For an $AR(p)$ process $ Y_t = a_1Y_{t-1}+a_2Y_{t-2}+...+a_qY_{t-q}$ :

Is having the coefficients $|a_1|,....,|a_p| < 1$ just a necessary condition for stationarity, or is it sufficient as well?


It is certainly not sufficient, try the AR(2) process $$ Y_t = 0.5Y_{t-1}+0.5Y_{t-2}+u_t $$ for which not all roots of the equation $1-0.5z-0.5z^2$ are outside the unit circle:

> polyroot(c(1, -0.5, -0.5))
[1]  1-0i -2+0i

Your condition is, moreover, also not necessary: try $$ Y_t = 1.1Y_{t-1}-0.9Y_{t-2}+u_t, $$ whose characteristic polynomial has all roots outside the unit circle:

> Re(polyroot(c(1, -1.1, 0.9)))^2+Im(polyroot(c(1, -1.1, 0.9)))^2
[1] 1.111111 1.111111

Here are the real and imaginary parts together with the unit circle:

enter image description here

plot(Re(polyroot(c(1, -1.1, 0.9))),Im(polyroot(c(1, -1.1, 0.9))),asp=1,ylim=c(-1,1),xlim=c(-1,1),col="red",pch=19)
draw.circle(0, 0, 1)

A necessary condition that I would argue is useful is given by $$\sum_{j=1}^pa_j<1,$$ see e.g. here, p. 54.

  • $\begingroup$ Are you saying $Y_t = 1.1Y_{t-1}-0.9Y_{t-2}+u_t$ is stationary? $\endgroup$ – Skander H. Jan 19 '18 at 8:04
  • $\begingroup$ Yes, because all roots are outside the unit circle. Try plot(arima.sim(list(ar=c(1.1,-0.9)),n=1000)). $\endgroup$ – Christoph Hanck Jan 19 '18 at 8:29
  • $\begingroup$ @Alex, effectively, the large negative coefficient on the second lag "undoes" the large positive on the first lag, so that, overall, the process does not become explosive. $\endgroup$ – Christoph Hanck Jan 22 '18 at 12:03

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.