Bounding the 3rd moment by the 4th moment

For a random variable X, the kth moment of X is the expected value of X^k.

For any random variable X with 0 mean, or negative mean, there's an inequality that bounds the 3rd moment, m₃ in terms of the 4th moment, m₄:

The following example shows that this bound is the best possible.

Define

u = (1 - a 3)/a 2
v = (1 + a 3)/a 2
p = (3 + a 3) / 6

and suppose X = u with probability p and X = v with probability q = 1 - p. Then X has mean 0, and you can show that exact equality holds in the inequality above.

Here's some Mathematica code to verify this claim.

 u = (1 - Sqrt[3])/Sqrt[2] v = (1 + Sqrt[3])/Sqrt[2] p = (Sqrt[3] + 1)/(2 Sqrt[3]) q = 1 - p m3 = p u^3 + q v^3 m4 = p u^4 + q v^4 Simplify[ (4/27)^(1/4) m4^(3/4) / m3 ]

As hoped, the code returns 1.

Reference: Iosif Pinelis, Relations Between the First Four Moments, American Mathematical Monthly, Vol 122, No 5, pp 479-481.