a*****k
发帖数: 704 | 1 【 以下文字转载自 Quant 讨论区 】
发信人: artwork (嘿嘿), 信区: Quant
标 题: how to show this
发信站: BBS 未名空间站 (Sun Jul 1 16:00:46 2007), 转信
lim_{\beta -> \infty} \sup_{0 \leq t\leq T}|\exp(-\beta t) \int^t_0 \exp(\
beta s)dW_s| = 0, a.s.
anybody has any idea? |
l****e
发帖数: 23 | 2 BDG inequality?
【在 a*****k 的大作中提到】
: 【 以下文字转载自 Quant 讨论区 】
: 发信人: artwork (嘿嘿), 信区: Quant
: 标 题: how to show this
: 发信站: BBS 未名空间站 (Sun Jul 1 16:00:46 2007), 转信
: lim_{\beta -> \infty} \sup_{0 \leq t\leq T}|\exp(-\beta t) \int^t_0 \exp(\
: beta s)dW_s| = 0, a.s.
: anybody has any idea?
|
a*****k
发帖数: 704 | 3 maybe it's true.But the things inside of the sup is not a martingle,
so I don't how to apply it...
could you elaborate?
【在 l****e 的大作中提到】
: BDG inequality?
|
l****e
发帖数: 23 | 4 Maybe BDG doesn't work. But I made it in a way similar to the proof of BDG:
Let M_{t}:=\int_{0}^{t}\exp{\beta s-t}dW_{s}.
Apply Ito Formula to |M_{t}|^{2} and integrate from 0 to T, we get
|M_{T}|^{2}=T-2\beta\int_{0}^{T}|M_{t}|^{2}dt+\int_{0}^{T}2M_{t}dW_{t}.
Then taking expectation gives
E|M_{T}|^{2}=T-2\beta E\int_{0}^{T}|M_{t}|^{2}dt
\geq T-2\beta TE\sup_{0\leq t\leq T}|M_{t}|^{2}.
That is
E\sup_{0\leq t\leq T}|M_{t}|^{2}
\leq (T-E|M_{T}|^{2})/(2\beta T)
=1/(2\beta)-(1-\exp{-2\beta T})/(4\be
【在 a*****k 的大作中提到】
: maybe it's true.But the things inside of the sup is not a martingle,
: so I don't how to apply it...
: could you elaborate?
|
a*****k
发帖数: 704 | 5 all the first several lines are right.
after "that is", when you move E\sup to the other side,
the inequality should be reversed.
That's where I think wrong.
correct me if I am wrong
【在 l****e 的大作中提到】
: Maybe BDG doesn't work. But I made it in a way similar to the proof of BDG:
: Let M_{t}:=\int_{0}^{t}\exp{\beta s-t}dW_{s}.
: Apply Ito Formula to |M_{t}|^{2} and integrate from 0 to T, we get
: |M_{T}|^{2}=T-2\beta\int_{0}^{T}|M_{t}|^{2}dt+\int_{0}^{T}2M_{t}dW_{t}.
: Then taking expectation gives
: E|M_{T}|^{2}=T-2\beta E\int_{0}^{T}|M_{t}|^{2}dt
: \geq T-2\beta TE\sup_{0\leq t\leq T}|M_{t}|^{2}.
: That is
: E\sup_{0\leq t\leq T}|M_{t}|^{2}
: \leq (T-E|M_{T}|^{2})/(2\beta T)
|
a*****k
发帖数: 704 | 6 but this could be a good idea..
【在 a*****k 的大作中提到】
: all the first several lines are right.
: after "that is", when you move E\sup to the other side,
: the inequality should be reversed.
: That's where I think wrong.
: correct me if I am wrong
|
l****e
发帖数: 23 | 7 OMG, you are right. (Blush ...)
【在 a*****k 的大作中提到】
: all the first several lines are right.
: after "that is", when you move E\sup to the other side,
: the inequality should be reversed.
: That's where I think wrong.
: correct me if I am wrong
|
a*****k
发帖数: 704 | 8 that's ok, baby.
even famous prof.s published wrong papers...
【在 l****e 的大作中提到】
: OMG, you are right. (Blush ...)
|
l****e
发帖数: 23 | 9 这里也错了，呵呵。实际上是：范数（L^{2}或其他）为零蕴涵函数为零值函数（几乎
必然或处处）。不是极限。
to 0 a.s..
【在 l****e 的大作中提到】
: OMG, you are right. (Blush ...)
|
l****e
发帖数: 23 | 10 又想了一下，好像结论不对啊，但是没有把握，请批评指正。：）
Problem: \lim_{\beta\to\infty}\sup_{t\in[0,T]}|\int_{0}^{t}e^{\beta(s-t)}dW_
{s}|=0, a.s.?
As before we denote M_{t}:=\int_{0}^{t}e^{\beta(s-t)}dW_{s}.
Integration by parts gives
M_{t}=W_{t}-\int_{0}^{t}\beta e^{\beta(s-t)}W_{s}ds.
Clearly there exists a subset \Omega_{1} of \Omega with probability 1 such
that:
1, for all \omega\in\Omega_{1}, W is continuous on [0,T] and
2, for all \omega\in\Omega_{1}, there exists a t*\in[0,T] such that W_{t*}\
neq0, namely, |W_{t*
【在 a*****k 的大作中提到】
: that's ok, baby.
: even famous prof.s published wrong papers...
|
|
|
B********e
发帖数: 10014 | 11 弱问一下： W_s啥意思？一种特殊测度？
【在 a*****k 的大作中提到】
: 【 以下文字转载自 Quant 讨论区 】
: 发信人: artwork (嘿嘿), 信区: Quant
: 标 题: how to show this
: 发信站: BBS 未名空间站 (Sun Jul 1 16:00:46 2007), 转信
: lim_{\beta -> \infty} \sup_{0 \leq t\leq T}|\exp(-\beta t) \int^t_0 \exp(\
: beta s)dW_s| = 0, a.s.
: anybody has any idea?
|
H****h
发帖数: 1037 | 12 布朗运动吧。
【在 B********e 的大作中提到】
: 弱问一下： W_s啥意思？一种特殊测度？
|
B********e
发帖数: 10014 | 13 哦，我去google学习下布朗运动的话dW_s的性质
【在 H****h 的大作中提到】
: 布朗运动吧。
|
B********e
发帖数: 10014 | 14 呵呵，放弃了
health有时间给讲一下这种积分有什么特别的地方，；）
【在 B********e 的大作中提到】
: 哦，我去google学习下布朗运动的话dW_s的性质
|
H****h
发帖数: 1037 | 15 我不是这方面的专家。统计版有很多高手。
【在 B********e 的大作中提到】
: 呵呵，放弃了
: health有时间给讲一下这种积分有什么特别的地方，；）
|
a*****k
发帖数: 704 | 16 结论应该是对的。这是Karatzs得continuous bm得课后习题，
在bdg那一节后面。
我觉得至少不能直接用bdg。但是不知道应该怎么证。
还有一个可能使这个跟前面一题有关系，关于shiryaev-equation,
但是我还是不知道怎么证。
maybe we should just send an email to Karatzs(wrong spelling) :)
dW_
【在 l****e 的大作中提到】
: 又想了一下，好像结论不对啊，但是没有把握，请批评指正。：）
: Problem: \lim_{\beta\to\infty}\sup_{t\in[0,T]}|\int_{0}^{t}e^{\beta(s-t)}dW_
: {s}|=0, a.s.?
: As before we denote M_{t}:=\int_{0}^{t}e^{\beta(s-t)}dW_{s}.
: Integration by parts gives
: M_{t}=W_{t}-\int_{0}^{t}\beta e^{\beta(s-t)}W_{s}ds.
: Clearly there exists a subset \Omega_{1} of \Omega with probability 1 such
: that:
: 1, for all \omega\in\Omega_{1}, W is continuous on [0,T] and
: 2, for all \omega\in\Omega_{1}, there exists a t*\in[0,T] such that W_{t*}\
|
l****e
发帖数: 23 | 17 呵呵，我翻看原题了，还是没想明白。如果你弄清楚了恳请告诉我，谢谢！：）
另外，前面第一章“停时”一节开头有个小问题，我们当初读的时候有些迷惑，一并请
教：
Let X be a stochastic process and T a stopping time of (the filtration
generated by X) {F^{X}_{t}}.
Suppose that for any \omega, \omega' in \Omega, we have X_{t}(\omega)=X_{t}(
\omega') for all t in the
intersection of [0,T(\omega)] and [0,\infty). Show that T(\omega)=T(\omega').
【在 a*****k 的大作中提到】
: 结论应该是对的。这是Karatzs得continuous bm得课后习题，
: 在bdg那一节后面。
: 我觉得至少不能直接用bdg。但是不知道应该怎么证。
: 还有一个可能使这个跟前面一题有关系，关于shiryaev-equation,
: 但是我还是不知道怎么证。
: maybe we should just send an email to Karatzs(wrong spelling) :)
:
: dW_
|
g******a
发帖数: 69 | 18 Just use the def of the stopping time.
\omega and \omega' are not distinguishable
up to time T(\omega).
}(
').
【在 l****e 的大作中提到】
: 呵呵，我翻看原题了，还是没想明白。如果你弄清楚了恳请告诉我，谢谢！：）
: 另外，前面第一章“停时”一节开头有个小问题，我们当初读的时候有些迷惑，一并请
: 教：
: Let X be a stochastic process and T a stopping time of (the filtration
: generated by X) {F^{X}_{t}}.
: Suppose that for any \omega, \omega' in \Omega, we have X_{t}(\omega)=X_{t}(
: \omega') for all t in the
: intersection of [0,T(\omega)] and [0,\infty). Show that T(\omega)=T(\omega').
|
