# Probability of two Points being divided by an high-Dimensional Hyperplane

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 $$\begin{equation} p:=P(x\cdot v<0 where $$\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 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, $$\begin{equation} p=2P(X<0-kX)=2P\big((X,Z)\in A\big), \end{equation}$$ 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.
• 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? – Alfred Feb 21 at 9:03
• 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. – Iosif Pinelis Feb 21 at 13:00