Dynamic Identification of DSGE Models:

Web Appendix

Ivana Komunjer and Serena Ng

October

This Appendix consists of details in ? along with two additional examples.

Details of Example 1

LetQ∗, C∗,K∗ be steady state values of output, consumption, and capital, and letR∗ = (1−δ)+αQ∗K∗ .The log-linearized model is

qt = αkt + zt

qt = (1− δK∗

Q∗)ct +

K∗

Q∗kt+1 − (1− δ)K

∗

Q∗kt

0 = Et

[ν(ct+1 − ct)− φβ

Q∗

K∗kt+2 + (α(1− α) + βφ(2− δ))βQ

∗

K∗kt+1

−φβ(1− δ)Q∗

K∗kt − αβ

Q∗

K∗zt+1

]zt = ψzt−1 + σεt

The structural model can be rewritten as

qt = γ1kt + zt

qt = γ2ct + γ3kt+1 + (1− γ2 − γ3)kt

= Et

[γ4(ct+1 − ct) + γ5kt+2 +

( γ1(1− γ1)γ1 + γ2 + γ3 − 1

+(1− γ2 − 2γ3)γ5

γ3

)kt+1

−(1− γ2 − γ3)γ5

γ3kt −

γ1

γ1 + γ2 + γ3 − 1zt+1

]zt = γ5zt−1 + γ6εt

where the structural parameters are

γ = (γ1, γ2, γ3, γ4, γ5, γ6) ≡(α, 1− δK

∗

Q∗,K∗

Q∗, ν, −φβQ

∗

K∗, ψ, σ

).

For this model, λkk is given by:

λkk =−B +

√B2 − 4AC2A

1

where

A ≡ −νK∗

C∗− φβ Y

∗

K∗, B ≡ ν

(1 +

1β

)K∗C∗

+ [α(1− α) + φ(2− δ)]β Y∗

K∗, C ≡ −ν

β

K∗

C∗− φβ(1− δ)Y

∗

K∗.

And λkz, λck, λcz equal:

λkz =(1− ψ)ν Y

∗

C∗ + αβψ Y ∗

K∗

ν[ 1β − λkk + 1− ψ]K∗

C∗ + β [α(1− α)− φ(2− δ)− φλkk − ψφ] Y ∗K∗

λck =K∗

C∗[1β− λkk]

λcz =K∗

C∗[Y ∗

K∗− λkz].

Solution Methods

θ0 = (α0, β0, δ0, φ0, ν0, ψ0, σ20) = (0.36, .95, 0.025, .16, 2, 0.85, 0.04).

For this model, direct the method of undetermined coefficients yields

kt = .9557kt−1 + .1521ztct = .5086kt−1 + .3334ztzt = .85zt−1 + εt.

Uhlig’s toolkit Let Wt = (yt, ct). The code

alpha=.36;beta=.95;delta=.025;nu=2;rho_Z=.85;phi=.16;sig2_Z=.04;Rstar=1/beta;kstar= (alpha/(Rstar-1+delta))^(1/(1-alpha));ystar=kstar^alpha;cstar=ystar-delta*kstar;AA=[ 0 ; kstar/ystar ; ];BB=[ alpha ; -((1-delta)/(1+g))*kstar/ystar ; ];CC=[-1 0; -1 1-delta*kstar/ystar];DD=[1 ; 0 ];FF=phi*beta*ystar/kstar;GG= -((alpha*(1-alpha)+phi*(2-delta))*beta*ystar/kstar);HH= phi*beta*(1-delta)*ystar/kstar;JJ= [0 -nu];KK= [ 0 nu];LL= alpha*beta*ystar/kstar;

2

MM=0;NN=rho_Z;Sigma=sig2_Z;[l_equ,m_states] = size(AA);[l_equ,n_endog ] = size(CC);[l_equ,k_exog ] = size(DD);

message = ’ ’;warnings = [];options;solve;disp(’PP’);mymprint(PP);disp(’QQ’);mymprint(QQ);disp(’RR’);mymprint(RR);disp(’SS’);mymprint(SS);

gives

P = .9557, Q = .1521, R =(.3600.5086

), S =

(1

.3334

). (1)

Sim’s code The code

n_k=1;n_w=2;n_z=1;n_f_k=1;n_f_w=2;n_y=n_f_k+n_f_w+ n_k+n_w+n_z;Jam0=zeros(n_y,n_y);Jam0(1:3,4:6)=eye(3);Jam0(4,end-1:end)=[1 -1];Jam0(5,4:6)=[-kstar/ystar -(1-delta*kstar/ystar) 1];Jam0(6,1:5)=[-phi*beta*ystar/kstar nu -alpha*beta*ystar/kstar ...

(alpha+phi*(2-delta))*beta*ystar/kstar -nu];Jam0(7,7)=1;Jam1=zeros(n_y,n_y);Jam1(1:3,1:3)=eye(3);Jam1(4,4)=alpha;Jam1(5,4)=-(1-delta)*kstar/ystar;Jam1(6,4)= phi*beta*(1-delta)*ystar/kstar;Jam1(end,end)=rho_Z;Jam2=zeros(n_y,1); Jam2(end)=1;

3

Jam3=zeros(n_y,n_f_k+n_f_w);Jam3(1:n_f_k+n_f_w,1:n_f_k+n_f_w)=-eye(n_f_k+n_f_w);;cons=zeros(ny,1);[G1,C,impact,fmat,fwt,ywt,gev,eu]=gensys(Jam0,Jam1,cons,Jam2,Jam3);

gives

Etkt+1

Etct+1

Etyt+1

ktctytzt

=

0 0 −0 0.9133 −0 −0 0.23340 0 −0 0.4860 −0 −0 0.30660 0 0 0.3440 −0 −0 0.76900 0 −0 0.9557 −0 −0 0.12930 0 −0 0.5086 −0 −0 0.28340 −0 −0 0.3600 −0 0 0.8500−0 0 0 −0 −0 0 0.8500

Et−1ktEt−1ctEt−1ytkt−1

ct−1

yt−1

zt−1

+

0.27460.36070.90470.15210.33341.00001.0000

εt

+

−1.3022 0.1681 0.2913−0.0213 1.0169 0.28910.0000 −0.0689 1.34500.0000 −0.1913 −0.04420.0000 1.0034 0.23160.0000 0.0000 −0.0000

0 −0.0000 0

∞∑s=1

−0.0000 0.3916 0.0904−0.0000 0.9081 0.2096−0.0000 −0.0000 −0.0000

s−1.0972.0956.7348

Etεt+s.

Thus

kt = .9557kt−1 + .1293zt−1 + .1521εtct = .5086kt−1 + .2834zt−1 + .3334εtyt = .3600kt−1 + .8500zt−1 + εt

zt = .85zt−1 + εt.

Note that since ψ = .85,

.1293zt−1 + .1521εt = .1521(ψzt−1 + εt) = .1521zt

.2834zt−1 + .3334εt = .3334(ψzt−1 + εt) = .3334zt.

Simple rearranging gives (??).

Klein’s alogirthm The code

A0=[ 1 0 0 0 0;0 1 0 0 0;0 0 0 0 0;0 -kstar/ystar 0 0 0;

0 (alpha+phi*(2-delta))*beta*ystar/kstar -phi*beta*ystar/kstar -alpha*beta*ystar/kstar\nu];B0=[rho_Z 0 0 0 0;

4

0 0 1 0 0;1 alpha 0 -1 0;0 -(1-delta)*kstar/ystar 0 -1 1-delta*kstar/ystar;0 phi*beta*(1-delta)*ystar/kstar 0 0 nu];

[f0,p0]=solab(A0,B0,2)

gives

F0 =

0.1521 0.95571.0000 0.36000.3334 0.5086

P0 =(

0.8500 00.1521 0.9557

)Thus, P,Q,R, S as defined in (??) can be easily extracted.

Details of Example 2

Uhlig’s Toolkit The code

tau = 2.0000;nu = 0.1000;kap = 0.3300;cyst = 0.8500;psi1 = 1.5000;psi2 = 0.1250;rhor = 0.7500;rhog = 0.9500;rhoz = 0.9000;rrst = 1.0000;pist = 3.2000;gamst= 0.5500;pistar2=exp(pist/400);rrstar2=exp(rrst/400);phi= tau*(1-nu)/nu/kap/pist^2;beta=1/rrstar2;gstar=1/cyst;cstar=(1-nu)^(1/tau);ystar=cstar*gstar;sig2r= .002^2;sig2g= .006^2;sig2z=.003^2;

par=[tau; kap; beta; psi1; psi2; rhor; rhog; rhoz; ];

Rstar=1/beta; gstar=.1; cstar=.1; ystar=.1; pistar=0.005; Zstar=0;ss0=[Rstar;gstar;cstar;ystar;pistar;;Zstar];[ss,fval,flag]=fsolve(’model_as_ss’,ss0);Rstar=ss(1);

5

gstar=ss(2);cstar=ss(3);ystar=ss(4);pistar=ss(5);Zstar=ss(6);disp(’Steady State’);disp(ss’);

VARNAMES=’R’;VARNAMES=strvcat(VARNAMES,’g’);VARNAMES=strvcat(VARNAMES,’c’);VARNAMES=strvcat(VARNAMES,’y’);VARNAMES=strvcat(VARNAMES,’pi’);VARNAMES=strvcat(VARNAMES,’e_R’);VARNAMES=strvcat(VARNAMES,’e_g’);VARNAMES=strvcat(VARNAMES,’e_Z’);

AA= [-1 -(1-rhor)*psi2 0;0 -1 0;0 -1 -1];

BB= [rhor 0 0;0 rhog 0;0 0 0];

CC= [ (1-rhor)*psi2 (1-rhor)*psi1;0 0;1 0];

DD= [1 0 0;0 1 0;0 0 0];

FF= [0 -1 0;0 0 0];

GG= [-1/tau 1 0;0 -kap 0];

HH = [0 0 0;0 0 0];

JJ= [ 1 1/tau;0 beta];

6

KK= [-1 0kap -1];

LL= [ 0 0 1/tau;0 0 0];

MM=[0 0 0;0 0 0];

NN= [ 0 0 0;0 0 0;0 0 rhoz];

Sigma=[sig2r 0 0;0 sig2g 0;0 0 sig2z];

[l_equ,m_states] = size(AA);[l_equ,n_endog ] = size(CC);[l_equ,k_exog ] = size(DD);

message = ’ ’;warnings = [];options;DISPLAY_IMMEDIATELY=1;solve;

gives Rtgtct

=

.5143 0 00 0 0

−1.1011 0 0

Rt−1

gt−1

ct−1

+

.6858 0 .60550 0 0

−1.1011 0 1.4863

εRtεgtzt

(ytπt

)=

(−.8528 .95 0−.5596 0 0

)Rt−1

gt−1

ct−1

+(−1.1011 1 1.4863−.7462 0 1.4909

)εRtεgtzt

(ztgt

)=

(.9 00 .95

)(zt−1

gt−1

)+(εztεgt

)Let Kt =

(Rt gt ct

)′ and Wt =(ytπt

). The solution is of the form P,Q,R,S with

P =

.5143 0 00 0 0

−1.1011 0 0

, Q =(−1.1011 1 1.4863−.7462 0 1.4909

), R =

(−.8528 .95 0−.5596 0 0

), S =

(−1.1011 1 1.4863−.7462 0 1.4909

)(2)

7

Dynare gives

Rt = .514326Rt−1 + .544980zt−1 + .605533εzt + .685769εrt= .514326Rt−1 + .606553zt + .685789εrt

gt = .95gt−1 + εgt

zt = .9zt−1 + εzt

yt = −.825828Rt−1 + .95gt−1 + εgt + 1.337764zt−1 + 1.486294εzt − 1.101104εrt= −.825828Rt−1+, 95gt−1 + 1.4868294zt + εgt − 1.101104εrt

ct = −.825828Rt−1 + 1.33764zt−1 + 1.486294εzt − 1.101104εrt= −.825828Rt−1 + 1.486294zt + 1.101104εrt

πt = −.559644Rt−1 + 1.341807zt−1 + 1.490897εzt − .746192εrt

Simple rearranging leads to the P,Q,R,S as defined in (??).

Sims’s Code Putting Gauss code provided by An and Schorfheide (2007) in Matlab:

tau = 2.0000;nu = 0.1000;kap = 0.3300;cyst = 0.8500;psi1 = 1.5000;psi2 = 0.1250;rhor = 0.7500;rhog = 0.9500;rhoz = 0.9000;rrst = 1.0000;pist = 3.2000;gamst= 0.5500;pistar2=exp(pist/400);rrstar2=exp(rrst/400);phi= tau*(1-nu)/nu/kap/pist^2;beta=1/rrstar2;gstar=1/cyst;cstar=(1-nu)^(1/tau);ystar=cstar*gstar;sig2r= .002^2;sig2g= .006^2;sig2z=.003^2;

Gam0=[ 1 0 0 0 0 0 0 0;0 1 0 0 0 0 0 0;

8

0 (1-rhor)*psi2 1 -(1-rhor)*psi2 -(1-rhor)*psi1 0 0 0;0 -kap 0 kap -1 0 beta 0 ;rhoz/tau (1-rhog) -1/tau -1 0 0 1/tau 1 ;0 0 0 0 0 0 0 0;0 0 0 0 1 0 0 00 0 0 1 0 0 0 0;];

Gam1=[ rhoz 0 0 0 0 0 0 0 ;0 rhog 0 0 0 0 0 0 ;0 0 rhor 0 0 0 0 0;0 0 0 0 0 0 0 0 ;g0 0 0 0 0 0 0 0 ;0 -1 0 1 0 -1 0 0 ;0 0 0 0 0 0 1 0;0 0 0 0 0 0 0 1;];

Gam2=zeros(8,3); Gam2(1:3,1:3)=eye(3);Gam3=zeros(8,2); Gam3(7:8,7:8)=eye(2);cons=zeros(8,1);[G1,C,impact,fmat,fwt,ywt,gev,eu]=gensys(Gam0,Gam1,cons,Gam2,Gam3);

gives

ztgtRtytπtct

Etπt+1

Etyt+1

=

0.9000 −0 0 0 −0 0 0 0−0 0.9500 −0 −0 0 −0 0 −0

0.5450 0 0.5143 0 0 −0 0 01.3377 0.9500 −0.8258 −0 −0 0 −0 −01.3418 0 −0.5596 −0 −0 0 −0 01.3377 0 −0.8258 −0 −0 0 −0 −00.9026 0 −0.2878 −0 −0 0 −0 00.7538 0.9025 −0.4247 −0 −0 0 −0 0

zt−1

gt−1

Rt−1

yt−1

πt−1

ct−1

Et−1πtEt−1yt

+

1.0000 −0 00 1.0000 −0

0.6055 0 0.68581.4863 1.0000 −1.10111.4909 0 −0.74621.4863 0 −1.10111.0029 0 −0.38380.8376 0.9500 −0.5663

εztεgtεRt

P,Q,R, S as defined in (??) can be easily extracted.

Example 3

An and Schorfheide (2007) considered two structural models with the same solution:

yt = ψyt−1 + ηt, ηt ∼ iid(0, 1) (3)

9

Model M1 has ψ = ρ, and is given by:

yt =1αEtyt+1 + ut, ut = ρut−1 + εt, εt ∼ iid(0, (1− ρ/α)2)

with θ ≡ (α, ρ). Model M2 has ψ = 12

(α−

√α2 − 4φα

)and is given by:

yt =1αEtyt+1 + εt, εt ∼ iid

(0,[α+

√α2 − 4φα2α

]2)with θ ≡ (α, φ).

The solution (1) is a special case of Equation (1) in ? obtained by setting Zt = ηt, Kt = yt, andleaving Wt empty. This implies nZ = 1, nK = 1, nW = 0. The solution parameters are P (θ) = ψ,Q(θ) = 1, Ψ(θ) = 0, and Σ(θ) = 1. They satisfy Assumptions 1 through whenever 0 < |ψ| < 1.Now note that nΛ = (nW + nK)(nK + nZ) + 2n2

Z = 4, nZ = 1, nθ = 2, so for both models we have:

nθ = 2 < (nW + nK)(nK + nZ) +nZ(nZ + 1)

2= 3.

Both models satisfy our order condition in Proposition 2. To examine the rank condition, the 4×3matrix ∆(θ) defined in Proposition 2 calculated for each of the two models is:

M1 : ∆(θ) =

0 1 00 0 10 0 00 0 −2

and M2 : ∆(θ) =

12

(1− α−2φ√

α2−4φα

)α√

α2−4φα0

0 0 10 0 00 0 −2

Hence, for both models the rank of ∆(θ) equals 2 < nθ + n2

Z = 3, so θ is not identifiable. This isnot surprising as in model M1, α can never be identified from (1), while in M2 the two componentsα and φ cannot be identified separately.

Indeed, we now confirm this intuitive observation by applying our partial and conditional iden-tification results. First, we show that the model M1 is partially identifiable in ρ. For this, set θi = ρand θ−i = α. Then, in Proposition 4 it holds that

rank(∂λ(θ,Id)∂θ−i

)= rank

0000

= 0 = rank∆(θ)− (nθ,i + n2Z) = 2− 2

which shows that model M1 is indeed partially identified in ρ.Next, we search for a priori restrictions that would allow for θ to be identified in each of

the models. For model M1, any scalar restriction such that ∂ϕ/∂α 6= 0 will be such that thecorresponding 5× 3 matrix ∆ϕ(θ) becomes of rank 3. Hence, θ will be identifiable under any suchrestriction. A simple example is the restriction α = α, i.e. ϕ(θ) = α − α. Of course, lettingϕ(θ) = ρ − ρ would fail to satisfy the rank condition. For model M2, the restrictions that willsatisfy the rank condition in Proposition 3 must be such that:

det

∂ϕ(θ)∂α

∂ϕ(θ)∂φ

12

(1− α−2φ√

α2−4φα

)α√

α2−4φα

6= 0.

For instance, setting α = α is sufficient to identify θ at any value θ0 = (α0, φ0) such that α0 6= 0.Similarly, setting φ = φ is also sufficient for identification provided φ0 6= 0.

10

Example 4

Chang- Doh and Schorfheide (2007) consider a one-sector stochastic growth model with technologyand labor supply shocks. Let Ct be consumption and Ht be hours. Utility is

Et

∞∑t=0

βt+s(

logCt+s −Ht+s/B

1+1/νt+s

1 + 1/ν

).

Resource constraint is

Yt = Ct +Kt+1 − (1− δ)Kt

= WtHt +RtKt

= (AtHt)αK1−αt

[1− φ

(Ht

Ht−1− 1)2]

.

Shocks are

logAt = γ + logAt−1 + εAt

logBt = ρBBt−1 + (1− ρB) logB0 + εBt

The parameter vector of interest is: θ = (α, β, γ, δ, φ, ν, ρB, lnA0, lnB0, σ2A, σ

2B)′ which is of

dimension nθ = 11. We focus on version 2 of the model (denoted A0) with stationary hours,0 6 ρB < 1, and with adjustment cost, φ > 0. The gensys code for this model is

1ν + 1− α+ 2φ(1 + β)/α 0 1 0 0 −2β φα −1− 1

ν

0 α(1− β (1−δ)exp(γ)) −1 0 1 −α(1− β 1−δ

exp(γ)) 0α −K∗

Y ∗ −C∗

Y ∗ 0 0 0 0−α 0 0 1 0 0 00 0 1 0 0 0 01 0 0 0 0 0 00 0 0 0 0 0 1

htkt+1

ctyt

Etct+1

Etht+1

Bt

=

2φα 1− α 0 0 0 0 00 0 0 0 0 0 00 −(1− α− I∗

Y ∗ + K∗

Y ∗ ) 0 0 0 0 00 1− α 0 0 0 0 00 0 0 0 1 0 00 0 0 0 0 1 00 0 0 0 0 0 ρB

ht−1

ktct−1

yt−1

Et−1ctEt−1htBt−1

+

−(1− α) 00 0

(1− α− I∗

Y ∗ + K∗

Y ∗ ) 0α− 1 0

0 00 00 1

(εAtεBt

)+

0 00 00 00 01 00 10 0

(ηctηht

)

11

Let Zt ≡ (at, bt)′, where at = ∆ lnAt − γ, bt = lnBt. Rearranging the ‘gensys’ output of Sims(2002), we obtain the P,Q,R, S representation of the model as follows:

Kt+1 =(htkt+1

)=

(Phh(θ) Phk(θ)Pkh(θ) Pkk(θ)

)(ht−1

kt

)+(Qha(θ) Qhb(θ)Qka(θ) Qkb(θ)

)(atbt

)Wt =

(ctyt

)=

(Ryh(θ) Ryk(θ)Rch(θ) Rck(θ)

)(ht−1

kt

)+(Sya(θ) Syb(θ)Sca(θ) Scb(θ)

)(atbt

)Zt =

(atbt

)=

(0 00 Ψb(θ)

)(at−1

bt−1

)+(εAtεBt

).

This model has two exogenous shocks εt ≡ (εAt, εBt)′ (nZ = 2), two observed state variablesKt ≡ (ht−1, kt)′ (nK = 2), and two endogenous variables Wt ≡ (yt, ct)′ (nW = 2). We have nΛ =(nK+nW )(nZ+nK)+2n2

Z = 24. Since nθ = 11 < (nK+nW )(nZ+nK)+nZ(nZ+1)/2 = 19 the ordercondition in Proposition 2 is satisfied. To check the rank condition, we use Table 2 of Chang, Doh,and Schorfheide (2009): θ0 = (0.658, 0.995, 0.004, 0.023, 11.36, 0.433, 0.8, 5.748, 3.171, 0.0112, 0.0342)′.Since rank∆(θ0) = 14 < nθ+n2

Z = 15, the parameter vector θ0 is not identifiable from the spectrumof {(ht−1, kt, yt, ct)′}.

To check for partial identifiability of each of the components of θ, we let θi be one among 11components of θ. The table below reports the ranks of ∂λ(θ0,Id)

∂θ−i. For the model to be partially

identifiable in θi, this rank should be equal to rank∆(θ0) − (nθ,i + n2Z) = 9. As can be seen from

Table 1, A0 is not partially identifiable in this model. Indeed, since lnAt is a unit root process withdrift, lnA0 cannot be identified from the stationary variables.

θi α β γ δ φ ν ρB A0 B0 σ2A σ2

B

9 9 9 9 9 9 9 10 9 9 9

Table 1: Rank of ∂λ(θ0,Id)∂θ−i

12