classwork soln

classwork soln - STA 4005 Classwork Solution 1. (a) Zt =...

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STA 4005 Classwork Solution 1. (a) Z t = Z t - 4 + a t - θ 1 a t - 1 - θ 2 a t - 2 = ( Z t - 8 + a t - 4 - θ 1 a t - 5 - θ 2 a t - 6 ) + a t - θ 1 a t - 1 - θ 2 a t - 2 So ψ 0 = 1 , ψ 1 = - θ 1 , ψ 2 = - θ 2 , ψ 3 = 0 and ψ 4 = 1. (b) Since ˆ Z t ( l ) = Z t + l - 4 l = 3 , 4 ˆ Z t (2) = Z t - 2 - θ 2 a t ˆ Z t (1) = Z t - 3 - θ 1 a t - θ 2 a t - 1 Thus ˆ Z t (1) = 25 - 0 . 5(3) + 0 . 25(2) = 24 ˆ Z t (2) = Z t - 2 - θ 2 a t = 20 + 0 . 25(3) = 20 . 75 ˆ Z t (3) = 25 ˆ Z t (4) = 40 (c) V ar ( e t ( l )) = σ 2 a l - 1 X j =0 ψ 2 j = 1 if l = 1 1 . 25 if l = 2 1 . 3125 if l = 3 1 . 3125 if l = 4 So 95% prediction limits are ( ˆ Z t ( l ) ± 2 q V ar ( e t ( l ))) = 24 ± 2 = (22 , 26) if l = 1 20 . 75 ± 2 . 2 = (18 . 51 , 22 . 99) if l = 2 25 ± 2 . 3 = (22 . 71 , 27 . 29) if l = 3 40 ± 2 . 3 = (37 . 71 , 42 . 29) if l = 4 Θ 2. AR characteristic polynomial is φ ( x ) = (1 - 1 . 6 x + 0 . 7 x 2 )(1 - 0 . 8 x 12 ). (a) Roots are 1 . 6 ± i 0 . 24 1 . 4 and the twelve roots of x 12 = 1 . 25 The moduli are 1 .
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This note was uploaded on 11/02/2011 for the course STAT 4005 taught by Professor Wu,kaho during the Spring '08 term at CUHK.

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classwork soln - STA 4005 Classwork Solution 1. (a) Zt =...

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