TSS module 9 ARI processes.
Cryer and Chan, chapter 5, show how to convert ARI (intergrated autoregressive) processes to stationary autoregressive process by taking differences. The ARI(1,1) format is shown in this illustrative test question, using equation 5.2.12 on page 97. Cryer and Chan provide the general expression for ARI(p,1) processes.
*Question 1.2: ARI(1,1) process
Which of the following is an ARI(1,1) process?
A. Yt = Yt-1 – (1 + ö) Yt-2 + et
B. Yt = Yt-1 + ö Yt-2 + et
C. Yt = Yt-1 – ö Yt-2 + et
D. Yt = (1 + ö) Yt-1 – ö Yt-2 + et
E. Yt = (1 + ö) Yt-1 + ö Yt-2 + et
Answer 1.2: D
Yt = (1 + ö) Yt-1 – ö Yt-2 + et
➾ Yt – Yt-1 = ö Yt-1 – ö Yt-2 + et
➾ ∇Yt = ö ∇Yt-1 + et
Final exam problems give ö1, ö2, … and derive the underlying AR(p) process. They also test variances of Yt given a starting point for the time series, covariances, and correlations.
**Question 1.3: ARI(1,1) process
An ARI(1,1) process Yt has first differences ∇ Yt that are an AR(1) process with parameter ö = 0.5.
Yt – Yt-1 = ö (Yt-1 – Yt-2) + åt
The ARI(1,1) process is written as a non-stationary ARMA process with weights øj applied to the error terms:
Yt = ø0 åt + ø1 åt-1 + ø2 åt-3 + ø3 åt-3 + …
What is the value of ø2?
A. 0.25
B. 0.5
C. 0.75
D. 1
E. 1.75
Answer 1.3: E
See equation 5.2.15 on page 97:
For ö = 0.5 and k = 2, this gives (1 – 0.52+1) / (1 – 0.5) = ⅞ / ½ = 1.75
Intuition: Expand the expression for the ARI(1,1) process:
Yt – Yt-1 = ö (Yt-1 – Yt-2) + åt
Yt = (1 + ö) Yt-1 – ö Yt-2 + åt
Yt = (1 + ö) Yt-1 – ö Yt-2 + åt