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Forum index » Science and Technology » Math » Probability
Understanding Proof of Markov Inequality
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Michael Zedeler
science forum beginner


Joined: 29 Nov 2005
Posts: 17

PostPosted: Fri May 19, 2006 9:53 am    Post subject: Re: Understanding Proof of Markov Inequality Reply with quote

Jason Simons wrote:
Quote:
The Markov inequality says the following:

If X is a non-negative 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.
--
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Visit my home page at http://michael.zedeler.dk/
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Jason Simons
science forum beginner


Joined: 03 Feb 2005
Posts: 8

PostPosted: Wed May 17, 2006 6:41 pm    Post subject: Understanding Proof of Markov Inequality Reply with quote

The Markov inequality says the following:

If X is a non-negative 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
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