I was given a problem to make a guess about the expected values of "l2-norm square" of random vectors of size n which are drawn from a uniform distribution of [-1,1] and standard normal distribution. One of the things I noticed is, since the variance of standard normal distribution > variance of uniform distribution [-1,1] (var=1/3), we have more spread in the data drawn from normal distribution most of the time compared to uniform. So the value of norms will be larger in case of the vectors when drawn from standard normal case compared to uniform as we square the result. Correct me if I am wrong and please let me know if there is anything to decipher? I am happy to know further are there anything interesting observation hidden behind this problem.
2026-03-26 20:40:45.1774557645
L2 norm of vector samples drawn from uniform distribution [-1,1] and standard normal distribution
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By linearity of expectation, $\mathbb{E}(||x||_2^2) = \sum_i \mathbb{E}(x_i^2)$ and $\mathbb{E}(x_i^2) = \operatorname{Var}(x_i)$ if $\mathbb{E}(x_i) = 0$, which is the case for both the standard normal and the uniform distribution supported on $[-1, 1]$. If $x_1, \dots, x_n$ are iid with mean $0$ and variance $\sigma^2$ then $\mathbb{E}(||x||_2^2) = n \sigma^2$.