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If $r \ge 1$ (not necessarily be an integer), then how to prove that the $r^\text{th}$ raw moment exists if and only if the $r^\text{th}$ central moment exists? I think I have to apply some inequality but I can't figure it out.

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    $\begingroup$ You should add the self-study tag and read its wiki. $\endgroup$ Dec 13 '16 at 21:40
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Two salient properties of real numbers $x, y,$ and $\mu$ are

  1. The triangle inequalities $$|x-\mu| \le |x|+|\mu|$$ and $$|x| \le |x-\mu| + |\mu|.$$

  2. Raising to the $r$ power for $r \gt 0$ preserves order: $$|x| \le |y| \text{ implies } |x|^r \le |y|^r$$ and (therefore) $$\max(|x|^r, |y|^r) = \max(|x|,|y|)^r.$$

Note, too, the simple relationship $$|x| + |y| \le 2\max(|x|, |y|).$$

Consequently for any random variable $X$ and real number $\mu$,

$$|X-\mu|^r \le \left(|X| + |\mu|\right)^r \le \left(2\max\left(|X|, |\mu|\right)\right)^r = 2^r\max\left(|X|^r, |\mu|^r\right)$$

and

$$|X|^r \le \left(|X-\mu| + |\mu|\right)^r \le \cdots \le 2^r\max\left(|X-\mu|^r, |\mu|^r\right).$$

Letting $\mu = \mathbb{E}(X)$ and taking expectations establishes the inequalities you need.

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  • $\begingroup$ In the RHS 2^r is included, so if we remove that then RHS can be small/large than LHS, can you please explain this a little more. $\endgroup$
    – Antora
    Dec 13 '16 at 23:14
  • $\begingroup$ You only need to show that the finiteness of one moment implies the finiteness of the other, not that one is actually smaller than the other. $\endgroup$
    – whuber
    Dec 13 '16 at 23:39

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