PS1_sol(3).pdf

# PS1_sol(3).pdf - Problem Set 1 ARMA Processes ECON 21200...

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Problem Set 1: ARMA Processes ECON 21200, Spring 2015 SOLUTIONS Problem 1 (Manipulating ARMA Processes) Suppose Y t = μ + φ 1 Y t - 1 + φ 2 Y t - 2 + t . Assume Var ( t ) = 1 . a) Solve for the roots of the lag polynomial, λ 1 and λ 2 , in terms of μ , φ 1 , and φ 2 . Solution: Write in lag-polynomial form and solve: ( 1 - φ 1 L - φ 2 L 2 ) Y t = μ + t λ 2 - φ 1 λ - φ 2 = 0 λ = φ 1 ± p φ 2 1 + 4 φ 2 2 1

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b) Assume Y t is stationary. Find the MA( ) form of Y t is Y t = α 0 + k =0 θ k t - k . Solution: If Y t is stationary, we can factor the lag polynomial and invert it term by term (1 - λ 1 L )(1 - λ 2 L ) Y t = μ + t (1 - λ 2 L ) Y t = (1 - λ 1 L ) - 1 μ + (1 - λ 1 L ) - 1 t (1 - λ 2 L ) Y t = μ 1 - λ 1 + X i =0 λ i 1 L i t Y t = μ (1 - λ 1 )(1 - λ 2 ) + (1 - λ 2 L ) - 1 X i =0 λ i 1 L i t = μ 1 - φ 1 - φ 2 + X j =0 λ j 1 L j ! X i =0 λ i 1 L i ! t = α 0 + X k =0 λ 0 1 λ k 2 + λ 1 1 λ k - 1 2 + · · · + λ k 1 λ 0 2 L k t = α 0 + X k =0 k X l =0 λ l 1 λ k - l 2 ! t - k c) Find the unconditional mean and variance of Y t . What is the unconditional distribution of Y t ? Solution: You can solve this in multiple ways, using either the AR(2) form of the process or the MA( ) form. I will use the AR(2) form. Set all the Y ’s equal to their means: μ Y = μ + φ 1 μ Y + φ 2 μ Y + 0 μ Y = μ 1 - φ 1 - φ 2 Note that you could also read this directly off the MA( ) form: μ Y equals the constant α 0 . To solve for the variance, first de-mean the process. Then square both sides and take expectations. Remember that we will need to solve multiple equations because there will be autocovariances in here, so we will also need to write the equation for γ 1 : γ 0 = φ 2 1 γ 0 + φ 2 2 γ 0 + 2 φ 1 φ 2 γ 1 + σ 2 γ 1 = φ 1 γ 0 + φ 2 γ 1 2
Use the second equation to solve for γ 1 in terms of γ 0 : (1 - φ 2 ) γ 1 = φ 1 γ 0 γ 1 = φ 1 1 - φ 2 γ 0 Now substitute γ 1 into the equation for γ 0 and solve: γ 0 = φ 2 1 γ 0 + φ 2 2 γ 0 + 2 φ 2 1 φ 2 1 - φ 2 γ 1 + σ 2 1 - φ 2 1 - φ 2 2 - 2 φ 2 1 φ 2 1 - φ 2 γ 0 = σ 2 ( 1 - φ 2 - φ 2 1 + φ 2 φ 2 1 - φ 2 2 + φ 3 2 - 2 φ 2 1 φ 2 ) γ 0 = (1 - φ 2 ) σ 2 ( 1 - φ 2 - φ 2 2 + φ 3 2 - φ 2 1 - φ 2 1 φ 2 ) γ 0 = (1 - φ 2 ) σ 2 (1 + φ 2 ) ( (1 - φ 2 2 ) - φ 2 1 ) γ 0 = (1 - φ 2 ) σ 2 γ 0 = 1 - φ 2 1 + φ 2 · σ 2 (1 - φ 2 ) 2 - φ 2 1 Therefore, the unconditional distribution of Y t is N ( μ Y , γ 0 ) . For parts (d) - (f), suppose μ = 0 . 5 , φ 1 = 1 . 2 and φ 2 = - 0 . 35 . d) Prove that this process is stationary. Solution: Plug the values of φ 1 and φ 2 into the answer for part (a) to get λ 1 = 0 . 7 and λ 2 = 0 . 5. Clearly, these are both smaller than 1 in magnitude. When both the roots are stable, the process is stationary. e) Use your answer to part (a) and write Y t = α 0 + k =0 θ k t - k (i.e. solve for α 0 and θ k explicitly). Solution: α 0 = μ 1 - φ 1 - φ 2 = 0 . 5 1 - 1 . 2+0 . 35 = 10 3 . The roots of the process are 0.7 and 0.5. The overall process is Y t = 10 3 + k =0 k j =0 0 . 5 j 0 . 7 k - j ! t - k . f) Calculate Var ( Y t ) and the first five autocovariances and autocorrelations in terms of σ 2 . (Remember, there are multiple ways to do this, but some ways will be much easier than others!) Solution: Here, we can substitute into the answer to part 3

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(c) to get γ 0 and γ 1 and then use the Yule-Walker equations to get the other parameters: γ 0 = 1 + 0 . 35
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