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help: proof of E(sup_n(Y_n)) < infinity
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Yecloud
science forum beginner


Joined: 18 Mar 2006
Posts: 17

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

Thanks a lot,
Cloud
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The World Wide Wade
science forum Guru


Joined: 24 Mar 2005
Posts: 790

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

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

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

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

PostPosted: Wed Jun 28, 2006 9:45 pm    Post subject: Re: help: proof of E(sup_n(Y_n)) < infinity Reply with quote
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 top-posting.

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

PostPosted: Wed Jun 28, 2006 10:19 pm    Post subject: Re: help: proof of E(sup_n(Y_n)) < infinity Reply with quote
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 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!
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Stephen Montgomery-Smith
science forum Guru


Joined: 01 May 2005
Posts: 487

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

PostPosted: Thu Jun 29, 2006 12:13 am    Post subject: Re: help: proof of E(sup_n(Y_n)) < infinity Reply with quote
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.
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Stephen Montgomery-Smith
science forum Guru


Joined: 01 May 2005
Posts: 487

PostPosted: Thu Jun 29, 2006 12:20 am    Post subject: Re: help: proof of E(sup_n(Y_n)) < infinity Reply with quote
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 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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Google
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