# markov inequality property question

#### yahiazr

##### New Member
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

Thanks
Yahia

#### BGM

##### TS Contributor
You just need to know the probability of the complementary event:

$$\Pr\{|X| \geq a\} \leq \frac {E[|X|]} {a}$$

$$\iff \Pr\{|X| < a\} = 1 - \Pr\{|X| \geq a\} \geq 1 - \frac {E[|X|]} {a}$$

You may google several different types of "extensions" of this inequality if you are seeking them. A famous one of course is the Chebyshev's inequality.

#### yahiazr

##### New Member
Hi BGM

Yes thanks a lot for your answer it really helped. I just wanted to get the upperbound of the random variable instead of the lowerbound.

Thanks again.