Prove the following using Chebyshev and Markov inequality.

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!

• You have stated the standard Chebyshev Inequality. Proofs can be found in many places, including the Wikipedia article. – André Nicolas Nov 28 '15 at 22:12
• 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. – Ian Nov 28 '15 at 22:17
• "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? – Did Nov 28 '15 at 22:35

The Chebyshev inequality is $$\mathbb{P}(|x - \mu| \geq a) \leq \frac{\sigma^2}{a^2}$$ .Substituting $$a = k\sigma$$gives the answer.
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*}