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Yecloud science forum beginner
Joined: 18 Mar 2006
Posts: 17

Posted: Wed Jun 28, 2006 5:20 pm Post subject:
help: proof of E(sup_n(Y_n)) < infinity



Hi, all,
I have a question on how to prove E(sup_n(Y_n)) < infinity.
Known: (1) Y_n >=0 is random variables and bounded for any n
(n=1,2,...);
(2) lim_{n>infinity} Y_n = 0;
Question: can we prove: E(sup_n(Y_n)) < infinity?
Thanks a lot,
Cloud 

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The World Wide Wade science forum Guru
Joined: 24 Mar 2005
Posts: 790

Posted: Wed Jun 28, 2006 7:15 pm Post subject:
Re: help: proof of E(sup_n(Y_n)) < infinity



In article
<1151515201.592454.316570@i40g2000cwc.googlegroups.com>,
"Yecloud" <yecloud@gmail.com> wrote:
Quote:  Hi, all,
I have a question on how to prove E(sup_n(Y_n)) < infinity.
Known: (1) Y_n >=0 is random variables and bounded for any n
(n=1,2,...);
(2) lim_{n>infinity} Y_n = 0;
Question: can we prove: E(sup_n(Y_n)) < infinity?

How could that be true? Can't we have E(Y_n) > oo? And isn't
E(sup_n(Y_n)) >= E(Y_m) for every m? 

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Yecloud science forum beginner
Joined: 18 Mar 2006
Posts: 17

Posted: Wed Jun 28, 2006 8:54 pm Post subject:
Re: help: proof of E(sup_n(Y_n)) < infinity



The World Wide Wade wrote:
Quote:  In article
1151515201.592454.316570@i40g2000cwc.googlegroups.com>,
"Yecloud" <yecloud@gmail.com> wrote:
Hi, all,
I have a question on how to prove E(sup_n(Y_n)) < infinity.
Known: (1) Y_n >=0 is random variables and bounded for any n
(n=1,2,...);
(2) lim_{n>infinity} Y_n = 0;
Question: can we prove: E(sup_n(Y_n)) < infinity?
How could that be true? Can't we have E(Y_n) > oo? And isn't
E(sup_n(Y_n)) >= E(Y_m) for every m?

Is it possible for Y_n >=0 and bouned (finite) for any n that its
expectation
E(Y_n) to be > oo?
Thanks,
Cloud 

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The World Wide Wade science forum Guru
Joined: 24 Mar 2005
Posts: 790

Posted: Wed Jun 28, 2006 9:15 pm Post subject:
Re: help: proof of E(sup_n(Y_n)) < infinity



In article
<1151528097.582178.19660@j72g2000cwa.googlegroups.com>,
"Yecloud" <yecloud@gmail.com> wrote:
Quote:  The World Wide Wade wrote:
In article
1151515201.592454.316570@i40g2000cwc.googlegroups.com>,
"Yecloud" <yecloud@gmail.com> wrote:
Hi, all,
I have a question on how to prove E(sup_n(Y_n)) < infinity.
Known: (1) Y_n >=0 is random variables and bounded for any n
(n=1,2,...);
(2) lim_{n>infinity} Y_n = 0;
Question: can we prove: E(sup_n(Y_n)) < infinity?
How could that be true? Can't we have E(Y_n) > oo? And isn't
E(sup_n(Y_n)) >= E(Y_m) for every m?
Is it possible for Y_n >=0 and bouned (finite) for any n that its
expectation
E(Y_n) to be > oo?

Sure. Let Y_n = n for every n. 

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Yecloud science forum beginner
Joined: 18 Mar 2006
Posts: 17

Posted: Wed Jun 28, 2006 9:19 pm Post subject:
Re: help: proof of E(sup_n(Y_n)) < infinity



But the second condition is lim_{n>infinity} Y_n = 0;, is it
enough to ensure the conclusion?
Thanks,
Zhenzhen
The World Wide Wade wrote:
Quote:  In article
1151528097.582178.19660@j72g2000cwa.googlegroups.com>,
"Yecloud" <yecloud@gmail.com> wrote:
The World Wide Wade wrote:
In article
1151515201.592454.316570@i40g2000cwc.googlegroups.com>,
"Yecloud" <yecloud@gmail.com> wrote:
Hi, all,
I have a question on how to prove E(sup_n(Y_n)) < infinity.
Known: (1) Y_n >=0 is random variables and bounded for any n
(n=1,2,...);
(2) lim_{n>infinity} Y_n = 0;
Question: can we prove: E(sup_n(Y_n)) < infinity?
How could that be true? Can't we have E(Y_n) > oo? And isn't
E(sup_n(Y_n)) >= E(Y_m) for every m?
Is it possible for Y_n >=0 and bouned (finite) for any n that its
expectation
E(Y_n) to be > oo?
Sure. Let Y_n = n for every n. 


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The World Wide Wade science forum Guru
Joined: 24 Mar 2005
Posts: 790

Posted: Wed Jun 28, 2006 9:45 pm Post subject:
Re: help: proof of E(sup_n(Y_n)) < infinity



In article
<1151529546.230282.254170@m73g2000cwd.googlegroups.com>,
"Yecloud" <yecloud@gmail.com> wrote:
Quote:  But the second condition is lim_{n>infinity} Y_n = 0;, is it
enough to ensure the conclusion?

No, but that wasn't the question you asked in the last post.
Here's an example to think about: Y_n(x) = n*x^n, x in [0,1).
PS: Please stop topposting.
Quote:  The World Wide Wade wrote:
In article
1151528097.582178.19660@j72g2000cwa.googlegroups.com>,
"Yecloud" <yecloud@gmail.com> wrote:
The World Wide Wade wrote:
In article
1151515201.592454.316570@i40g2000cwc.googlegroups.com>,
"Yecloud" <yecloud@gmail.com> wrote:
Hi, all,
I have a question on how to prove E(sup_n(Y_n)) < infinity.
Known: (1) Y_n >=0 is random variables and bounded for any n
(n=1,2,...);
(2) lim_{n>infinity} Y_n = 0;
Question: can we prove: E(sup_n(Y_n)) < infinity?
How could that be true? Can't we have E(Y_n) > oo? And isn't
E(sup_n(Y_n)) >= E(Y_m) for every m?
Is it possible for Y_n >=0 and bouned (finite) for any n that its
expectation
E(Y_n) to be > oo?
Sure. Let Y_n = n for every n. 


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Yecloud science forum beginner
Joined: 18 Mar 2006
Posts: 17

Posted: Wed Jun 28, 2006 10:19 pm Post subject:
Re: help: proof of E(sup_n(Y_n)) < infinity



The World Wide Wade wrote:
Quote:  In article
1151529546.230282.254170@m73g2000cwd.googlegroups.com>,
"Yecloud" <yecloud@gmail.com> wrote:
But the second condition is lim_{n>infinity} Y_n = 0;, is it
enough to ensure the conclusion?
No, but that wasn't the question you asked in the last post.
Here's an example to think about: Y_n(x) = n*x^n, x in [0,1).
PS: Please stop topposting.

I see. In this case, sup_n(Y_n) is still n with x to be a rv in [0,1),
is that right? How about if we change the second condition to
be limsup_{n>infinity} Y_n = 0? As Y_n >=0, the original second
condition still holds. Can the conclusion be hold?
Thanks! 

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Stephen MontgomerySmith1 science forum Guru
Joined: 01 May 2005
Posts: 487

Posted: Wed Jun 28, 2006 10:26 pm Post subject:
Re: help: proof of E(sup_n(Y_n)) < infinity



Yecloud wrote:
Quote:  Hi, all,
I have a question on how to prove E(sup_n(Y_n)) < infinity.
Known: (1) Y_n >=0 is random variables and bounded for any n
(n=1,2,...);
(2) lim_{n>infinity} Y_n = 0;
Question: can we prove: E(sup_n(Y_n)) < infinity?

No. Let Y_n be defined on [0,1] to be n^2 I_(0,1/n) ("I" means
indicator function). Then Y_n>0 but E(sup Y_n) >= sup E(Y_n) = infinity. 

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The World Wide Wade science forum Guru
Joined: 24 Mar 2005
Posts: 790

Posted: Thu Jun 29, 2006 12:13 am Post subject:
Re: help: proof of E(sup_n(Y_n)) < infinity



In article <YlDog.804568$084.43536@attbi_s22>,
Stephen MontgomerySmith <stephen@math.missouri.edu> wrote:
Quote:  Yecloud wrote:
Hi, all,
I have a question on how to prove E(sup_n(Y_n)) < infinity.
Known: (1) Y_n >=0 is random variables and bounded for any n
(n=1,2,...);
(2) lim_{n>infinity} Y_n = 0;
Question: can we prove: E(sup_n(Y_n)) < infinity?
No. Let Y_n be defined on [0,1] to be n^2 I_(0,1/n) ("I" means
indicator function). Then Y_n>0 but E(sup Y_n) >= sup E(Y_n) = infinity.

And thus another multipost dialog ends with a splat. 

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Stephen MontgomerySmith1 science forum Guru
Joined: 01 May 2005
Posts: 487

Posted: Thu Jun 29, 2006 12:20 am Post subject:
Re: help: proof of E(sup_n(Y_n)) < infinity



Yecloud wrote:
Quote:  The World Wide Wade wrote:
In article
1151529546.230282.254170@m73g2000cwd.googlegroups.com>,
"Yecloud" <yecloud@gmail.com> wrote:
But the second condition is lim_{n>infinity} Y_n = 0;, is it
enough to ensure the conclusion?
No, but that wasn't the question you asked in the last post.
Here's an example to think about: Y_n(x) = n*x^n, x in [0,1).
PS: Please stop topposting.
I see. In this case, sup_n(Y_n) is still n with x to be a rv in [0,1),
is that right? How about if we change the second condition to
be limsup_{n>infinity} Y_n = 0? As Y_n >=0, the original second
condition still holds. Can the conclusion be hold?
Thanks!

If Y_n>0 then limsup Y_n = 0. How can weakening the hypothesis help?
But in general you are trying to deduce "convergence or boundedness in
L_1" from "convergence or boundedness in measure" and this just isn't
going to happen. 

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