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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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Are $X_i$ independent? –  Davide Giraudo Nov 4 '12 at 18:14
    
The case $p=1$ is the easiest :) –  Hans Engler Nov 4 '12 at 18:43

2 Answers 2

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

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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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