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:
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 method of moments is something different.
The moments don’t require standardization. They actually tell you the skewness and kurtosis numbers, aka the statistics. Only when those values return the “standard” values would the normal have achieved actual normality.
Normalizing the data lets you achieve before you would have otherwise. But, that has nothing to do with the actual achievement of normality. It is an approximation that hides much.