3 points are randomly selected from a multinormal distribution $\mathcal{N}(\vec{0},\Sigma)$ in $\mathbb{R^3}$ with $\Sigma=\begin{pmatrix}\sigma^2&0 &0 \\0&\sigma^2&0\\0&0&\sigma^2 \end{pmatrix}$ and variance $\sigma^2$.

How to prove that the expected absolute triangle area of the 3 random points is presumably $\mathbb{E}(A)=\sqrt{3}\sigma^2$? The conjecture was found by simulations.

  • 2
    $\begingroup$ start by defining how the area of the triangle can be expressed with the (x,y,z) random variables that are jointly distributed as Normal (note the marginals are also normally distributed, and by nature of the covariance matrix, they're all independent as well). Once you have an expression for that, find the distribution, then find the expected value. $\endgroup$ – creutzml Jan 31 at 3:29

This problem can be solved through a series of simplifications and then looking things up.

First, $\sigma$ merely establishes a unit of measurement: in a system where $\sigma$ is one unit, the covariance matrix is the identity and the unit of area is $\sigma^2:$ that's why the result is a multiple of $\sigma^2.$ So from now on we may take $\sigma=1.$

Second, let the three (independent) random points (each with coordinates from this trivariate standard Normal distribution) be $X,$ $Y,$ and $Z.$ Let $i$ denote one of the three components of these vectors. The triangle in question can be translated to the origin (without changing its area) by subtracting $Z,$ where it is determined by the vectors $U = X-Z$ and $V = Y-Z.$ The components of these vectors are Normal with zero means and covariances

$$\operatorname{Cov}(U_i,V_i) = \operatorname{Cov}(X_i-Z_i, Y_i-Z_i) = 1$$

and variances

$$\operatorname{Var}(V_i) = \operatorname{Var}(U_i) = \operatorname{Var}(X_i-Z_i) = 2.$$

Consequently the correlation of $U_i$ and $V_i$ is $\rho = 1/2.$

Third, we may exploit properties of Normal distributions to describe the distribution of $U,V$ in an equivalent way. Define $\rho^\prime = \sqrt{1-\rho^2}$ so that $\rho^2 + (\rho^\prime)^2 = 1.$

An equivalent description of the distribution of $(U,V)$ begins with independent components $U_i,W_i$ (all with zero mean and variance of $2.$) If we set

$$V = \rho^\prime\,W + \rho\,U$$


$$\operatorname{Var}(V) = (\rho^2 + (\rho^\prime)^2)(2) = 2$$


$$\operatorname{Cov}(U,V) = \rho\,(2) = 2\rho.$$


This version of $(U,V),$ which (in $n=3$ dimensions) also is $2n$-variate Normal, has exactly the same first and second moments as the original description: thus the distributions are the same.

Fourth, geometry tells us the area of the triangle $OVU$ is the same as the area of the triangle $O(\rho^\prime W)U$ and that, in turn, is $\rho^\prime$ times the area of triangle $OWU,$ which trigonometry tells us is

$$\operatorname{Area}(OWU) = \frac{1}{2} |W|\,|U|\,\sin(\theta_{UW}).$$ Here, $\theta_{UW}$ is the angle made between vectors $U$ and $W.$

Now we may call on well-known (simple) results:

  1. $|U|/\sqrt{2}$ and $|W|/\sqrt{2}$ have $\chi(n)$ distributions.

  2. $t = (1 + \cos(\theta_{UW}))/2$ has a Beta$((n-1)/2, (n-1)/2)$ distribution..

  3. $|U|,|W|,$ and $\theta_{UW}$ are independent. (This follows directly from the spherical symmetry of the $n$-variate standard Normal distribution.)

This information is enough to work out the distribution of the area. (When $n=3$ it happens to have a Gamma distribution but in other dimensions its PDF is proportional to a modified Bessel $K$ function.)

The expected area is particularly easy to find. We can look up (or readily) compute the $\chi(n)$ expectation,

$$E\left[\frac{|U|}{\sqrt{2}}\right] = E\left[\frac{|W|}{\sqrt{2}}\right] = \sqrt{2} \frac{\Gamma((n+1)/2)}{\Gamma(n/2)},$$

and with almost no work we can find the expectation of $\sin(\theta_{UV}) = 2\sqrt{t(1-t)}$ as

$$\eqalign{ E\left[2t^{1/2}(1-t)^{1/2}\right] &= \frac{1}{B((n-1)/2,(n-1)/2)} \int_0^1 2t^{1/2}(1-t)^{1/2} t^{(n-1)/2-1}(1-t)^{(n-1)/2-1}\, \mathrm{d}t \\ &= \frac{2}{B((n-1)/2,(n-1)/2)} \int_0^1 t^{n/2-1}(1-t)^{n/2-1}\, \mathrm{d}t \\ &= \frac{2\,B(n/2,n/2)}{B((n-1)/2,(n-1)/2)}. } $$

Plug everything into the area formula for triangle $OWU$ to obtain

$$\eqalign{ E[\operatorname{Area}(OWU)] &= E\left[\frac{1}{2} |W|\,|U|\,\sin(\theta_{UW})\right] \\ & = \frac{1}{2} \left((\sqrt{2})(\sqrt{2}) \frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n}{2}\right)}\right)^2\ \frac{2\,B\left(\frac{n}{2},\frac{n}{2}\right)}{B\left(\frac{n-1}{2},\frac{n-1}{2}\right)} \\ & = 4\frac{\Gamma\left(\frac{n+1}{2}\right)^2 \Gamma(n-1)}{\Gamma\left(\frac{n-1}{2}\right)^2 \Gamma(n)} \\ &= 4 \frac{\left(\frac{n-1}{2}\right)^2}{n-1} = n-1. }$$

(The third line expanded the Beta functions in terms of Gamma functions and the last line used the defining relation $\Gamma(z+1) = z\Gamma(z)$ several times.)

We must remember the other two factors dropped along the way: this area has to be multiplied by $\rho^\prime$ (lost at step 4) and then by $\sigma^2$ (lost at step 1).

We have thereby obtained a general formula for the expectation of a triangular area in any number of dimensions and even when the components of the vectors $U$ and $V$ are correlated with correlation coefficient $\rho.$ (Bear in mind these components have variances of $2,$ not $1.$) It is

$$E[\operatorname{Area}(OVU)] = \rho^\prime\, (n-1)\, \sigma^2.$$

Earlier we saw that $\rho=1/2,$ so $\rho^\prime = \sqrt{3}/2$ (there is where the square root of $3$ comes from!) and for $n=3$ this yields

$$E[\operatorname{Area}(XYZ)] = \sqrt{3}\, \sigma^2.$$

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  • $\begingroup$ This post is related to the expected volume of normal distributed tetrahedron coordinates: math.stackexchange.com/questions/3545749 $\endgroup$ – granular bastard Feb 14 at 17:53
  • $\begingroup$ I figure you might go there--and then on to 4D, 5D, and so on. Although the analysis gets more complicated, the simplifying approaches I used here also work in higher dimensions. Ultimately it comes down to the product of independent chi-distributed variables and a variable for the spherical angle they determine:finding the expectation of the latter is the crux of the question. $\endgroup$ – whuber Feb 14 at 18:07
  • $\begingroup$ Is this method somehow applicable if the expectations of the original distributions do not agree? $\endgroup$ – granular bastard Feb 14 at 19:29
  • $\begingroup$ Most of it is: use a suitable rotation to make the expectations zero for all but one variable. The result gets complicated because a non-central chi distribution is involved and (I suspect) it may be difficult to evaluate the expectation of the spherical sector. $\endgroup$ – whuber Feb 14 at 19:51
  • $\begingroup$ In the third step the variable $W$ emerges from nowhere. Could you explain? $\endgroup$ – granular bastard Feb 15 at 0:38

Rather than an answer I want to extend your speculation: The distribution of the area with $\sigma=1$ has a Gamma distribution with parameters 2 and $\sqrt{3}/2$.

Why? First, a histogram of random samples looks very much like a Gamma distribution. (I'm using Mathematica here because I know the OP also uses Mathematica.)

(* Define the area of the triangle of 3 points in 3-space *)
x1 = {x[1], x[2], x[3]};
x2 = {x[4], x[5], x[6]};
x3 = {x[7], x[8], x[9]};
area = Area[Polygon[{x1, x2, x3}]]

$$\frac{1}{2} \sqrt{(x_2 (x_4-x_7)+x_5 x_7-x_4 x_8+x1 (x_8-x_5))^2+(x_3 (x_4-x_7)+x_6 x_7-x_4 x_9+x_1 (x_9-x_6))^2+(x_3 (x_5-x_8)+x_6 x_8-x_5 x_9+x_2 (x_9-x_6))^2}$$

(* Look at the distribution of some random samples of area *)
n = 10000;
a = ConstantArray[0, n];
Do[a[[j]] = area /. Thread[Table[x[i], {i, 9}] -> 
     RandomVariate[NormalDistribution[0, 1], 9]], {j, n}]
Histogram[a, Automatic, "PDF"]

Histogram of random samples of area

Fortunately all of the even moments of the random variable area are readily determined. So we'll match the 2nd and 4th moments of area with that of a Gamma distribution and determine the parameters of the Gamma distribution.

(* Expectation of 2nd and 4th moments of area *)
m2 = Expectation[area^2, Table[x[i] \[Distributed] NormalDistribution[0, 1], {i, 9}]]
(* 9/2 *)
m4 = Expectation[area^4, Table[x[i] \[Distributed] NormalDistribution[0, 1], {i, 9}]]
(* 135/2 *)

(* Expectation of 2nd and 4th moments of a gamma distribution *)
g2 = Expectation[z^2, z \[Distributed] GammaDistribution[a, b]]
(* a (1+a) b^2 *)
g4 = Expectation[z^4, z \[Distributed] GammaDistribution[a, b]]
(* a (1+a) (2+a) (3+a) b^4 *)

(* Get solution(s) for a and b where a > 0 and b > 0 *)
Select[{a, b} /. Solve[{m2 == g2, m4 == g4}, {a, b}], #[[1]] > 0 && #[[2]] > 0 &][[1]]
(* {2,Sqrt[3]/2} *)

So we have a Gamma distribution with parameters $2$ and $\sqrt{3}/2$ which has a mean of $2 \times \sqrt{3}/2=\sqrt{3}$.

But do higher order moments now match? Yes.

 Table[{2 k, Expectation[area^(2 k), 
    Table[x[i] \[Distributed] NormalDistribution[0, 1], {i, 9}]],
   Expectation[z^(2 k), z \[Distributed] GammaDistribution[2, Sqrt[3]/2]]}, {k, 1, 5}],
 TableHeadings -> {None, {"\nk", "\nE[area^k]", "k-th moment of a\nGamma(2,3^(1/2)/2)"}}]

Table of matching even moments

It looks like a Gamma and we can match (eventually) many even moments.

This doesn't get you a proof but if the distribution of the area is really a multiple of a Gamma distribution, then that might suggest some avenues to go about getting a proof to others. (This approach will almost certainly apply to your cross-product questions on several of these forums.)

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