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I gather from this question and answer, a seasonal series is non-stationary as its mean depends on which month it is.

Suppose I have a series which possibly has a unit root (stochastic trend) but also exhibit strong seasonality (see the chart below).

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Now I think I can remove seasonality first and then use a suitable unit root test. My doubt is, would it be theoretically sound if I directly perform unit root test on the original series.

My feeling is that a strong seasonality, such that a large part of variance in series is coming from seasonal fluctuation, can hide the underlying stochastic trend, causing the unit root test to misidentify the series as having no unit root.

Does this make sense? Further, what if we try to handle the seasonality within the unit root test. For example, in ADF test, we include seasonal dummies (just like we include a trend variable to handle deterministic trend)?. But even if this is correct, we wouldn't have the right critical values for it, right?

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  • $\begingroup$ Comment: A negative unit root produces cycles of period 2 unless the error is very large. $\endgroup$
    – Alexis
    Mar 7 '21 at 17:11
  • $\begingroup$ @Alexis: what do you mean when you say negative unit root? $\endgroup$
    – Dayne
    Mar 8 '21 at 11:42
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    $\begingroup$ For example, $y_t = -y_{t-1} + \varepsilon_{t}$ has a unit root (memory is infinite, perturbations always affect the future state), however the root is negative which induces period 2 cycles. It is easy to simulate a see why in R. $\endgroup$
    – Alexis
    Mar 8 '21 at 17:33
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    $\begingroup$ And let me retract my "unless the error is very large" clause: A negative (1 lag as in my example in the previous comment) unit root produces cycles of period 2. $\endgroup$
    – Alexis
    Mar 8 '21 at 19:12
  • $\begingroup$ @Alexis: this is interesting. I haven't simulated to check this but if we expand this we will get $y_t = \sum_{i=0}^t (-1)^{i}\varepsilon_i$. Besides iid assumption, if we add an assumption of symmetric distribution on $\varepsilon_t$ then wouldn't it be same as $\sum_{i=0}^t \varepsilon_i$? $\endgroup$
    – Dayne
    Mar 9 '21 at 15:33
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A few days after I posted this question, I attempted simulations to test my intuition that strong seasonality can hide the underlying stochastic trend in the usual ADF test. Consider the following process:

\begin{align} y_t &= S_t+y_{t-1} + \varepsilon_t \\ \text{where, } \,\, &S_t \sim N(\mu_i,\sigma_s^2) \,\,\,\text{for }t=4n+i;\tag{$i=1,2,3,4;n \in \mathbb N$}\\ &\sum\limits_i \mu_i = 0; \\ &\varepsilon_t \sim IN(0,\sigma_{\varepsilon}^2) \end{align}

We can think of this process as a quarterly time series exhibiting stable seasonality. Intuitively, the strongness of seasonality can be measured by $(\text{range}(\mu_i))/\sigma_{\varepsilon}^2$ (further logic in the end). Higher this ratio, more likely it would be that seasonality will hide unit root in the usual test.

Chart below shows the results of the sumulation. 10,000 series of length 1000 are generated from the above model and a pure random walk model. Values for seasonal term are taken as $\mu = (1.7,0.8,-1.0,-1.5)$. DF-statistic is calculated for each series using urca::ur.df(., lag = 0) and density estimates are plotted.

enter image description here

From above chart, it is interesting to see that when $\sigma_{\varepsilon}$ is small the usual unit root test can be very wrong.

Theoretical justification: In this paper, the authors have derived the analogous DF distribution in presence of an additive outlier. The distribution is similar to the usual DF distribution with an additional term as function of $\theta/\sigma_{\varepsilon}^2$; where $\theta$ is the coefficient of the outlier dummy. The above process is somewhat a special case with four additive type outliers and perhaps that's why we are getting the similar results.

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  • $\begingroup$ So in other words, it is highly dependent of values (outliers) that can foul the test, which would result in a pre screening of outliers before conducting unit-root tests? (upvote for you) $\endgroup$ Mar 6 '21 at 11:24
  • $\begingroup$ @PatrickBormann: Yes it appears that presence of outlier can potentially influence the asymptomatic distribution of the t-statistic (i found in my simulations that the distortion is more for smaller samples). But I don't know how to detect for outliers in the series has unit root. Most algos for detection are designed for stationary series afaik. $\endgroup$
    – Dayne
    Mar 6 '21 at 11:29
  • $\begingroup$ A friend of mine once wrote an algorithm for outlier detection, a so called streaming algorithm: towardsdatascience.com/… maybe you want to have a look at it, perhaps it can help you, while it is not especially for unit root, maybe you can test if it handles your artifical case. $\endgroup$ Mar 6 '21 at 11:31
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I am no expert in this topic, but there seem to be dedicated unit root tests developed for seasonal data.

See for example:

Ghysels, Eric, Hahn S. Lee, and Jaesum Noh. "Testing for unit roots in seasonal time series: some theoretical extensions and a Monte Carlo investigation." Journal of Econometrics 62, No. 2 (1994): 415-442.

While we find the procedure proposed by Hylleberg, Engle, Granger, and Yoo (1990) the most useful among the alternative procedures, we caution users of many remaining serious obstacles when testing for unit roots in seasonal time series.

The Hylleberg, Engle, Granger, and Yoo (1990) approach is also included in the uroot package in R, see here for the Vignette.

According to the ADF test, Dickey et al. (1984) obtained the critical values of $\phi$ under the following data generating process. $(1 − L^S)\,y_t = \Phi y_{t−S} + \epsilon_t$, $\epsilon_t ∼ iid (0, \sigma_{\epsilon^2})$ where $S$ is peridocity of the data: 4 in the case of quarterly series and 12 for monthly series. Under the null hypothesis $\Phi$ is equal to 1 and the process contains all the roots in Table (2), page 7. In practice, however, it may be some, but not all, seasonal unit roots. Hence, it would be convenient to specify a model in which the regressors allow to test for individual roots. That is what Hylleberg et al. (1990) achieve in the case of quarterly series.The seasonal operator can be decomposed into the following polynomials: $(1 − L^4) = (1 − L)(1 + L)(1 − iL)(1 + iL)$, where the roots are $(±1, ±i)$. Assuming $y_t$ is generated by an AR(p) process, Hylleberg et al. (1990) show that $y_t$ can be represented as $\phi (L) \Delta^4 y_t = \pi_1 y_{1,t−1} + \pi_2 y_{2,t−1} + \pi_3 y_{3,t−2} + \pi_4 y_{3,t−1} + \epsilon_t$ ,

Hylleberg, Svend, Robert F. Engle, Clive WJ Granger, and Byung Sam Yoo. "Seasonal integration and cointegration." Journal of Econometrics 44, No. 1-2 (1990): 215-238.

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    $\begingroup$ Thanks for links to these useful papers. I haven't got thru both the papers in detail yet but it appears that they have a lot to do with seasonal unit roots. I am rather interested in case when there may be a usual unit root plus a stable seasonality. In fact, few days after posting this question, I ran simulations and found that the limiting distribution shifts left of actual DF distribution in presence of seasonality - thereby misidentifying series as stationary. $\endgroup$
    – Dayne
    Mar 3 '21 at 0:51
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A series is not automatically non-stationary if it has a seasonal pattern. For example, Rob Hyndman shows that a cyclic pattern may exist that makes it unpredictable. There is also a good explanation on towarddatascience.com on that topic.

Instead of directly performing the ADF test I would first consider ACF/PACF plots to see if a single series may be stationary or not, that normally clears out a lot of questions. The correlogram is the first plot to be conducted to get a first glance. Normally this is the starting point in checking if you need additional tests. See here (you have a lot of scrolling to do).

However, I believe there is no harm on using the ADF on a non transformed series, because you should already have insights from your ACF. If you don't have any insights, well I once developed a genetic algorithm which fits on a VAR/VECM in R. These approaches normally need a unified level of variables in the model. Thus if there is one variable that is stationary and one that is not, you should log transform and difference all variables, or difference them a second time, you need a unified level. This is also a sign for me in no harm, as we try to measure before and after logging and differencing with the ADF. In addition, believe in Stata it is also one of the first tests conducted on non transformed variables before the KPSS as this comes after the ADF: If your original variable already has a unit root, then it is non stationary. In the end you want to look at the real data.

Further, what if we try to handle the seasonality within the unit root test. For example, in ADF test, we include seasonal dummies (just like we include a trend variable to handle deterministic trend)?. But even if this is correct, we wouldn't have the right critical values for it, right?

It is possible to do seasonal differencing, search for it with the tag: Rob Hyndman, the master of time series ;-) I do not believe you do it with dummies, you do seasonal differencing and then you can normally interpret ADF.

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  • $\begingroup$ Hi, welcome to CV! If possible, please add references for your links, in case they die in the future $\endgroup$
    – Antoine
    Mar 2 '21 at 20:34
  • $\begingroup$ The titles you added to the links are not informative enough to be helpful in case the links die. $\endgroup$ Mar 6 '21 at 9:02
  • $\begingroup$ I filled in the original links $\endgroup$ Mar 6 '21 at 11:03
  • $\begingroup$ Welcome to CV, Patric Bormann! When you wrote "For example Rob hyndman, shows that a cyclic pattern may exist that makes it unpredictable." did you mean cyclic (up and down and up again without regard to intervals between peaks and valleys) or periodic (cycles with regular or nearly regular temporal intervals)? $\endgroup$
    – Alexis
    Mar 7 '21 at 17:05
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    $\begingroup$ You are totally right, in addition I forgot to mention that the cyclic pattern I referenced to could be seen on the first image on that link on towarddatascience with the letter (g) , it is cyclic. $\endgroup$ Mar 7 '21 at 19:25

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