Let a point on the plane be randomly chosen via $(\sqrt{\frac{t}{1-t}}\cos(2\pi\theta),\sqrt{\frac{t}{1-t}}\sin(2\pi\theta))$, where $t$ and $\theta$ are uniformly randomly chosen on $[0,1]$ (equivalently, choose a point uniformly randomly on the surface of the sphere and then project stereographically). Then, what is the probability that two random line segments (determined by their endpoints) will intersect?

This is a repost of a subproblem in a previous post that never got answered. Monte Carlo simulation suggests that the answer is precisely $1/5$, but I have no fruitful ideas left how to prove it.

  • $\begingroup$ @MvG That is correct. $\endgroup$ – Feryll Jul 14 '18 at 10:13
  • $\begingroup$ It seems that your radius $\sqrt{\frac t{1-t}}$ is based on an erroneous calculation in one of my comments to the original question. I'd missed a factor $2$ in your stereographic projection, so the radius $r^2=\frac{1+z}{1-z}$ (with $z=2t+1$) corresponds to the point $(x/(1-z),y/(1-z))$. The difference is that between stereographic projection with a sphere cut in half by the plane or sitting on the plane. Here the median circle is the unit circle; there it's the circle with radius $2$. I'm sorry about the error. I think it would be good to make the two questions use the same projection. $\endgroup$ – joriki Jul 14 '18 at 13:39
  • $\begingroup$ Perhaps the cleanest solution would be to take out the $2$s in the other question; then both MvG's answer here and my comments under the other question would be correct. $\endgroup$ – joriki Jul 14 '18 at 13:40
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    $\begingroup$ I derived the $\sqrt{\frac{t}{1-t}}$ myself, indeed based on the projection without the factor of $2$. It doesn't ultimately matter which projection you use since they're equivalent up to multiplication, which doesn't affect whether line segments intersect, but to possibly save confusion, I'll go ahead and modify the previous question's projection. $\endgroup$ – Feryll Jul 14 '18 at 15:40
  • $\begingroup$ It's perhaps worth pointing out that the convex hull of 4 points is either a triangle or a quadrilateral with probability 1, and in the former case no pair of line segments between them can intersect and in the latter case there's a probability of $1/3$ that the line segments intersect. So this problem is equivalent to showing there's a $3/5$ probability that the convex hull is a quadrilateral. $\endgroup$ – Contravariant Jul 23 '18 at 16:18

This is not a finished solution, just a collection of ideas, but with a bit of luck it will get you there.

  1. Switch to Cartesian coordinates. Expressing intersections there will be easier. To achieve this, you need a probability density function $p(x,y)$. It should be proportional to the ratio of sphere surface area per plane surface area after stereographic projection, for infinitesimally small areas. It should only depend on the (squared) radius $x^2+y^2$. And of course it should sum up to one, as in $$\iint_{-\infty}^{+\infty}p(x,y)\,\mathrm dx\,\mathrm dy=1$$ Unless I made a mistake, the probability density function you want should be $$p(x,y)=\frac1{\pi\left(x^2+y^2+1\right)^2}$$ This is based not on your formula for $t$ but on my considerations for stereographic projection of the unit sphere onto the equatorial plane. Please double-check this.

  2. With probability $1$ any three random points do not lie on a line. In that case you can express the fourth point as a linear combination of these, namely $$P_4=\lambda_1P_1+\lambda_2P_2+\lambda_3P_3\qquad\text{with }\lambda_1+\lambda_2+\lambda_3=1$$ Then segment $(P_1,P_2)$ will intersect segment $(P_3,P_4)$ iff $\lambda_1>0,\lambda_2>0,\lambda_3<0$.

  3. Combine these. Three points are random in the plane, the fourth random but satisfying these constraints. \begin{align*} f_1&=\iint_{-\infty}^{+\infty}f_2\,p(x_1,y_1)\,\mathrm dx_1\,\mathrm dy_1 \\ f_2&=\iint_{-\infty}^{+\infty}f_3\,p(x_2,y_2)\,\mathrm dx_2\,\mathrm dy_2 \\ f_3&=\iint_{-\infty}^{+\infty}f_4\,p(x_3,y_3)\,\mathrm dx_3\,\mathrm dy_3 \\ f_4&=\int_0^1\int_{1-\lambda_1}^{+\infty} q(x_4,y_4)\,\mathrm d\lambda_2\,\mathrm d\lambda_1 +\int_1^{+\infty}\int_0^{+\infty} q(x_4,y_4)\,\mathrm d\lambda_2\,\mathrm d\lambda_1 \\ x_4 &= \lambda_1x_1+\lambda_2x_2+(1-\lambda_1-\lambda_2)x_3 \\ y_4 &= \lambda_1y_1+\lambda_2y_2+(1-\lambda_1-\lambda_2)y_3 \end{align*}

  4. The formulation above is using a different probability density function $q$ in the last step, due to the different parametrization. You will need to express $q$ in terms of $p$, using regular rules for integration by substitution. Essentially $\mathrm dx\,\mathrm dy$ describes a rectangular area in the plane. The corresponding area $\mathrm d\lambda_1\,\mathrm d\lambda_2$ projects onto the plane as a parallelogram-shaped area which you can compute as a the absolute value of a determinant $$q(x,y)=\left\lvert\det\begin{pmatrix}x_1-x_3&x_2-x_3\\y_1-y_3&y_2-y_3\end{pmatrix}\right\rvert\,p(x,y)$$ You can of course move that determinant in front of the integral.

  5. Now you only have to hope that some combination of brain power and computer algebra system can compute these integrals without building up too much complexity.

By now I gave that a try, and the results are rather discouraging. Seems the terms will become already quite complicated at the innermost integral, and Sage asks me for complicated case distinctions. So I'm no longer optimistic this will be a suitable approach without any fundamental insights into the computation of the integrals.

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    $\begingroup$ Please see my comments under the original question (linked to in this question). After first going down this Cartesian route, I now believe that a solution on the sphere is more promising. $\endgroup$ – joriki Jul 14 '18 at 13:26
  • $\begingroup$ @joriki: I concur that computing these integrals appears harder than I would have hoped. Do you want to expand on the computation on the sphere, detailing in a different “answer” how far you got with that? Reading comments is not ideal for that. I feel bad for causing this question to have a formal answer when I'm no longer sure it really helps answering, but I guess I'll try to make up for that with a bounty when I can, and having all available information around might help readers to build on that. $\endgroup$ – MvG Jul 15 '18 at 22:22

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