d*e
发帖数: 843 | 1 Consider a stochastic integral I(t) = int_0^t c(s)a(t,s)dW(s),
where
1) c(s) (>=0) is a function such that int_0^\infty c(s)ds = \infty and
int_0^\infty c^2(s)ds < \infty
2) a(t,s) is a deterministic function satisfying
|a(t,s)| <= e^{-\int_s^t c(r)dr}
Q: show that I(t)->0 almost surely as t -> \infty
Note: if a(t,s) = e^{-\int_s^t c(r)dr}, the problem can be solved by the
method given by "MathFi" in this post
http://www.mitbbs.com/article/Quant/31266465_0.html
But it seems that the same approach cannot be applied to the case
|a(t,s)| <= e^{-\int_s^t c(r)dr}
Any comments will be greatly appreciated!
* MathFi, if you see this, I would especially appreciate your comments. | d*e
发帖数: 843 | 2 re
【在 d*e 的大作中提到】
: Consider a stochastic integral I(t) = int_0^t c(s)a(t,s)dW(s),
: where
: 1) c(s) (>=0) is a function such that int_0^\infty c(s)ds = \infty and
: int_0^\infty c^2(s)ds < \infty
: 2) a(t,s) is a deterministic function satisfying
: |a(t,s)| <= e^{-\int_s^t c(r)dr}
: Q: show that I(t)->0 almost surely as t -> \infty
: Note: if a(t,s) = e^{-\int_s^t c(r)dr}, the problem can be solved by the
: method given by "MathFi" in this post
: http://www.mitbbs.com/article/Quant/31266465_0.html
| E*****T
发帖数: 1193 | | d*e
发帖数: 843 | 4 随机还夹逼？你给个证明吧
【在 E*****T 的大作中提到】
: 难道不是一夹逼就出来了？
| z****g
发帖数: 1978 | 5 换个模不就有了？
【在 d*e 的大作中提到】
: 随机还夹逼？你给个证明吧
| d*e
发帖数: 843 | 6 这么简单？给我写个三行的证明吧，多谢了
【在 z****g 的大作中提到】
: 换个模不就有了？
| d*e
发帖数: 843 | 7 换哪个模？
【在 z****g 的大作中提到】
: 换个模不就有了？
| z****g
发帖数: 1978 | 8 Normally, for this kind of proof, you try to prove
|I(t)|-> 0 in some sense.
For non-stochastic integral, you just use try to prove |I(t)|^2 -> 0, then
loose the left side to L^2.
For stochastic integral, I(t) is essentially a random variable if given t,
so I guess E(I(t)^2) would work, and luckily you can loose E(I(t)^2) using
The It^o isometry | d*e
发帖数: 843 | 9 yes, no problem to prove E(I(t)^2)->0, using ito isometry
I had my problem in proving almost sure convergence though
the original post is here
http://www.mitbbs.com/article_t/Quant/31266211.html
【在 z****g 的大作中提到】
: Normally, for this kind of proof, you try to prove
: |I(t)|-> 0 in some sense.
: For non-stochastic integral, you just use try to prove |I(t)|^2 -> 0, then
: loose the left side to L^2.
: For stochastic integral, I(t) is essentially a random variable if given t,
: so I guess E(I(t)^2) would work, and luckily you can loose E(I(t)^2) using
: The It^o isometry
| z****g
发帖数: 1978 | 10 Then you just try to see if E(A^2) = 0 means a.s. zero. It is a linear space
with good property, should not be difficult.
【在 d*e 的大作中提到】
: yes, no problem to prove E(I(t)^2)->0, using ito isometry
: I had my problem in proving almost sure convergence though
: the original post is here
: http://www.mitbbs.com/article_t/Quant/31266211.html
| | | d*e
发帖数: 843 | 11 does it have anything to do with "E(A^2) = 0 means a.s. zero."?
f with L^2 norm =0 on [0,1] implies f=0 a.e.
yet f_n->0 a.e. and f_n->0 in L^2 are two different concepts..
space
【在 z****g 的大作中提到】
: Then you just try to see if E(A^2) = 0 means a.s. zero. It is a linear space
: with good property, should not be difficult.
| z****g
发帖数: 1978 | 12 probably you don't see the big picture.
you have |I(t)| -> 0 with some given norm, so the last step you should do
is to see if this norm you are using implies the convergence you are
looking for. Ito isometry is merely a tool to prove |I(t)| -> 0 in your
case.
Since it is a linear space, the only thing you need to verify is:
for a given sequence of random variables {A_n} if E(A_n^2) -> 0 means
{A_n}
-> 0 a.s.(or a.e)
【在 d*e 的大作中提到】
: does it have anything to do with "E(A^2) = 0 means a.s. zero."?
: f with L^2 norm =0 on [0,1] implies f=0 a.e.
: yet f_n->0 a.e. and f_n->0 in L^2 are two different concepts..
:
: space
| d*e
发帖数: 843 | 13 thanks. true, i didn't see your picture at this level.
but I am still not clear about what you meant
1) why is linearity relevant?
2) what space you refer to?
3) how to verify "a given sequence of random variables
{A_n} if E(A_n^2) -> 0 means {A_n} -> 0 a.s.(or a.e)"
could you please elaborate? it seems to me the last statement 3) is just a
re-statement of my problem and does not simplify it in any sense?
thanks again!
【在 z****g 的大作中提到】
: probably you don't see the big picture.
: you have |I(t)| -> 0 with some given norm, so the last step you should do
: is to see if this norm you are using implies the convergence you are
: looking for. Ito isometry is merely a tool to prove |I(t)| -> 0 in your
: case.
: Since it is a linear space, the only thing you need to verify is:
: for a given sequence of random variables {A_n} if E(A_n^2) -> 0 means
: {A_n}
: -> 0 a.s.(or a.e)
| z****g
发帖数: 1978 | 14 Ok, forget 1) and 2), you don't need them here.
For 3), clearly you see it converges by measure. So if you wanna
almost everywhere convergence you should add some conditions.
I can't remember all the results, but there are definitely some
conditions you should meet to make the result a.e converge. One that I
can remember
seems to be: the random variable is defined on bounded interval on real
axis, which does not apply in your case. You probably need to check real
analysis or measure theory. I check the wiki, seems these may work:
1. If μ is σ-finite and (fn) converges (locally or globally) to f in
measure, there is a subsequence converging to f almost everywhere.
2. If μ is σ-finite, (fn) converges to f locally in measure if and only
if every subsequence has in turn a subsequence that converges to f
almost
everywhere.
Try to verify the above because for Ito process is adapted. Or prove by
contradiction using c(s)'s property because c(s)'s property looks
special.
just a
【在 d*e 的大作中提到】
: thanks. true, i didn't see your picture at this level.
: but I am still not clear about what you meant
: 1) why is linearity relevant?
: 2) what space you refer to?
: 3) how to verify "a given sequence of random variables
: {A_n} if E(A_n^2) -> 0 means {A_n} -> 0 a.s.(or a.e)"
: could you please elaborate? it seems to me the last statement 3) is just a
: re-statement of my problem and does not simplify it in any sense?
: thanks again!
| d*e
发帖数: 843 | 15 thanks I was aware of this result but I'm not sure how a.s convergence of
a subsequence can be used here. I also tried to apply BC lemma somehow,
but no success..
【在 z****g 的大作中提到】
: Ok, forget 1) and 2), you don't need them here.
: For 3), clearly you see it converges by measure. So if you wanna
: almost everywhere convergence you should add some conditions.
: I can't remember all the results, but there are definitely some
: conditions you should meet to make the result a.e converge. One that I
: can remember
: seems to be: the random variable is defined on bounded interval on real
: axis, which does not apply in your case. You probably need to check real
: analysis or measure theory. I check the wiki, seems these may work:
: 1. If μ is σ-finite and (fn) converges (locally or globally) to f in
| z****g
发帖数: 1978 | 16 Try the second result, locally convergence by measure. It is a complete
and sufficient criteria. Or, the question itself is wrong since it is very
rare to prove a.s. convergence for stochastic process.
of
【在 d*e 的大作中提到】
: thanks I was aware of this result but I'm not sure how a.s convergence of
: a subsequence can be used here. I also tried to apply BC lemma somehow,
: but no success..
| z****g
发帖数: 1978 | 17 Also, you may need to check the basic analytically form of Ito integral. I
remember the definition is quite loose as it does not require the summation
converges on 'arbitrary' partition of t, in this sense you can choose a sub-
sequence of partition that will converge almost everywhere. Also, I remember
the integral itself converges only in terms of measure.
【在 d*e 的大作中提到】
: thanks I was aware of this result but I'm not sure how a.s convergence of
: a subsequence can be used here. I also tried to apply BC lemma somehow,
: but no success..
| d*e
发帖数: 843 | 18 多谢，我再试试，我相信是对的，但也有可能本来就不成立
有谁能想到简单的证明请告诉我啊
summation
sub-
remember
【在 z****g 的大作中提到】
: Also, you may need to check the basic analytically form of Ito integral. I
: remember the definition is quite loose as it does not require the summation
: converges on 'arbitrary' partition of t, in this sense you can choose a sub-
: sequence of partition that will converge almost everywhere. Also, I remember
: the integral itself converges only in terms of measure.
:
| M****i
发帖数: 58 | 19 You could try the following idea to see whether it helps.
Let b(t,s)=c(s)a(t,s), (d/ds)b(t,s)=f(t,s), h(t)=(2tlnlnt)^(1/2) and T>e.
Then integration by parts gives for t>T,
I(t)=b(t,t)W(t)-\int_0^t W(s)f(t,s) ds
=(b(t,t)h(t))*(W(t)/h(t))-\int_0^T W(s)f(t,s) ds-\int_T^t (W(s)/h(s))*(f(t,s)h(s)) ds.
Observe that for t>=T, W(t)/h(t) is bounded a.s. by iterated logarithm of BM, hence a
sufficient condition to ensure that I(t) tends to zero a.s. is that
b(t,t)h(t), \int_0^T |f(t,s)| ds and \int_T^t |f(t,s)h(s)| ds tend to zero. It should be aware that this method is essentially derministic not stochastic because we work path
by path. |
|
