I commonly see people doing trend analysis of (monthly) timeseries data which show a strong inter-annual cycle following this scheme:

  1. compute climatological means ("mean January", "mean February", ..., "mean December")
  2. subtract climatological means from actual data, to yield an "anomaly timeseries"
  3. perform linear regression on this "anomaly timeseries"

Climatological in this case means multi-year average of individual months, e.g., an average of the 10 Januaries from 2000 to 2009. As Nick Cox points out in his comment, anomaly just means deviation from a reference level; there is no implication of anything pathological or very unusual.

While user31264's answer makes sense for processes where the seasonal component is truly purely additive. However, in atmospheric science we often have processes where the amplitude of the seasonal variation depends on the base level, i.e., is somewhat multiplicative.

Even in these scenarios, people often use the approach I outlined above. However, I could nowhere find a rigorous statistical explanation of why this approach is actually valid. Why is the linear regression of these anomalies a reasonable solution to the regression of the original timeseries data? Can you give me any justification why this could be reasonable to do? Those people I asked mostly say "everyone's doing it" ...

  • $\begingroup$ For others I will add that in climatology "anomaly" just means deviation from a reference level; there is no implication of anything pathological or very unusual. In terms of the question, who is claiming that this is "optimal"? What is optimal will depend at least tacitly on a model for the time series, quite apart from any other considerations? $\endgroup$ – Nick Cox Oct 18 '13 at 12:26
  • $\begingroup$ True, my choice of the word "optimal" was not optimal. Replaced with "reasonable". $\endgroup$ – andreas-h Oct 18 '13 at 13:28
  • $\begingroup$ (1) Is the multiplicative part negligibly small? After all, the base level changed by maybe 1 degree of Kelvin, while the base temperature is about 300 K (roughly). (2) Is there any statistical evidence that the seasonal variation shows long term trends, or there is a long term change in seasonal patterns? $\endgroup$ – user31264 Oct 18 '13 at 13:47
  • $\begingroup$ In my case (tropospheric NO2 pollution levels), the amplitude of the seasonal variation is clearly changing with time (clearly meaning by visual inspection). Which in my opinion shows that the multiplicative part cannot be neglected. As to (2), I'm not sure how to provide this statistical evidence. Any suggestions? $\endgroup$ – andreas-h Oct 18 '13 at 14:18
  • $\begingroup$ You may be interested in this post, which proposes a local-level model to fit trends in Alaska temperature data and compares it with a linear trend. Details are given in this paper. The analysis is done for the annual series (seasonal paths) but the model can be extended to incorporate a seasonal component. $\endgroup$ – javlacalle Oct 3 '14 at 11:57

The underlying model is considered to be $T(t)=f(t)+g(t)+h(t)$, where $T$ is temperature, $t$ is time, $f(t)$ is a function without seasonality, $g(t)$ is a periodic function, $h(t)$ is non-autocorrelative noise. The underlying assumption is that the movement of Earth around the orbit, which is periodical and independent on anything else, determines $g(t)$. There is also a small correction due to increase of concentration of greenhouse gases (and other things which are not seasonal in nature), which is . Note that the concentration of greenhouse gases showls little seasonality (unlike its derivative). In order to estimate $g(t)$, we calculate the mean temperature for each month.

In order to obtain the trends, the autoregression of $f(t)+h(t)$ is better than the autoregression of the whole $T(t)$, because $g(t)$, being periodic, has its own strong autocorrelation.

  • $\begingroup$ Thanks, this makes sense for purely additive processes. However, I often see people use this approach when the underlying process is clearly not purely additive. See my updated question. $\endgroup$ – andreas-h Oct 18 '13 at 13:35

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