2

This is actually two questions, that have a similar premise:

  1. If points in $\mathbb{R}^2$ are chosen stereographically randomly (i.e. chosen uniformly randomly on the surface of the unit sphere and then projected via $(x,y,z) \mapsto (\dfrac{x}{1-z}, \dfrac{y}{1-z})$), then what is the probability that two random line segments (determined by their endpoints) in the plane will intersect?

  2. If points on the surface of the unit sphere are connected by a geodesic, then what is the probability that two random geodesics on the sphere will intersect?

The first problem was easy enough for me to run a Monte Carlo simulation to obtain the answer—though I have no idea how to tackle it theoretically—and I believe I have an argument that the answer to #2 should be $\dfrac{1}{8}$ (just looking for verification). It is as follows:

If $P(g)$ is the probability that a random geodesic will intersect the given geodesic $g$, then we observe that $P(g) = \dfrac{1}{2} \cdot \dfrac{\theta}{2\pi}$, where $\theta$ is the length of $g$ (probability of $\dfrac{1}{2}$ that the random geodesic will have endpoints on opposite hemispheres of the great circle induced by $g$, multiplied by the proportion of this great circle that $g$ covers). As can be computed, the probability that a random geodesic will have length $\leq \theta$ (for $0 \leq \theta \leq \pi$) is $\dfrac{2\pi(1-\cos(\theta))}{4\pi} = \dfrac{1-\cos(\theta)}{2}$; differentiating, we get the probability density function $D(\theta) = \dfrac{\sin(\theta)}{2}$. We combine these results with Fubini's theorem to get

$$\int_0^\pi \dfrac{\theta D(\theta)}{4\pi}~~d\theta = \dfrac{1}{8}$$

In short, is my answer for #2 correct, and how do I solve #1?

Edit: Problem 1 reposted here.

Feryll
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  • You could provide the result of your simulation for 1. so that we could check any theoretical results we might obtain. – joriki Jul 08 '18 at 20:41
  • @joriki The answer, provided there were no flaws in my implementation, appeared to be exactly 1/5. – Feryll Jul 09 '18 at 02:45
  • That's very interesting! I didn't expect it to have such a nice answer. It seems very difficult to derive that, but a first step would be to determine the density for $r$ in the plane. We have

    $$ r^2=\frac{1+z}{1-z} $$

    and thus

    $$ z=\frac{r^2-1}{r^2+1}=1-\frac2{r^2+1};. $$

    Since on a sphere slices of equal height have equal area, $z$ is uniformly distributed over $[-1,1]$, so the density for $r$ is

    $$ \frac12\frac{\mathrm dz}{\mathrm dr}=\frac{2r}{(r^2+1)^2};. $$

     – joriki Jul 09 '18 at 05:52
  • (In case you simulated points on the sphere and projected them to the plane, you could use this density to make the simulation a bit more efficient; note that it has a definite integral in closed form, so $r$ can be sampled by inverting the cumulative distribution function.) – joriki Jul 09 '18 at 05:56
  • I also get $\frac15\pm10^{-5}$ from simulations. It's a bit unfortunate that you asked these two questions in one; there's now an answer to the question, which will cause it to get less attention, even though it only answers the easier part of it. You could remove the first part from the question and post it as a separate question. – joriki Jul 09 '18 at 06:35
  • Actually, I think the approach with the density for $r$ was probably misguided; it seems more promising to stay on the sphere. – joriki Jul 09 '18 at 06:45
  • Simulations show that the density of the distance $d$ of the plane spanned by three uniformly random points on the unit sphere from the origin is $\frac32(1-d^2)$. This implies the $\frac15$ result, so we just have to figure out how to prove this. – joriki Jul 09 '18 at 10:12
  • Interesting. I've yet to sit down and think about the geometrical interpretation of #1 as closely as I maybe could, but how does that plane-distance density result imply the overall answer of 1/5? (At any rate, I'll let you know if I happen to repost this) – Feryll Jul 09 '18 at 10:49
  • The lines in the plane are circles through the projective point on the sphere. Two line segments intersect iff the corresponding circle arcs intersect. I actually got it a bit wrong above; the density $\frac32(1-d^2)$ implies that the intersection probability of a line and a segment is $\frac25$; I'm missing an argument for why the intersection probability of two segments should be exactly half that. The $\frac25$ is obtained by integrating the density for $d$ with the probability $2(\frac12+\frac d2)(\frac12-\frac d2)$ of choosing two points on different sides of the circle. – joriki Jul 09 '18 at 11:29
  • @joriki I think the following argument can be made: Given a circle on the sphere and two points on said circle, the probability densities of the line segments of either arc are equal. As well, a line segment formed by two points (one on the inside of the circle, one on the outside) will cross some uniformly random point on the circle. If we integrate the product of the probability densities of (probability of length of arc = $L$) and (probability of line segment crossing arc of length = $L$), we get exactly $1/2$, since generally $\int_0^1 f(L)g(L) = 1/2$ when $f(L)=f(1-L)$ and $g(L)+g(1-L)=1$ – Feryll Jul 14 '18 at 10:32
  • and also $\int_0^1 f(L)=1$. At any rate, I have since reposted the problem in its own question. – Feryll Jul 14 '18 at 10:34
  • I'd thought about that argument, too, but I don't think it's sound (at least not in that form). We're not uniformly choosing points on these circles. We're uniformly choosing points, then determining the circles they lie on, and they must have a preference towards lying further away from the projection point, since they're uniformly distributed on the meridians whereas these circles lie closer to the projection point on average than the meridians. I'd seen the repost (and upvoted it :-). I'd suggest adding a link here, too (even though it's already in the sidebar, that's easily overlooked). – joriki Jul 14 '18 at 10:37

1 Answers1

1

This is a variation of a well known Putnam exam problem, A6 from 1992. I will have to double check what you did, I’m on my phone, but a version can found in many places online, like this