I am reading this paper https://arxiv.org/abs/1304.8019. In introduction, it said

In many applications, even simple estimation problems involving angular data are often considered as traditional linear or nonlinear estimation problems and handled with classical techniques such as the Kalman Filter [1], the extended Kalman Filter (EKF), or the unscented Kalman Filter (UKF) [2]. In a circular setting, most traditional approaches to filtering suffer from assuming a Gaussian probability density at a certain point. They fail to take into account the periodic nature of the problem and assume a linear vector space instead of a curved manifold. This shortcoming can cause poor results, in particular when the angular uncertainty is large. In certain cases, the filter may even diverge.

My background is engineering, and I understand what Gaussian distribution in N-dimension means and the role of Gaussian distribution in Kalman filter. But I don't understand why Gaussian distribution fail to estimate angular data in its nature. It will be great if there are mathematical explanation and reference. Thank you.


1 Answer 1


The traditional approach was that the models produced a probability distribution for all possible values of an angle. In most cases, the distribution is assumed to be Gaussian since it is easier to deal with and reason about, also because the mean of the Gaussian distribution (where the distribution is centered at) always has the highest probability and the probabilities around it diminish symmetrically. So imagine that you have a model that outputs for each possible angle (0-359) a probability (0-1).

The problem with this approach is twofold. The first is that the Gaussian distribution is unimodal which means the highest probability only occurs at one point which is the mean as discussed. However, if an object is symmetric (think of a double-ended arrow) then when rotating it 180 degrees it will remain the same. This means that if it was initially placed at 0 degrees or 180 degrees then there wouldn't be any difference. So the distribution must be multimodal and must have, in this case, two identical and high probabilities at 0 and 180 degrees.

The second problem is that the traditional approach doesn't reflect the geometry of the problem. Particularly, the periodicity of angles, i.e., when you add 360 degrees to an angle it remains the same. However, $\theta$ and $\theta + 2\pi$ are different values on the real line. This is what the author means by using a "Linear Geometry", i.e., using a vector space such as the real line. Instead, we must use a "Circular Geometry". For example, instead of looking at the real line, we look at the unit circle. When you walk on the real line you keep finding new values indefinitely; however, when you walk on the unit circle, after some time, you return to where you started. So instead of attaching a probability to $\theta$, we attach a probability to $(\cos \theta, \sin \theta)$.

  • $\begingroup$ Nice answer. But it seems to me that these two problems can easily be overcame by manual modification. E.g., the first problem can be solved just be adding 180 degrees on the output of a Gaussian model. The second problem can be solved by dividing the value by 2*pi, and taking remainder as result. So, I don't understand why Gaussian model will fail, it just requires one more trick. So, I think my question now is, if there were experiment or reference show that using these two tricks is not enough to solve the problem of Gaussian model. Therefore, other distribution method must be used. $\endgroup$
    – Lion Lai
    Jul 16, 2020 at 7:14
  • $\begingroup$ @LionLai The problem is that you cannot always add 180 degrees. The has to always output a probability for each angle for ANY shape. What if the shape didn't have any symmetry? Forget that you know how the shape is posed and whether you know anything else about it like its symmetry. You want the model to find that out for you. If it is not symmetric then the highest probability would be at only one of the degrees between 0-359. If it is symmetric then the distribution should have two high values, etc. $\endgroup$ Jul 17, 2020 at 10:18
  • $\begingroup$ On the other concern, dividing by 2*pi and taking the remainder works for periodicity, but the Gaussian distribution does not do that for you. If you look at a typical bell-shaped curve, then you will see that it streches to infinity on both ends and it doesn't stop to just $2\pi$ that is the problem. You want a distribution that is only defined on the interval $[0,2\pi)$ that respects the geometry of the manifold. $\endgroup$ Jul 17, 2020 at 10:21
  • $\begingroup$ Your answer is highly appreciated. I would like to know if you could show some materials and reference regard this question, so I could understand it more thoroughly and methametically. Thank you. $\endgroup$
    – Lion Lai
    Jul 23, 2020 at 6:03
  • 1
    $\begingroup$ Your welcome. You will need good probability and statistics books. Introduction to Probability and Statistics for Engineers and Scientists by Sheldon Ross is an example. If you want to learn more about Manifolds, first you have to have a good background in Calculus and Linear Algebra, and ideally read a book on Real Analysis, then read a book on Multivariable Calculus that includes material on Manifold such as Advanced Calculus of Several Variables by C. H. Edwards Jr. $\endgroup$ Jul 23, 2020 at 11:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .