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  • STA256 Lecture 11

Chebyshev's Inequality

STA256

• $P(|X-\mu |\ge c)\le \frac{{\sigma }^2}{c^2}$
• $P(|X-\mu |\ge k\sigma )\le \frac{1}{k^2}$
• $P(|X-\mu |<k\sigma )\ge 1-\frac{1}{k^2}$
• Gives the bound of the probability of an observations being within k standard deviations from the mean.
• Need that the $var(x)<\mathrm{\infty }$
• Proof:
  • By Markov's Inequality
  • For a function $g(x)$ and a constant $c$
  • The $P[g(x)\ge c]\le \frac{E[g(x)]}{c}$
  • Let $g(x)=(x-\mu {)}^2$
  • $c=(k\sigma {)}^2$
  • By Markov's
  • You can also see that $ϵ=k\sigma$ and use $ϵ$ for the proof
  • $P[(x-\mu {)}^2\ge (k\sigma {)}^2]\le \frac{E[(x-\mu {)}^2]}{(k\sigma {)}^2}$
    • $E[(x-\mu {)}^2]$ is the Variance of $X$
  • $P[(x-\mu {)}^2\ge (k\sigma {)}^2]\le \frac{{\sigma }^2}{k^2{\sigma }^2}$
  • $P[(x-\mu {)}^2\ge (k\sigma {)}^2]\le \frac{1}{k^2}$
  • $P[|x-\mu |\ge k\sigma ]\le \frac{1}{k^2}$
  • Done

STA260

• So let a random variable $X_n$ have a finite mean and variance $\mathrm{E}[X]={\mu }_X$ and $\mathrm{Var}(X)={\sigma }_X^2$.
• For $k>0$
• $P(|X_n-{\mu }_X|>k{\sigma }_X)\le \frac{1}{k^2}$
• $⟺P(|X_n-{\mu }_X|\le k{\sigma }_X)>1-\frac{1}{k^2}$
• $⟺P(|X_n-{\mu }_X|>k)\le \frac{{\sigma }_X^2}{k^2}$
• $⟺P(|X_n-{\mu }_X|\le k)\ge 1-\frac{{\sigma }_X^2}{k^2}$