\f(a) X t 0.3X t 1 Z t is an ARMA (p, q) process with p = 1, q = 0, i.e., it is AR(1). In B notation it can be written as 1 0.3B X t Z t . The root of 10.3B = 0 is 1/0.3, which lies outside the unit circle, so Xt is stationary. AR process is always invertible. The MA representation of this model is X t 0.3 j j j 0 (b) X t Zt 1.3Z t 1 0.4 Z t 2 is an ARMA(p, q) process with p = 0, q = 2, i.e., it is MA(2). In B notation it can be written as The roots of 1 1.3z 0.4z 0 2 X t 1 1.3ZB 0.4B Zt 2 . are 2 and 1.25. They all lie outside of unit circle, so it's invertible. MA process is always stationary. (c) X t 0.5X t 1 Z t 1.3Z t 1 0.4Z t 2 1 can be written as 0.5B X t 1 1.3B 0.4B 2 Z t First, however, we need to check if there are common factors. The associated polynomial z 1 1.3z 0.4 z 2 can be written as 1 1.25 z 1 2 z 1 0.8z 1 0.5z 1 1 Hence, there is a common factor 1 0.5z and the model can be simplified. Dividing both sides of the model by 1 0.5z we obtain X t 1 0.8B Z t The root of 1 0.5B = 0 is 2, which lies outside the unit circle, so Xt is stationary. The roots of 1 1.3B + 0.4B2 = 0 are 2 and 1.25, which lie outside the unit circle, so Xt is invertible. (a) X t 0.3X t 1 Z t is an ARMA (p, q) process with p = 1, q = 0, i.e., it is AR(1). In B notation it can be written as 1 0.3B X t Z t . The root of 10.3B = 0 is 1/0.3, which lies outside the unit circle, so Xt is stationary. AR process is always invertible. The MA representation of this model is X t 0.3 j j j 0 (b) X t Zt 1.3Z t 1 0.4 Z t 2 is an ARMA(p, q) process with p = 0, q = 2, i.e., it is MA(2). In B notation it can be written as The roots of 1 1.3z 0.4z 0 2 X t 1 1.3ZB 0.4B Zt 2 . are 2 and 1.25. They all lie outside of unit circle, so it's invertible. MA process is always stationary. (c) X t 0.5X t 1 Z t 1.3Z t 1 0.4Z t 2 1 can be written as 0.5B X t 1 1.3B 0.4B 2 Z t First, however, we need to check if there are common factors. The associated polynomial z 1 1.3z 0.4 z 2 can be written as 1 1.25 z 1 2 z 1 0.8z 1 0.5z 1 1 Hence, there is a common factor 1 0.5z and the model can be simplified. Dividing both sides of the model by 1 0.5z we obtain X t 1 0.8B Z t The root of 1 0.5B = 0 is 2, which lies outside the unit circle, so Xt is stationary. The roots of 1 1.3B + 0.4B2 = 0 are 2 and 1.25, which lie outside the unit circle, so Xt is invertible