Cumulative distrubution function question - Math Help Forum

$Math Help Forum$

$Go Back$		Math Help Forum > University Math Help > Advanced Probability and Statistics
$Reload this Page$ Cumulative distrubution function question

Download MHF Flyer

Mark Forums Read

$Reply$

#1

$Old$ November 5th, 2009, 10:19 AM

ggurtin $ggurtin is offline$

Newbie

Join Date: Oct 2009

Posts: 2

Thanks: 0

Thanked 0 Times in 0 Posts

$ggurtin is on a distinguished road$

$Default$ Cumulative distrubution function question

Can anyone provide any insight in how I could show the following (attached below)

Attached Thumbnails

$cumulative-distrubution-function-question-q3.jpg$

$Reply With Quote$

Advertisement

#2

$Old$ November 5th, 2009, 11:50 PM

$matheagle's Avatar$

matheagle $matheagle is online now$

MHF Contributor

Join Date: Feb 2009

Posts: 1,367

Country:

United States

Thanks: 99

Thanked 560 Times in 503 Posts

$matheagle is a splendid one to behold$ $matheagle is a splendid one to behold$ $matheagle is a splendid one to behold$ $matheagle is a splendid one to behold$ $matheagle is a splendid one to behold$ $matheagle is a splendid one to behold$ $matheagle is a splendid one to behold$

$Default$

$0\le nP(X>n) =n\int_n^{\infty}f(x)dx =\int_n^{\infty}nf(x)dx\le \int_n^{\infty}xf(x)dx\to 0$

as $n\to \infty$ since the first moment is finite.

Last edited by matheagle; November 7th, 2009 at 12:30 AM.

$Reply With Quote$

#3

$Old$ November 6th, 2009, 11:54 AM

$Moo's Avatar$

$Moo is offline$

A Cute Angle

Join Date: Mar 2008

Location: P(I'm here)=1/3, P(I'm there)=t+1/3

Posts: 5,049

Country:

United Nations

Thanks: 506

Thanked 2,914 Times in 2,397 Posts

$Moo has a reputation beyond repute$ $Moo has a reputation beyond repute$ $Moo has a reputation beyond repute$ $Moo has a reputation beyond repute$ $Moo has a reputation beyond repute$ $Moo has a reputation beyond repute$ $Moo has a reputation beyond repute$ $Moo has a reputation beyond repute$ $Moo has a reputation beyond repute$ $Moo has a reputation beyond repute$ $Moo has a reputation beyond repute$

$Default$

Hello,

$1-F(x)=P(X>n)=P(X\cdot \bold{1}_{X>n}>n)=P(Y_n>n)\leq \frac{E(Y_n)}{n}$ by Markov's inequality.
where $Y_n=X\cdot \bold{1}_{X>n}$
As n goes to infinity, $Y_n$ obviously goes to 0 (X is almost surely finite since it's integrable)
and $|Y_n|$ is bounded by |X|, which is integrable.
So we can apply the dominated convergence theorem, and we have $n(1-F(x))\leq E(Y_n) \to 0 \quad \quad \square$

__________________
Everything is possible. The impossible just takes longer.

To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

shinhidora production

To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

$Reply With Quote$

$Reply$

« Quantiles/Order Statistics | Top | Gaussian Probability Density Function »

Thread Tools
$Show Printable Version$ Show Printable Version $Email this Page$ Email this Page
Display Modes
$Linear Mode$ Linear Mode $Hybrid Mode$ Switch to Hybrid Mode $Threaded Mode$ Switch to Threaded Mode

Posting Rules
You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off Trackbacks are Off Pingbacks are Off Refbacks are Off

Forum Jump

All times are GMT -7. The time now is 10:32 PM.

your link here Sponsors

rv solar, e-learning,

Math Help Forum is a community of maths forums with an emphasis on maths help in all levels of mathematics.
Register to post your math questions or just hang out and try some of our math games or visit the arcade.