Bounding the 3rd moment by the 4th moment

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

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:

m_3 \leq \left(\frac{4}{27} \right )^{1/4} m_4^{3/4}

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

Define

u = (1 – √ 3)/√ 2
v = (1 + √ 3)/√ 2
p = (3 + √ 3) / 6

and suppose Xu with probability p and Xv 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.

5 thoughts on “Bounding the 3rd moment by the 4th moment

  1. Hi,
    Small mistake in your post:
    According to the cited paper [Pinelis 2015], the inequality is valid (assuming the 4th moment do exist) *only if* the mean of X is non-positive.

    In the general case, for any random variable X, the inequality is not true, and the best (tight) bound is then the obvious m_3 ≤ m_4^{3/4}.

  2. That’s very interesting that you can improve on Holder’s inequality in such a way just because you have a non-positive mean. But of course it’s because the left hand side is not the L-3 norm, but rather simply the expectation of the X^3.

  3. Hi John
    In using the method of moments in general to define the expectation of a random variable, X, are you assuming a normal distribution or a standard normal distribution?

  4. Not sure what you mean. The moments of a random variable are defined as in this post. The method of moments is something different.

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