Show that $\mathbb{P}(|X| \geq a) \geq \frac{\mathbb{E}|X|-a}{M-a}$ Let X be a random variable that takes values in the interval $[-M,M]$ only. Show that
\begin{equation}
\mathbb{P}(|X| \geq a) \geq \frac{\mathbb{E}|X|-a}{M-a} 
\end{equation}
if $0 \leq a <M$.
I'm having difficulties proving this for a homework assignment. I tried using Markov's inequality, with contradiction, but I can't find one. Could anybody give me some tips on how to approach this problem and maybe even state the (in)equality's I will have to use. Thanks in advance!
 A: By the fact that $X \in [-M, M]$, the following inequality holds: $|X| \leq M$ and therefore $|X| - a \leq M - a$. Introduce the indicator function $1_{|X| \geq a}$ which is $1$ when $|X| \geq a$ and $0$ when $|X| < a$.
Check whether or not $|X| - a \leq (M - a)1_{|X| \geq a}$ holds:
1) If $|X| < a$, then $|X|-a < 0$ and $1_{|X| \geq a} = 0$. Therefore $(M-a)1_{|X| \geq a} = (M-a) \cdot 0 = 0$ and thus $|X|-a \leq (M-a)1_{|X| \geq a}$.
2) If $|X| \geq a$, then $1_{|X| \geq a} = 1$ and thus $(M-a)1_{|X| \geq a} = (M-a) \cdot 1 = (M-a)$. Because it is already shown that $|X|-a \leq (M-a)$, we can conclude that $|X|-a \leq (M-a)1_{|X| \geq a}$.
From 1) and 2) it follows that $X - a \leq (M - a)1_{|X| \geq a}$ for all possible values of $|X|$. If we take the expected value of both sides, we get $\mathbb{E}(|X|-a) \leq \mathbb{E}((M-a)1_{|X| \geq a})$. By the fact that $\mathbb{E}(1_{|X| \geq a}) = \mathbb{P}(|X| \geq a)$ and linearity of $\mathbb{E}$, we get $\mathbb{E}|X| - a \leq (M-a)\mathbb{P}(|X| \geq a)$. Because $M-a > 0$, this rewrites to $\mathbb{P}(|X| \geq a) \geq \frac{\mathbb{E}|X|-a}{M-a}$, which was to be proved.
