Author 
Message 
Michael Zedeler science forum beginner
Joined: 29 Nov 2005
Posts: 17

Posted: Fri May 19, 2006 9:53 am Post subject:
Re: Understanding Proof of Markov Inequality



Jason Simons wrote:
Quote:  The Markov inequality says the following:
If X is a nonnegative random variable and a is a positive constant,
then
P( X >= a ) <= E[X]/a
[...] Let I be the indicator random variable that is 1 when X >= a and
0 otherwise. Then
I <= X/a, so
E[I] = P( I = 1 ) = P( X >= a ) <= E[ X/a ] = E[X]/a
My question concerns the motivation behind the first move. Why should I
set up an indicator variable "I" equal to 1 when X >= a and zero
otherwise. We're concerned about the "probability" of X >= a not the
value of X itself. I never would have come up with this idea (at least
at my current level of understanding). It's apparent that there is some
relationship between X >= a and the theorem, but I don't see it.

It is a general rule that for any indicator variable I[A] (A being an
event where I[A] takes the value 1, 0 otherwise), E(I[A]) = P(A). This
can be used as a shortcut in some places.
If you have questions regarding some of the expressions above, please
post them here.
Regards,
Michael.

Which is more dangerous? TV guided missiles or TV guided families?
Visit my home page at http://michael.zedeler.dk/
Get my vcard at http://michael.zedeler.dk/vcard.vcf 

Back to top 


Jason Simons science forum beginner
Joined: 03 Feb 2005
Posts: 8

Posted: Wed May 17, 2006 6:41 pm Post subject:
Understanding Proof of Markov Inequality



The Markov inequality says the following:
If X is a nonnegative random variable and a is a positive constant,
then
P( X >= a ) <= E[X]/a
I tried some examples and can see that, for my examples, this is true.
The following was given in class as a proof of this.
Proof: Let I be the indicator random variable that is 1 when X >= a and
0 otherwise. Then
I <= X/a, so
E[I] = P( I = 1 ) = P( X >= a ) <= E[ X/a ] = E[X]/a
My question concerns the motivation behind the first move. Why should I
set up an indicator variable "I" equal to 1 when X >= a and zero
otherwise. We're concerned about the "probability" of X >= a not the
value of X itself. I never would have come up with this idea (at least
at my current level of understanding). It's apparent that there is some
relationship between X >= a and the theorem, but I don't see it.
Anyone help?
thanks,
Jason 

Back to top 


Google


Back to top 



The time now is Thu Jan 17, 2019 1:22 pm  All times are GMT

