quiz2_sol - 1. Since X t − ΦX t −1 = Z t + θ Z t −1...

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Unformatted text preview: 1. Since X t − ΦX t −1 = Z t + θ Z t −1 1) , we can get X t − ΦBX t = Z t + θ BZ t , B is a backwards operator. Xt = 1+θ B 1+θ B Z t = φ ( B ) Z t , where φ ( B) = , and EX t = 0 1 − ΦB 1 − ΦB ,then we get We set φ ( B) = 1 + ϕ1 B + ϕ2 B 2 + ϕ3 B 3 + (1 − ΦB)(1 + ϕ1 B + ϕ 2 B 2 + ϕ3 B3 + ) = 1+θ B We expand the left side and match the coefficients of each power B k , we can get ϕ1 = Φ + θ {ϕ2 = Φϕ1 = Φ (Φ + θ ) ϕk = Φϕk −1 = Φ 2ϕk − 2 = =Φ k −1 (Φ + θ ) So, we get X t = (1 + ϕ1 B + ϕ 2 B 2 + ϕ3 B 3 + ) Z t = Z t + ϕ1Z t −1 + ϕ2 Z t − 2 2), ϕk = Φ (Φ + θ ) k −1 We multiply by X t − k in eq 1) and take the expectation on both sides, E ( X t X t − k ) − E (ΦX t −1 X t − k ) = E ( Z t X t − k ) + E (θ Z t −1 X t − k ) = E[ Z t ( Z t − k + ϕ1Z t − k −1 + ϕ2 Z t − k − 2 We can say that i) if k>1, E ( Z t X t − k ) = E (θ Z t −1 X t − k ) = 0 , so )] + θ E[ Z t −1 ( Z t − k + ϕ1Z t − k −1 + ϕ 2 Z t − k − 2 )] γ (k ) = Φγ (k − 1) ii) iii) For k=o γ (0) − Φγ (−1) = σ z2 + θϕ1 = σ z2 + θ (Φ + θ )σ z2 For k=1 γ (1) − Φγ (0) = σ z2θ We can get, γ (0) = 1 + θ 2 + 2Φθ 2 (1 + θΦ )(Φ + θ 2 σ z , γ (1) = σ z , γ (k ) = Φγ (k − 1) , k>1 2 1− Φ 1 − Φ2 So, (1 + θΦ )(Φ + θ # { 1 + θ 2 + 2Φθ ρ (k ) = Φρ (k − 1) k > 1 ρ (1) = 2. 5 X t = 4 X t −1 − X t − 2 + Z t ⇒ X t = 4 1 ⇒ 1 − B − B 2 = 0 , 5 5 Set, Y = 4 1 1 X t −1 − X t − 2 + Z t 5 5 4 1 B 1 4 1 ⇒ Y2 − Y + = 0 5 5 5 We get roots, Y1 = 21 21 + i , Y2 = − i , 55 55 Here, we set ρ (k ) = A1 ( + i ) k + A2 ( − i ) k Here, 2 5 1 5 21 55 ρ (0) = A1 + A2 = 1 { ρ (1) = A1 ( + i ) + A2 ( − i ) = 2 5 1 5 21 55 4 2 = 63 We got A1 = So, 12 12 − i , A2 = + i 23 23 ρ (k ) = ( − i)( + i ) k + ( + i )( − i) k # 3. 1 2 2 3 2 5 1 5 1 2 2 3 21 55 6 X t = X t −1 + X t − 2 + Z t 1 1 1 X t −1 + X t − 2 + Z t 6 6 6 1 12 ⇒ 1− B − B = 0 6 6 ⇒ Xt = We set Y = 1 , B 1 1 ⇒Y2 − Y − = 0 6 6 We got roots Y1 = 1 1 , Y2 = − 2 3 Since, ρ (k ) = A1 ( ) k + A2 (− ) k S0, 1 2 1 3 ρ (0) = A1 + A2 = 1 { ρ (1) = A1 ( ) + A2 (− ) = 1 2 1 3 1 5 We can get A1 = So, 16 9 , A2 = 25 25 ρ (k ) = 16 1 k 9 1k ( ) + (− ) 25 2 25 3 4. 1) X t − α 2 X t − 2 = Z t − θ Z t −1 ⇒ (1 − α 2 B 2 ) X t = (1 − θ B ) Z t 1 − α 2 B 2 = 0 ⇒ B1 = 1 − θ B = 0 ⇒ B3 = 1 1 (*) α2 , B2 = − 1 α2 θ If B1 > 1, and B2 > 1 ⇒ α 2 < 1 , then X t is stationary. If B3 > 1 ⇔ θ < 1 , then X t is invertible. 2) Since X t is stationary, EX t = 0 Since, X t = (1 − θ B ) Z t = φ ( B ) Z t , set φ ( B) = 1 + ϕ1 B + ϕ 2 B 2 + ϕ3 B 3 + 2 (1 − α 2 B ) = (1 − θ B ) (1 − α 2 B 2 ) We can get (1 − α 2 B 2 )(1 + ϕ1 B + ϕ2 B 2 + ϕ3 B 3 + ) = 1 − θ B , expand the right‐hand items and match the coefficients of each power B k , we get ϕ1 = −θ , ϕ 2 = α 2 , ϕ k = α 2ϕ k − 2 , so X t = (1 − θ B + α 2 B 2 + ϕ3 B 3 + ) Z t , hence from eq (*) E ( X t X t − k ) − E (α 2 X t −1 X t − k ) = E ( Z t X t − k ) − E (θ Z t −1 X t − k ) = E[ Z t ( Z t − k − θ Z t − k −1 + α 2 Z t − k − 2 If k=0, we get )] − θ E[ Z t −1 ( Z t − k − θ Z t − k −1 + α 2 Z t − k − 2 )] γ (0) − α 2γ (2) = E ( Z t X t ) − θ E ( Z t −1 X t ) = E[ Z t ( Z t − θ Z t −1 + α 2 Z t − 2 = σ +θ σ 2 z 2 2 z )] − θ E[ Z t −1 ( Z t − θ Z t −1 + α 2 Z t − 2 )] If k=2,we get γ (2) − α 2γ (0) = E ( Z t X t − 2 ) − θ E ( Z t −1 X t − 2 ) = E[ Z t ( Z t − 2 − θ Z t −3 + α 2 Z t − 4 =0 From 2 equations above, we get )] − θ E[ Z t −1 ( Z t − 2 − θ Z t −3 + α 2 Z t − 4 )] γ (0) = # 1 + θ 2 2 σz 2 1− α2 ...
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