Perturbation Methods II: General Case

Transcription

1 Perturbation Methods II: General Case (Lectures on Solution Methods for Economists VI) Jesús Fernández-Villaverde 1 and Pablo Guerrón 2 February 21, University of Pennsylvania 2 Boston College

2 The General case Most of arguments in the previous set of lecture notes are easy to generalize. The set of equilibrium conditions of many DSGE models can be written using recursive notation as: E t H(y, y, x, x ) = 0, where y t is a n y 1 vector of controls and x t is a n x 1 vector of states. n = n x + n y. H maps R ny R ny R nx R nx into R n. 1

3 Partitioning the state vector The state vector x t can be partitioned as x = [x 1 ; x 2 ] t. x 1 is a (n x n ɛ ) 1 vector of endogenous state variables. x 2 is a n ɛ 1 vector of exogenous state variables. Why do we want to partition the state vector? 2

4 Exogenous stochastic process I x 2 = Ax 2 + λη ɛ ɛ Process with 3 parts: 1. The deterministic component Ax 2, where A is a n ɛ n ɛ matrix, with all eigenvalues with modulus less than one. 2. The scaled innovation η ɛɛ, where: 2.1 η ɛ is a known n ɛ n ɛ matrix. 2.2 ɛ is a n ɛ 1 i.i.d. innovation with bounded support, zero mean, and variance/covariance matrix I. 3. The perturbation parameter λ. 3

5 Exogenous stochastic process II We can accommodate very general structures of x 2 through changes in the definition of the state space: i.e. stochastic volatility. More general structure: x 2 = Γ(x 2 ) + λη ɛ ɛ where Γ is a non-linear function satisfying that all eigenvalues of its first derivative evaluated at the non-stochastic steady state lie within the unit circle. Note we do not impose Gaussanity. 4

6 The perturbation parameter The scalar λ 0 is the perturbation parameter. If we set λ = 0, we have a deterministic model. Important: there is only ONE perturbation parameter. The matrix η ɛ takes account of relative sizes of different shocks. Why bounded support? Samuelson (1970), Jin and Judd (2002). 5

7 Solution of the model The solution to the model is of the form: y = g(x; λ) x = h(x; λ) + ληɛ where g maps R nx R + into R ny and h maps R nx R + into R nx. The matrix η is of order n x n ɛ and is given by: [ ] η = η ɛ 6

8 Perturbation We wish to find a perturbation approximation of the functions g and h around the non-stochastic steady state, x t = x and λ = 0. We define the non-stochastic steady state as vectors ( x, ȳ) such that: H(ȳ, ȳ, x, x) = 0. Note that ȳ = g( x; 0) and x = h( x; 0). This is because, if λ = 0, E t H = H. 7

9 Plugging-in the proposed solution Substituting the proposed solution, we define: F (x; λ) E t H(g(x; λ), g(h(x; λ) + ηλɛ, λ), x, h(x; λ) + ηλɛ ) = 0 Since F (x; λ) = 0 for any values of x and λ, the derivatives of any order of F must also be equal to zero. Formally: F x k λj (x; λ) = 0 x, λ, j, k, where F x k λj (x, λ) denotes the derivative of F with respect to x taken k times and with respect to λ taken j times. 8

10 First-order approximation We are looking for approximations to g and h around (x, λ) = ( x, 0) of the form: g(x; λ) = g( x; 0) + g x ( x; 0)(x x) + g λ ( x; 0)λ h(x; λ) = h( x; 0) + h x ( x; 0)(x x) + h λ ( x; 0)λ As explained earlier, g( x; 0) = ȳ and h( x; 0) = x. The remaining four unknown coefficients of the first-order approximation to g and h are found by using the fact that: F x ( x; 0) = 0 and F λ ( x; 0) = 0 Before doing so, we need to introduce the tensor notation. 9

11 Tensors General trick from physics. An n th -rank tensor in a m-dimensional space is an operator that has n indices and m n components and obeys certain transformation rules. [H y ] i α is the (i, α) element of the derivative of H with respect to y: 1. The derivative of H with respect to y is an n n y matrix. 2. Thus, [H y ] i α is the element of this matrix located at the intersection of the i-th row and α-th column. 3. Thus, [H y ] i α[g x] α β [h x] β j = n y nx H i g α h β α=1 β=1. y α x β x j [H y y ]i αγ: 1. H y y is a three dimensional array with n rows, ny columns, and ny pages. 2. Then [H y y ]i αγ denotes the element of H y y located at the intersection of row i, column α and page γ. 10

12 Solving the system I g x and h x can be found as the solution to the system: [F x ( x; 0)] i j = [H y ] i α[g x ] α β[h x ] β j + [H y ] i α[g x ] α j + [H x ] i β[h x ] β j + [H x ] i j = 0; i = 1,..., n; j, β = 1,..., n x ; α = 1,..., n y Note that the derivatives of H evaluated at (y, y, x, x ) = (ȳ, ȳ, x, x) are known. Then, we have a system of n n x quadratic equations in the n n x unknowns given by the elements of g x and h x. We can solve with a standard quadratic matrix equation solver. 11

13 Solving the system II g λ and h λ are the solution to the n equations: Then: [F λ ( x; 0)] i = E t {[H y ] i α[g x ] α β[h λ ] β + [H y ] i α[g x ] α β[η] β φ [ɛ ] φ + [H y ] i α[g λ ] α +[H y ] i α[g λ ] α + [H x ] i β[h λ ] β + [H x ] i β[η] β φ [ɛ ] φ } i = 1,..., n; α = 1,..., n y ; β = 1,..., n x ; φ = 1,..., n ɛ. [F λ ( x; 0)] i = [H y ] i α[g x ] α β[h λ ] β + [H y ] i α[g λ ] α + [H y ] i α[g λ ] α + [f x ] i β[h λ ] β = 0; i = 1,..., n; α = 1,..., n y ; β = 1,..., n x ; φ = 1,..., n ɛ. Certainty equivalence: linear and homogeneous equation in g λ and h λ. Thus, if a unique solution exists, it satisfies: h λ = 0 g λ = 0 12

14 Second-order approximation I The second-order approximations to g around (x; λ) = ( x; 0) is [g(x; λ)] i = [g( x; 0)] i + [g x ( x; 0)] i a[(x x)] a + [g λ ( x; 0)] i [λ] [g xx( x; 0)] i ab[(x x)] a [(x x)] b [g xλ( x; 0)] i a[(x x)] a [λ] [g λx( x; 0)] i a[(x x)] a [λ] [g λλ( x; 0)] i [λ][λ] where i = 1,..., n y, a, b = 1,..., n x, and j = 1,..., n x. 13

15 Second-order approximation II The second-order approximations to h around (x; λ) = ( x; 0) is [h(x; λ)] j = [h( x; 0)] j + [h x ( x; 0)] j a[(x x)] a + [h λ ( x; 0)] j [λ] [h xx( x; 0)] j ab [(x x)] a[(x x)] b [h xλ( x; 0)] j a[(x x)] a [λ] [h λx( x; 0)] j a[(x x)] a [λ] [h λλ( x; 0)] j [λ][λ], where i = 1,..., n y, a, b = 1,..., n x, and j = 1,..., n x. 14

16 Second-order approximation III The unknowns of these expansions are [g xx ] i ab, [g xλ] i a, [g λx ] i a, [g λλ ] i, [h xx ] j ab, [h xλ ] j a, [h λx ] j a, [h λλ ] j. These coefficients can be identified by taking the derivative of F (x; λ) with respect to x and λ twice and evaluating them at (x; λ) = ( x; 0). By the arguments provided earlier, these derivatives must be zero. 15

17 Solving the system I We use F xx ( x; 0) to identify g xx ( x; 0) and h xx ( x; 0): [F xx ( x; 0)] i jk = ( [Hy y ]i αγ[g x ] γ δ [h x] δ k + [H y y ] i αγ[g x ] γ k + [H y x ]i αδ[h x ] δ k + [H y x] i ) αk [gx ] α β[h x ] β j +[H y ] i α[g xx ] α βδ[h x ] δ k[h x ] β j + [H y ] i α[g x ] α β[h xx ] β jk + ( [H yy ] i αγ[g x ] γ δ [h x] δ k + [H yy ] i αγ[g x ] γ k + [H yx ]i αδ[h x ] δ k + [H yx ] i αk) [gx ] α j +[H y ] i α[g xx ] α jk + ( [H x y ]i βγ[g x ] γ δ [h x] δ k + [H x y ] i βγ[g x ] γ k + [H x x ]i βδ[h x ] δ k + [H x x] i βk) [hx ] β j +[H x ] i β[h xx ] β jk +[H xy ] i jγ[g x ] γ δ [h x] δ k + [H xy ] i jγ[g x ] γ k + [H xx ]i jδ[h x ] δ k + [H xx ] i jk = 0; i = 1,... n, j, k, β, δ = 1,... n x ; α, γ = 1,... n y. 16

18 Solving the system II We know the derivatives of H. We also know the first derivatives of g and h evaluated at (y, y, x, x ) = (ȳ, ȳ, x, x). Hence, the above expression represents a system of n n x n x linear equations in then n n x n x unknowns elements of g xx and h xx. 17

19 Solving the system III Similarly, g λλ and h λλ can be obtained by solving: [F λλ ( x; 0)] i = [H y ] i α[g x ] α β[h λλ ] β +[H y y ]i αγ[g x ] γ δ [η]δ ξ[g x ] α β[η] β φ [I ]φ ξ +[H y x ]i αδ[η] δ ξ[g x ] α β[η] β φ [I ]φ ξ +[H y ] i α[g xx ] α βδ[η] δ ξ[η] β φ [I ]φ ξ + [H y ]i α[g λλ ] α +[H y ] i α[g λλ ] α + [H x ] i β[h λλ ] β +[H x y ]i βγ[g x ] γ δ [η]δ ξ[η] β φ [I ]φ ξ +[H x x ]i βδ[η] δ ξ[η] β φ [I ]φ ξ = 0; i = 1,..., n; α, γ = 1,..., n y ; β, δ = 1,..., n x ; φ, ξ = 1,..., n ɛ a system of n linear equations in the n unknowns given by the elements of g λλ and h λλ. 18

20 Cross-derivatives The cross derivatives g xλ and h xλ are zero when evaluated at ( x, 0). Why? Write the system F λx ( x; 0) = 0 taking into account that all terms containing either g λ or h λ are zero at ( x, 0). Then: [F λx ( x; 0)] i j = [H y ] i α[g x ] α β[h λx ] β j + [H y ] i α[g λx ] α γ [h x ] γ j + [H y ] i α[g λx ] α j + [H x ] i β[h λx ] β j = 0; i = 1,... n; α = 1,..., n y ; β, γ, j = 1,..., n x. This is a system of n n x equations in the n n x unknowns given by the elements of g λx and h λx. The system is homogeneous in the unknowns. Thus, if a unique solution exists, it is given by: g λx = 0 h λx = 0 19

21 Structure of the solution The perturbation solution of the model satisfies: g λ ( x; 0) = 0 h λ ( x; 0) = 0 g xλ ( x; 0) = 0 h xλ ( x; 0) = 0 Standard deviation only appears in: 1. A constant term given by 1 g 2 λλλ 2 for the control vector y t. 2. The first n x n ɛ elements of 1 h 2 λλλ 2. Correction for risk. Quadratic terms in endogenous state vector x 1. Those terms capture non-linear behavior. 20

22 Higher-order approximations We can iterate this procedure as many times as we want. We can obtain n-th order approximations. Problems: 1. Existence of higher order derivatives (Santos, 1992). 2. Numerical instabilities. 3. Computational costs. 21