Linearization. (Lectures on Solution Methods for Economists V: Appendix) Jesús Fernández-Villaverde 1 and Pablo Guerrón 2 February 21, 2018

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1 Linearization (Lectures on Solution Methods for Economists V: Appendix) Jesús Fernández-Villaverde 1 and Pablo Guerrón 2 February 21, University of Pennsylvania 2 Boston College

2 Basic RBC Benchmark set up: max E 0 t=0 β t {log c t + ψ log (1 l t )} c t + k t+1 = k α t (e zt l t ) 1 α + (1 δ) k t, t > 0 z t = ρz t 1 + ε t, ε t N (0, σ) This is a dynamic optimization problem. The previous problem does not have a paper and pencil solution. Traditional solution: linearization. 1

3 Equilibrium conditions From the household problem+firms s problem+aggregate conditions: { 1 1 ( ) } = βe t 1 + αk α 1 t+1 c t c (ezt+1 l t+1 ) 1 α δ t+1 c t ψ = (1 α) kt α (e zt l t ) 1 α lt 1 1 l t c t + k t+1 = k α t (e zt l t ) 1 α + (1 δ) k t z t = ρz t 1 + ε t Do we substitute first? 2

4 Steady state I If σ = 0, the equilibrium conditions are: 1 = β 1 ( 1 + αk α 1 t+1 c t c l 1 α t+1 δ) t+1 c t ψ = (1 α) kt α lt α 1 l t c t + k t+1 = k α t l 1 α t + (1 δ) k t The equilibrium conditions imply a steady state: 1 c = β 1 ( 1 + αk α 1 l 1 α δ ) c ψ c = (1 α) k α l α 1 l c + δk = k α l 1 α 3

5 Steady state II Solution: k = µ Ω + ϕµ l = ϕk c = Ωk y = k α l 1 α where ϕ = ( 1 α ( )) 1 1 β 1 + δ 1 α, Ω = ϕ 1 α δ, and µ = 1 ψ (1 α) ϕ α. 4

6 Linearization I Loglinearization or linearization? Loglinearization: 1. Take variable x t and substitute by xe xt where: x t = log xt x 2. A variable x t represents the log-deviation with respect to the steady state. 3. Linearize with respect to x t. Advantages and disadvantages. We can linearize and perform later a change of variables. 5

7 Linearization II We linearize: { 1 1 ( ) } = βe t 1 + αk α 1 t+1 c t c (ezt+1 l t+1 ) 1 α δ t+1 c t ψ = (1 α) kt α (e zt l t ) 1 α lt 1 1 l t c t + k t+1 = k α t (e zt l t ) 1 α + (1 δ) k t z t = ρz t 1 + ε t around l, k, and c with a First-order Taylor Expansion. 6

8 Linearization III We get: 1 c (c t c) = E t { 1 c (c t+1 c) + α (1 α) β y k z t+1+ α (α 1) β y k 2 (k t+1 k) + α (1 α) β y kl (l t+1 l) 1 c (c t c) + 1 (1 l) (l t l) = (1 α) z t + α k (k t k) α (l t l) l { ( ) } y (1 α) z t + α (c t c) + (k t+1 k) = k (k t k) + (1 α) l (l t l) + (1 δ) (k t k) z t = ρz t 1 + ε t } 7

9 Rewriting the system I α 1 (c t c) = E t {α 1 (c t+1 c) + α 2 z t+1 + α 3 (k t+1 k) + α 4 (l t+1 l)} (c t c) = α 5 z t + α k c (k t k) + α 6 (l t l) (c t c) + (k t+1 k) = α 7 z t + α 8 (k t k) + α 9 (l t l) z t = ρz t 1 + ε t α 1 = 1 c α 2 = α (1 α) β y k α 3 = α (α 1) β y k 2 α 4 = α (1 ( α) β y kl ) α α 5 = (1 α) c α 6 = l + 1 (1 l) c α 7 = (1 α) y α 8 = y α k + (1 δ) α 9 = y (1 α) l y = k α l 1 α 8

10 Rewriting the system II After some algebra the system is reduced to: A (k t+1 k) + B (k t k) + C (l t l) + Dz t = 0 E t ( G (k t+1 k) + H (k t k) + J (l t+1 l) +K (l t l) + Lz t+1 + Mz t ) = 0 E t z t+1 = ρz t We have eliminated one control: c t. This is not necessary in general: 1. Policy functions that we find. 2. Numerical differences. How do we solve this system of equations? Different yet equivalent approaches. 9

11 Undetermined coefficients We guess policy functions of the form (k t+1 k) = P 1 (k t k) + P 2 z t (l t l) = R 1 (k t k) + R 2 z t Plug them in, use linearity of expectation and E t z t+1 = ρz t to get: A (P 1 (k t k) + P 2 z t ) + B (k t k) + C (R 1 (k t k) + R 2 z t ) + Dz t = 0 G (P 1 (k t k) + P 2 z t ) + H (k t k) + J (R 1 (P 1 (k t k) + P 2 z t ) + R 2 Nz t ) +K (R 1 (k t k) + R 2 z t ) + (LN + M) z t = 0 10

12 Solving the system I Since these equations need to hold for any value (k t+1 k) or z t, we need to equate each coefficient to zero. Coefficients on (k t k): AP 1 + B + CR 1 = 0 GP 1 + H + JR 1 P 1 + KR 1 = 0 Coefficients on z t : AP 2 + CR 2 + D = 0 (G + JR 1 ) P 2 + JR 2 N + KR 2 + LN + M = 0 11

13 Solving the system II We have a system of four equations on four unknowns. To solve it, first note that R 1 = 1 C (AP 1 + B) = 1 C AP 1 1 C B Then: a quadratic equation on P 1. ( B P1 2 + A + K J GC ) KB HC P 1 + JA JA = 0 12

14 Solving the system III We have two solutions: P 1 = 1 2 B A K J + GC JA ± one stable and another unstable. ( (B A + K J GC JA ) 2 4 ) 0.5 KB HC JA If we pick the stable root and find R 1 = 1 C (AP 1 + B), we have to a system of two linear equations on two unknowns with solution: P 2 = D (JN + K) + CLN + CM AJN + AK CG CJR 1 R 2 = ALN AM + DG + DJR 1 AJN + AK CG CJR 1 13

15 Practical implementation How do we do this in practice? Solving quadratic equations: A Toolkit for Analyzing Nonlinear Dynamic Stochastic Models Easily by Harald Uhlig. Using dynare. 14

16 General structure of linearized system Given m states s t, n controls y t, and k exogenous stochastic processes z t+1, we have: As t + Bs t 1 + Cy t + Dz t = 0 E t (Fs t+1 + Gs t + Hs t 1 + Jy t+1 + Ky t + Lz t+1 + Mz t ) = 0 E t z t+1 = Nz t where C is of size l n, l n and of rank n, F is of size (m + n l) n, and that N has only stable eigenvalues. 15

17 Policy functions I We guess policy functions of the form: s t = Ps t 1 + Qz t y t = Rs t 1 + Uz t where P, Q, R, and U are matrices such that the computed equilibrium is stable. 16

18 Policy functions II For simplicity, suppose l = n (standard case, see Uhlig s chapter for the general case). Then: 1. P satisfies the matrix quadratic equation: ( F JC 1 A ) P 2 ( JC 1 B G + KC 1 A ) P KC 1 B + H = 0 The equilibrium is stable iff max (abs (eig (P))) < R is given by: R = C 1 (AP + B) 3. Q satisfies: N ( F JC 1 A ) + I k ( JR + FP + G KC 1 A ) vec (Q) = vec (( JC 1 D L ) N + KC 1 D M ) 4. U satisfies: U = C 1 (AQ + D) 17

19 How to solve quadratic equations To solve for the m m matrix P in ΨP 2 ΓP Θ = 0 1. Define the 2m 2m matrices: [ Γ Θ Ξ = I m 0 m ], and = [ Ψ 0 m I m 0 m ] 2. Let s be the generalized eigenvector and λ be the corresponding generalized eigenvalue of Ξ w.r.t.. Then, we can write s = [λx, x ] for some x R m. 3. If m generalized eigenvalues λ 1, λ 2,..., λ m with generalized eigenvectors s 1,..., s m of Ξ w.r.t., written as s = [λx i, x i ] for some x i R m and if (x 1,..., x m ) is linearly independent, then: P = ΩΛΩ 1 is a solution to the matrix quadratic equation where Ω = [x 1,..., x m ] and Λ = [λ 1,..., λ m ]. The solution of P is stable if max λ i < 1. Conversely, any diagonalizable solution P can be written in this way. 18