Arima process - factorising (1-B)

[John Lee]

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).

 

[jm_kinuthia]

Assignment X4 - Questions X4.5

(ii)b - How did they arrive at p=1 ? I thought p=2. ie ARIMA(2,1,1).

 

[Julie Lewis]

The p in ARIMA is the number of roots that are left after you have eliminated all the roots of 1 (provided getting rid of the 1's means the process is stationary, ie all the remaining roots are bigger than 1 in modulus).

If you start with AR(2) but it's not stationary, say because the roots of the char poly are 1 and 3, then differencing once will eliminate the root of 1 and the only remaining root is 3. So this would be ARIMA(1,1,0). Note that the ARIMA p+d must add to the original p (1+1=2 in this case).