Consider a random vector $X\sim\textrm{Unif}\left[B\left(0,\sqrt{n}\right)\right]$, that is $X$ is uniformly distributed on the Euclidean ball $B(0,\sqrt{n})$ in $\mathbb{R}^n$ centered at the origin with radius $\sqrt{n}$. $$B\left(0,\sqrt{n}\right)=\left\{x\in\mathbb{R}^n:x_1^2+x_2^2+...+x_n^2\leq n\right\}.$$ Show that $X$ is sub-Gaussian and $\|X\|_{\Psi_2}\leq C$, where $C$ is an absolute constant.
From the book "High dimensional probability" by Vershynin, it said that the above argument is an extension of uniform distribution on the sphere in section 3.4.3. While I am not sure how to connect the argument that uniform distribution on the sphere is sub-Gaussian to this? Thank you!
Consider the following way to generate $X$: first, generate $Y$ uniformly distributed on the sphere centered at the origin with radius $\sqrt{n}$, and an independent uniform random variable $U\sim\mathrm{U}_{[0,1]}$. Then set $X=U^{1/n}Y$.
Then you can use the same proof as the one you mention from the book; by rotational invariance, it suffices to bound the tail probability $$ \mathbb{P}\{|X_1|\geq t\} $$ for all $t\geq 0$. But now, $\mathbb{P}\{|X_1|\geq t\} = \mathbb{P}\{|U^{1/n}Y_1|\geq t\} \leq \mathbb{P}\{|Y_1|\geq t\}$ (since $0\leq U^{1/n}\leq 1$), and since $\mathbb{P}\{|Y_1|\geq t\}$ has the desired subgaussian bound, so has $\mathbb{P}\{|X_1|\geq t\}$.