Dynare
Wouter J. Den Haan
University of Amsterdam
July 26, 2010
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Introduction
• What is the objective of perturbation?
• Peculiarities of Dynare
• Some examples
Introduction
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Objective of 1st-order perturbation
• Obtain linear approximations to the policy functions that satisfy the first-order conditions
• state variables: xt = [x\/t Xijt %3,t • • • xn,t]'
• result:
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Underlying theory
• Model:
Et f (g(x))] = 0,
• f (x) is completely known
• g(x) is the unknown policy function.
• Perturbation: Solve sequentially for the coefficients of the Taylor expansion of g(x).
• More info:
• notes and slides on perturbation
• slides on Blanchard-Kahn conditions
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Neoclassical growth model
• xt = [kt-i, zt]
• yt = [ct, kt, zt ]
• linearized solution:
ct = C + flc,k(kt-i - k)+ flc,z(zt - z)
kt = k + flk,k(kt-i - k) + flk,z(zt - z)
zt = pzt_i + £t
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Linear in what variables?
• Dynare does not understand what a is.
• could be level of consumption
• could be log of consumption
• could be rainfall in Scotland
• Dynare simply generates a linear solution in what you specify as the variables
• More on this below
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Peculiarities of Dynare
• Variables known at beginning of period t must be dated t — 1.
• Thus,
• kt: the capital stock chosen in period t
• kt—1: the capital stock available at beginning of period t
Introduction Do it yourself Tricks IRFs & Simulations Pruning
Peculiarities of Dynare
The solution
ct = c + aCfk(kt-i - k) + ac,z(zt - z) kt = k + ak,k(kt-i - k) + akiz (zt - z)
Zt = pZt-i + £t can of course be written (less conveniently) as
ct = C + (kt-i - k)+ ac,z_i (zt-i - z) + aCiz£t kt = k + akik(kt-i - k)+ akiz_x (zt-i - z) + akiz£t
zt = pzt-i + £t
with aCiz_x = pac,z and akiz_i = pakiz
Practical
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Peculiarities of Dynare
• Dynare gives the solution in the less convenient form:
ct = c + aC/k(kt-i - k) + ac,z_t (zt-\ - z) + aCiZ£t kt = k + aKk(kt-i ~ k) + akfZ_r (zt-i ~ z) + ak,z£t
Zt = pZt-1 + £t
• Since the Dynare solution satisfies
ac,z_x = pac,z and a^fz_x = pak,z
one could always rewrite the Dynare solution in the more convenient form
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Dynare program blocks
• Labeling block: indicate which symbols indicate what
• variables in "var"
• exogenous shocks in "varexo"
• parameters in "parameters"
• Parameter values block: Assign values to parameters
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Dynare program blocks
• Model block: Between "model" and "end" write down the n equations for n variables
• note that dynare has no conditional expectations but if an equation has a (+1) variable, then Dynare knows there is a conditional expectation
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Dynare program blocks
• Initialization block: Dynare has to solve for the steady state. This can be the most difficult 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
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Dynare program blocks
• Solution & Properties block:
• Solve the model with the command
• 1st-order: stoch_simul(order=1,nocorr,nomomentsJRF=0)
• 2nd-order: stoch_simul(order=2,nocorr,nomomentsJRF=0)
• Dynare can calculate IRFs and business cycle statistics. E.g.,
• stoch_simul(order=1,IRF=30),
• but I would suggest to program this yourself (see below)
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Running Dynare
• In Matlab change the directory to the one in which you have your *.mod files
• In the Matlab command window type
dynare programname
• This will create and run several Matlab files
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Model with productivity in levels (FOCs A)
Specif cation of the problem
c}-v
max{ct,ktg E £r=i s.t.
ct + kt = ztk*_x + (1 - S)kt-i zt = (1 - p)+ pzt-i + £t ko given
Et[et+i] = 0 & Et[£2+i] = a2
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Distribution of innovation
• 1st-order approximations:
• the distribution of et does not matter, except that Ef [et+i] has to be zero.
• 2nd-order approximations:
• u matters (it affects the mean)
• higher-order moments do not
• Also see notes and slides on perturbation theory
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Everything in levels: FOCs A
Model equations:
ct + kt zt
Dynare equations: c~(-nu)
=beta*c(+1)~(-nu)*(alpha*z(+1)*k~(alpha-1)+1-delta); c+k=z*k(-1)~alpha+(1-delta)k(-1); z=(1-rho)+rho*z(-1)+e;
ztkU + (1 - 5)kt-i (1 - p)+ pzt-1 + ^
+1
6)
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Policy functions reported by Dynare
• 5 = 0.025,v = 2, a = 0.36, ß = 0.99, and p = 0.95
POLICY AND TRANSITION FUNCTIONS
constant k(-1)
e
k
37.989254 0.976540 2.597386 2.734091
z
1.000000
-0.000000
0.950000
1.000000
c
2.754327 0.033561 0.921470 0.969968
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
!!!! You have to read output as
constant k(-l)-kss z(-1)-zss e
k
37.989254 0.976540 2.597386 2.734091
z
1.000000 -0.000000 0.950000 1.000000
c
2.754327 0.033561 0.921470 0.969968
• That is, explanatory variables are relative to steady state.
• (Note that steady state of e is zero by definition)
• If explanatory variables take on steady state values, then choices are equal to the constant term, which of course is simply equal to the corresponding steady state value
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Changing amount of uncertainty
Suppose <j = 0.1 instead of 0.007 POLICY AND TRANSITION FUNCTIONS
constant
k(-1)
z(-1)
e
k
37.989254 0.976540 2.597386 2.734091
z
1.000000 -0.000000 0.950000 1.000000
c
2.754327 0.033561 0.921470 0.969968
• Any change?
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Model with productivity in logs
Specification of the problem
c}~v - 1
max E V B*-1^
{ct,kt}    ttl 1 - V
s.t.
ct + kt = exp(zt+ (1 - 5)kt-i
Zt = pZt-l + £t
ko given, Et[et+i] = 0
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Variables in levels & prod. in logs - FOCs B
Dynare equations: c~(-nu)
=beta*c(+1)~(-nu)*(alpha*exp(z(+1))*k~(alpha-1)+1-delta);
c+k=exp(z)*k(-1)~alpha+(1-delta)k(-1);
z=rho*z(-1)+e;
Model equations:
—V
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Policy functions reported by Dynare
• 5 = 0.025,v = 2, a = 0.36 and ff = 0.99 POLICY AND TRANSITION FUNCTIONS
constant
k(-1) z(-1)
e
k
37.989254 0.976540 2.597386 2.734091
z
0.000000 -0.000000 0.950000 1.000000
c
2.754327 0.033561 0.921470 0.969968
• What does z stand for here?
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Linear solution in what?
Dynare gives a linear system in what you specify the variables to be
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
All variables in logs - FOCs C
Model equations:
(exp(ct ))~v = = Et [/Kexp(cm))-17(a exp(zm)(exp(kt))a"1 + 1 - 5)
exp(ct) + exp(kt) = exp(Zt)(exp(kt_i))a + (1 - 5) exp(kt_i)
zt = pzt_i + et
The variables ct and kt are the log of consumption and capital.
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
All variables in logs - FOCs C
Model equations (rewritten a bit)
exp (-vct)
= Et [ßexp(-vCf+i)(aexp(Zt+i + (a - 1)kt) + 1 - 5)
exp(Ct) + exp(kt)  = exp(Zt + akt-i) + (1 - 5) exp(kt_i)
Z t  = pz t-i + et
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
All variables in logs - FOCs C
Dynare equations:
exp(-nu*lc)=beta*exp(-nu*lc(+1))* (alpha*exp(lz(+1)+(alpha-1)*lk))+1-delta); exp(lc)+exp(lk)
=exp(lz+alpha*lk(-1))+(1-delta)exp(lk(-1)); lz=rho*lz(-1)+e;
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
All variables in logs - FOCs C
• This system gives policy functions that are linear in the variables lc, i.e., ln(ct), lk,i.e., ln(kt), and lz, i.e., ln(zt),
• Programmers often do not make clear that a variable is a log. That is, they would simply use c, k, and z in the dynare equations above instead of lc, lk, and lz
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
All variables in logs - FOCs C
Dynare equations (with different notation):
exp(-nu*c)=beta*exp(-nu*c(+1))*
(alpha*exp(z(+1)+(alpha-1)*k))+1-delta);
exp(c)+exp(k)=exp(z+alpha*k(-1))+(1-delta)exp(k(-1));
z=rho*z(-1)+e;
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
POLICY AND TRANSITION FUNCTIONS for foc B
k z c
constant 37.989254 0.000000 2.754327 k(-1)       0.976540    -0.000000 0.033561
z(-1) 2.597386 0.950000 0.921470 e        2.734091   1.000000 0.969968
POLICY AND TRANSITION FUNCTIONS for foc C kzc
constant 3.637303 0.000000 1.013173
k(-1) 0.976540 0.000000 0.462887
z(-1) 0.068372 0.950000 0.334554
e 0.071970 1.000000 0.352162
Introduction
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IRFs & Simulations
Pruning
These are not the same solutions
Suppose that k0 = 49.3860 & zt = 0 Vt
49.5
49 48.5
48 47.5
47 46.5
46 45.5
45 44.5
0
Tricks
Practical
5
10
15
20
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Example with analytical solution
• If 5 = v = 1 then we know the analytical solution. It is
kt = af exp(zt )kat_1 ct = (1 - af) exp(zt)kcl_x
or
ln kt  =  ln(af) + a ln kt_1 + zt ln ct  =  ln(1 — af) + a ln kt_1 + zt
• That is, the policy rules are linear in the logs
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Dynare solutions I
Dynare equations (with consumption and capital in logs): exp(-c)
=beta*exp(-c(+1))*(alpha*exp(z(+1)+(alpha-1)*k)));
exp(c)+exp(k)=exp(z+alpha*k(-1));
z=rho*z(-1)+e;
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Dynare solutions I
• Dynare solution (a = 0.36 and ft = 0.99)
k z c
constant   -1.612037 0.000000 -1.021009
k(-1)       0.359999 -0.000000 0.360000
z(-1)        0.949998 0.950000 0.950000
e 0.999998 1.000000 1.000000
• Note that c, k, and z are logs
• Check yourself that this is correct
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Dynare solutions II
Dynare equations (with consumption and capital in levels):
1/c=beta*(1/c(+1))*(alpha*exp(z(+1))*k~(alpha-1)); c+k=exp(z)*k(-1)"alpha; z=rho*z(-1)+e;
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Dynare solutions II
• Dynare solution (a = 0.36 and ft = 0.99)
k z c
constant   0.199482 0.000000 0.360231
k(-1)       0.360000 0.000000 0.650101
z(-1)        0.189507 0.950000 0.342219
e        0.199482 1.000000 0.360231
• Note that c and k indicate levels and z logs
• This is not the same !!!
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Substitute out consumption- FOCs D
Model equations:
[zt exp(akf_i) + (1 - 5) exp(fct-i) - exp(kt)] v
E J a\  [zt+1 exp(akt) + (1 - 5) exp(kt) - exp(fct+i)] ~V x  \ \ tY\ (azt+iexp((a - 1)kt) + 1 - 5) )\
zt = (1 - p) + pzt-i + et
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Dynare solution
Dynare equations (with capital in logs):
(z*exp(alpha*lk(-1)+(1-delta)*exp(lk(-1))-exp(lk))~(-nu) =beta*(z(+1)*exp(alpha*lk+(1-delta)*exp(lk)-exp(lk(+1))~( *(alpha*exp(z(+1)+(alpha-1)*lk))+(1-delta));
z=1-rho+rho*z(-1)+e;
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Dynare solution
• Dynare solution (a = 0.36 and f = 0.99)
lk z
constant   3.637303 0.000000
lk(-1)       0.976540 0.000000
z(-1)        0.068372 0.950000
e        0.071970 1.000000
• Given this law of motion for ln(kt) you can solve ct using the non-linear equation
ct = exp(zt) exp(a * lnkt_1) + (1 — 5) exp(lnkt_1) — exp(lnkt)
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Do it yourself!
• Try to do as much yourself as possible
Introduction Do it yourself Tricks IRFs & Simulations Pruning
What (not) to do your self
• Policy functions:
• can be quite tricky so let Dynare do it.
• IRFs, business cycle statistics, etc:
• easy to program yourself
• you know exactly what you are getting
Practical
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Why do things yourself?
• Dynare linearizes everything
• Suppose you have an RBC in log of capital
• Add the following equation to introduce investment
exp(if) = exp(kt) - (1 - 3) exp(kf_i)
• Dynare will approximate this linear equation.
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Why do things yourself?
• Now suppose you have an approximation in levels
• Add the following equation to introduce output
• Dynare will take a first-order condition of this equation to get a first-order approximation for yt
• But you already have solutions for kt and ht
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Why do things yourself?
• Getting the policy rules requires a bit of programming
• Thus, it makes sense to use Dynare for this
• But the more you program yourself, the better you understand the results
• Try, therefore, to program the simpler things, like IRFs, simulated time paths, and business cycle statistics yourself, that is, simply use
• stoch_simul(order=1,nocorr,nomoments,IRF=0)
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Tricks
• Incorporating Dynare in other Matlab programs
• Reading parameter values in *.mod file from external file
• Reading Dynare policy functions as they appear on the screen
• How to get good initial conditions (to solve for steady state)
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Keeping variables in memory
• Dynare clears all variables out of memory
• To overrule this, use
dynare program.mod noclearall
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Saving solution to a file
• Replace the file "disp_dr.m" with the provided file
• I made two changes:
• The original Dynare file only writes a coefficient to the screen if it exceeds 10-6 in absolute value. I eliminated this condition
• I save the policy functions, exactly the way Dynare now writes them to the screen
To load the policy rules into the matrix "decision" simply type
load dynarerocks
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Saving solution to a file
• Note that Dynare also saves policy functions, but for second-order this is not what you see on the screen
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Saving solution to a file
• Note that Dynare also saves policy functions, but for second-order this is not what you see on the screen
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Loops
• This trick allows you to run the same dynare program for different parameter values
• Suppose your Dynare program has the command
nu=3;
• You would like to run the program twice; once for nu=3, and once for nu=5.
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Loops
[1 In your Matlab program, loop over the different values of nu. In each iteration, first save the current value of nu (and the associated name) to the file wouterrocks with
"save parametersle nu
and then run Dynare
^ In your Dynare program file, replace the command "nu = 3" with
load parameterfle set_param_value('nu',nu);
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Using loop to get good initial conditions
With a loop you can update the initial conditions used to solve for steady state
ft Use parameters to definite initial conditions ft Solve model for simpler case ft Gradually change parameter
ft You can even gradually change models using weighting coefficients
ft Alternative: (also) use different algorithm to solve for steady state
O solve_algo=1,2, or 3
[2 solve for coefficients instead of variables
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Simple model with endogenous labor
= Et [jSc^(a exp(zt+i) (kt/ht+1)a~1 + 1 - <5)~
ct + kt = exp^k^h1^ + (1 - ^)kf_i C"v(1 - a) exp(zt)(kt_1/ht)a = h?
Zt = pZt-1 + £t
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Simple model with endogenous labor
ft Solve for c,k,h using
1 = p(cc (k/h)a~x + 1 - 5) c + k = kahx~a + (1 - 5)k c~v(1 - a)(k/h)a = *hK
* = 1
ft Or solve for c,k, * using
1 = p(cc (k/h)a~x + 1 - 5) c + k = kahx~a + (1 - 5)k c~v(1 - a)(k/h)a = *hK h = 0.3
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Impulse Response functions
Definition: The effect of a one-standard-deviation shock
• Take as given A0, z0, and time series for et, fet}J=i
• Let fkt}J=i be the corresponding solutions
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Impulse Response functions
• Consider the time series ejf such that
4 = £t for t = T e*t = et + a for t = T
• Let {k"[}J= 1 be the corresponding solutions
• Impulse response functions are calculated as
IRFk = k*T+j - kT+j   for j > 0
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
IRFs for linear systems
• Value of kT-\ and values of original shock {gf}J=t irrelevant for IRFs
• Thus, make your life easy by setting
• t = 1
• gT+j = 0 for j > OTake as given ko, Zo, and time series for Et,
{gf }J=1
• If k is in logs then subtract k and you have the IRF
• If k is in levels calculate (kT+j — k)/k or ln(kT+j/k)
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Impulse Response functions
1st-order case: Dynare gives you
kt = k + akik(kt-\ - k) + akiZ_l (zt_i - z) + aki££t
• Start at k0 = k and z0 = z (= 0)
• Let e\ = ae and et = 0 for t > 1
• Calculate time path for zt
• Calculate time path for kt
• Calculate time path for other variables
• Calculate % change (subtract steady-state value if variables are in logs)
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Impulse Response functions
2nd-order case:
• One could repeat procedure described in last slide
• But with a non-linear law of motion results do depend on initial value of k, realizations of shocks in the original series, and whether eT = eT + u or z% = eT — u
• For example, IRF can be different when initial capital stock is low than when it is high
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
How to calculate a simulated data set
Dynare gives you
kt = k + ak/k(kt-i - k) + ak/Z_l (zt_i - Z) + akAzt
• Start at k0 = k and z0 = z (= 0)
• Use a random number generator to get a series for £t for t = 1 to t = T
• Calculate time path for Zt
• Calculate time path for kt
• Calculate time path for other variables
• Discard an initial set of values
• Note that procedure is the same for first and second-order solutions
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Simulate higher-order & pruning
• first-order solutions are by construction stationary
• simulation cannot be problematic
• simulation of higher-order can be problematic
• simulation of 2nd-order will be problematic for large shocks
• trick proposed: Pruning
• pruning:
• is a trick to ensure stability
• it uses a distorted numerical approximation
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Pruning
k(n)(k _1,z): the nth-order perturbation solution for k as a function of k_i and z.
• k(n): the value of kt generated with k(n)(-).
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Pruning
For n > 1, the regular perturbation solution k(n can be written
• For n
as
kin) - kss
a(n) + 4n) (k^ - kss) + a? (zt - zss) + ~k(n)(k(% zt)
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practic;
Pruning
• With pruning one would simulate two series
- kss   = - kss) + az1}(zt - Zss)
+k(n)(k(l)1,zt)
k(n) - kss
(1)
• k\   is stationary as long as BK conditions are satisfied
k(n) (k(1_)1,Zt) is then also stationary
(n)
< 1 then ensures that k\   is stationary
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Pruning
• The pruned simulated series,       is NOT a function of the corresponding state variables k)   and zt
Introduction Do it yourself Tricks IRFs & Simulations Pruning Practical
Practical
• Dynare expects files to be in a regular path like e:\... and cannot deal with subdirectories like //few.eur.nl/.../...
• The solution is to put your *.mod files on a memory stick