For a random variable *X*, the *k*th 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, *m*_{3} in terms of the 4th moment, *m*_{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 *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.

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}.

Thanks. I edited the post to put in the condition on the first moment that I’d left out.

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.

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?

Not sure what you mean. The moments of a random variable are defined as in this post. The

methodof moments is something different.