0.1
Exercises
1. A function,
f
, is said to be
Lipschitzcontinuous
on [0
, T
] if there exists a constant
C >
0 such that for any
t, t
0
∈
[0
, T
],

f
(
t
)

f
(
t
0
)

< C

t

t
0

.
Show that a Lipschitzcontinuous function has bounded variation and zero 2variation
over [0
, T
].
Proof.
Let the partition
π
n
: 0
≤
t
0
< t
1
< ... < t
n
=
T
. Then,
sup
π
n
n
X
i
=1

f
(
t
i
)

f
(
t
i

1
)
 ≤
sup
π
n
C
n
X
i
=1

t
i

t
i

1

=
cT <
∞
and
n
X
i
=1

f
(
t
i
)

f
(
t
i

1
)

2
≤
C
2
n
X
i
=1

t
i

t
i

1

2
≤
C
2
ˆ
sup
1
≤
i
≤
n

t
i

t
i

1

n
X
i
=1

t
i

t
i

1

!
≤
C
2
T
sup
1
≤
i
≤
n

t
i

t
i

1
 →
0
2.
(a) Show directly that
tB
t
is not a martingale.
(b) Using the Doob’s decomposition theorem, it can be show:
if
X
t
is a martingale
and has a bounded variation, then
X
t
must be a constant in
t
.
Use the above result to show that
R
t
0
B
s
ds
can not be a martingale.
(
Remark
:
By Doob’s decomposition theorem, any sub, supermartingale
X
t
has the unique decomposition
X
t
=
M
t
+
A
t
, where
M
t
is a martingale and
A
t
is
of bounded variation with
A
0
= 0. So if
X
t
is already a martingale, then
A
t
= 0,
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '11
 D
 Xt, TBT, Vasicek model, Bs ds

Click to edit the document details