There is the following exercise in [Vershynin's book on High-Dimensional Probability](https://www.math.uci.edu/~rvershyn/papers/HDP-book/HDP-book.html).


**Exercise 3.1.6:**
Let $X = (X\_1, \dots, X\_n) \in \mathbb{R}^n$ be a random vector with independent coordinates $X\_i$ that satisfy $\mathbb{E}[X\_i^2] = 1$ and $\mathbb{E}[X\_i^4] \leq K^4$. Show that $\mathrm{Var}(\|X\|\_2) \leq CK^4$.


What if we remove the condition that the coordinates are independent? Suppose instead that we have only the condition that the coordinates are uncorrelated (i.e., $X$ has covariance $I$), and we additionally have some bound on the fourth moment (e.g., $\mathbb{E} [ \langle X, v\rangle^4]\leq K^4$ for every unit vector $v$). What is the best bound we can prove on $\mathrm{Var}(\|X\|\_2)$?



Variance of the norm of a random variable under finite-moment assumptions
$\newcommand\R{\mathbb R}
\renewcommand{\P}{\operatorname{\mathsf P}}\newcommand\E{\operatorname{\mathsf{E}}}\newcommand\Var{\operatorname{\mathsf{Var}}}$The best bound on $\Var \|X\|\_2$ is about $n$.


Indeed,
\begin{equation\*}
 \Var \|X\|\_2\le \E \|X\|\_2^2=\sum\_{i=1}^n \E X\_i^2=n. \tag{1}\label{1}
\end{equation\*}


On the other hand, suppose that
\begin{equation\*}
 \mu\_X=q\,\mu\_{aG}+p\,\mu\_{bG},
\end{equation\*}
where $\mu\_Y$ denotes the distribution of a random vector $Y$, $G$ is a standard Gaussian random vector in $\R^n$, and
\begin{equation\*}
 a:=0,\quad b\in[1,\infty),\quad p:=\frac1{b^2},\quad q:=1-p. 
\end{equation\*}
Then the covariance matrix of $X$ is $pb^2I=I$.


Also, for all unit vectors $v\in\R^n$,
\begin{equation\*}
 \E (X\cdot v)^4=pb^4\E (G\cdot v)^4=3b^2=K^4
\end{equation\*}
if $b=K^2/\sqrt3$ (assuming that $K^4\ge3$).


So, all your conditions hold.


However, $\E\|X\|\_2^2=pb^2\E\|G\|\_2^2=n$ and $\E\|X\|\_2=pb\E\|G\|\_2\le pb\sqrt{\E\|G\|\_2^2}=pb\sqrt n=\sqrt n/b$ and hence
\begin{equation\*}
 \Var \|X\|\_2=\E\|X\|\_2^2-(\E\|X\|\_2)^2\ge\Big(1-\frac1{b^2}\Big)n=\Big(1-\frac3{K^4}\Big)n; \tag{2}\label{2}
\end{equation\*}
the inequality $\Var \|X\|\_2\ge\big(1-\frac3{K^4}\big)n$ trivially holds if $K^4\not\ge3$.



> 
> **Conclusion 1:** The trivial upper bound $n$ on $\Var \|X\|\_2$ in \eqref{1} is optimal, in the sense that it cannot be replaced by $cn$ for any real $c<1$, if $K$ is allowed to be large enough.
> 
> 
> 




---


Nonetheless, the upper bound $n$ on $\Var \|X\|\_2$ in \eqref{1} can be "second-order" improved as follows. Note that
\begin{equation\*}
 \E\|X\|\_2^4=\E\Big(\sum\_{i=1}^n X\_i^2\Big)^2=\sum\_{i,j=1}^n \E X\_i^2 X\_j^2
 \le\sum\_{i,j=1}^n \frac{\E X\_i^4+\E X\_j^4}2
 =n\sum\_{i=1}^n \E X\_i^4\le n^2K^4. 
\end{equation\*}
So, by the log-convexity of $\E\|X\|\_2^p$ in $p>0$,
\begin{equation\*}
 (\E\|X\|\_2)^2\ge\frac{(\E\|X\|\_2^2)^3}{\E\|X\|\_2^4}\ge\frac{n^3}{n^2K^4}
 =\frac n{K^4}. 
\end{equation\*}
So,
\begin{equation\*}
 \Var \|X\|\_2=\E\|X\|\_2^2-(\E\|X\|\_2)^2\le\Big(1-\frac1{K^4}\Big)n, \tag{3}\label{3}
\end{equation\*}
which is an improvement of \eqref{1}.



> 
> **Conclusion 2:** In view of \eqref{2} and \eqref{3}, the best upper bound on $\Var \|X\|\_2$ is of the form
> \begin{equation\*}
>  \Big(1-\frac C{K^4}\Big)n 
> \end{equation\*}
> for some $C\in[1,3]$.
> 
> 
> 


This refines Conclusion 1.


