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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!

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  • $\begingroup$ Mimic the proof of Markov's inequality by finding a lower bound on $(M-a)1_{|X|\geq a}$. $\endgroup$ Apr 28 at 18:48
  • $\begingroup$ Ehm, I understand where you want to go to, but when mimicing, I seem to get stuck with the -a... $\endgroup$ Apr 28 at 19:10
  • $\begingroup$ $E[|X|]-a = E[ |X| - a]$. Does a lower bound still elude you ? $\endgroup$ Apr 28 at 19:13
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    $\begingroup$ If I can show that $|X|-a \leq (M-a)1_{|X| \geq a}$, I am of course done. I understand that $\endgroup$ Apr 28 at 19:22
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    $\begingroup$ 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. $\endgroup$ Apr 28 at 19:43
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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.

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  • $\begingroup$ Perfect, well done ! $\endgroup$ Apr 30 at 13:37

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