# Distribution of the individual coordinates of a uniform random vector on a high-dimensional sphere

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 $$R$$.

Without loss of generality, $$R=1$$. Let $$Z_1,\ldots,Z_n$$ be iid standard normal random variables (r.v.'s). Then $$$$\sqrt n\, X_1\overset{\text{D}}=\frac{\sqrt n\,Z_1}{\sqrt{Z_1^2+\cdots+Z_n^2}} \overset{\text{D}}= \frac{Z_1+\cdots+Z_n}{\sqrt{Z_1^2+\cdots+Z_n^2}}=:T_1,$$$$ where $$\overset{\text{D}}=$$ denotes the equality in distribution. By the top display on page 20 (you may also want to see the published version), $$$$d_{Ko}(T_1,Z_1)\le d_{Ko}(T,Z_1)+\frac{0.24}n,$$$$ where $$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 $$n\ge 5$$ $$$$d_{Ko}(T,Z_1)<\frac{0.16}{n-1},$$$$ so that $$$$\sup_{x\in\mathbb R}|P(\sqrt n\,X_1\le x)-\Phi(x)| =d_{Ko}(T_1,Z_1)\le\frac{0.24}n+\frac{0.16}{n-1}\sim\frac{0.4}n.$$$$

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.

• OK great. This is better than my imagined $n^{-1/2}$ rate. The references are even a bigger treasure. Thanks! – dohmatob Nov 13 '18 at 14:27
• Thanks. I have added refs. to the published versions of the papers. – Iosif Pinelis Nov 13 '18 at 14:29
• 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. – dohmatob Nov 13 '18 at 14:36
• 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. – Iosif Pinelis Nov 13 '18 at 14:42
• 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. – dohmatob Nov 13 '18 at 14:54

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).

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). $$\quad\quad\Box$$