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quiz3_sol - 1 8 X t = 2 X t 1 X t 2 Z t 1 1 1 1 B B 2...

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1. 1 2 8 2 t t t t X X X Z = + + Using backwards operator, 2 1 1 1 4 8 B B and set 1 Y B = , 2 1 1 0 4 8 Y Y = , 1 1 4 Φ = , 2 1 8 Φ = we can get roots of eq, 1 2 1 1 , 2 4 Y Y = = − Set ACF 1 2 1 1 ( ) ( ) ( ) 2 4 k k k A A ρ = + , we can get 1 2 (0) 1 A A ρ = + = and 1 1 2 2 1 1 2 (1) ( ) 2 4 1 7 A A ρ Φ = + − = = − Φ So, 1 2 5 2 , 7 7 A A = 5 1 2 1 ( ) ( ) ( ) 7 2 7 4 k k k ρ = + 2. a) 1 1 2 2 t t t t X X X Z α α = + + 1 1 2 1 1 n n n n X X X Z α α + + = + + 1 1 1 2 1 (1) ( | ) n n n n n X E X X X X X α α + = = + " 1 1 1 (1) ( | ) n n n n X E X X X Z + + = = " So, 2 1 ( (1)) ( ) n n Var e Var Z σ + = = 2 ˆ . .( (1)) n s e e σ =
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b) 2 1 1 1 2 2 1 1 2 2 2 1 1 2 1 2 (2) ( | ) ( | ) (1) ( ) n n n n n n n n n n n n n X E X X X E X X Z X X X X E Z X X X α α α α α α α α + + + + = = + + = + + = + + " " 2 2 1 1 2 2 1 1 2 1 2 2 1 1 2 1 1 2 2 1 1 2 1 2 1 1 2 (2) (2) ( ) ( ) ( ) n n n n n n n n n n n n n n n n n n n e X X X X Z X X X X X Z X Z X X X Z Z α α α α α α α α α α α α α α α + + + + + + + = = + + + + = + + + + + + = + 2 2 2 1 ( (2)) n Var e α σ σ = + 3) For k>2 1 1 1 2 2 1 1 2 1 2 ( ) ( | ) ( | ) ( 1) ( 2) ( ) ( 1) ( 2) n n k n n k n k n k n n n n k n n X k E X X X E X X Z X X X k
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