hw3sol - AMS316 Homework 3 Solution 3.1 Solution(k = cov(Xt...

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AMS316 Homework 3 Solution 3.1 Solution: γ ( k ) = cov ( X t , X t + k ) = cov ( Z t + 0 . 7 Z t - 1 - 0 . 2 Z t - 2 , Z t + k + 0 . 7 Z t + k - 1 - 0 . 2 Z t + k - 2 ) when k = 0, γ (0) = V ar ( X t ) = V ar ( Z t ) + 0 . 7 V ar ( Z t ) - 0 . 2 V ar ( Z t ) = 1 . 5 V ar ( Z t ); when k = 1, γ (1) = cov ( Z t + 0 . 7 Z t - 1 - 0 . 2 Zt - 2 , Z t - 1 + 0 . 7 Z t - 0 . 2 Z t - 1 ) = 0 . 7 V ar ( Z t ) - 0 . 14 V ar ( Z t ) = 0 . 56 V ar ( Z t ) Hence ρ (1) = γ (1) γ (0) = 0 . 37. When k = 2, γ (2) = cov ( Z t + 0 . 7 Z t - 1 - 0 . 2 Z t - 2 , Z t +2 + 0 . 7 Z t +1 - 0 . 2 Z t ) = - 0 . 2 cov ( Z t , Z t ) = - 0 . 2 V ar ( Z t ) Hence ρ (2) = γ (2) γ (0) = - 0 . 2 1 . 5 = - 0 . 13. It’s easy to find out ac.f when k = - 1 , - 2, therefore, ρ ( k ) = 1 : k = 0 0 . 37 : k = ± 1 - 0 . 13 : k = ± 2 0 : otherwise (1) 3.2 Solution: For X t = m k =0 Z t - k m +1 , we have E ( X t ) = E m X k =0 Z t - k m + 1 ! = 0 . (2) 1
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Since ρ ( k ) = γ ( k ) γ (0) = corr ( X t , X t + k ), γ (0) = V ar ( X t ) = V ar ( m X k =0 Z t - k m + 1 ) = 1 ( m + 1) 2 m X k =0 var ( Z t - k ) = σ 2 1 m + 1 , when m k 1, γ ( k ) = cov ( X t , X t + k ) = cov m X k =0 Z t - k m + 1 , m X k =0 Z t - k + i m + 1 ! = E " ( m - k X t =0 Z t - k m + 1 )( m - k X i =0 Z t - k + i m + 1 ) # = σ 2 m - k X i =0 1 ( m + 1) 2 ! = σ 2 m - k + 1 ( m + 1) 2 when k m , γ ( k ) = 0. Hence ρ ( k ) = γ ( k ) γ (0) = ( σ 2 ( m - k +1) ( m +1) 2 / σ 2 ( m +1) ( m +1) 2 : k = 0 , 1 , ..., m 0 : k m (3)
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