# Proof that if higher moment exists then lower moment also exists

The $r$-th moment of a random variable $X$ is finite if $$\mathbb E(|X^r|)< \infty$$

I am trying to show that for any positive integer $s<r$, then the $s$-th moment $\mathbb E[|X^s|]$ is also finite.

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Is this homework? If so, what have you tried so far? Also, I've tried to make your question more readable, please let me know if I've made a mistake. –  Gschneider Apr 14 '12 at 15:07
I read billingsley textbook and searched internet but no exact proof exists. What I found is just a clue maybe jensen's inequality can be used. –  nona Apr 14 '12 at 15:09
Consider rewriting $|X^r|$ as $|X^s \cdot X^{r-s}|$ and see if that gets you anywhere. –  Gschneider Apr 14 '12 at 15:16
There is a difference between a moment existing and being finite. In particular, a moment can exist, but be infinite. The terminology you're being introduced to is a bit imprecise. In any event, this is a standard result about $L_p$ spaces; it is not true that "no exact proof exists". :) –  cardinal Apr 14 '12 at 15:16
thanks for your answer, both of you. –  nona Apr 14 '12 at 15:21
$0<s<r \Longrightarrow \forall X \, |X|^s \le \max(1, |X|^r)$
(+1) I like this because it relies on only the most basic properties of expectation, namely monotonicity. In case one is worried about what to do with the right-hand side, they can note that $\max(1,|X|^r) \leq 1 + |X|^r$. If one prefers an application of Jensen, they can write $|X|^r = (|X|^s)^{r/s}$ and note that $r/s \geq 1$. –  cardinal Apr 14 '12 at 16:54
@cardinal: (+1) I prefer your inequality as it directly involves $|X|^r$... –  Xi'an Apr 15 '12 at 7:21