(Forecasting an ARMA(2,2) process) Consider the ARMA(2,2) process: y_t phi_1y_t-
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(Forecasting an ARMA(2,2) process) Consider the ARMA(2,2) process: y_t phi_1y_t-1 + phi_2y_t-2 + euler_t + theta_1 euler_t-1 + theta_2 euler_t_2. Verify that the optimal 1-step ahead forecast made at time T is y_T+1,T = phi_1y_T + phi_2y_T-1 + theta_1 euler_T + theta_2 euler_T-1. Verify that the optimal 2-step ahead forecast made at time T is y_T+2,T = phi_1y_T+1,T + phi_2y_T + theta_2 euler_T, and express it purely in terms of elements of the time- T information set. Verify that the optimal 3-step ahead forecast made at time T is y_T+3,T = phi_1y_T+2,T + phi_2y_T+1,T, and express it purely in terms of elements of the time-T information set. Show that for any forecast horizon h greater than or equal to three, y_T+h,T = phi_1y_T+h-1,T + phi_2y_T+h-2,T.Explanation / Answer
a) et follows Normal Distribution with Mean 0 and variance (sigma t square).
E(Yt)=Phi1Yt-1 + Phi2Yt-2 +et+theta1 et-1 + theta2 et-1
E(Yt+1)= Phi1Y(t+1)-1 + Phi2Y(t+1)-2 +e(t+1)+theta1 e(t+1)-1 + theta2 e(t+1)-2
= Phi1Yt + Phi2Y(t-1) +e(t+1)+theta1 et + theta2 et-1
Note: e(t+1) = 0 Future error
= Phi1Yt + Phi2Y(t-1) +theta1 et + theta2 et-1
(b) et follows Normal Distribution with Mean 0 and variance (sigma t square).
E(Yt)=Phi1Yt-1 + Phi2Yt-2 +et+theta1 et-1 + theta2 et-1
E(Yt+2)=Phi1Y(t+2)-1 + Phi2Y(t+2)-2 +e(t+2)+theta1 e(t+2)-1 + theta2 e(t+2)-2
=Phi1Y(t+1) + Phi2Yt +e(t+2)+theta1 e(t+1) + theta2 et
Note: e(t+2)+theta1 e(t+1) = 0 Future error
= Phi1Y(t+1) + Phi2Yt +theta2 et
(C)
(b) et follows Normal Distribution with Mean 0 and variance (sigma t square).
E(Yt)=Phi1Yt-1 + Phi2Yt-2 +et+theta1 et-1 + theta2 et-1
E(Yt+2)=Phi1Y(t+2)-1 + Phi2Y(t+2)-2 +e(t+2)+theta1 e(t+2)-1 + theta2 e(t+2)-2
=Phi1Y(t+1) + Phi2Yt +e(t+2)+theta1 e(t+1) + theta2 et
Note: e(t+2)+theta1 e(t+1) = 0 Future error
E(Yt+2) = Phi1Y(t+1) + Phi2Yt +theta2 et
E(Yt+2+1) = Phi1Y(t+1+1) + Phi2Y(t+1) +theta2 e(t+1)
E(Yt+3) = Phi1Y(t+2) + Phi2Y(t+1) +theta2 e(t+1)
Note: theta2 e(t+1) = 0 Future error
E(Yt+3) = Phi1Y(t+2) + Phi2Y(t+1)
(d) E(Yt+3) = Phi1Y(t+2) + Phi2Y(t+1)
E(Yt+b) = Phi1Y(t+b-1) + Phi2Y(t+b-2)
Hence proved
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