For a stationary Vector autoregressive process of order 1, eigenvalues of A should be smaller than one. However, I am getting some eigenvalues as a complex number after the estimation. however, the real components of the complex eigenvalues, as well as their norm both, are less than unity. Does that imply that process is stationary??


1 Answer 1


The process is stationary when all the complex eigenvalues are within the complex unit circle. This implies you are correct in checking that the norm is less than 1.

An AR process with complex eigenvalues will tend to behave a little differently than one with only real eigenvalues, but it is still stationary.

  • $\begingroup$ "An AR process with complex eigenvalues will tend to behave a little differently" Would it be possible for you to elaborate on that? If its problematic for the empirical estimation, I always have the option of putting the restriction of real eigenvalue in the optimization $\endgroup$ Commented Jul 6, 2019 at 7:17
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    $\begingroup$ What I meant is that if you look at the PACF of AR(p) processes with real roots vesus those with complex roots, then they tend to look a bit different. It's been a while since I've looked at this area specifically, but normally I would just simulate from the different series and see how they behave. $\endgroup$
    – John
    Commented Jul 7, 2019 at 3:29
  • $\begingroup$ @SudarshanKumar I don't know if I had noticed your comment on the restrictions before. I'm not sure how easy that is or not. I only know that calculating all the eigenvalues of potentially large matrices at each iteration can significantly slow down your algorithm. You may be better off finding out what about the process is causing the eigenvalues to be non-real and seeing if you can model that. $\endgroup$
    – John
    Commented Jul 15, 2019 at 18:39

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