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(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 $\cos{\theta}$?

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    $\begingroup$ "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? $\endgroup$ Jul 28, 2020 at 16:02
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    $\begingroup$ @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 $\cos\theta$. $\endgroup$ Jul 28, 2020 at 16:10
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    $\begingroup$ 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. $\endgroup$ Jul 28, 2020 at 17:10
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    $\begingroup$ 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

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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.

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  • $\begingroup$ 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. $\endgroup$ 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

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