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:
$$\begin{align*} E(\|X\|_2^2 -n)^2 &= E[(\sum_{i=1}^n X_i^2)^2 ]-n^2 \\ &=E[\sum_{i=1}^n X_i^4]+E[\sum_{i<j}X_i^2X_j^2] -n^2 \\ &\leq nK^4 + 2{{n}\choose {2}}-n^2 \\ &\leq n(K^4-1) \\ & \leq nk^4 \end{align*}$$

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?

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

1 Answer 1


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, $$Var(\|X\|_2) \leq E (\|X\|_2 - \sqrt{n})^2 \leq K^4 .$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .