Thread: 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

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.

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

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.

Originally Posted by yahiazr

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

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