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!



Show the coordinate distribution has a very large sub-gaussian norm
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. 


