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The 3 points of a triangle in $\mathbb{R}3$ are not precisely known. The uncertainty of the points shall be described by three-dimensional Gaussians where the expectation values $(x_i,y_i,z_i), i=1,2,3$ and standard deviations are known. What is the expected area and area distribution? (here a 2d sketch)

The area can be calculated via the cross product and is given in terms of $(x_i,y_i,z_i)$ as

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

Following simplifications shall be made:

  • standard deviations are equal for all points and all directions
  • random distributions are mutually independent
  • points do not interfere, i.e. standard deviation $\ll$ point distances
granular_bastard
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    Maybe see: https://math.stackexchange.com/questions/516219/finding-out-the-area-of-a-triangle-if-the-coordinates-of-the-three-vertices-are/516223, https://math.stackexchange.com/questions/92255/probability-distribution-for-the-perimeter-and-area-of-triangle-with-fixed-circu – Alessio K Dec 25 '19 at 22:08
  • the given link considers the case of exact known points but in the problem the points are randomly distributed – granular_bastard Dec 25 '19 at 23:19

1 Answers1

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You don't need a square root; the area is given by $A=|x_1y_2-x_1y_3-x_3y_2-x_2y_1+x_2y_3+x_3y_1|$. By your third assumption, we can assume that the argument of the absolute value doesn't change sign, so the area is linear in each coordinate. Then by your second assumption the problem factorizes and you can replace each coordinate by its expected value: The expected area is just the area of the triangle formed by the expected vertices. You don't need the first assumption for that.

joriki
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  • I specified my original question. Does the square root change everything? – granular_bastard Dec 26 '19 at 14:30
  • @granularbastard: You didn't "specify" it, you turned it into a completely different question. That's very bad style, changing a question so that a valid answer is no longer valid. You should ask a new question instead. – joriki Dec 26 '19 at 20:13
  • Actually I tried to simplify the problem from 3d to 2d but now it seems that the simplification was not valid and a triangle with random points in 2d is not the same as a triangle in 3d. That is surprising. – granular_bastard Dec 26 '19 at 20:41
  • the problem was specified here:

    https://math.stackexchange.com/questions/3490253/area-of-triangle-with-normal-distributed-points

     – granular_bastard Dec 28 '19 at 18:11