Dynamic Macroeconomic Modeling with Matlab

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2 An Outline of the Theory are equivalent. 2. There exists locally smooth manifolds W u (x ∗ ) and W s (x ∗ ), called a local unstable manifold and a local stable manifold, respectively, tangent to the linear spaces N u and N s , respectively. 3. W s (x ∗ ) is characterized by ||φ(y)−φ(x ∗ )|| → 0 exponentially as t → ∞ for any y ǫW s (x ∗ ), and W u (x ∗ ) is characterized by ||φ(y) − φ(x ∗ )|| → 0 exponentially as t → −∞ for any y ǫW u (x ∗ ). For a well-defined solution the number of initial boundary conditions has to equal the di- mension of the stable manifold and the number of eigenvalues with negative real part of the Jacobian matrix evaluated at the steady state. What is a numerical solution of the problem? A solution is a time vector T = {t0,t1,...,tM} with t0 = 0 and tM = ∞, and a corresponding solution vector (x,y), such that (x(ti),y(ti)) denote the solution at time ti. 2.4 The Relaxation Algorithm “The relaxation method determines the solution by starting with a guess and improving it iteratively. As the iteration improves, the result is said to relax to the true solution.” (Press et al. p. 765) Relaxation can be seen in Figure 4. An initial (time-dependent) guess is made, which is represented in the (k,c) phase diagram. In every iteration, the initial guess improves until it is sufficiently close to the true solution. The principle of relaxation is to construct a large set of non-linear equations, whose solution represents the desired trajectory. This can be achieved by a discretisation of the differential equations on a mesh of points in time. This set of equations is augmented by additional algebraic equations representing equilibrium conditions or (static) no-arbitrage conditions at each mesh point. Finally, equations representing the initial and final boundary conditions are appended. The set of equations as a whole is solved simultaneously. Constructing the set of equations We discretize the differential equation according to ˙x = f(x,t) ⇒ ∆x ∆t = xk+1 − xk xk+1 + xk ≈ f , tk+1 − tk 2 tk+1 + tk . (18) 2
2 An Outline of the Theory c 0.68 0.66 0.64 0.62 0.6 0.58 0.56 saddle path initial guess saddle point 0.54 0 1 2 3 4 5 6 7 8 Figure 4: Relaxation in the Ramsey model These are (M − 1) · nd equations. The set of equations is augmented by ni initial boundary conditions and nf final boundary conditions, such that the set of equations expands to M · nd. Finally, algebraic equations are required to be fulfilled at every mesh point. Hence, appending M · na equations gives M · (nd + na) = M · n equations in total. We derived a square system of non-linear equations. Example 5 (Ramsey model) We construct a mesh of 3 points T = {0, 50, 100} and solve for c0,c50,c100,k0,k50,k100 numerically. For simplicity, we write the differential equations as k ˙k = f(c,k) ˙c = g(c,k) Then, the difference equations take the form k50 − k0 c50 − c0 = f , 50 2 k50 − k0 2
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