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Let $a>0$, $X$ some random variable and $\mathbb{E}[X]=\mu<t$. I was trying to prove the following simple inequality: $$ \textrm{Pr}\left[|X-t|\geq a\right]\leq\textrm{Pr}\left[|X-\mu|\geq a\right] $$ but being unsuccessful now I suspect it is not true. Can anyone provide a simple counterexample?

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  • $\begingroup$ You should use $P(|..| ) \le \frac{E(|..|)}{a} \le \frac{\mu}{a}$ and $P(|X-\mu|\ge a) \le \frac{Var(X)}{a^{2}}$ $\endgroup$ – openspace May 11 '16 at 12:30
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The inequality is not valid. Consider the following counterexample. Let $X \sim \mbox{Bernoulli}(1/2)$ and $\alpha = 1$. By definition, $X \in \{0,1\}$ and $\mathbb{E}[X] = \mu = 1/2$. Choosing $t = 1>\mu$, we have \begin{align} \mathbb{P}(|X-t| \geq \alpha) &= \mathbb{P}(|X-1|\geq 1) = \mathbb{P}(X = 0) = 1/2, \\ \mathbb{P}(|X-\mu| \geq \alpha) &= \mathbb{P}\left(\left|X -\frac{1}{2}\right| \geq 1\right) = 0. \end{align}

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