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Consider a real random variable $X$ with zero mean. Does the following inequality hold in general?

$$\langle X^4\rangle \ge 3 \langle X^2\rangle^2$$

I'm not sure how to prove this or if a counter-example exists. If the inequality is true, it is also tight because for a standard normal this is an equality.

I found a related inequality, valid for symmetric distributions (all odd moments vanish):

$$\langle X^4\rangle \ge 2\langle X^2\rangle$$

This is proved by

Dreier, Ilona. "Inequalities between the second and fourth moments." Statistics: A Journal of Theoretical and Applied Statistics 32.2 (1998): 189-198.

But I'm actually not sure if this is connected to the inequality above.

Update: The inequality from Dreier 1998 is stated under the assumption of a distribution with a non-negative characteristic function.

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  • $\begingroup$ This notation $\langle \cdot \rangle$ reminds me of the notation used in my quantum mechanics textbook. $\endgroup$
    – Galen
    Mar 17, 2023 at 19:43
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    $\begingroup$ You second inequality obviously does not hold, because it changes when you rescale $X.$ There must be a typographical error in it. I suspect you mean to square the moment on the rhs--but it would still be incorrect. You can show $E[X^4]\ge E[X^2]^2$ in many ways, with equality attained for a Rademacher variable. See stats.stackexchange.com/questions/512568 for some applicable techniques for addressing moment inequalities. $\endgroup$
    – whuber ♦
    Mar 17, 2023 at 21:06
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    $\begingroup$ It would be good to define your notation explicitly to avoid any confusion. If $\langle \cdot\rangle$ is intended to be mean-corrected (I have seen it used that way in relation to random variables), this is basically asking "is kurtosis always at least 3?" to which the answer is "definitely not". For example, all the symmetric beta distributions are counterexamples, including the uniform for a very simple case. $\endgroup$
    – Glen_b
    Mar 18, 2023 at 0:23
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    $\begingroup$ @Glen_b What does "mean-corrected" mean? Note that I assume at the outset that the mean is zero, $\langle X \rangle = 0$. $\endgroup$
    – a06e
    Mar 18, 2023 at 9:51
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    $\begingroup$ Mean-correction is to subtract the mean so you have mean 0. If the mean is zero, it's already mean corrected. When it's mean corrected, which it is here, you're talking about kurtosis. $\endgroup$
    – Glen_b
    Mar 18, 2023 at 23:37

2 Answers 2

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Both of the inequalities you assert are false

For a random variable with zero mean, the moment quantity $\langle X^4 \rangle /\langle X^2 \rangle^2$ is the kurtosis of the distribution, which has a lower bound of one, not three. Thus, you can find abundant counter-examples to your first inequality by choosing any distribution that is platykurtic (i.e., has a kurtosis less than three). You can also find counter-examples to your second inequality by choosing any symmetric distribution that is sufficiently playkurtic to have a kurtosis less than two.

The counter-example given in the other answer here is the most platykurtic distribution there is (and it is also symmetric), so it gives a simple counter-example to both purported inequalities. Other examples of symmetric platykurtic distributions include the discrete uniform distribution, the continuous uniform distribution, the Wigner semicircle distribution, the raised cosine distribution, certain parameterisations of the beta distribution, and many others.

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An easy counterexample is a two point distribution $P(X = \pm 1) = 1/2$, for which $E[X] = 0$, $E[X^2] = E[X^4] = 1$. Hence $1 = E[X^4] < 3(E[X^2])^2 = 3$.

This example also showed the related inequality you claimed does not hold either.

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