Suppose $X$ is a isotropic sub-Gaussian $n$-dimensional random vector (i.e. $EXX^T=I\_n$, and for any unit vector $u$,$\|\left<X,u\right>\|\_{\psi\_2}\le K$). It is said that $\|X\|\_2-\sqrt n$ may not be sub-Gaussian with bounded norm $CK$ which does not depend on $n$. But I havn't found a counter example. 


When $X$ is a uniform ball distribution or a uniform hypercube distribution, it can both be proved that $\|X\|\_2-\sqrt n$ is sub-Gaussian. Moreover, if $X\_i$ are independent, the proposition is also true.


Can someone show a counter example? Thank you!



The norm of isotropic sub-Gaussian random vector may not be sub-Gaussian
Let 
\begin{equation}
 \mu\_X=\tfrac12\,\mu\_{aZ}+\tfrac12\,\mu\_{bZ}, 
\end{equation}
where $\mu\_U$ denotes the probability distribution of a random vector $U$, $Z\sim N(0,I\_n)$, 
and $a,b$ are constants such that  

\begin{equation}
 0<a<1<b\quad\text{and}\quad \tfrac12\,a^2+\tfrac12\,b^2=1. 
\end{equation} 


Then $EXX^T=I\_n$. Also, for any unit vector $u$ and real $s>0$
\begin{equation}
 E\exp\{\left<X,u\right>^2/s^2\}=\frac1{2\sqrt{1-2a^2/s^2}}+\frac1{2\sqrt{1-2b^2/s^2}}<2 
\end{equation}
if $s$ is large enough (depending only on $a,b$), 
so that, by the [definition](https://en.wikipedia.org/wiki/Sub-Gaussian_distribution) of $\|\cdot\|\_{\psi\_2}\|$, we have $\|\left<X,u\right>\|\_{\psi\_2}\le s$. For instance, here we can take $a=1/5,b=7/5,s=3$. 


On the other hand, for
\begin{equation}
 t:=(b-1)\sqrt{n}/2, 
\end{equation}
\begin{multline}
 2\,Ee^{(\|X\|-\sqrt n)^2/t^2}>Ee^{(\|bZ\|-\sqrt n)^2/t^2}
 >Ee^{(\|bZ\|-\sqrt n)^2/t^2}1\_{\|Z\|^2>n} \\ 
 >e^{(b\sqrt n-\sqrt n)^2/t^2}\,P(\|Z\|^2>n)=e^4\,P(\|Z\|^2>n)\to e^4/2>4, 
\end{multline}
because, by the central limit theorem, $P(\|Z\|^2>n)\to1/2$. 
So, for all large enough $n$, 
\begin{equation}
 \|\|X\|-\sqrt n\|\_{\psi\_2}\ge t=(b-1)\sqrt{n}/2\to\infty,
\end{equation}
as desired. 


