THis question refers to the last unanswered comment in this question:

Decomposing a time series with some zero values

I see, thank you for the explanation. Let's suppose that the time series follows a multiplicative model. Wouldn't it be helpful to add a positive constant to the time series (so that all the values become greater than zero), perform the decomposition on the transformed data and then go back to the original scale by a back-trasformation? – John M

I'm doing this right now with my model, but I'm not sure if this holds as a solution and I can't seem to find anything on the internet. I know that 0 values and negative values can result in wrong decomposition with imaginary numbers.

  • $\begingroup$ The quotation, as given, is obviously wrong, because when a series follows a multiplicative model, adding a constant to it guarantees the result cannot be multiplicative. I suspect it might have been intended to say that sometimes after adding a constant a series might then follow a multiplicative model. Could you please explain your interpretation and state clearly what you mean by "holds as a solution"? $\endgroup$
    – whuber
    Feb 4, 2022 at 15:57
  • $\begingroup$ I just dont understand why this: d = seasonal_decompose(df, period=6, model="multiplicative", two_sided=False) is so different to this: d = seasonal_decompose(df + constant, period=6, model="multiplicative", two_sided=False) and why the model is not multiplicative anymore $\endgroup$
    – Fenrir
    Feb 4, 2022 at 16:21
  • $\begingroup$ It would violate laws of arithmetic if those models were exactly the same. Statistically, both might be equally good fits to the data provided the constant is sufficiently small. $\endgroup$
    – whuber
    Feb 4, 2022 at 17:59


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