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If I have a time series that has got seasonality, does that automatically make the series non stationary? My intuition (probably off) is that it does not.

Seasonality means that the series goes up and down around a constant value....something like a sine wave. So by this logic a time series with seasonality is a (weakly) stationary series (constant mean).

Is this wrong? Why?

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    $\begingroup$ Given the votes and the discussions in the comments, have you considered accepting a different answer? $\endgroup$ May 25, 2021 at 20:23

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Seasonality does not make your series non-stationary. The stationarity applies to the errors of your data generating process, e.g. $y_t=sin(t)+\varepsilon_t$, where $\varepsilon_t\sim\mathcal{N}(0,\sigma^2)$ and $Cov[\varepsilon_s,\varepsilon_t]=\sigma^21_{s=t}$ is a stationary process, despite having a periodic wave in it, because the errors are stationary.

Seasonality does not make your process stationary either. Consider the same process but $\varepsilon_t\sim\mathcal{N}(0,t\sigma^2)$, in this case the error variance is non-stationary and seasonality has nothing to do with it.

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    $\begingroup$ I disagree with this answer. The series is not even weakly stationary (a.k.a. wide-sense stationary) because $E[Y_t] = \sin(t)$ is not a constant. It is what is sometimes referred to as covariance-stationary because the covariance $\operatorname{cov}(Y_{t_1},Y_{t_2})$ depends only on the difference $t_1-t_2$ between the time instants. The series is, of course, not strictly stationary in any sense of the word. $\endgroup$ Oct 16, 2015 at 12:57
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    $\begingroup$ Determinism, that is, lack of randomness, is not what is relevant here; it is the definition of stationarity (or weak stationarity since time series folks seem to use stationary to mean weakly stationary or wide-sense stationary) that is relevant, and by the usual definitions, your answer is incorrect. See, for example this more recent question where the issue is discussed in detail and the accepted answer there (by @Silverfish) is a contradiction of your answer here. $\endgroup$ Oct 16, 2015 at 13:58
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    $\begingroup$ @Aksakal You are not reading what I am writing correctly. I don't claim that random walk is stationary. I said you cannot claim that a process is stationary because a modified version of it is stationary. Random walk is non-stationary because its unconditional variance is growing, however if we follow your logic of conditioning it has constant conditional variance. In general you are wrong in the definition of WSS. $\endgroup$ Oct 16, 2015 at 15:13
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    $\begingroup$ You are side tracking. You can call a process trend stationary, difference stationary, etc., but that process is not stationary considering the formal definition of stationarity. You are wrong and turning this into a pissing contest. Open any signal processing book you will find the definition as it is used in academy. Just suck it up. $\endgroup$ Oct 16, 2015 at 15:18
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    $\begingroup$ @Aksakal Nowhere in your answer do you mention trend stationarity or covariance stationarity or any other form of stationarity (weak sense, to order $n$, ot whatever. If $t$ is assumed to be an integer, then $\sin(0), \sin(1), \sin(2), \sin(3),\cdots$ is not even a periodic sequence (in the sense that $x_{n+N} = x_n$ for all integer $n$) or an ultimately periodic sequence either. Thus, your example, acceptable though it is to the OP, is not by any means an answer to the question asked. I am downvoting your answer: it in not just "not useful", it is downright misleading. Poor OP. $\endgroup$ Oct 16, 2015 at 18:48
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IMHO, persistent seasonality, by definition, is a type of non-stationarity: the mean of a seasonal process varies with the season, E[z(t*s+j)] = f(j), where s is the number of seasons, j is a particular season (j=1,...,s), and t is specific period (typically a year). Thus, E[y(t)] = E[sin(t)+u(t)] = sin(t) is not a stable mean, although it is deterministic: you could group observations with different means.

Luis

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    $\begingroup$ +1 I agree with your statement that seasonality is a type of non-stationarity. $\endgroup$ Oct 16, 2015 at 13:11
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A seasonal pattern that remains stable over time does not make the series non-stationary. A non-stable seasonal pattern, for example a seasonal random walk, will make the data non-stationary.

Edit (after new answer and comments)

A stable seasonal pattern is not stationary in the sense that the mean of the series will vary across seasons and, hence, depends on time; but it is stationary in the sense that we can expect the same mean for the same month in different years.

A stable seasonal pattern may therefore fit in the concept of a cyclostationary process, i.e., a process with a periodic mean and a periodic autocorrelation function.

The above does not apply to a non-stable seasonal pattern.

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    $\begingroup$ +1 for bringing up the concept of cyclostationary processes. $\endgroup$ Oct 17, 2015 at 13:58
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I don`t agree that seasonality is a type of non-stationarity because the concept of stationarity in natural systems already incorporates the idea of fluctuation within an unchanging envelope of variability (Milly et al., 2008). Speaking about hydrological time-series, even though they are stochastic (random process) and commonly have seasonality, that is, they contain wet and dry periods, they will always be stationary if the mean and the variance do not change over time. So, ignoring the uncertainties of the effects of climate change, a hydrological time series should normally be stationary even though it still has seasonality. This is why stationarity is a widely accepted concept for civil engineering design, and because of this concept hydrologists are able to calculate the recurrence time of floods, for example.

Link for Milly et al. (2008): "Stationarity Is Dead: Whither Water Management?"

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  • $\begingroup$ While your link is interesting, the definition it uses is not the same as the one used in the mathematical theories of stationary processes. Mathematics needs precise definitions that can be used in proofs, and fluctuate within an unchanging envelope of variability is not so. But see the answer by @javlacalle $\endgroup$ Dec 10, 2021 at 21:49

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