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# solutions3 - Matlab Solutions Time Series(2DD23 Exercise...

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Matlab Solutions Time Series (2DD23) Week 3 Exercise 3.1 The formula of the autocorrelationfunction is: ρ( k ) = γ ( k ) γ ( 0 ) . γ ( k ) = Cov ( X t , X t + k ) = Cov ( Z t + 0 . 7 Z t - 1 - 0 . 2 Z t - 2 , Z t + k + 0 . 7 Z t + k - 1 - 0 . 2 Z t + k - 2 ) Now use the formula: Cov ( X 1 + X 2 , Y ) = Cov ( X 1 , Y ) + Cov ( X 2 , Y ) . γ ( k ) = Cov ( Z t , Z t + k ) + Cov ( Z t , 0 . 7 Z t + k - 1 ) + Cov ( Z t , - 0 . 2 Z t + k - 2 )) + Cov ( 0 . 7 Z t - 1 , Z t + k ) + Cov ( 0 . 7 Z t - 1 , 0 . 7 Z t + k - 1 ) + Cov ( 0 . 7 Z t - 1 , - 0 . 2 Z t + k - 2 )) + Cov ( - 0 . 2 Z t - 2 , Z t + k ) + Cov ( - 0 . 2 Z t - 2 , 0 . 7 Z t + k - 1 ) + Cov ( - 0 . 2 Z t - 2 , - 0 . 2 Z t + k - 2 )) Now use Cov ( aX , Y ) = a Cov ( X , Y ), Cov ( X , X ) = Var ( X ) and for k 6= 0 : Cov ( Z t , Z t + k ) = 0 . The case k = 0 : γ ( k ) = Cov ( Z t , Z t ) + 0 . 7 Cov ( Z t , Z t - 1 ) - 0 . 2 Cov ( Z t , Z t - 2 )) + 0 . 7 Cov ( Z t - 1 , Z t ) + 0 . 7 2 Cov ( Z t - 1 , Z t - 1 ) - 0 . 14 Cov ( Z t - 1 , Z t + k - 2 )) - 0 . 2 Cov ( Z t - 2 , Z t ) - 0 . 14 Cov ( Z t - 2 , Z t + k - 1 ) + ( - 0 . 2 ) 2 Cov ( Z t - 2 , Z t + k - 2 )) = ( 1 + 0 . 7 2 + ( - 0 . 2 ) 2 2 Z = 1 . 53 σ 2 Z The case k = 1 : γ ( k ) = ( 0 . 7 + 0 . 7 ( - 0 . 2 ))σ 2 Z = 0 . 56 σ 2 Z The case k = 2 : γ ( k ) = - 0 . 2 σ 2 Z The case k > 2 : γ ( k ) = 0 And in general: γ ( k ) = γ ( - k ) . 1

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ρ( k ) = γ ( k ) γ ( 0 ) = γ ( k ) 1 . 53 σ 2 Z = 1 k = 0 0 . 56 1 . 53 k = ± 1 - 0 . 2 1 . 53 k = ± 2 0 otherwise Of course we could have used the formulas from paragraph 3 . 4 . 3 immediately. Exercise 3.2 γ ( k ) = Cov ( X t , X t + k ) = Cov 1 m + 1 m X j = 0 Z t - j , 1 m + 1 m X j = 0 Z t - j + k = 1 ( m + 1 ) 2 Cov m X j = 0 Z t - j , m X j = 0 Z t - j + k = ( 0 k > m m - k + 1 ( m + 1 ) 2 σ 2 Z k = 0 , . . . , m The autocorrelation function is: ρ( k ) = γ ( k ) γ ( 0 ) . We have: γ ( 0 ) = m + 1 ( m + 1 ) 2 σ 2 Z = σ 2 Z m + 1 . ρ( k ) = ( 0 k > m m - k + 1 m + 1 k = 0 , . . . , m Exercise 3.3 E ( X t ) = E ( Z t + C ( Z t - 1 + Z t - 2 + . . . )) = 0 + C ( 0 + 0 + . . . ) = 0 Var ( X t ) = Var ( Z t + C ( Z t - 1 + Z t - 2 + . . . )) = σ 2 Z + C 2 2 Z + σ 2 Z + . . . ) = ∞ So X t is not stationary. Now regards the first order differences of
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