Let $X=(X_1,\ldots,X_n)$ be a random vector uniformly distributed on the $n$-dimensional sphere of radius $R > 0$. Intuitively, i think that for large $p$ every coordinate $X_i$ is normally distributed with variance $R^2/n$, but I'm not quite sure.
Question
More formaly, if $\Phi$ is the CDF of the standard Guassian $\mathcal N(0, 1)$, what is a good upper bound for the quantity $\alpha_n := \sup_{z \in \mathbb R}|P(X_1 \le nR^{-2}z) – \Phi(z)|$ ?
Observations
My wild guess is that $\alpha_n \le Cn^{-1/2}$ for some absolute constant $C$ independent of $n$ and $R$.
Best Answer
Without loss of generality, $R=1$. Let $Z_1,\ldots,Z_n$ be iid standard normal random variables (r.v.'s). Then \begin{equation} \sqrt n\, X_1\overset{\text{D}}=\frac{\sqrt n\,Z_1}{\sqrt{Z_1^2+\cdots+Z_n^2}} \overset{\text{D}}= \frac{Z_1+\cdots+Z_n}{\sqrt{Z_1^2+\cdots+Z_n^2}}=:T_1, \end{equation} where $\overset{\text{D}}=$ denotes the equality in distribution. By the top display on page 20 (you may also want to see the published version), \begin{equation} d_{Ko}(T_1,Z_1)\le d_{Ko}(T,Z_1)+\frac{0.24}n, \end{equation} where $d_{Ko}(X,Y):=\sup_{x\in\mathbb R}|P(X\le x)-P(Y\le x)|$ is the Kolmogorov distance between r.v.'s $X,Y$, and $T$ is a r.v. with the Student distribution $t_{n-1}$ with $n-1$ degrees of freedom.
By Theorem 1.2 (you may also want to see the published version), for $n\ge 5$ \begin{equation} d_{Ko}(T,Z_1)<\frac{0.16}{n-1}, \end{equation} so that \begin{equation} \sup_{x\in\mathbb R}|P(\sqrt n\,X_1\le x)-\Phi(x)| =d_{Ko}(T_1,Z_1)\le\frac{0.24}n+\frac{0.16}{n-1}\sim\frac{0.4}n. \end{equation}
I think the latter constant factor $0.4$ can be improved to about $0.16$ by using directly the method of proof of Theorem 1.2.