The norm of isotropic sub-Gaussian random vector may not be sub-Gaussian 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!
 A: 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 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.  
A: 
It is said that $\|X\|_2-\sqrt n$ may not be sub-Gaussian. 

Who says that? It is sub-Gaussian.
Let $a_i = \|\langle X,e_i\rangle\|_{\psi_2}$, $i=1,\dots,n$. Then for $a > \sqrt{n}\max_{1\le i\le n} a_i$ we have
$$
\mathrm E[e^{\|X\|^2_2/a^2}] = \mathrm E\biggl[\prod_{i=1}^n e^{\langle X,e_i\rangle^2_2/a^2}\biggr] \le \biggl(\prod_{i=1}^n \mathrm E [e^{n\langle X,e_i\rangle^2_2/a^2}] \biggr)^{1/n}\le \biggl(\prod_{i=1}^n \mathrm E [e^{\langle X,e_i\rangle^2_2/a_i^2}] \biggr)^{1/n}  \le 2,
$$
whence 
$$
\bigl\|\|X\|_2\bigr\|_{\psi_2}\le \sqrt{n}\max_{1\le i\le n} \|\langle X,e_i\rangle\|_{\psi_2}.
$$
