
{
s
≥
0,
R
s
=
x
}
.
To compute the law of this last expression, we shall first
consider the whole perpetuity
∫
+
∞
0
ds
(
R
2
s
+
2
aR
s
+
b
)
2
.
From the occupation time formula and Ray-Knight’s third
theorem (see Theorem 10):
∫
+
∞
0
ds
(
R
2
s
+
2
aR
s
+
b
)
2
=
∫
+
∞
0
L
y
∞
(
R
)
(
y
2
+
2
ay
+
b
)
2
dy
(law)
=
∫
+
∞
0
Z
(
2
)
y
(
y
2
+
2
ay
+
b
)
2
dy
where
(
Z
(
2
)
y
,
y
≥
0
)
denotes a two-dimensional squared
Bessel process starting from 0. Then, from Theorem 8, the
Laplace transform of the right-hand side equals :
E
exp
−
λ
∫
+
∞
0
Z
(
2
)
y
(
y
2
+
ay
+
b
)
2
dy
=
F
(
∞
)
where
F
is the unique solution on
[
0,
+
∞
[
of :
F
′′
=
2
λ
(
x
2
+
2
ax
+
b
)
2
F
such that
F
is positive, non increasing, and
F
(
0
) =
1.
Therefore, from Lemma 3, there exist two constants
α
and
β
such that :
F
(
x
) =
√
x
2
+
2
ax
+
b
(
α
cosh
(
η
(
x
)
√
2
λ
+
∆
)
+
β
sinh
(
η
(
x
)
√
2
λ
+
∆
))
.
Since
F
(
0
) =
1, we deduce that:
α
=
1
√
b
.
Next, as
F
is positive and non increasing, the limit
F
(
∞
)
necessarily exists, so we must have :
1
√
b
cosh
(
η
(+
∞
)
√
2
λ
+
∆
)
+
β
sinh
(
η
(+
∞
)
√
2
λ
+
∆
)
=
0
which yields
β
=
−
1
√
b
cosh
(
η
(+
∞
)
√
2
λ
+
∆
)
sinh
(
η
(+
∞
)
√
2
λ
+
∆
)
.
Finally, thanks to the additivity formula of sinh:
F
(
x
) =
√
x
2
+
2
ax
+
b
√
b
sinh
(
√
2
λ
+
∆
(
η
(
∞
)
−
η
(
x
))
)
sinh
(
η
(+
∞
)
√
2
λ
+
∆
)
−−−−→
x
→
+
∞
√
2
λ
+
∆
√
b
sinh
(
η
(+
∞
)
√
2
λ
+
∆
)
and we recover the result of Remark 4. Now, to obtain the
result for any
x
>
0 we shall use a decomposition of the
paths of the three-dimensional Bessel process at its last pas-
sage time
G
x
=
sup
{
t
≥
0,
R
t
=
x
}
. From Theorem 9, we
may write :
∫
+
∞
0
ds
(
R
2
s
+
2
aR
s
+
b
)
2
=
∫
G
x
0
ds
(
R
2
s
+
2
aR
s
+
b
)
2
+
∫
+
∞
G
x
ds
(
R
2
s
+
2
aR
s
+
b
)
2
=
∫
G
x
0
ds
(
R
2
s
+
2
aR
s
+
b
)
2
+
∫
+
∞
0
ds
(
R
2
G
x
+
s
+
2
aR
G
x
+
s
+
b
)
2
(law)
=
∫
G
x
0
ds
(
R
2
s
+
2
aR
s
+
b
)
2
+
∫
+
∞
0
ds
(
(
x
+
e
R
s
)
2
+
2
a
(
x
+
e
R
s
) +
b
)
2
where
( e
R
s
,
s
≥
0
)
is an independent copy of
(
R
s
,
s
≥
0
)
.
Taking the Laplace transform of both sides, we obtain, from
Analítika,
Revista de análisis estadístico
, 4 (2014), Vol. 7(1): 7-19
11