Optimal policy computation with Dynare MONFISPOL workshop, Stresa Michel Juillard 1 March 12, 2010 1 Bank of France and CEPREMAP
Optimal policy computation with DynareMONFISPOL workshop, Stresa
Michel Juillard1
March 12, 2010
1Bank of France and CEPREMAP
Introduction
Dynare currently implements two manners to compute optimalpolicy in DSGE models
◮ optimal rule under commitment (Ramsey policy)◮ optimal simple rules
A New Keynesian model
Thanks to Eleni Iliopulos!
Utility function:
Ut = ln Ct − φN1+γ
t
1 + γ
Uc,t =1Ct
UN,t = −φNγt
Recursive equilibrium
1Ct
=βEt
{1
Ct+1
Rt
πt+1
}
πt
Ct(πt − 1) =βEt
{πt+1
Ct+1(πt+1 − 1)
}
+ εAt
ω
Nt
Ct
(φCtN
γt
At−
ε− 1ε
)
AtNt =Ct +ω
2(πt − 1)2
ln At =ρ ln At−1 + ǫt
Ramsey problem
max E0
∞∑
t=0
βt
{ln Ct − φ
N1+γt
1 + γ
− µ1,t
(1Ct
− βEt
{1
Ct+1
Rt
πt+1
})
− µ2,t
(πt
Ct(πt − 1)− βEt
{πt+1
Ct+1(πt+1 − 1)
}
−εAt
ω
Nt
Ct
(φCtN
γt
At−
ε− 1ε
))
− µ3,t
(AtNt − Ct −
ω
2(πt − 1)2
)
− µ4,t (ln At − ρ ln At−1 − ǫt)
}
µi ,t = λi ,tβt where λi ,t is a Lagrange multiplier.
F.O.C.
µ1,t−1CtRt−1
π2t
+1Ct
[2πt − 1](µ2,t − µ2,t−1
)= µ3,tω (
1Ct
−µ1,t
C2t
+ µ1,t−1Rt−1
C2t πt
− µ2,t1
C2t
[πt (πt − 1) +
AtNt
ω(ε− 1)
]
+µ2,t−11
C2t
[πt (πt − 1)] = µ3,t
φNγt + φ
ε
ωNγ
t µ2,t (γ + 1)− µ2,tAt
Ctω(ε− 1) = µ3,tAt
µ2,t
ω
Nt
Ct(ε− 1) + µ3,tNt +
µ4,t
At−
ρβµ4,t+1
At= 0
µ1,tβEt
{1
Ct+1πt+1
}= 0
Dynare code
var pai, c, n, r, a;varexo u;parameters beta, rho, epsilon, omega, phi, gamma;
beta=0.99;gamma=3;omega=17;epsilon=8;phi=1;rho=0.95;
Dynare code (continued)
model;a = rho*a(-1)+u;1/c = beta*r/(c(+1)*pai(+1));pai*(pai-1)/c = beta*pai(+1)*(pai(+1)-1)/c(+1)
+epsilon*phi*n^(gamma+1)/omega-exp(a)*n*(epsilon-1)/(omega*c);
exp(a)*n = c+(omega/2)*(pai-1)^2;end;
Dynare code (continued)
initval;pai=1;r=1/beta;c=0.96717;n=0.96717;a=0;end;
shocks;var u; stderr 0.008;end;
planner_objective(ln(c)-phi*((n^(1+gamma))/(1+gamma)));
ramsey_policy(planner_discount=0.99,order=1);
Ramsey policy: General nonlinear case
max{yτ}∞τ=0
Et
∞∑
τ=t
βτ−tU(yτ )
s.t.Eτ f (yτ+1, yτ , yτ−1, ετ ) = 0
yt ∈ Rn : endogenous variables
εt ∈ Rp : stochastic shocks
andf : R3n+p
→ Rm
There are n − m free policy instruments.
Lagrangian
The Lagrangian is written
L (yt−1, εt) = Et
∞∑
τ=t
βτ−tU(yτ )− [λτ ]η [f (yτ+1, yτ , yτ−1, ετ )]η
where λτ is a vector of m Lagrange multipliers We adopt tensornotations because later on we will have to deal with the secondorder derivatives of [λ (st)]η [F (st)]
ηγ .
It turns out that it is the discounted value of the Lagrangemultipliers that are stationary and not the multipliersthemselves. It is therefore handy to rewrite the Lagrangian as
L (yt−1, εt ) = Et
∞∑
τ=t
βτ−t(
U(yτ )− [µτ ]η [f (yτ+1, yτ , yτ−1, ετ )]η)
whith µt = λτ/βτ−t .
Optimization problem reformulated
The optimization problem becomes
max{yτ}∞τ=t
min{λτ}
∞
τ=t
Et
∞∑
τ=t
βτ−t(
U(yτ )− [µτ ]η [f (yτ+1, yτ , yτ−1, ετ )]η)
for yt−1 given.
First order necessary conditions
Derivatives of the Lagrangian with respect to endogenousvariables:[∂L∂yt
]
α
=Et{[U1(yt )]α − [µt ]η [f2(yt+1, yt , yt−1, εt )]
ηα
− β [µt+1]η [f3(yt+2, yt+1, yt , εt+1)]ηα
}τ = t
[∂L∂yτ
]
α
=Et{[U1(yτ )]α − [µτ ]η [f2(yτ+1, yτ , yτ−1, ετ )]
ηα
− β [µτ+1]η [f3(yτ+2, yτ+1, yτ , ετ+1)]ηα
− β−1 [µτ−1]η [f1(yτ , yτ−1, yτ−2, ετ−1)]α}
τ > t
First order conditions
The first order conditions of this optimization problem are
Et{[U1(yτ )]α − [µτ ]η [f2(yτ+1, yτ , yτ−1, ετ )]
ηα
−β [µτ+1]η [f3(yτ+2, yτ+1, yτ , ετ+1)]ηα
−β−1 [µτ−1]η [f1(yτ , yτ−1, yτ−2, ετ−1)]ηα
}= 0
[f (yτ+1, yτ , yτ−1, ετ )]η = 0
with µt−1 = 0 and where U1() is the Jacobian of function U()with respect to yτ and fi() is the first order partial derivative off () with respect to the i th argument.
Cautionary remark
The First Order Conditions for optimality are only necessaryconditions for a maximum. Levine, Pearlman and Pierse (2008)propose algorithms to check a sufficient condition. I still need toadapt it to the present framework where the dynamicconditions, f () are given as implicit functions and variables canbe both predetermined and forward looking.
Nature of the solution
The above system of equations is nothing but a larger systemof nonlinear rational expectation equations. As such, it can beapproximated either to first order or to second order. Thesolution takes the form
[yt
µt
]= g (yt−2, yt−1, µt−1, εt−1, εt)
The optimal policy is then directly obtained as part of the set ofg() functions.
The steady state problem
The steady state is solution of
U1(y)− µ′[
f2(y , y , y ,0) + βf3(y , y , y ,0)
+β−1f1(y , y , y ,0)]
= 0
f (y , y , y ,0) = 0
Computing the steady stateFor a given value y , it is possible to use the first matrix equationabove to obtain the value of µ that minimizes the sum of squareresiduals, e:
M = f2(y , y , y , x ,0) + βf3(y , y , y , x ,0)
+β−1f1(y , y , y ,0)
µ′ = U1(y)M′(MM ′
)−1
e′ = U1(y)− µ′M
Furthermore, y must satisfy the m equations
f (y , y , y ,0) = 0
It is possible to build a sytem of equations whith only n unkownsy , but we must provide n − m independent measures of theresiduals e. Independent in the sense that the derivatives ofthese measures with respect to y must be linearly independent.
A QR trick
At the steady state, the following must hold exactly
U1 (y) = µ′M
This can only be if
M⋆ =
[M
U1 (y)
]
is of rank m The reordered QR decomposition of M⋆ is suchthat
M⋆ E = Q R(m + 1)× n n × n (m + 1)× (m + 1) (m + 1) × n
where E is a permutation matrix, Q an orthogonal matrix and Ra triangular matrix with diagonal elements ordered indecreasing size.
A QR trick (continued)
◮ When U1 (y) = µ′M doesn’t hold exactly M⋆ is full rank(m + 1) and the n − m last elements of R may be differentfrom zero.
◮ When U1 (y) = µ′M holds exactly M⋆ has rank m and then − m last elements of R are zero.
◮ The last n − m elements of the last row of R provide then − m independent measures of the residuals e
◮ In practice, we build a nonlinear function with y as inputand that returns the n − m last elements of the last row ofR and f (y , y , y ,0). At the solution, when y = y , thisfunction must return zeros.
When an analytical solution is available
Let’s write yx the n − m policy instruments and yy , the mremaining endogenous variables, such that
y⋆ =
[y⋆
xy⋆
y
].
The analytical solution is given by
y⋆y = f (y⋆
x ) .
We have then
f (y⋆, y⋆, y⋆,0) = 0
Combining this analytical solution with the n − m residualscomputed above, we can reduce the steady state problem to asystem of n − m nonlinear equations that is much easier tosolve.
First order approximation of the FOCs
Et
{[U11]αγ
[yt]γ
− [µt ]η [f2]η
α − β [µt+1]η [f3]η
α − β−1 [µt−1]η [f1]
η
α
− [µ]η
(β [f13]
η
γα
[yt+2
]γ+ β
−1 [f13]η
αγ
[yt−2
]γ+
([f12]
η
γα + β [f23]η
γα
) [yt+1
]γ
+([f23]
η
αγ + β−1 [f12]
η
αγ
) [yt−1
]γ+
([f33]
η
αγ + β [f22]η
αγ + β−1 [f11]
η
αγ
) [yt]γ
+ [f24]η
αδ [εt ]δ + β [f34]
η
αδ [εt+1]δ + β
−1 [f14]η
αδ [εt−1]]δ
)}= 0
Et{[f1]ηγ
[yt+1
]γ+ [f2]ηγ
[yt]γ
+ [f3]ηγ[yt−1
]γ+ [f4]ηδ [εt ]
δ}
= 0
where yt = yt − y , µt = µt − µ and fij the second orderderivatives corresponding to the i th and the j th argument of thef () function.
The first pitfall
A naive approach ot linear-quadratic appoximation that wouldconsider a linear approximation of the dynamics of the systemand a second order approximation of the objective function,ignores the second order derivatives fij that enter in the firstorder approximation of the dynamics of the model underoptimal policy.
The approximated solution function
yt = y + g1yt−2 + g2yt−1 + g3µt−1 + g4εt−1 + g5εt
µt = µ+ h1yt−2 + h2yt−1 + h3µt−1 + h4εt−1 + h5εt
Dynare implementation
When the ramsey_policy instruction is given, Dynare forms thematrices of coefficients of the first order expansion of the firstorder necessary conditions, without writing explicitely theequations for the first order necessary conditions. In the future,we will write a *.mod file containing all the equations for theRamsey problem.
Evaluating welfare under optimal policy
◮ Optimal policy under commitment is mostly interesting as abenchmark to evaluate other, more practical policies.
◮ The welfare evaluation must be consistent with theapproximated solution. Idealy, if the approximated solutionis the exact solution of an approximated problem,approximated welfare should be the objective function ofthe approximated problem.
◮ A first order approximation of the dynamics under optimalpolicy is the exact solution of some quadratic problem.
◮ Plugging policy rule in original model and compute anapproximation of welfare should provide the samemeasure and avoid paradoxical rankings.
◮ We should aim at a second order approximation of welfare.
Conjectures
1. Under optimal policy, the second order approximation ofwelfare is equal to a second order approximation of theLagrangian
2. Second order derivatives of the policy function drop fromthe second order approximation of welfare
Intuition from the static case
Considermax
yU(y)
s.t.f (y , s) = 0
The Lagrangian is
L(s) = U(y)− [λ]η [f (y , s)]η
The solution takes the form
y = g(s)
λ = h(s)
First order condition for optimality
[U1]α +− [λ]η [f1(y , s)]η = 0
f (y , s) = 0
Second order approximation of the Lagrangian
Taking the Taylor expansion around s:
U −
[λ]η
[f]η
−
[f]η[h1]
ηγ [s] +
(([U1]α −
[λ]η[f1]
ηα
)[g1]
αγ −
[λ]η[f2]
ηγ
)[s]γ
+12
(([U1]α −
[λ]η[f1]
ηα
)[gss]
αγθ +
([U11]αδ −
[λ]η[f11]
ηαδ
)[gs]
αγ [gs]
δθ
−
[λ]η[f22]
ηγδ −
[f]η[h11]
ηγδ − 2
( [λ]η[f21]
ηαδ [g1]
αγ
+ [h1]ηγ [f1]
ηα [g1]
αδ
))[s]γ [s]δ
Simplifying
Taking the Taylor expansion of the Lagrangian around s:
U + [U1]α [g1]αγ [s]
γ
+12
(([U11]αδ −
[λ]η[f11]
ηαδ
)[gs]
αγ [gs]
δθ −
[λ]η[f22]
ηγδ
− 2([
λ]η[f21]
ηαδ [g1]
αγ + [h1]
ηγ [f1]
ηα [g1]
αδ
))[s]γ [s]δ
In addition . . .
From steady state:
[U1]α −
[λ]η[f1]
ηα = 0
when we compute a second order approximation of the FOCs,we have also
[f1]ηα [gss]
αγθ + [f11]
ηαδ [gs]
αγ [gs]
δθ + [f22]
ηγδ + 2 [f21]
ηαδ [g1]
αγ = 0
So,
[U1]α [gss]αγθ =−
[λ]η
([f11]
ηαδ [gs]
αγ [gs]
δθ + [f22]
ηγδ + 2 [f21]
ηαδ [g1]
αγ
)
Extending the intuition to DSGE models
Steady state condition:
[U1]α − [µ]η([f2]
ηα + β [f3]
ηα + β−1 [f1]
ηα
)= 0
◮ One needs to properly select timing of variables in welfaresummation over time.
◮ One needs to use the fact that initial value of Lagrangemultipliers is zero.
Timeless perspective
◮ Time inconsistency problem stems from the fact that the initialvalue of the lagged Lagrange multipliers is set to zero.
◮ If, later on, the authorities re–optimize, they reset the Lagrangemultipliers to zero.
◮ This mechanism reflects the fact that authorities make theirdecision after that private agents have formed their expectations(on the basis of the previous policy).
◮ When private agents expect the authorities to reoptimize theeconomy switch to a Nash equilibrium (discretionary solution)
◮ For optimal policy in a timeless perspective, Svensson andWoodford suggest that authorities relinquish their first periodadvantage and act as if the Lagrange mulitpliers had beeninitialized to zero in the far away past.
◮ What should be the initial value of the Lagrange multipliers foroptimal policy in a timeless perspective?
Eliminating the Lagrange multipliers
A linear approximation of the solution to a Ramsey problemtakes the form:
[yt
µt
]=
[A11 A12
A21 A22
] [yt−1
µt−1
]+
[B1
B2
]εt
= A1yt−1 + A2µt−1 + Bεt
where A1 and A2 are the conforming sub-matrices.The problem is to eliminate µt−1 from the expression for yt bysubstituting values of yt−1, yt−2, εt−1.
A QR algolrithm (I)
A2E = QR =
[Q11 Q12
Q21 Q22
] [R1 R2
0 0
]
where E is a permutation matrix, R1 is upper triangular and R2
is empty if A2 is full (column) rank.Replacing A2 by its decomposition, one gets
Q′
[yt
µt
]= Q′A1yt−1 + RE ′µt−1 + Q′Bεt
The bottom part of the above system can be written
Q′12yt + Q′
22µt =[
Q′12 Q′
22
] (A1yt−1 + Bεt
)
This system in turn contains necessarily more equations thanelements in µt .
A QR algolrithm (II)
A new application of the QR decomposition to Q′22 gives:
Q′22E = QR =
[Q11 Q12
Q21 Q22
][R1 R2
0 0
]
where E is a permutation matrix. When R2 isn’t empty, thesystem is undetermined and it is possible to choose some ofthe multipliers µt , the ones corresponding to the columns of R2.We set them to zero, following a minimal state space type ofarguments. Note however, that the QR decomposition isn’tunique. Once the economy is managed according to optimalpolicy, all choices of decomposition would be equivalent, butthe choice may matter for the initial period. This is an issue thatdeserves further studying.
A QR algolrithm (III)
Then, we have
µt = E
[R−1
1
[Q′
11 Q′
21
] ([Q′
12 Q′
22
] (A1yt−1 + Bεt
)− Q′
12yt)
0
]
and
µt−1 = E
[R−1
1
[Q′
11 Q′
21
] ([Q′
12 Q′
22
] (A1yt−2 + Bεt−1
)− Q′
12yt−1)
0
]
A QR algolrithm (IV)
Replacing, µt−1 in the original equation for yt , one obtainsfinally
yt = M1yt−1 + M2yt−2 + M3εt + M4εt−1
where
M1 = A11 − A12E
[R−1
1
[Q′
11 Q′21
]Q′
12
0
]
M2 = A12E
[R−1
1
[Q′
11 Q′21
] [Q′
12 Q′22
]A1
0
]
M3 = B1
M4 = A12E
[R−1
1
[Q′
11 Q′21
] [Q′
12 Q′22
]B
0
]
A linear–quadratic example (I)
var y inf r dr;varexo e_y e_inf;
parameters delta sigma alpha kappa gamma1 gamma2;
delta = 0.44;kappa = 0.18;alpha = 0.48;sigma = -0.06;
A linear–quadratic example(II)
model(linear);y = delta*y(-1)+(1-delta)*y(+1)+sigma *(r-inf(+1))+e_yinf = alpha*inf(-1)+(1-alpha)*inf(+1)+kappa*y+e_inf;dr = r - r(-1);end;
shocks;var e_y;stderr 0.63;var e_inf;stderr 0.4;end;
A linear–quadratic example(III)
planner_objective y^2 + inf^2 + 0.2*dr^2;
ramsey_policy(planner_discount=1);
Warning: the solution isn’t necessarily minimum state!
Optimal simple rule
Exemple (Clarida, Gali, Gertler)
yt = δyt−1 + (1 − δ)Et yt+1 + σ(rt − Et inft+1) + eyt
inft = αinf−1 + (1 − α)Et inft+1 + κyt + einft
rt = γ1inft + γ2yt
Objectifarg min
γ1,γ2var(y) + var(inf )
= arg minγ1,γ2
limβ→1
E0
∞∑
t=1
(1 − β)βt (y2t + inf 2
t )
DYNARE example
var y inf r;varexo e_y e_inf;
parameters delta sigma alpha kappa gamma1 gamma2;
delta = 0.44;kappa = 0.18;alpha = 0.48;sigma = -0.06;
(continued)
model(linear);y = delta*y(-1)+(1-delta)*y(+1)+sigma *(r-inf(+1))+e_y;inf = alpha*inf(-1)+(1-alpha)*inf(+1)+kappa*y+e_inf;r = gamma1*inf+gamma2*y;end;
shocks;var e_y;stderr 0.63;var e_inf;stderr 0.4;end;
(continued)
optim_weights;inf 1;y 1;end;
gamma1 = 1.1;gamma2 = 0;
osr_params gamma1 gamma2;
osr;
Another example
yt = δyt−1 + (1 − δ)Et yt+1 + σ(rt − Et inft+1) + eyt
inft = αinf−1 + (1 − α)Et inft+1 + κyt + einft
rt = γ1inft + γ2yt
drt = rt − rt−1
Objectifminγ1,γ2
var(y) + var(inf ) + 0.2var(dr)