Expected triangle area of normal distributed vertices with colinear expectations

Question

For the bounty the already answered problem was reformulated

This question was already answered for random variables in $\mathbb{R}^3$. Now I am looking for the solution in $\mathbb{R}^2$ that could maybe be extracted from the solution for $\mathbb{R}^3$ below. (In case the solution for $\mathbb{R}^n$ automatically appears then this would be also of interest.)

Due to reduction of the dimension $\mathbb{R}^3\to\mathbb{R}^2$ the variables from the original problem change:

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

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

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$\Large\text{Original problem}$

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

Cross post

This question was already posted on StackExchange Mathematics and there is no answer after 2 bounties within a year. 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.

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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).

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Answer 1

How was your formula derived?

You can do most of the work in your head without touching pen or paper. First, project to a random plane. Then the area will drop twice on average (Archimedes), so the answer is twice the expectation (in $F_3$) of the mean of the triangle area in the planar problem with data $A=F_3a, B=F_3b$ where $F_3$ is the "shrinking factor" for the projection of a fixed 3D vector to a random 2D plane. Now, in the plane, choose the horizontal axis along the line joining the point means. Let $P_a=(H_a,V_a),P_b=\dots,P_o=\dots$ be the random points ($P_o$ is the "central" one). Then the planar area is one half (which neatly cancels with the above $2$) times $|V_o(H_b-H_a)+V_a(H_o-H_b)+V_b(H_a-H_o)|$ (from geometry or from the cross product, whichever you like more). Since $V_{o,a,b}$ are independent $N(0,\sigma)$, we can take the expectation in them to get $\sqrt{\frac 2\pi}\sigma$ times the expectation of $\sqrt{(H_b-H_a)^2+(H_o-H_b)^2+(H_a-H_o)^2}$.

Now $H_o=\sigma N_o$, $H_a=-A+\sigma N_a$, $H_b=B+\sigma N_b$ where $N_{o,a,b}$ are independent $N(0,1)$. Plugging in, we see that we need the expectation of the distance from the point $(B+A,-B,-A)$ to $\sigma(N_a-N_b,N_b-N_o,N_o-N_a)$ in the plane $\Pi=\{x+y+z=0\}$. The second point has a rotationally symmetric normal distribution in $\Pi$, so only the absolute value of the first point matters. Computing this absolute value and the expectation of the square of the second point, we get $E|\rho e-\sqrt 3\sigma N_2|$ where $\rho=\sqrt{2(A^2+B^2+AB)}$ and $N_2$ is the standard normal on the plane and $e$ is any unit vector on the plane.

Now project to a random line. This will give a factor $\frac \pi 2$ (one over the average of $|\cos|$ over the period and a factor $F_2$ (shrinking factor for the projection from the plane to a random line) on $\rho$. The ultimate result will be $$ \sqrt{\frac \pi 2}\sigma E\left|F_3F_2\sqrt{3\sigma^2 r^2}+\sqrt 3\sigma N_1\right| $$ where $N_1$ is $N(0,1)$, and $r^2=2(a^2+b^2+ab)/3\sigma^2$.

Next note that $F_3F_2$ can be viewed as the shrinking factor for the projection of a $3D$ vector to a random line, so, by Archimedes again, it is uniformly distributed on $[0,1]$. Carrying out $\sqrt 3\sigma$ and using the symmetry of $N_1$, we get $$ \sqrt 3\sigma^2\sqrt{\frac \pi 2}E|Ur+N_1| $$ where $U$ is uniform on $[-1,1]$ and $N_1$ is $N(0,1)$ and they are independent.

Setting up the double integral, using symmetry of $N_1$ again, and recalling the formula for the Gaussian density, we get $$ \sqrt 3\sigma^2\int_0^\infty (E_U|Ur+x|)e^{-x^2/2}\,dx $$ Now we can easily compute $E_U|Ur+x|$, which is $\frac r2+\frac{x^2}{2r}$ for $0\le x\le r$ and $x$ for $x\ge r$. Integration by parts (here you finally need to write something) finishes the story.