Show the coordinate distribution has a very large sub-gaussian norm Consider a random vector X with the coordinate distribution is uniformly distributed in the set $\{\sqrt{n}e_i : i = 1,..., n\}$, where $e_i$ denotes the n-element set of the canonical basis vectors in $R^n$. Show that $ \parallel X \parallel_{ \psi 2}\asymp  \sqrt{\frac{n}{{ log n}}}.$
By the definition of the sub-gaussian norm of a random vector, $\parallel X\parallel_{ \psi_2}=\sup_{x∈S^{n−1}}\parallel <X, x>\parallel_{\psi 2} $, for all $x ∈ R^n$.
I tried to consider $\sup_{x∈S^{n−1}}\parallel <X, x>\parallel_{\psi 2}=\sup_{x∈S^{n−1}} \parallel \sum_{i=1}^n x_iX_i\parallel_{\psi_2}$, while I am not sure how to precede next? Thank you!
 A: The subgaussian norm of a real-valued random variable $Y$ is 
$$\|Y\|:=\|Y\|_{\psi_2}:=\inf\{t>0\colon Ee^{Y^2/t^2}\le2\}.$$
If $Y$ is such that $P(Y=0)<1$ and $Ee^{Y^2/t^2}<\infty$ for all real $t>0$, then 
$Ee^{Y^2/t^2}$ continuously decreases in real $t>0$ from $\infty$ to $0$, so that $\|Y\|$ is the unique positive root of the equation
$$Ee^{Y^2/\|Y\|^2}=2.$$
So, for $x=(x_1,\dots,x_n)\in S^{n-1}$ and $\nu_x:=\|X\cdot x\|$, where $X\cdot x:=\langle X,x\rangle$, we have 
$$2=E\exp\frac{(X\cdot x)^2}{\|X\cdot x\|^2}
=\frac1n\,\sum_{i=1}^n\exp\frac{nx_i^2}{\nu_x^2}
\le\frac1n\,\Big(n-1+\exp\frac n{\nu_x^2}\Big), \tag{1}$$
because $\sum_{i=1}^n\exp\frac{nu_i}{\nu_x^2}$ is convex in $u=(u_1,\dots,u_n)$ in the simplex that is the image of $S^{n-1}$ under the map $(x_1,\dots,x_n)\mapsto(x_1^2,\dots,x_n^2)$, and hence $\sum_{i=1}^n\exp\frac{nu_i}{\nu_x^2}$ attains its maximum on this simplex at one of the vertices of this simplex, which latter are the standard basis vectors $e_1,\dots,e_n$ in $\mathbb R^n$. 
It follows from (1) that 
$$\nu_x\le\sqrt{\frac n{\ln(n+1)}}$$
for all $x\in S^{n-1}$, and this upper bound on $\nu_x$ is attained if $x$ is one of the $e_i$'s. Thus, 
$$\|X\|=\max_{x\in S^{n-1}}\nu_x=\sqrt{\frac n{\ln(n+1)}}\sim \sqrt{\frac n{\ln n}}$$
as $n\to\infty$, as desired. 
