A0094709U Problem Set 2.docx - ECA5103 Quantitative and...

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ECA5103 Quantitative and Computing Methods Name : Lin Jia He (Monday Group) Student Number : A0094709U Question 1 Part 1 Question 1A Part 1. Given an AR (1) process Y t = 2 + 0.5 Y t 1 + U t ,U t iid N ( 0,6 ) Y E ( ¿¿ t )= E ( 2 + 0.5 Y t 1 + U t ) ¿ Y E ( ¿¿ t )= E [ 2 + 0.5 ( 2 ) + 0.5 2 Y t 2 + U t + 0.5 U t 1 ] ¿ Y E ( ¿¿ t )= E [ 2 + 0.5 ( 2 ) + 0.5 2 ( 2 )+ 0.5 3 Y t 3 + U t + 0.5 U t 1 + 0.5 2 U t 2 ] ¿ ……………… Y E ( ¿¿ t )= lim n→∞ E [ 2 + 0.5 ( 2 ) + + 0.5 n ( 2 )+ 0.5 n + 1 Y t n 1 + U t + 0.5 U t 1 + 0.5 n U t n ] ¿ Since 0.5 E [ ¿¿ n + 1 Y t n 1 ]= 0 lim n→∞ ¿ when E [ Y t n 1 ] has a finite mean Y E ( ¿¿ t )= lim n→∞ E [ 2 + 0.5 ( 2 ) + + 0.5 n ( 2 ) + U t + 0.5 U t 1 + 0.5 n U t n ] ¿
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Y E ( ¿¿ t )= j = 0 0.5 j ( 2 ) + i = 0 0.5 i ( U t i ) ¿ Y E ( ¿¿ t )= j = 0 0.5 j ( 2 ) ¿ since E [ U z ] = 0 for all z R Y E ( ¿¿ t )= 2 1 1 2 ¿ Y E ( ¿¿ t )= 4 ¿ Y Var ( ¿¿ t )= Cov ( Y t ,Y t ) ¿ Y Var ( ¿¿ t )= Cov [ j = 0 0.5 j ( 2 ) + i = 0 0.5 i ( U t i ) , j = 0 0.5 j ( 2 ) + i = 0 0.5 i ( U t i ) ] ¿ Since Cov [ a, x ] = 0 for any a being a constant Y Var ( ¿¿ t )= Cov [ i = 0 0.5 i ( U t i ) , i = 0 0.5 i ( U t i ) ] ¿
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Y Var ( ¿¿ t )= k = 0 0.5 2 k Var ( U t k ) ¿ , since U t iid N ( 0,6 ) implies Cov ( U t ,U t s ) = 0 for s≠ 0 Y Var ( ¿¿ t )= k = 0 0.5 2 k ( 6 ) ¿ , since Var ( U z ) = 6 for all z R Y Var ( ¿¿ t )= 6 ( 1 + 0.5 2 + 0.5 4 + ) ¿ Y Var ( ¿¿ t )= 6 ¿ ( 1 1 0.5 2 ¿ Y Var ( ¿¿ t )= 8 ¿ Question 1B Part 1. In similar vein to the argument from above in Y Var ( ¿¿ t )= Cov ( Y t ,Y t ) ¿ Cov ( Y t ,Y t k ) = Cov [ i = 0 0.5 i ( U t i ) , i = 0 0.5 i ( U t k i ) ] Cov ( Y t ,Y t k ) = 0.5 k Cov [ U t 0.5 k + U t 1 0.5 k 1 + i = 0 0.5 i ( U t k i ) , i = 0 0.5 i ( U t k i ) ] Since U t iid N ( 0,6 ) implies Cov ( U t ,U t s ) = 0 for s≠ 0 Cov ( Y t ,Y t k ) = 0.5 k Cov [ i = 0 0.5 i ( U t k i ) , i = 0 0.5 i ( U t k i ) ] Cov ( Y t ,Y t k ) = 0.5 k [ var ( U t k ) + 0.5 2 ( 1 ) var ( U t k 1 ) + 0.5 2 ( 2 ) var ( U t k 2 ) + ]
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Cov ( Y t ,Y t k ) = 0.5 k [ i = 0 0.5 2 i ( 6 ) ] , since Var ( U z ) = 6 for all z R Cov ( Y t ,Y t k ) = 0.5 k ( 6 1 0.5 2 ¿ = 0.5 k ( 8 ) By the above Cov ( Y t ,Y t 1 ) = 0.5 1 ( 8 ¿ = 4 Cov ( Y t ,Y t 2 ) = 0.5 2 ( 8 ¿ = 2 Question 1C Part 1. Note that by 1A, we can show in the same way Var ( Y z ) = 8 for all z R Hence Corr ( Y t ,Y t 1 ) = Cov ( Y t ,Y t 1 ) Var ( Y t ) Var ( Y t 1 ) Corr ( Y t ,Y t 1 ) = 4 ( 8 )( 8 ) = 1 2 Corr ( Y t ,Y t 2 ) = 2 ( 8 )( 8 ) = 1 4 Question 1D Part 1. Note that by 1A, we can show in the same way E [ Y z ]= 4 for all z R And since Var ( Y z ) = 8 for all z R By 1B, Cov ( Y t ,Y t k ) = 0.5 k ( 8 ) , Cov ( Y t ,Y t k ) is a function of k (time distance between data points) and not t (time at which data occurred)
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Hence, Y t fulfill the condition of being a stationary time series. As per above lim k →∞ Cov ( Y t ,Y t k ) = lim k → ∞ 0.5 k ( 8 ) = 0 The autocovariance of the series tends to zero the further the data points are apart, and by such Y t is almost “independent” of Y t h when h grows to infinity. Hence, Y t is weakly dependent .
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