Unit Roots, Stationarity, and Deterministic Trends
Summary
The discussion separates unit-root testing from the different meanings of stationarity. An augmented Dickey–Fuller test evaluates evidence for a unit root under its model assumptions; a conclusion about that test does not by itself establish that a series is stationary in every sense. The answers illustrate the distinction with an autoregressive process and explain that deterministic components and stochastic components must be considered separately.
A process with a deterministic trend can have a stationary random component while its overall level changes over time. The example also distinguishes finite variance at a given time from a stable long-run distribution. These points help prevent treating mean, variance, covariance, and unit-root properties as interchangeable. The short discussion is conceptual rather than a full treatment of test specification, power, or alternative stationarity tests, and one answer’s example appears to contain a parameterization inconsistency that readers should check before relying on its conclusion.
Key ideas
- A unit-root test does not establish every form of stationarity by itself.
- Stationarity definitions concern the stochastic component and may exclude deterministic trends.
- A process can have finite variance at each time while its level changes over time.
- The treatment of deterministic components affects how stationarity should be interpreted.
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Full text
# Does unit root stationary imply mean stationary and variance stationary?
# Does unit root stationary imply mean stationary and variance stationary?
Newbie question. I am reading about stationary series and understand that it has many forms:
- mean stationary
- variance stationary
- covariance stationary
My question is does unit root stationary imply all of the above? If I run an augmented dicky fuller test and find that the series is unit root stationarity is that sufficient to say that the series is stationary in all respects?
## Answer by Bryce (score 2, accepted)
https://quant.stackexchange.com/a/16070
Consider the following AR(1) process with i.i.d. normal errors that have zero mean and finite variance $\sigma^2>0$,
$$ x_t = (1-\rho)\mu + \rho x_{t-1} + \epsilon _t$$
Now suppose $ \rho = 1/2$ and $\mu = 1$. This process does not have a unit root, and it is not mean stationary. At any point in time, the process has finite variance, although as time diverges to infinity, the level of $x_t$ diverges to infinity as well.
## Answer by user1483 (score 1)
https://quant.stackexchange.com/a/16128
Write the series in the answer as
$(x_t - \mu) = \rho (x_{t-1} - \mu) + \varepsilon_t$
then if $\rho=.5$ and $\varepsilon_t$ is $N(0,\sigma^2)$, $(x_t - \mu)$ is stationary with mean $0$ and variance $\frac{\sigma^2}{1-\rho}$.
A time series process can have a deterministic part and a pure random part. The definition of stationarity (strict or strong or second order or \dots) refers only to the pure random part. If the deterministic part is a trend and the random part is stationary the series is trend stationary.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.