# Uniform distribution in Euclidean ball is sub-gaussian [closed]

Consider $$n-$$dimensional Euclidean ball centred at 0 with radius $$\sqrt{n}$$. We want to show that the uniform distribution $$X$$ in this ball is sub-gaussian and $$||X||_{\psi_2} where $$C$$ is absolute constant.

Clarify: $$X$$ is subgaussian if $$\langle X,x \rangle$$ is subgaussian for any $$x \in \mathbb{R}^n$$ and $$||X||_{\psi_2}=||\sup_x\langle X,x\rangle||_{\psi_2}$$ where sup is over all unit vector $$x$$.

Attempt: Uniform distribution on ball can be represented by $$R,\varphi_1,..,\varphi_{n-1}$$ jointly where $$R$$ is a uniform distribution on $$[0, \sqrt{n}]$$ representing radius, $$\varphi_i$$ representing the angles in spherical coordinates and they are uniform on $$[0,\pi]$$. All these variables are independent.

By symmetry, I only need to show $$||\langle X, (1,0,0,0,...)\rangle||_{\psi_2}=||X_1||_{\psi_2}=||R\cos\varphi_1||_{\psi_2}. Then it is not clear to me how to proceed

• This sounds like a homework question... MO is for research-level questions only, perhaps Math.SE will be better suited. Jun 14, 2020 at 21:35

$$\newcommand\Ga\Gamma$$ For each unit vector $$x$$, the random variable (r.v.) $$\langle X,x\rangle$$ equals $$V:=\sqrt n\,W_nR$$ in distribution, where $$W_n:=\frac{Z_1}{\sqrt{Z_1^2+\dots+Z_n^2}},$$ $$Z_1,\dots,Z_n$$ are iid $$N(0,1)$$ r.v.'s, and $$R$$ is a r.v. (independent of $$V$$ and) such that $$0\le R\le1$$.
So, it suffices to show that for some real $$c>0$$ $$\sup_{n\ge2}Ee^{cnW_n^2}<\infty. \tag{1}$$ Note that $$W_n^2$$ has the beta distribution with parameters $$1/2,(n-1)/2$$. So, for any $$c\in(0,1/2)$$ and $$n\ge3$$ \begin{align} Ee^{cnW_n^2} &=\frac{\Ga(n/2)}{\sqrt\pi\,\Ga((n-1)/2)}\int_0^1 e^{cnw^2}w^{-1/2}(1-w)^{(n-3)/2}\,dw \\ &\le\frac{\Ga(n/2)}{\sqrt\pi\,\Ga((n-1)/2)}\int_0^1 e^{cnw}w^{-1/2}e^{-(n-3)w/2}\,dw \\ &=O(\sqrt n)O(1/\sqrt n)=O(1). \end{align} Also, clearly $$Ee^{cnW_n^2}<\infty$$ for $$n=2$$. Thus, (1) holds, as desired.
• Thank you! But would you mind explaining how in the last step the integral is of order $O(1/n)$. I tried to upper bound it by changing upper limit to infinity and write it as gamma function but this only gives me $O(1/\sqrt{n})$. Jun 15, 2020 at 14:31
• @user135520 : That the expectation under the $\sup$ in (1) is finite is trivial. That the supremum of this expectation over all $n\ge2$ is finite is not so trivial. Feb 6 at 2:43