Proving probability inequality

[Lily@pie]

Homework Statement

Prove the following
a>0, X is a non-negative function

Ʃ_{n\in N} P(X>an)≥\frac{1}{a}(E[X]-a)

Ʃ_{n\in N} P(X>an)≤\frac{E[X]}{a}

The Attempt at a Solution

I know that
\sum_{n\in N} P(X>an)=\sum_{k \in N} kP((k+1)a≥X>ka)=\sum_{k \in N} E[k1_{[(k+1)a,ka)}(X)]
where 1_{[(k+1)a,ka)} is the indicator function.

I'm not sure how to preceed from here.

All I know is that for all ε>0, E[ε1_{[ε,∞)}(X)]≤E[X]

 

[haruspex]

The second is not true. Is there a factor 1/a missing on the right?

 

[Lily@pie]

The second is not true. Is there a factor 1/a missing on the right?

Sorry, my bad... I've edited it...

 

[haruspex]

You can eliminate a by defining Y = X/a. Does that help?