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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.

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  • $\begingroup$ 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. $\endgroup$
    – user65203
    Commented Jul 24, 2020 at 7:39

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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.

This approximation breaks down for large variances, of course, so you may want to experiment a bit to see what the limitations are.

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  • $\begingroup$ Small variance assumption is not a terrible one in my case, so this approximation is quite satisfactory. I'm adding 3 use case examples that compare sampled distribution (density histogram) with your approximation (blue line), if you're interested. Last one is about the biggest variance I expect and the approximation is not terrible. imgur.com/a/kXJbQHg $\endgroup$
    – simejanko
    Commented Jul 31, 2020 at 7:55
  • $\begingroup$ @simejanko Great. As you can see, the last two histograms are, in fact, noticeably not normal, so any approximation of them by a normal distribution would have its limitations. If you want to catch that skew, I can think of it, but if what we have now is already "not perfect but good enough", then let's stop here. $\endgroup$
    – fedja
    Commented Jul 31, 2020 at 14:35
  • $\begingroup$ should be plenty good enough for my use case, much appreciated. I'm aware that normal assumption breaks completely when the three points are relatively close and/or their variances are large, but that shouldn't happen in my case. The fact that I get to approximate with normal distribution also greatly simplifies the rest of my solution. $\endgroup$
    – simejanko
    Commented Jul 31, 2020 at 17:57

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