3.1. The Case of Call Option
In this section, we derive the formula of European call option with the Hull-White interest rate using the Mellin transform. However, since the European call option has the payoff function
P
(
T
,
s
,
r
)
=
h
(
s
)
=
(
s
-
K
)
+
, the Mellin transform of the payoff function does not exist. Therefore, a somewhat modified form of
h
(
s
)
=
(
s
-
K
)
+
is needed to guarantee the existence of the integral, and we define the sequence of the payoff function
h
n
(
s
)
such that
lim
n
→
∞
h
n
(
s
)
=
h
(
s
)
as follows:
(10)
h
n
(
S
)
=
{
S
T
-
K
,
if
K
<
S
T
<
n
0
,
if
S
T
<
K
.
If we define the call option price with the payoff function
h
n
(
s
)
=
P
n
(
T
,
s
,
r
)
as
(11)
P
n
(
t
,
s
,
r
)
=
E
*
{
exp
(
-
∫
t
T
r
t
*
d
t
*
)
h
n
(
S
T
)
∣
S
t
=
s
,
r
t
=
r
}
,
P
n
(
t
,
s
,
r
)
satisfies the PDE given by (4). Then, if we find the solution
P
n
(
t
,
s
,
r
)
by using the Mellin transform, we can obtain the formula of the option price
P
(
t
,
s
,
r
)
from
P
(
t
,
s
,
r
)
=
lim
n
→
∞
P
n
(
t
,
s
,
r
)
.
If we define
p
^
n
(
t
,
w
,
r
)
as the Mellin transform of
P
n
(
t
,
s
,
r
)
, then the inverse of the Mellin transform is given by
(12)
P
n
(
t
,
s
,
r
)
=
1
2
π
i
∫
c
-
i
∞
c
+
i
∞
p
^
n
(
t
,
w
,
r
)
s
-
w
d
w
.
By substituting (12) into the PDE
L
^
P
n
=
0
mentioned above, the PDE is transformed by
(13)
-
∂
p
^
n
∂
τ
+
(
1
2
σ
2
w
(
w
+
1
)
-
r
w
-
r
)
p
^
n
+
(
b
(
T
-
τ
)
-
a
r
-
ρ
s
r
σ
σ
ˇ
w
)
∂
p
^
n
∂
r
+
1
2
σ
ˇ
2
∂
2
p
^
n
∂
r
2
=
0
,
where
τ
=
T
-
t
and the terminal condition is given by
p
^
n
(
0
,
w
,
r
)
=
∫
0
∞
P
n
(
0
,
s
,
r
)
s
w
-
1
d
s
=
∫
K
n
(
s
-
K
)
s
w
-
1
d
s
=
n
w
(
n
/
(
w
+
1
)
-
K
/
w
)
+
K
w
+
1
/
w
(
w
+
1
)
.
To simplify PDE (13), let us assume
p
^
n
(
τ
,
w
,
r
)
can be expressed by the form
(14)
p
^
n
(
τ
,
w
,
r
)
=
exp
{
1
2
σ
2
w
(
w
+
1
)
τ
}
g
^
n
(
τ
,
w
,
r
)
,
and then
g
^
n
must satisfy
(15)
-
∂
g
^
n
∂
τ
+
(
-
a
r
+
(
b
(
T
-
τ
)
-
ρ
s
r
σ
σ
ˇ
w
)
)
∂
g
^
n
∂
r
+
1
2
σ
ˇ
2
∂
2
g
^
n
∂
r
2
-
r
(
w
+
1
)
g
^
n
=
0
with the final condition
g
^
n
(
0
,
w
,
r
)
=
p
^
n
(
0
,
w
,
r
)
=
n
w
(
n
/
(
w
+
1
)
-
K
/
w
)
+
K
w
+
1
/
w
(
w
+
1
)
.
Now, to solve PDE (15), we set
g
^
n
(
τ
,
w
,
r
)
=
E
n
(
w
)
H
(
τ
,
w
)
e
-
G
(
τ
)
(
w
+
1
)
r
. Here,
g
^
n
(
0
,
w
,
r
)
=
E
n
(
w
)
and by substituting this functional form of the solution into PDE (15), we have two ordinary differential equations (ODEs) with respect to
H
(
τ
,
w
)
and
G
(
τ
)
as follows:
(16)
-
H
τ
-
(
1
1
(
b
(
T
-
τ
)
-
ρ
s
r
σ
σ
ˇ
w
)
G
(
τ
)
(
w
+
1
)
h
h
h
h
h
h
h
-
1
2
σ
ˇ
2
G
2
(
τ
)
(
w
+
1
)
2
)
H
=
0
,
G
′
(
τ
)
+
a
G
(
τ
)
-
1
=
0
with
H
(
0
,
w
)
=
1
and
G
(
0
)
=
0
. ODEs (16) yield
(17)
H
(
τ
,
w
)
=
exp
{
∫
0
τ
L
(
τ
*
,
w
)
d
τ
*
}
,
G
(
τ
)
=
1
-
e
-
a
τ
a
,
where
L
(
τ
,
w
)
=
(
1
/
2
)
σ
ˇ
2
G
2
(
τ
)
(
w
+
1
)
2
-
(
b
(
T
-
τ
)
-
ρ
s
r
σ
σ
ˇ
w
)
G
(
τ
)
(
w
+
1
)
.
Therefore, from the solution of
g
^
n
(
τ
,
w
,
r
)
and (14), we obtain the solution of
p
^
n
(
τ
,
w
,
r
)
as follows:
(18)
p
^
n
(
τ
,
w
,
r
)
=
E
n
(
w
)
exp
{
M
1
(
τ
)
(
w
+
1
)
2
+
1
2
σ
2
w
(
w
+
1
)
τ
777777777777
i
77
+
η
w
(
w
+
1
)
(
τ
-
G
(
τ
)
)
-
M
2
(
τ
)
(
w
+
1
)
77777777777
i
777
-
G
(
τ
)
r
1
2
}
,
where
(19)
M
1
(
τ
)
=
σ
ˇ
2
2
a
2
(
τ
+
2
a
(
e
-
a
τ
-
1
)
-
1
2
a
(
e
-
2
a
τ
-
1
)
)
M
2
(
τ
)
=
∫
0
τ
G
(
τ
*
)
b
(
T
-
τ
*
)
d
τ
*
,
η
=
ρ
s
r
σ
σ
ˇ
a
.
From the definition of
M
1
(
τ
)
,
M
2
(
τ
)
, and
G
(
τ
)
, the bond price stated in Section 2.1 is
(20)
B
(
T
-
τ
,
r
;
T
)
=
exp
{
M
1
(
τ
)
-
M
2
(
τ
)
-
G
(
τ
)
r
}
.
Finally, we have the price of the European call option:
(21)
P
n
(
t
,
w
,
r
)
=
1
(
2
π
i
)
×
∫
c
-
i
∞
c
+
i
∞
E
n
(
w
)
exp
{
M
1
(
τ
)
(
w
+
1
)
2
+
1
2
σ
2
w
(
w
+
1
)
τ
ggggggghhhhhhhhhhihhhg
+
η
w
(
w
+
1
)
(
τ
-
G
(
τ
)
)
ggggghhhhhhhgghhihhhhg
-
M
2
(
τ
)
(
w
+
1
)
-
G
(
τ
)
r
1
2
}
f
f
f
f
f
f
f
f
f
f
f
f
×
s
-
w
d
w
,
and to compute the above integral (21), we define the following equation:
(22)
γ
(
s
)
=
1
(
2
π
i
)
∫
c
-
i
∞
c
+
i
∞
e
φ
(
τ
,
w
,
r
)
s
-
w
d
w
,
where
φ
(
τ
,
w
,
r
)
=
{
M
1
(
τ
)
(
w
+
1
)
2
+
(
1
/
2
)
σ
2
w
(
w
+
1
)
τ
+
η
w
(
w
+
1
)
(
τ
-
G
(
τ
)
)
-
M
2
(
τ
)
(
w
+
1
)
-
G
(
τ
)
r
}
. Then
γ
(
s
)
leads to the following equation:
(23)
γ
(
s
)
=
1
(
2
π
i
)
∫
c
-
i
∞
c
+
i
∞
exp
{
(
1
2
σ
2
τ
+
M
1
(
τ
)
+
η
(
τ
-
G
(
τ
)
)
)
w
2
hhhhhhhhhhhhhhhhh
+
(
1
2
σ
2
τ
+
2
M
1
(
τ
)
-
M
2
(
τ
)
hhhhhhhhhhhhhhhhhhhhh
+
η
(
τ
+
G
(
τ
)
)
-
G
(
τ
)
r
1
2
)
w
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
+
M
1
(
τ
)
-
M
2
(
τ
)
-
G
(
τ
)
r
1
2
}
s
-
w
d
w
=
exp
{
-
1
4
(
(
1
/
2
)
σ
2
τ
+
η
(
τ
-
G
(
τ
)
)
+
M
2
(
τ
)
+
G
(
τ
)
r
)
2
(
1
/
2
)
σ
2
τ
+
M
1
(
τ
)
+
η
(
τ
-
G
(
τ
)
)
}
·
1
(
2
π
i
)
×
∫
c
-
i
∞
c
+
i
∞
exp
{
1
2
2
1
(
1
2
σ
2
τ
+
M
1
(
τ
)
+
η
(
τ
-
G
(
τ
)
)
)
s
s
s
s
s
g
g
h
h
g
g
i
i
g
g
s
s
s
h
s
×
{
w
+
1
2
(
1
2
1
1
+
(
M
1
(
τ
)
-
M
2
(
τ
)
-
G
(
τ
)
r
)
g
g
g
g
g
g
g
g
g
g
g
g
g
g
g
g
h
i
g
h
i
g
d
i
v
g
×
(
1
2
σ
2
τ
+
M
1
(
τ
)
f
f
f
f
f
f
f
f
f
f
f
f
h
h
h
h
f
f
i
h
h
i
h
h
h
h
h
i
f
f
f
+
η
(
τ
-
G
(
τ
)
)
1
2
)
-
1
)
}
2
}
h
h
h
h
h
h
h
h
h
h
h
h
×
s
-
w
d
w
.
To compute the integral of (23), we use the following lemma.
Lemma 1.
Let
α
and
β
be complex numbers satisfying
Re
(
α
)
≥
0
. Then,
(24)
1
2
π
i
∫
c
-
i
∞
c
+
i
∞
e
α
(
w
+
β
)
2
x
-
w
d
w
=
1
2
(
π
α
)
-
1
/
2
x
β
e
-
(
1
/
4
α
)
(
ln
x
)
2
holds.
Proof.
Let
f
(
x
)
=
(
1
/
2
π
i
)
∫
c
-
i
∞
c
+
i
∞
e
α
(
w
+
β
)
2
x
-
w
d
w
and
s
=
w
+
β
. Then
f
(
x
)
=
(
x
β
/
2
π
i
)
∫
c
*
-
i
∞
c
*
+
i
∞
e
α
s
2
x
-
s
d
s
, where
c
*
=
c
+
β
. If
s
=
c
*
+
i
z
, then
f
(
x
)
becomes
(25)
f
(
x
)
=
x
β
2
π
∫
-
∞
∞
e
α
(
c
*
+
i
z
)
2
x
-
(
c
*
+
i
z
)
d
z
=
x
β
2
π
∫
-
∞
∞
e
-
α
(
z
-
i
(
c
*
-
ln
x
/
2
α
)
)
2
e
-
(
1
/
4
α
)
(
ln
x
)
2
d
z
=
1
2
(
π
α
)
-
1
/
2
x
β
e
-
(
1
/
4
α
)
(
ln
x
)
2
(
since
∫
-
∞
∞
e
-
z
2
d
z
=
π
)
.
However, to apply Lemma 1 to (23), the following lemma is also required.
Lemma 2.
For
M
1
(
τ
)
and
G
(
τ
)
given above,
(
1
/
2
)
σ
2
τ
+
M
1
(
τ
)
+
η
(
τ
-
G
(
τ
)
)
≥
0
holds.
Proof.
From the definition of
M
1
(
τ
)
and
G
(
τ
)
,
(26)
1
2
σ
2
τ
+
M
1
(
τ
)
+
η
(
τ
-
G
(
τ
)
)
=
1
2
∫
0
τ
(
σ
2
+
2
ρ
s
r
σ
σ
ˇ
G
(
t
*
)
+
σ
ˇ
2
G
2
(
t
*
)
)
d
t
*
is satisfied. However, since
(27)
0
≤
(
σ
-
σ
ˇ
G
(
t
*
)
)
2
≤
σ
2
+
2
ρ
s
r
σ
σ
ˇ
G
(
t
*
)
+
σ
ˇ
2
G
2
(
t
*
)
≤
(
σ
+
σ
ˇ
G
(
t
*
)
)
2
,
we can prove
(
1
/
2
)
σ
2
τ
+
M
1
(
τ
)
+
η
(
τ
-
G
(
τ
)
)
≥
0
, where
-
1
≤
ρ
s
r
≤
1
.
In (23), if we set
(28)
α
=
1
2
σ
2
τ
+
M
1
(
τ
)
+
η
(
τ
-
G
(
τ
)
)
,
β
=
1
2
(
1
+
M
1
(
τ
)
-
M
2
(
τ
)
-
G
(
τ
)
r
(
1
/
2
)
σ
2
τ
+
M
1
(
τ
)
+
η
(
τ
-
G
(
τ
)
)
)
,
then, from Lemmas 1 and 2,
γ
(
s
)
yields the following equation:
(29)
γ
(
s
)
=
exp
{
-
(
(
1
/
2
)
σ
2
τ
+
η
(
τ
-
G
(
τ
)
)
+
M
2
(
τ
)
+
G
(
τ
)
r
)
2
4
α
}
×
1
2
α
π
s
β
e
-
(
1
/
4
α
)
(
ln
s
)
2
.
Now, we are trying to use relation to multiplicative convolution of Mellin transform and find
P
n
(
t
,
s
,
r
)
. The Mellin convolution of
f
and
g
is given by the inverse Mellin transform of
f
^
(
w
)
g
^
(
w
)
as follows:
(30)
f
(
x
)
*
g
(
x
)
=
M
w
-
1
[
f
^
(
w
)
g
^
(
w
)
;
x
]
=
∫
0
∞
u
-
1
f
(
x
u
)
g
(
u
)
d
u
,
where
f
(
x
)
*
g
(
x
)
is the symbol of the Mellin convolution of
f
and
g
and
M
w
-
1
is the symbol of the inverse Mellin transform. It is referred to in [11] with more details.
In (21) and (22), since
e
φ
(
τ
,
w
,
r
)
is the Mellin transform of
γ
(
s
)
and
E
n
(
w
)
is the Mellin transform of the payoff function
h
n
(
s
)
, we have the following formula by using the Mellin convolution property mentioned above:
(31)
P
n
(
τ
,
s
,
r
)
=
∫
0
∞
h
n
(
u
)
γ
(
s
u
)
1
u
d
u
=
1
2
α
π
[
∫
K
n
(
u
-
K
)
exp
{
-
(
1
2
σ
2
τ
+
η
(
τ
-
G
(
τ
)
)
hhhhhhhhhhhhhhhhhhhhhhhhhhih
+
M
2
(
τ
)
+
G
(
τ
)
r
1
2
)
2
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
×
(
4
α
)
-
1
1
2
}
hhhhhhhhhhhihh
×
(
s
u
)
β
e
-
(
1
/
4
α
)
(
ln
(
s
/
u
)
)
2
1
u
d
u
]
.
Therefore, to find the European option price
P
(
t
,
s
,
r
)
, if we take
n
→
∞
in both sides of (31), then
(32)
P
(
t
,
s
,
r
)
=
lim
n
→
∞
P
n
(
τ
,
s
,
r
)
=
1
2
α
π
×
[
∫
K
∞
(
u
-
K
)
exp
{
(
1
2
2
-
(
1
2
σ
2
τ
+
η
(
τ
-
G
(
τ
)
)
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
g
h
h
h
h
h
h
h
+
M
2
(
τ
)
+
G
(
τ
)
r
1
2
)
2
(
4
α
)
-
1
}
gggggggggg
×
(
s
u
)
β
e
-
(
1
/
4
α
)
(
ln
(
s
/
u
)
)
2
1
u
d
u
]
=
1
2
α
π
×
[
∫
K
∞
u
exp
{
(
1
2
2
-
(
1
2
σ
2
τ
+
η
(
τ
-
G
(
τ
)
)
+
M
2
(
τ
)
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
i
h
h
+
G
(
τ
)
r
1
2
)
2
(
4
α
)
-
1
}
h
h
h
h
h
h
h
h
h
h
h
×
(
s
u
)
β
e
-
(
1
/
4
α
)
(
ln
(
s
/
u
)
)
2
1
u
d
u
]
-
1
2
α
π
[
∫
K
∞
K
exp
{
(
1
2
2
-
(
1
2
σ
2
τ
+
η
(
τ
-
G
(
τ
)
)
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
+
M
2
(
τ
)
+
G
(
τ
)
r
1
2
)
2
(
4
α
)
-
1
}
hhhhhhhhhhhhhh
×
(
s
u
)
β
e
-
(
1
/
4
α
)
(
ln
(
s
/
u
)
)
2
1
u
d
u
]
=
P
1
(
t
,
s
,
r
)
-
P
2
(
t
,
s
,
r
)
.
Theorem 3.
Under the payoff function
P
(
T
,
S
,
r
)
=
(
S
T
-
K
)
+
, the formula of European call option with Hull-White interest rate is given by
(33)
P
(
t
,
s
,
r
)
=
P
(
T
-
τ
,
s
,
r
)
=
s
Φ
(
d
1
)
-
K
B
(
T
-
τ
,
r
;
T
)
Φ
(
d
2
)
,
d
1
=
ln
(
s
/
K
)
-
ln
B
(
t
,
r
;
T
)
+
(
1
/
2
)
∫
0
τ
σ
^
2
(
t
*
)
d
t
*
∫
0
τ
σ
^
2
(
t
*
)
d
t
*
,
d
2
=
d
1
-
∫
0
τ
σ
^
2
(
t
*
)
d
t
*
,
where
B
(
T
-
τ
,
r
;
T
)
is the price of zero-coupon bond mentioned in Section 2.1,
σ
^
(
t
)
=
σ
2
+
2
ρ
s
r
σ
σ
ˇ
X
(
t
)
+
σ
ˇ
2
X
2
(
t
)
with
X
(
t
)
=
-
(
1
/
B
)
(
∂
B
/
∂
r
)
=
G
(
t
)
, and
Φ
is the normal cumulative distribution function defined by
Φ
(
z
)
=
(
1
/
2
π
)
∫
-
∞
z
e
-
z
2
/
2
d
z
.
Proof.
Most of all, let
y
=
ln
(
s
/
u
)
/
2
α
. By applying change of variable
(
u
→
y
)
,
P
1
(
t
,
s
,
r
)
of (32) leads to
(34)
P
1
(
t
,
s
,
r
)
=
1
2
α
π
[
∫
K
∞
u
exp
{
(
1
2
2
-
(
1
2
σ
2
τ
+
η
(
τ
-
G
(
τ
)
)
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
+
M
2
(
τ
)
+
G
(
τ
)
r
1
2
)
2
(
4
α
)
-
1
}
h
h
h
h
h
h
h
h
h
h
f
h
h
×
(
s
u
)
β
e
-
(
1
/
4
α
)
(
ln
(
s
/
u
)
)
2
1
u
d
u
]
=
exp
{
-
(
(
1
/
2
)
σ
2
τ
+
η
(
τ
-
G
(
τ
)
)
+
M
2
(
τ
)
+
G
(
τ
)
r
)
2
4
α
}
j
j
j
j
j
j
j
×
1
2
π
s
∫
-
∞
ln
(
s
/
K
)
/
2
α
e
2
α
(
β
-
1
)
y
-
(
1
/
2
)
y
2
d
y
=
exp
{
-
(
(
1
/
2
)
σ
2
τ
+
η
(
τ
-
G
(
τ
)
)
+
M
2
(
τ
)
+
G
(
τ
)
r
)
2
4
α
h
h
h
h
h
h
h
h
h
+
α
(
β
-
1
)
2
(
(
s
N
(
τ
)
)
)
2
d
}
1
2
π
s
h
h
h
h
h
×
∫
-
∞
ln
(
s
/
K
)
/
2
α
e
-
(
1
/
2
)
(
y
-
2
α
(
β
-
1
)
)
2
d
y
=
exp
{
-
(
(
1
/
2
)
σ
2
τ
+
η
(
τ
-
G
(
τ
)
)
+
M
2
(
τ
)
+
G
(
τ
)
r
)
2
4
α
h
h
h
h
h
h
h
h
h
+
α
(
β
-
1
)
2
(
(
s
N
(
τ
)
)
)
2
d
}
1
2
π
s
∫
-
∞
d
1
e
-
(
1
/
2
)
z
2
d
z
,
where
d
1
=
(
ln
(
s
/
K
)
-
2
α
(
β
-
1
)
)
/
2
α
. From the definition of
α
,
β
, and
B
(
T
-
τ
,
r
;
T
)
mentioned before, we have
(35)
-
1
4
(
(
1
/
2
)
σ
2
τ
+
η
(
τ
-
G
(
τ
)
)
+
M
2
(
τ
)
+
G
(
τ
)
r
)
2
α
+
α
(
β
-
1
)
2
=
0
,
2
α
=
∫
0
τ
(
σ
2
+
2
ρ
s
r
σ
σ
ˇ
G
(
t
*
)
+
σ
ˇ
2
G
2
(
t
*
)
)
d
t
*
,
2
α
(
β
-
1
)
=
(
M
1
(
τ
)
-
M
2
(
τ
)
-
G
(
τ
)
r
)
-
(
1
2
σ
2
τ
+
M
1
(
τ
)
+
η
(
τ
-
G
(
τ
)
)
)
.
Therefore, the solution of
P
1
(
t
,
s
,
r
)
is
(36)
P
1
(
t
,
s
,
r
)
=
s
1
2
π
∫
-
∞
d
1
e
-
(
1
/
2
)
z
2
d
z
,
d
1
=
ln
(
s
/
K
)
-
ln
B
(
t
,
r
;
T
)
+
(
1
/
2
)
∫
0
τ
σ
^
2
(
t
*
)
d
t
*
∫
0
τ
σ
^
2
(
t
*
)
d
t
*
,
where
σ
^
2
(
t
)
=
σ
2
+
2
ρ
s
r
σ
σ
ˇ
G
(
t
)
+
σ
ˇ
2
G
2
(
t
)
.
Also, by setting
y
=
ln
(
s
/
u
)
/
2
α
and using the similar way stated above,
P
2
(
t
,
s
,
r
)
of (32) is transformed into
(37)
P
2
(
t
,
s
,
r
)
=
1
2
α
π
[
∫
K
∞
K
exp
{
1
2
2
-
(
1
2
σ
2
τ
+
η
(
τ
-
G
(
τ
)
)
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
i
h
h
+
M
2
(
τ
)
+
G
(
τ
)
r
1
2
)
2
(
4
α
)
-
1
}
k
k
k
k
k
k
k
k
k
k
k
i
h
v
k
k
×
(
s
u
)
β
e
-
(
1
/
4
α
)
(
ln
(
s
/
u
)
)
2
1
u
d
u
]
=
exp
{
-
(
(
1
/
2
)
σ
2
τ
+
η
(
τ
-
G
(
τ
)
)
+
M
2
(
τ
)
+
G
(
τ
)
r
)
2
4
α
g
g
g
g
g
g
i
g
g
+
α
β
2
(
(
s
N
(
τ
)
)
)
2
d
}
1
2
π
K
∫
-
∞
ln
(
s
/
K
)
/
2
α
e
-
(
1
/
2
)
(
y
-
2
α
β
)
2
d
y
=
exp
{
-
(
(
1
/
2
)
σ
2
τ
+
η
(
τ
-
G
(
τ
)
)
+
M
2
(
τ
)
+
G
(
τ
)
r
)
2
4
α
h
h
h
h
h
i
h
g
i
h
+
α
β
2
(
(
s
N
(
τ
)
)
)
2
d
}
1
2
π
K
∫
-
∞
ln
(
s
/
K
)
/
2
α
-
2
α
β
e
-
(
1
/
2
)
z
2
d
z
.
Similarly,
(38)
exp
{
-
(
(
1
/
2
)
σ
2
τ
+
η
(
τ
-
G
(
τ
)
)
+
M
2
(
τ
)
+
G
(
τ
)
r
)
2
4
α
h
h
h
i
h
+
α
β
2
(
(
s
N
(
τ
)
)
)
2
d
}
=
exp
{
M
1
(
τ
)
-
M
2
(
τ
)
-
G
(
τ
)
r
}
=
B
(
T
-
τ
,
r
;
T
)
,
ln
(
s
/
K
)
2
α
-
2
α
β
=
ln
(
s
/
K
)
-
ln
B
(
t
,
r
;
T
)
-
(
1
/
2
)
∫
0
τ
σ
^
2
(
t
*
)
d
t
*
∫
0
τ
σ
^
2
(
t
*
)
d
t
*
.
Therefore, the solution of
P
2
(
t
,
s
,
r
)
is
(39)
P
2
(
t
,
s
,
r
)
=
K
B
(
T
-
τ
,
r
;
T
)
1
2
π
∫
-
∞
d
2
e
-
(
1
/
2
)
z
2
d
z
,
d
2
=
ln
(
s
/
K
)
-
ln
B
(
t
,
r
;
T
)
-
(
1
/
2
)
∫
0
τ
σ
^
2
(
t
*
)
d
t
*
∫
0
τ
σ
^
2
(
t
*
)
d
t
*
.
Finally, by combining (32), (36), and (39), we obtain the following result:
(40)
P
(
t
,
s
,
r
)
=
P
(
T
-
τ
,
s
,
r
)
=
s
∫
-
∞
d
1
e
-
(
1
/
2
)
z
2
d
z
-
K
B
(
T
-
τ
,
r
;
T
)
×
∫
-
∞
d
2
e
-
(
1
/
2
)
z
2
d
z
,
d
1
=
ln
(
s
/
K
)
-
ln
B
(
t
,
r
;
T
)
+
(
1
/
2
)
∫
0
τ
σ
^
2
(
t
*
)
d
t
*
∫
0
τ
σ
^
2
(
t
*
)
d
t
*
,
d
2
=
d
1
-
∫
0
τ
σ
^
2
(
t
*
)
d
t
*
.
The proof is completed.