# Distribution of an angle between 3 normally distributed 2D points

Example image

Given 3 normally distributed points in 2D space, what would the distribution of the angle $$\alpha$$ between the three points be & can it be reasonably well approximated with a (circular) normal distribution.

In my particular case, I have an additional assumption that the covariance matrices are limited to the multipliers of the identity matrix $$I_2$$, which may simplify the problem.

$$A \sim N(\mu_A, \sigma_A * I_{2})$$

$$B \sim N(\mu_B, \sigma_B * I_{2})$$

$$C \sim N(\mu_C, \sigma_C * I_{2})$$

$$\alpha \sim ?$$

I'd be happy with an approximate/engineered solution, could some additional assumptions simplify the problem further?

I don't have much of a maths background, so my first engineered approximation was to sample the said distribution and infer circular normal distribution parameters that way. The problem with using this approach in practice is its poor computational performance.

• If I am right the vectors $AB$ and $AC$ also follow a normal law, and maybe finding the distribution of the dot product is tractable. The length of the vectors must follow a Raighley law.
– user65203
Commented Jul 24, 2020 at 7:39

Let us treat $$a=\mu_A,b=\mu_B,c=\mu_C$$ as complex numbers. Then we are interested in the random variable that is $$\Im\log\frac {(a+\sigma_AX_A)-(b+\sigma_BX_B)}{(c+\sigma_C X_C)-(b+\sigma_B X_B)}$$ where $$X_A,X_B,X_C$$ are independent standard complex Gaussians (mean $$0$$, covariance $$I_2$$). The general case looks rather hopeless, but when $$\sigma$$'s are small compared to the distances, we can linearize and get $$\Im\left\{\log\frac{a-b}{c-b}+\frac{\sigma_A}{a-b}X_A-\frac{\sigma_C}{c-b}X_C -\sigma_B\left[\frac1{a-b}-\frac1{c-b}\right]X_B\right\}$$ The first term is just the angle $$\alpha_0$$ the means make and the sum of the last three terms is a complex Gaussian with $$0$$ mean and the covariance $$\sigma I_2$$ where $$\sigma^2=\sigma_A^2\frac{1}{|a-b|^2}+\sigma_C^2\frac1{|c-b|^2}+\sigma_B^2\left|\frac 1{a-b}-\frac 1{c-b}\right|^2$$ (everything is measured in radians, of course). The projection to the imaginary axis yields the $$N(\alpha_0,\sigma)$$ normal law for the angle you are interested in.