3 Exotic Options
137
implies
S >
1 +
rT
1 +
D
0
αT
A
, i.e.,

(1 +
D
0
αT
)
S
+ (1 +
rT
)
A <
0. Thus
if
1 +
rT
1 +
D
0
αT
<
1
α
, then when
S >
A
α
, the PDE cannot be used. If
1 +
rT
1 +
D
0
αT
>
1
α
, then when
S >
1 +
rT
1 +
D
0
αT
A
, the PDE cannot be
used; while
A
α
< S <
1 +
rT
1 +
D
0
αT
A
, we have

(1 +
D
0
αT
)
S
+ (1 +
rT
)
A >
0 and the PDE can be used; Putting those cases together, we
have that when
S <
max
A
α
,
1 +
rT
1 +
D
0
αT
A
¶
, the PDE can be used,
and otherwise the PDE cannot be used. Consequently
S
f
(
A, T
) =
A
max
1
α
,
1 +
rT
1 +
D
0
αT
¶
.
b) When
A < E
, max(
A

E,
0) = 0, so the PDE always can be used.
When
A > E
, max(
A

E
) =
A

E
. Thus
∂
(
A

E
)
∂t
+
L
SAT
(
A

E
) =
S

A
T
·
1

r
(
A

E
)
=
1
T
[
S
+
rTE

(1 +
rT
)
A
]
.
The last expression is greater than zero if
S
+
rTE
1 +
rT
> A
, and less
than zero if
S
+
rTE
1 +
rT
< A
. If
S
+
rTE
1 +
rT
< E
, then
A > E
implies
A >
S
+
rTE
1 +
rT
, i.e.,
S
+
rTE

(1+
rT
)
A <
0. Thus if
S
+
rTE
1 +
rT
< E
, then
when
A > E
, the PDE cannot be used. If
S
+
rTE
1 +
rT
> E
, then when
A >
S
+
rTE
1 +
rT
, we have
S
+
rTE

(1+
rT
)
A <
0 and the PDE cannot
be used; while
E < A <
S
+
rTE
1 +
rT
, we have
S
+
rTE

(1 +
rT
)
A >
0
and the PDE can be used. Putting those cases together, we have that
when
A <
max
E,
S
+
rTE
1 +
rT
¶
, the PDE can be used, and otherwise
the PDE cannot be used. Consequently
A
f
(
S, T
) = max
E,
S
+
rTE
1 +
rT
¶
.
9. Suppose that sampling is done discretely at
t
=
t
1
, t
2
,
· · ·
, t
K
, where
0
≤
t
1
< t
2
<
· · ·
< t
K
≤
T
. Let
H
(
t
) = max
(
S
(
t
1
)
,
· · ·
, S
(
t
i
*
(
t
)
))
,
where
i
*
(
t
) is the number of samplings before time
t.
Assume
dS
=
μSdt
+
σSdX
and the dividends are paid continuously with dividend yield
D
0
.