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This may be a silly question, but, for a random variable $X$ defined on reals, I am wondering if the existence of the finite fourth central moment $E[(X-E[X])^4]$ implies the existence of the finite third absolute moment $E[|X-E[X]|^3]$? Formally, does the following statement hold:

$$E[(X-E[X])^4]<\infty\Rightarrow E[|X-E[X]|^3]<\infty$$

The reason for my question is that, while trying to apply the Berry-Esseen Theorem to bound the total variation distance between the distribution of the (appropriately normalized) mean of $n$ i.i.d. random variables and the standard normal distribution, I am finding that the fourth central moment of the i.i.d. random variables the distribution of whose average is being approximated is much easier to compute than their third absolute moment...

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  • $\begingroup$ This works because the measure of the space is finite. When integrating over spaces of infinite measure, this doesn't hold. $\endgroup$ – Michael Hardy Aug 10 '13 at 17:28
  • $\begingroup$ I am not that familiar with measure theory, but all probability measures are finite, correct? And thanks for fixing the typo in my question. $\endgroup$ – M.B.M. Aug 10 '13 at 17:41
  • $\begingroup$ Probability measures are finite. But in some other contexts than probability theory one integrates over spaces of infinite measure. $\endgroup$ – Michael Hardy Aug 10 '13 at 23:45
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Yes.

Proof: Apply the inequality $|x|^3\leqslant1+x^4$ to $x=X-E[X]$ and integrate.

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  • $\begingroup$ This is a very pretty answer, thanks! $\endgroup$ – M.B.M. Aug 10 '13 at 16:25
  • $\begingroup$ Is there a way to bound $|x|^3$ by a polynomial involving $x^4$ and $x$? I just wonder how you come up with these inequalities. $\endgroup$ – Glassjawed Dec 7 '13 at 0:04
  • $\begingroup$ @Glassjawed $|x|^3\leqslant x^4+|x|$. Just draw the graphs of all these functions on, say, $(-5,5)$. $\endgroup$ – Did Dec 7 '13 at 7:12
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Split the random variable into the following regions

  1. $ |X - E[X]| > 1 $. In this region, $|X-E[X]|^3 < |X-E[X]|^4$

  2. $|X - E[X]| \leq 1$. In this region, $|X-E[X]|^3 \leq 1 $.

Hence conclude that $E\left[|X-E[X]|^3\right] < E\left[|X-E[X]|^4\right] + 1$

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