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Unformatted text preview: 1 Xt = ∑ 1 Zt −k k =0 m + 1 1 = ( Z t + Z t −1 + m +1 m + Zt −m ) So, E( X t ) = 1 ( E ( Z t ) + E ( Z t −1 ) + m +1 =0 + E ( Z t − m )) Then, γ (h) = Cov( X t , X t + h ) = E ( X t , X t + h )
1 1 ( Z t + Z t −1 + + Z t − m ), ( Z t + h + Z t + h −1 + + Z t + h − m )] m +1 m +1 1 = E[( Z t + Z t −1 + + Z t − m ), ( Z t + h + Z t + h −1 + + Z t + h − m )] (m + 1) 2 = E[
i) If, h>m, this means t+k‐m>t, there is no overlap between ( Z t + Z t −1 + + Z t − m ) and ( Z t + h + Z t + h −1 + i ,i , d → + Z t + h − m ) , since Z k ⎯⎯⎯ N (0, σ z ) , γ ( h) = 0 ii) If, h<m, this means t+k‐m<t, the overlap between ( Z t + Z t −1 + + Z t − m ) and ( Z t + k + Z t + k −1 + + Z t + k − m ) is ( Z t + h − m + Z t + h − m+1 + + Zt ) + Z t )] γ (k )= 1 E[( Z t + k − m + Z t + k − m +1 + (m + 1) 2 + Z t ), ( Z t + k − m + Z t + k − m +1 + The correlation of rest items is equal to 0. γ ( h) =
= 1 Var[( Z t + h − m + Z t + h − m +1 + (m + 1) 2 + Z t )] m−h 1 ∑ Var (Zt +h−m+k ) (m + 1) 2 k =0 m − h +1 2 σz = (m + 1) 2 Also, Var ( X t ) = Var[∑ = 1 Zt −k ] k =0 m + 1 m m 1 ∑Var (Zt −k ) (m + 1) 2 k =0 1 = σ z2 m +1 So, ρ ( h) =
So, γ ( h)
Var ( X t ) = m + h −1 , If h<m m +1 0 ρ ( h) = { m + h − 1 m +1 h>m h<m # 2 X t = Z t + C ( Z t −1 + Z t − 2 + a) ) E ( X t ) = E ( Z t ) + E[C ( Z t −1 + Z t − 2 + =0 )] Var ( X t ) = Var ( Z t ) + Var[C ( Z t −1 + Z t − 2 + )] (Because of independence of Z t ) = σ + C (σ + σ +
2 z 2 2 z 2 z ) =∞ So, the infinite order MA process { X t } has a constant mean, but it has infinite variance. We can conclude that the process { X t } is not stationary. b) Yt = X t − X t −1 = ( Z t + CZ t −1 + CZ t − 2 + = Z t + CZ t −1 − Z t −1
So, ) − ( Z t −1 + CZ t − 2 + CZ t −3 + ) E (Yt ) = E ( Z t + CZ t −1 − Z t −1 ) =0 γ (0) = Var (Yt ) = Var ( Z t + CZ t −1 − Z t −1 )
= Var ( Z t ) + Var[(C − 1) Z t ] = σ z2 + (C − 1) 2 σ z2 γ (1) = Cov(Yt , Yt +1 ) = Cov[( Z t + CZ t −1 − Z t −1 ), ( Z t +1 + CZ t − Z t )]
= Cov( Z t , CZ t − Z t ) = Cov( Z t , CZ t ) − Cov( Z t , Z t ) = (C − 1)σ z2 γ (2) = Cov(Yt , Yt + 2 ) = Cov[( Z t + CZ t −1 − Z t −1 ), ( Z t + 2 + CZ t +1 − Z t +1 )]
=0 ρ (1) = γ (1) γ (0) C −1 = 1 + (C − 1) 2 So,{ Yt } is a first‐order MA process. {Y }t is stationary. ACF of {Y }t is , 1 k =0 k =1 O.W . # ρ (1){ C −1 1 + (C − 1) 2 0 3 X t = λ1 X t −1 + λ2 X t − 2 + Z t We set backwards operator B, so X t = λ1 BX t + λ2 B 2 X t + Z t ⇒ Z t = (1 − λ1 B − λ2 B 2 ) X t We can get characteristic equation, 1 − λ1 B − λ2 B 2 = 0 Set B = 1 , we can get standard form, Y Y 2 − λ1Y − λ2 = 0 To assure we have real root, we have first condition i) λ12 + 4λ2 ≥ 0 Then, we get roots Y= λ1 ± λ12 + 4λ2
2 To assure our results are stationary, we have second condition ii) Y= λ1 ± λ12 + 4λ2
2 <1 which means −1 < λ1 + λ2 < 1 Combining with first condition, we get the range of λ1 and λ2 , λ12 + 4λ2 ≥ 0 { −1 < λ1 + λ2 < 1 If λ1 = 1 2 , λ2 = 3 9 We get characteristic equation, 1 2 Y2 − Y − = 0 3 9
So, Y = Since 2 1 or Y = − 3 3 ρ (k ) = AY1 k + A2Y2k 1
So, ρ (0) = A1 + A2 = 1 By, Y.W. eq, ρ (1) = λ1 ρ (0) + λ2 ρ (−1) = λ1 ρ (0) + λ2 ρ (1)
So, 1 λ 3 ρ (1) = 1 = 3 = 27 1 − λ2 1 − 9 ⇒{ ρ (0) = A1 + A2 = 1 ρ (1) =
2 1 3 A1 − A2 = 3 3 7 16 5 , A2 = , 21 21 We get, A1 = So, ρ (k ) = 16 2 k 5 1 k ( ) + (− ) # 21 3 21 3 4 Xt = 1 1 X t −1 + X t − 2 + Z t 12 12 So, we get characteristic equation, 1− 1 1 B − B2 = 0 12 12 Transform to standard equitation, Y= 1 , B 1 1 Y − = 0 12 12 1 1 , Y2 = − 3 4 Y2 − We get roots, Y1 = ρ (0) = A1 + A2 = 1
⇒{ ρ (1) = A1 − A2 =
45 32 , A2 = 77 77 1 3 1 4 1 1 1 1 ρ (0) + ρ (−1) = 12 = 12 12 11 1− 1 12 So, A1 = ρ (k ) = 45 1 k 32 1 k ( ) + (− ) # 77 3 77 4 ...
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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.
 Fall '09

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