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2.6. COINTEGRATION 65in q t−1 because ∆q t is stationary and this can be true only if q t−1 dropsout from the right side of (2.92).By analogy, suppose that in the bivariate case the vector (q t ,f t )isgenerated according to"qtf t#="a11 a 12a 21#" #qt−1+a 22 f t−1"b11 b 12b 21#" #qt−2+b 22 f t−2" #uqtu ft, (2.93)where (u qt ,u ft ) 0 iid ∼ N(0, Σ u ). Rewrite (2.93) as the vector analog of theaugmented Dickey—Fuller test equation" #∆qt=∆f twhere"r11 r 12r 21#" #qt−1−r 22 f t−1"b11 b 12b 21#" #∆qt−1+b 22 ∆f t−1" # "r11 r 12 a11 + b=11 − 1 a 12 + b 12r 21 r 22 a 21 + b 21 a 22 + b 22 − 1#≡ R." #uqtu ft,(2.94)If {q t } and {f t } areunitrootprocesses,theirÞrst differences are stationary.This means the terms on the right hand side of (2.94) are stationary.Linear combinations of levels of the variables appear in the system.r 11 q t−1 + r 12 f t−1 appears in the equation for ∆q t and r 21 q t−1 + r 22 f t−1appears in the equation for ∆f t .If {q t } and {f t } do not cointegrate, there are no values of the r ijcoefficients that can be found to form stationary linear combinationsof q t and f t . The level terms must drop out. R is the null matrix, and(∆q t , ∆f t ) follows a vector autoregression.If {q t } and {f t } do cointegrate, then there is a unique combinationof the two variables that is stationary. The levels enter on the right sidebut do so in the same combination in both equations. This means thatthe columns of R are linearly dependent and the R, which is singular,can be written as" #r11 −βrR =11.r 21 −βr 21(2.94) can now be written as" #∆qt∆f t=" #r11r 21(q t−1 − βf t−1 ) −"b11 b 12b 21#" #∆qt−1+b 22 ∆f t−1" #uqtu ft