The Ramsey model - linearization, backward integration, relaxation (complete)

Transcription

1 The Ramsey model - linearization, backward integration, relaxation (complete) (Quantitative Dynamic Macroeconomics, Lecture Notes, Thomas Steger, University of Leipzig) Ramsey Model: linearized version 1. Model setup, first-order conditions, and dynamic system à 1.1. Model setup Consider a very simple neoclassical economy. There is no technical change and no population growth. The production function for final output is Y = AK a L 1-a 0 <a<1 (1) The utility function of the representative individual is of the CIES type u C = C1-s s with C, s>0 (2) The economy's resource constraint is given by Y -dk = K + C or K = Y -dk - C (3) à 1.2. The social planner's problem The social planner's dynamic problem reads C 1-s - 1 max C 0 1 -s -rt t s.t. K = AK a L 1-a -dk - C K 0 = K 0 (4) à 1.3. First-order conditions and dynamic system The current-value Hamiltonian of the above-stated problem reads H = C1-s s +l AKa L 1-a -dk - C (5) and the necessary first-order conditions are given by H C = C -s -l=0 (6)

2 2 Ramsey_complete.nb l =-H K +rl=- A a K a-1 L 1-a -d l+rl (7) à 1.4. The Keynes-Ramsey rule of optimal consumption C -s =l C -sc -s-1 =l C -sc -s-1 = l C -s l -C s C = l l Using the second first-order condition yields - s C C =- A a K a-1 L 1-a -d +r C = C s A a Ka-1 L 1-a -d-r (8) (9) (10) (11) (12) (13) à 1.5. The complete dynamic system The complete dynamic system can be expressed as C = C s A a Ka-1 L 1-a -d-r K = AK a L 1-a -dk - C K 0 = K 0 ; C = C è (14) (15) (16) where C è is the steady state level of consumption (see below). 2. Some conceptual remarks Consider the following (non-linear) system of differential equations (think of this system as the reduced form of the Ramsey model) C = g C, K K = f C, K K 0 = K 0 ; C = C è Linearizing this system around the steady state C = C è and K = K è by means of a first-order Taylor approximation gives (border conditions suppressed) C = è g C è, K è g C, K + C - C è g C, K + K - K è C K (17) (18) (19) (20)

3 Ramsey_complete.nb 3 K = è f C è, K è f C, K + C - C è f C, K + K - K è C K (21) Notice that the first terms on the RHS are zero by definition g C è, K è = 0 and f C è, K è = 0 remember that the steady state is defined by C = 0 and K = 0. Moreover, since C = C - C è and K = K - K è we can write C - C è K - K è = è g C, K C - C è g C, K + K - K è C K = è f C, K C - C è f C, K + K - K è C K (22) (23) This is a system of linear and homogenous differential equations in the variables C - C è and K - K è. The coefficient matrix J = g C,K C f C,K C g C,K K f C,K K (24) is denoted as Jacobian matrix. We know that the solution to such a system may be expressed as C - C è = è B 1 v 11 e l 1 t + B 2 v 12 e l 2 t K - K è = è B 1 v 21 e l 1 t + B 2 v 22 e l 2 t (25) (26) where B 1 and B 2 are "arbitrary constants of integration" (determined by initial conditions), l 1, l 2 the eigenvalues resulting from J - I l = 0 (with J denoting the Jacobian matrix of the differential equation system and I the identity matrix), and v 11 v 12, v 21 v 22 are the eigenvectors associated with the eigenvalues l 1, l 2 (see, for instance, Chiang, 2004). 3. The dynamic system (linearized form) The reduced form of the Ramsey model is restated for convenience C = C AL1-a a K a-1 -d-r s K = AK a L 1-a - C -dk K 0 = K 0 ; C = C è (27) (28) (29) Linearizing the above dynamic system around the steady state (K è, C è ) by means of a first-order Taylor approximation gives C - C è K - K è = è a A K a-1 L 1-a -d-r s C - C è + C a a -1 A K a-2 L 1-a s = è -1 C - C è + A a K a-1 L 1-a -d K - K è K - K è (30) (31) Evaluating the coefficients of this dynamic system at C = C è and K = K è gives a li near DES with constant coefficients

4 4 Ramsey_complete.nb C - C è K - K è = è a C - C è + b K - K è = è c C - C è + d K - K è (32) (33) where a = a AKè a-1 L 1-a -d-r s, b = Cè a a-1 AK è a-2 L 1-a, c =-1, and d = A a K è a-1 L 1-a -d. s 4. Numerical considerations Set of parameters (notice the implied shock: a once and for all increase in TFP by 30 percent) paraminitial A 1, L 1, 0.35, 0.01, 0.02, 2 ; paramfinal A 1.3, L 1, 0.35, 0.01, 0.02, 2 ; Steady state CSS, KSS A AL 1 1 L 1 AL1 1 1, AL ; CSS, KSS. paraminitial CSS, KSS. paramfinal , , Jacobian matrix, eigenvalues and eigenvectors The Jacobian matrix of the above DES jacobian D C AK 1 L 1, AK L 1 K C, C, K TraditionalForm AK a-1 a L 1-a -d-r ACK a-2 L 1-a a-1 a s s -1 AK a-1 L 1-a a-d Next we evaluate the Jacobian matrix at the steady state jacobianss jacobian. C CSS, K KSS ; Here we use Mathematica to compute the eigenvalues and eigenvectors

5 Ramsey_complete.nb 5 eigen Eigensystem jacobianss. paramfinal ; A nice summary of the result can be produced as follows TableForm eigen, TableHeadings "Eigenvalues", "Eigenvectors", Automatic, C, K, TableSpacing 4, 3, Eigenvalues Eigenvectors C K After having determined the eigenvalues and eigenvectors, the solution to the linearized system can be expressed as follows solutionlinearized B1 eigen 2, 1 Exp eigen 1, 1t B2 eigen 2, 2 Exp eigen 1, 2t CSS, KSS ColumnForm B t B t AL B t B t AL AL 1 AL1 The final step consists in determining B 1 and B 2, which is what we consider next. 6. Determining arbitrary constants of integration To determine the two "arbitrary constants of integration" B 1 and B 2, let us consider the solution to the linearized system evaluated at t = 0 (notice that e 0 1) C 0 = B 1 v 11 + B 2 v 12 + C è K 0 = B 1 v 21 + B 2 v 22 + K è (34) (35) Remember that B 1 is associated with the unstable root (l 1 > 0) and B 2 is associated with the stable root (l 2 < 0). The impact of the unstable eigenvalue on the dynamic evolution can only be eliminated by setting C 0 such that B 1 = 0! Technically this means we set B 1 = 0 and solve the above system for C 0 and B 2. sys solutionlinearized C0, K0. t 0, K0 KSS. paraminitial, B1 0 ;

6 6 Ramsey_complete.nb solintialconditions Solve sys, C0, B2 ; solcomplete solutionlinearized. solintialconditions; 7. Plotting time paths Plotting the result C t and K t gives pc Plot solcomplete 1, 1. Append paramfinal, B1 0, CSS. paraminitial, CSS. paramfinal, t, 0, 300, AxesLabel t, C, PlotRange 3, 5 C t

7 Ramsey_complete.nb 7 pk Plot solcomplete 1, 2. Append paramfinal, B1 0, KSS. paraminitial, KSS. paramfinal, t, 0, 300, AxesLabel t, K, PlotRange 35, 70 K t COMMENTS [1] The (asymptotic) ROC is given by the absolute value of the negative eigenvalue. [2] A more intuitive measure for the speed of convergence is the half life which can be determined as follows Solve Abs eigen 1, 2 Log 0.5 t, t t The saving rate At first, we define the instantaneous saving rate and the steady state saving rate savingt 1 solcomplete 1, 1 A solcomplete 1, 2 ; savingss 1 CSS A KSS ; Notice that the steady state saving rate is independent of A

8 8 Ramsey_complete.nb savingss PowerExpand FullSimplify 1 L 1 1 The saving rate is then plotted Plot savingt. Append paramfinal, B1 0, savingss. paraminitial, t, 0, 200, AxesLabel t, "saving rate", PlotStyle Thickness 0.01, PlotRange 0, 0.25 saving rate t Ramsey model: backward integration 1. Simulation (numerical analysis) à 1.1. Steady state and phase diagram At first, one needs to determine the steady state (k è, c è ) which is defined by k = 0 and c = 0. Off General::spell ; Off General::spell1 ;

9 Ramsey_complete.nb 9 sol Solve Ak 1 0, A k c k 0, c, k FullSimplify c A 1 1 A A 1 1,k A 1 1 For future uses, let us label the stationary solution as csol and ksol csol, ksol A 1 1 A A 1 1, A 1 1 ; à 1.2. Simulation of the adjustment process At first, we define the differential equation system to be solved numerically in Mathemtica Syntax. Notice that [1] the time index must appear explicitly and [2] we multiply the RHS of each differential equation by -1 since the system is solved backwards in time (backward integration). Clear dec, dek ; dec c' t c t A k t 1 ; dek k' t A k t c t k t ; The set of parameters must be specified numerically Clear,,, ; 0.3; 1; 0.04; 0.02; A 1; Now the NDSolve command can be applied to the dynamic system under study. Notice that we start nearby the steady state. Clear nsba, cb, kb, cf, kf ; nsba NDSolve dec, dek, c 0 csol ,k 0 ksol , c t, k t, t, 0, 255 NDSolve::ndsz : At t `, step size is effectively zero; singularity or stiff system suspected. c t InterpolatingFunction0., , t, k t InterpolatingFunction0., , t The numerical solutions are given as interpolating functions. In the next step we have to extract and evaluate these interpolating functions to be able to visualize the result.

10 10 Ramsey_complete.nb cb t : Evaluate nsba 1, 1, 2 kb t : Evaluate nsba 1, 2, 2 Now since we solved the system backwards in time but would like to see the solution in forward-looking time we have to reverse the direction of the time index. Clear tende ; tende 164; cf t : cb tende t kf t : kb tende t To receive a first impression, we plot the solution trajectory in k, c -space tra ParametricPlot kf t, cf t, t, 0, tende, AxesLabel k, c, PlotStyle Thickness 0.01, AspectRatio 1 2 c k Here we combine the plots of the isoclines together with the trajectory in k, c -space

11 Ramsey_complete.nb 11 pkdot0 Plot A k k, k, 0, 15, AxesLabel k, c ; cdot0 Table ksol, i, i, 0, 2, ; pcdot0 ListPlot cdot0, AxesLabel k, c ; phase Show pcdot0, pkdot0 2.0 c k Combine both plots to receive a kind of phase diagram Show tra, phase, PlotRange All, AxesOrigin 0, c k à 1.3. Ploting time paths First, we plot the time path of capital and consumption

12 12 Ramsey_complete.nb Plot kf t, t, 0, tende, AxesLabel t, "k t ", PlotStyle Thickness 0.01 Plot cf t, t, 0, tende, AxesLabel t, "c t ", PlotStyle Thickness 0.01 k t t c t t Next consider the saving rate and the growth rate of output. Here we need at first some definitions Clear s ; s t : Akf t cf t Akf t ;

13 Ramsey_complete.nb 13 Plot s t, s tende, t, 0, 150, AxesLabel t, "s t ", PlotRange 0, 1, PlotStyle Thickness 0.01 s t t 2. Empirical implications [1] The growth rate of output decreases implying b-convergence. [2] The fact that the saving rate decreases with income is clearly at odds with empirical regularities. Notice that this depends crucially on the numerical value for s. Ramsey Model: relaxation 1. Relaxation: a sketch à The Relaxation Algorithm: a sketch "The relaxation method determines the solution by starting with a guess and improving it, iteratively. As the iteration improves the solution, the result is said to relax to the true solution." (Press et al., p. 765)

14 14 Ramsey_complete.nb à Starting point: a general ODE Consider the following autonomous, ordinary differential equation system (ODE), expressed in vector notation x t = f x t with f : N Ø N B x 0 = 0 with B : N Ø n 1 F x = 0 with F : N Ø n 2 ; n 1 + n 2 = N (1) (2) (3) A solution is a function x = x t which maps each tœ into a unique x N such that (10), (11), and (12) are satisfied. Typical solution procedures [1] Multiple shooting; [2] backward integration; [3] time elimination. à Discretization: corresponding algebraic problem The ODE x = f x can then be approximated by a "finite difference equation" (FDE) x v+1 - x v = t v+1 - t v f x v for all v œ 1, 2,..., M (4) where x n denotes the value of x at t v for v œ 1, 2,..., M, t v œ T = t 1, t 2,..., t M (the time mesh, equal to set of natural numbers in the simplest case), and x v = x v + x v+1 2. A solution of this equation (plus boundary conditions) consists of a set of points x v with v œ 1, 2,..., M in the state space of dimension N (approximating the solution trajectory). Notice: The initial problem of solving an ODE has been transformed into a problem of solving a set of N M N = NM algebraic equations (the above set of equations plus the Ò of finite difference equation Ò of boundary conditions boundary conditions)!

15 Ramsey_complete.nb 15 An illustrative example Consider the Ramsey model (N=2) with three mesh points (M=3) x = c, k ; c = g c, k ; k = f c, k ; v œ 1, 2, 3 (5) Between the two mesh points v = 2 and v = 1 we have c 2 - c 1 = 2-1 g c 1 + c 2 k 2 - k 1 = 2-1 f 2 c 1 + c 2 2, k 1 + k 2 2, k 1 + k 2 2 Between the two mesh points v = 3 and v = 2 we have c 3 - c 2 = 3-2 g c 2 + c 3 k 3 - k 2 = 3-2 f 2 c 2 + c 3 2, k 3 + k 2 2, k 3 + k 2 2 Boundary conditions can be stated as: [1] k 1 k initial and [2] c 3 c final. (6) (7) (8) (9) This represents an algebraic system comprising 6 equations in 6 unknowns. In what follows, we solve such a system for {(c 1, k 1, (c 2, k 2, (c 3, k 3 } using the FindRoot command. 2. Preliminaries nx 2 number of dynamic variables: state and costate variables ; ny 1 number of static variables: control variables ; The model is specified in the form of the (current-value) Hamiltonian of the social planner problem ham Y 1, j X 2, j AX 1, j L 1 X 1, j Y 1, j ; dynequstate D ham, X 2, j; dynequcostate D ham, X 1, j X 2, j ; dynequ Join dynequstate, dynequcostate ; statequcontrol D ham, Y 1, j; Set of parameters (shock specification) Initial steady state paraminitial A 1, L 1, 0.35, 2, 0.01, 0.02, 0.5 ; paramfinal A 1.3, L 1, 0.35, 2, 0.01, 0.02, 0.5 ;

16 16 Ramsey_complete.nb XX 1; YY 1; j 0; vars Join Table X i, j, XX, i, 1, nx 2, Table X i, j, 1 XX, i, nx 2 1, nx, Table Y i, j, YY, i, 1, ny; sysinitial Join dynequstate, dynequcostate, statequcontrol. paraminitial; solinitial FindRoot sysinitial, vars sysinitial. solinitial Clear j ; X 1, , X 2, , Y 1, , , Final steady state XX 1; YY 1; j n; vars Join Table X i, j, XX, i, 1, nx 2, Table X i, j, 1 XX, i, nx 2 1, nx, Table Y i, j, YY, i, 1, ny; sysfinal Join dynequstate, dynequcostate, statequcontrol. paramfinal; solfinal FindRoot sysfinal, vars sysfinal. solfinal Clear j ; X 1, n , X 2, n , Y 1, n , ,

17 Ramsey_complete.nb Discretization: algebraic system n 300 ; number of mesh points subslist1 Join Table X i, j X i, j 1 X i, j 2, i, 1, nx, Y i, j 1 Y i, j Table Y i, j, i, 1, ny; 2 equmain Table Table X i, j X i, j 1, i, 1, nx dynequ. subslist1, j, 1, n ; discretization equstatic Table statequcontrol, j, 0, n ; equborder Join Table X i, 0 solinitial i, 2, i, 1, nx 2, Table X i, n solfinal i, 2, i, nx 2 1, nx; equations Join Flatten equmain, Flatten equstatic, equborder ; 4. Rootfinding start1 X i, n X i, 0 Join Table X i, j, X i, n n j, i, 1, nx. n Join solinitial, solfinal, Y i, n Y i, 0 Table Y i, j, Y i, n n j, i, 1, ny. n Join solinitial, solfinal ; startvalues Flatten Table start1, j, 0, n, 1 ; Timing sol2 FindRoot equations. paramfinal, startvalues ; Max equations.paramfinal.sol , Null

18 18 Ramsey_complete.nb 5. Plots p1 ListPlot Join Table solinitial 1, 2, z, 1, 0.1 n, Table sol2 i, 2, i, 1, nx ny n, nx ny, PlotRange 30, 75, AxesLabel t, K ; p2 ListPlot Join Table solinitial 2, 2, z, 1, 0.1 n, Table sol2 i, 2, i, 2, nx ny n, nx ny, PlotRange 0, 0.1, AxesLabel t, ; p3 ListPlot Join Table solinitial 3, 2, z, 1, 0.1 n, Table sol2 i, 2, i, 3, nx ny n, nx ny, PlotRange 2, 6, AxesLabel t, C ; GraphicsArrayp1, p2, p3, p K t t C 6 C t t à The saving rate The initial and final steady state saving rate

19 Ramsey_complete.nb 19 savinginitial A solinitial 1, 2 L 1 solinitial 3, 2. A solinitial 1, 2 L 1 paraminitial savingfinal A solfinal 1, 2 L 1 solfinal 3, 2. paramfinal A solfinal 1, 2 L The saving rate along the adjustment process outputadjustment Table A sol2 i, 2 L 1. paramfinal, i, 1, nx ny n, nx ny ; consumptionadjustment Table sol2 i, 2, i, 3, nx ny n, nx ny ; savingadjustment outputadjustment consumptionadjustment ; outputadjustment ListPlot Join Table savinginitial, z, 1, 0.1 n, savingadjustment, PlotRange 0, 0.25, AxesLabel t, "saving rate" saving rate t

20 20 Ramsey_complete.nb Project #4: economic development with subsistence needs Consider the following economy. There is no technical change and no population growth. The production function for final output (per capita) is y = k a 0 <a<1 The utility function of the representative individual is characterized by subsistence consumption (10) u c = c - c min 1-s s with c min, s>0 (11) The economy's resource constraint is given by y -dk = k + c or k = y -dk - c (12) Notation: y: per capita output; k: per capita capital; c: per capita consumption; c min : subsistence consumption needs (per capita); a: elasticity of y w.r.t. k; d: depreciation rate; s: preference parameter. à Exercises (1) Derive the Keynes-Ramsey rule (KRR) of optimal consumption. Provide an economic interpretation of this form of the KRR. (2) Simulate the transitional dynamics of the resulting dynamic system. Assume that the economy starts (in forward looking time) "marginally above k min ", where k min solves c min = k min a. The set of parameters should be employed 0.3; 1; 0.04; 0.02; cmin 0.5; (3) Plot the time paths of the saving rate and the growth rate of per capita income. Give an economic interpretation of the results.