A random variable has mean 0 and variance 1. Prove that E[|X|] is at most 1, identify every distribution attaining equality, and state how small E[|X|] can be.

A random variable has mean 0 and variance 1. Prove that E[|X|] is at most 1, identify every distribution attaining equality, and state how small E[|X|] can be.

Approach: Apply Cauchy-Schwarz to the pair |X| and 1, then examine the equality case, and search the low end by pushing mass toward zero while keeping the variance fixed.

E[|X|] is at most 1, with equality exactly when X takes the values +1 and -1 with probability 1/2 each, and the infimum over all such distributions is 0. Cauchy-Schwarz gives E[|X| * 1] <= sqrt(E[X^2]) sqrt(E[1]) = sqrt(1) = 1. The equality condition requires |X| to be a constant almost surely, and the mean zero and unit variance conditions then force that constant to be 1 with symmetric signs. For the low end, take X equal to plus or minus 1/sqrt(p) with probability p/2 each and 0 with probability 1 - p. The variance is 1 for every p, while E[|X|] = p / sqrt(p) = sqrt(p), which tends to 0 as p tends to 0, so the infimum is 0 and no minimising distribution exists. The same gap is why a heavy tailed series can have a large standard deviation and a small mean absolute deviation, and why the ratio of the two is used as a tail diagnostic.

Follow-up: For a normal variable the ratio E[|X|]/sigma is sqrt(2/pi). What does a sample ratio of 0.5 suggest about the data?

Key concepts: cauchy-schwarz inequality, equality condition, absolute moment, infimum.