4
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Can you cite the source of your claim? My guess is that $\text{Var}\|X\|$ will grow with $n$. $\endgroup$ – nemo Nov 30 '18 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 at 4:13
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.