An Introduction to Perturbation Methods in Macroeconomics. Jesús Fernández-Villaverde University of Pennsylvania

Transcription

1 An Introduction to Perturbation Methods in Macroeconomics Jesús Fernández-Villaverde University of Pennsylvania 1

2 Introduction Numerous problems in macroeconomics involve functional equations of the form: H (d) =0 Examples: Value Function, Euler Equations. Regular equations are particular examples of functional equations. How do we solve functional equations? 2

3 Two Main Approaches 1. Projection Methods: d n (x, θ) = nx i=0 θ i Ψ i (x) We pick a basis {Ψ i (x)} i=0 and project H ( ) against that basis. 2. Perturbation Methods: d n (x, θ) = nx i=0 θ i (x x 0 ) i We use implicit-function theorems to find coefficients θ i. 3

4 Intuition of Perturbation Methods Most complicated problems have particular cases that are easy to solve. Often, we can use the solution to the particular case as a building block of the general solution. Very successful in physics. Judd and Guu (1993) showed how to apply it to economic problems. 4

5 ASimpleExample Imaginewewanttofind the (possible more than one) roots of: such that x<0. x 3 4.1x +0.2 =0 This a tricky, cubic equation. How do we do it? 5

6 Main Idea Transform the problem rewriting it in terms of a small perturbation parameter. Solve the new problem for a particular choice of the perturbation parameter. Use the previous solution to approximate the solution of original the problem. 6

7 Step 1: Transform the Problem Write the problem into a perturbation problem indexed by a small parameter ε. This step is usually ambiguous since there are different ways to do so. A natural, and convenient, choice for our case is to rewrite the equation as: where ε 0.1. x 3 (4 + ε) x +2ε =0 7

8 Step 2: Solve the New Problem Index the solutions as a function of the perturbation parameter x = g (ε): g (ε) 3 (4 + ε) g (ε)+2ε =0 andassumeeachofthissolutionissmooth(thiscanbeshowntobe the case for our particular example). Note that ε = 0 is easy to solve: x 3 4x =0 that has roots g (0) = 2, 0, 2. g (0) = 2. Since we require x < 0, we take 8

9 Step 3: Build the Approximated Solution By Taylor s Theorem: x = g (ε) ε=0 = g (0) + X n=1 g n (0) n! ε n Substitute the solution into the problem and recover the coefficients g (0) and gn (0) n! for n =1,... in an iterative way. Let s do it! 9

10 Zeroth -Order Approximation We just take ε =0. Before we found that g (0) = 2. Is this a good approximation? x 3 4.1x +0.2 = =0.4 It depends! 10

11 First -Order Approximation Take the derivative of g (ε) 3 (4 + ε) g (ε)+2ε =0withrespectto ε: 3g (ε) 2 g 0 (ε) g (ε) (4 + ε) g 0 (ε)+2=0 Set ε =0 3g (0) 2 g 0 (0) g (0) 4g 0 (0) + 2 = 0 Butwejustfoundthatg (0) = 2, so: that implies g 0 (0) = g 0 (0) + 4 = 0

12 First -Order Approximation By Taylor: x = g (ε) ε=0 ' g (0) + g1 (0) 1! ε 1 or x ' ε For our case ε 0.1 x = = 2.05 Is this a good approximation? x 3 4.1x +0.2 = =

13 Second -Order Approximation Takethederivativeof3g (ε) 2 g 0 (ε) g (ε) (4 + ε) g 0 (ε)+2 = 0 with respect to ε: 6g (ε) ³ g 0 (ε) 2 +3g (ε) 2 g 00 (ε) g 0 (ε) g 0 (ε) (4 + ε) g 00 (ε) =0 (1) Set ε =0 6g (0) ³ g 0 (0) 2 +3g (0) 2 g 00 (0) 2g 0 (0) 4g 00 (0) = 0 Since g (0) = 2 andg 0 (0) = 1 2, we get: 8g 00 (0) 2=0 that implies g 00 (0) =

14 Second -Order Approximation By Taylor:x = g (ε) ε=0 ' g (0) + g1 (0) 1! ε 1 + g2 (0) 2! ε 2 or x ' ε ε2 For our case ε 0.1 x = = Is this a good approximation? x 3 4.1x +0.2 = = e

15 Some Remarks The exact solution (up to machine precession of 14 decimal places) is x = A second-order approximation delivers: x = Relative error: Yes, this was a rigged, but suggestive, example. 15

16 A Couple of Points to Remember 1. We transformed the original problem into a perturbation problem in such a way that the zeroth-order approximation has an analytical solution. 2. Solving for the first iteration involves a nonlinear (although trivial in our case) equation. All further iterations only require to solve a linear equation in one unknown. 16

17 An Application in Macroeconomics: Basic RBC max E 0 X t=0 β {log c t } c t + k t+1 = e z t kt α +(1 δ) k t, t>0 z t = ρz t 1 + σε t, ε t N (0, 1) Equilibrium Conditions 1 1 ³ = βe t 1+αe z t+1 k c t c t+1 α 1 δ t+1 c t + k t+1 = e z t k α t +(1 δ) k t z t = ρz t 1 + σε t 17

18 Computing the RBC The previous problem does not have a known paper and pencil solution. Oneparticularcasethemodelhasaclosedformsolution:δ =1. Why? Because,theincomeandthesubstitutioneffect from a productivity shock cancel each other. Not very realistic but we are trying to learn here. 18

19 Solution By Guess and Verify c t =(1 αβ) e z t k α t k t+1 = αβe z t k α t How can you check? Plug the solution in the equilibrium conditions. 19

20 Another Way to Solve the Problem Now let us suppose that you missed the lecture where Guess and Verify was explained. You need to compute the RBC. What you are searching for? A policy functions for consumption: c t = c (k t,z t ) and another one for capital: k t+1 = k (k t,z t ) 20

21 Equilibrium Conditions We substitute in the equilibrium conditions the budget constraint and the law of motion for technology. Then, we have the equilibrium conditions: 1 c (k t,z t ) = βe αe ρz t+σε t+1 k (k t,z t ) α 1 t c (k (k t,z t ), ρz t + σε t+1 ) c (k t,z t ) + k (k t,z t ) = e z t kt α The Euler equation is the equivalent of x 3 4.1x +0.2 =0inour simple example, and c (k t,z t )andk (k t,z t ) are the equivalents of x. 21

22 A Perturbation Approach You want to transform the problem. Which perturbation parameter? standard deviation σ. Why σ? Set σ =0 deterministic model, z t =0ande z t =1. 22

23 Taylor s Theorem We search for policy function c t = c (k t,z t ; σ) andk t+1 = k (k t,z t ; σ). Equilibrium conditions: Ã 1 E t c (k t,z t ; σ) β αeρz t+σε t+1k (k t,z t ; σ) α 1! c (k (k t,z t ; σ), ρz t + σε t+1 ; σ) = 0 c (k t,z t ; σ)+k (k t,z t ; σ) e z t kt α = 0 We will take derivatives with respect to k t,z t, and σ. 23

24 Asymptotic Expansion c t = c (k t,z t ; σ) k,0,0 c t = c (k, 0; 0) +c k (k, 0; 0) (k t k)+c z (k, 0; 0) z t + c σ (k, 0; 0) σ c kk (k, 0; 0) (k t k) c kz (k, 0; 0) (k t k) z t c kσ (k, 0; 0) (k t k) σ c zk (k, 0; 0) z t (k t k) c zz (k, 0; 0) zt c zσ (k, 0; 0) z t σ c σk (k, 0; 0) σ (k t k)+ 1 2 c σz (k, 0; 0) σz t c σ 2 (k, 0; 0) σ

25 Asymptotic Expansion k t+1 = k (k t,z t ; σ) k,0,0 k t+1 = k (k, 0; 0) +k k (k, 0; 0) k t + k z (k, 0; 0) z t + k σ (k, 0; 0) σ k kk (k, 0; 0) (k t k) k kz (k, 0; 0) (k t k) z t k kσ (k, 0; 0) (k t k) σ k zk (k, 0; 0) z t (k t k) k zz (k, 0; 0) zt k zσ (k, 0; 0) z t σ k σk (k, 0; 0) σ (k t k)+ 1 2 k σz (k, 0; 0) σz t k σ 2 (k, 0; 0) σ

26 Comment on Notation From now on, to save on notation, I will just write F (k t,z t ; σ) =E t 1 c(k t,z t ;σ) β αeρz t +σε t+1k(k t,z t ;σ) α 1 c(k(k t,z t ;σ),ρz t +σε t+1 ;σ) c (k t,z t ; σ)+k (k t,z t ; σ) e z tk α t = " 0 0 # Note that: F (k t,z t ; σ) =H (c (k t,z t ; σ),c(k (k t,z t ; σ),z t+1 ; σ),k(k t,z t ; σ),k t,z t ; σ) I will use H i to represent the partial derivative of H with respect to the i component and drop the evaluation at the steady state of the functions when we do not need it. 26

27 Zeroth -Order Approximation First, we evaluate σ =0: F (k t, 0; 0) = 0 Steady state: or, 1 βαkα 1 c = c 1=αβk α 1 27

28 Steady State Then: c = c (k, 0; 0) = (αβ) α 1 α (αβ) 1 1 α k = k (k, 0; 0) = (αβ) 1 1 α Howgoodisthisapproximation? 28

29 First -Order Approximation We take derivatives of F (k t,z t ; σ) aroundk, 0, and 0. With respect to k t : F k (k, 0; 0) = 0 With respect to z t : F z (k, 0; 0) = 0 With respect to σ: F σ (k, 0; 0) = 0 29

30 Solving the System Remember that: F (k t,z t ; σ) =H (c (k t,z t ; σ),c(k (k t,z t ; σ),z t+1 ; σ),k(k t,z t ; σ),k t,z t ; σ) Then: F k (k, 0; 0) = H 1 c k + H 2 c k k k + H 3 k k + H 4 =0 F z (k, 0; 0) = H 1 c z + H 2 (c k k z + c k ρ)+h 3 k z + H 5 =0 F σ (k, 0; 0) = H 1 c σ + H 2 (c k k σ + c σ )+H 3 k σ + H 6 =0 30

31 Solving the System I Note that: F k (k, 0; 0) = H 1 c k + H 2 c k k k + H 3 k k + H 4 =0 F z (k, 0; 0) = H 1 c z + H 2 (c k k z + c k ρ)+h 3 k z + H 5 =0 is a quadratic system of four equations on four unknowns: c k, c z, k k, and k z. Procedures to solve quadratic systems: Uhlig (1999). Why quadratic? Stable and unstable manifold. 31

32 Solving the System II Note that: F σ (k, 0; 0) = H 1 c σ + H 2 (c k k σ + c σ )+H 3 k σ + H 6 =0 is a linear, and homogeneous system in c σ and k σ. Hence c σ = k σ =0 32

33 Comparison with Linearization After Kydland and Prescott (1982) a popular method to solve economic models has been the use of a LQ approximation. Close relative: linearization of equilibrium conditions. When properly implemented linearization, LQ, and first-order perturbation are equivalent. Advantages of linearization: 1. Theorems. 2. Higher order terms. 33

34 Second -Order Approximation We take second-order derivatives of F (k t,z t ; σ) around k, 0, and 0: F kk (k, 0; 0) = 0 F kz (k, 0; 0) = 0 F kσ (k, 0; 0) = 0 F zz (k, 0; 0) = 0 F zσ (k, 0; 0) = 0 F σσ (k, 0; 0) = 0 Remember Young s theorem! 34

35 Solving the System We substitute the coefficients that we already know. A linear system of 12 equations on 12 unknowns. Why linear? Cross-terms kσ and zσ are zero. Conjecture on all the terms with odd powers of σ. 35

36 Correction for Risk We have a term in σ 2. Captures precautionary behavior. We do not have certainty equivalence any more! Important advantage of second order approximation. 36

37 Higher Order Terms We can continue the iteration for as long as we want. Often, a few iterations will be enough. The level of accuracy depends on the goal of the exercise: Fernández- Villaverde, Rubio-Ramírez, and Santos (2005). 37

38 AComputer In practice you do all this approximations with a computer. Burden: analytical derivatives. Why are numerical derivatives a bad idea? More theoretical point: do the derivatives exist? (Santos, 1992). 38

39 Code This afternoon there will be some demonstrations by Juilliard and Levin. First and second order: Matlab and Dynare. Higher order: Mathematica, Fortran code by Jinn and Judd. 39

40 An Example Let me run a second order approximation. Our choices Calibrated Parameters Parameter β α ρ σ Value

41 Computation Steady State: c =(αβ) 1 α α (αβ) 1 α 1 = k =(αβ) 1 α 1 = First order components. c k (k, 0; 0) = k k (k, 0; 0) = c z (k, 0; 0) = k k (k, 0; 0) = 0.33 c σ (k, 0; 0) = 0 k k (k, 0; 0) = 0 41

42 Comparison c t =0.6733e z t k 0.33 t c t ' (k t k) z t and: k t+1 =0.3267e z t k 0.33 t k t+1 ' (k t k) +0.33z t 42

43 Second-Order Terms c kk (k, 0; 0) = k kk (k, 0; 0) = c kz (k, 0; 0) = k kz (k, 0; 0) = c kσ (k, 0; 0) = 0. k kσ (k, 0; 0) = 0 c zz (k, 0; 0) = k zz (k, 0; 0) = c zσ (k, 0; 0) = 0 k zσ (k, 0; 0) = 0 c σ 2 (k, 0; 0) = 0 k σ 2 (k, 0; 0) = 0 43

44 Zeroth-Order Approximation First-Order Approximation Second-Order Approximation

45 Non Local Accuracy test (Judd, 1992, and Judd and Guu, 1997) Given the Euler equation: we can define: 1 c i (k t,z t ) = E t αez t+1k i (k t,z t ) α 1 c i ³ k i (k t,z t ),z t+1 EE i (k t,z t ) 1 c i (k t,z t ) E t αez t+1k i (k t,z t ) α 1 c i ³ k i (k t,z t ),z t+1 44

46 Figure : Euler Equation Errors at z = 0, τ = 2 / σ = Linear Log-Linear FEM Chebyshev Perturbation 2 Perturbation 5 Value Function -5 Log10 Euler Equation Error Capital

47 Changes of Variables We approximated our solution in levels. You may have read about loglinearization. Note that: x t = xe log x t x = xe bx t We can substitute all variables x t by bx t. Why can loglinearization be a good idea? 45

48 Judd (2002) Why stop there? Why not in powers of the state variables? Judd (2002) has provided methods for changes of variables. We apply and extend ideas to the stochastic neoclassical growth model. 46

49 A General Transformation We look at solutions of the form: c µ c µ 0 = a ³ k ζ k ζ 0 + cz ϕ k 0γ k γ 0 = c ³ k ζ k ζ 0 + dz ϕ where ϕ 1. Note that: 1. If γ, ζ, µ and ϕ are 1 we get the linear representation. 2. As γ, ζ and µ tend to zero and ϕ is equal to 1 we get the loglinear approximation. 47

50 Theory The first order solution can be written as f (x) ' f (a)+(x a) f 0 (a) Expand g(y) =h (f (X (y))) around b = Y (a), where X (y) isthe inverse of Y (x). Then: g (y) =h (f (X (y))) = g (b)+g α (b)(y α (x) b α ) where g α = h A fi AXi α comes from the application of the chain rule. From this expression it is easy to see that if we have computed the values of f A i, then it is straightforward to find the value of g α. 48

51 Theory The first order solution can be written as f (x) ' f (a)+(x a) f 0 (a) Expand g(y) =h (f (X (y))) around b = Y (a), where X (y) isthe inverse of Y (x). Then: g (y) =h (f (X (y))) = g (b)+g α (b)(y α (x) b α ) where g α = h A fi AXi α comes from the application of the chain rule. From this expression it is easy to see that if we have computed the values of f A i, then it is straightforward to find the value of g α. 49

52 Figure : Euler Equation Errors at z = 0, τ = 2 / σ = Linear Optimal Change Log10 Euler Equation Error Capital

53 Figure : Euler Equation Errors at z = 0, τ = 2 / σ = Log10 Euler Equation Error Linear Log-Linear FEM Chebyshev Perturbation 2 Perturbation 5 Value Function Optimal Change Capital

54 En Savoir Plus I General Perturbation theory: Advanced Mathematical Methods for Scientists and Engineers: Asymptotic Methods and Perturbation Theory by Carl M. Bender, Steven A. Orszag. Perturbation in Economics: 1. Perturbation Methods for General Dynamic Stochastic Models by Hehui Jin and Kenneth Judd. 2. Perturbation Methods with Nonlinear Changes of Variables by Kenneth Judd. 50

55 En Savoir Plus II A gentle introduction: Solving Dynamic General Equilibrium Models Using a Second-Order Approximation to the Policy Function by Martín Uribe and Stephanie Schmitt-Grohe. Solving quadratic equations: A Toolkit for Analysing Nonlinear Dynamic Stochastic Models Easily by Harald Uhlig. 51