# 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 $$$$\mathbb{P}(|X| \geq a) \geq \frac{\mathbb{E}|X|-a}{M-a}$$$$

if $$0 \leq a .

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!

• Mimic the proof of Markov's inequality by finding a lower bound on $(M-a)1_{|X|\geq a}$. Apr 28 at 18:48
• Ehm, I understand where you want to go to, but when mimicing, I seem to get stuck with the -a... Apr 28 at 19:10
• $E[|X|]-a = E[ |X| - a]$. Does a lower bound still elude you ? Apr 28 at 19:13
• If I can show that $|X|-a \leq (M-a)1_{|X| \geq a}$, I am of course done. I understand that Apr 28 at 19:22
• No, I don't understand the reasoning. If $|X| < a$, then the inequality rewrites as $|X|-a\leq 0$, which is true since $|X| -a <0$ . If $|X| \geq a$, the inequality rewrites as $|X|-a \leq M-a$, which is the same as $|X|\leq M$, which is true by a hypothesis of the problem. So in each case, the inequality $|X|-a \leq (M-a)1_{|X| \geq a}$ holds. Apr 28 at 19:43

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.