Probability that $n$ randomly chosen spherical belt points are linearly independent Let $S = \{x : x \in \mathbb{R}^d, \|x\| = 1, |x_1| \leq k\}$ be a spherical belt defined by some constant $k \in (0, 1)$ that bounds the size of the first coordinate. Let $y_1, \dots, y_n$ be $n$ points randomly sampled from the uniform distribution over $S$. What is the probability that $y_1, \dots, y_n$ are linearly independent?
Context: I'm interested in understanding for which conditional distributions over the unit sphere can we say that $d$ random samples are w.h.p linearly independent. Belt seems like a natural conditional distribution to think about (conditioning on the value of the first coordinate).
 A: The probability that $\ y_1, y_2,\dots, y_n\ $ are linearly independent is $0$ if $\ n>d $ or $1$ if $\ n\le d\ $. The first assertion is trivial, since any $\ d+1\ $ or more vectors in $\ \mathbb{R}^d\ $ are always linearly dependent.
If $\ n\le d\ $, let $\ V_i\ $ be the vector space spanned by $\ \big\{y_1,y_2,\dots,y_n\big\}\setminus$$\big\{y_i\big\}\ $.  Then $\ y_1,y_2,\dots,y_n\ $ are linearly dependent if and only if $\ y_i\in V_i\ $ for some $\ i\ $.  Therefore
$$
\big\{\,y_1,y_2,\dots,y_n\ \text{ linearly dependent}\ \big\}=\bigcup_{i=1}^n\big\{\,y_i\in V_i\big\}\ .
$$
However, given $\ y_1,y_2, \dots,y_{i-1},y_{i+1},\dots,y_n\ $, $\ V_i\ $ is a linear space of dimension at most $\ n-1\le d-1\ $, intersecting the $\  d-1\ $ dimensional manifold $\ S\ $ in a submanifold of dimension at most $\ d-2\ $, and $\ y_i\ $ being uniformly distributed over $\ S\ $, and independent of $\ y_1,y_2,$$ \dots,$$y_{i-1},$$y_{i+1},$$\dots,y_n\ $, it therefore has probability zero of lying in $\ V_i\ $.  That is
$$
\mathbb{P}\big(y_i\in V_i\,\big|\,y_1,y_2, \dots,y_{i-1},y_{i+1},\dots,y_n\,\big)=0\ ,
$$
and taking expectations gives $\ \mathbb{P}\big(y_i\in V_i\,\big)=0\ $. Therefore,
\begin{align}
\mathbb{P}\big(\,y_1,y_2,\dots,y_n\ \text{ linearly dependent}\big)&=\mathbb{P}\Big(\bigcup_{i=1}^n\big\{\,y_i\in V_i\big\}\Big)\\
&\le\sum_{i=1}^n\mathbb{P}\big(y_i\in V_i\,\big)\\
&=0\ .
\end{align}
Therefore $\ \mathbb{P}\big(\,y_1,y_2,\dots,y_n\ \text{ linearly independent}\big)=1\ $
