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solutions9 - ω cos ω k d ω = Z π ω k d sin ω k = h ω...

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Matlab Solutions Time Series (2DD23) Week 9 Exercise 6.2b X t = Z t + 0 . 5 Z t - 1 - 0 . 3 Z t - 2 The power spectral density function is: f (ω) = 1 π 2 k = 1 γ ( k ) cos ω k + γ ( 0 ) = 1 π 2 k = 0 γ ( k ) cos ω k - γ ( 0 ) The autocovariance function of a MA ( q ) process has a cut-off after lag q . γ ( 0 ) = Var ( X t ) = σ 2 Z 2 i = 0 β 2 i = ( 1 2 + 0 . 5 2 + ( - 0 . 3 ) 2 2 Z = 1 . 34 σ 2 Z γ ( 1 ) = σ 2 Z 0 β 1 + β 1 β 2 ) = ( 1 × 0 . 5 + 0 . 5 × - 0 . 3 2 Z = 0 . 35 σ 2 Z γ ( 2 ) = σ 2 Z 0 β 2 ) = - 0 . 3 σ 2 Z So f (ω) = σ 2 Z π ( 1 . 34 + 2 ( 0 . 35 cos ω - 0 . 3 cos 2 ω) ) . Exercise 6.4 f * (ω) = 2 - ω) π 2 Use formula 6.9: ρ( k ) = γ ( k ) γ ( 0 ) = γ ( k ) σ 2 X = π 0 cos ω k f (ω) σ 2 X d ω = π 0 f * (ω) cos ω k d ω = π 0 2 - ω) π 2 cos ω k d ω = 2 π π 0 cos ω k d ω - 2 π 2 π 0 ω cos ω k d ω Intermediate results: π 0 cos
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Unformatted text preview: ω cos ω k d ω = Z π ω k d sin ω k = h ω k sin ω k i π-1 k Z π sin ω k d ω = + ² 1 k 2 cos ω k ³ π = 1 k 2 cos k π-1 k 2 = -1 k 2 + (-1 ) k k 2 1 So we get: ρ( k ) = -2 π 2 ±-1 k 2 + (-1 ) k k 2 ² = ³ k is even 4 ( k π) 2 k is odd Only if k = , we have ρ( k ) = 1 . 2...
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