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If $X_1,X_2$ have finite second moments then Cauchy-Schwarz gives $\langle |X_1||X_2|\rangle^2 \leq \langle |X_1|^2\rangle \langle |X_2|^2\rangle $

If $(X_n)_{n=1}^N$ have their $N$th moments is it so that $\langle\prod_n|X_n|\rangle^N \leq \prod_n\langle |X_n|^N\rangle $?

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    $\begingroup$ I think this is probably what you are looking for: en.wikipedia.org/wiki/… $\endgroup$ – b00n heT Jan 2 at 8:33
  • $\begingroup$ Thanks, it does appear much closer to Holder's inequality. I will think about it in the morning. $\endgroup$ – enthdegree Jan 2 at 8:36
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    $\begingroup$ Cauchy-Schwarz is indeed a special case of Hölder's inequality. Good luck :) $\endgroup$ – b00n heT Jan 2 at 8:37
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This can be proven using Holder's inequality. Let's do it by induction, obviously this is true for $N=1,2$ so I focus on the induction step. Suppose we are given $(X_n)_{n=1}^{N+1}$ and have finite $N+1$'th moments, then \begin{align*} \left\langle \prod_{n=1}^{N+1} |X_n|\right\rangle&\leq \left\langle \prod_{n=1}^N |X_n|^{\frac{N+1}{N}} \right\rangle^{\frac{N}{N+1}}\cdot\langle |X_{N+1}|^{N+1} \rangle^{\frac{1}{N+1}}\\ &\leq \prod_{n=1}^{N+1} \langle |X_n|^{N+1} \rangle^{\frac{1}{N+1}} \cdot \langle|X_{N+1}|^{N+1} \rangle^{\frac{1}{N+1}}\\ \end{align*} where the first inequality is Holder's inequality with $p=\frac{N+1}{N}$ and $q=N+1$, the second one is induction hypothesis.

Taking all this to the power $N+1$ you get your desired result.

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