# 2 since every solution of θ z 0 lies outside the

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Since every solution of θ ( z ) = 0 lies outside the unit circle, { X t } is invertible. Problem 3.4 Let us compute the ACF and PACF of the AR(2) process. a. For the calculations of ACF, we can use both Methods in BD. Here, we present Method 2. The solutions of φ ( z ) = 1 - 0 . 8 z 2 = 0 are z 1 = 0 . 8 - 1 / 2 and z 2 = - 0 . 8 - 1 / 2 . Thus the roots are differents and are out side the unit cirle. The process is then causal. If we use the notation in BD, we have m = max ( p, q + 1) = max (2 , 0 + 1) = 2 and the general solution is given by γ X ( h ) = α 1 ξ - h 1 + α 2 ξ - h 2 , h 0 where ξ 1 and ξ 2 are the roots of the equation φ ( z ) = 0 ( ξ 1 = - ξ 2 = 0 . 8 - 1 / 2 ). The initial conditions, gives ( m = 2 equations k = 0 , ...m - 1) ( k = 0) γ X (0) - 0 . 8 γ X (2) = σ 2 ( k = 1) γ X (1) = 0 . Now, if we replace γ X (0), γ X (1), and γ X (2) from general solution in these two equations we obtain ( α 1 + α 2 ) - 0 . 8( α 1 ξ 2 + α 2 ξ 2 ) = σ 2 α 1 = α 2 Therefore, α 1 = α 2 = σ 2 2(1 - (0 . 8) 2 ) . This implies that γ X ( h ) = 0 ifh is odd σ 2 1 - (0 . 8) 2 (0 . 8) h/ 2 if h is even . b. The calculations of PACF in AR(2). In the AR(p) case, we have that PACF at lags more than h are 0 and α ( p ) = φ p . Then, by the definition of PACF we obtain α X ( h ) = 1 h = 0 γ X (1) X (0) = 0 h = 0 φ 2 = 0 . 8 h = 2 0 h > 2 . Problem 3.8 Because X t is the unique stationary solution of the AR(1) equation ( | φ | > 1), X t is X t = - j =1 φ - j Z t + j . 3
On the other hand, for h 0, γ X ( h ) = Cov ( X t + h , X t ) = Cov ( - i =1 φ - i Z t + h + i , - j =1 φ - j Z t + j ) = σ 2 j =1 φ - j φ - ( h + j ) = σ 2 φ - h φ - 2 1 - φ - 2 = σ 2 1 φ h
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