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State Chebyshev's inequality.

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Answer

  • P(∣X  −  μ∣  ≥  kσ)P( \mid X\; - \;{\mu} \mid \; \ge \;k{\sigma}) ≤ 1/k21 / k^{2}.
  • Distribution-free bound on tail probabilities — at most 1/k21 / k^{2} of the mass is more than k SDs from the mean, for any distribution with finite variance.
  • Consequence: at least 75% within 2σ, 89% within 3σ.
  • Much looser than the 95/99.7% for normal — general bounds are wide.
  • Applied in concentration arguments, generalization bounds in learning theory.
Check yourself — multiple choice
  • Only for normal distributions
  • P(∣X  −  μ∣  ≥  kσ)P( \mid X\; - \;{\mu} \mid \; \ge \;k{\sigma}) ≤ 1/k21 / k^{2} for any distribution with finite variance — distribution-free tail bound
  • P = 1/k
  • Only equality

Chebyshev: 1/k21 / k^{2} tail bound holds for any distribution with finite variance.

#probability#theory

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