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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 $O(1)$ (a concenquence of tail standard tail bounds on chi-squared distribution).

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Here is my solution without the reduction trick to $1$D gaussian.


Let $U := X/\|X\|$. Since $U$ is uniformly distributed on the unit $n$-sphere, it follows that the random variable $U^Tz$ has the same distribution as $U_1$ (the first coordinate of the random vector $U$), which in turn (by the Archimedean projection property) has the same distribution as the first coordinate of a point draw uniformly in the unit ball in $\mathbb R^{n-1}$. Thus, $P(U_1 > \delta)$ is the probability that a random point in the unit ball in $\mathbb R^{n-2}$ lies in on given side of an equatorial hyperplane, we have

$$ \begin{split} P(|U^Tz| > \delta) &= P(|U_1| > \delta)= 2P(U_1 > \delta) = 1-I\left(\delta;\frac{1}{2}, \frac{n-1}{2}\right)\\ &= I\left(1-\delta;\frac{n-1}{2},\frac{1}{2}\right), \end{split} \tag{2} $$

where $I(t; a, b)$ is the normalized incomplete beta function, defined by $I_t(t; a, b) := B(t;a,b) / B(1; a, b)$, with $B(t; a, b):= \int_{0}^t s^{a-1}(1-s)^{b-1}ds$.

Theorem ($U^Tz$ is sub-exponential! ). Let $U$ be uniformly distributed on the unit $n$-sphere and let $z$ be a fixed vector on this sphere. If $n$ is large enough, then for every $\delta \in [0, 1]$, it holds that $$ P(|U^Tz| > \delta) \le e^{-\frac{n-1}{4}\delta}. \tag{3} $$

Proof. Let $p = I(1-\delta; 1/2, (n-1)/2)$. It is known since Temme (1992) that for $p \in (0, 1)$ and large $a > 0$, the solution of the equation $p = I(t; a,b)$ is given (approximately) by

$$ t=t_p(a, b) \approx e^{-(1/a)Q_{1-p}(\Gamma(b,1))}, \tag{4} $$

where $Q_{1-p}(\Gamma(b,1))$ is the $1-p$ quantile of the unit-scale gamma distribution with shape parameter $b$. Now by standard concentration results (e.g see Boucheron et al. textbook),

$$ Q_{1-p}(\Gamma(b,1)) \le \log(1/p) + \sqrt{2b\log(1/p)}. \tag{5} $$

In particular, for $a=(n-1)/2$ and $b=1/2$ we get

$$ Q_{1-p}(\Gamma(1/2,1)) \le \log(1/p) + \sqrt{\log(1/p)} \le 2\log(1/p). \tag{6} $$

Putting (2), (4), and (6) together and using the basic inequality $e^{-t} \ge 1-t\;\forall t > -1$, we see that $$ \begin{split} 1-\delta &\ge t_{2p}\left((n-1)/2,1/2\right) \ge e^{-\frac{2Q_{1-2p}(\Gamma(1/2,1))}{n-1}} \ge e^{-\frac{2}{n-1}\left(\log\left(\frac{1}{2p}\right) + \sqrt{\log\left(\frac{1}{2p}\right)}\right)}\\ & \ge 1 - \frac{2\left(\log\left(\frac{1}{2p}\right) + \sqrt{\log\left(\frac{1}{2p}\right)}\right)}{n-1} \ge 1-\frac{4\log\left(\frac{1}{2p}\right)}{n-1}, \end{split} $$

from which (3) follows upon combining with (2). $\quad\quad\Box$

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