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?
$\begingroup$
$\endgroup$
$\begingroup$
$\endgroup$
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:
library(plotrix)
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.
answered Jan 19 '18 at 7:06
-
$\begingroup$ 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
