I am searching for an example of an unstable VAR($p$) process (its reverse characteristic polynomial has no roots inside and on the complex unit circle) which is stationary. I come up with this question as it is often stated in (multivariate) time series analysis that stable (VAR($p$)) processes are stationary, but the reverse is not necessarily true. So, what kind of process can be unstable but remains stationary?

  • 1
    $\begingroup$ Are you sure that it should not be the other way round? For example, an AR(1) process with coefficient l.t. 1 in absolute value whose error variance changes at some point is stable (which means mean-reverting for me), but not stationary, because its distribution changes over time. $\endgroup$ – Christoph Hanck Nov 3 '15 at 15:17
  • $\begingroup$ @ChristophHanck, OP might be correct. it's from Lutkepohl's "Introduction to Multiple Time Series Analysis". I found the citation here. $\endgroup$ – Richard Hardy Nov 3 '15 at 16:00
  • $\begingroup$ Thanks @ChristophHanck for your comment. I was not aware that there could be a confusion regarding the definition of stability. I refer to stable VAR($p$) processes defined trough the norm of the roots of the characteristic polynomial. Sorry, I will edit this in my question. As far as I understand your point, this assumption is not fulfilled by an AR(1) process with unit root? $\endgroup$ – muffin1974 Nov 3 '15 at 16:00
  • $\begingroup$ What I was referring to was the idea that $y_t = \rho y_{t-1}+u_t$ is not stationary if $u_t\sim(0,\sigma_1^2)$ for e.g. $t=1,...,T/2$ and $u_t\sim(0,\sigma_2^2)$ afterwards, because the variance change implies that the unconditional distribution is not independent of $t$. That should be true whether we have an AR or VAR process, and irrespective of whether $\rho$ is less than one in absolute value or not. $\endgroup$ – Christoph Hanck Nov 3 '15 at 16:05
  • $\begingroup$ @RichardHardy, thanks for the link, which indeed points in that direction. Lütkepohl rules out my type of counterexample at the beginning of Section 2 by assuming the variance of $u_t$ not to depend on $t$. $\endgroup$ – Christoph Hanck Nov 3 '15 at 16:07

Your Answer

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

Browse other questions tagged or ask your own question.