Suppose that $X,Y$ are independent random $d$-dimensional vectors each uniformly distributed on the unit sphere, and let $Z=Y\cdot\sqrt{1-\alpha}$ be a uniformly selected vector on a slightly smaller sphere.


I'm interesting in calculating (or lower-bounding) the probability that $X$ and $Z$ are far from each other, namely, what is:
$$
\Pr[\lVert X-Z \rVert^2\le \alpha]
$$
as a function of $d,\alpha$?


This quantity appears to play a key role in a vector quantization paper I'm trying to understand (<https://arxiv.org/pdf/2010.03246.pdf>, Section 4.3), which they unfortunately didn't analyze.



Let $\alpha\in(0,1),d\in\mathbb N^+$ and $X,Y\in\mathbb S^d$ be uniform, what is $\Pr[\lVert X-Y\cdot\sqrt{1-\alpha} \rVert^2\le \alpha]$?
$\newcommand\al\alpha$By spherical symmetry, the conditional distribution of $\|X-Y\,\sqrt{1-\al}\|$ given $Y$ does not depend on $Y$. So, letting $e\_1:=(1,0,\dots,0)$ and writing $X=(X\_1,\dots,X\_d)$, we have
$$\begin{aligned}
&P(\|X-Y\,\sqrt{1-\al}\|^2\le\al) \\ &=P(\|X-e\_1\,\sqrt{1-\al}\|^2\le\al) \\ 
&=P((X\_1-\sqrt{1-\al})^2+X\_2^2+\cdots+X\_d^2\le\al) \\ 
&=P((X\_1-\sqrt{1-\al})^2+1-X\_1^2\le\al) \\ 
&=P(X\_1\ge\sqrt{1-\al}) \\
&=\tfrac12\,P(X\_1^2\ge1-\al).
\end{aligned}$$
Noting that $X\_1^2$ has the beta distribution with parameters $1/2,(d-1)/2$, we conclude that
$$\begin{aligned}
P(\|X-Y\,\sqrt{1-\al}\|^2\le\al)
&=\tfrac12\,(1-F\_{1/2,\,(d-1)/2}(1-\al)) \\ 
&=\tfrac12\,F\_{(d-1)/2,\,1/2}(\al),
\end{aligned}$$
where $F\_{a,b}$ is the cdf of the beta distribution with parameters $a,b$.


