Chebyshev's inequality

Description:

• If $X$ is a random variable with finite mean $\mu$ and variance ${\sigma }^2$, then for any value $k>0,P(|X-\mu |\ge k)\le \frac{{\sigma }^2}{k^2}$
• Proof:
  • $P((X-\mu {)}^2\ge k^2)\le \frac{E[(X-\mu {)}^2]}{k^2}$
    • By Markov’s inequality
  • $P((X-\mu {)}^2\ge k^2)=P(|X-\mu |\ge k)$
  • Proof complete