Chebyshev's inequality bounds P(|X - mu| >= 3 sigma) by 1/9. Construct a distribution that attains the bound exactly, explain why no distribution can do worse, and compare the bound with the normal value.

Chebyshev's inequality bounds P(|X - mu| >= 3 sigma) by 1/9. Construct a distribution that attains the bound exactly, explain why no distribution can do worse, and compare the bound with the normal value.

Approach: Work out where the proof of the inequality wastes anything, then place mass only at the points where the two sides of the underlying comparison agree.

Put mass 1/18 at -3, mass 1/18 at +3 and mass 8/9 at 0, with mu = 0 and sigma = 1. The variance is (1/18)*9 + (1/18)*9 + 0 = 1, and P(|X| >= 3) = 1/18 + 1/18 = 1/9, so this three point distribution attains the sharp bound with equality. The proof of the Chebyshev inequality replaces X^2 by 9 * 1{|X| >= 3}, which loses nothing only when X takes the values 0 and plus or minus 3 alone, and this construction places mass exactly there. No distribution can exceed 1/9, since for any mean zero unit variance random variable the second moment is at least 9 times the probability that |X| reaches 3. For a normal tail the true probability is 0.0027, about 41 times smaller, which is the price of a bound that uses only two moments and holds for every distribution.

Follow-up: If you also know the distribution is symmetric and unimodal, what replaces 1/9, and how large is the improvement at three standard deviations?

Key concepts: chebyshev inequality, sharp bound, three point distribution, normal tail.