To check for stationarity of ARIMA would be to:
- show that d=0
- or solve the characteristic polynomial of AR or MA portion or BOTH?
Same way as usual - you check the characteristic equation of the (differenced) AR part to check stationarity.
Using your example from earlier (but now with RHS):
X(t) - 0.6X(t-1) - 0.3X(t-2) - 0.1X(t-3) = e(t) + 0.8e(t-1)
The LHS can be written as:
[X(t) - X(t-1)] + 0.4[X(t-1) - X(t-2)] + 0.1[X(t-2) - X(t-3)]
Using D for the difference operator, we can write this as:
DX(t) + 0.4DX(t-1) + 0.1DX(t-2)
The sum of these coefficients is 1 + 0.4 + 0.1 which is not zero, so we cannot difference this any more.
So we now check for stationarity.
The characteristic polynomial of the final differenced equation is:
1 + 0.4z + 0.1z² = 0
Unfortunately solving this gives complex roots of -2 ± i×root(6).
The magnitude of these roots are both root(2² + 6) = 3.162 which is greater than 1 so the differenced equation is stationary.
So we have an ARIMA(2, 1, 1).