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Must a time series be stationary if it has no unit root? I am not quite sure.

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    $\begingroup$ Close-voters: I'm unclear on why this is being voted to close as "unclear". It seems like a straightforward time-series question to me. I'd ask anyone not familiar with time-series to skip this in the voting queue. Thank you! $\endgroup$ – Stephan Kolassa Mar 2 '18 at 6:42
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    $\begingroup$ It is possible that this was being voted as "unclear" because if fails to specify the type of stationarity (e.g., strict, weak, asymptotic, etc.) at issue. However, I agree that it is a straightforward question even without specification of this. $\endgroup$ – Ben Mar 2 '18 at 22:17
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No, roots outside the unit circle gives asymptotic stationarity, not strict stationarity: For a standard ARMA time-series model, the recursive formula for a model locks in the autocorrelation of the process, but does not lock in the marginal/joint distribution for the process. To get the latter you also need to specify the marginal distribution at some point in time, and this specification then locks in the full joint distribution of the process.

If all of the roots of an AR process are outside the unit circle (no unit roots and no explosive roots), and the error terms are IID, then the process will have mean-reverting behaviour and will converge asymptotically to a stationary distribution. This asymptotic convergence to a stationary distribution is a weaker property than strict stationarity. It is possible for such a process to be non-stationary by specifying a "starting distribution" that is not the asymptotic stationary distribution.

Stationarity for an AR(1) process: Consider a standard first-order auto-regressive process defined by the recursive equation:

$$\begin{matrix} X_{t+1} = \mu + \alpha (X_t - \mu) + \varepsilon_t & & \varepsilon_t \sim \text{ IID N}(0, \sigma^2). \end{matrix}$$

Suppose that $| \alpha | <1$ so that this has a single root $1/\alpha$ outside the unit circle. This model form defines a broad set of possible time-series processes --- i.e., all processes that obey the requisite recursive equation and have the specified noise distribution. However, with this recursive equation you have not yet specified a "starting distribution" for any particular point in the series, and so this presently encompasses some stationary and some non-stationary processes. Now, it can be shown that (regardless of the starting distribution) this process has the asymptotic stationary distribution:

$$X_{\infty} \sim \text{N} \Big( \mu, \frac{\sigma^2}{1 - \alpha^2} \Big).$$

In order to form a strictly stationary process you would need to impose the requirement that the marginal distribution at some arbitrary time is equal to this stationary distribution. Alternatively, if you want to specify a non-stationary process, you could impose a different marginal distribution at a particular "starting time" (perhaps with a different mean or variance, or maybe even a different distributional form).

Non-stationary AR(1) model (with root still inside the unit circle): Suppose you specify the "starting distribution" that is still a normal distribution, but with some arbitrary mean and variance:

$$X_0 \sim \text{N} ( \mu_0, \sigma_0^2 ).$$

With this "starting distribution" it can be shown that the series of marginal distributions is:

$$X_t \sim \text{N} \Big( (1- \alpha^t) \mu + \alpha^t \mu_0, \alpha^{2t} \sigma_0^2 +\frac{1-\alpha^{2t}}{1-\alpha^2} \sigma^2 \Big).$$

For $\mu_0 \neq \mu$ this series has non-stationary mean, though it converges to the asymptotic mean $\mu$. For $\sigma_0 \neq \sigma$ this series has non-stationary variance, though it converges to the asymptotic variance $\sigma^2 / (1-\alpha^2)$. In this case the process is non-stationary, but it gets closer and closer to being stationary the further you get from the prescribed "starting point" of the process.

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    $\begingroup$ Can we say it is stationary when a time series passes both the adf test and the kpss test. $\endgroup$ – inaMinute Feb 27 '18 at 7:16
  • $\begingroup$ These are both unit-root tests, so they will test the roots of the AR part. In cases where you have observed time-series data it is usual to assume a model form that is stationary, so long as you have roots inside the unit circle. $\endgroup$ – Ben Feb 27 '18 at 7:26
  • $\begingroup$ however ,some articles said we use adf test to check unit root and use kpss test to check trend stationary.Is that correct. $\endgroup$ – inaMinute Feb 27 '18 at 8:06
  • $\begingroup$ You would have to look into the details of the test, but my understanding is that this test assumes the stationary form in the case where the root is within the unit circle (or else it appeals to the asymptotic distribution, which is the stationary form). $\endgroup$ – Ben Feb 27 '18 at 11:50

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