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I am working on a problem where I have to first classify whether a time-series is periodic or not and then, if it is periodic identify its period(s) (the time-series could have multiple periodicities, also).

I need a reliable approach to say that a time-series is not periodic. I have been searching about this topic and these are some leads:

Are there some other properties of time-series that can be used?

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  • $\begingroup$ I would love some of the promising citations you have seen for your two bullet points. :) Also: Welcome to CV, Hari. $\endgroup$ – Alexis Mar 12 at 7:46
  • $\begingroup$ Please, see the updates. $\endgroup$ – Hari Mar 12 at 7:54
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I would approach the problem "check if a time-series is not periodic" as follows

Maybe this is a bit naive approach but for, at least a simple case, it seems to work (see example below).

Example

As minimal test for this 2-stages approach we could see how it works if we have the following 2 series:

  • Periodic series: $y_{i}^{(1)} = \sin(x_{i}) + \epsilon_{i}$
  • Non periodic series: $y_{i}^{(2)} = \alpha + \beta x_{i} + \epsilon_{i}$

where $x_{i} = [-2\pi, 2\pi]$ and $\epsilon_{i} \sim N(0, \sigma^{2})$ and $i = 1,...,N$.

Below you will find a R code for these settings. The harmonic test is performed for the two series using the PML package.

library(PML)
N  = 100
x  = seq(-2*pi, 2*pi, len = N)
ep = rnorm(N) * 0.25
y1 = sin(x) + ep
y2 =  x + 0.5 + ep 

re1 = test.harmonic(y1, p = 0.025/(N - 1)) # p: correct for multiple freq 
re1$sig
       
        frequency     prop (g)  p-value     p-threshold
      1 50.000000 0.3565748 3.148508e-08 0.0002525253
      2  2.857143 0.0316902 1.000000e+00 0.0002525253

re2 = test.harmonic(y2, p = 0.025/(N-1))
re2$sig 
      
      frequency  prop (g)      p-value  p-threshold
      1 100     0.2077127   0.0006863734 0.0002525253

From the results above we notice that for $y^{(1)}$ we have one significant frequency while for the linear function we do not find any.

I hope this answer helps (maybe it does not solve your problem but suggests a possible line of investigation).

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