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Working on an exercise from Shorack's Probability for Statisticians, Ex 3.4.3 (Littlewood's inequality). Can't seem to find any material for this. Help is appreciated.

Prove:Exercise 4.3 (Littlewood's inequality) Let $m_r \equiv \mathbb{E}|X|^r$ denote the $r$th absolute moment. Then for $r \ge s \ge t \ge 0$, we have $m^{r-s}_t m^{s-t}_r \ge m^{r-t}_s$.

Thank you very much!

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It's a consequence of Hölder's inequality: we can write $s=\alpha t+(1-\alpha)r$, where $\alpha\in [0,1]$ (you can deal with the case $\alpha\in \{0,1\}$ by hand). We have $$m_s^{r-t}=(E|X|^s)^{r-t}=(E[|X|^{t\alpha}|X|^{r(1-\alpha)}])^{r-t}.$$ We apply Hölder's inequality to $f=|X|^t$, $g=|X|^r$ and $p=\frac 1{\alpha}>1$, $q=\frac 1{1-\alpha}$. We get $$m_s^{r-t}\leq (E|X|^t)^{(r-t)\alpha}(E|X|^r)^{(r-t)(1-\alpha)}=m_t^{(r-t)\alpha}m_r^{(r-t)(1-\alpha)}.$$ But $\alpha$ can be computed to see the exponents are what we expected.

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Awesome! Thank you! – RVC Sep 18 '12 at 9:48

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