Dynamic Identification of DSGE Models

Transcription

1 Dynamic Identification of DSGE Models Ivana Komunjer and Serena Ng UCSD and Columbia University All UC Conference 2009 Riverside

2 1 Why is the Problem Non-standard? 2 Setup Model Observables 3 Identification Analysis Observational Equivalence Rank and Order Conditions 4 Illustration Example 1: An and Schorfheide (2007) Example 2: Stochastic Growth Model 5 Conclusion

3 Why is the Problem Non-standard? Motivation In estimation of DSGE models, some parameters are always fixed How many (if any) should we fix? Does fixing the parameters guarantee identification of the remaining parameters? Why is the Problem Non-standard? Models are dynamic with non-iid shocks. Likelihood may not exist because of stochastic singularity. Reduced form parameters are generally not identified.

4 Why is the Problem Non-standard? Literature 1 multivariate ARMA models Hannan (1971), Zellner & Palm (1974), Hatanaka (1975), Wallis (1977) rule out stochastically singular models 2 classical rank conditions Fisher (1966), Rothenberg (1971), Iskrev (2009) require the reduced form parameters to be identified 3 objective function approaches: likelihood or GMM Rothenberg (1971), Iskrev (2007, 2009), Canova & Sala (2009) depend on the population moments of the data 4 approaches based on observational equivalence linear simultaneous equations systems: Cowles (1950, 1953), linear SVARs: Rubio-Ramirez, Waggoner & Zha (2009) nonlinear models: this paper

5 Why is the Problem Non-standard? Summary of the Results 1 show that the reduced form parameters of DSGE models are generally not identifiable 2 provide rank and order conditions for identifying the structural parameters from the observable spectrum exploits all information in the autocovariances (spectrum) of the observed variables exploits restrictions on observationally equivalent transformations of the model depends only on the system matrices in the solution equations should be evaluated prior to collecting data 3 analyze several examples

6 Setup Model 1 Why is the Problem Non-standard? 2 Setup Model Observables 3 Identification Analysis Observational Equivalence Rank and Order Conditions 4 Illustration Example 1: An and Schorfheide (2007) Example 2: Stochastic Growth Model 5 Conclusion

7 Setup Model Setup Reduced form Solution: K t+1 (n K 1) W t (n W 1) Z t+1 (n Z 1) = P(θ) K t + Q(θ) Z t (n K n K ) (n K n Z ) = R(θ) K t + S(θ) Z t (n W n K ) (n W n Z ) = Ψ(θ) Z t + ɛ t+1 (n Z n Z ) (n Z 1) K t : endogenous (state) variables W t : other endogenous ( jump ) variables Z t : latent exogenous shocks θ: parameter vector of interest (dim n θ )

8 Setup Model Assumptions 1 For every θ Θ, {ɛ t } WN(0, Σ(θ)) and Σ(θ) nonsingular. 2 For every θ Θ and any z C, det(id Ψ(θ)z) det(id P(θ)z) = 0 implies z > 1. Moreover, for every (θ, θ) Θ 2, P(θ) and Ψ( θ) have no eigenvalues in common. 3 n K n Z n K + n W. 4 For every θ Θ, we have: (i) Q(θ) full rank; (ii) ( Q(θ), S(θ) ) full rank. 5 P( ), Q( ), R( ), S( ), Ψ( ), Σ( ) continuously differentiable on Θ.

9 Setup Observables Observables 1 Lemma 1 shows that Y t is weakly stationary Y t (n Y n K +n W ) ( Kt W t ) = j=0 h(j; θ)ɛ t j = H(L; θ) ɛ t (n Y n Z )(n Z 1) in general, n Z n Y and h(0; θ) = Id ny 2 Lemma 2 shows that Y t is invertible: ɛ t = (n Z 1) j=0 g(j; θ)y t j = G(L; θ) Y t (n Z n Y )(n Y 1) ɛ t is fundamental for Y t rank of H(z; θ) is n Z and a left inverse G(z; θ) exists

10 Setup Observables Transfer (impulse response) function: ( z[id P(θ)z] H(z; θ) = { 1 Q(θ)[Id Ψ(θ)z] 1 ) R(θ)z[Id P(θ)z] 1 Q(θ) + S(θ) } [Id Ψ(θ)z] 1 Spectrum: Ω(z; θ) (n Y n Y ) + j= E(Y t Y t+j )z j = H(z; θ) Σ(θ) (n Y n Z )(n Z n Z ) H(z 1 ; θ) (n Z n Y ) Ω(z; θ) only depends on the reduced form parameter Λ(θ) (P(θ), Q(θ), R(θ), S(θ), Ψ(θ), Σ(θ))

11 Identification Analysis Observational Equivalence 1 Why is the Problem Non-standard? 2 Setup Model Observables 3 Identification Analysis Observational Equivalence Rank and Order Conditions 4 Illustration Example 1: An and Schorfheide (2007) Example 2: Stochastic Growth Model 5 Conclusion

12 Identification Analysis Observational Equivalence Identification Problem given θ 0, we know Λ(θ 0 ), from which we can compute Ω(z; θ 0 ) the identification problem: given Ω(z; θ 0 ), can we recover the θ 0 that generated it? mathematically, θ 0 is identifiable from the spectrum if for every z C: H(z; θ 0 )Σ(θ 0 )H(z 1 ; θ 0 ) = H(z; θ 1 )Σ(θ 1 )H(z 1 ; θ 1 ) implies θ 1 = θ 0

13 Identification Analysis Observational Equivalence Two sources of observational equivalence: 1 Equivalence of transfer functions: each H(z; θ) can be obtained by many combinations of (P(θ), Q(θ), R(θ), S(θ)) in the transfer function all such combinations are characterized in Lemma 3 ( minimum state ) Lemma 3 2 Equivalence of spectral densities: there are many couples (H(z; θ), Σ(θ)) that can jointly produce the same spectrum all such couples are characterized in Lemma 4 ( minimum phase ) Lemma 4

14 Identification Analysis Observational Equivalence Proposition 1 Let Assumptions 1 through 4 hold. Then θ 0 and θ 1 generate the same spectrum for Y t if and only if: P(θ 1 ) = P(θ 0 ) and R(θ 1 ) = R(θ 0 ) and there exists a full rank n Z n Z matrix U such that: Q(θ 1 ) = Q(θ 0 )U S(θ 1 ) = S(θ 0 )U Ψ(θ 1 ) = U 1 Ψ(θ 0 )U Σ(θ 1 ) = U 1 Σ(θ 0 )U 1.

15 Identification Analysis Observational Equivalence Implications: 1 the reduced form parameter Λ(θ) is generally not identifiable (only its components P(θ) and R(θ) are) 2 Proposition 1 defines a system of (n K + n W )(n K + n Z ) + 2n 2 Z equations (components of Λ(θ)) in (components of θ and U) n θ + n 2 Z unknowns 3 identification at θ 0 requires that the unique solution to this system be θ 1 = θ 0 and U = Id 4 this system is finite so we can write necessary and sufficient conditions for local uniqueness (local identifcation)

16 Identification Analysis Rank and Order Conditions Rank and Order Conditions Let (θ) vecp(θ) θ vecq(θ) θ vecr(θ) θ vecs(θ) θ vecψ(θ) θ vecσ(θ) 0 n 2 K n 2 Z Id nz Q(θ) 0 nk n W n 2 Z Id nz S(θ) Id nz Ψ(θ) Ψ(θ) Id nz θ (Id n 2 + T nz,n Z Z )(Σ(θ) Id nz )

17 Identification Analysis Rank and Order Conditions Proposition 2 Let Assumptions 1 through 5 hold. If the rank of (θ) remains constant in a neighborhood of θ 0, then a necessary and sufficient rank condition for θ 0 to be locally identified from the spectrum of {Y t } is: rank (θ 0 ) = n θ + n 2 Z. Moreover, a necessary order condition is: n θ (n K + n W )(n Z + n K ) + n Z(n Z + 1). 2

18 Identification Analysis Rank and Order Conditions Identification Under A Priori Restrictions When the rank or order conditions fail we need: Let ϕ (θ) n ϕ a priori restrictions ϕ(θ) = 0 ϕ(θ) θ vecp(θ) θ vecq(θ) θ vecr(θ) θ vecs(θ) θ vecψ(θ) θ vecσ(θ) 0 nϕ n 2 Z 0 n 2 K n 2 Z Id nz Q(θ) 0 nk n W n 2 Z Id nz S(θ) Id nz Ψ(θ) Ψ(θ) Id nz θ (Id n 2 + T nz,n Z Z )(Σ(θ) Id nz )

19 Identification Analysis Rank and Order Conditions Proposition 3 Let Assumptions 1 through 5 hold and assume that ϕ is continuously differentiable on Θ. If the rank of ϕ (θ) remains constant in a neighborhood of θ 0, then a necessary and sufficient rank condition for θ 0 to be locally identified from the spectrum of {Y t } and the a priori restrictions ϕ(θ 0 ) = 0 is: rank ϕ (θ 0 ) = n θ + n 2 Z. Moreover, a necessary order condition for identification is: [ n ϕ n θ (n K + n W )(n Z + n K ) + n ] Z(n Z + 1). 2

20 Illustration Example 1: An and Schorfheide (2007) 1 Why is the Problem Non-standard? 2 Setup Model Observables 3 Identification Analysis Observational Equivalence Rank and Order Conditions 4 Illustration Example 1: An and Schorfheide (2007) Example 2: Stochastic Growth Model 5 Conclusion

21 Illustration Example 1: An and Schorfheide (2007) Example 1: An and Schorfheide (2007) y t = E t y t+1 + g t E t g t+1 1 ( ) rt E t π t+1 E t z t+1 τ π t = τ(1 ν) βe t π t+1 + νπ 2 φ (y t g t ) c t = y t g t r t = ρ r r t 1 + (1 ρ r )ψ 1 π t + (1 ρ r )ψ 2 (y t g t ) + ɛ rt g t = ρ g g t 1 + ɛ gt z t = ρ z z t 1 + ɛ zt with ɛ rt WN(0, σ 2 r ), ɛ gt WN(0, σ 2 g ), ɛ zt WN(0, σ 2 z ) mutually uncorrelated π is steady state inflation rate

22 Illustration Example 1: An and Schorfheide (2007) P, Q, R, S representation of An and Schorfheide s (2007) model is obtained by letting: ɛ rt ɛ rt Z t g t, ɛ t ɛ gt (n Z = 3) z t ɛ zt K t r t 1 (n K = 1) y t W t π t (n W = 3) c t

23 Illustration Example 1: An and Schorfheide (2007) Version I: n θ = 13 θ = (τ, β, ν, φ, π, ψ 1, ψ 2, ρ r, ρ g, ρ z, σ 2 r, σ 2 g, σ 2 z ) order condition: n θ = 13 < (n K + n W )(n Z + n K ) + n Z(n Z + 1) 2 = 31 so the order condition in Proposition 2 holds rank condition: rank (θ 0 ) = 21 < n θ + n 2 Z = 22 not surprising as An and Schorfheide (2007) noted that the components ν, φ, and π are not separately identifiable

24 Illustration Example 1: An and Schorfheide (2007) Version II: n θ = 11 τ(1 ν) κ νπ 2 φ θ = (τ, β, κ, ψ 1, ψ 2, ρ r, ρ g, ρ z, σr 2, σg, 2 σz 2 ) order condition: n θ = 11 < (n K + n W )(n Z + n K ) + n Z(n Z + 1) 2 = 31 rank condition: rank (θ 0 ) = 20 = n θ + n 2 Z

25 Illustration Example 2: Stochastic Growth Model Example 2: Stochastic Growth Model [ max E t β {C t,k t } s=0 t+s( C1 ν t+s 1 ν ) ] production technology: Q t = Z t K α t (1 Φ t) feasibility: Q t = C t + K t+1 (1 δ)k t exogenous technology process: Z 0 given Z t = Z 0 exp(z t ) where z t = ψz t 1 + ɛ t and ɛ t WN(0, σ 2 ) adjustment cost: Φ t (K t+1, K t ) = φ 2 [ K t+1 K t 1] 2

26 Illustration Example 2: Stochastic Growth Model Reduced Form Solution: k t = λ kk k t 1 + λ kz z t c t = λ ck k t 1 + λ cz z t z t+1 = ψz t + ɛ t+1, ɛ t WN(0, σ 2 ) The parameters of interest are: θ (α, β, δ, φ, ν, ψ, σ 2) The reduced form parameters are: λ (λ kk, λ kz, λ ck, λ cz, ψ, σ 2) each λ is a nonlinear function of θ analytic expressions are provided (Appendix D)

27 Illustration Example 2: Stochastic Growth Model We consider 3 versions of the model, all with Model I: ν = 1, φ = 0 θ 0 = (0.36, 0.95, 0.025, φ, ν, 0.85, 0.04). n θ = 5 = (n K + n W )(n Z + n K ) + n Z(n Z + 1) 2 order holds rank (θ 0 ) = 6 = n θ + n 2 Z = rank condition holds Model I is identifiable at θ 0 from the spectrum of {(k t 1, c t ) }.

28 Illustration Example 2: Stochastic Growth Model Model II: ν > 0 unrestricted and φ = 0 Here n θ = 6, so n θ (n K + n W )(n Z + n K ) n Z(n Z + 1) 2 = 1 need n ϕ = 1 Let ν 0 = 2. Then ϕ(θ) = β 0.95, rank ϕ (θ 0 ) = 6 < n θ + n 2 Z = 7 ϕ(θ) = δ 0.025, rank ϕ (θ 0 ) = 7 θ 0 is conditionally identifiable by the restriction δ =

29 Illustration Example 2: Stochastic Growth Model Model III: ν > 0 and φ > 0 unrestricted Here n θ = 7, so n θ (n K + n W )(n Z + n K ) n Z(n Z + 1) 2 = 2 need n ϕ = 2 Let ν 0 = 2 and φ 0 = For the rank condition to hold we need: rank ϕ (θ 0 ) = n θ + n 2 Z = 8

30 Illustration Example 2: Stochastic Growth Model Fixing 2 components of θ gives the following ranks of ϕ (θ 0 ): θ j β δ φ ν ψ σ θ i α β δ φ ν ψ - 7

31 Conclusion 1 Why is the Problem Non-standard? 2 Setup Model Observables 3 Identification Analysis Observational Equivalence Rank and Order Conditions 4 Illustration Example 1: An and Schorfheide (2007) Example 2: Stochastic Growth Model 5 Conclusion

32 Conclusion Conclusion Existing Identification Analysis: check Hessian of the likelihood compare properties of the VAR implied by the DSGE model with a factor augmented VAR gradient of the mapping from the deep to the reduced form parameters assumed to be identified Drawbacks: the methods depend on parameters that are possibly not consistently estimated the reduced parameters themselves are not identifiable We provide simple to evaluate order and rank conditions that only depend on the functional form of the solution matrices

33 Lemma 3 Lemma 3: Similarity Transform Let Assumptions 1 to 4 hold. Consider the realizations (P(θ 0 ), Q(θ 0 ), R(θ 0 ), S(θ 0 ), Ψ(θ 0 )) and (P(θ 1 ), Q(θ 1 ), R(θ 1 ), S(θ 1 ), Ψ(θ 1 )) of the transfer functions H(z; θ 0 ) and H(z; θ 1 ), respectively, with (θ 0, θ 1 ) Θ 2. Then the two realizations are equivalent, i.e. H(z; θ 0 ) = H(z; θ 1 ) for every z C, if and only if P(θ 1 ) = P(θ 0 ), R(θ 1 ) = R(θ 0 ) and there exists a full rank n Z n Z matrix T such that: Q(θ 1 ) = Q(θ 0 )T 1, S(θ 1 ) = S(θ 0 )T 1, Ψ(θ 1 ) = TΨ(θ 0 )T 1. Back

34 Lemma 4 Lemma 4: Spectral Decomposition Let Assumptions 1 to 4 hold. Consider two couples (H(z; θ 0 ), Σ(θ 0 )) and (H(z; θ 1 ), Σ(θ 1 )) with spectral densities Ω(z; θ 0 ) and Ω(z; θ 1 ), respectively, where (θ 0, θ 1 ) Θ 2. Then the two couples are equivalent, i.e. Ω(z; θ 0 ) = Ω(z; θ 1 ) for every z C, if and only if there exists a full rank n Z n Z matrix U such that: for every z C H(z; θ 1 ) = H(z; θ 0 )U and UΣ(θ 1 )U = Σ(θ 0 ). Back