Triangle whose corners are N(0,1) variables

Question

A friend of mine and I have been exchanging and solving math puzzles and this is the last one:

A triangle is formed by three points on a plane, whose $x$ and $y$ coordinates are $N(0,1)$ random variables. Prove that the area of the triangle is also normally distributed and find the mean and variance.

Any help greatly appreciated!

Answer 1

Let the vertices of the triangle be random variables

$x = (X_1,X_2)$

$y = (Y_1,Y_2)$

$z = (Z_1,Z_2)$

If we translate the triangle to put $x$ at the origin the area doesn't change. We can do this in general, and it means finding the area of the random triangle with vertices. . .

$x' = (0,0)$

$y' = (Y_1-X_1,\ Y_2-X_2) = (Y'_1,Y'_2)$

$z' = (Z_1-X_1,\ Z_2-X_2) = (Z'_1, Z'_2)$

Then each of the random variables $Y'_1,Y'_2,Z'_1,Z'_2$ is $N(0, \sqrt 2)$ because of how normal distributions behave under sums/differences.

Then you could apply the formula for the area of a triangle with vertices at $(0,0), (a,b), (c,d)$. The area is given by $\Delta = \frac{1}{2} |ad-cb|$. So the area of the random triangle has distribution $\frac{1}{2} |Y_1' Z_2' - Y_2' Z'_1| = \frac{1}{2} |(Y_1-X_1) (Z_2-X_2) - (Y_2-X_2) (Z_1-X_1)|$.

Can you compute the mean and variance of that?