hw3sol - AMS316 Homework 3 Solution 3.1 Solution: γ ( k )...

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Unformatted text preview: AMS316 Homework 3 Solution 3.1 Solution: γ ( k ) = cov ( X t , X t + k ) = cov ( Z t + 0 . 7 Z t- 1- . 2 Z t- 2 , Z t + k + 0 . 7 Z t + k- 1- . 2 Z t + k- 2 ) when k = 0, γ (0) = V ar ( X t ) = V ar ( Z t ) + 0 . 7 V ar ( Z t )- . 2 V ar ( Z t ) = 1 . 5 V ar ( Z t ); when k = 1, γ (1) = cov ( Z t + 0 . 7 Z t- 1- . 2 Zt- 2 , Z t- 1 + 0 . 7 Z t- . 2 Z t- 1 ) = 0 . 7 V ar ( Z t )- . 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- . 2 Z t- 2 , Z t +2 + 0 . 7 Z t +1- . 2 Z t ) =- . 2 cov ( Z t , Z t ) =- . 2 V ar ( Z t ) Hence ρ (2) = γ (2) γ (0) =- . 2 1 . 5 =- . 13. It’s easy to find out ac.f when k =- 1 ,- 2, therefore, ρ ( k ) = 1 : k = 0 . 37 : k = ± 1- . 13 : k = ± 2 : 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 !...
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This note was uploaded on 02/28/2012 for the course AMS 316 taught by Professor Xing during the Fall '09 term at SUNY Stony Brook.

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hw3sol - AMS316 Homework 3 Solution 3.1 Solution: γ ( k )...

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