e*******a
发帖数: 514 | 1 Suppose b(t) is a standard wiener process on [0,1], then
w(t) := b(t) + \int_0^1 {log(1 - min(t, s))}db(s) is also a standard wiener
process.
不知哪位朋友能帮忙指点一下，或给个参考文献。
万分感谢！ | n******t
发帖数: 4406 | 2 Just use the fact that w(t) is a continuous martingale and check its
quadratic variation is t.
wiener
【在 e*******a 的大作中提到】
: Suppose b(t) is a standard wiener process on [0,1], then
: w(t) := b(t) + \int_0^1 {log(1 - min(t, s))}db(s) is also a standard wiener
: process.
: 不知哪位朋友能帮忙指点一下，或给个参考文献。
: 万分感谢！
| Q***5
发帖数: 994 | 3 But you need to change the filtration first. w(t) is not even adaptive to
the filtration of b(t), let along a martingale.
It seems that this problem be proved directly by definition of B.M.
【在 n******t 的大作中提到】
: Just use the fact that w(t) is a continuous martingale and check its
: quadratic variation is t.
:
: wiener
| n******t
发帖数: 4406 | 4 I do not see it is a big problem.
What I actually mean is just splitting the integral into two part,
from 0 up to time t, and from t to 1.
【在 Q***5 的大作中提到】
: But you need to change the filtration first. w(t) is not even adaptive to
: the filtration of b(t), let along a martingale.
: It seems that this problem be proved directly by definition of B.M.
| Q***5
发帖数: 994 | 5 But the t to 1 part involves `information' that is not available at t in the
original filtration. -- for example, there should be a b(1) term from the t
to 1 integration.
If I understand the question correctly, we need to prove that w(t) is B.M.
with respect to filtration \sigma (w(s), s<=t), not \sigma(b(s), s<=t).
Since the process is a Gaussian process, we only need to check the
independent incremental property and the std of w(t_2)-w(t_1), using the 2
part integration you mentioned.
【在 n******t 的大作中提到】
: I do not see it is a big problem.
: What I actually mean is just splitting the integral into two part,
: from 0 up to time t, and from t to 1.
| n******t
发帖数: 4406 | 6
the
t
~~~~~~~~~~~~~~~~~~~~
Yes, we can not use the filtration generated by b(t), but I think
in this case there is no problem to enlarge the filtration.
Basically, if Let the first part be X(t) and the second part be Y(t),
then from the definition, we could see that we can actually add
their QVs together which gives out t.
【在 Q***5 的大作中提到】
: But the t to 1 part involves `information' that is not available at t in the
: original filtration. -- for example, there should be a b(1) term from the t
: to 1 integration.
: If I understand the question correctly, we need to prove that w(t) is B.M.
: with respect to filtration \sigma (w(s), s<=t), not \sigma(b(s), s<=t).
: Since the process is a Gaussian process, we only need to check the
: independent incremental property and the std of w(t_2)-w(t_1), using the 2
: part integration you mentioned.
| e*******a
发帖数: 514 | 7 谢谢两位的回复，我最后用了以下方法证出：
Let a(t) be a function in L^2[0,1].
A(t) := \int_t^1 a(t) dt, and Z := \int_0^1 a(t)w(t) dt.
Then we could show
E(Z) = 0, and Var(Z) = \int_0^1 [A(t)]^2 dt.
This implies w(t) is a standard wiener process.
q.e.d. | Q***5
发帖数: 994 | 8 Why "E(Z) = 0, and Var(Z) = \int_0^1 [A(t)]^2 dt implies w(t) is a standard
wiener process."? Is that from some theroem?
Thanks
【在 e*******a 的大作中提到】
: 谢谢两位的回复，我最后用了以下方法证出：
: Let a(t) be a function in L^2[0,1].
: A(t) := \int_t^1 a(t) dt, and Z := \int_0^1 a(t)w(t) dt.
: Then we could show
: E(Z) = 0, and Var(Z) = \int_0^1 [A(t)]^2 dt.
: This implies w(t) is a standard wiener process.
: q.e.d.
| e*******a
发帖数: 514 | 9 E(Z^2) = E{[\int_0^1 a(t) w(t)dt]^2}
=E{[\int_0^1 a(t) (\int_0^t dw(s)) dt]^2}
=E{[\int_0^1 (\int_s^1 a(t)dt)dw(s) ]^2}
=\int_0^1 [A(s)]^2 d_s
standard
【在 Q***5 的大作中提到】
: Why "E(Z) = 0, and Var(Z) = \int_0^1 [A(t)]^2 dt implies w(t) is a standard
: wiener process."? Is that from some theroem?
: Thanks
|
|
