Let x=<x1,x2,...,xd>x = <x_1, x_2, ..., x_d> be an dd dimension random unit vector, i.e., a vector generated uniformly at random from the sphere of the unit ball. What is the probability that ∣x1∣≥1n|x_1| \ge \frac{1}{n} for some given n≥2n \ge 2?

When d=1d = 1, then x1=1x_1 = 1 or x1=−1x_1 = -1. Therefore P[∣x1∣≥1n]=1P[|x_1| \ge \frac{1}{n}] = 1.

When d=2d = 2,

P[∣x1∣≥1n]=42π∫1/n11−t2dt=2πarccos⁡1n P[|x_1| \ge \frac{1}{n}] = \frac{4}{2 \pi} \int_{1/n}^1 \sqrt{1 - t^2} d_t = \frac{2}{\pi} \arccos \frac{1}{n}

By

arcsin⁡(z)=z+(12)z33+(1⋅32⋅4)z55+(1⋅3⋅52⋅4⋅6)z77+⋯=∑n=0∞(2n−1)!!(2n)!!⋅z2n+12n+1=∑n=0∞(2nn)z2n+14n(2n+1)∣z∣≤1 \begin{aligned} \arcsin(z) &= z + \left({\frac {1}{2} } \right) {\frac {z^{3} } {3} } +\left({\frac {1\cdot 3}{2\cdot 4} } \right){\frac {z^{5} } {5} } +\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6} } \right){\frac {z^{7} } {7} } +\cdots \\\\ &=\sum _{n=0}^{\infty }{\frac {(2n-1)!!}{(2n)!!} } \cdot {\frac {z^{2n+1} } {2n+1} } \\\\ &=\sum _{n=0}^{\infty }{\frac { {\binom {2n}{n} } z^{2n+1} } {4^{n}(2n+1)} } \qquad |z|\leq 1 \end{aligned}

We get

arccos⁡1n=π2−arcsin⁡1n=π2−(1n+(12)1n33+(1⋅32⋅4)1n55+(1⋅3⋅52⋅4⋅6)1n77+⋯ )≥π2−1n−12n3(∑i=0∞1/2i)since n≥2≥π2−54n \begin{aligned} \arccos \frac{1}{n} &= \frac{\pi}{2} - \arcsin \frac{1}{n} \\\\ &= \frac{\pi}{2} - (\frac{1}{n} + \left({\frac {1}{2} } \right) {\frac {\frac{1}{n}^{3} } {3} } +\left({\frac {1\cdot 3}{2\cdot 4} } \right){\frac {\frac{1}{n}^{5} } {5} } +\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6} } \right){\frac {\frac{1}{n}^{7} } {7} } +\cdots ) \\\\ &\ge \frac{\pi}{2} - \frac{1}{n} - \frac{1}{2 n^3}\left( \sum_{i = 0}^\infty 1 / 2^i \right) \qquad since \ n \ge 2 \\\\ &\ge \frac{\pi}{2} - \frac{5}{4n} \end{aligned}

It follows that

P[∣x1∣≥1n]=2πarccos⁡1n≥1−1n P[|x_1| \ge \frac{1}{n}] = \frac{2}{\pi} \arccos \frac{1}{n} \ge 1 - \frac{1}{n}

For general values of dd, we would like to give an easier estimation by upper bounding the following integration

∫x1=01/n∫x2=01∫x3=01...∫xd=01⏟1 dxddxd−1...dx1≤1n∫x2=01∫x3=01...∫xd=01⏟1 dxddxd−1...dx2x22+x32+...+xd2≤1−x12x22+x32+...+xd2=1 \begin{aligned} \begin{matrix} \int_{x_1 = 0}^{1 / n} & \underbrace{\int_{x_2 = 0}^{1} \int_{x_3 = 0}^{1} ... \int_{x_d = 0}^{1} } & 1 \ d_{x_d} d_{x_{d - 1} } ... d_{x_1} \le \frac{1}{n} &\underbrace{\int_{x_2 = 0}^{1} \int_{x_3 = 0}^{1} ... \int_{x_d = 0}^{1} } & 1 \ d_{x_d} d_{x_{d - 1} } ... d_{x_2} \\\\ & x_2^2 + x_3^2 + ... + x_d^2 \le 1 - x_1^2 & & x_2^2 + x_3^2 + ... + x_d^2 = 1 & \end{matrix} \end{aligned}

Using the result that the surface area of a d−1d - 1 dimension sphere is given by

2πd2Γ(d2) {\frac {2\pi ^{\frac {d}{2} } }{\Gamma \left({\frac {d}{2} } \right)} }

We get

P[∣x1∣≥1n]=1−1n2πd2Γ(d2)2πd+12Γ(d+12)=1−1nΓ(d+12)Γ(d2)π12≥1−d2πen \begin{aligned} P[|x_1| \ge \frac{1}{n}] &= 1 - \frac{1}{n} \frac{ {\frac {2\pi ^{\frac {d}{2} } }{\Gamma \left({\frac {d}{2} } \right)} } }{ {\frac {2\pi ^{\frac {d + 1}{2} } }{\Gamma \left({\frac {d + 1}{2} } \right)} } } \\ &= 1 - \frac{1}{n} \frac{ \Gamma \left({\frac {d + 1}{2} } \right)}{ \Gamma \left({\frac {d}{2} } \right) \pi^{\frac {1}{2} } } \\ &\ge 1 - \sqrt{ \frac{d}{2\pi} } \frac{e}{n} \end{aligned}

where by Stirling’s formula

Γ(d+12)Γ(d2)≤(d+12)d+12−1/2e1/(12(d+12))/(d2)d2−1/2=(1+1d)d/2e1/(6(d+1))(d2)1/2≤e(d2)1/2 \begin{aligned} \frac{ \Gamma \left({\frac {d + 1}{2} } \right)}{ \Gamma \left({\frac {d}{2} } \right)} &\le (\frac{d + 1}{2})^{\frac {d + 1}{2} - 1 / 2} e^{1 / (12 (\frac {d + 1}{2}))} / (\frac{d}{2})^{\frac{d}{2} - 1 / 2} \\ &= (1 + \frac{1}{d})^{d/2} e^{1 / (6(d + 1))} (\frac{d}{2})^{1 / 2} \\ &\le e (\frac{d}{2})^{1 / 2} \end{aligned}