2
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The above example will pass ADF test as trend-stationary. $\endgroup$ – Wisentgenus Jan 7 '15 at 3:06
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.