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quiz2_sol

# quiz2_sol - 1 Since X t X t 1 = Z t Z t 1 1,wecanget X t BX...

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1. Since 1 1 t t t t X X Z Z θ − Φ = + 1) , we can get t t t t X BX Z BZ θ − Φ = + , B is a backwards operator. 1 ( ) 1 t t t B X Z B Z B θ φ + = = − Φ , where 1 ( ) 1 B B B θ φ + = − Φ , and 0 t EX = We set 2 3 1 2 3 ( ) 1 B B B B φ ϕ ϕ ϕ = + + + + " ,then we get 2 3 1 2 3 (1 )(1 ) 1 B B B B B ϕ ϕ ϕ θ − Φ + + + + = + " We expand the left side and match the coefficients of each power k B , we can get 1 2 1 2 1 1 2 { ( ) ( ) k k k k ϕ θ ϕ ϕ θ ϕ ϕ ϕ θ = Φ + = Φ = Φ Φ + = Φ = Φ = = Φ Φ + " So, we get 2 3 1 2 3 1 1 2 2 (1 ) t t t t t X B B B Z Z Z Z ϕ ϕ ϕ ϕ ϕ = + + + + = + + " " 2), 1 ( ) k k ϕ θ = Φ Φ + We multiply by t k X in eq 1) and take the expectation on both sides, 1 1 1 1 2 2 1 1 1 2 2 ( ) ( ) ( ) ( ) [ ( )] [ ( )] t t k t t k t t k t t k t t k t k t k t t k t k t k E X X E X X E Z X E Z X E Z Z Z Z E Z Z Z Z θ ϕ ϕ θ ϕ ϕ − − − − − − − − Φ = + = + + + + + " " We can say that i) if k>1, 1 ( ) ( ) 0 t t k t t k E Z X E Z X θ = = , so ( ) ( 1) k k γ γ = Φ ii) For k=o 2 2 2 1 (0) ( 1) ( ) z z z γ γ σ θϕ σ θ θ σ − Φ = + = + Φ + iii)

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