Talk:Circular mean

message: Statements like ""the arithmetic mean of 0° and 360° is 180°, although 0° would be clearly the better choice" are very misleading. 
Consider the case of a process in which the resulting angle is the sum of many small angular displacements about a mean which we can assume is 0 degrees. As the number of displacements N grows larger and the angular displacements δ grow smaller (with Nδ remaining finite)  the distribution tends towards a normal distribution of angles about 0 degrees where the range of angles is from negative infinity to positive infinity. The problem is that we cannot measure the "true" angle \theta, we can only measure the "measured angle" \theta_m which is constrained to lie inside, lets say, 0 to 2\pi. A measured angle of 358 degrees could result from a small number of displacements in the negative direction (true angle=-2 degrees), or it could result from a large number of displacements in the positive direction (true angle=358 degrees) or even a very large number of displacements in the positive direction (true angle = 360+358=718 degrees) etc. etc. To say that for two measured angles of 2 and 358 degrees, 0 degrees is a better estimator of the mean of the underlying distribution rather than 2+358=180 is unjustified. It could happen that the distribution had a mean of 180 degrees. Because the two measurements are bunched around 0 degrees, its more likely that the mean will be some multiple of 360 degrees, but not for certain, and you cannot say its simply zero either. The average of the true angles is simply unknown, and the average of the measured angles is misleading, no matter how you do it.
This is why z=e^{i\theta_m}=e^{i\theta} is used as the working variable. It is unambiguous, despite the ambiguity in the true angle. It's what we actually measure. When we calculate the average z value, we obtain Ave(z)=cos(2 degrees), and thats a valid estimate for e^{i\mu} where \mu is the mean. But when we calulate arg(Ave(z)) we do not get zero. We re-introduce the uncertainty. arg(Ave(z))=2n\pi where n is some unknown integer. In other words, an estimate for the mean being 360 degrees (true value) is just as valid as an estimate of 0 degrees.
I don't have any references for this, but I really believe that the article is glossing over the true nature of the problem in favor of some hand-waving and incorrect arguments. Does anyone agree, and are there any references that can be found for this? PAR (talk) 19:08, 1 January 2010 (UTC)
replies: text: If you have "true" angles, and -2° is distinct from 358°, then you haven't really got a circular topology, so the sort of mean described by this article is not appropriate (I suppose). --catslash (talk) 13:55, 19 January 2011 (UTC)
thread_title:  This article is misleading 

message: PAR, the case is question is a a set of points on the rim of a circle, not a set of angle measurements. A point can be assigned the angle of 1, 361, or -719, but these angles all refer to the same point. If you map those points in some way to a distribution on R1, the average of that distribution might not correspond to what you would naturally call the average on the circle. The article was a little more relaxed and non-rigorous than you are trying to make it. 
James M, 07:00ish, 5 June 2010 (UTC)
: Is there anything in what I wrote that gives the impression I am confusing the two? I agree with your last sentence. Is there anything wrong with that? PAR (talk) 19:34, 6 June 2010 (UTC)
thread_title:  Don't mistake the angle for the actual point 

message: The Cartesian average of the points would NOT be on the circle's rim, except in the trivial case of every point coinciding. The average converted to polar coordinates would have a radius of less than one. That angle with the radius set to one would give a point on the circle's rim. That error probably added to the confusion. 
James M, 07:16, 5 June 2010 (UTC)
thread_title:  However the article had a serious error 

message: I think the definition should be discussed more. I must say, I never found a proper discussion on this problem of mean averaging.
For example, given the formula in this article, the mean of 0°, 10° and 180° would be .... 10°, which is quite counter-intuitive. I would rather have a mean of 85°.
Also, isn't it a problem that the mean of a distribution would not be in the domain of this distribution? The argument of the euclidean distance works only if the circle is a meaningful representation of the circular quantities and the distance from one point to any other is a meaningful measure for the circular quantity considered.
I usually use this definition for the discrete average of circular quantities:
 \hat{\theta} = \mathop{\rm arg\,min}_{\tilde{\theta} \in D} \sum_{i=1}^{N} d^2(\theta_i, \tilde{\theta}) 
with d(x,y) the cyclic distance (i.e. d(350,10) = 20, d^2(350,10) = 400), and D the cyclic domain.
For some reason, \mathop doesn't behave as usual for subscripts ... but that's the idea.
This definition has the advantage to behaves exactly as the usual average when all the data is confined in an area where the circularity has no impact (i.e. if the spread of the data is less than half the total size of the domain). However, it is difficult to find this argmin.
What do you think?
--PierreBarbierDeReuille (talk) 14:52, 16 December 2010 (UTC)
replies: text: You have to ask, what is the point of taking a mean? Its because it is an estimator of one of the parameters of the underlying population, and that population has a distribution which YOU assume. For example, in linear statistics, you assume a normal distribution with unknown mean and variance, and then you have a sample and you want to know the best estimate of the mean of the normal population. Thats just the sum of the sample values divided by the number of samples. If you assume a different distribution, then maybe thats not the way to go. Its the same with circular statistics - you assume that some underlying population gave you the sample, and you want to estimate the parameters of that population using that sample. Usually one assumes a von Mises distribution, or a wrapped normal distribution, and then the mean taken in the prescribed way will be an estimator of the mean of the von Mises or wrapped normal distribution. Approach it this way and the mean as described makes more sense. PAR (talk) 07:28, 6 February 2013 (UTC)
thread_title:  Problem with the definition 

message: Isn't \operatorname{atan2}\left(\frac{1}{n}\cdot\sum_{j=1}^n \sin\alpha_j, \frac{1}{n}\cdot\sum_{j=1}^n \cos\alpha_j\right)  the same as \operatorname{atan2}\left(\sum_{j=1}^n \sin\alpha_j, \sum_{j=1}^n \cos\alpha_j\right) ?
And \arg\left(\frac{1}{n}\cdot\sum_{j=1}^n \exp(i\cdot\alpha_j)\right)  the same as \arg\left(\sum_{j=1}^n \exp(i\cdot\alpha_j)\right) ?
Multiplying or dividing by a positive real number should not affect the angle. --208.80.119.68 (talk) 17:43, 10 July 2012 (UTC)
replies: text: I was thinking just the same thing. --Amble (talk) 02:42, 22 July 2015 (UTC)
replies: text: You are right, the scaling in the formulas as they stand is unnecessary. I guess I wrote it this way because I explained above the relation to the arithmetic mean of points on the circle and the meaning of the distance from the origin. I have rewritten the formulas and added a remark on the omitted scaling. HenningThielemann (talk) 17:38, 13 February 2016 (UTC)
thread_title:  Don't need to divide by n