Advanced Monetary Theory and Policy
EPOS 2012/13
Using Dynare
Giovanni Di Bartolomeo
[email protected]
The RBC model
• The Neo-classical growth model
u2 (ct , lt )
 zt f 2 (kt , nt )
u1 (ct , lt )
ct  kt 1  zt f (kt , nt )  (1   )kt
u1 (ct , lt )  E t  u1 (ct 1 , lt 1 )  zt f1 (kt , nt )  (1   ) 
The RBC model
• The Neo-classical growth model
u2 (ct , lt )
 zt f 2 (kt , nt )
u1 (ct , lt )
ct  kt 1  zt f (kt , nt )  (1   )kt
u1 (ct , lt )  E t  u1 (ct 1 , lt 1 )  zt f1 (kt , nt )  (1   ) 
How to solve it?
Dynare
• Dynare is preprocessor and a collection of Matlab
routines, which solves DSGE models
• A suite of programs for the simulation and estimation of
rational expectation models, developed by a group of
leading applied DSGE researchers headed by Michel
Juillard since 1994
• Widely used by central banks, IMF, academics research
and in teaching graduate students
• The Dynare home page is http://www.dynare.org/, here
you can download it
• Instructions how to install Dynare can be found in the
manual (Chapter 2, Installation and configuration)
• Dynare is open source!!!!
What you do, what dynare do!!!
• You tell Dynare what the variables, shocks and
parameters of your model
• You write down the model equations
• You tell Dynare to find the steady state
• Dynare (log)linearizes the model around the steady state
... and solves the recursive equilibrium laws of motion of
the linearized model
• Finally, the model is analyzed via impulse responses,
stochastic simulations and moments
Features
•
•
•
•
•
•
•
•
•
Computes the steady state of the D(S)GE model
Computes the solution of deterministic models
Computes the first and second order approximation of
linear/non-linear stochastic models
Estimates the parameters of DSGE models using MLE
or Bayesian methods
Computes optimal policy for LQ economies
Simple regression tool
Useful checking tool
No or little programming skills required
The general model
• The model
• Et{f( yt+1, yt, yt−1, vt; θ)} = 0
– y: vector of endogenous (state and jump) variables
– v: vector of exogenous shocks
– θ: vector of model parameters
– f(.): linear or non-linear function
• Steady state: y = {y : [f(yt+1, yt, yt-1, vt; θ)] = 0, t →∞}
• Solution: yt = g(yt-1,vt; ψ)
• Compute statistics of interest: zt = h(vt; ψ, y0, ybar)
Labeling and parameter value blocks
• Labeling block: indicate (list) which symbols indicate
what
– (endogenous) variables in "var"
– exogenous shocks in "varexo"
– parameters in "parameters"
• Parameter values block: Assign values to parameters
• You may also find it useful to give some parameter
transformations. e.g. beta is a parameter, and you
define steady state real interest rate: rr = 1/beta −1
Model block
• The model block starts with
model;
... and ends with
end;
• Between ’model’ and ’end’ you write down the necessary
equations
– defining the dynamic equilibrium of the model firstorder conditions
– constraints
• Remember: you need as many equations as there are
variables
Notation: Time conventions
• Time indices are given in parenthesis
Xt+1 is written X(+1),
Xt−1 is written X(−1)
Xt is written X (no time index needed)
• Note In Dynare, the time index refers to the period when
the value of the variable is determined
Example
• The value of the capital stock, which is used in
production in period t, is determined in period t − 1
• In period t − 1, the agents decide how much to consume
and invest, and they simultaneously determine the size
of the capital stock that will be available in period t.
• In a theory model we often write the period t production
function
Yt= ZtKtαLt1−α
• In Dynare we need to write
Y = Z*K(−1)^(alpha)*L^(1−alpha)
• Note that we now use L for labor (do not make confusion
with leisure)
The RBC model (dynare notation)
• The model is non-linear. Difficult to solve!!!!
 Ct  1    Z t Kt1 Lt 
Ct  Kt  Z t Kt1 L1t  (1   ) Kt 1
 Ct 1 
  Zt 1 Kt 1 
Et 
  Et 
  (1   )
 1
  Ct 
 Lt 1

State-space form
•
Generalised state-space form (matrix notation)
A0 Et X t 1  A1 X t  B0vt 1
•
Many techniques available to solve this class of models
Dynare uses standard: Blanchard-Kahn
•
We will discuss this issue later on (other classes)
•
Linearization
• In the long-run the model in on the steady state.We want
to evaluate what occurs after a shock
– If the economy goes back to the steady state
(stability)
– and how (path followed by the variables after the
shock in backing to the steady state
• Linearize the model, consider a linear approximation
(Taylor expansion) of the non-linear model around the
steady state and compute the effects of small
perturbations around steady state
• Our results are good is we are in the neighborhood of
the steady state (small perturbations)
• After we will go back on this issue
Linearization and log-linearization
• Dynare linearizes the model around the steady state
• It does not log-linearize the model
• Advantages of log-linearization: We are dealing with
percentage deviations from the steady state (or the
balanced growth path)
• Note:
log(Y)log(Yss) = % deviation from the steady state
• Log-deviations (or percentage deviations) are easy to
interpret (is a deviation ’small’ or ’large’?)
• Similar measures are used in empirical examination of
the data (e.g. applying HP filter to log(GDP))
Log-linearization: A useful trick
• A useful trick: Write down the model in terms of
logarithmic transformations of the original variables
• Adopt the notation
ly = log(Y), lc = log(C), lk = log(K) etc.
• Then use
Y = exp(ly), C = exp(lc), K=exp(lk) etc.
• When Dynare linearizes the model in terms of the
logarithmic transformations, it log-linearizes the model in
terms of the original variables
Log-linearization in practice
• Adopt the notation
ly = log(Y), lc = log(C), lk = log(K) etc.
• Note: For employment we use the notation
lh = log(L)
• where h refers to hours worked
• Then the resource constraint
Yt = Ct + Kt+1 − (1− δ)Kt
• can be written as
exp(ly) = exp(lc) + exp(lk) − (1− delta) ∗ exp(lk(−1))
• Remember the Dynare conventions pertaining to time
indexation
Conditional expectations
• The expectation operator Et is not used in Dynare code.
– Dynare ’knows’ when one has to take expectations,
we do not have to tell this explicitly.
• E.g., the consumption Euler equation
 Ct 1 
 Yt 1 
Et 
   Et 
  (1   )
  Ct 
 Kt 
1
• is written as
(1/beta)∗ exp(lc(+1) − lc) = alfa ∗ (exp(y(+1) − lk) + 1− δ
Dynare program blocks
• Initialization block: Dynare has to solve for the steady
state. This can be the most diffcult part (since it is a true
non-linear problem). So good initial conditions are
important
• Random shock block: Indicate the standard deviation for
the exogenous innovation
Initialization block
• It is often useful to give Dynare an initial guess of the
steady state values, after the command
initval;
• Dynare finds the steady state with the command
steady;
Random shock block
shock;
var e; we shock only one ’variable’ (the TFP shock e)
stderr 0.007; the standard error of the TFP shock is 0.007
– If there are several shocks, say e_z and e_c, you
have to be more careful: stderr e_z 0.007; stderr e_c
0.002;
– More generally, if there are several shocks, you may
want to give the variance-covariance matrix of the
shocks;
end;
Stochastic simulation (IRF)
stoch_simul(order=1,irf=100) ly lh lc li lk z;
– order=1 means that Dynare takes a first order Taylor
approximation around the steady state (order=2 =>
second order approximation, order=3 …)
– irf=100 means that you want Dynare to compute the
impulse responses for 100 periods
• Dynare computes the moments of the endogenous
variables mean, standard deviation, variance, skewness,
kurtosis (contemporaneous) cov-matrix autocorrelations
• The list of variables, we want to Dynare to analyze is ly
lh lc li lk z
• Notice that z ≡ log(Zt)−log(Z) = log (Zt), since Z=1 and
log(Z)=0 (a useful normalization).
Stochastic simulation (IRF)
• Alternatively we can write:
stoch_simul(order=1,periods=1000,irf=100);
• Now Dynare simulates the model for 1000 periods, and
computes the moments based on the simulated data
The structure of the code in a nutshell
•
•
•
•
•
•
•
•
•
•
•
•
•
var (list of endogenous variables );
varexo (list of shocks) ;
parameters (parameters + parameter values + transformations);
model;
model equations;
end;
initval; (initial guesses for computing the steady state)
steady; (compute the steady state)
shocks; (the shock structure of the model )
var (what variables are shocked);
stderr (the standard error of the shocks);
end;
stoch_simul(order=1,irf=100) ly lh lc li lk z; (analyzing the model)
The RBC DSGE model (dynare notation)
• Our RBC model (social planner)
 Ct  1    Z t K t1Lt 
Ct  K t  Z t K t1L1t  (1   ) K t 1
 1
 Ct 1 
 Z t 1 K t 
Et 

   Et 
  Ct 
 Lt 1 
 (1   )
Now recall that
• Competitive markets imply
Wt  MPL
rt 1  MPK  (1   )
• From the production function definition
MPL  1    Zt K L


t 1 t
MPK  Zt 1  Kt / Lt 1 
 1    Yt / Lt
 1
  Yt 1 / Kt 
Full micro-founded RBC model
1/  Ct  E t  rt 1 / Ct 1 
[Euler equation, SD ]
 Ct  Wt
[Labor supply]
K t  I t  (1   ) K t 1
[Capital accumulation]
Yt  Z t K t1 Lt 
[Production function]
Yt  I t  Ct
[Market equilibrium]
Wt  1    Yt / Lt
[Market equilibrium]
rt 1   E t Yt 1 / K t   (1   ) [Real net interest rate, SS ]
log Z t  1    log( Z )   log Z t 1   t ,  t
iid [TFP]
Full micro-founded RBC model
1/  Ct  E t  rt 1 / Ct 1 
[Euler equation, SD ]
 Ct  Wt
[Labor supply]
K

I

(1


)
K
[Capital
accumulation]
t
t
t

1
Check that the social planner model and
 
Ythe

Z
K
[Producti
function]
t
t t 1 Lt
micro-foundend
one
areonthe
same
Yt  I t  Ct First
Wt  1    Yt / Lt
[Market equilibrium]
welfare theorem!!!
[Market equilibrium]
rt 1   E t Yt 1 / K t   (1   ) [Real net interest rate, SS ]
log Z t  1    log( Z )   log Z t 1   t ,  t
iid [TFP]
RBC DSGE model (after some algebra)
 Ct 1 
 Yt 1 
Et 
  (1   )
   Et 
  Ct 
 Kt 
 
 Ct  1    Z t K t 1 Lt

1
t 1 t
Ct  K t  Z t K L
 (1   ) K t 1
Yt  Z t K t1 Lt 
I t  Yt  Ct
Wt  1    Yt / Lt
log Z t  1    log( Z )   log Z t 1   t ,  t
iid
1. Definitions
// following are the ’endogenous’ variables:
var ly, lh, lc, li, lw, lk, z ;
// following are the shocks
varexo e;
parameters beta, fi, alpha, delta, psi;
alpha = 0.33;
fi = 0.74;
beta = 0.99;
delta = 0.024;
psi =0.262;
Steady state (solved)
1/     K ss / Lss 
 1
 (1   )
 Css  1    K ss Lss
 1
Css  K ss Lss   K ss
Yss  K ss Lss
I ss  Yss  Css
Wss  1    Yss / Lss
log Z ss  0, Z  1  ss  0
Steady state (solved)
// Steady state expressed in terms of parameters
rr = 1/beta-1;
zss = 1;
css = (1alpha)*(zss/psi)*(alpha*zss/(rr+delta))^(alpha/(1alpha));
kss = css/((rr+delta)/alpha-delta);
hss = (1/psi)*(1-alpha*rr/(rr+(1-alpha)*delta));
yss = zss*kss^(alpha)*hss^(1-alpha);
iss = delta*kss;
wss = (1-alpha)*yss/hss;
2. The model
 Ct 1 
 Yt 1 
Et 
  (1   )
   Et 
  Ct 
 Kt 
 Ct  1   Yt / Lt 
Ct  K t  Z t K t1 L1t  (1   ) K t 1
Yt  Z t K t1 Lt 
I t  Yt  Ct
Wt  1    Yt / Lt
log Z t  1    log( Z )   log Z t 1   t ,  t
iid
(a) Euler equation
(1/beta)*exp(lc(+1)-lc)= alpha*exp(ly(+1)-lk) + 1 - delta;
 Ct 1 
 Yt 1 
Et 
  (1   )
   Et 
  Ct 
 Kt 
rt 1
Inter-temporal marginal rate of substitution between
consumption today and tomorrow is equal to the real
interest rate (marginal product of capital) [dynamic]
(b) Labor market equilibrium
psi*exp(lc)= (1-alpha)*exp(ly-lh);
 Yt 
 Ct  1     
 Lt 
The marginal rate of substitution between consumption and
leisure should be equated to the wage rate (marginal
product of labor) [static]
(c) Capital accumulation
exp(ly) = exp(lc) + exp(lk) - (1-delta)*exp(lk(-1));

1
t 1 t
Zt K L
 Ct  Kt  (1   ) Kt 1
Yt
Note that
Yt  Ct  Kt  (1   ) Kt 1
Yt  Ct  Kt  (1   ) K t 1
I t  Kt  (1   ) Kt 1
Kt  I t  (1   ) K t 1
(d) Production function
exp(ly) = exp(z+alpha*lk(-1)+(1-alpha)*lh);
Yt  Zt Kt1Lt 
Cobb-Douglas production function
(e) Investment
exp(li)=exp(ly)-exp(lc);
It  Yt  Ct
Maybe, more familiar identity: Y = C + I
(f) Real wage
exp(lw)=(1-alpha)*exp(ly-lh);
Wt  1    Yt / Lt
Real wage is equal to the marginal product of labor
Recall that

 Kt 1 
Yt
MPL  1    Zt 
  1   
Lt
 Lt 
(g) Law of motion of TFP
z = fi*z(-1) + e;
log Zt  1    log( Z )   log Zt 1   t ,  t
iid
• Note for later use: we assume that the steady state value
of the total factor productivity is one, its log is zero
Model: Summary
model;
(1/beta)*exp(lc(+1)-lc)= alpha*exp(ly(+1)-lk) + 1 delta; //Euler equation
psi*exp(lc)= (1-alpha)*exp(ly-lh); //Labour market
equilibrium
exp(ly) = exp(lc) + exp(lk) - (1-delta)*exp(lk(-1));
//Resource constraint
exp(ly) = exp(z+alpha*lk(-1)+(1-alpha)*lh);
//Production function
exp(li)=exp(ly)-exp(lc); // Investment
exp(lw)=(1-alpha)*exp(ly-lh); //Real wage
z = fi*z(-1) + e; //Law of motion of TFP
end;
4. Initial values
//Initial guesses for the computation of steady
state
initval;
lc=log(css);
lh=log(hss);
lk=log(kss);
ly=log(yss);
li=log(iss);
lw=log(wss);
z=zss;
end;
steady;
5. Stochastic block and simulation
shocks;
var e;
stderr 0.007;
end;
stoch_simul(order=1,irf=100) ly lh lc li lk z;
To get the results
• Save the file as [Name of the file].mod
• For example rbc.mod
• You run the model by writing to the Matlab command
window
dynare rbc.mod
Dynare output
• The output is shown on the screen, and in separate
figures (impulse responses).
• Output includes
–
–
–
–
–
–
Policy and transition functions
Moments of the endogenous variables
mean, variance, standard deviation, skewness, kurtosis
matrix of contemporaneous correlations
coefficients of autocorrelation
Impulse responses
• If you have included the periods option in stoch_simul
the output also includes the results (time paths of
endogenous variables) from the stochastic simulation.
Output storing
• The output is shown on the screen, and in separate
figures (impulse responses).
• Output is also stored in a separate structure, called oo_.
The structure oo_ contains (for example)
– The steady state (oo_.steady_state )
– The variance-covariance matrix (oo_.var)
– The autocorrelations (oo_.autocorr)
– The impulse responses (oo_.irfs)
– The coefficients of the policy and transition functions
(oo_.dr)
– Results (time paths of endogenous variables) from
stochastic
simulations (oo_.endo_simul)
Dynare output
Type dynare rbc.mod
Starting Dynare (version 4.3.0).
Starting preprocessing of the model file ...
Found 7 equation(s).
Evaluating expressions...done
Computing static model derivatives:
- order 1
Computing dynamic model derivatives:
- order 1
Processing outputs ...done
Preprocessing completed.
Starting MATLAB/Octave computing.
Steady state
STEADY-STATE RESULTS:
ly
lh
lc
li
lw
lk
z
0.02
-1.10
-0.24
-1.44
0.72
2.23
0.00
• The term constant is the steady state (of log
transformation)
• E.g. lh = -1.1 and thus L = exp(-1.1) = 0.33
Summary
MODEL SUMMARY
Number of variables:
Number of stochastic shocks:
Number of state variables:
Number of jumpers:
Number of static variables:
7
1
2
2
3
Shock co-variance
MATRIX OF COVARIANCE OF EXOGENOUS SHOCKS
Variables
e
e
0.000049
Policy and transition functions
POLICY AND TRANSITION FUNCTIONS
ly
lh
lc
li
Const 0.018 -1.099 -0.245
-1.441
lk(-1) -0.007 -0.503 0.496
-1.672
z(-1)
1.868
1.683 0.184
7.433
e
2.524
2.275 0.249
10.045
lk
2.288
0.935
0.178
0.241
z
0.000
0.000
0.740
1.000
• Transition functions: how the period t values of the state variables (lk
and z) depend on t-1 values of the state variables, and the shock
• Policy functions: how the period t values of the other variables
depend on t-1 values of the state variables, and the shock
– Rows: period t values of the variables
– Columns: period t-1 values of the state variables + the shock (e)
• The term constant is the steady state (of log transformation)
Moments
THEORETICAL MOMENTS
VARIABLE
ly
lh
lc
li
lk
MEAN
0.0186
-1.0994
-0.2457
-1.4414
2.2884
STD. DEV.
0.0262
0.0221
0.0095
0.0980
0.0167
VARIANCE
0.0007
0.0005
0.0001
0.0096
0.0003
z
0.0000
0.0104
0.0001
Correlations
MATRIX OF CORRELATIONS
Variables
ly
lh
lc
li
lk
z
ly
1.0000
0.9363
0.5824
0.9652
0.4846
1.0000
lh
0.9363
1.0000
0.2597
0.9956
0.1465
0.9347
lc
0.5824
0.2597
1.0000
0.3494
0.9933
0.5860
li
0.9652
0.9956
0.3494
1.0000
0.2388
0.9640
lk
0.4846
0.1465
0.9933
0.2388
1.0000
0.4885
z
1.0000
0.9347
0.5860
0.9640
0.4885
1.0000
Autocorrelation
COEFFICIENTS OF AUTOCORRELATION
Order
ly
lh
lc
li
lk
z
1
0.7389
0.6860
0.9811
0.6931
0.9901
0.7400
2
0.5458
0.4571
0.9517
0.4690
0.9668
0.5476
3
0.4030
0.2910
0.9155
0.3060
0.9345
0.4052
4
0.2974
0.1711
0.8751
0.1880
0.8966
0.2999
5
0.2192
0.0852
0.8326
0.1031
0.8553
0.2219
Impulse response functions (IRF)
How to write the solution in a VAR form
POLICY AND TRANSITION FUNCTIONS
Dynare output
Log(y)
Log(h)
Log(c)
Log(i)
Log(k)
z
Constant
0,019
-1,099
-0,246
-1,441
2,288
0,000
Log(k(-1))
-0,008
-0,504
0,496
-1,673
0,936
0,000
z(-1)
1,868
1,684
0,184
7,434
0,178
0,740
e
2,524
2,275
0,249
10,046
0,241
1,000
0,333
0,782
0,237
9,859
1,000
Levels Exp(Constant)
Steady state
1,019
We rewrite the table in equations
Trasfor
m
Dynare output
ly
lh
lc
li
lk
z
ly = 0,019 + -0,008 lk(-1)- 2,288
+ 1,868 z(-1)
+ 2,524
e
Constant 0,019 -1,099 -0,246 -1,441 2,288 0,000
lh= -1,099 + -0,504 lk(-1)- 2,288
+ 1,684 z(-1)
+ 2,275
e
lk(-1)
-0,008 -0,504 0,496 -1,673 0,936 0,000
lc= -0,246 + 0,496
lk(-1)- 2,288
+ 0,184 z(-1)
+ 0,249
e
z(-1)
1,868 1,684 0,184 7,434 0,178 0,740
li= -1,441 + -1,673 lk(-1)- 2,288
+ 7,434 z(-1)
+ 10,046
e
e
2,524 2,275 0,249 10,046 0,241 1,000
lK= 2,288 + 0,936
lk(-1)- 2,288
+ 0,178 z(-1)
+ 0,241
e
z= 0,000 + 0,000
lk(-1)- 2,288
+ 0,740 z(-1)
+ 1,000
e
We rewrite the table in equations
Trasfor
m
Dynare output
ly
lh
lc
li
lk
z
ly = 0,019 + -0,008 lk(-1)- 2,288
+ 1,868 z(-1)
+ 2,524
e
Constant 0,019 -1,099 -0,246 -1,441 2,288 0,000
lh= -1,099 + -0,504 lk(-1)- 2,288
+ 1,684 z(-1)
+ 2,275
e
lk(-1)
-0,008 -0,504 0,496 -1,673 0,936 0,000
lc= -0,246 + 0,496
lk(-1)- 2,288
+ 0,184 z(-1)
+ 0,249
e
z(-1)
1,868 1,684 0,184 7,434 0,178 0,740
li= -1,441 + -1,673 lk(-1)- 2,288
+ 7,434 z(-1)
+ 10,046
e
e
2,524 2,275 0,249 10,046 0,241 1,000
lK= 2,288 + 0,936
lk(-1)- 2,288
+ 0,178 z(-1)
+ 0,241
e
z= 0,000 + 0,000
lk(-1)- 2,288
+ 0,740 z(-1)
+ 1,000
e
We rewrite the table in equations
Trasfor
m
Dynare output
ly
lh
lc
li
lk
z
ly = 0,019 + -0,008 lk(-1)- 2,288
+ 1,868 z(-1)
+ 2,524 e
Constant 0,019 -1,099 -0,246 -1,441 2,288 0,000
lh= -1,099 + -0,504 lk(-1)- 2,288
+ 1,684 z(-1)
+ 2,275 e
lk(-1)
-0,008 -0,504 0,496 -1,673 0,936 0,000
lc= -0,246 + 0,496
lk(-1)- 2,288
+ 0,184 z(-1)
+ 0,249 e
z(-1)
1,868 1,684 0,184 7,434 0,178 0,740
li= -1,441 + -1,673 lk(-1)- 2,288
+ 7,434 z(-1)
+ 10,046 e
e
2,524 2,275 0,249 10,046 0,241 1,000
lK= 2,288 + 0,936
lk(-1)- 2,288
+ 0,178 z(-1)
+ 0,241 e
z= 0,000 + 0,000
lk(-1)- 2,288
+ 0,740 z(-1)
+ 1,000 e
We rewrite the table in equations
Trasfor
m
Dynare output
ly
lh
lc
li
lk
z
ly = 0,019 + -0,008 lk(-1)- 2,288
+ 1,868 z(-1)
+ 2,524 e
Constant 0,019 -1,099 -0,246 -1,441 2,288 0,000
lh= -1,099 + -0,504 lk(-1)- 2,288
+ 1,684 z(-1)
+ 2,275 e
lk(-1)
-0,008 -0,504 0,496 -1,673 0,936 0,000
lc= -0,246 + 0,496
lk(-1)- 2,288
+ 0,184 z(-1)
+ 0,249 e
z(-1)
1,868 1,684 0,184 7,434 0,178 0,740
li= -1,441 + -1,673 lk(-1)- 2,288
+ 7,434 z(-1)
+ 10,046 e
e
2,524 2,275 0,249 10,046 0,241 1,000
lK= 2,288 + 0,936
lk(-1)- 2,288
+ 0,178 z(-1)
+ 0,241 e
z= 0,000 + 0,000
lk(-1)- 2,288
+ 0,740 z(-1)
+ 1,000 e
Now we manipulate our equations
Given
ly =
0,019
+
-0,008
lk(-1)- 2,288
+
1,868
z(-1)
+
2,524
e
lh=
-1,099
+
-0,504
lk(-1)- 2,288
+
1,684
z(-1)
+
2,275
e
lc=
-0,246
+
0,496
lk(-1)- 2,288
+
0,184
z(-1)
+
0,249
e
li=
-1,441
+
-1,673
lk(-1)- 2,288
+
7,434
z(-1)
+
10,046
e
lK=
2,288
+
0,936
lk(-1)- 2,288
+
0,178
z(-1)
+
0,241
e
+
0,000
lk(-1)- 2,288
+
0,740
z(-1)
+
1,000
e
z
We move the steady state on the left
ly -
0,019
=
-0,008
lk(-1)- 2,288
+
1,868
z(-1)
+
2,524
e
lh -
-1,099
=
-0,504
lk(-1)- 2,288
+
1,684
z(-1)
+
2,275
e
lc -
-0,246
=
0,496
lk(-1)- 2,288
+
0,184
z(-1)
+
0,249
e
li -
-1,441
=
-1,673
lk(-1)- 2,288
+
7,434
z(-1)
+
10,046
e
lk -
2,288
=
0,936
lk(-1)- 2,288
+
0,178
z(-1)
+
0,241
e
=
0,000
lk(-1)- 2,288
+
0,740
z(-1)
+
1,000
e
z
Obtaining the log deviations
y=
-0,008
k(-1)
+
1,868
z(-1)
+
2,524
E
h=
-0,504
k(-1)
+
1,684
z(-1)
+
2,275
E
c=
0,496
k(-1)
+
0,184
z(-1)
+
0,249
E
i=
-1,673
k(-1)
+
7,434
z(-1)
+
10,046
E
k=
0,936
k(-1)
+
0,178
z(-1)
+
0,241
E
z=
0,000
k(-1)
+
0,740
z(-1)
+
1,000
e
Looking at the shock
Form log deviations (y=log(y)-log(yss) …)
y = -0,008
k(-1)
h = -0,504
k(-1)
c=
0,496
k(-1)
i = -1,673
k(-1)
k=
0,936
k(-1)
z=
0,000
k(-1)
+
+
+
+
+
+
1,868
1,684
0,184
7,434
0,178
0,740
Note
z =
Multiply
2,524
2,275
0,249
10,046
0,241
*z
*z
*z
*z
*z
=
=
=
=
=
phi
0,740
z(-1)
+
e
1,868
1,684
0,184
7,434
0,178
z(-1)
z(-1)
z(-1)
z(-1)
z(-1)
2,52
2,27
0,24
10,0
0,24
+
+
+
+
+
e
e
e
e
e
z(-1)
z(-1)
z(-1)
z(-1)
z(-1)
z(-1)
+ 2,524 e
+ 2,275 e
+ 0,249 e
+ 10,046 e
+ 0,241 e
+ 1,000 e
Looking at the shock
Form log deviations
y=
h=
c=
i=
k=
z=
-0,008
-0,504
0,496
-1,673
0,936
0,000
k(-1)
k(-1)
k(-1)
k(-1)
k(-1)
k(-1)
+
+
+
+
+
+
1,868
1,684
0,184
7,434
0,178
0,740
z(-1)
z(-1)
z(-1)
z(-1)
z(-1)
z(-1)
2,524
2,275
0,249
10,046
0,241
e
z
z
z
z
z
Note that
z
=
phi
0,740
z(-1)
+
e
2,524
*z
=
1,868
z(-1)
2,524
+ e
2,275
*z
=
1,684
z(-1)
2,275
+ e
0,249
*z
=
0,184
z(-1)
0,249
+ e
10,046
*z
=
7,434
z(-1)
10,05
+ e
0,241
*z
=
0,178
z(-1)
0,241
+ e
k(-1)
k(-1)
k(-1)
k(-1)
k(-1)
z(-1)
+
+
+
+
+
+
Multiply by
Hence we can write
y=
h=
c=
i=
k=
z=
-0,008
-0,504
0,496
-1,673
0,936
0,740
+
+
+
+
+
+
2,524
2,275
0,249
10,046
0,241
1,000
e
e
e
e
e
e
Compact form
X t  AX t 1  BZt
Zt  Zt 1  St
y = -0,008 k(-1)
h = -0,504 k(-1)
c = 0,496 k(-1)
i = -1,673 k(-1)
k = 0,936 k(-1)
z = 0,740 z(-1)
+
+
+
+
+
+
2,524 z
2,275 z
0,249 z
10,046 z
0,241 z
e