# Variance of the Euclidean norm under finite moment assumptions

Let $$X = (X_1,X_2 \cdots X_n)$$ be random vector in $$R^n$$ with independent coordinate $$X_i$$ that satisfy $$E[X_i^2]=1$$ and $$E[X_i^4] \leq K^4$$. Then show that $$\operatorname{Var}(\| X\|_2) \leq CK^4$$ where $$C$$ is a absolute constant and $$\| \ \|_2$$ denotes euclidian norm.

Here is my attempt:
\begin{align*} E(\|X\|_2^2 -n)^2 &= E[(\sum_{i=1}^n X_i^2)^2 ]-n^2 \\ &=E[\sum_{i=1}^n X_i^4]+E[\sum_{i

since $$E(\|X\|_2^2 -n)^2 \leq nk^4 \rightarrow E\left(\frac{\|X\|_2^2}{n} -1\right)^2 \leq \frac{K^4}{n}$$
and since $$(\forall z \geq 0 \ \ |z-1|\leq |z^2-1|) \rightarrow E(\frac{\|X\|_2}{\sqrt n} -1)^2\leq E(\frac{\|X\|_2^2}{n} -1)^2$$

thus: $$E(\frac{\|X\|_2}{\sqrt n} -1)^2 \leq K^4/n \rightarrow E(\|X\|_2-\sqrt n)^2\leq K^4$$

by Jensen inequality: $$(E[\|X\|_2] - \sqrt n)^2 \leq K^4$$

which is equivalence to $$|E[\|X\|_2] - \sqrt n)| \leq K^2$$

then when I am trying to bound $$Var(\| X\|_2)$$ I meet some problem :

$$\operatorname{Var}(\| X\|_2)=E[\|X\|_2^2] -(E[\|X\|_2])^2 \leq n- (K^2-\sqrt n)^2 \leq -K^4+2K^2\sqrt n$$ which is not bound by constant , how can I bound that?

• Can you cite the source of your claim? My guess is that $\text{Var}\|X\|$ will grow with $n$.
– nemo
Nov 30, 2018 at 9:19
• For example, this is Exercise 3.1.6 in the HDP book by Vershynin. Book's draft is freely available online. Feb 16, 2019 at 4:13

You had most of the steps correct. As you argued correctly, $$E (\|X\|_2 - \sqrt{n})^2 \leq K^4 .$$
Note that the mean minimizes the squared error, i.e., for any $$c \in \mathbb{R}$$, $$Var(X) \leq E(X-c)^2$$. Therefore, $$Var(\|X\|_2) \leq E (\|X\|_2 - \sqrt{n})^2 \leq K^4 .$$