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 Y_t = 1.4Y_t-1 + 0.1Y_t-2 – 0.5Y_t-3 + e_t + 0.3e_t-1 + 0.2e_t-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

Y_t – Y_t-1 = 0.4Y_t-1 + 0.1Y_t-2 – 0.5Y_t-3 + e_t + 0.3e_t-1 + 0.2e_t-2

= 0.4Y_t-1 – 0.4Y_t-2 + 0.4Y_t-2 + 0.1Y_t-2  – 0.5Y_t-3 + e_t + 0.3e_t-1 + 0.2e_t-2

= 0.4Y_t-1 – 0.4Y_t-2 + 0.5Y_t-2 – 0.5Y_t-3 + e_t + 0.3e_t-1 + 0.2e_t-2

➾ W_t = 0.4 W_t-1 + 0.5W_t-2 + e_t + 0.3e_t-1 + 0.2e_t-2

See equation 5.2.2 on page 92.

Intuition: The ARIMA(p,1,q) process is

Y_t – Y_t-1 = ö₁ (Y_t-1 – Y_t-2) + ö₂ (Y_t-2 – Y_t-3) + … + ö_p (Y_t-p – Y_t-p-1) + å_t – è₁ å_t-1 – è₂ å_t=2 – … – è_q å_t-q

Rewrite this as

Y_t = (1 + ö₁) Y_t-1 + (ö₂ – ö₁) Y_t-2 + (ö₃ – ö₂) Y_t-3 + … + (ö_p – ö_p-1) Y_t-p + + å_t – è₁ å_t-1 – è₂ å_t=2 – … – è_q å_t-q

The AR(1) process for ∇ Y_t = W_t has ö₁ = 0.4 and ö₂ = 0.5, which give

           1 + ö₁ = 1.4

           ö₂ – ö₁ = 0.1

           –ö₂ = –0.5