d*e
发帖数: 843 | 1 什么样的函数c(t)能够使得下面这个积分概率1收敛到0，当t趋无穷大？
$I(t)=\int_0^t c(s)*exp(-\int_s^t c(r)dr) dB(s)$？
想了半天也没想出来 |
c****o
发帖数: 1280 | 2 you should first see that I(t) is normal distribution with mean 0 and
variance \int_0^t c(s)^2*exp(-2\int_s^t c(r)dr) ds, and in order to make it
goes to 0 with probability 1, I suggest variance goes to 0, and for the
variance, one such function I come up with is c(s) such that
(1) c(s)<=1
(2)c(\infty)=0
(3) \int_0^\infty c(s)=\infty
I might did the calculation wrong.......
【在 d*e 的大作中提到】
: 什么样的函数c(t)能够使得下面这个积分概率1收敛到0，当t趋无穷大？
: $I(t)=\int_0^t c(s)*exp(-\int_s^t c(r)dr) dB(s)$？
: 想了半天也没想出来
|
M****i
发帖数: 58 | 3 My idea:
If the function c(t) verifies:
i) c(t)>=0;
ii) \int_0^s c(t) dt<\infty, for every s>=0;
iii) \int_0^\infty c(t) dt=\infty;
iv) \int_0^\infty c(t)^2 dt<\infty.
Then the integral
I(t)=\int_0^t c(s)*exp(-\int_s^t c(r)dr) dB(s)
converges to 0 in probability.
Example: c(t) = 1/(t+1).
Proof: The idea is to show that I(t) is L^2 convergent to 0, which implies
convergence in probability. For this purpose,
by Ito isometry (since iv) holds) we get
E(I(t)^2)
=\int_0^t c(s)^2*\exp(-2\int_s^t c(r) dr) ds
=\int_0^\infty 1_[0,t](s)*c(s)^2*\exp(-2\int_s^t c(r) dr) ds
=\int_0^\infty 1_[0,t](s)*c(s)^2*\exp(-2\int_0^t c(r) dr)*\exp(2\int_0^s c(r
) dr) ds.
Then use Lebesgue dominated convergece theorem to the above integral to
conclude that E(I(t)^2) converges to 0 as t tends to infinity. (Observe that
i) and iv) give us the right to choose the dominating function to be c(s)^2
, ii) and iii) implies that the function in the integral tend to 0). CQFD
|
d*e
发帖数: 843 | 4 You got the same answer as me.
Convergence in L^2 is OK, which implies convergence in prob, OK too.
But how about convergence in probability 1 (almost sure)?
【在 M****i 的大作中提到】
: My idea:
: If the function c(t) verifies:
: i) c(t)>=0;
: ii) \int_0^s c(t) dt<\infty, for every s>=0;
: iii) \int_0^\infty c(t) dt=\infty;
: iv) \int_0^\infty c(t)^2 dt<\infty.
: Then the integral
: I(t)=\int_0^t c(s)*exp(-\int_s^t c(r)dr) dB(s)
: converges to 0 in probability.
: Example: c(t) = 1/(t+1).
|
d*e
发帖数: 843 | 5 (2) might not be enough, we might need \int_0^\infty c^2(s)ds<\infty?
Given variance goes to 0, and I(t) is normally distributed, how to show I(t)
->0 in almost surely (in prob. 1)?
I appeciated all the above answers... thanks
it
【在 c****o 的大作中提到】
: you should first see that I(t) is normal distribution with mean 0 and
: variance \int_0^t c(s)^2*exp(-2\int_s^t c(r)dr) ds, and in order to make it
: goes to 0 with probability 1, I suggest variance goes to 0, and for the
: variance, one such function I come up with is c(s) such that
: (1) c(s)<=1
: (2)c(\infty)=0
: (3) \int_0^\infty c(s)=\infty
: I might did the calculation wrong.......
|
M****i
发帖数: 58 | 6 For this question, the simplest method that I can find is to use a
representation theorem of continuous local martingale and the theorem of
iterated logarithm of Brownian motion. An advantage of this method is that
c(t) can be allowed to be stochastic. More precisely,
Let c(t) be a stochastic process and we write
f(t)=\int_0^t c(s)^2 exp(2*\int_0^s c(r)dr) ds,
g(t)=exp(-\int_0^t c(r)dr).
Assume that
i)f(t)<\infty, a.s. for every t>=0;
ii)f(t) tends to infinity when t tends to infinity.
iii)g(t)*(2*f(t)*lnlnf(t))^(1/2) tends to 0 a.s.
when t tends to infinity.
Then I(t) tends to 0 a.s.
Example: c(t)=1/(1+t). Then
g(t)*(2*f(t)*lnlnf(t))^(1/2)=(2*t*lnlnt)^(1/2)/(1+t), which tends to 0. So
that I(t) tends to 0 a.s.
Proof: By i), M(t)=I(t)/g(t) is a continuous local martingale with quadratic
variation f(t). Then there exists a Brownian motion W such that M(t)=W(f(t)
). Hence I(t)=g(t)*W(f(t))
=(g(t)*(2*f(t)*lnlnf(t))^(1/2))*(W(f(t))/(2*f(t)*lnlnf(t))^(1/2)).
Then ii) ensures that the theorem of iterated logarithm of Brownian motion
can be applied. CQFD |
d*e
发帖数: 843 | 7 Thanks.
What I thought was using Martingale Convergence Theorem.
Since I(t) is L^2-bounded, it should converge both in L^2 and a.s. sense.
Since the L^2 limit is 0 (under conditions given in previous posts),
the a.s. limit must be 0 as well.
Correct me if I'm wrong.
【在 M****i 的大作中提到】
: For this question, the simplest method that I can find is to use a
: representation theorem of continuous local martingale and the theorem of
: iterated logarithm of Brownian motion. An advantage of this method is that
: c(t) can be allowed to be stochastic. More precisely,
: Let c(t) be a stochastic process and we write
: f(t)=\int_0^t c(s)^2 exp(2*\int_0^s c(r)dr) ds,
: g(t)=exp(-\int_0^t c(r)dr).
: Assume that
: i)f(t)<\infty, a.s. for every t>=0;
: ii)f(t) tends to infinity when t tends to infinity.
|
M****i
发帖数: 58 | 8 You are welcome.
Generally speaking, the martingale convergence theorem can not be applied
here because I(t) is only a semimartingale, not a martingale:
I(t)=g(t)*M(t),
where
g(t)=exp(-\int_0^t c(r)dr)
M(t)=\int_0^t c(s)*exp(\int_0^s c(r)dr) dB(s).
Note that g(t) is of finite variation and only M(t) could be a martingale.
This can also be verified by my previous example
c(t)=1/(1+t). In this case I(t)=B(t)/(1+t), which is not a martingale, but
it converges to 0 a.s. by iterated logarithm.
For your idea, maybe a possible way is to begin with Girsanov theorem to find another probability measure Q under which I(t) is a martingale. Since dI(t)=c(t)*(dB(t)-I(t)dt), if we use Girsanov theorem directly, then dI(t)=c(t)dW(t) under Q, with W(t) a Q-BM. But its L^2 norm under Q is E_Q(int_0^t c(t)^2 dt) which can not tends to 0 unless c(t) is identically 0. Have you another idea on this way?
Another solution to your problem is to estimate the probability P{sup{|I(t)|
the Borel-Cantelli lemma can be used. If you do so, you will find that
everything goes well. But it seems that this method does not work if c(t) is stochastic. So that in my opinion, my posted solution seems a little better. |
d*e
发帖数: 843 | 9 Good point, it's not a martingale, and thanks a lot for all your
thoughts on this question. I'm not very familiar with Girsanov Theorem
though.
A second thought: $I(t)$ is still a supermartingale if we choose $c(s)$
non-negative and satisfyingall the conditions such that $I(t)$
converge in $L^2$ to 0.
Now that $I(t)$ is $L^2$-bounded implies that $I(t)$ is uniformly
integrable. By the MCT, a uniformly integrable supermartingale can be
closed and so $I(t)$ has an almost sure limit.
This limit must be 0, since we have shown $I(t)->0$ in $L^2$.
Again, corrent me if I'm wrong. Many thanks.
applied
martingale.
but
【在 M****i 的大作中提到】
: You are welcome.
: Generally speaking, the martingale convergence theorem can not be applied
: here because I(t) is only a semimartingale, not a martingale:
: I(t)=g(t)*M(t),
: where
: g(t)=exp(-\int_0^t c(r)dr)
: M(t)=\int_0^t c(s)*exp(\int_0^s c(r)dr) dB(s).
: Note that g(t) is of finite variation and only M(t) could be a martingale.
: This can also be verified by my previous example
: c(t)=1/(1+t). In this case I(t)=B(t)/(1+t), which is not a martingale, but
|
M****i
发帖数: 58 | 10 Without any other condition, I don't think that I(t) can be a
supermartingale even if c(t)>=0. Because M(t)=\int_0^t c(s)*exp(\int_0^s c(r
)dr) dB(s) is not necessarily positive. This will cause a problem when
checking the definition of supermartingale.
By the way, in order to ensure an a.s. limite exists for a supermartingale X
(t), it suffices to verify that its negative part max{-X(t),0} is L^1
bounded. Or, more practically and very frequently,
to check an in fact equivalent but seems a little stronger condition: X(t) is itself L^1 bounded. |
|
|
a********e
发帖数: 508 | 11 actually the neccessary and sufficient condition is
c(inf)=0, \lim \int c(s)ds=inf
by applying the rule \lim f/g=\lim f'/g'on var(I)
【在 d*e 的大作中提到】
: 什么样的函数c(t)能够使得下面这个积分概率1收敛到0，当t趋无穷大？
: $I(t)=\int_0^t c(s)*exp(-\int_s^t c(r)dr) dB(s)$？
: 想了半天也没想出来
|
d*e
发帖数: 843 | 12 var(I) goes to zero is easy. How about almost sure convergence?
【在 a********e 的大作中提到】
: actually the neccessary and sufficient condition is
: c(inf)=0, \lim \int c(s)ds=inf
: by applying the rule \lim f/g=\lim f'/g'on var(I)
|
a********e
发帖数: 508 | 13 概率收敛means convergence in probability.....
【在 d*e 的大作中提到】
: var(I) goes to zero is easy. How about almost sure convergence?
|
d*e
发帖数: 843 | 14 conv. in prob 1 is not conv. in prob
【在 a********e 的大作中提到】
: 概率收敛means convergence in probability.....
|
a********e
发帖数: 508 | 15 sorry.
I suspect the neccessary and sufficient condition should be the ones
in the case of convergence in probability plus some boundary conditions
on c(t)e^{\int c(r)dr}.
the proof should be similar to that of the strong law of large number
【在 d*e 的大作中提到】
: conv. in prob 1 is not conv. in prob
|
d*e
发帖数: 843 | 16 thanks.. hehe.
if you can write a proof down using strong LLN
that'll be even nicer..
【在 a********e 的大作中提到】
: sorry.
: I suspect the neccessary and sufficient condition should be the ones
: in the case of convergence in probability plus some boundary conditions
: on c(t)e^{\int c(r)dr}.
: the proof should be similar to that of the strong law of large number
|
d*e
发帖数: 843 | 17 原题应该可以用MathFi的方法解决了
同主题链接：http://mitbbs.com/article_t/Quant/31266211.html
但是我遇到的问题其实是这样的
考虑积分
I(t)=\int_0^t c(s)*a(t,s)dB(s)
其中a(t,s)是依赖于t和s的确定函数
满足以下估计：
|a(t,s)|<=exp(-\int_s^t c(r)dr)
问题还是：
什么样的函数c(t)能够I(t) a.s. 收敛到0，当t趋无穷大？
问题是随机积分好像没有对diffusion项的比较定理？
L^2收敛是比较好办的，不知道额外需要什么条件才可以保证a.s.收敛。。。
谢谢！
【在 d*e 的大作中提到】
: 什么样的函数c(t)能够使得下面这个积分概率1收敛到0，当t趋无穷大？
: $I(t)=\int_0^t c(s)*exp(-\int_s^t c(r)dr) dB(s)$？
: 想了半天也没想出来
|
