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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
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
"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?
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

 Quote: In article 1151515201.592454.316570@i40g2000cwc.googlegroups.com>, "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? 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
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
"Yecloud" <yecloud@gmail.com> wrote:

 Quote: The World Wide Wade wrote: In article 1151515201.592454.316570@i40g2000cwc.googlegroups.com>, "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? 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.
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

 Quote: In article 1151528097.582178.19660@j72g2000cwa.googlegroups.com>, "Yecloud" wrote: The World Wide Wade wrote: In article 1151515201.592454.316570@i40g2000cwc.googlegroups.com>, "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? 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.
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
"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).

 Quote: The World Wide Wade wrote: In article 1151528097.582178.19660@j72g2000cwa.googlegroups.com>, "Yecloud" wrote: The World Wide Wade wrote: In article 1151515201.592454.316570@i40g2000cwc.googlegroups.com>, "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? 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.
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

 Quote: In article 1151529546.230282.254170@m73g2000cwd.googlegroups.com>, "Yecloud" 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 top-posting.

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!
Stephen Montgomery-Smith1
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.
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 Montgomery-Smith <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 multi-post dialog ends with a splat.
Stephen Montgomery-Smith1
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" 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 top-posting. 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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