HW1_Solutions - Solutions to HW 1 Exercise 3.1 (h) = Cov(X...

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Solutions to HW 1 Exercise 3.1 ( )( ) 12 1 () C o v ( , ) ( , ) . 7 . 2 . 7 . 2 th t t t t t hX X E X X E Z ZZ Z Z Z γ ++ + + + 2 == = + + , i.e., since 2 ,, (Z , ) 0, , σ = = ts st EZ () 22 2 1 2 ( 0 ) . 7. 2 . 2 ( ) . 4 9 ( ) . 0 4 ( ) 1 . 5 3 tt ttt t t t t Z Z Z Z Z E Z E Z E Z 2 −− ⎡⎤ =+ + + = ⎣⎦ , 11 1 2 1 (1) .7 .2 ( ) .14 ( ) .56 t t t t Z Z Z Z Z E Z E Z 2 +− + −=− = , 21 (2) .7 .2 .7 .2 .2 ( ) .2 t t t t Z Z Z Z Z E Z −+ −= = , ( 321 (3) .7 .2 .7 .2 0 t Z Z Z Z Z +++ + ) = , and / ( 0 ) . ρ γγ = hh Problem 1 { } {} 2 1 2 1 1 2 1 2 121 2 (, ) ( ) ()( ) ( ) { } ( ) { } ( ) ( ) { } ()() ()() . t t t t E X t X t EXX tX X t t t t EX t t t t t t t t t t γμ μ μμ μμμ =− + = += =−−+= Problem 2 For a strictly stationary process, shifting the time origin has no effect on the joint distributions, so that all the joint moments also remain the same. Therefore, in particular, [ ] [ ] 0 t EX , a constant independent of t , and [ ] [ ] 0 h h + = is independent of t for each h . Then 2 o v [ , ] ] h h X E X X (see Problem 1) also depends only on the lag . Problem 3 t X is not stationary since t t α β is not constant. Problem 4 2 2 4 1 1 2 1 1 () ( ) () ( )0 , ( 0 )( ) ( 1 )() ( ) 0 ( 1 ) , , σσ + = = = = = t t t t t t t t t t t t EZZ EZ EZ EZZ ZZ EZ Z
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HW1_Solutions - Solutions to HW 1 Exercise 3.1 (h) = Cov(X...

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