Solution Methods Jesús Fernández-Villaverde University of Pennsylvania March 16, 2016 Jesús Fernández-Villaverde (PENN) Solution Methods March 16, 2016 1 / 36 Functional equations A large class of problems in macroeconomics search for a function d that solves a functional equation: (d) = 0 H More formally: 1 Let J1 and J2 be two functional spaces and let : J1 J2 be an operator between these two spaces. H ! 2 Let W Rl . 3 Then, we need to find a function d : W Rm such that (d) = 0. ! H Notes: 1 Regular equations are particular examples of functional equations. 2 0 is the space zero, different in general that the zero in the reals. Jesús Fernández-Villaverde (PENN) Solution Methods March 16, 2016 2 / 36 Example I Let’sgo back to our basic stochastic neoclassical growth model: ¥ t max E0 ∑ b u (ct , lt ) t=0 zt a 1 a ct + kt+1 = e Ak l + (1 d) kt , t > 0 t t 8 zt = rzt 1 + s#t , #t (0, 1) N 1 2 2 log st = (1 r ) log s + r log st 1 + 1 r hut s s s The first order condition: zt+1 a 1 1 a u0 (ct , lt ) = bEt u0 (ct+1, lt+1) 1 + ae Ak l d t+1 t+1 Where is the stochastic volatility? Jesús Fernández-Villaverde (PENN) Solution Methods March 16, 2016 3 / 36 Example II Define: xt = (kt , zt 1, log st 1, #t , ut ) There is a decision rule (a.k.a. policy function) that gives the optimal choice of consumption and capital tomorrow given the states today: 1 d (xt ) = lt d = 2 d (xt ) = kt+1 From these two choices, we can find: zt a 1 1 a 2 ct = e Ak d (xt ) + (1 d) kt d (xt ) t Jesús Fernández-Villaverde (PENN) Solution Methods March 16, 2016 4 / 36 Example III Then: 1 = u0 ct , d (xt ) H 1 u0 ct+1, d (xt+1) bEt z 2 a 1 1 1 a = 0 1 + ae t+1 Ad (xt ) d (xt+1) d ( ) If we find g, and a transversality condition is satisfied, we are done! Jesús Fernández-Villaverde (PENN) Solution Methods March 16, 2016 5 / 36 Example IV There is a recursive problem associated with the previous sequential problem: V (xt ) = max u (ct , lt ) + bEt V (xt+1) kt+1,lt f g zt a 1 a ct + kt+1 = e Ak l + (1 d) kt , t > 0 t t 8 zt = rzt 1 + s#t , #t (0, 1) N 1 2 2 log st = (1 r ) log s + r log st 1 + 1 r hut s s s Then: d (xt ) = V (xt ) and (d) = d (xt ) max u (ct , lt ) + bEt d (xt+1) = 0 H kt+1,lt f g Jesús Fernández-Villaverdee (PENN) Solution Methods March 16, 2016 6 / 36 How do we solve functional equations? General idea: substitute d (x) by dn (x, q) where q is an n dim vector of coeffi cients to be determined. Two Main Approaches: 1 Perturbation methods: n n i d (x, q) = ∑ qi (x x0) i=0 We use implicit-function theorems to find qi . 2 Projection methods: n n d (x, q) = ∑ qi Yi (x) i=0 ¥ We pick a basis Yi (x) and “project” ( ) against that basis. f gi=0 H · Jesús Fernández-Villaverde (PENN) Solution Methods March 16, 2016 7 / 36 Comparison with traditional solution methods Linearization (or loglinearization): equivalent to a first-order perturbation. Linear-quadratic approximation: equivalent (under certain conditions) to a first-order perturbation. Parameterized expectations: a particular example of projection. Value function iteration:it can be interpreted as an iterative procedure to solve a particular projection method. Nevertheless, I prefer to think about it as a different family of problems. Policy function iteration: similar to VFI. Jesús Fernández-Villaverde (PENN) Solution Methods March 16, 2016 8 / 36 Advantages of the functional equation approach Generality: abstract framework highlights commonalities across problems. Large set of existing theoretical and numerical results in applied math. It allows us to identify more clearly issue and challenges specific to economic problems (for example, importance of expectations). It allows us to deal effi ciently with nonlinearities. Jesús Fernández-Villaverde (PENN) Solution Methods March 16, 2016 9 / 36 Perturbation: motivation Perturbation builds a Taylor-series approximation of the exact solution. Very accurate around the point where the approximation is undertakes. Often, surprisingly good global properties. Only approach that handle models with dozens of state variables. Relation between uncertainty shocks and the curse of dimensionality. Jesús Fernández-Villaverde (PENN) Solution Methods March 16, 2016 10 / 36 References General: 1 A First Look at Perturbation Theory by James G. Simmonds and James E. Mann Jr. 2 Advanced Mathematical Methods for Scientists and Engineers: Asymptotic Methods and Perturbation Theory by Carl M. Bender, Steven A. Orszag. 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. 3 A gentle introduction: “Solving Dynamic General Equilibrium Models Using a Second-Order Approximation to the Policy Function” by Martín Uribe and Stephanie Schmitt-Grohe. Jesús Fernández-Villaverde (PENN) Solution Methods March 16, 2016 11 / 36 Our RBC-SV model Let me come back to our RBC-SV model. Three changes: 1 Eliminate labor supply and have a log utility function for consumption. 2 Full depreciation. 3 A = 1. Why? Jesús Fernández-Villaverde (PENN) Solution Methods March 16, 2016 12 / 36 Environment ¥ t max E0 ∑ b log ct t=0 zt a s.t. ct + kt+1 = e k , t > 0 t 8 zt = rzt 1 + s#t 1 2 2 log st = (1 r ) log s + r log st 1 + 1 r hut s s s Equilibrium conditions: 1 1 zt+1 a 1 = bEt ae kt+1 ct ct+1 zt a ct + kt+1 = e kt zt = rzt 1 + s#t 1 2 2 log st = (1 r ) log s + r log st 1 + 1 r hut s s s Jesús Fernández-Villaverde (PENN) Solution Methods March 16, 2016 13 / 36 Solution Exact solution (found by “guess and verify”): zt a ct = (1 ab) e k t zt a kt+1 = abe kt Note that this solution is the same than in the model without stochastic volatility. Intuition. However, the dynamics of the model is affected by the law of motion for zt . Jesús Fernández-Villaverde (PENN) Solution Methods March 16, 2016 14 / 36 Steady state Steady state is also easy to find: 1 1 a k = (ab) a 1 c = (ab) 1 a (ab) 1 a z = 0 log s = log s Steady state in more general models. Jesús Fernández-Villaverde (PENN) Solution Methods March 16, 2016 15 / 36 The goal Define xt = (kt , zt 1, log st 1, #t , ut ; l). Role of l: perturbation parameter. We are searching for decision rules: c = c (x ) d = t t kt+1 = k (xt ) Then, we have a system of functional equations: 1 1 zt+1 a 1 = bEt ae k (xt ) c (xt ) c (xt+1) zt a c (xt ) + k (xt ) = e kt zt = rzt 1 + st l#t 1 2 2 log st = (1 r ) log s + r log st 1 + 1 r hlut s s s Jesús Fernández-Villaverde (PENN) Solution Methods March 16, 2016 16 / 36 Taylor’stheorem We will build a local approximation around x = (k, 0, log s, 0, 0; l). Given equilibrium conditions: 1 1 zt+1 a 1 Et b ae k (xt ) = 0 c (xt ) c (xt+1) zt a c (xt ) + k (xt ) e k = 0 t We will take derivatives with respect to (k, z, log s, #, u; l) and evaluate them around x = (k, 0, log s, 0, 0; l). Why? Apply Taylor’stheorem and a version of the implicit-function theorem. Jesús Fernández-Villaverde (PENN) Solution Methods March 16, 2016 17 / 36 Taylor series expansion I ct = c (kt , zt 1, log st 1, #t , ut ; 1) k,0,0 = c (x) j +ck (x)(kt k) + cz (x) zt 1 + clog s (x) log st 1 + +c# (x) #t + cu (x) ut + cl (x) 1 2 1 + ckk (x)(kt k) + ckz (x)(kt k) zt 1 + 2 2 1 + ck log s (x)(kt k) log st 1 + ... 2 Jesús Fernández-Villaverde (PENN) Solution Methods March 16, 2016 18 / 36 Taylor series expansion II kt = k (kt , zt 1, log st 1, #t , ut ; 1) k,0,0 = k (x) j +kk (x)(kt k) + kz (x) zt 1 + klog s (x) log st 1 + +k# (x) #t + ku (x) ut + kl (x) 1 2 1 + kkk (x)(kt k) + kkz (x)(kt k) zt 1 + 2 2 1 + kk log s (x)(kt k) log st 1 + ... 2 Jesús Fernández-Villaverde (PENN) Solution Methods March 16, 2016 19 / 36 Comment on notation From now on, to save on notation, I will write 1 1 z a 1 b ae t+1 k (xt ) 0 F (x ) = E c(xt ) c(xt+1 ) = t t z a c (xt ) + k (xt ) e t k 0 " t # I will take partial derivatives of F (xt ) and evaluate them at the steady state Do these derivatives exist? Jesús Fernández-Villaverde (PENN) Solution Methods March 16, 2016 20 / 36 First-order approximation We take first-order derivatives of F (xt ) around x. Thus, we get: DF (x) = 0 A matrix quadratic system. Why quadratic? Stable and unstable manifold. However, it has a nice recursive structure that we can and should exploit.