quiz3_sol - 1. 8 X t = 2 X t −1 + X t − 2 + Z t Using...

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Unformatted text preview: 1. 8 X t = 2 X t −1 + X t − 2 + Z t Using backwards operator, 1 1 1 1 − B − B 2 and set Y = , 4 8 B 1 1 1 1 Y 2 − Y − = 0 , Φ1 = , Φ 2 = 4 8 4 8 we can get roots of eq, Y1 = 1 1 , Y2 = − 2 4 1 4 k Set ACF ρ (k ) = A1 ( ) + A2 ( − ) , we can get k 1 2 ρ (0) = A1 + A2 = 1 and ρ (1) = So, A1= , A2 = 1 1 Φ1 2 = A1 + (− ) A2 = 2 4 1 − Φ2 7 5 7 2 7 2 7 1 4 ρ (k ) = ( ) k + (− ) k 2. a) X t = α1 X t −1 + α 2 X t − 2 + Z t 51 72 X n +1 = α1 X n + α 2 X n −1 + Z n +1 X n (1) = E ( X n +1 | X 1 X n (1) = E ( X n +1 | X 1 X n ) = α1 X n + α 2 X n −1 X n ) = Z n +1 So, Var (en (1)) = Var ( Z n +1 ) = σ 2 ˆ s.e.(en (1)) = σ 2 b ) X n (2) = E ( X n + 2 | X 1 Xn) Xn) = E (α1 X n +1 + α 2 X n + Z n + 2 | X 1 = α1 X n (1) + α 2 X n + E ( Z n + 2 ) = α12 X n + α1α 2 X n −1 + α 2 X n en (2) = X n + 2 − X n (2) = α1 X n +1 + α 2 X n + Z n + 2 − (α12 X n + α1α 2 X n −1 + α 2 X n ) = α1 (α1 X n + α 2 X n −1 + Z n +1 ) + α 2 X n + Z n + 2 − (α12 X n + α1α 2 X n −1 + α 2 X n ) = α1Z n +1 + Z n + 2 Var (en (2)) = α12σ 2 + σ 2 3) For k>2 X n (k ) = E ( X n+ k | X 1 Xn) Xn) = E (α1 X n + k −1 + α 2 X n + k − 2 + Z n + k | X 1 = α1 X n (k − 1) + α 2 X n (k − 2) + E ( Z n + k ) = α1 X n (k − 1) + α 2 X n (k − 2) 3. a) X n +1 = α X n + Z n +1 − θ Z n X n (1) = E ( X n +1 | X 1 = α X n − θ Zn en (1) = X n +1 − X n (1) = Z n +1 Var (en (1)) = σ 2 b) Xn) Xn) = E (α X n + Z n +1 − θ Z n | X 1 X n + 2 = α X n +1 + Z n + 2 − θ Z n +1 = α (α X n + Z n +1 − θ Z n ) + Z n + 2 − θ Z n +1 X n (2) = E (α X n +1 + Z n + 2 − θ Z n +1 | X 1 = α E ( X n +1 | X 1 Xn) = α X n (1) = α (α X n − θ Z n ) en (2) = X n + 2 − X n (2) Xn) = α (α X n + Z n +1 − θ Z n ) + Z n + 2 − θ Z n +1 − α (α X n − θ Z n ) = (α − θ ) Z n +1 + Z n + 2 Var (en (2)) = σ 2 + σ 2 (α − θ ) 2 c) X n (k ) = E (α X n + k −1 + Z n + k − θ Z n + k −1 | X 1 = α E ( X n + k −1 | X 1 = α X n (k − 1) = =α =α k −1 k −1 k Xn) Xn) X n (1) (α X n − θ Z n ) = α X n − α k −1θ Z n When k → ∞ , α k → 0 ,so X n (k ) → 0 . 4) a) (1 − α B )(1 − B) X t = (1 − θ B) Z t , so {1 − (α + 1) B + α B 2 } X t = Z t − θ Z t −1 , We can get, X n +1 = (α + 1) X n − α X n −1 + Z n +1 − θ Z n X n (1) = E ( X n +1 | X 1 Xn) Xn) = E ((α + 1) X n − α X n −1 + Z n +1 − θ Z n | X 1 = (α + 1) X n − α X n −1 − θ Z n en (1) = X n +1 − X n (2) = Z n +1 Var (en (1)) = σ 2 b) X n + 2 = (α + 1) X n +1 − α X n + Z n + 2 − θ Z n +1 = (α + 1){(α + 1) X n − α X n −1 + Z n +1 − θ Z n } − α X n + Z n + 2 − θ Z n +1 X n (2) = E ( X n + 2 | X 1 Xn) Xn) = E ((α + 1) X n +1 − α X n + Z n + 2 − θ Z n +1 | X 1 = (α + 1) X n (1) − α X n = (α + 1){(α + 1) X n − α X n −1 − θ Z n } − α X n en (2) = X n + 2 − X n (2) = (α + 1 − θ ) Z n +1 + Z n + 2 Var (en (2)) = (α + 1 − θ ) 2 σ 2 + σ 2 ...
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This note was uploaded on 08/08/2010 for the course AMS 316 taught by Professor Xing during the Fall '09 term at SUNY Stony Brook.

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