# Removing outliers from circular average data

I'm trying to find the average from a set of circular data and am using the following which is doing what I'm expecting.

$$a = \arctan\left(\frac{\sum\limits_{i=1}^N \sin(a_i)}{\sum\limits_{i=1}^N\cos(a_i)}\right).$$

However I would love to be able to remove results which are furthest from the average but not sure what my approach should be.

Essentially I want to hone in on the cluster of most alike results in order to give the value which represents the average of the data in the highest density around the circle.

I'd love to just hear of any strategies I could investigate to help move this problem forward.

• I don't understand the down vote here. The question makes sense to me and seems to be encountered in statistical/mathematical modeling. – Iosif Pinelis Dec 7 '17 at 17:45
• Thanks all for the commentary so far. Getting my head around the answers at the moment – trusted Dec 11 '17 at 14:55

## 3 Answers

An alternative approach is just to choose some real trigonometric polynomial $p(x)=\sum_{k=-m}^m c_ke^{ikx}$ that is like $ct^2$ near the origin and like $1$ far from it and to minimize $q(x)=\sum_j p(a-a_j)$. What you are currently doing is just using $p(x)=1-\cos x$. So all you need to start ignoring the outliers to a higher extent is to localize more near $0$. For instance, you can consider using $1-(\frac{1+\cos x}2)^m$ with $m=2,3,\dots$ or something else like that and see how that works. The cost is that you'll need to compute $m$ complex Fourier coefficients of $\sum_j\delta_{a_j}$ instead of just one and the corresponding minimization problem for the resulting trigonometric polynomial $q(x)$ of degree $m$ will no longer have a simple closed form solution, so you'll have to resort to numerical minimization techniques, but you'll most likely get what you want if you choose $m$ in a good way. Notice that for large $m$, if you have two equal size tight clusters with positive distance between them, the minimum will be attained with $a$ near one of them rather than at the midpoint, but that is exactly what you wanted, isn't it?

Here is a family of alternatives. Let us identify the unit circle $C$ with the interval $[0,2\pi)$. For any two points $a$ and $b$ on the circle, consider the circular distance \begin{equation*} d(a,b):=\min(|a-b|,2\pi-|a-b|); \end{equation*} this is the geodesic (shortest) distance on the circle. For any real $p\ne0$, any $J\subseteq[N]:=\{1,\dots,N\}$, and any point $a\in C$, let \begin{equation*} d_p(a,A_J):=\Big(\sum_{j\in J}d(a,a_j)^p\Big)^{1/p}, \end{equation*} where $A_J:=\{a_j\colon j\in J\}$, so that $d_p(a,A_J)$ is an "$L^p$ summary distance" from $a$ to the subset $A_J$ of your data set. Note that $d_p(a,A_J)\to\min_{j\in J}d(a,a_j)$ as $p\to-\infty$ and $d_p(a,A_J)\to\max_{j\in J}d(a,a_j)$ as $p\to\infty$, provided that $J\ne\emptyset$.

Let then $a_{p,J}$ be any minimizer of the distance $d_p(a,A_J)$ in $a\in C$, so that $d_p(a_{p,J},A_J)\le d_p(a,A_J)$ for all $a\in C$; we may call $a_J$ an $L^p$-average of the set $A_J$. If the data set is not well enough clustered, then this average may be not unique. However, if the circle were replaced by a straight line with the usual metric on it and if $p=2$, then the $L^p$-average would be unique and it would coincide with the usual arithmetic mean. Anyhow, the minimal distance $d_p(a_J,A_J)$ is uniquely determined for each $J\subseteq[N]$.

Now take a real $q$ and re-enumerate your data points $a_i$ in $[0,2\pi)$ so that \begin{equation*} d_q(a_1,A_{[N]\setminus\{1\}})\le d_q(a_2,A_{[N]\setminus\{2\}})\le \dots \le d_q(a_N,A_{[N]\setminus\{N\}}); \tag{1} \end{equation*} here one may want to use a (finite) negative $q$, so that, in accordance with a previous remark, $d_q(a_j,A_{[N]\setminus\{j\}})$ is somewhat similar to $\min_{i\in [N]\setminus\{j\}}d(a_j,a_i)$. Then, for some natural $n<N$, the points $a_{N-n+1},\dots,a_N$ are the $n$ data points with the largest "$L^q$ summary distance" to your data set, and so, these $n$ points may be considered the outliers; you may use the values of $d_q$ in (1) to choose your number $n$ of such outliers.

Letting then $J_n:=[N-n]=\{1,\dots,N-n\}$ and choosing a presumably positive real $p$, you will have $a_{p,J_n}$ as a "value which represents the truncated average of the data in the highest density around the circle", after the removal of $n$ "farthest outliers".

To illustrate this approach, a sample of size $40$ from the normal distribution with $\mu=0.6$ and $\sigma=0.1$ was simulated, and also another sample, of size $10$ from the normal distribution with the same $\mu=0.6$ and the greater $\sigma=0.15$. Then the two samples were combined into one, of size $50$; the fractional part of each of the $50$ values was taken; and all the results were multiplied by $2\pi$. This will be our set of the $a_i$'s. Then the approach described above was used with $q=-5$, $p=2$, and $n=2$ (outliers). The points on the unit circle obtained from the sample of size $40$ are the red points here in the picture; the blue points come from the more spread-out sample of size $10$; the two outliers (in this case they both come from the more spread-out, blue'' sample) are represented by the two blue points of a greater size; the big black point represents the truncated average $a_{p,J_n}$. Remark: The subtle point in this approach is the recommendation that usually one should choose a finite negative value for $q$. Indeed, outliers may be understood, informally, as data points that lie far from the rest of data. So, one can try to measure the distances from potential outliers $a_j$ to the rest of the data as the minimum distance $\min_{i\in [N]\setminus\{j\}}d(a_j,a_i)$, which would correspond to $q=-\infty$, as mentioned before. But this choice of $q$ may be not the best sometimes. Indeed, usually outliers may be expected to be far away from one another. However, sometimes two outliers may turn out to be very close to each other, and then the minimum distance may not detect such outliers. To cover for such cases, one can use negative but finite values of $q$, to account also for the distances from the two (or more) such close-to-each-other outliers to the rest of the data. It seems to be a good idea to consider several negative values of $q$ and see how the outliers respond to such variation of the value of $q$.

In the first version of this answer, it was suggested to consider summary distances from potential outliers to subsets of the rest of the data of a certain size $m:=N-n<N$. This would take care of the problem of outliers very close to one another. However, then the volume of calculations would be much larger, on the order of $N^{n-1}$ times as large. Choosing a negative but finite $q$ seems to be a computationally much cheaper way to attain comparable quality of the results.

• I have modified the previous approach. The new approach is logically simpler. It is also substantially lighter computationally, and it likely has similar (quasi-)optimality properties. – Iosif Pinelis Dec 7 '17 at 22:10
• A disadvantage of this definition is that it depends on the choice of the x-axis. I thought so too until I started typing my answer. :-) The real problem is that $\tan x=\tan(\pi+x)$, so sometimes the OP gets the diametrically opposite point to the one he's looking for. However, up to this funny effect the formula is rotation invariant (unbelievable, isn't it?) – fedja Dec 7 '17 at 22:44
• Good point! Yes, this looks really unbelievable, until one looks at it a bit more closely. :-) – Iosif Pinelis Dec 7 '17 at 23:16
• I have added a remark (highlighted) explaining why one should choose a finite negative $q$ to measure the distances from potential outliers to the rest of the data. – Iosif Pinelis Dec 8 '17 at 14:10

First, in case you haven't run across this before, a classic reference for this area of statistics is Mardia and Jupp's Directional Statistics, which includes a number of measures capturing circular analogues of mean, variance, "median direction", etc.

Second, here is one concrete suggestion (which I think is a particular instance of fedja's answer). Interpret your circular data as points on the unit circle in $\mathbb{R}^2$. One common approach is to find the point in $\mathbb{R}^2$ that minimizes the $L_2$ error between it and your observations. This point happens to be equal to the mean in each dimension. This point lies on or within the unit circle, so (assuming the point does not happen to be the origin), project back it back to the circle to get your estimate. (The distance from the inner point to the unit circle also provides a measure of variance.)

You are interested in a more robust statistic. Recall that the mean is an $L_2$ minimizer (as above), whereas the median is an $L_1$ minimizer. So, take your data, select a point that minimizes the $L_1$ error to the observations, and project back to the circle. (Sadly, the "median direction" mentioned above is not, I believe, the same as this method.)