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Suppose $X$ is a random variable with mean $\mu$ and variance $\sigma^2$. Show that $$P(|X-\mu| \geq k\sigma) \leq \frac{1}{k^2}$$

Question: I know the exercise wants me to use Markov inequality and Chebyshev inequality, but I can't reach the same answer. If someone can help me, it will be appreciated.

Thanks!

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    $\begingroup$ You have stated the standard Chebyshev Inequality. Proofs can be found in many places, including the Wikipedia article. $\endgroup$ – André Nicolas Nov 28 '15 at 22:12
  • $\begingroup$ Usually the Chebyshev inequality is either written that way or as $P(|X-\mu| \geq \varepsilon) \leq \frac{\sigma^2}{\varepsilon^2}$. Starting from the latter way of writing it, you can identify $k$ such that $\varepsilon=k \sigma$, and then the result follows. $\endgroup$ – Ian Nov 28 '15 at 22:17
  • $\begingroup$ "I know the exercise want me to use Markov inequality and chebyshev inequality, but I can't reach the same answer." Which answer can you reach? $\endgroup$ – Did Nov 28 '15 at 22:35
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The Chebyshev inequality is $$\mathbb{P}(|x - \mu| \geq a) \leq \frac{\sigma^2}{a^2}$$ .Substituting $$a = k\sigma $$gives the answer.

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Thanks André, I found it.

For any event $A$, let $I_A$ be the indicator random variable of $A$, i.e. $I_A$ equals $1$ if $A$ occurs and $0$ otherwise. Then $\newcommand{\E}{\operatorname{E}}$ \begin{align*} \Pr(|X-\mu| \geq k\sigma) &=\E(I_{|X-\mu| \geq k\sigma}) \\ &= \E\left(I_{\left(\frac{X-\mu}{k\sigma}\right)^2 \geq 1}\right) \\ &\leq \E\left(\left(\frac{X-\mu}{k\sigma}\right)^2\right) \\ &=\frac{1}{k^2} \cdot \frac{\E((X-\mu)^2)}{k\sigma} \\ &=\frac{1}{k^2}.\end{align*}

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