Simple inequality of tails of random variables

Question

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?

Answer 1

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}