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INTRODUCTION TO DYNAREICE 2005
Michel Juillard
CEPREMAP, Paris Sciences Economics, University Paris 8
INTRODUCTION TO DYNARE – p. 1/13
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Acknowledgments
DYNARE started at CEPREMAP in 1994.
DYNARE development: S. Adjemian, O. Kamenik
Built on work of: R. Boucekkine, F. Collard, J.P. Laffargue,M. Ratto, F. Schorfheide, C. Sims, R. Wouters
Public domain software: cygwin, gnumex, lapack, styxbox,asamin, asa.
INTRODUCTION TO DYNARE – p. 2/13
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DYNARE
1. computes the steady state of the model
2. computes the solution of deterministic models(arbitrary accuracy)
3. computes first and second order approximation tosolution of stochastic models
4. estimates (maximum likelihood or Bayesian approach)parameters of DSGE models
INTRODUCTION TO DYNARE – p. 3/13
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Solution of deterministic models
based on work of Laffargue, Boucekkine and myself
approximation: impose return to equilibrium in finitetime instead of asymptotically
computes the trajectory of the variables numerically
uses a Newton–type method
usefull to study full implications of non–linearities
INTRODUCTION TO DYNARE – p. 4/13
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An example
The effect of a change in tax rate in a model withmonopolistic competition (adapted from Hairault, Langotand Portier, 2001)
Wt = ln ct + η ln(1 − ht) + βWt+1
ct + it = Akα
t−1h1−α
t
it = kt − (1 − δ)kt−1
1
ct
= βEt
�1
ct+1
(zt+1 + 1 − δ)
�
η
1 − ht
=wt
ct
α
�kt−1
ht
�α−1
= (1 + µ)(1 + τt)zt
(1 − α)
�kt−1
ht
�α
= (1 + µ)(1 + τt)wt
INTRODUCTION TO DYNARE – p. 5/13
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DYNARE implementation – Preambule
var Welf w c h i k z;varexo tau;
parameters beta delta alpha mu eta rho Abar;delta = 0.025;eta = 2;mu = 0.1;alpha = 0.36;rho = 0.95;beta = 0.988;Abar = 1;
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DYNARE implementation – Model
model;Welf = log(c)+eta * log(1-h)+beta * Welf(+1);c+i = Abar * k(-1)ˆalpha * hˆ(1-alpha);i = k - (1-delta) * k(-1);1/c = beta * (1/c(+1)) * (z(+1)+1-delta);eta/(1-h) = w/c;alpha * (k(-1)/h)ˆ(alpha-1) = (1+mu) * (1+tau) * z;(1-alpha) * (k(-1)/h)ˆalpha = (1+mu) * (1+tau) * w;end;
INTRODUCTION TO DYNARE – p. 7/13
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DYNARE implementation – Initialization
initval;Welf = -100;w = 0.5;c = 0.6;h = 0.3;i = 0.4;k = 3;z = 0.1;tau = 0;end;
steady;
INTRODUCTION TO DYNARE – p. 8/13
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DYNARE implementation – Initialization
endval;Welf = -100;w = 0.5;c = 0.6;h = 0.3;i = 0.4;k = 3;z = 0.1;tau = -mu/(1+mu);end;
steady;
INTRODUCTION TO DYNARE – p. 9/13
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DYNARE implementation – Computations
simul(periods=300);
dsample 0 50;rplot Welf;rplot k;rplot c;rplot h;
INTRODUCTION TO DYNARE – p. 10/13
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Stochastic models: First order approximation
In a a stochastic framework, the unknowns are the decisionfunctions.For a large class of DSGE models, DYNARE computesapproximated decision rules and transition equations of theform
yt = y + Ayt−1 + But
with yt = yt − y.Method proposed by Klein (2000) and Sims (2002).DYNARE computes also theoretical moments and IRFs.
INTRODUCTION TO DYNARE – p. 11/13
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Example
In the previous example, one introduces stochasticproductivity according to
ln At = (1 − ρ) ∗ ln A + ρ ∗ ln At−1 + et
and one considers the case of no tax.New instructions:shocks;var e; stderr 0.072;end;
stoch_simul(order=1) Welf h c i w z;
INTRODUCTION TO DYNARE – p. 12/13
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Second order approximation
Two features:
decision rules and transition functions are 2nd orderpolynomials
departure from certainty equivalence: the variance offuture shocks matters
Decision rules and transition equations of the form
yt = y+Ayt−1+But+0.5(
y′t−1Cyt−1 + u′
tDut
)
+y′t−1Fut+∆ (Σu)
Method suggested by K. Judd, developped by C. Sims(2002), S. Schmitt-Grohe and M. Uribe (2003), F. Collardand M. Juillard (2000).
INTRODUCTION TO DYNARE – p. 13/13
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A k-order Perturbation Approach toSolve Complete Market RBC Models
July 2004
prepared for the conference “Computational Methods forDynamic Stochastic Economic Models”, SITE 2004.
Michel JuillardCEPREMAP and University Paris 8
– p. 1/19
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Solving DSGE models
Let’s consider the following model:
Et (f(yt−1, yt, yt+1, ut)) = 0
with
ut = σεt, E(εt) = 0, E(
[εt]β1 . . . [εt]
βk
)
= [Σ]β1...βk
The solution takes the form:
yt = g(yt−1, ut, σ)
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The perturbation method
Computes a Taylor expansion for g() from thecoefficients of the Taylor expansion of f().
The Taylor expansion is generaly computed around thedeterministic equilibrium of the model:
f(y, y, y, 0, 0) = 0
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The state variables
The state variables are yt−1, and ut.Then,
yt+1 = g(yt, ut+1, σ)
= g(g(yt−1, ut, σ), ut+1, σ)
and
F (yt−1, ut, σ, ut+1) = f(
yt−1, g(yt−1, ut, σ),
g(g(yt−1, ut, σ), ut+1, σ), ut
)
F (yt−1, ut, σ) = Et
(
F (yt−1, ut, σ, ut+1))
= 0
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The first order approximation
First order Taylor expansion of the structural model:
F (1)(yt−1, ut, σ) = Et
{
f(y, y, y, 0, 0) + fy−
y
+fy(gyy + guu + gσσ)
+fy+gy(gyy + guu + gσσ)
+ fy+guu′ + fy+
gσσ + fuu}
= 0
where y = yt−1 − y, u = ut, u′ = ut+1. The partial derivatives are taken at the
deterministic equilibrium and aren’t stochastic.
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for this to hold . . .
(
fy−
+ fygy + fy+gygy
)
y = 0
(fygu + fy+gygu + fu) u = 0
(fygσ + fy+gygσ + fy+
gσ) σ = 0
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k order approximation
Let’s write
s = [yt−1, ut]′
s = [y, 0]′
s = s − s
u = ut
u′ = ut+1
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Tensor notation
∂jF i
∂sα1. . . ∂sαj
=[
F i
sj
]
α1...αj
and
n∑
α1
. . .
n∑
αj
∂jF i
∂sα1. . . ∂sαj
sα1. . . sαj
=[
F i
sj
]
α1...αj
[s]α1 . . . [s]αj
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Taylor expansion of the model
F(p)
i(s, σ, u′)
= Fi(s, 0, 0) +
pX
j=1
1
j!
��
F i
sj
�
α1...αj[s]α1 . . . [s]αj
+
j−1X
k=1
�
F i
skσj−k
�
α1...αk[s]α1 . . . [s]αk σj−k
+
j−1X
k=1
h
F i
sku′j−k
i
α1...αkβ1...βj−k
[s]α1 . . . [s]αk
�u′
�β1 . . .
�u′
�βj−k
+
j−1X
k=1
h
F i
u′kσj−k
iβ1...βk
�u′
�β1 . . .
�u′
�βk σj−k
+
j−2X
k=1
j−1Xm=k+1
hF i
sku′m−kσj−m
iα1...αkβ1...βm−k
[s]α1 . . . [s]αk
�
u′
�
β1 . . .
�
u′
�
βm−k σj−m
+
�F i
σj
�σj +
hF i
u′ji
β1...βj
�u′
�β1 . . .
�u′
�βj
�
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Reminding
u = σε
E(
[u]β1 . . . [u]βk
)
= σk [Σ]β1...βk
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Taking the expectation
E
�
F(p)
i(s, σ)
�
= Fi(s, 0, 0) +
pX
j=1
1
j!
��
F i
sj
�
α1...αj[s]α1 . . . [s]αj
+
j−1X
k=1
�
F i
skσj−k
�
α1...αk[s]α1 . . . [s]αk σj−k
+
j−1X
k=1
h
F i
sku′j−k
i
α1...αkβ1...βj−k
[s]α1 . . . [s]αk [Σ]β1...βj−k σj−k
+
j−1X
k=1
h
F i
u′kσj−k
iβ1...βk
[Σ]β1...βk σj
+
j−2Xk=1
j−1Xm=k+1
hF i
sku′m−kσj−m
iα1...αkβ1...βm
[s]α1 . . . [s]αk [Σ]β1...βm−k σj−k
+
�F i
σj
�σj +
hF i
u′j
iβ1...βj
[Σ]β1...βj σj
�
= 0.
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Constraints on the partial derivatives
(1)[
F isj
]
α1...αj
= 0
(2)[
F iskσj−k
]
α1...αk
+[
F isku′
j−k
]
α1...αkβ1...βj−k
[Σ]β1...βj−k
+∑j−1
m=k+1
[
F isku′
m−kσj−m
]
α1...αkβ1...βm
[Σ]β1...βm−k = 0
(3)[
F iσj
]
+[
F iu′j
]
β1...βj
[Σ]β1...βj
+∑j−1
k=1
[
F iu′kσj−k
]
β1...βk
[Σ]β1...βk = 0
j = 1, . . . , p k = 1, . . . , j − 1
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F as a composition of functions
Let’s define z as
z =
yt−1
yt
yt+1
ut
= z(y, u, σ, u′) =
y
g(y, u, σ)
g(g(y, u, σ), u′, σ)
u
andF (y, u, σ, u′) = f
(
z(y, u, σ, u′))
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kth order derivatives of a composition
Faa Di Bruno formula: if y = f(z(s)), then
[
f i
sj
]
α1...αj
=
j∑
l=1
[
f i
zl
]
β1...βl
∑
c⊂Ml,j
l∏
m=1
[zsN (cm) ]
βm
αcm
where Ml,j is the set of all partitions of the set of j indiceswith l classes and N (cm) is the cardinality of class cm.Note that M1,j = {1, . . . , j} and Mj,j = {{1}, {2}, . . . , {j}}.Good news: The highest order derivatives appear only onceand are multiplied by first order derivatives.
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Recoveringgyj
Back into matrix notation. The partial derivatives unfoldalong the columns.
Fyj = fy+
(
gyj
j⊗
k=1
gy + gygyj
)
+ fygyj + D = 0
where D is a term depending on partial derivatives of g() oforder lower than j and therefor already computed. Thisrequires solving the generalized Sylvester equation
(fy+gy + fy) gyj + fy+gyj
j⊗
k=1
gy = −D,
using Kamenik’s algorithm.– p. 15/19
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Recovering other terms ingsj
Fsj = fy+
(
gyj
j⊗
k=1
gs + gygsj
)
+ fygsj + D = 0
where D is a term depending on partial derivatives of g() oforder lower than j and therefor already computed. Thisrequires solving the linear system
(fy+gy + fy) gsj = −D − fy+gyj
j⊗
k=1
gs,
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Recoveringgykσj−k
Must be solved in decreasing order of k.
Fykσj−k = fy+
(
gykσj−k
k⊗
`=1
gy + gygykσj
)
+ fygyj + D + E = 0
where D is a term not depending on gykσj−k , but on gyrσj−r ,for r > k and
[E]iα1...αk
=[
F i
sku′j−k
]
α1...αkβ1...βj−k
[Σ]β1...βj−k
+
j−1∑
m=k+1
[
F i
sku′m−kσj−m
]
α1...αkβ1...βm
[Σ]β1...βm−k
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Recoveringgykσj−k (continued)
This requires solving the generalized Sylvester equation
(fy+gy + fy) gykσj−k + fy+gykσj−k
k⊗
`=1
gy = −D − E.
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Recoveringgσj
Fσj = fy+gygσj + fy+gσj + fygσj + D + E = 0
where D is a term not depending on gσj and
[E]iβ1...βj
=[
F i
u′j
]
β1...βj
[Σ]β1...βj +
j−1∑
k=1
[
F i
u′kσj−k
]
β1...βk
[Σ]β1...βk
This requires solving the linear system
(fy+gy + fy+ + fy) gσj = −D − E
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