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Given $X_i\sim \mathcal{N}(0,1)$ what is the behaviour of $ ||X||_{l^p}=(\sum_{i=1}^n|X_i|^p )^{1/p}$ as $n\rightarrow \infty$? For $p=2$ results about $\chi$-distribution tell us that $\mathbb{P}(||X||_{l^2}\le 2n^\frac{1}{2} )\rightarrow 1.$

I am interested in analgous statments for $p\ne1$,i.e.

$\mathbb{P}(||X||_{l^p}\le Cn^{e(p)} ),$ where $C$ is allowed to depend on $p$.

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    The case $p=1$ is the easiest :)2012-11-04

2 Answers 2

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By the law of large numbers for i.i.d. random variables, $ \frac1n\sum_{i=1}^n|X_i|^p\to E[|X_1|^p], $ almost surely, hence $ \frac1{n^{1/p}}\|X\|_p\to E[|X_1|^p]^{1/p}, $ almost surely. In particular, for every positive $x$, $ P[(1-x)n^{1/p}E[|X_1|^p]^{1/p}\leqslant\|X\|_p\leqslant(1+x)n^{1/p}E[|X_1|^p]^{1/p}]\to1. $

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Hint: Assuming $X_i$'s are independent you can apply the Central Limit Theorem on $\frac{1}{n}\left\Vert X\right\Vert_{\ell^p}^p=\frac{1}{n}\sum_{i=1}^n\lvert X_i\rvert ^p.$