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 $||X||_{\psi_2}<C$ 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 $||\langle X, (1,0,0,0,...)\rangle||_{\psi_2}=||X_1||_{\psi_2}=||R\cos\varphi_1||_{\psi_2}<C$. Then it is not clear to me how to proceed

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

1 Answer 1


$\newcommand\Ga\Gamma$ 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.

  • $\begingroup$ 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 $O(1/\sqrt{n})$. $\endgroup$
    – Daniel Li
    Jun 15, 2020 at 14:31
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    $\begingroup$ @DanielLi : That was a mistake, which is now corrected. Thank you for your comment. $\endgroup$ Jun 15, 2020 at 18:11
  • $\begingroup$ OK, sure. I read your answer in a hurry. Please disregard my comment (deleted). $\endgroup$
    – dohmatob
    Jul 5, 2020 at 9:05
  • $\begingroup$ This maybe a dumb question, but isn't the uniform distribution on the ball bounded? $\endgroup$
    – user135520
    Feb 5, 2022 at 21:52
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    $\begingroup$ @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. $\endgroup$ Feb 6, 2022 at 2:43

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