Macroeconometric modelling

Transcription

1 Macroeconometric modelling 2 Background Gunnar Bårdsen CREATES November 2009

2 Models with steady state We are interested in models with a steady state They need not be long-run growth models, but they need to be stable so lim t f (Z t) = f ( Z ), where Z is a vector of suitably transformed stationary variables. Otherwise we cannot do policy analysis. This is Samuelsons (1941, 1942) correspondence principle see Evans and Honkapohja (2007) for a recent reappraisal. To motivate, some classical and simple model examples follow.

3 Keynesian multiplier model Foundation of policy and disequilibrium macro No growth, discrete time: with constant steady state Y t = C t + Ī C t = c 0 + c 1 Y t 1 Ȳ = c Ī. 1 c 1 1 c 1 if c 1 < 1. The model is cast in Equilibrium Correction form (EqCM) as: Y t Y t 1 = c 0 (1 c 1 ) Y t 1 + Ī = (1 c 1 ) Y c 0 t Ī 1 c 1 1 c }{{ 1 } Ȳ Y t = (1 c 1 ) ( Y t 1 Ȳ ).

4 Solow-Swan growth model I Foundation of growth literature In discrete time, with population growth and technological progress, see f. ex. Blanchard and Fischer (1989): A t A t 1 N t N t 1 = a = n } Y t = C t + I t (1) Y t = F (A t N t, K t 1 ) (2) C t = (1 s) Y t (3) I t = K t K t 1 + δk t 1 (4) (AN) t (AN) t 1 = (1 + a) (1 + n) 1 = g (5) If the production function satisfies the Inada conditions, the stability condition (1 δ) 1+g < 1 of s f ( Kt 1 ) = (1 + g) (AN) t ( Kt (AN) t+1 ) ( ) Kt 1 (1 δ) (AN) t (6)

5 Solow-Swan growth model II Foundation of growth literature gives per capita effi ciency level of capital of lim t K t 1 = g + δ (AN) t s.

6 Ramsey growth model in continous time Foundation of the RBC literature 1. order Taylor expansion of model in per capita terms, no technological progress or depreciation, see Blanchard and Fischer (1989): ċ (t) = β (k k ) k (t) = (c c ) + θ (k k ) The stability conditon is now θ (4β+θ 2 ) 2 < 0.

7 Phillips (1954, 1957) proportional control model Foundation of policy control literature Continous time with no growth, see Turnovsky (1977, p ): ( ) ( ) ( ) Ẏ (t) α (c 1) α Y Y = Ġ (t) βγ β G G The stability conditions are α (c 1) β < 0 αβ (γ + 1 c) > 0,

8 New Keynesian Model with Taylor rule Presently Discrete time, no growth, see Wickens (2008, p ): p t = µ + βe t p t+1 + γx t + e πt x t = E t x t+1 α (R t E t p t+1 θ) + e xt R t = θ + π + µ ( p t p ) + νx t + e Rt where p is inflation, x is the output gap and R is the interest β rate and the e-s are shocks. If 1 > 1+α(ν+µγ) > 0, the solutions for inflation and output gap are p t = p + (1 + αν) e πt + γ (e xt αe Rt ) 1 + α (ν + µγ) x t = αµe πt e xt + αe Rt 1 + α (ν + µγ) so on average inflation p is equal to the target p and the output gap x is zero.

9 Vector Autoregressive Representations (VARs) I We follow the exposition in Lütkepohl (2005) see there for further details We are interested in analyzing the k-dimensional VAR(p) process y t y t = p A i y t i + u t, u t (0,Σ u ) i=1 The system is stable if Π = det (I k A 1 z A p z p ) 0 for z 1.

10 Structural VARs (SVARs) I The system has a so-called Wold Moving Average representation (intuitively, "solving" by recursive substitution): y t = Φ i u t i, Φ s = i=0 s Φ s j A j, s = 1, 2,..., Φ o = I k j=1 The elements of Φ j are the responses of the variables to changes in the errors. Since the errors most likely are correlated, the responses will not reflect shocks to the variables. Solution: make errors uncorrelated: innovations the new response functions should now reflect unique impulses to the variables: impulse response functions. Example: the "A-model"

11 Structural VARs (SVARs) II 1. Assume the underlying identified model A 0y t = A 0 p A i y t i + A 0u t. i=1 A 0u t = ε t (0,Σ ε), Σ ε = A 0Σ ua 0 where Σ ε = A 0Σ ua 0 is the diagonal covariance matrix of the model errors the innovations. 2. The MA-representation is now y t = i=0 Φ i A 1 0 A 0u t i = Θ i ε t i i=0 3. The elements of Θ i are the unique impulse responses. We will return to this under the heading of dynamic multipliers. The fundamental paper on SVARs with common trends and cointegration is King, Plosser, Stock, and Watson (1991).

12 Structural VARs (SVARs) III See Anders Warne s lecture notes at for a brilliant introduction to this literature.

13 VAR representations of DSGE models I VAR solution This follows Fernandez-Villaverde, Rubio-Ramirez, Sargent, and Watson (2007). In compact notation the (log-linearised) equilibrium conditions of a large class of models, including DSGE models, can be written FE t ξ t+1 + Gξ t + Hξ t 1 = Ju t, (7) where ξ t is a vector of state variables and u t is a vector of uncorrelated white noise shocks (e.g., shocks to technology and preferences) and the elements in the coeffi cient matrices are non-linear functions of the underlying structural parameters in the DSGE model. If the Blanchard-Kahn conditions (see Blanchard and Kahn (1980)) are satisfied, the model has a unique stable solution ξ t = Aξ t 1 + Bu t.

14 VAR representations of DSGE models II VAR solution Adding a set of measurement equations relating the elements of ξ t to a vector of observable variables y t gives the state-space representation ξ t = Aξ t 1 + Bu t (8) y t = Cξ t 1 + Du t. (9) If D is non-singular we get from (9) u t = D 1 (y t Cξ t 1 ), which substituted into (8) and rearranging gives ξ t = Aξ t 1 + BD 1 (y t Cξ t 1 ) ( I ( A BD 1 C ) L ) ξ t = BD 1 y t = ( A BD 1 C ) ξ t 1 + BD 1 y t

15 VAR representations of DSGE models III VAR solution The solution can be approximated by a vector autoregression (VAR) (or a vector equilibrium correction model (VEqCM)) if the moving average (MA) representation is invertible (that is, it must be possible to recover the shocks u t from the current and lagged values of the observables (see e.g., Watson (1994)). If the number of observables equals the number of shocks and D 1 exists, a necessary and suffi cient condition for invertibility is that the eigenvalues of A BD 1 C are strictly less than one in modulus (see Fernandez-Villaverde, Rubio-Ramirez, Sargent, and Watson (2007)).

16 VAR representations of DSGE models IV VAR solution If this condition is satisfied, then ξ t = ( A BD 1 C ) j BD 1 y t j. j=0 Shifting back one period and substituting into (9) then gives the VAR representation y t = C ( A BD 1 C ) j BD 1 y t j 1 + Du t. (10) j=0 If all the endogenous state variables are observable and included in the VAR, the VAR representation is of finite order (e.g., Ravenna (2007)). In general, however, the VAR is of infinite order (corresponding to a VARMA representation).

17 An example: the New Keynesian Phillips Curve The model is p t = b f p1e t p t+1 + b b p1 p t 1 + b p2 x t + ε pt (11) x t = b x x t 1 + ε xt (12) where all coeffi cients are assumed to be between zero and one. The solution (the statistical system) is ( ) p x ) ( up u x t t = = ( α1 b p2 b x b f p1 (α 2 b x ) 0 b x ( 1 b f p1 α 2 b p2 b f p1 (α 2 b x ) 0 1 ) ( ) p x ) ( ) εp ε x t 1 t ( up + u x ) t, Derivation of the solution

18 Hard to find I Standard dynamic price-wage model ( ) ( ) ( 1 γ2 p 1 γ1 = δ 2 1 x δ t 1 1 ( γ δ 3 ) ( p x ) ( gap u ) ) t 1 ( ep t + e x ) t with statistical system (reduced form) ( ) ( ) p γ 2 δ 1 +1 γ = 2 δ 2 1 γ 1+γ 2 ( ) γ 2 δ 2 1 p x t δ 1+δ 2 γ 2 δ 2 1 γ 1δ 2 +1 x γ 2 δ 2 1 ( ) ( ) vp γ 3 δ γ = 2 δ 2 1 γ 3 ( 2 γ 2 δ 2 1 gap v δ x γ 2 3 γ 2 δ 2 1 δ 3 u γ 2 δ 2 1 t t 1 ( vp + ) t + v x ) t, ( e p+γ 2 e x γ 2 δ 2 1 ex +δ 2e p γ 2 δ 2 1 ) t

19 Hard to find II could be observationally equivalent same with conditional models. See Bårdsen, Jansen, and Nymoen (2004) for testing with conditional models.

20 Estimation and identification I Errors in variables Remember that the model is p t = b f p1e t p t+1 + b b p1 p t 1 + b p2 x t + ε pt which can be rewritten as π = γe t π t+1 + δx t + v pt. where π t = p t α 1 p t 1 and α 1 is the backward stable root of the solution. The model is often estimated by means of instrumental variables, using the errors in variables method (evm) where expected values are replaced by actual values and the expectational errors: π t = γπ t+1 + δx t + v pt γη t+1. (13)

21 Estimation and identification II Errors in variables The implications of estimating the model by means of the errors in variables method is to induce moving average errors. Following Blake (1991), this can be readily seen using the expectational errors as follows.

22 Estimation and identification III Errors in variables 1. Lead (18) one period and subtract the expectation to find the RE error: η t+1 = γe t π t+2 + δx t+1 + v pt+1 E t π t+1 ( ) ( δbx δbx = γ x t+1 + δx t+1 + v pt+1 1 γb x 1 γb x ( ) δ = (x t+1 b x x t ) + v pt+1 1 γb x ( ) δ = ε xt+1 + v pt+1 1 γb x ) x t

23 Estimation and identification IV Errors in variables 2. Substitute into (13): ( ) γδ π t = γπ t+1 + δx t + v pt γv pt+1 ε xt+1. 1 γb x 3. Finally, reexpress in terms of original variables: ( ) ( ) 1 b p2 p t α 1 p t 1 = ( p t+1 α 1 p t ) + α 2 bp1 f α x t 2 ( ) ( ) ( ( ) ( ) ) 1 b p α2 bp1 + bp1 f α ε pt 2 α 2 bp1 f α ε pt+1 f α 2 ( ) ε xt+1. b α2 x

24 Estimation and identification V Errors in variables Giving p t = bp1 p f t+1 + bp1 p b t 1 + b p2 x t ( ) ( 1 bp2 + ε pt ε pt+1 α 2 α 2 b x ) ε xt+1, where we have exploited the two well-known relationships between the roots: α 1 + α 2 = 1 b f p1 α 1 α 2 = bb p1 bp1 f.

25 Estimation and identification VI Errors in variables So even though the original model has white noise errors, the estimated model will have first order moving average errors.

26 Does the MA(1) process prove that the forward solution applies? I Assume that the true model is p t = b p1 p t 1 + ε pt, b p1 < 1 and the the following model is estimated by means of instrumental variables p t = b f p1 p t+1 + ε f pt What are the proprties of ε f pt? ε f pt = p t b f p1 p t+1 Assume, as is common in the literature, that we find that b f p1 1. Then ε f pt p t p t+1 = 2 p t+1 = [ε pt+1 + (b p1 1)ε pt +...].

27 Does the MA(1) process prove that the forward solution applies? II So we get a model with a moving average residual, but this time the reason is not forward looking behaviour but misspecification.

28 References I Bårdsen, G., Ø. Eitrheim, E. S. Jansen, and R. Nymoen (2005). The Econometrics of Macroeconomic Modelling. Oxford: Oxford University Press. Bårdsen, G., E. S. Jansen, and R. Nymoen (2004). Econometric evaluation of the New Keynesian Phillips curve. Oxford Bulletin of Economics and Statistics 66(s1), Blake, D. (1991). The estimation of rational expectations models: A survey. Journal of Economic Studies 18(3), Blanchard, O. J. and S. Fischer (1989). Lectures on Macroeconomics. Cambridge, Massachusetts: The MIT Press.

29 References II Blanchard, O. J. and C. M. Kahn (1980). The solution of linear difference models under rational expectations. Econometrica 48(5), Chiang, A. C. (1984). Fundamental Methods of Mathematical Economics (3. ed.). McGraw-Hill. Davidson, J. (2000). Econometric Theory. Oxford: Blackwell. Dejong, D. N. and C. Dave (2007). Structural Macroeconometrics. Princeton University Press.

30 References III Evans, G. W. and S. Honkapohja (2007, 02). The e-correspondence principle. Economica 74(293), Fernandez-Villaverde, J., J. Rubio-Ramirez, T. Sargent, and M. Watson (2007). ABCs (and Ds) of understanding VARs. American Economic Review 97, King, R. G., C. I. Plosser, J. H. Stock, and M. W. Watson (1991). Stochastic trends and economic fluctuations. American Economic Review 81, Lütkepohl, H. (2005). New Introduction to Multiple Time Series Analysis. Springer.

31 References IV McCandless, G. (2008). The ABCs of RBCs. Harvard University Press. Pesaran, M. H. (1987). The Limits to Rational Expectations. Oxford: Basil Blackwell Inc. Phillips, A. W. (1954). Stabilization policy in a closed economy. Economic Journal 64, Phillips, A. W. (1957). Stabilization policy and the time form of lagged responses. Economic Journal 67,

32 References V Ravenna, F. (2007). Vector autoregressions and reduced form representations of DSGE models. Journal of Monetary Economics 54, Samuelson, P. A. (1941). The stability of equilibrium: Comparative statics and dynamics. Econometrica 9(2), Samuelson, P. A. (1942). The stability of equilibrium: Linear and nonlinear systems. Econometrica 10(1), Turnovsky, S. J. (1977). Macroeconomic Analysis and Stabilization Policy. Cambridge: Cambridge University Press.

33 References VI Watson, M. W. (1994). Vector autoregressions and cointegration, Volume 4 of Handbook of Econometrics, Chapter 47, pp Amsterdam: Elsevier Science. Wickens, M. (2008). Macroeconomic Theory. Princeton University Press.

34 Appendix: Solution of the New Keynesian Phillips Curve Method of repeated substitution All of the Rational expectations techniques rely on the law of iterated expectations, saying E t E t+k x t+j = E t x t+j, that your average revisions of expectations given more information will be zero. The method of repeated substitution is the brute force solution, cumbersome and not very general. But since it s also instructive to see exactly what goes on, we use with this method. See Bårdsen, Eitrheim, Jansen, and Nymoen (2005, Appendix A1) for alternative analytical methods and McCandless (2008) and Dejong and Dave (2007) for general methods based on matrix decompositions.

35 We start by using a trick to get rid of the lagged dependent variable, following Pesaran (1987, p ), by defining p t = π t + α p t 1 (14) where α will turn out to be the backward stable root of the process of p t. We take expectations one period ahead E t p t+1 = E t π t+1 + αe t p t E t p t+1 = E t π t+1 + απ t + α 2 p t 1.

36 Next, we substitute for E t p t+1 into original model: π t + α p t 1 = bp1 f ( Et π t+1 + απ t + αt 1 p 2 ) t 1 +bp1 p b t 1 + b p2 x t + ε pt ( ) ( ) b f p1 b f p1 α 2 α + bp1 b π t = 1 bp1 f α E t π t bp1 f ( ) ( ) α b p bp1 f α x t + 1 bp1 f α ε t. The parameter α is defined by b f p1α 2 α + b b p1 = 0 p t 1 or α 2 1 bp1 f α + bb p1 bp1 f = 0 (15)

37 with the solutions α 1 α 2 } = 1 ± 1 4bp1 f bb p1. (16) 2b f p1 The model will typically have a saddle point behaviour with one root bigger than one and one smaller than one in absolute value. In the following we will use the backward stable solution, defined by: 1 1 4b α p1 f bb p1 1 = < 1. 2b f p1

38 In passing might be noted that the restriction b b p1 = 1 bf p1 often imposed in the literature implies the roots α 1 = 1 bf p1 b f p1 α 2 = 1, as given in (16) as before. We choose α 1 < 1 in the following. So we now have a pure forward-looking model ( ) ( ) ( ) b f p1 b p2 1 π t = 1 bp1 f α E t π t bp1 f α x t bp1 f α ε pt. 1

39 Finally, using the relationship α 1 + α 2 = 1 b f p1 between the roots, see f.ex. Chiang (1984, p. 506), so: 1 b f p1α 1 = b f p1α 2, (17) the model becomes ( ) ( ) ( ) 1 b p2 1 π t = E t π t+1 + α 2 bp1 f α x t + 2 bp1 f α ε pt (18) 2 π = γe t π t+1 + δx t + v pt (19)

40 Following Davidson (2000, p ), we now derive the solution in two steps: 1. Find E t π t+1 2. Solve for π t. Find E t π t+1 : Define the expectational errors as: η t+1 = π t+1 E t π t+1. (20) We start by reducing the model to a single equation: π t = γπ t+1 + δb x x t 1 + δε xt + v pt γη t+1.

41 Solving forwards then produces: π t = γ (γπ t+2 + δb x x t + δε xt+1 + v pt+1 γη t+2 ) + δb x x t 1 + δε xt + v pt γη t+1 = (δb x x t 1 + δε xt + v pt γη t+1 ) + γ (δb x x t + δε xt+1 + v pt+1 γη t+2 ) + (γ) 2 π t+2 n = (γ) j (δb x x t+j 1 + δε xt+j + v pt+j γη t+j+1 ) + (γ) n+1 π t+n+1. j=0 By imposing the transversality condition: lim n (γ)n+1 π t+n+1 = 0

42 and then taking expectations conditional at time t, we get the discounted solution : E t π t+1 = = (γ) j (δb x E t x t+j + δe t ε xt+j+1 + E t v pt+j+1 γe t η t+j+2 ) j=0 (γ) j (δb x E t x t+j ). j=0 However, we know the process for the forcing variable, so: E t 1 x t = b x x t 1 E t x t = x t E t x t+1 = b x x t E t x t+2 = E t (E t+1 x t+2 ) = E t b x x t+1 = bx 2 x t E t x t+j = bx j x t.

43 We can therefore substitute in: E t π t+1 = (γ) j ( δb x bx j ) x t j=0 = δb x (γb x ) j x t j=0 ( δbx = 1 γb x ) x t. and substitute back the expectation into the original equation: π t = γe t π t+1 + δx t + v pt ( ) δbx = γ x t + δx t + v pt. 1 γb x

44 Finally, using (14) and (19) we get the complete solution: ( ) ( ) b p2 b f p1 b p t α 1 p t 1 = p1 f α b x 2 bp1 f α ( ) 2 bp1 1 f x t bp1 f α b x 2 ( ) ( ) b p2 1 + bp1 f α x t + 2 bp1 f α ε pt (21) 2 ( ) ( ) ( ) ( ) 1 b p2 b x b p2 1 = α 2 bp1 f (α x t + 2 b x ) bp1 f α x t + 2 bp1 f α ε pt 2 ( ) ( ) b p2 1 p t = α 1 p t 1 + bp1 f (α x t + 2 b x ) bp1 f α ε pt (22) 2 Return