Assignment_1_solutions_JX

Assignment_1_solutions_JX - Problem Set 1 ARMA Models...

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Problem Set 1: ARMA Models Solutions Question 1 Consider the process: y t = μ + θ 1 u t - 1 + θ 2 u t - 2 + θ 3 u t - 3 + u t Compute: 1. E [ y t ] E [ y t ] = E [ μ + θ 1 u t - 1 + θ 2 u t - 2 + θ 3 u t - 3 + u t ] = μ + θ 1 E [ u t - 1 ] + θ 2 E [ u t - 2 ] + θ 3 E [ u t - 3 ] + E [ u t ] = μ 2. V [ y t ] V [ y t ] = γ 0 = E [( y t - E [ y t ])( y t - E [ y t ])] = E [( y t - μ )( y t - μ )] = E ( u t + θ 1 u t - 1 + θ 2 u t - 2 + θ 3 u t - 3 ) 2 = E u 2 t + θ 2 1 u 2 t - 1 + θ 2 2 u 2 t - 2 + θ 2 3 u 2 t - 3 + E [2 θ 1 u t u t - 1 + 2 θ 2 u t u t - 2 + 2 θ 3 u t u t - 3 ] + E [2 θ 1 θ 2 u t - 1 u t - 2 + 2 θ 1 θ 3 u t - 1 u t - 3 + 2 θ 2 θ 3 u t - 2 u t - 3 ] = ( 1 + θ 2 1 + θ 2 2 + θ 2 3 ) σ 2 1

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3. γ 1 = E [( y t - E [ y t ])( y t - 1 - E [ y t - 1 ])] γ 1 = E [( y t - E [ y t ])( y t - 1 - E [ y t - 1 ])] = E [( y t - μ )( y t - 1 - μ )] = E [( u t + θ 1 u t - 1 + θ 2 u t - 2 + θ 3 u t - 3 )( u t - 1 + θ 1 u t - 2 + θ 2 u t - 3 + θ 3 u t - 4 )] = E [ u t u t - 1 + u t θ 1 u t - 2 + u t θ 2 u t - 3 + u t θ 3 u t - 4 ] + E θ 1 u 2 t - 1 + θ 2 1 u t - 1 u t - 2 + θ 1 u t - 1 θ 2 u t - 3 + θ 1 u t - 1 θ 3 u t - 4 + E θ 2 u t - 2 u t - 1 + θ 2 θ 1 u 2 t - 2 + θ 2 2 u t - 2 u t - 3 + θ 2 u t - 2 θ 3 u t - 4 + E θ 3 u t - 3 u t - 1 + θ 3 u t - 3 θ 1 u t - 2 + θ 3 θ 2 u 2 t - 3 + θ 2 3 u t - 3 u t - 4 = E θ 1 u 2 t - 1 + θ 2 θ 1 u 2 t - 2 + θ 3 θ 2 u 2 t - 3 = ( θ 1 + θ 1 θ 2 + θ 2 θ 3 ) σ 2 4. γ 2 = E [( y t - E [ y t ])( y t - 2 - E [ y t - 2 ])] γ 2 = E [( y t - E [ y t ])( y t - 2 - E [ y t - 2 ])] = E [( y t - μ )( y t - 2 - μ )] = E [( u t + θ 1 u t - 1 + θ 2 u t - 2 + θ 3 u t - 3 )( u t - 2 + θ 1 u t - 3 + θ 2 u t - 4 + θ 3 u t - 5 )] = E [ u t u t - 2 + u t θ 1 u t - 3 + u t θ 2 u t - 4 + u t θ 3 u t - 5 ] + E θ 1 u t - 1 u t - 2 + θ 2 1 u t - 1 u t - 3 + θ 1 u t - 1 θ 2 u t - 4 + θ 1 u t - 1 θ 3 u t - 5 + E θ 2 u 2 t - 2 + θ 2 u t - 2 θ 1 u t - 3 + θ 2 2 u t - 2 u t - 4 + θ 2 u t - 2 θ 3 u t - 5 + E θ 3 u t - 3 u t - 2 + θ 3 θ 1 u 2 t - 3 + θ 3 θ 2 u t - 3 u t - 4 + θ 2 3 u t - 3 u t - 5 = E θ 2 u 2 t - 2 + θ 1 θ 3 u 2 t - 3 = ( θ 1 + θ 1 θ 3 ) σ 2 2
5. γ 3 = E [( y t - E [ y t ])( y t - 3 - E [ y t - 3 ])] γ 3 = E [( y t - E [ y t ])( y t - 3 - E [ y t - 3 ])] = E [( y t - μ )( y t - 3 - μ )] = E [( u t + θ 1 u t - 1 + θ 2 u t - 2 + θ 3 u t - 3 )( u t - 3 + θ 1 u t - 4 + θ 2 u t - 5 + θ 3 u t - 6 )] = E [ u t u t - 3 + u t θ 1 u t - 4 + u t θ 2 u t - 5 + u t θ 3 u t - 6 ] + E θ 1 u t - 1 u t - 3 + θ 2 1 u t - 1 u t - 4 + θ 1 u t - 1 θ 2 u t - 5 + θ 1 u t - 1 θ 3 u t - 6 + E θ 2 u t - 2 u t - 3 + θ 2 u t - 2 θ 1 u t - 4 + θ 2 2 u t - 2 u t - 5 + θ 2 u t - 2 θ 3 u t - 6 + E θ 3 u 2 t - 3 + θ 3 θ 1 u t - 3 u t - 4 + θ 3 θ 2 u t - 3 u t - 5 + θ 2 3 u t - 3 u t - 6 = E θ 3 u 2 t - 3 = θ 3 σ 2 6. γ 4 = E [( y t - E [ y t ])( y t - 4 - E [ y t - 4 ])] γ 4 = E [( y t - E [ y t ])( y t - 4 - E [ y t - 4 ])] = 0 7. γ 5 = E [( y t - E [ y t ])( y t - 5 - E [ y t - 5 ])] γ 5 = E [( y t - E [ y t ])( y t - 5 - E [ y t - 5 ])] = 0 3

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8. γ - 1 = E [( y t - E [ y t ])( y t +1 - E [ y t +1 ])] γ - 1 = E [( y t - E [ y t ])( y t +1 - E [ y t +1 ])] = E [( y t - μ )( y t +1 - μ )] = E [( u t + θ 1 u t - 1 + θ 2 u t - 2 + θ 3 u t - 3 )( u t +1 + θ 1 u t + θ 2 u t - 1 + θ 3 u t - 2 )] = E u t u t +1 + θ 1 u 2 t + u t θ 2 u t - 1 + u t θ 3 u t - 2 + E θ 1 u t - 1 u t +1 + θ 2 1 u t - 1 u t + θ 1 θ 2 u 2 t - 1 + θ 1 u t - 1 θ 3 u t - 2 + E θ 2 u t - 2 u t +1 + θ 2 θ 1 u t - 2 u t + θ 2 2 u t - 2 u t - 1 + θ 2 θ 3 u 2 t - 2 + E θ 3 u t - 3 u t +1 + θ 3 u t - 3 θ 1 u t + θ 3 θ 2 u t - 3 u t - 2 + θ 2 3 u t - 3 u t - 2 = E θ 1 u 2 t + θ 1 θ 2 u 2 t - 1 + θ 2 θ 3 u 2 t - 2 = ( θ 1 + θ 1 θ 2 + θ 2 θ 3 ) σ 2 4
Question 2 Note: because the calculations are simple, I want you to compute the results by hand for this question.

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• Fall '16
• ROSSI
• Random walk, Mean squared error, Yt

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