# Must a time series be stationary if it has no unit root?

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

• 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! Mar 2 '18 at 6:42
• 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.
– Ben
Mar 2 '18 at 22:17

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. (Strictly, for an ARMA model, you need to specify the joint distribution of $$p$$ elements, where $$p$$ is the degree of the autoregressive characteristic polynomial.)

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 an "anchoring 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 an "anchoring 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 if $$\sigma>0$$ then (regardless of the anchoring 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 anchoring distribution. Alternatively, if you want to specify a non-stationary process, you could impose a different marginal distribution at a particular 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 an "anchoring distribution" for the element $$X_0$$ 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 "anchoring 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 "anchoring point" of the process.

• Can we say it is stationary when a time series passes both the adf test and the kpss test. Feb 27 '18 at 7:16
• 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.
– Ben
Feb 27 '18 at 7:26
• however ,some articles said we use adf test to check unit root and use kpss test to check trend stationary.Is that correct. Feb 27 '18 at 8:06
• 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).
– Ben
Feb 27 '18 at 11:50
• Does the ADF and KPSS capture cases where the variance changes? Or deterministic non-stationarity. A second issue is that (I think) the ADF and KPSS does not capture non-stationarity due to structural breaks. Aug 24 '20 at 13:46

A process could be nonstationary in the sense of its distribution not being constant over time even without invoking any kind of autoregressive structure when, e.g., its variance changes over time (in a permanent fashion, so not like in a GARCH situation).

E.g., a permanent break in the variance at point in time $$c$$ would imply nonstationarity: $$X_t\sim\begin{cases}N(0,\sigma_1^2)&\quad t=-\infty,\ldots,c\\N(0,\sigma_2^2)&\quad t=c+1,\ldots,\infty\\\end{cases}$$

• While I agree with this answer (+1), and it is a useful perspective, the OP asks about the absence of a unit root, which suggests the context of an ARMA model, or at least a model that has a root that may or may not be a unit root.
– Ben
Aug 24 '20 at 22:30
• That is true, and your answer (+1, too :-)) surely is more pertinent for the OP. I just wanted to recall that there are other types of nonstationarities that are not related to autoregressive dynamics. Aug 25 '20 at 4:31