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Recently I was studying something about random matrix theory, and class of sub-guassian / sub-exponential random variables is of interest. In the literature it gave an inequality as following:

$\sup_{p\geq 1} \frac{\|X^2\|_p}{p} \leq 2\sup_{p\geq 1} (\frac{\|X\|_p}{\sqrt{p}})^2$

which gives a sufficient condition such that:

$\|X\|_{2p} \leq \sqrt{2}\|X\|_p$

The proof of this inequality is not provided in the literature. Suppose the random variable $X\in\mathbb{L}^\infty(\mathbb{R})$ therefore the previous defined sub-gaussian norm and sub-exponential norm both exist, then how do we proof the sufficient condition?

Any comment is greatly appreciated :)

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1 Answer 1

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This is a direct consequence of the general fact that $\|X^2\|_p=E[|X|^{2p}]^{1/p}=\left(\|X\|_{2p}\right)^2$ for every $p\gt0$ and every random variable $X$ in $L^{2p}$. Hence, $$ \sup_{p\geqslant 1} \frac{\|X^2\|_p}{p}=2\,\sup_{p\geqslant 2} \left(\frac{\|X\|_p}{\sqrt{p}}\right)^2\leqslant 2\,\sup_{p\geqslant 1} \left(\frac{\|X\|_p}{\sqrt{p}}\right)^2. $$ The rôle of the (not always true) inequality $\|X\|_{2p} \leqslant \sqrt{2}\,\|X\|_p$ is not clear.

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Yeh...This actually come from this neat consequence. Thanks a lot!! –  chooingbobo May 17 '13 at 20:43

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