solutions2 - ( x t-3 + x t-4 + x t-5 + x t-6 + x t-7 ))-3 4...

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Matlab Solutions Time Series (2DD23) Week 2 Exercise 2.5 a. 12 X t = X t - X t - 12 = ( a + bt + S t + ε t ) - ( a + b ( t - 12 ) + S t - 12 + ε t - 12 ) = a - a + bt - bt + 12 b + S t - S t - 12 + ε t - ε t - 12 = 0 + 0 + 12 b + 0 + ε t - ε t - 12 We can see that 12 b is constant, and ε t - ε t - 12 is a linear combination of { ε t } , so it is stationary. b. 12 X t = X t - X t - 12 = (( a + bt ) S t + ε t ) - (( a + b ( t - 12 )) S t - 12 + ε t - 12 ) = a ( S t - S t - 12 ) + bt ( S t - S t - 12 ) + 12 bS t - 12 + ε t - ε t - 12 = 0 + 0 + 12 bS t - 12 + ε t - ε t - 12 This series is not stationary because of the seasonal term 12 bS t - 12 . The operator 2 12 will make the series stationary: 2 12 X t = ∇ 12 ( 12 X t ) = ∇ 12 ( X t - X t - 12 ) = ( 12 bS t - 12 + ε t - ε t - 12 ) - ( 12 bS t - 24 + ε t - 12 - ε t - 24 ) = 12 b ( S t - 12 - S t - 24 ) + ε t - 2 ε t - 12 ) + ε t - 24 ) = 0 + ε t - 2 ε t - 12 ) + ε t - 24 ) This is a linear combination of { ε t } , so it is stationary. Extra Exercise Solution First compute the convolution of ( 1 5 , 1 5 , 1 5 , 1 5 , 1 5 ) * ( - 3 4 , 3 4 , 1 , 3 4 , - 3 4 ) : y t = X j a j x t - j = 1 5 x t + 1 5 x t - 1 + 1 5 x t - 2 + 1 5 x t - 3 + 1 5 x t - 4 = 1 5 ( x t + x t - 1 + x t - 2 + x t - 3 + x t - 4 ) 1
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z t = X j b j y t - j = - 3 4 y t + 3 4 y t - 1 + y t - 2 + 3 4 y t - 3 + - 3 4 y t - 4 = - 3 4 ( 1 5 ( x t + x t - 1 + x t - 2 + x t - 3 + x t - 4 )) + 3 4 ( 1 5 ( x t - 1 + x t - 2 + x t - 3 + x t - 4 + x t - 5 )) + 1 5 ( x t - 2 + x t - 3 + x t - 4 + x t - 5 + x t - 6 ) + 3 4 ( 1 5
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Unformatted text preview: ( x t-3 + x t-4 + x t-5 + x t-6 + x t-7 ))-3 4 ( 1 5 ( x t-4 + x t-5 + x t-6 + x t-7 + x t-8 )) = 1 20 (-3 x t + 4 x t-2 + 7 x t-3 + 4 x t-4 + 7 x t-5 + 4 x t-6-3 x t-7 ) This can be denoted as (-3 20 , , 1 5 , 7 20 , 1 5 , 7 20 , 1 5 , ,-3 20 ) . In the same way the convolution of the other lters can be computed. The order in which you convolute the lters is not important. E.g. ( 1 4 , 1 4 , 1 4 , 1 4 ) * ( 1 4 , 1 4 , 1 4 , 1 4 ) = 1 16 ( 1 , 2 , 3 , 4 , 3 , 2 , 1 ) . 2...
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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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solutions2 - ( x t-3 + x t-4 + x t-5 + x t-6 + x t-7 ))-3 4...

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