... Solving and Estimating Dynamic General Equilibrium Models Using Log Linear Approximation

... Solving and Estimating Dynamic General Equilibrium Models Using Log Linear Approximation 1

Overview To Estimate and Analyze Dynamic General Equilibrium Models, Need Solution. Will Review a Linearization Strategy for Doing This. 2

Model Solution Methods 1. Example #1: A Simple RBC Model. Define a Model Solution Motivate the Need to Somehow Approximate Model Solutions Describe Basic Idea Behind Linear Approximations Some Strange Examples to be Prepared For 2. Example #2: Putting the Stochastic RBC Model into General Canonical Form 3. Example #3: Stochastic RBC Model with Hours Worked (Matrix Generalization of Previous Results) 4. Example #4: Example #3 with Exotic Information Sets. 5. Summary so Far. 6. Example #5: Sticky price model with no capital - log linearizing about flexible price equilibrium. 7. Computing Impulse Response Functions. 9

Example #1: Nonstochastic RBC Model subject to: Maximize {ct,k t+1 } X t=0 β t Ct 1 σ 1 σ, First order condition: C t + K t+1 (1 δ)k t = K α t,k 0 given C σ t βct+1 σ αk α 1 t+1 +(1 δ), or, after substituting out resource constraint: v(k t,k t+1,k t+2 )=0,t=0, 1,..., with K 0 given. 10

Example #1: Nonstochastic RBC Model... Solution : a function, K t+1 = g(k t ), such that v(k t,g(k t ),g[g(k t )]) = 0, for all K t. Problem: This is an Infinite Number of Equations (one for each possible K t ) in an Infinite Number of Unknowns (a value for g for each possible K t ) With Only a Few Rare Exceptions this is Very Hard to Solve Exactly Easy cases: If σ =1,δ=1 g(k t )=αβk α t. If v is linear in K t,k t+1,k t+1. Standard Approach: Approximate v by a Linear Function. 14

Approximation Method Based on Linearization Three Steps Compute the Steady State Do a Linear Expansion About Steady State Solve the Resulting Linearized System Step 1: Compute Steady State - Steady State Value of K, K - C σ βc σ αk α 1 +(1 δ) =0, αk α 1 +(1 δ) = 1 β " K α = (1 δ) 1 β # 1 1 α. K satisfies: v(k,k,k )=0. 16

Approximation Method Based on Linearization... Step 2: Replace v by First Order Taylor Series Expansion About Steady State: v 1 (K t K )+v 2 (K t+1 K )+v 3 (K t+2 K )=0, Here, v 1 = dv u(k t,k t+1,k t+2 ) dk t, at K t = K t+1 = K t+2 = K. Conventionally, do Log-Linear Approximation: (v 1 K) ˆK t +(v 2 K) ˆK t+1 +(v 3 K) ˆK t+2 =0, ˆK t K t K K. Write this as: α 2 ˆKt + α 1 ˆKt+1 + α 0 ˆKt+2 =0, α 2 = v 1 K, α 1 = v 2 K, α 0 = v 3 K 20

Approximation Method Based on Linearization... Step 3: Solve Posit the Following Policy Rule: ˆK t+1 = A ˆK t, Where A is to be Determined. Compute A : α 2 ˆKt + α 1 A ˆK t + α 0 A 2 ˆKt =0, or α 2 + α 1 A + α 0 A 2 =0. A is the Eigenvalue of Polynomial In General: Two Eigenvalues. Can Show: In RBC Example, One Eigenvalue is Explosive. The Other Not. There Exist Theorems (see Stokey-Lucas, chap. 6) That Say You Should Ignore the Explosive A. 23

Some Strange Examples to be Prepared For Other Examples Are Possible: Both Eigenvalues Explosive Both Eigenalues Non-Explosive What Do These Things Mean? 25

Some Strange Examples to be Prepared For... Example With Two Explosive Eigenvalues Preferences: X t=0 β t C γ t γ,γ <1. Technology: Production of Consumption Goods C t = k α t n 1 α t Production of Capital Goods k t+1 =1 n t. 26

Some Strange Examples to be Prepared For... Planning Problem: Euler Equation: max X t=0 β t hk α t (1 k t+1 ) 1 αi γ γ v(k t,k t+1,k t+2 )= (1 α)k αγ t (1 k t+1 ) [(1 α)γ 1] + βαk (αγ 1) t+1 (1 k t+2 ) (1 α)γ =0, t =0, 1,... Steady State: k = αβ 1 α + αβ. 27

Some Strange Examples to be Prepared For... Log-linearize Euler Equation: α 0ˆkt+2 + α 1ˆkt+1 + α 2ˆkt =0 With β =0.58, γ =0.99, α =0.6, Both Roots of Euler Equation are explosive: 1.6734, 1.0303 Other Properties: Steady State: 0.4652 Two-Period Cycle: 0.8882, 0.0870 28

Some Strange Examples to be Prepared For... Meaning of Stokey-Lucas Example Illustrates the Possibility of All Explosive Roots Economics: If Somehow You Start At Single Steady State, Stay There If You are Away from Single Steady State, Go Somewhere Else If Linearized Euler Equation Around Particular Steady State Has Only Explosive Roots All Possible Equilibria Involve Leaving that Steady State Linear Approximation Not Useful, Since it Ceases to be Valid Outside a Neighborhood of Steady State Could Linearize About Two-Period Cycle (That s Another Story...) The Example Suggests That Maybe All Explosive Root Case is Unlikely 34

Some Strange Examples to be Prepared For... Another Possibility: Both Roots Stable Many Paths Converge Into Steady State: Multiple Equilibria Can Happen For Many Reasons Strategic Complementarities Among Different Agents In Private Economy Certain Types of Government Policy This is a More Likely Possibility Avoid Being Surprised by It By Always Thinking Through Economics of Model. 35

Example #2: RBC Model With Uncertainty Model subject to Maximize E 0 X t=0 β t Ct 1 σ 1 σ, C t + K t+1 (1 δ)k t = K α t ε t, where ε t is a stochastic process with Eε t = ε, say. Let and suppose ˆε t = ε t ε, ε ˆε t = ρˆε t 1 + e t,e t N(0,σ 2 e). First Order Condition: E t C σ t βct+1 σ αk α 1 t+1 ε t+1 +1 δ ª =0. 36

Example #2: RBC Model With Uncertainty... First Order Condition: E t v(k t+2,k t+1,k t,ε t+1,ε t )=0, where v(k t+2,k t+1,k t,ε t+1,ε t ) =(K α t ε t +(1 δ)k t K t+1 ) σ β (K α t+1ε t+1 +(1 δ)k t+1 K t+2 ) σ αk α 1 t+1 ε t+1 +1 δ. Solution: a g(k t,ε t ), Such That E t v (g(g(k t,ε t ),ε t+1 ),g(k t,ε t ),K t,ε t+1,ε t )=0, For All K t,ε t. Hard to Find g, Except in Special Cases One Special Case: v is Linear. 38

Example #2: RBC Model With Uncertainty... Linearization Strategy: Step 1: Compute Steady State of K t when ε t is Replaced by Eε t Step2: Replace v By its Taylor Series Expansion About Steady State. Step 3: Solve Resulting Linearized System. Logic: If Actual Stochastic System Remains in a Neighborhood of Steady State, Linear Approximation Good 39

Example #2: RBC Model With Uncertainty... Step 1: Steady State: Step 2: Linearize - K = " 1 β αε (1 δ) # 1 1 α. v(k t+2,k t+1,k t,ε t+1,ε t ) ' v 1 (K t+2 K )+v 2 (K t+1 K )+v 3 (K t K ) +v 3 (ε t+1 ε)+v 4 (ε t ε) µ = v 1 K Kt+2 K µ + v K 2 K Kt+1 K µ + v K 3 K Kt K µ µ K εt+1 ε εt ε +v 3 ε + v 4 ε ε ε = α 0 ˆKt+2 + α 1 ˆKt+1 + α 2 ˆKt + β 0ˆε t+1 + β 1ˆε t. 40

Example #2: RBC Model With Uncertainty... Step 3: Solve Linearized System Posit: ˆK t+1 = A ˆK t + Bˆε t. Pin Down A and B By Condition that log-linearized Euler Equation Must Be Satisfied. Note: ˆK t+2 = A ˆK t+1 + Bˆε t+1 = A 2 ˆKt + ABˆε t + Bρˆε t + Be t+1. Substitute Posited Policy Rule into Linearized Euler Equation: h i E t α 0 ˆKt+2 + α 1 ˆKt+1 + α 2 ˆKt + β 0ˆε t+1 + β 1ˆε t =0, so must have: E t {α 0 h A 2 ˆKt + ABˆε t + Bρˆε t + Be t+1 i +α 1 h A ˆK t + Bˆε t i + α 2 ˆKt + β 0 ρˆε t + β 0 e t+1 + β 1ˆε t } =0 41

Example #2: RBC Model With Uncertainty... Then, h i E t α 0 ˆKt+2 + α 1 ˆKt+1 + α 2 ˆKt + β 0ˆε t+1 + β 1ˆε t h i = E t {α 0 A 2 ˆKt + ABˆε t + Bρˆε t + Be t+1 h +α 1 A ˆK i t + Bˆε t + α 2 ˆKt + β 0 ρˆε t + β 0 e t+1 + β 1ˆε t } where = α(a) ˆK t + F ˆε t =0 α(a) =α 0 A 2 + α 1 A + α 2, F = α 0 AB + α 0 Bρ + α 1 B + β 0 ρ + β 1 Find A and B that Satisfy: α(a) =0,F=0. 42

Example #3 RBC Model With Hours Worked and Uncertainty Maximize subject to E t X t=0 β t U(C t,n t ) C t + K t+1 (1 δ)k t = f(k t,n t,ε t ) and Eε t = ε, ˆε t = ρˆε t 1 + e t,e t N(0,σ 2 e) ˆε t = ε t ε. ε 43

Example #3 RBC Model With Hours Worked and Uncertainty... First Order Conditions: E t v K (K t+2,n t+1, K t+1,n t,k t,ε t+1,ε t )=0 and v N (K t+1,n t,k t,ε t )=0. where v K (K t+2,n t+1, K t+1,n t,k t,ε t+1,ε t ) = U c (f(k t,n t,ε t )+(1 δ)k t K t+1,n t ) βu c (f(k t+1,n t+1,ε t+1 )+(1 δ)k t+1 K t+2,n t+1 ) [f K (K t+1,n t+1,ε t+1 )+1 δ] and, v N (K t+1,n t,k t,ε t ) = U N (f(k t,n t,ε t )+(1 δ)k t K t+1,n t ) +U c (f(k t,n t,ε t )+(1 δ)k t K t+1,n t ) f N (K t,n t,ε t ). Steady state K and N such that Equilibrium Conditions Hold with ε t ε. 46

Example #3 RBC Model With Hours Worked and Uncertainty... Log-Linearize the Equilibrium Conditions: v K (K t+2,n t+1, K t+1,n t,k t,ε t+1,ε t ) = v K,1 K ˆKt+2 + v K,2 N ˆNt+1 + v K,3 K ˆKt+1 + v K,4 N ˆNt + v K,5 K ˆKt +v K,6 εˆε t+1 + v K,7 εˆε t v K,j Derivative of v K with respect to j th argument, evaluated in steady state. v N (K t+1,n t,k t,ε t ) = v N,1 K ˆKt+1 + v N,2 N ˆNt + v N,3 K ˆKt + v N,4 εˆε t+1 v N,j Derivative of v N with respect to j th argument, evaluated in steady state. 48

Example #3 RBC Model With Hours Worked and Uncertainty... Representation Log-linearized Equilibrium Conditions Let µ ˆKt+1 z t =,s t =ˆε t, t = e t. ˆN t Then, the linearized Euler equation is: E t [α 0 z t+1 + α 1 z t + α 2 z t 1 + β 0 s t+1 + β 1 s t ]=0, s t = Ps t 1 + t, t N(0,σ 2 e), P= ρ. Here, vk,1 K α 0 = v K,2 N vk,3 K,α 0 0 1 = v K,4 N v N,1 K v N,2 N, vk,5 K α 2 = 0 v N,3 K, 0 µ µ vk,6 ε vk,7 ε β 0 =,β 0 1 =. v N,4 ε Previous is a Canonical Representation That Essentially All Linearized Models Can be Fit Into (See Christiano (2002).) 52

Example #3 RBC Model With Hours Worked and Uncertainty... Again, Look for Solution z t = Az t 1 + Bs t, where A and B are pinned down by log-linearized Equilibrium Conditions. Now, A is Matrix Eigenvalue of Matrix Polynomial: α(a) =α 0 A 2 + α 1 A + α 2 I =0. Also, B Satisfies Same System of Linear Equations as Before: F =(β 0 + α 0 B)P +[β 1 +(α 0 A + α 1 )B] =0. Go for the 2 Free Elements of B Using 2 Equations Given by 0 F =. 0 53

Example #3 RBC Model With Hours Worked and Uncertainty... Finding the Matrix Eigenvalue of the Polynomial Equation, α(a) =0, and Determining if A is Unique is a Solved Problem. See Anderson, Gary S. and George Moore, 1985, A Linear Algebraic Procedure for Solving Linear Perfect Foresight Models, Economic Letters, 17, 247-52 or Articles in Computational Economics, October, 2002. See also, the program, DYNARE. 54

Example #3 RBC Model With Hours Worked and Uncertainty... Solving for B Given A, Solve for B Using Following (Linear) System of Equations: F =(β 0 + α 0 B)P +[β 1 +(α 0 A + α 1 )B] =0 To See this, Use vec(a 1 A 2 A 3 )=(A 0 3 A 1 ) vec(a 2 ), to Convert F =0Into vec(f 0 )=d + qδ =0,δ= vec(b 0 ). Find B By First Solving: δ = q 1 d. 57

Example #4: Example #3 With Exotic Information Set Suppose the Date t Investment Decision is Made Before the Current Realization of the Technology Shock, While the Hours Decision is Made Afterward. Now, Canonical Form Must Be Written Differently: E t [α 0 z t+1 + α 1 z t + α 2 z t 1 + β 0 s t+1 + β 1 s t ]=0, where E [X1t ˆε E t X t = t 1 ]. E [X 2t ˆε t ] Convenient to Change s t : Adjust β i s: s t = β 0 = µ ˆεt ˆε t 1,P= µ vk,6 ε 0 0 0 ρ 0 1 0,β 1 =, t = µ vk,7 ε 0 v N,4 ε 0 µ et 0,. 58

Example #4: Example #3 With Exotic Information Set... Posit Following Solution: z t = Az t 1 + Bs t. Substitute Into Canonical Form: E t [α 0 z t+1 + α 1 z t + α 2 z t 1 + β 0 s t+1 + β 1 s t ] = α(a)z t 1 + E t Fs t + E t β 0 t+1 = α(a)z t 1 + E t Fs t =0, Then, F11 F E t Fs t = E 12 F11ˆε t s F 21 F t = E t + F 12ˆε t 1 t 22 F 21ˆε t + F 22ˆε t 1 0 F12 + ρf = 11 s F 21 F t = Fs t. 22 Equations to be solved: α(a) =0, F =0. F Only Has Three Equations How Can We Solve for the Four Elements of B? Answer: Only Three Unknowns in B Because B Must Also Obey Information Structure: 0 B12 B =. B 21 B 22 60

Summary so Far Solving Models By Linear Approximation Involves Three Steps: a. Compute Steady State b. Log-Linearize Equilibrium Conditions c. Solve Linearized Equations. Step 3 Requires Finding A and B in: z t = Az t 1 + Bs t, to Satisfy Log-Linearized Equilibrium Conditions: E t [α 0 z t+1 + α 1 z t + α 2 z t 1 + β 0 s t+1 + β 1 s t ] s t = Ps t 1 + t, t iid We are Led to Choose A and B so that: α(a) =0, (standard information set) F =0, (exotic information set) F =0 and Eigenvalues of A are Less Than Unity In Absolute Value. 63

Example #5: A Sticky Price Model (Gali, Lopez-Salido, Valles) Technology grows forever: model has no steady state. Absence of capital implies flexible price ( natural ) equilibrium can be computed analytically. Equilibrium quantities determined by static labor market conditions ( MRS = MPL ) Interest rate determined by requirement that people have no incentive to smooth away equilibrium fluctuations in consumption. Model is approximately log-linear around flexible price equilibrium allocations. Exploit this to apply standard linearization strategy. 64

Example #5: A Sticky Price Model (Gali, Lopez-Salido, Valles)... Model: Households choose consumption and labor. Monopolistic firms produce and sell output using labor, subject to sticky prices Monetary authority obeys a Taylor rule. 65

Preferences: E 0 X t=0 Household efficiency conditions: Household Ã! β t log C t exp (τ t ) N 1+ϕ t,τ t = λτ t 1 + ε τ t. 1+ϕ C 1 t = βe t C 1 t+1 R t/ π t+1,mrs t =exp(τ t ) N ϕ t C t = W t P t. Take logs of intertemporal Euler equation: c t = logβ + r t +log E t Ct+1 1 / π t+1 = logβ + r t +log[e t exp ( c t+1 π t+1 )] ' log β + r t +log[exp( E t c t+1 E t π t+1 )] = logβ + r t E t c t+1 E t π t+1 Note the approximation that was used here! 66

Household... Household efficiency conditions (repeated): C 1 t = βe t C 1 t+1 R t/ π t+1,mrs t =exp(τ t ) N ϕ t C t = W t P t. Log-linear approximation of intertemporal Euler equation: c t = [r t E t π t+1 rr]+e t c t+1 rr log β, c t log C t,π t log π t,r t log R t Log of intratemporal Euler equation: w t p t = c t + ϕn t + τ t (= log MRS t ) 67

Firms Final Good Firms: Buy intermediate goods Y t (i),i [0, 1] at prices P t (i) and sell Y t at price P t Technology: Y t = µz 1 0 Y t (i) ε 1 ε ε ε 1 di,ε 1. Demand for intermediate good (fonc for optimization of Y t (i)): µ ε Pt (i) Y t (i) =Y t P t 68

Firms... Intermediate Good Firms Technology: Y t (i) =A t N t (i), a t =loga t, a t = ρ a t 1 + ε t. Optimal price setting under flexible prices (constant markup over marginal cost, MC t ): P t (i) = ε ε 1 MC t,mc t = W t A t Implication: under flexible prices P t (i) =P t (j),alli, j, so P t (i) =P t. Then, if mc t log MC t P t, under flexible prices: mc t = w t p t a t = log ε ε 1. 69

Flexible Price Equilibrium ( Natural Equilibrium) Employment in all sectors the same N t (i) =N t all i this is because all firms set the same price Aggregate resource constraint: y t = a t + n t,y t =logy t Goods market clearing y t = c t. 70

Flexible Price Equilibrium ( Natural Equilibrium)... Combine household and firm first order conditions for labor: ε mc t = c t + ϕn t + τ t a t = log. ε 1 Using resource constraint and market clearing: ε y t + ϕ (y t a t )+τ t a t = log, ε 1 Natural rate of output, yt : yt = γ + a t 1 1+ϕ τ t, γ = log ε ε 1 1+ϕ. Natural rate of employment: n t = y t a t = γ 1 1+ϕ τ t Note: income and substitution effects of a t on n t cancel! 72

Flexible Price Equilibrium ( Natural Equilibrium)... Now, seek interest rate needed to reconcile households with intertemporal consumption pattern implied by efficient allocations Intertemporal Euler equation: where rr t is the real interest rate: y t = [rr t rr]+e t y t+1, rr t = r t E t π t+1. To determine natural real interest rate, rrt, substitute in natural output, yt : yt z } yt+1 { γ + a t 1 µ z } { 1+ϕ τ t = [rrt rr]+e t γ + a t+1 1 1+ϕ τ t+1 or, rr t = rr + ρ a t + 1 1+ϕ (1 λ) τ t. 73

Flexible Price Equilibrium ( Natural Equilibrium)... Natural rate: rr t = rr + ρ a t + 1 1+ϕ (1 λ) τ t. a t jumps a t will keep rising in future (if ρ>0) rise in c t smaller than rise in c t+1 people would like to use financial markets to smooth away from this discourage this by having a high interest rate. τ t jumps τ t will be less high in the future (unless λ>1) c t falls more than c t people want to smooth away discourage this by having a high interest rate. 75

Sticky Price Equilibrium A fraction, 1 θ, of firms permitted to optimize their prices. A fraction, θ, not permitted. Prices of different intermediate good firms will be different The price level now will not be a fixed markup over marginal cost real marginal cost: mc t = c t + ϕn t + τ t a t. output will deviate from natural rate (output gap, x t, will not be zero): x t = y t yt Write deviation of marginal cost from its natural rate : cmc t = mc t mc =(1+ϕ) y t (ϕ +1)a t + τ t (1 + ϕ) yt +(ϕ +1)a t τ t =(1+ϕ) x t. Calvo reduced form inflation equation: (1 θ)(1 βθ)(1+ϕ) π t = βe t π t+1 + κx t,κ=. θ 76

Sticky Price Equilibrium... Intertemporal equation: y t = [r t E t π t+1 rr]+e t y t+1 y t = [rr t rr]+e t y t+1 Subtract: x t = [r t E t π t+1 rr t ]+E t x t+1 We now have two equations ( IS curve and Phillips curve ) in three unknowns: π t,r t,x t. Need another equation! 78

Monetary Authority Target interest rate: ˆr t = φ π π t + φ x x t Actual interest rate choice when there is some inertia: r t = αr t 1 +(1 α)ˆr t + u t u t = δu t 1 + η t. Writing out policy rule: r t = αr t 1 + u t +(1 α)φ π π t +(1 α)φ x x t 79

Collection Equations of Equilibrium βe t π t+1 + κx t π t =0(Calvo pricing equation) [r t E t π t+1 rr t ]+E t x t+1 x t =0(intertemporal equation) αr t 1 + u t +(1 α)φ π π t +(1 α)φ x x t r t =0(policy rule) rrt ρ a t 1 1+ϕ (1 λ) τ t =0(definition of natural rate) Note: Preference and technology shocks enter system through rrt Flexible price equilibrium with π t =0can be supported by setting r t = rrt. Practical issue: how to measure rrt??? 80

Solving the Sticky Price Model Exogenous shocks: s t = a t u t = ρ 0 0 0 δ 0 a t 1 u t 1 + ε t η t τ t 0 0 λ τ t 1 ε τ t s t = Ps t 1 + t Equilibrium conditions: β 000 π t+1 1 σ 100 x t+1 0000 r t+1 0000 + 0000 0000 00α 0 0000 π t 1 x t 1 r t 1 rr t 1 1 κ 0 0 + 0 1 1 1 σ σ (1 α)φ π (1 α)φ x 1 0 0 0 0 1 000 0 0 0 + 000 000 s t+1 + 0 0 0 0 1 0 000 σψρ 0 1 σ+ϕ rr t+1 E t [α 0 z t+1 + α 1 z t + α 2 z t 1 + β 0 s t+1 + β 1 s t ]=0 π t x t r t rr t (1 λ) s t 81

Solving the Sticky Price Model... Collecting: E t [α 0 z t+1 + α 1 z t + α 2 z t 1 + β 0 s t+1 + β 1 s t ]=0 Solution: As before, want A such that z t = Az t 1 + Bs t s t Ps t 1 t =0. α 0 A 2 + α 1 A + α 2 =0, Want B such that: F =(β 0 + α 0 B)P +[β 1 +(α 0 A + α 1 )B] =0 Note: if α =0,A=0. 83

Examples with Sticky Price Model Baseline parameters: φ x =0,φ π =1.5, β=0.99, ϕ=1,ρ=0.2, θ=0.75, α=0,δ=0.2, λ=0.5. 0.03 0.02 Dynamic Response to a Technology Shock inflation output gap 0.2 0.15 0.15 0.1 0.1 nominal rate natural nominal rate actual nominal rate 0.01 0.05 0.05 0 2 4 6 natural real rate 0.2 0.15 0.1 0.05 0 2 4 6 employment 0.2 0.15 0.1 0.05 0 0 5 10 1 0.5 0 2 4 6 log, technology 0 0 2 4 6 natural employment actual employment 1.2 1.15 1.1 0 2 4 6 output natural output actual output 1.05 1 0 2 4 6 Under policy rule, interest rate not increased enough, so consumption and employment increase and inflationisdrivenup. 84

Examples with Sticky Price Model... Dynamic Response to a Preference Shock 0.1 0.08 0.06 0.04 0.02 inflation 0.25 0.2 0.15 0.1 0.05 output gap 0.25 0.2 0.15 0.1 0.05 nominal rate natural real rate actual nominal rate 0 2 4 6 natural real rate 0.25 0.2 0.15 0.1 0.05 0 2 4 6 employment 0 2 4 6 preference shock 1 0.5 0 0 2 4 6 0.1 0.2 0.3 0.4 0 2 4 6 output natural output actual output 0.5 0 2 4 6 0.1 0.2 0.3 0.4 0.5 0 2 4 6 natural employment actual employment Under policy rule, interest rate not increased enough. This encourages consumption above what is needed for the zero-inflation equilibrium. The extra demand drives up output gap, inflation 85

Examples with Sticky Price Model... Dynamic Response to a Monetary Policy Shock inflation output gap nominal rate 0.05 0.1 0.15 0 2 4 6 natural real rate 1 0.5 0 0.5 1 0 2 4 6 employment 0 0.2 0.4 0.6 0.8 0 2 4 6 0.2 0.4 0.6 0.8 0 2 4 6 monetary policy shock 1 0.5 0 0 2 4 6 natural employment actual employment 0.6 0.4 0.2 0 0 2 4 6 output 0 0.2 0.4 0.6 0.8 0 2 4 6 natural real rate actual nominal rate natural output actual output Monetary policy shock drives up the interest rate High interest rate discourages current consumption Output, output gap and employment fall Fall in costs causes inflationtodrop. 86

Estimation of Model Parameters Limited information methods Matching model impulse response functions with VAR model response functions Rotemberg and Woodford, Christiano, Eichenbaum and Evans, Del Negro, Schorfheide and others. Full information methods Maximum likelihood Bayesian maximum likelihood 87

Computing Impulse Response Functions For Model Impulse Response Function (IRF): Suppose System is in Steady State IRF Is Response of System to a One-Time Innovation in Exogenous Variables, Relative to What Trajectory Would have Been, Absent a Shock. One-Time Shock: 1 6=0, t =0for t>1. Simulate Exogenous Variables: s t = Ps t 1 + t,t=1, 2,...,T, s 0 =0. Simulate Endogenous Variables: z t = Az t 1 + Bs t,t=1, 2,..., T, z 0 =0. Note: z t = µ Kt+1 K K N t N N Which are Percentage Deviations From Unshocked, Steady State Path., 94

Computing Impulse Response Functions For Model... Other Endogenous Variables Not Included in z t Suppose We Also Would Like Output, y t so that y t = ε t K α t N 1 α t, ŷ t =ˆε t + α ˆK t +(1 α) ˆN t, a Linear Function of z t (where ˆN t is), z t 1 (where ˆK t is) and s t. CanObtainOtherVariables,SayY t, In the Same Way: Y t = α z z t +ᾱ z z t 1 + α s s t. Use This Expression and Previous Results to Simulate the Impulse Response to Y t 95

A Little Technical Note We have defined ˆx t = x t x x, where x ~ steady state. Then, So, x t x =ˆx t +1 log x t x = log(ˆx t +1)' ˆx t log x t ' log x +ˆx t 96

Suppose we have: Maximum Likelihood Estimation data, Yt data,t=1,..., T. For example, Y data t = µ log yt log N t,, where y t is (detrended) output, N t is employment. Next, we obtain a compact summary of the model and its implications for the observed data. The compact summary is called, in Kalman filter language, the state-space, observer representation of the system. 97

Maximum Likelihood Estimation... Model: solution: z t = Az t 1 + Bs t,z t = µ ˆkt ˆN t law of motion, shocks: s t = Ps t 1 + t Model implication for observables: Y t = = µ µ log yt log y = log N t log N µ log y + log N + µ ŷt ˆN t µ αz z t +ᾱ z z t 1 + α s s t z 2,t = a 0 + a 1 z t + a 2 z t 1 + b 0 s t. Here a 0 = µ log y log N,a 1 = α z 01,a 2 = ᾱ z 00,b 0 = αs 0. 99

Maximum Likelihood Estimation... Kalman filter language: The state, ξ t : what is required to compute the model implication for observables ξ t = z t z t 1. s t Law of motion for the state: z t+1 z t = A 0 BP I 0 0 z t z t 1 + B 0 t+1, s t+1 0 0 P s t I or ξ t+1 = Fξ t + D t+1,e t+1 0 t+1 = I Q = BDD0 B 0 0 BDD 0 0 0 0 DD 0 B 0 DD 0,F= A 0 BP I 0 0 0 0 P. 100

Maximum Likelihood Estimation... Observer equation: Y data t = a 0 + Hξ t + w t,ew t w 0 t = R, H = a 1 a 2 b 0 b 1 Here, w t denotes iid measurement error (could be autocorrelated, see Anderson, Hansen, McGrattan, Sargent (1996)). 101

Maximum Likelihood Estimation... State space-observer system: ξ t+1 = Fξ t + D t+1 Y data t = a 0 + Hξ t + w t,ew t w 0 t = R Parameters of system: (F, D, a 0,H,R). These are functions of model parameters, γ. In Hamilton s textbook, formulas for computing likelihood are provided. L Y data 1,...,Y data T ; γ. 102

Bayesian Maximum Likelihood Place a prior on each of the N parameters, γ, p i (γ i ),i=1,..., N. Typically, priors across different parameters independent: p (γ) = NY p i (γ i ). i=1 Example, p(γ 3 ) could be normal density function centered on γ 3 =6, with variance σ 2 =3. Bayesian maximum likelihood: max γ L Y data 1,..., Y data T ; γ + NX log [p i (γ i )] i=1 103

Bayesian Maximum Likelihood... By-product of formulas to compute likelihood: ˆξ t T E [ξ t Ω T ],t=1, 2,..., T, where Ω T = {Y 1,...,Y T }. So, can compute estimate of shocks, s t Simulation of model in response to estimated shocks is provided by mod el Yt = a 0 + Hˆξ t T. Can use these and estimated shocks to obtain visual assessment of model fitto data. 104

Figure 4a: EA, Actual (solid line) and Fitted (dotted line) Data 50 0-50 -100 Growth, Real Net Worth (annual, percent) GDP Growth (annual, percent) 5 0-5 15 10 5 0-5 -0.5 M 1 Growth (annual, percent) Risk Premium (annual rate, basis points) 1 0.5 0 Growth, Relative Price of Invest. (annual, %) 2 0-2 1985 1990 1995 2000 2005 4 2 0 5 0-5 Inflation (APR) Real Wage Growth (annual, percent) 4 2 0-2 -4 M 3 Growth (annual, percent) Re (annual rate, basis points) 8 6 4 2 0 100-100 Growth, Real Price of Oil (annual, percent) 0-200 1985 1990 1995 2000 2005 10 0-10 10 0-10 Log, Hours Investment Growth (annual, percent) 20 Consumption Growth (annual, percent) 4-2 02-4 -6-8 Government Consump. Growth (annual, percent) 4 2 0-2 -4-6 1985 1990 1995 2000 2005 1 Notes: (i) Model fitted values, produced by two-sided Kalman smoothing. 0 0

Figure 5a: EA, Estimated Economic Shocks %dev. from s.s. 10 0-10 Firm Markup, λ f,t Level 6 4 2 Inflation Objective (APR) Level 300 200 100 Banking Technology Shock Level Investment Specific Shock, µ U,t 1.02 1 0.98 Level 0.9 0.8 0.7 0.6 0.5 Money Demand Shock, χ % dev. from s.s. 1 0-1 Government Consumption, g t Level Persistent Component of Technology 1.015 1.02 1.01 1 1.005 0.98 1 0.96 0.995 0.94 Level Financial Wealth Shock Level 1.02 1 0.98 Technology Shock, ε t Level Monetary Policy (basis points) 50 0-50 Level Riskiness of Entrepreneurs, σ t 0.5 0.4 0.3 1985 1990 1995 2000 2005 Ratio to s.s. 1.4 1.2 Consumption Preference, ζ c,t 1 1985 1990 1995 2000 2005 Ratio to s.s. Marginal Investment Efficiency, ζ i,t 1.4 1.2 1 0.8 1985 1990 1995 2000 2005 Ratio to s.s. 2 1.5 1 Energy shock, τ oil t 0.5 1985 1990 1995 2000 2005

Fernandez-Villaverde, Rubio-Ramirez, Sargent Result Use the state space, observer representation to derive DSGE implication for VAR. System (ignoring constant terms and measurement error): ξ t = Fξ t 1 + D t, Substituting: Y t = Hξ t. Y t = HFξ t 1 + HD t Suppose HD is square and invertible. Then t =(HD) 1 Y t (HD) 1 HFξ t 1. 105

Fernandez-Villaverde, Rubio-Ramirez, Sargent Result... Substitute latter into the state equation: ξ t = Fξ t 1 + D (HD) 1 Y t D (HD) 1 HFξ t 1 h i = I D (HD) 1 H Fξ t 1 + D (HD) 1 Y t, so ξ t = Mξ t 1 + D (HD) 1 Y t,m= h i I D (HD) 1 H F. If eigenvalues of M are less than unity, ξ t = D (HD) 1 Y t + MD(HD) 1 Y t 1 + M 2 D (HD) 1 Y t 2 +... Then, t =(HD) 1 Y t (HD) 1 HF h i D (HD) 1 Y t 1 + MD(HD) 1 Y t 2 + M 2 D (HD) 1 Y t 3 +... 106

Fernandez-Villaverde, Rubio-Ramirez, Sargent Result... or, where Y t = B 1 Y t 1 + B 2 Y t 2 +... + u t, u t = HD t B j = HFM j 1 D (HD) 1,j=1, 2,... The latter is the VAR representation. Note: t is invertible because it lies in space of current and past Y t s Note: VAR is infinite-ordered. 107