2
$\begingroup$

I am proposing here a multi-d version of the question I proposed here and from which I have really learnt a lot.

Problem. Given three points "on the Earth", i.e. three couples of longitude/latitude $p=(\alpha_1, \beta_1), q=(\alpha_2, \beta_2)$ and $s=(\alpha_3, \beta_2)$, consider the regular tetrahedron $T$ "spanned" by these three points (I am normalizing the radius of the Earth to be equal to 1).

Consider now a probability measure $\mu$ and suppose the points $p,q,s$ (or, better to say: their coordinates $(\alpha_i, \beta_i)$) are i.i.d. random variables w.r.t. to this probability measure.

Q. Which is the distribution/probability measure $\mu$ on $[0,2\pi] \times [0,\pi]$ which maximizes the mean volume (without sign) of the tetrahedron, i.e. $$ \int_{\mathbb ([0,2\pi] \times [0,\pi])^3} \vert \text{Vol }(T) \vert d\mu(\alpha_1, \beta_1)d\mu(\alpha_2,\beta_2)d\mu(\alpha_3,\beta_3)? $$

Alternatively, we can directly consider probability distributions on the unit sphere $S^2 \subset \mathbb R^3$ and in this case the functional becomes $$ \int_{\mathbb (S^2)^3} \vert \text{Vol }(T) \vert d\mu(p)d\mu(q) d\mu(s) $$ and I honestly do not know which one is easier (if it is).

$\endgroup$
6
  • $\begingroup$ It actually could make quite a difference whether you have three iid random variables describing the locations of the points, or six iid random variables giving the locations in spherical coordinates. For example, if the coordinates are iid then it is impossible for the distribution of the point $p$ to be uniform over the surface area of the sphere. (And there is the fact that in conventional spherical coordinates, the difference of maximum and minimum values is $\pi$ for one coordinate and $2\pi$ for the other.) $\endgroup$
    – David K
    Feb 19, 2019 at 14:49
  • $\begingroup$ @DavidK Uh, interesting, I did not notice it. Well, I believe the interesting case is when the random variables are describing the locations of points on the sphere. How can I formulate this precisely? Should we consider three iid random variables on the unit sphere and maximize the expectation w.r.t. their common distribution? Is this the correct way of stating the problem? I apologize but I am not so confident with language of probability theory... Thanks. $\endgroup$
    – Romeo
    Feb 19, 2019 at 15:22
  • $\begingroup$ The question would be more natural with $\mu$ as a probability measure on $[0,2\pi]\times[0,\pi]$, so that the integrand is $|\text{Vol}(T)|\, d\mu(\alpha_1,\beta_1)\, d\mu(\alpha_2,\beta_2)\, d\mu(\alpha_3,\beta_3)$. $\endgroup$
    – user210229
    Feb 19, 2019 at 15:37
  • $\begingroup$ @MattF. Nice to hear from you again :) Well, is a probability measure $\mu$ on $[0,2\pi] \times [0,\pi]$ the same as a probability measure on $S^2$? If that's true then probably it is the best way of asking the question. Do you agree? $\endgroup$
    – Romeo
    Feb 19, 2019 at 15:49
  • $\begingroup$ That’s good too $\endgroup$
    – user210229
    Feb 19, 2019 at 16:13

1 Answer 1

2
+50
$\begingroup$

This is not an answer, but an amusing observation. If I didn't screw up my calculus, for the uniform distribution on the spherethe integral has a value of $\frac{\pi}{48} \approx 0.06545...$. On the other hand, since the volume is maximized when the points are orthogonal, let us try a measure concentrated at the vertices of an orthonormal basis, i.e. $\mu = \frac13 \big( \delta_{e_1}+\delta_{e_2}+ \delta_{e_3} \big)$. The expected value of the volume for this distribution will be $\frac1{27} =0.037...$, since $6$ out of $27$ choices yield a tetrahedron of volume $\frac16$, and the rest give volume $0$. We see that the uniform distribution wins.

Notice that in 2 dimensions (for the area of triangle Apex angle of a triangle as a random variable) this difference is: $\frac1{\pi} \approx 0.3183...$ vs. $\frac14= 0.25$.

I suspect that for the analogous problem in $d$ dimensions ($d$-dimensional volume of a simplex generated by random points on $S^{d-1}$) the difference would go to zero as $d\rightarrow \infty$ (due to the phenomenon of concentration of measure), but the uniform measure would be the maximizer.

1) Here is the computation of the integral. I'll work with the expected volume of parallelepiped vs tetrahedron, so that I don't have to carry $\frac16$ around. We need to compute \begin{equation} \int\limits_{S^{d-1}}\int\limits_{S^{d-1}}\int\limits_{S^{d-1}} | \det ( x\, y\, z) | d\sigma(x)\, d\sigma(y) \, d\sigma(z) = \int\limits_{S^{d-1}} \int\limits_{S^{d-1}} | \det ( x\, y\, z) | d\sigma(x)\, d\sigma(y), \end{equation} where by rotational invariance we can fix one of the variables (e.g. $z$), thus reducing to a double integral above. Since this integral is independent of the value of $z$, let us choose $z= e_3 = (0,0,1)$. Then $$| \det ( x\, y\, z) | = |x_1 y_2 - x_2 y_1 | = | \sin (\phi_x - \phi_y) | \cdot \sin \theta_y \cdot \sin \theta_y,$$ where we switched to polar coordinates. Recalling that $$ d\sigma = \frac1{4\pi} \sin \theta d\theta d\phi,$$ (we divide by $4\pi$ since $\sigma $ is a normalized measure) we reduce our integral to: \begin{align*} \int\limits_{S^{d-1}} \int\limits_{S^{d-1}} | \det ( x\, y\, e_3) | d\sigma(x)\, d\sigma(y)& = \frac1{16\pi^2} \bigg(\int_0^\pi \sin^2 \theta \, d\theta \bigg)^2 \cdot \bigg( \int_0^{2\pi}\int_0^{2\pi} | \sin (\phi_x - \phi_y ) | \, d\phi_x d\phi_y \bigg)\\ & = \frac1{16\pi^2} \cdot \bigg( \frac{\pi}2 \bigg)^2 \cdot 8\pi = \frac{\pi}{8}. \end{align*} (The integrals above are straightforward to compute.) Therefore, the expected area of the tetrahedron is $\frac{1}{6} \cdot \frac{\pi}{8} = \frac{\pi}{48} \approx 0.06545...$.

2) Here is the derivation of the recursive formula for the expected volume in $d$ dimensions. Let $E_d$ be the expected $d$-dimensional volume of the parallelepiped, spanned by $d$ independent uniformly distributed random points on the sphere $S^{d-1}$. Then the volume of the tetrahedron is simply $V_d = \frac{1}{d!} E_d$.

For any $x\in S^{d-1}$ let us write $x= (\sqrt{1-t^2} u, t)$, where $t\in [-1,1]$ is its $d^{th}$ coordinate, and $u\in \mathbb S^{d-2}$. It is a classical fact that under this change of variables $$ d\sigma (x) = \frac{\omega_{d-2}}{\omega_{d-1}} (1-t^2)^{\frac{d-3}2} dt \, d\sigma_{d-2} (u) , $$ where $\omega_{d-1} = \frac{2\pi^{d/2}}{\Gamma (d/2)}$ is the Lebesgue surface measure of the sphere $S^{d-1}$, and $\sigma_{d-2}$ is the normalized surface measure on $S^{d-2}$. Notice that this implies, in particular, that $$\int_{-1}^1 (1-t^2)^{\frac{d-3}2} dt = \frac{\omega_{d-1}}{\omega_{d-2}}. $$ We then use the same trick as above: fix $x_d = e_d = (0,0,\dots,0,1)$ to reduce the number of variables: \begin{align*} \mathbb E_{x_1,\dots, x_d \in S^{d-1}} | \det (x_1 \, \dots \, x_d ) | & = \mathbb E_{x_1,\dots, x_{d-1} \in S^{d-1}} | \det (x_1 \, \dots \, e_d ) |\\ & = E_{x_1,\dots, x_{d-1} \in S^{d-1}} \prod_{i=1}^{d-1} (1-t_i^2)^{1/2} \cdot \det (u_1 \, \dots \, u_{d-1} ) |\\ & = E_{t_1,\dots, t_{d-1} \in [-1,1] } \prod_{i=1}^{d-1} (1-t_i^2)^{1/2} \cdot E_{u_1,\dots, u_{d-1} \in S^{d-2}} | \det (u_1 \, \dots \, u_{d-1} ) |\\ & = \bigg( \frac{\omega_{d-2}}{\omega_{d-1}} \int_{-1}^1 (1-t^2)^{\frac{d-2}2} dt \bigg)^{d-1} \cdot E_{d-1}\\ & = \bigg( \frac{\omega_{d-2}\omega_{d}}{\omega_{d-1}^2} \bigg)^{d-1} E_{d-1} . \end{align*} Therefore $$ E_d = \bigg( \frac{\omega_{d-2}\omega_{d}}{\omega_{d-1}^2} \bigg)^{d-1} E_{d-1} = \left( \frac{\big( \Gamma \big(\frac{d}{2}\big)\big)^2 }{\Gamma \big(\frac{d-1}{2}\big) \Gamma \big(\frac{d+1}{2}\big)} \right)^{d-1} E_{d-1}.$$

Let us check the case $d=3$. We have $$ E_{d-1}= E_2= \frac{1}{4\pi^2} \int_0^{2\pi}\int_0^{2\pi} |sin (\phi_1 - \phi_2) | d\phi_1 d\phi_2 = \frac{2}{\pi} ,$$ also $\omega_{d-1} = \omega_2 = 4\pi$, $\omega_{d-2} = \omega_1 = 2\pi$, $\omega_d = \omega 3 = 2\pi^2$. Therefore, $$ E_3 = \bigg( \frac{\omega_{1}\omega_{3}}{\omega_{2}^2} \bigg)^2 E_{2} = \bigg( \frac{2\pi \cdot 2\pi^2}{(4\pi)^2} \bigg)^2 \cdot \frac{2}{\pi} = \bigg( \frac{\pi}{4} \bigg)^2 \cdot \frac{2}{\pi} = \frac{\pi}{8}, $$ as we have obtained above.

$\endgroup$
15
  • $\begingroup$ This is also interesting, and I want to ask for confirmation what happens in the $d$-dimensional case: it seems to me we have $d^d$ choices (i.e. the entrances of each vector of the canonical base) and of these $d(d-1)$ (?) will yield a volume of $1/(d+1)!$ and the others zero. So the expected volume for the distribution $1/d (\delta_1 + \ldots + \delta_d)$ is $1/d^d \cdot 1/((d+1)(d-2)!)$. Am I right? Thanks. $\endgroup$
    – Romeo
    Feb 26, 2019 at 17:51
  • $\begingroup$ I think you are slightly off here: we are indeed, choosing $d$ vectors out of the $d$ basis elements (with repetitions), so $d^d$ choices. We are interested in the choices where none of them repeat and all elements of the basis are taken (otherwise the volume is zero): there are $d!$ such choices, and in each case the volume of the tetrahedron spanned by the basis is $1/{d!}$, hence the expected value will just be $1/{d^d}$. $\endgroup$ Feb 27, 2019 at 13:02
  • $\begingroup$ That was easy, you are right. Sorry for messing things up :-( and thanks for clarification. Do you know if there is a close formula for the integral wrt the uniform distribution in $\mathbb R^d$ (so to compare the asymptotic behavior with $d^{-d}$)? Thanks. $\endgroup$
    – Romeo
    Feb 27, 2019 at 18:28
  • $\begingroup$ I don't know one, but I'll try to compute it at some point. $\endgroup$ Feb 27, 2019 at 18:45
  • $\begingroup$ How did you set up the calculus to get .0398 for the uniform distribution? I got .12 for that by the code in my answer. $\endgroup$
    – user210229
    Mar 2, 2019 at 3:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .