In circular statistics, the expectation value of a random variable $Z$ with values on the circle $S$ is defined as $$ m_1(Z)=\int_S z P^Z(\theta)\textrm{d}\theta $$ (see wikipedia). This is a very natural definition, as is the definition of the variance $$ \mathrm{Var}(Z)=1-|m_1(Z)|. $$ So we didn't need a second moment in order to define the variance!

Nonetheless, we define the higher moments $$ m_n(Z)=\int_S z^n P^Z(\theta)\textrm{d}\theta. $$ I admit that this looks rather natural as well at first sight, and very similar to the definition in linear statistics. But still I feel a little bit uncomfortable, and have the following


1. What is measured by the higher moments defined above (intuitively)? Which properties of the distribution can be characterized by their moments?

2. In the computation of the higher moments we use multiplication of complex numbers, although we think of the values of our random variables merely as vectors in the plane or as angles. I know that complex multiplication is essentially addition of angles in this case, but still: Why is complex multiplication a meaningful operation for circular data?


1 Answer 1


The moments are the Fourier coefficients of the probability measure $P^Z$. Suppose (for the sake of intuition) that $Z$ has a density. Then the argument (angle from $1$ in the complex plane) of $Z$ has a density on $[0,2\pi)$, and the moments are the coefficients when that density is expanded in a Fourier series. Thus the usual intuition about Fourier series applies -- these measure the strengths of frequencies in that density.

As for your second question, I think you already gave the answer: "complex multiplication is essentially addition of angles in this case".

  • $\begingroup$ Thank you, this is really helpful. (Shame on me for not recognizing a Fourier series even when hurtling towards it...) $\endgroup$
    – Rasmus
    Aug 6, 2010 at 12:55
  • $\begingroup$ Does this mean that the moments of a circular distribution should be compared to the characteristic function of a linear distribution rather than to its moments? $\endgroup$
    – Rasmus
    Aug 6, 2010 at 17:53
  • $\begingroup$ @Rasmus: I guess that depends on exactly what you want to do with the information, but in general I'd say yes. $\endgroup$ Aug 6, 2010 at 19:13

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