# solutions4 - Matlab Solutions Time Series (2DD23) Exercise...

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Matlab Solutions Time Series (2DD23) Week 4 Exercise 3.4 X t - μ = 0 . 7 ( X t - 1 - μ) + Z t = 0 . 7 ( 0 . 7 ( X t - 2 - μ) + Z t - 1 ) + Z t = 0 . 7 ( 0 . 7 ( 0 . 7 ( X t - 3 - μ) + Z t - 2 ) + Z t - 1 ) + Z t = 0 . 7 3 ( X t - 3 - μ) + 0 . 7 2 Z t - 2 + 0 . 7 Z t - 1 + Z t = X i = 0 0 . 7 i Z t - i Now we can compute the expectation and the variance of X t - μ : E ( X t - μ) = 0 Var ( X t - μ) = σ 2 Z X i = 0 ( 0 . 7 2 ) i = 1 1 - 0 . 7 2 σ 2 Z The autocovariance function: γ ( k ) = Cov ( X t , X t + k ) = Cov ( X t - μ, X t + k - μ) = Cov ( X i = 0 0 . 7 i Z t - i , X i = 0 0 . 7 i Z t + k - i ) = Cov ( X i = 0 0 . 7 i Z t - i , X j =- k 0 . 7 j + k Z t - j ) = σ 2 Z X i = 0 0 . 7 i 0 . 7 i + k = σ 2 Z 0 . 7 k X i = 0 ( 0 . 7 2 ) i = σ 2 Z 0 . 7 k 1 - 0 . 7 2 The autocorrelation function: ρ( k ) = γ ( k ) γ ( 0 ) = 0 . 7 k 1

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Exercise 3.7 X t = 1 12 X t - 1 + 1 12 X t - 2 + Z t X t - 1 12 X t - 1 - 1 12 X t - 2 = Z t ( 1 - 1 12 B - 1 12 B 2 ) X t = Z t X t = Z t 1 - 1 12 B - 1 12 B 2 The auxiliary equation becomes: y 2 - 1 12 y - 1 12 = 0 ( y - 1 24 ) 2 = 1 12 + 1 24 2 = ( 7 24 ) 2 y - 1 24 = 7 24 y - 1 24 = - 7 24 y = 1 3 y = - 1 4 The roots lie within the unit circle, so X t is stationary. Now we will compute the autocorrelation- function: ρ( k ) = A 1 π | k | 1 + A 2 π | k | 2 , where π 1 = 1
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## This note was uploaded on 09/23/2009 for the course MATH 200 taught by Professor Gur during the Spring '09 term at Accreditation Commission for Acupuncture and Oriental Medicine.

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solutions4 - Matlab Solutions Time Series (2DD23) Exercise...

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