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.      Y_t = Y_t-1 – (1 + ö) Y_t-2 + e_t

B.      Y_t = Y_t-1 + ö Y_t-2 + e_t

C.      Y_t = Y_t-1 – ö Y_t-2 + e_t

D.     Y_t = (1 + ö) Y_t-1 – ö Y_t-2 + e_t

E.      Y_t = (1 + ö) Y_t-1 + ö Y_t-2 + e_t

Answer 1.2: D

Y_t = (1 + ö) Y_t-1 – ö Y_t-2 + e_t

➾ Y_t – Y_t-1 = ö Y_t-1 – ö Y_t-2 + e_t

➾ ∇Y_t = ö ∇Y_t-1 + e_t

Final exam problems give ö₁, ö₂, … and derive the underlying AR(p) process. They also test variances of Y_t given a starting point for the time series, covariances, and correlations.

**Question 1.3: ARI(1,1) process

An ARI(1,1) process Y_t has first differences ∇ Y_t that are an AR(1) process with parameter ö = 0.5.

Y_t – Y_t-1 = ö (Y_t-1 – Y_t-2) + å_t

The ARI(1,1) process is written as a non-stationary ARMA process with weights øj applied to the error terms:

Y_t = ø₀ å_t + ø₁ å_t-1 + ø₂ å_t-3 + ø₃ å_t-3 + …

What is the value of ø₂?