hw4_solution

# hw4_solution - X j =1 3 . 4 j w t + h-j ) = ( 1 , h = 0 3 ....

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Homework 4 solutions 1. Note that k X t = k i =0 ( - 1) i ( k i ) X t - i and let a i = ( - 1) i ( k i ) . We have E ( k X t ) = 0 and Cov ( k X t + h , k X t ) = i j a i a j Cov ( X t + h ,X t ) which will depend only h since X t is stationary and so k X t is stationary. 2. (a) ARIMA (0 , 0 , 1) 2 (b) x t = (1 + Φ B 2 ) w t w t = 1 (1 + Φ B 2 ) x t = X j =0 ( - Φ B 2 ) j x t π k = ± 0 , k = 2 n - 1 ( - Φ) n , k = 2 n, n = 0 , 1 , ··· 3. (1 - 0 . 8 B 12 ) x t = (1 + 0 . 5 B ) w t ρ ( h ) = ± 0 . 8 n , h = 12 n, n = 0 , 1 , ··· 0 . 4 × 0 . 8 n , h = 12 n ± 1 , n = 0 , 1 , ··· 0 10 20 30 40 50 0.0 0.2 0.4 0.6 0.8 1.0 Lag ACF 4. (a) ν ( h ) = 0 , h = 0 1 , h = 1 0 . 8 h - 1 , h = 2 , 3 , ··· (b) ν ( h ) = ± 3 , h = 0 - ( - 0 . 6) h / 3 , h = 1 , 2 , ··· 1

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0 5 10 15 20 0.0 0.2 0.4 0.6 0.8 1.0 (a) lag Impulse Weights 0 5 10 15 20 0.5 1.5 2.0 2.5 3.0 (b) lag 5. x t = 1 + 0 . 8 B 1 - 0 . 4 B w t = (1 + 0 . 8 B ) X j =0 0 . 4 j w t - j = w t + X j =1 3 × 0 . 4 j w t - j γ wx ( h ) = cov ( w t ,w t + h +
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Unformatted text preview: X j =1 3 . 4 j w t + h-j ) = ( 1 , h = 0 3 . 4 h , h = 1 , 2 , w = 1 2 x = V ar w t + X j =1 3 . 4 j w t-j = 1 + X j =1 9 . 16 j = 2 . 28 / . 84 wx ( h ) = wx ( h ) w x = q . 84 2 . 28 , h = 0 q . 84 2 . 28 3 . 4 h , h = 1 , 2 , 2...
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## This note was uploaded on 05/01/2009 for the course PSTAT 120A taught by Professor Mackgalloway during the Spring '08 term at UCSB.

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hw4_solution - X j =1 3 . 4 j w t + h-j ) = ( 1 , h = 0 3 ....

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