TSS module 9 Identifying ARIMA processes.
Cryer and Chan show how to write an ARIMA(p,1,q) process as a non-stationary ARMA process. Some final exam problems ask you to identify the proper ARIMA process and its parameters.
**Question 1.2: ARIMA Process
A time series is Yt = 1.4Yt-1 + 0.1Yt-2 – 0.5Yt-3 + et + 0.3et-1 + 0.2et-2
What is the process followed by this time series?
A. ARIMA(1,1,1)
B. ARIMA(2,1,2)
C. ARIMA(2,1,1)
D. ARIMA(1,2,1)
E. ARIMA(2,2,1)
Answer 1.2: B
Rewrite the ARIMA process as
Yt – Yt-1 = 0.4Yt-1 + 0.1Yt-2 – 0.5Yt-3 + et + 0.3et-1 + 0.2et-2
= 0.4Yt-1 – 0.4Yt-2 + 0.4Yt-2 + 0.1Yt-2 – 0.5Yt-3 + et + 0.3et-1 + 0.2et-2
= 0.4Yt-1 – 0.4Yt-2 + 0.5Yt-2 – 0.5Yt-3 + et + 0.3et-1 + 0.2et-2
➾ Wt = 0.4 Wt-1 + 0.5Wt-2 + et + 0.3et-1 + 0.2et-2
See equation 5.2.2 on page 92.
Intuition: The ARIMA(p,1,q) process is
Yt – Yt-1 = ö1 (Yt-1 – Yt-2) + ö2 (Yt-2 – Yt-3) + … + öp (Yt-p – Yt-p-1) + åt – è1 åt-1 – è2 åt=2 – … – èq åt-q
Rewrite this as
Yt = (1 + ö1) Yt-1 + (ö2 – ö1) Yt-2 + (ö3 – ö2) Yt-3 + … + (öp – öp-1) Yt-p + + åt – è1 åt-1 – è2 åt=2 – … – èq åt-q
The AR(1) process for ∇ Yt = Wt has ö1 = 0.4 and ö2 = 0.5, which give
1 + ö1 = 1.4
ö2 – ö1 = 0.1
–ö2 = –0.5
These are the coefficients of Yt-k in the original equation.
Jacob: What about the moving average terms?
Jacob: The coefficients of the error terms åt-k remain unchanged; they are the negatives of the moving average parameters.
Jacob: What is the procedure for this transformation?
Rachel: The sum of the coefficients for the Y terms are equal on both sides of the equation.