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Math Help - markov inequality property question

  1. #1
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    Cairo
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    markov inequality property question

    Dear All,


    As you know that markov inequality property is:


    My question is that valid that to use invariant of this property in the reverse way, I mean is the following is true ? :


    P(X<=a) >= E(X)/a


    I appreciate your help.


    Thanks
    Yahia
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  2. #2
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    Re: markov inequality property question

    No this doesn't hold. Here is a simple counter example:

    Assume P(X<=a) >= E(X)/a

    consider X~Exp(k)

    then P(X<=a) = 1-exp(ka)

    and E(X) = 1/k

    but 1-exp(ka) >= 1/ka doesn't hold.

    To get a feeling for why the Markov inequality can't be reversed I suggest looking up how the proof of the original inequality works. Essentially it relies on the fact that I(X>=a)a <= X where I is the indicator function. But I(X<=a)a >= X does not hold.
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  3. #3
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    Re: markov inequality property question

    Thanks a lot StaryNight for your answer. What about substituting P(X>=t) by 1-P(X<t). Then we will have P(X<t) > 1 - E(X)/t

    is that right ?

    Thanks
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  4. #4
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    Re: markov inequality property question

    Yes, that looks fine to me. Sorry for omitting the modulus operator in my last answer - the equality only works for |X|, since I(|X|>=a)a <= |X| but not I(X>=a)a <= X. So to clarify P(|X|<t) > 1-E(|X|)/t. Hope this clears things up.


    Quote Originally Posted by yahiazr View Post
    Thanks a lot StaryNight for your answer. What about substituting P(X>=t) by 1-P(X<t). Then we will have P(X<t) > 1 - E(X)/t

    is that right ?

    Thanks
    Follow Math Help Forum on Facebook and Google+

  5. #5
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    Re: markov inequality property question

    Yes it is clear now. Thanks a lot for your help.
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