Inequalities between Probability and expectation I would like to prove the following
When $a>0$ and X is a non-negative and measurable function
$\frac{E[X]}{a}\ge \sum _{ n\in N  }^{  }{ P(X>an) } \ge \frac { 1 }{ a } \left( E[X]-a \right) $ 
I know that 
$\sum _{ n\in N }^{  }{ P(X>an) } $
$=\sum _{ k\in N }^{  }{ kP\left( ka<X\le (k+1)a \right)}$
$=\sum _{ k\in N }^{  }{ kE\left( { 1 }_{ (ka,(k+1)a] }\left( X \right)  \right)  } $
$=\sum _{ k\in N }^{  }{ E\left( k{ 1 }_{ (ka,(k+1)a] }\left( X \right)  \right)  } $
If it is without the summation, I know that 
$E\left( k{ 1 }_{ (k,\infty ) }\left( X \right)  \right) \le E\left[ X \right] $
because $k{ 1 }_{ (k,\infty ) }\left( x \right) \le x$
I'm quite stuck here. It seems that I'm kind of there but I'm missing something important.
Thanks a lot!!
 A: You’re quite near indeed.
Starting where you stopped : with
$$
A=\sum _{ k\in N }^{  }{ E\left( k{ 1 }_{  (ka,(k+1)a] }\left( X \right)  \right)  },
$$
we have
$$
A=\sum _{ k\in N }^{  }{ E\left( k{ 1 }_{X \in  (ka,(k+1)a] } \right)  },
$$
so
$$
A \leq \sum _{ k\in N }  E\left( \frac{X}{a} { 1 }_{X \in  (ka,(k+1)a] } \right)  =
E\left( \frac{X}{a}\sum_{k\in N}{ 1 }_{X \in  (ka,(k+1)a] } \right)=\frac{E\left( X \right)}{a}
$$
The second inequality is similar, you use $(k+1)a \geq X$ instead of $X \geq ka$.
A: $$\begin{align}
E[X] &= \sum_{n=0}^\infty \int_{an}^{a(n+1)}P\{X>x\}\,\mathrm dx\\
&\leq \sum_{n=0}^\infty a\cdot P\{X > an\}&\scriptstyle{\text{because}~P\{X>x\}~ \text{is bounded above by}~P\{X>an\}~\text{for}~ x\in [an,a(n+1)]}\\
&= \sum_{n=0}^\infty a\cdot P\{X > an\}
\end{align}$$ 
giving $\displaystyle \sum_{n=0}^\infty P\{X > an\} \geq \frac{1}{a}E[X]$
and so since $P\{X > 0\} \leq 1$, we have that
$$ \sum_{n=1}^\infty P\{X > an\} 
\geq \left.\left.\frac{1}{a}\right(E[X] - a\right)$$
which is the second inequality, but the first one doesn't seem quite right.
In fact, the result that the OP wants to prove implies that
$$E[X] \geq \left.\left.\frac{1}{a}\right(E[X] - a\right) = \frac{1}{a}E[X] - 1
~ \text{for all}~ a > 0$$
which cannot be true.  Bounding $P\{X > x\}$ below by $P\{X > a(n+1)\}$ for
$x \in [an,a(n+1)]$ gives
$$E[X] \geq \sum_{n=1}^\infty a\cdot P\{X > an\} \Rightarrow \frac{E[X]}{a} \geq \sum_{n=1}^\infty P\{X > an\}$$ which, as Did points out in his comment
on the question, is really what the question should be asking.
