Time Series Exam, 2009: Solutions - STAT

and the cross-covariancefunction (substituting also the given parametervalues) isγ XY (0) = σ 2 ǫǫ θ 11θ 21 +σ 2 ǫη (1+θ 12θ 21 +θ 11 θ 22 )+σ 2 ηη θ 12θ 22 = 10,γ XY (1) = σ 2 ǫη θ 11 +σ 2 ηη θ 12 = 0,γ XY (−1) = σ 2 ǫǫ θ 21 +σ 2 ǫη θ 22 = 2,γ XY (h) = 0 otherwise.(b) Equivalent forms for the two bivariate series areX t = θV t−1 +U t ,Y t = V t ,where we used the second equation in the first, and denoted θ 12 byθ, andX ′ t = θV t +U t−1 ,Y ′t = V t−1 .The cross-covariance functions areγ XY (0) = Ω 12 ,γ XY (1) = θΩ 22 ,γ XY (−1) = 0,γ XY (h) = 0 otherwiseandγ XY (0) = Ω 12 ,γ XY (1) = 0,γ XY (−1) = θΩ 22 ,γ XY (h) = 0 otherwise.Since γ XY (h) = γ YX (−h), the two structures are indeed equivalent:a simple exchange of names transforms one model into the other.Alternative solution: For the AR model we have (I −AB)Z t = W t ,where B is the usual operator, and A is a 2×2 matrix. On invertingthis we have Z t = (I − AB) −1 W t = (I + AB + A 2 B 2 + ...)W t =(I+AB)W t because A 2 = 0. Hence we can’t distinguish these AR(1)and MA(1) models.5