W4437 HW 02 Answer keys.pdf - 1 Assignment 2 Textbook...

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Unformatted text preview: 1 Assignment 2 Textbook problems: 1.4, 1.5, 1.7, 1.10. 2 Answer keys: = ; independent of 1.4 (30 points; 5 point per part) (a) , (c) (d) (e) 0 is stationary. , Therefore, = , Therefore, = , Therefore, = , = , =* ℎ=0 ℎ = ±1 ℎ = ±2 ℎ is stationary. ! + 1 "# $ is NOT stationary except in the trivial case when is an integer multiple of %. = ; independent of ; independent of = 0; independent of is stationary. &' ( ! 2 +ℎ "+ ℎ ) is NOT stationary except in the trivial case when is an integer multiple of %. = 0; independent of Therefore, ; independent of ℎ ; independent of = 0; independent of Therefore, (f) = = 0; independent of Therefore, (b) + 0 0 + ℎ=0 ; independent of |ℎ| > 0 is stationary. Moreover, 1.5 (20 points; 10+5+5) 1+4 12 ℎ = 34 0 When 4 = 0.8, (a) The ACVF and ACF of 1.64 12 ℎ = 30.8 0 ~/0 0, + . 1 5 ℎ = 0 5 ℎ = ±2 and 72 ℎ = 34/ 1 + 4 ℎ 0 are, respectively 5 ℎ = 0 1 and 7 ℎ = 3 5 ℎ = ±2 0.49 2 ℎ 0 5 ℎ = 0 5 ℎ = ±2 . ℎ 5 ℎ = 0 5 ℎ = ±2 . ℎ 3 (b) When 4 = 0.8, > ! ? + + (c) When 4 = −0.8, @ + + + + 1 /4" = AA 16 ! B, CD? BD? C" = 1 E412 0 + 412 2 F = 0.61. 16 1 E412 0 + 412 2 F = 0.21. 16 The negative lag-2 correlation in means that positive deviations of from zero tend to be followed two time units later by a compensating negative deviation, resulting in smaller variability in the sample mean than in part (b). > ! ? + + @ + /4" = + + H = I2 + IJ , which is independent of . 1.7 (5 points) Since K , H = 0 for all independent of . +H and , , Therefore, + H is weakly stationary by definition, and 12 shown above. For M = ∑PODQ 1.10 (5 points) O O for some constants , we have P ∇M = A ODQ Q , ⋯ , PT O O P −A , since ODQ O −1 That is, ∇M is a polynomial of degree S − 1. P −1 = P O =S −S P PT? + H = 12 ℎ + 1J ℎ , which is J PT? ℎ = 12 ℎ + 1J ℎ , ∀ ℎ, as is PT +A + ⋯. ODQ O O By successive application of the difference operator ∇, we deduce that ∇P ? M = 0. ...
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