Chapter 5: The Cointegrated VAR model Katarina Juselius July 1, 2012 Katarina Juselius () Chapter 5: The Cointegrated VAR model July 1, 2012 1 / 41
An intuitive interpretation of the Pi matrix Consider the VAR(2) model in ECM form with m = 1 : x t = Γ 1 x t 1 + Πx t 1 + µ + ε t (1) If x t I (1), then x t I (0) and Π cannot have full rank as a stationary variable x t cannot be equal to a nonstationary variable, x t 1, plus other stationary terms. Either Π = 0, or it must have reduced rank: Π = αβ 0 where α and β are p r matrices, r p. Thus, under the I(1) hypothesis, the cointegrated VAR model is given by: x t = Γ 1 x t 1 + αβ 0 x t 1 + µ + ε t (2) where β 0 x t 1 is an r 1 vector of stationary cointegration relations. Katarina Juselius () Chapter 5: The Cointegrated VAR model July 1, 2012 2 / 41
Decomposing the Pi martix The estimate of the unrestricted Π matrix: 2 Πx t = 6 4 0.26 0.24 1.48 5.04 4.99 0.02 0.12 0.33 2.11 0.51 0.02 0.01 0.77 0.64 0.24 0.00 0.00 0.01 0.11 0.04 0.00 0.00 0.03 0.01 0.09 3 2 7 6 5 4 mt r 1 yt r 1 p t 1 R m,t 1 R b,t 1 By setting α 1 = 0.26 and β 0 1 = [1, 0.9, 5.7, 19.3, 19.3], we can rougly reproduce the rst row of Π as α 11 β 0 1. Assuming that Γ 1 = 0, the cointegrated VAR model can now be written as: 2 6 4 m r t y r t 2 p t R m,t R b,t 3 2 7 = 6 5 4 0.26 α 21 α 31 α 41 α 51 3 7 fm r 0.9y r + 5.7 p 19.3(R m R b )g t 1 + 5 Katarina Juselius () Chapter 5: The Cointegrated VAR model July 1, 2012 3 / 41 3 7 5
Decomposing the Pi with r=2 By setting α 2 = 0.12 and β 0 2 = [0, 1, 0, 17.5, 0] we can roughly reproduce the second row in Π. The cointegrated VAR with r = 2 becomes: 2 6 4 m r t y r t 2 p t R m,t R b,t 3 7 = 5 2 6 4 +µ + ε t 0.26 0 0 0.12 0 0 0 0 0 0 3 f(m r 0.9y r ) 5.7 p 19.3(R m 7 5 fy r + 17.5R m g t 1 Can we reproduce any of the other rows with r = 3? Katarina Juselius () Chapter 5: The Cointegrated VAR model July 1, 2012 4 / 41
Decomposing the Pi with r=3 The coe cients in third row of Π shows that only in ation rate seems signi cant suggesting that in ation rate is stationary by itself. Setting α 33 = 0.77 and β 0 3 = [0, 0, 1, 0, 0] we an roughly reproduce the third row. However, p t I (0) implies that the money demand relation can be speci ed as the sum of two stationary components: 0.26f(m r 0.9y r ) 5.7 p 19.3(R m R b )g t 1 = 0.26f(m r 0.9y r ) 19.3(R m R b )g t 1 1.5 p t 1 The CVAR model for r = 3 can be represented as: 2 m r 3 2 3 t 0.26 0 1.5 yt r 2 6 2 p t 7 = 0 0.12 0 fm r 0.9y r 19.3(R m R b ) 6 0 0 0.77 4 7 fy r + 17.5R m g t 1 4 R m,t 5 4 0 0 0 5 p t 1 R b,t 0 0 0 Katarina Juselius () Chapter 5: The Cointegrated VAR model July 1, 2012 5 / 41
A theory consistent scenario for the Pi matrix The choice of α and β, in addition to reproducing the Π matrix, should ideally describe an interpretable economic structure. The above tentatively identi ed cointegration relations are quite di erent from the monetary theory consistent cointegration relations in Chapter 2. Here we add a hypothetical adjustment structure of α coe cients that are roughly consistent with the theory of monetary in ation. 2 6 4 m r t y r t 2 p t R m R b,t 3 2 7 = 6 5 4 a 11 a 12 a 13 0 a 22 a 23 a 31 0 a 33 a 41 0 0 0 a 52 0 3 2 4 7 5 (m p y r ) t 1 (R b R m ) t 1 ( p R m ) t 1 2 3 5 + 6 4 µ 1 µ 2 µ 3 µ 4 µ 5 3 2 7 + 6 5 4 A row in the Π matrix is often a linear combination of several cointegration relations. Hence, a structure of interpretable long-run relations may not always be easy to uncover by exclusively inspecting the Π matrix, but it often Katarina Juselius helps. () Chapter 5: The Cointegrated VAR model July 1, 2012 6 / 41
Figure: A cross plot of real money stock, m r t, and real aggregate income, y r t and the regression line. Katarina Juselius () Chapter 5: The Cointegrated VAR model July 1, 2012 7 / 41 The pulling and pushing forces Consider the bivariate system with xt 0 = [mt r,yt r ], where mt r is real money stock and yt r is real income and a crossplot of the two variables. 6.2 Lrm LYr 6.1 6.0 5.9 5.8 5.7 5.6 5.5 6.55 6.60 6.65 6.70 6.75 6.80 6.85 6.90 6.95 7.00
Interpretation Assume that the equilibrium position corresponds to a constant money velocity m r y r = β 0, so that β 0 = [1, 1] and the attractor set is β? = [1, 1] indicated by the 45 0 line. If β 0 x t = (m r t y r t ) β 0 6= 0, then the adjustment coe cient, α, will force the process back towards the attractor set with a speed of adjustment that depends on the length of α and the size of the equilibrium error β 0 x t β 0. The (long-run) equilibrium position β 0 x t = β 0 describes a system at rest and hence no economic force (incentive) to move the system to a new position. When new (exogenous) shocks hit the system, causing β 0 x t β 0 6= 0, the adjustment forces are activated and pull the process back towards the long-run equilibrium point. The common trend, measured by α 0? Σt i=1 ε i, has pushed money and income along the line de ned by β?, the attractor set. Thus, positive shocks to the system will push the process higher up along β?, whereas negative shocks will move it down. Katarina Juselius () Chapter 5: The Cointegrated VAR model July 1, 2012 8 / 41
The geometry of the pulling and pushing forces m t 6 @ @@ β 0 x t @R α sp(β? ) α 0? t i=1 ε i Katarina Juselius () Chapter 5: The Cointegrated VAR model July 1, 2012 9 / 41 - y t
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