Q***5
发帖数: 994 | 1 Give an example of a uniformly integrable martingale which is not in H^1.
( a martingle (X_t) is H^1 if E(X^*) is finite, where X^*(w) = sup_t |X_t(w)
|)
A hint can be found in ex 3.15 on page 74 of the following book. The result
seems unintuitive (therefore interesting). I still can not figure out how to
construct such an example.
http://books.google.com/books?id=1ml95FLM5koC&lpg=PP1&dq=Revuz%20Yor&pg=PA74#v=onepage&q=&f=false | c****o
发帖数: 1280 | 2 For martingale, I am not that sure, but for general functions, f in H^1 can
imply that the integral of f=0, h^1 is a proper subset of L^1.
Maybe you can generalize the result somehow.
w)
result
to
【在 Q***5 的大作中提到】
: Give an example of a uniformly integrable martingale which is not in H^1.
: ( a martingle (X_t) is H^1 if E(X^*) is finite, where X^*(w) = sup_t |X_t(w)
: |)
: A hint can be found in ex 3.15 on page 74 of the following book. The result
: seems unintuitive (therefore interesting). I still can not figure out how to
: construct such an example.
: http://books.google.com/books?id=1ml95FLM5koC&lpg=PP1&dq=Revuz%20Yor&pg=PA74#v=onepage&q=&f=false
| Q***5
发帖数: 994 | 3 Thanks.
From the hint, I guess we need construct a function f>0 on [0,1), such that,
\int_0^1 f < \infty, and \int_0^1 g = \infty, where
g(t) =(1/(1-t)) * \int_t^1 f(x)dx for any t in [0 1)
I was tempted to consider f = 1/(1-x)^a, where 0
, because the correspond g has finite integeral.
I guess we need something like: f = h(x)/(1-x), where h(x) goes to 0 as x
goes to 1, fast enough to make f finitely integerable, but slow enough so as
to make the corresponding g to
【在 c****o 的大作中提到】
: For martingale, I am not that sure, but for general functions, f in H^1 can
: imply that the integral of f=0, h^1 is a proper subset of L^1.
: Maybe you can generalize the result somehow.
:
: w)
: result
: to
| p*****k
发帖数: 318 | 4 QL365, i just noticed your question.
seems in the note at the end of chapter II, Reuvz and Yor cited
the following reference for this problem:
Sur certaines proprietes des espaces de Banach H1 et BMO
C. Dellacherie, P. A. Meyer and M. Yor (1978)
unfortunately it's in french and i could only recognize the relevant
pages which i have uploaded here:
http://ifile.it/51fur9c
i need to think a little more about this, but seems they were
saying the example is:
X_t(w) = MX(1-w) * I(t<1-w) + X(w) * I(t>= | Q***5
发帖数: 994 | 5 Many thanks, glad someone else is also interested in this kind of questions.
Thanks for your kindness in uploading the paper, unfortunately, I can not
read French either, it seems that they constructed a martingale in the same
way as suggested in Reuvz and Yor's exerciese, but they did not seem to
provide a example of X(t) (or f(t) in Reuvz and Yor's) that leads to the
desirable counter example.
In the passage right after equation 17, 18, they seem to say something about
the integerability of X^
【在 p*****k 的大作中提到】
: QL365, i just noticed your question.
: seems in the note at the end of chapter II, Reuvz and Yor cited
: the following reference for this problem:
: Sur certaines proprietes des espaces de Banach H1 et BMO
: C. Dellacherie, P. A. Meyer and M. Yor (1978)
: unfortunately it's in french and i could only recognize the relevant
: pages which i have uploaded here:
: http://ifile.it/51fur9c
: i need to think a little more about this, but seems they were
: saying the example is:
| Q***5
发帖数: 994 | 6 How about this:
Let f(x) (or X(x) in that French paper) to be
f(x) = 1/((1-x) * log(1-x)^2)
Let g(x) = (1/(1-x)) * \int_x^1 f(s)ds = - 1/((1-x)log(1-x))
Then for any 0<=t<1, X_t(w) is:
X_t(w) = f(w)*1_{w=t},
X_t seems to be uniformly integrable -- I still have some trouble with the
proof of this, my idea is that X_1 = f seems to close the martingale X_t ( E
(f|F_t) = X_t), and E|f| is finite -- only that f is infinity at 1, will
this be an issue?
Now,
X^*(w) = g(w), but E|X^*| = | Q***5
发帖数: 994 | 7 To make f(0), g(0) finite, we should replace the x in the right hand of
f(x) = 1/((1-x) * log(1-x)^2)
with (x+a)/(1+a),with a>0 a constant; Otherwise, E|f| is NOT finite.
E
【在 Q***5 的大作中提到】
: How about this:
: Let f(x) (or X(x) in that French paper) to be
: f(x) = 1/((1-x) * log(1-x)^2)
: Let g(x) = (1/(1-x)) * \int_x^1 f(s)ds = - 1/((1-x)log(1-x))
: Then for any 0<=t<1, X_t(w) is:
: X_t(w) = f(w)*1_{w=t},
: X_t seems to be uniformly integrable -- I still have some trouble with the
: proof of this, my idea is that X_1 = f seems to close the martingale X_t ( E
: (f|F_t) = X_t), and E|f| is finite -- only that f is infinity at 1, will
: this be an issue?
| p*****k
发帖数: 318 | 8 no progress on my side (analysis is not my bag), but i tried
google translator which seems pretty good :P
one thing is that they mentioned X(w)=log(w) in the last paragraph,
but this alone does not seem to be the counterexample. easy to verify
X-MX is indeed 1, then they mentioned some function Y to approach
X with Y-MY in the neighborhood of 1, then Z=MY gives contradiction...
if you understand what they say, plz let me know.
i also asked some math major and was pointed to some counterexamples | Q***5
发帖数: 994 | 9 Many thanks,
"i also asked some math major " -- I had automatically assume that you are
math major yourself, giving the depth of math knowledge you showed in
previous posts. You are not? Pei(4) Fu(2)!
【在 p*****k 的大作中提到】
: no progress on my side (analysis is not my bag), but i tried
: google translator which seems pretty good :P
: one thing is that they mentioned X(w)=log(w) in the last paragraph,
: but this alone does not seem to be the counterexample. easy to verify
: X-MX is indeed 1, then they mentioned some function Y to approach
: X with Y-MY in the neighborhood of 1, then Z=MY gives contradiction...
: if you understand what they say, plz let me know.
: i also asked some math major and was pointed to some counterexamples
|
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