# Expected triangle area of normal distributed vertices with colinear expectations

Situation

Given are 3 independent multinormal distributions $$X_i=\mathcal{N}(\vec\mu_i,\Sigma)_{i=1,2,3}$$ in $$\mathbb{R^3}$$.

For simplification the expectations are colinear:

$$\vec\mu_1=\begin{pmatrix}-a\\0\\0\end{pmatrix}, \vec\mu_2=\begin{pmatrix}0\\0\\0\end{pmatrix}, \vec\mu_3=\begin{pmatrix}b\\0\\0\end{pmatrix}$$ with $$a\ge0, b\ge0$$.

The covariance matrix is:

$$\Sigma=\begin{pmatrix}\sigma^2&0&0\\0&\sigma^2&0\\0&0&\sigma^2\end{pmatrix}$$.

Question

What is the expected absolute area $$\mathbb{E}(A)$$ of triangle $$x_1,x_2,x_3$$ with $$x_i\sim~X_i$$?

Sketch No right solution was given in two bounties

The answerer of the 1st bounty solved the easier problem $$\mathbb{E}(A^2)$$ but not $$\mathbb{E}(A)$$. No answer was given in the 2nd bounty.

What is known?

Simplified cases: solutions and approximations

1. $$\,\,\,\,\,\,\text{max}(a,b)=0 \rightarrow \mathbb{E}(A)=\sqrt{3}\sigma^2$$ (proof)

2. $$\,\,\,\,\,\,a \gg \sigma \land b=0 \rightarrow \mathbb{E}(A)=\frac{\sigma}{2}\sqrt{\pi}a$$ (proof below)

3. $$\,\,\,\,\,\,\text{max}(a,b) \ll \sigma \rightarrow \mathbb{E}(A) \approx \sqrt{3}\sigma^2$$ (presumed by simulations)

4. $$\,\,\,\,\,\,\text{min}(a,b) \gg \sigma \rightarrow \mathbb{E}(A)\approx \frac{\sigma}{2}\sqrt{\pi(a^2+b^2+ab)}$$ (presumed by simulations)

Proof for case 2:

As $$a\gg \sigma$$ the triangle can be assumed as a right triangle and the expected area is $$\mathbb{E}(A)=\frac{1}{2}ac$$ with length $$c=pq$$. The length $$p$$ is derived from the expectation of a shifted central chi-distribution with 3 degrees of freedom: $$\frac{p}{\sqrt{2}\sigma}= \mathbb{E}(\chi_3)=\sqrt{\frac{8}{\pi}}$$. The length must be corrected by $$q=\frac{\pi}{4}$$, the perpendicular component of $$p$$ to the line $$\overline{\mu_1 \mu_2}$$ (see here).

• I would immediately simplify your problem and set $\sigma = 1$ (without loss of generality). Then I'd use Heron's formula for the area of a triangle of sides $a,b,c$. I'd plug in these lengths given random points from the three distributions, then compute the expectation of the area.... It might prove computationally tricky, but that's the principled way to do it. Jan 23 '20 at 21:28

I'll try to give a partial answer since the bounty is expiring soon.

Let me call the 3 components of the $$X_i$$ to be $$x_i$$, $$y_i$$, and $$z_i$$

The norm of the cross product of two vectors is the area of the parallelogram swept by them. So half of that is the triangle you're looking for.

$$\text{area}=\frac{1}{2} \left\| \left(\left\{x_2,y_2,z_2\right\}-\left\{x_1,y_1,z_1\right\}\right)\times \left(\left\{x_3,y_3,z_3\right\}-\left\{x_1,y_1,z_1\right\}\right)\right\|$$

Simplifying:

$$\text{area}=\frac{1}{2} \sqrt{\left(x_3 \left(y_2-y_1\right)+x_2 \left(y_1-y_3\right)+x_1 \left(y_3-y_2\right)\right){}^2+\left(x_3 \left(z_2-z_1\right)+x_2 \left(z_1-z_3\right)+x_1 \left(z_3-z_2\right)\right){}^2+\left(y_3 \left(z_2-z_1\right)+y_2 \left(z_1-z_3\right)+y_1 \left(z_3-z_2\right)\right){}^2}$$

The square of the area is much nicer to work with:

$$\text{areaSq}=\frac{1}{4} \left(\left(x_3 \left(y_2-y_1\right)+x_2 \left(y_1-y_3\right)+x_1 \left(y_3-y_2\right)\right){}^2+\left(x_3 \left(z_2-z_1\right)+x_2 \left(z_1-z_3\right)+x_1 \left(z_3-z_2\right)\right){}^2+\left(y_3 \left(z_2-z_1\right)+y_2 \left(z_1-z_3\right)+y_1 \left(z_3-z_2\right)\right){}^2\right)$$

So now that we know the area in terms of the random variables, the expectation becomes a definite integral over the domains of the three random variables. Mathematica can easily integrate this. which works out to:

$$\frac{1}{2} \left(2 a^2 \sigma ^2+2 a b \sigma ^2+2 b^2 \sigma ^2+9 \sigma ^4\right)$$

and then we can apply Jensen's inequality which supports your assertions about asymptotic cases.

Unfortunately, trying to find the expectation of the absolute value of area rather than the square of area results in a super nasty integral with an analytic solution that evidently exceeds my patience and my computer's 32 GB of memory.

I'm not surprised that you see a $$\sqrt{\pi }$$ term in your empirically observed asymptotic limits since in the very simple case of looking at the expected value of the absolute value of a normally distributed random variable:

$$\text{Expectation}[\left| x\right| ,x\text{~NormalDistribution}[\mu ,\sigma ]]$$

evaluates to the rather ugly: $$\mu \text{ Erf}\left(\frac{\mu }{\sqrt{2} \sigma }\right)+\sqrt{\frac{2}{\pi }} \sigma e^{-\frac{\mu ^2}{2 \sigma ^2}}$$

Hopefully this is enough to inspire you to get the answer you need!

• The answer is a nine-dimensional improper integral with a radical in the integrand. I thought about spending hours to work on substitutions that might allow this integral to be solved in terms of elementary and special functions. There is research that solves the expected absolute value of a bivariate product. This is way beyond that. It seems like the current state of the art is to use numerical integration. Jan 31 '20 at 19:50
• Actually this problem is already a simplification. The real problem is to calculate the expected area of 3 points in R3, not necessarily laying on a line. This is identical to calculate the expectation of a cross product stats.stackexchange.com/questions/445185 Another simplification was posted here stats.stackexchange.com/questions/447196 and there is was conjectured that the distribution of the triangle area follows a Gamma distribution. Integration might be not the right solution and simulations are of course always possible and these were already made but it is to slow. Jan 31 '20 at 20:04
• It does look like a gamma distribution ($\left\{2,\frac{\sqrt{3}}{2}\right\}$) is a good fit for the case where all the means are zero. If not, then I think you need a generalized gamma which needs 4 parameters. BTW, it seems like this would be a very useful and novel result if it were proven. In any event I think you could have a pretty good approximation. Jan 31 '20 at 23:08