Uniform distribution in Ball with radius $\sqrt{n}$ is sub-gaussian

Question

I have to show that a random vector $X$ who ist uniformly distributed on the Ball with Radius $\sqrt{n}$ is sub-gaussian with $$\lVert X \rVert_{\psi_2}\leq C$$ I already know that the same result does hold for a random vector on the sphere with radius $\sqrt{n}$ (1). I tried to show that $r Y$ is uniformly distributed on the Ball with Radius 1 for r uniformly distributed in $[0,1]$ and $Y$ uniformly distributed on the sphere with radius 1. Then I could use this to prove the claim in a similar way (1) is proved. However I am not sure if this is the right way?

Answer 1

Let $X$ be uniformly distributed on the ball $B_{\sqrt n}$ of radius $\sqrt n$ in $\mathbb R^n$. Then $$X=RY,$$ where $R:=|X|/\sqrt n$ and $Y:=\sqrt n\,X/|X|$ is uniformly distributed on the sphere $S_{\sqrt n}$ of radius $\sqrt n$ in $\mathbb R^n$.

Note that $0\le R\le1$ and hence $E\exp\{c(X\cdot t)^2\}=E\exp\{cR^2(Y\cdot t)^2\}\le E\exp\{c(Y\cdot t)^2\}$ for all real $c>0$ and $t\in\mathbb R^n$. So, $\|X\|_{\psi_2}\le\|Y\|_{\psi_2}$.

Also, you know that $\|Y\|_{\psi_2}\le C$. Thus, $$\|X\|_{\psi_2}\le C,$$ as desired.