I have two points $x_1,x_2 \in \mathbb S^n $ which are distant $d$ from each other, where $d<<1$. I also have a vector $v$ sampled uniformly at random from $\mathbb S^n$.

What is the probability that $x_1$ and $x_2$ lie on different sides of the hyperplane perpendicular to $v$?

Thank you!


Let us change notation somewhat: Let $x=(x_1,\dots,x_n)$ and $y=(y_1,\dots,y_n)$ be points in $\mathbb R^n$ such that $|x|=|y|=1$ and $|x-y|=d\in(0,1)$, where $|\cdot|$ is the Euclidean norm. Let $v$ be a random vector uniformly distributed on the unit sphere $S^{n-1}$ in $\mathbb R^n$. The probability in question is \begin{equation} p:=P(x\cdot v<0<y\cdot v)+P(y\cdot v<0<x\cdot v)=2P(x\cdot v<0<y\cdot v), \end{equation} where $\cdot$ is the dot product.

The key note is that the random vector $v$ equals in distribution the random vector $(Z_1,\dots,Z_n)/\sqrt{\sum_1^n Z_i^2}$, where $Z_1,\dots,Z_n$ are independent standard normal random variables (r.v.'s). Hence, \begin{equation} p=2P(X<0<Y), \end{equation} where $X:=\sum_1^n x_i Z_i$ and $Y:=\sum_1^n y_i Z_i$. The r.v.'s $X$ and $Y$ are jointly normal with zero means, unit variances, and correlation \begin{equation} r=EXY=x\cdot y=\tfrac12\,(|x|^2+|y|^2-|x-y|^2)=1-d^2/2. \end{equation} So, the pair $(X,Y)$ equals $(X,rX+\sqrt{1-r^2}\,Z)$ in distribution, where $Z$ is a standard normal r.v. independent of the standard normal r.v. $X$.

So, \begin{equation} p=2P(X<0<rX+\sqrt{1-r^2}\,Z)=2P(X<0,Z>-kX)=2P\big((X,Z)\in A\big), \end{equation} where \begin{equation} k:=r/\sqrt{1-r^2} \end{equation} and $A$ is the angle between the rays $\{(x,0)\colon x\le0\}$ and $\{(x,-kx)\colon x\le0\}$, emanating from the origin. Since the distribution of the random vector $(X,Z)$ in $\mathbb R^2$ is rotation invariant, we conclude that the probability in question is \begin{equation} p=\frac\theta{\pi}, \end{equation} where \begin{equation} \theta:=\text{arccot}\,k=\arccos r=\arccos(1-d^2/2) \end{equation} is the measure of the angle $A$.

In particular, it follows that $p=2d/\pi+O(d^3)\sim 2d/\pi$ as $d\downarrow0$, which agrees with the intuition.

  • $\begingroup$ Thank you very much! Just one more question: why does it not depend on the dimension $n$? Intuitively, the bigger the dimension is, the smaller the probability should be, or am I wrong? $\endgroup$ – Alfred Feb 21 at 9:03
  • $\begingroup$ I don't see an intuition (or an argument) that the probability should decrease (or increase) with the dimension $n$. The best and shortest explanation that I have at the moment of why the probability does not depend on $n$ is in the first half of this answer: that the joint distribution of the signs of $x\cdot v$ and $y\cdot v$ does not depend on $n$ (whereas the joint distribution of $x\cdot v$ and $y\cdot v$ themselves likely does depend on $n$). So, the problem is essentially two dimensional, and the answer depends only on the distance between the points $x,y$ on the unit sphere. $\endgroup$ – Iosif Pinelis Feb 21 at 13:00

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