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Perturbation Methods Jesús Fernández-Villaverde University of Pennsylvania May 28, 2015 Jesús Fernández-Villaverde (PENN) Perturbation Methods May 28, 2015 1 / 91

Introduction Introduction Remember that we want to solve a functional equation of the form: H ( d ) = 0 for an unknown decision rule d . Perturbation solves the problem by specifying: n d n ( x , θ ) = θ i ( x − x 0 ) i ∑ i = 0 We use implicit-function theorems to find coefficients θ i ’s. Inherently local approximation. However, often good global properties. Jesús Fernández-Villaverde (PENN) Perturbation Methods May 28, 2015 2 / 91

Introduction Motivation Many complicated mathematical problems have: either a particular case 1 or a related problem. 2 that is easy to solve. Often, we can use the solution of the simpler problem as a building block of the general solution. Very successful in physics. Sometimes perturbation is known as asymptotic methods. Jesús Fernández-Villaverde (PENN) Perturbation Methods May 28, 2015 3 / 91

Introduction The World Simplest Perturbation √ What is 26? Without your Iphone calculator, it is a boring arithmetic calculation. But note that: � √ √ 26 = 25 ( 1 + 0 . 04 ) = 5 ∗ 1 . 04 ≈ 5 ∗ 1 . 02 = 5 . 1 Exact solution is 5.099. √ We have solved a much simpler problem ( 25) and added a small coefficient to it. More in general � √ √ y = x 2 ( 1 + ε ) = x 1 + ε where x is an integer and ε the perturbation parameter. Jesús Fernández-Villaverde (PENN) Perturbation Methods May 28, 2015 4 / 91

Introduction Applications to Economics Judd and Guu (1993) showed how to apply it to economic problems. Recently, perturbation methods have been gaining much popularity. In particular, second- and third-order approximations are easy to compute and notably improve accuracy. A first-order perturbation theory and linearization deliver the same output. Hence, we can use much of what we already know about linearization. Jesús Fernández-Villaverde (PENN) Perturbation Methods May 28, 2015 5 / 91

Introduction Regular versus Singular Perturbations Regular perturbation: a small change in the problem induces a small change in the solution. Singular perturbation: a small change in the problem induces a large change in the solution. Example: excess demand function. Most problems in economics involve regular perturbations. Sometimes, however, we can have singularities. Example: introducing a new asset in an incomplete markets model. Jesús Fernández-Villaverde (PENN) Perturbation Methods May 28, 2015 6 / 91

Introduction References General: A First Look at Perturbation Theory by James G. Simmonds and 1 James E. Mann Jr. Advanced Mathematical Methods for Scientists and Engineers: 2 Asymptotic Methods and Perturbation Theory by Carl M. Bender, Steven A. Orszag. Economics: Perturbation Methods for General Dynamic Stochastic Models” by 1 Hehui Jin and Kenneth Judd. Perturbation Methods with Nonlinear Changes of Variables” by 2 Kenneth Judd. A gentle introduction: “Solving Dynamic General Equilibrium Models 3 Using a Second-Order Approximation to the Policy Function” by Martín Uribe and Stephanie Schmitt-Grohe. Jesús Fernández-Villaverde (PENN) Perturbation Methods May 28, 2015 7 / 91

A Baby Example A Baby Example: A Basic RBC Model: ∞ β t log c t ∑ max E 0 t = 0 s.t. c t + k t + 1 = e z t k α t + ( 1 − δ ) k t , ∀ t > 0 z t = ρ z t − 1 + σε t , ε t ∼ N ( 0 , 1 ) Equilibrium conditions: � � 1 1 1 + α e z t + 1 k α − 1 = β E t t + 1 − δ c t c t + 1 c t + k t + 1 = e z t k α t + ( 1 − δ ) k t z t = ρ z t − 1 + σε t Jesús Fernández-Villaverde (PENN) Perturbation Methods May 28, 2015 8 / 91

A Baby Example Computing a Solution The previous problem does not have a known “paper and pencil” solution except when (unrealistically) δ = 1 . Then, income and substitution effect from a technology shock cancel each other (labor constant and consumption is a fixed fraction of income). Equilibrium conditions with δ = 1: α e z t + 1 k α − 1 1 t + 1 = β E t c t c t + 1 c t + k t + 1 = e z t k α t z t = ρ z t − 1 + σε t By “guess and verify”: c t = ( 1 − αβ ) e z t k α t k t + 1 = αβ e z t k α t Jesús Fernández-Villaverde (PENN) Perturbation Methods May 28, 2015 9 / 91

A Baby Example Another Way to Solve the Problem Now let us suppose that you missed the lecture when “guess and verify” was explained. You need to compute the RBC. What you are searching for? A decision rule for consumption: c t = c ( k t , z t ) and another one for capital: k t + 1 = k ( k t , z t ) Note that our d is just the stack of c ( k t , z t ) and k ( k t , z t ) . Jesús Fernández-Villaverde (PENN) Perturbation Methods May 28, 2015 10 / 91

A Baby Example Equilibrium Conditions We substitute in the equilibrium conditions the budget constraint and the law of motion for technology. And we write the decision rules explicitly as function of the states. Then: α e ρ z t + σε t + 1 k ( k t , z t ) α − 1 1 c ( k t , z t ) = β E t c ( k ( k t , z t ) , ρ z t + σε t + 1 ) c ( k t , z t ) + k ( k t , z t ) = e z t k α t System of functional equations. Jesús Fernández-Villaverde (PENN) Perturbation Methods May 28, 2015 11 / 91

A Baby Example 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. This step is usually ambiguous since there are different ways to do so. Use the previous solution to approximate the solution of original the problem. Jesús Fernández-Villaverde (PENN) Perturbation Methods May 28, 2015 12 / 91

A Baby Example A Perturbation Approach Hence, we want to transform the problem. Which perturbation parameter? Standard deviation σ . Why σ ? Discrete versus continuous time. Set σ = 0 ⇒ deterministic model, z t = 0 and e z t = 1. We know how to solve the deterministic steady state. Jesús Fernández-Villaverde (PENN) Perturbation Methods May 28, 2015 13 / 91

A Baby Example A Parametrized Decision Rule We search for decision rule: c t = c ( k t , z t ; σ ) and k t + 1 = k ( k t , z t ; σ ) Note new parameter σ . We are building a local approximation around σ = 0 . Jesús Fernández-Villaverde (PENN) Perturbation Methods May 28, 2015 14 / 91

A Baby Example Taylor’s Theorem Equilibrium conditions: � � α e ρ z t + σε t + 1 k ( k t , z t ; σ ) α − 1 1 E t c ( k t , z t ; σ ) − β = 0 c ( k ( k t , z t ; σ ) , ρ z t + σε t + 1 ; σ ) c ( k t , z t ; σ ) + k ( k t , z t ; σ ) − e z t k α t = 0 We will take derivatives with respect to k t , z t , and σ . Apply Taylor’s theorem to build solution around deterministic steady state. How? Jesús Fernández-Villaverde (PENN) Perturbation Methods May 28, 2015 15 / 91

A Baby Example Asymptotic Expansion I c t = c ( k t , z t ; σ ) | k , 0 , 0 = 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 + 1 2 c kk ( k , 0 ; 0 ) ( k t − k ) 2 + 1 2 c kz ( k , 0 ; 0 ) ( k t − k ) z t + 1 2 c k σ ( k , 0 ; 0 ) ( k t − k ) σ + 1 2 c zk ( k , 0 ; 0 ) z t ( k t − k ) + 1 t + 1 2 c zz ( k , 0 ; 0 ) z 2 2 c z σ ( k , 0 ; 0 ) z t σ + 1 2 c σ k ( k , 0 ; 0 ) σ ( k t − k ) + 1 2 c σ z ( k , 0 ; 0 ) σ z t + 1 2 c σ 2 ( k , 0 ; 0 ) σ 2 + ... Jesús Fernández-Villaverde (PENN) Perturbation Methods May 28, 2015 16 / 91

A Baby Example Asymptotic Expansion II k t + 1 = k ( k t , z t ; σ ) | k , 0 , 0 = k ( k , 0 ; 0 ) + k k ( k , 0 ; 0 ) ( k t − k ) + k z ( k , 0 ; 0 ) z t + k σ ( k , 0 ; 0 ) σ + 1 2 k kk ( k , 0 ; 0 ) ( k t − k ) 2 + 1 2 k kz ( k , 0 ; 0 ) ( k t − k ) z t + 1 2 k k σ ( k , 0 ; 0 ) ( k t − k ) σ + 1 2 k zk ( k , 0 ; 0 ) z t ( k t − k ) + 1 t + 1 2 k zz ( k , 0 ; 0 ) z 2 2 k z σ ( k , 0 ; 0 ) z t σ + 1 2 k σ k ( k , 0 ; 0 ) σ ( k t − k ) + 1 2 k σ z ( k , 0 ; 0 ) σ z t + 1 2 k σ 2 ( k , 0 ; 0 ) σ 2 + ... Jesús Fernández-Villaverde (PENN) Perturbation Methods May 28, 2015 17 / 91

A Baby Example Comment on Notation From now on, to save on notation, I will write � � � 0 � c ( k t , z t ; σ ) − β α e ρ zt + σε t + 1 k ( k t , z t ; σ ) α − 1 1 F ( k t , z t ; σ ) = E t = c ( k ( k t , z t ; σ ) , ρ z t + σε t + 1 ; σ ) 0 c ( k t , z t ; σ ) + k ( k t , z t ; σ ) − e z t k α t Note that: F ( k t , z t ; σ ) = H ( c t , c t + 1 , k t , k t + 1 , z t ; σ ) = H ( c ( k t , z t ; σ ) , c ( k ( k t , z t ; σ ) , z t + 1 ; σ ) , k t , k ( k t , z t ; σ ) , 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. Jesús Fernández-Villaverde (PENN) Perturbation Methods May 28, 2015 18 / 91

A Baby Example Zeroth-Order Approximation First, we evaluate σ = 0: F ( k t , 0 ; 0 ) = 0 Steady state: c = βα k α − 1 1 c or 1 = αβ k α − 1 Then : α 1 1 − α − ( αβ ) c = c ( k , 0 ; 0 ) = ( αβ ) 1 − α 1 k = k ( k , 0 ; 0 ) = ( αβ ) 1 − α Jesús Fernández-Villaverde (PENN) Perturbation Methods May 28, 2015 19 / 91