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I have a sequence of $N$ dependent random variables

$$y_i = \frac{x_i}{||\vec x||_2} \quad \mathrm{for} \quad \vec x \sim \mathcal N(0,\mathbb{1}_N),$$

where the $x_i$ are the iid elements of $\vec x$ and $||.||_2$ is the Euclidean norm. Note that $\sum_i y_i^2 = 1$.

For large $N$, by the CLT the distribution for $||\vec x||_2$ is closely approximated by $\mathcal N(\sqrt{N},1/\sqrt{2})$ and the distribution for $y_i$ goes to $\mathcal N(0,1/\sqrt{N})$, although this clearly does not enforce the constraint.

enter image description here

Because this fits pretty well and is independent of the correlation between $x_i$ and $||\vec x||_2$, I thought this meant I could approximate the $y_i$ as if they were independent as $N \to \infty$. However, when I try to use the iid approximation, $y_i \sim \mathcal N(0,1/\sqrt{N})$, and the CLT to compute things like $\sum_i^N y_i^4$, I get a value for the variance which is too large. This obviously results from neglecting the constraint, since if I draw the $y_i \sim \mathcal N(0,1/\sqrt{N})$, the fit is good although the two $y_i$ histograms are nearly identical.

enter image description here

My question: Can I get a better approximation for the variance of these sums $\sum_i^N |y_i|^m$ as $N\to \infty$, since the results appear to be normally distributed? In particular, I don't understand why

$$\langle | y|^m \rangle \sim \int | y|^m \mathcal N(0,1/\sqrt{N}) dy $$

gives a bad approximation to the moments when the distributions for $y_i$ are very closely approximated by $\mathcal N(0,1/\sqrt{N})$.

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  • $\begingroup$ Just a remark: the CLT doesn't apply here. $\|x\|_2$ will not be normally distributed but rather $\chi$-distributed. $\endgroup$
    – lemon
    Aug 20, 2014 at 1:57
  • $\begingroup$ @lemon, You're right that $||x||$ will be $\chi$-distributed, but this is almost-Gaussian for large $N$. Then the ratio distribution for Gaussians gives $\mathcal N(0,1/\sqrt{N})$. $\endgroup$
    – user162190
    Aug 20, 2014 at 2:03

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