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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 $R$.

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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.

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  • $\begingroup$ OK great. This is better than my imagined $n^{-1/2}$ rate. The references are even a bigger treasure. Thanks! $\endgroup$ – dohmatob Nov 13 '18 at 14:27
  • $\begingroup$ Thanks. I have added refs. to the published versions of the papers. $\endgroup$ – Iosif Pinelis Nov 13 '18 at 14:29
  • $\begingroup$ Great. Would you mind throwing in 1 or two details hinting the "hidden" computation "$d_{Ko}(T1, T) \le 0.24/n$" ? I guess this follows from your delta-method, but a sentence saying what's going on (e.g "one can take the function $f=...$", etc.) might be really useful. $\endgroup$ – dohmatob Nov 13 '18 at 14:36
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    $\begingroup$ No, the inequality $d_{Ko}(T_1,Z_1)\le d_{Ko}(T,Z_1)+\frac{0.24}n$ does not use the delta method results at all. Rather, it follows immediately from elementary formula (4.24) in the delta-method paper, since the cdf's of $T_1$ and $T$ are easy to express in terms of each other. $\endgroup$ – Iosif Pinelis Nov 13 '18 at 14:42
  • $\begingroup$ Off-topic: I wonder if you would mind helping on this mathoverflow.net/questions/314409/… or this math.stackexchange.com/questions/2976654/…. Thanks in advance. $\endgroup$ – dohmatob Nov 13 '18 at 14:54
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We may assume $R=1$. A useful trick is to realize the uniform measure on the unit sphere as the distribution of $$ \left(\frac{G_1}{|G|},\dots,\frac{G_n}{|G|} \right), $$ where $G=(G_1,\dots,G_n)$ is a Gaussian vector with independant $N(0,1)$ coordinates, and $|G|=\sqrt{G_1^2+\cdots+G_n^2}$. With this in hand you can now write $$ P(X_1 \leq \frac{z}{\sqrt{n}}) = P(G_1 \leq \frac{|G|}{\sqrt{n}} z) \approx P(G_1 \leq z) ,$$ where in the last step you have to argue that $|G|$ concentrates around $\sqrt{n}$ with fluctuations $O(1)$ (google "chi-squared concentration").

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