A2 soln - STA4005B Assignment 2 (Solution) 1) 3 2 1 4 . 3 ....

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Unformatted text preview: STA4005B Assignment 2 (Solution) 1) 3 2 1 4 . 3 . 3 . − − − − + − = t t t t t a a a a Z 6 . 1 ) 4 . ( 68 . 1 )) 4 . )( 3 . ( 3 . ( 04 . 2 )) 3 . )( 4 . ( ) 3 . )( 3 . ( 3 . ( 36 . 5 ) 4 . 3 . 3 . 1 ( 2 3 2 2 2 1 2 2 2 2 − = − = = − − + = − = − + − + − = = + + + = a a a a σ γ σ γ σ γ σ γ = k γ for 4 ≥ k ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∑ = 100 1 100 1 t t Z Var [ ] [ ] 015096 . ) ) 6 . 1 ( 97 ) 68 . 1 ( 98 ) 04 . 2 ( 99 ( 2 ) 36 . 5 ( 100 10000 1 ) 2 97 98 99 ( 2 100 10000 1 ) , ( 100 1 99 98 3 2 1 100 2 1 100 2 1 2 = + − + + − × + = + + + + + × + = + + + + + + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = γ γ γ γ γ γ L L L Z Z Z Z Z Z Cov 2a) Fitting model (a) in S-plus, we get ) 12 2 sin( 7818 . 14 ) 12 2 cos( 0034 . 22 266 . 64 ˆ t t Z t π π − − = Call: lm(formula = Zt ~ cos + sin, data = STA4005B.asg2, na.action = na.exclude) Residuals: Min 1Q Median 3Q Max -9.184 -2.448 -0.008571 2.361 8.716 Coefficients: Value Std. Error t value Pr(>|t|) (Intercept) 64.2660 0.2837 226.5299 226....
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A2 soln - STA4005B Assignment 2 (Solution) 1) 3 2 1 4 . 3 ....

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