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!
Best Answer
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.