Linearization (Lectures on Solution Methods for Economists V: Appendix), Lecture notes of Algebra

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Linearization

(Lectures on Solution Methods for Economists V: Appendix)

Jes´us Fern´andez-Villaverde^1 and Pablo Guerr´on^2 November 21, 2021 (^1) University of Pennsylvania

(^2) Boston College

Basic RBC

• Benchmark set up: max E 0

∑^ ∞

t=

βt^ {log ct + ψ log (1 − lt )}

ct + kt+1 = k tα (ezt^ lt )^1 −α^ + (1 − δ) kt , ∀ t > 0 zt = ρzt− 1 + εt , εt ∼ N (0, σ)

• This is a dynamic optimization problem.
• The previous problem does not have a “paper and pencil” solution.
• Traditional solution: linearization.

Steady state I

• If σ = 0, the equilibrium conditions are: 1 ct^ =^ β^

ct+

1 + αkα t+1−^1 l t^1 +1−α − δ

ψ ct 1 − lt = (1 − α) kα t l− tα

ct + kt+1 = k tα l t^1 −α+ (1 − δ) kt

• The equilibrium conditions imply a steady state: 1 c = β 1 c

1 + αkα−^1 l^1 −α^ − δ

ψ c 1 − l = (1 − α) kαl−α

c + δk = kαl^1 −α

Steady state II

Solution:

k = μ Ω + ϕμ l = ϕk c = Ωk y = kαl^1 −α

where ϕ =

1 α

1 β −^ 1 +^ δ

)) (^1) −^1 α , Ω = ϕ^1 −α^ − δ, and μ = (^) ψ^1 (1 − α) ϕ−α.

Linearization II

We linearize:

1 ct^ =^ βEt

ct+

1 + αkα t+1−^1 (ezt+1^ lt+1)^1 −α^ − δ

ψ ct 1 − lt = (1 − α) k tα (ezt^ lt )^1 −α^ l t−^1

ct + kt+1 = k tα (ezt^ lt )^1 −α^ + (1 − δ) kt zt = ρzt− 1 + εt

around l, k, and c with a First-order Taylor Expansion.

Linearization III

We get:

c (ct^ −^ c) =^ Et

− (^1) c (ct+1 − c) + α (1 − α) β yk zt+1+ α (α − 1) β (^) ky 2 (kt+1 − k) + α (1 − α) β (^) kly (lt+1 − l)

c (ct − c) + 1 (1 − l) (lt − l) = (1 − α) zt + α k (kt − k) − α l (lt − l)

(ct − c) + (kt+1 − k) =

y

(1 − α) zt + α k (kt − k) + (1− l α)(lt − l)

• (1 − δ) (kt − k)

zt = ρzt− 1 + εt

Rewriting the system II

• After some algebra the system is reduced to: A (kt+1 − k) + B (kt − k) + C (lt − l) + Dzt = 0

Et

G (kt+1 − k) + H (kt − k) + J (lt+1 − l) +K (lt − l) + Lzt+1 + Mzt

Et zt+1 = ρzt

• We have eliminated one control: ct. 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.

Undetermined coefficients

• We guess policy functions of the form (kt+1 − k) = P 1 (kt − k) + P 2 zt (lt − l) = R 1 (kt − k) + R 2 zt
• Plug them in, use linearity of expectation and Et zt+1 = ρzt to get: A (P 1 (kt − k) + P 2 zt ) + B (kt − k) + C (R 1 (kt − k) + R 2 zt ) + Dzt = 0

G (P 1 (kt − k) + P 2 zt ) + H (kt − k) + J (R 1 (P 1 (kt − k) + P 2 zt ) + R 2 Nzt )

+K (R 1 (kt − k) + R 2 zt ) + (LN + M) zt = 0

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) = − (^) C^1 AP 1 − (^) C^1 B
• Then: P 12 +

B

A

K

J

GC

JA

P 1 +

KB − HC

JA

a quadratic equation on P 1.

Solving the system III

• We have two solutions:

P 1 = −

− B

A −^

K

J +^

GC

JA ±

B

A +^

K

J −^

GC

JA

KB − HC

JA

one stable and another unstable.

• If we pick the stable root and find R 1 = − (^) C^1 (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

General structure of linearized system

Given m states st , n controls yt , and k exogenous stochastic processes zt+1, we have:

Ast + Bst− 1 + Cyt + Dzt = 0 Et (Fst+1 + Gst + Hst− 1 + Jyt+1 + Kyt + Lzt+1 + Mzt ) = 0 Et zt+1 = Nzt

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.

Policy functions I

We guess policy functions of the form:

st = Pst− 1 + Qzt yt = Rst− 1 + Uzt

where P, Q, R, and U are matrices such that the computed equilibrium is stable.

How to solve quadratic equations

To solve for the m × m matrix P in ΨP^2 − ΓP − Θ = 0

1. Define the 2m × 2 m matrices:

Ξ =

[

Im 0 m

]

, and ∆ =

[

Ψ (^0) m (^0) m Im

]

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 ∈ Rm.
3. If ∃ m generalized eigenvalues λ 1 , λ 2 , ..., λm with generalized eigenvectors s 1 , ..., sm of Ξ w.r.t. ∆, written as s′^ = [λx i′ , x i′ ] for some xi ∈ Rm^ and if (x 1 , ..., xm) is linearly independent, then: P = ΩΛΩ−^1 is a solution to the matrix quadratic equation where Ω = [x 1 , ..., xm] and Λ = [λ 1 , ..., λm]. The solution of P is stable if max |λi | < 1. Conversely, any diagonalizable solution P can be written in this way. (^18)