Let $\alpha\in(0,1),d\in\mathbb N^+$ and $X,Y\in\mathbb S^d$ be uniform, what is $\Pr[\lVert X-Y\cdot\sqrt{1-\alpha} \rVert^2\le \alpha]$?

Question

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.

Answer 1

$\newcommand\al\alpha$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 $a,b$.