Asymptotic scaling of mean and variance for the norm of random vector (non gaussian components)

Question

The norm of a vector whoose components $X_i$ are normally distributed follows the Non-central chi distribution and it can be shown that, increasing the number of components $k$ (i.e. the dimension of the vector) the mean of the norm scale as $M= O(k^{\frac{1}{2}})$ while its variance scale as $V= O(1) $ (see for example here).

Now let's assume generic non Gaussian independent r.v. $X_i$, such that

• $\mu_i, \, \sigma_i < \infty \, \forall i$
• $\lim_{k \rightarrow \infty} \frac{1}{k} \sum_i^k \mu_i = \mu$, $\lim_{k \rightarrow \infty} \frac{1}{k} \sum_i^k \sigma^2_i = \sigma^2$
• $X_i$ follows the Central limit theorem $\forall i$
• $X_i$ has fast decaying tails, i.e. $P_X(X_i=x) = o(x^{-n}) \, \forall n \in \mathbb{N}$ for $x \rightarrow \pm \infty$

How do the mean and variance of the norm distribution scale with $k$ in the limit as $k \rightarrow \infty$ ?

Can we extend the result presented here?

Answer 1

The first two moments of $X_i$ are given, $\mathbb{E}[X_i]=\mu_i$, $\mathbb{E}[X_i^2]=\sigma_i^2+\mu_i^2$. We also need the fourth moment, $\mathbb{E}[X_i^4]=\tau_i^4$.

Then with $Y =\sqrt{\sum_{i=1}^k(X_i/\sigma_i)^2}$ we have, in view of the independence of the $X_i$'s, $$\mathbb{E}[Y^2]=M=\sum_{i=1}^k\bigl[1+(\mu_i/\sigma_i)^2\bigr],$$ $$\text{var}\,[Y^2]=V=\sum_{i=1}^k \bigl[(\tau_i/\sigma_i)^4-\bigl(1+(\mu_i/\sigma_i)^2\bigr)^2\bigr].$$ Both $M$ and $V$ are assumed to scale linearly with $k$. For large $k$ the distribution of $Y^2$ tends to a Gaussian with mean $M$ and variance $V$. We now follow the same steps as in https://mathoverflow.net/a/453690/11260 : to leading order in $1/k$ one has $$\mathbb{E}[Y] =M^{1/2}+{\cal O}(k^{-1/2}),$$ $$\text{var}\,[Y]=\frac{V}{4 M}+{\cal O}(k^{-1}).$$ So the mean of $Y$ scales as $\sqrt{k}$ while the variance tends to a constant in the limit $k\rightarrow\infty$.