5
$\begingroup$

(edited) Consider the unit sphere $\mathbb{S}^2\subset \mathbb{R}^3$, and its upper $(z>0)$ and lower $(z<0)$ hemispheres.

Draw two independent, uniformly distributed points $X,Y$ on $\mathbb{S}^2$. Given $\theta\in[0,\pi/2]$, what is the probability that $X$ and $Y$ belong to different hemispheres among the above two, conditioning by the event that the chord $[X,Y]$ makes an angle $\theta$ with the $z$-axis?

Numerically I find that this probability is $\cos{\theta}$.

Could anyone help me justify this $\cos{\theta}$?

Improve this question
$\endgroup$
14
  • 5
    $\begingroup$ "consider a random chord which makes an angle $\theta$ with the $z$ axis". This definition of the probability space is too vague to allow for any more meaningful answer than "something between $0$ and $1$"). Also, I suspect that MSE is a more appropriate place for this question unless your probability distribution is very fancy. $\endgroup$
    – fedja
    Jul 28, 2020 at 15:39
  • $\begingroup$ Can you describe your numerical work in detail? $\endgroup$
    – Iosif Pinelis
    Jul 28, 2020 at 16:02
  • 1
    $\begingroup$ @EmilJeřábek : As I understand it, the question is whether the conditional probability that the random chord touches both hemispheres given that the chord makes an angle $\theta$ with the $z$ axis is $\cos\theta$. $\endgroup$
    – Iosif Pinelis
    Jul 28, 2020 at 16:10
  • 1
    $\begingroup$ I took the liberty to integrate the edit to the question itself. It should be clearer this way, but I might have been heavy-handed so let me know if you prefer me to reverse part or all of the edit. $\endgroup$
    – Benoît Kloeckner
    Jul 28, 2020 at 17:10
  • 3
    $\begingroup$ With the new formulation it is, indeed, true. The straightforward computation results in the integral $\frac 2\pi\int_0^{\frac\pi 2}\frac {d\alpha}{1+\tan^2\theta\cos^2\alpha}$, which is elementary but somewhat unpleasant to compute (the usual trig substitution $z=\tan\alpha$). However such a simple answer should have an equally simple explanation, so I'd rather wait for someone to find a "no pen or paper solution" $\endgroup$
    – fedja
    Jul 28, 2020 at 20:00

1 Answer 1

Reset to default
7
$\begingroup$

This is not a true "no pen or paper" solution requested by fedja, but at least it avoids integrals. :-)

Let $X$ and $Y$ be independent random vectors on the unit sphere. Write $E = (X - Y) / |X - Y|$ for the unit vector parallel to the chord $XY$, and $Z = \tfrac{1}{2} (X + Y)$.

notation

Claim: Conditionally on $E = e$, the projection $Z = \tfrac{1}{2}(X + Y)$ of the chord $XY$ onto its perpendicular bisector plane $\pi_e = \{v : v \perp e\}$ (the entire chord projects onto a single point) is uniformly distributed over the unit disk in $\pi_e$.

Given the above claim, the proof is straightforward. Indeed: given any unit vector $e$ such that $\theta = \arcsin |e \cdot (0,0,1)|$ as in the question, and conditionally on $E = e$, $X$ and $Y$ belong to different hemispheres if and only if $Z$ belongs to an ellipse, which is the projection of the equator of the unit sphere onto $\pi_e$. This ellipse has semi-axes $1$ and $\cos \theta$ (this becomes pretty clear if one draws a picture). The area of this ellipse is equal to $\cos \theta$ times the area of the unit disk, and the desired result follows. Thus, it remains to prove the claim.

projection

Proof of the claim: Since the random variable $X \cdot Y$ is uniformly distributed over $[-1, 1]$ (Archimedes's theorem!), the random variable $$ \|Z\|^2 = \|\tfrac{1}{2} (X + Y)\|^2 = \tfrac{1}{2} (1 - X \cdot Y) $$ is uniformly distributed over $[0, 1]$. By rotational symmetry, $\|Z\|^2$ and $E$ are independent. It follows that conditionally on $E = e$, $\|Z\|^2$ is uniformly distributed over $[0, 1]$. Again by symmetry, the conditional distribution of $Z$ (given $E = e$) is invariant under rotations of $\pi_e$, and so it follows that this conditional distribution is uniform over the unit disk on $\pi_e$, as desired.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Thanks Mateusz, there are many things I don't get: - What do you exactly mean by "the projection Z of the chord XY"? I get what projecting a point or a vector means, but a chord? If it's the same Z as defined before, then I would say that is the middle of the chord, is it not? - I'm not sure what plane $\pi_e$ you're talking about, there's an infinity perpendicular to e, do you implicitly mean the one passing by the origin? - Shouldn't it be arccos and not arcsin? $\endgroup$
    – user655870
    Jul 30, 2020 at 21:02
  • $\begingroup$ @user655870: Since $e$ is parallel to the chord $XY$, the plane $\pi_e$ is perpendicular to it, and hence the entire chord projects onto a single point $Z$. And indeed $Z = \tfrac{1}{2} (X + Y)$: since $|X| = |Y| = 1$, the line passing through the origin and the midpoint $\tfrac{1}{2} (X + Y)$ of $XY$ is perpendicular to $XY$, and hence it lies in the plane $\pi_e$. I'll try to add some pictures to my answer in a few minutes. And please do ask if anything is unclear. $\endgroup$
    – Mateusz Kwaśnicki
    Jul 30, 2020 at 21:19
  • $\begingroup$ Thanks for the clarifications and pictures! Now I see. I'm still puzzled about how you define your $\theta$, in my case it's the one formed with the z axis (perp to the equatorial plane), did I miss something basic again? $\endgroup$
    – user655870
    Jul 31, 2020 at 9:22
  • $\begingroup$ I remember going in the direction of your claim in the following way: Imagine you distribute random points on the surface of the sphere with uniform density, then you link them by pairs in a given direction so as to make all possible parallel chords in that given direction. Then I thought the $\cos{\theta}$ would be obvious if the density of chords crossing $\pi_e$ was uniform across $\pi_e$, but I thought it couldn't be uniform, because on the edge for example, isn't it infinite? $\endgroup$
    – user655870
    Jul 31, 2020 at 9:26
  • $\begingroup$ I'm trying to reconcile your claim with my intuition, when I look at a (transparent) spherical shell, the density of protected matter is infinite on the edge. So there should be an infinite density of chords there, no? I can't picture how it can be uniform. $\endgroup$
    – user655870
    Jul 31, 2020 at 9:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.