Let $X=(X\_1,\ldots,X\_n)$, where $X\_i \sim P\_{p\_i}(0,\frac{1}{\lambda})$ are iid, $P\_{p\_i}$ is sub gaussian distribution for $i^\text{th}$ element, and 0 and $1/\lambda$ are mean and variance.


I'm looking for a result on the concentration of $\|X\|\_2^2$ something of the form $E \|X\|\_2^2 \leq c$ with $P(\|X\|\_2^2 \geq c+\epsilon)\leq f(\epsilon)$. What happens when all the distribution are normals?


I have asked a similar question on <https://math.stackexchange.com/questions/3798100/concentration-of-l2-norm-of-a-vector-sampled-from-a-distribution>



Concentration of $\ell_2$ norm of a vector sampled from a distribution
WLOG, let $\lambda = 1$ (rescale your problem appropriately, if necessary). Then, it is well-known consequence of [Bernstein's inequality](https://en.wikipedia.org/wiki/Bernstein_inequalities_(probability_theory)) (e.g see theorem 3.1.1 of "High-dimensional Probability" book by R. Vershynin) that


$$
\mathbb P\left(\left|\frac{\|X\|^2}{n}-1\right| \le \epsilon\right) \ge 1 - 2e^{-Cn\min(\epsilon,\epsilon^2)},\;\forall \epsilon \ge 0.
$$


Here, $C$ is a constant which is independent of $n$. In other words, $\|X\|^2$ has good (exponential / Gaussian) concentration around the value $n$.


