Business Cycles Problem Set 4, Exercise 2econ.sciences-po.fr/sites/default/files/file/yann...
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Business Cycles Problem Set 4, Exercise 2
Business CyclesProblem Set 4, Exercise 2
F. Ferroni1
1Banque de France,
February 2, 2011
Business Cycles Problem Set 4, Exercise 2
Preliminary on DSGE models
Agents in the economy optimize. Compute FOC
we obtain a set of optimality equations
Etf (xt+1, xt, zt+1, zt) = 0
zt = g(zt−1, εt) εt ∼ N(0, σ)
where xt is a vector of endogenous variables and zt is a vector ofexogenous processes
We want to express the endogenous variables as a function ofexogenous ones, i.e. solve the model
xt = h(zt)
Analytical solution of h are not implementable. Log linearize theequilibrium conditions around a point, the steady state, and solvenumerically the linearized sistem
Business Cycles Problem Set 4, Exercise 2
Preliminary on DSGE models
Agents in the economy optimize. Compute FOC
we obtain a set of optimality equations
Etf (xt+1, xt, zt+1, zt) = 0
zt = g(zt−1, εt) εt ∼ N(0, σ)
where xt is a vector of endogenous variables and zt is a vector ofexogenous processes
We want to express the endogenous variables as a function ofexogenous ones, i.e. solve the model
xt = h(zt)
Analytical solution of h are not implementable. Log linearize theequilibrium conditions around a point, the steady state, and solvenumerically the linearized sistem
Business Cycles Problem Set 4, Exercise 2
Preliminary on DSGE models
Agents in the economy optimize. Compute FOC
we obtain a set of optimality equations
Etf (xt+1, xt, zt+1, zt) = 0
zt = g(zt−1, εt) εt ∼ N(0, σ)
where xt is a vector of endogenous variables and zt is a vector ofexogenous processes
We want to express the endogenous variables as a function ofexogenous ones, i.e. solve the model
xt = h(zt)
Analytical solution of h are not implementable. Log linearize theequilibrium conditions around a point, the steady state, and solvenumerically the linearized sistem
Business Cycles Problem Set 4, Exercise 2
Preliminary on DSGE models
Agents in the economy optimize. Compute FOC
we obtain a set of optimality equations
Etf (xt+1, xt, zt+1, zt) = 0
zt = g(zt−1, εt) εt ∼ N(0, σ)
where xt is a vector of endogenous variables and zt is a vector ofexogenous processes
We want to express the endogenous variables as a function ofexogenous ones, i.e. solve the model
xt = h(zt)
Analytical solution of h are not implementable. Log linearize theequilibrium conditions around a point, the steady state, and solvenumerically the linearized sistem
Business Cycles Problem Set 4, Exercise 2
The RBC model
max E0
∞∑t=0
βt [ln ct + θ ln(1− ht)]
s.t. ezt kαt h1−αt + (1− δ)kt ≥ ct + kt+1
where 0 < β < 1, 0 < α < 1, 0 < δ < 1, θ is the Frish elasticity, and k0 isgiven. Assume further
zt = ρzt−1 + εt εt ∼ N(0, σ)
Business Cycles Problem Set 4, Exercise 2
The RBC model
max E0
∞∑t=0
βt [ln ct + θ ln(1− ht)]
s.t. ezt kαt h1−αt + (1− δ)kt ≥ ct + kt+1
where 0 < β < 1, 0 < α < 1, 0 < δ < 1, θ is the Frish elasticity, and k0 isgiven. Assume further
zt = ρzt−1 + εt εt ∼ N(0, σ)
Business Cycles Problem Set 4, Exercise 2
The RBC model
max E0
∞∑t=0
βt [ln ct + θ ln(1− ht)]
s.t. ezt kαt h1−αt + (1− δ)kt ≥ ct + kt+1
where 0 < β < 1, 0 < α < 1, 0 < δ < 1, θ is the Frish elasticity, and k0 isgiven. Assume further
zt = ρzt−1 + εt εt ∼ N(0, σ)
Business Cycles Problem Set 4, Exercise 2
Question 1
Derive the first order condition for ct, ht, kt+1
Write the Lagrangian
E0
∞∑t=0
βt[ln ct + θ ln(1− ht) + λt(ezt kαt h1−α
t + (1− δ)kt − ct − kt+1)]=
E0([ln c0 + θ ln(1− h0) + λt(ez0 kα0 h1−α
0 + (1− δ)k0 − c0 − k1)]+ ...
βt[ln ct + θ ln(1− ht) + λt(ezt kαt h1−α
t + (1− δ)kt − ct − kt+1)]+
βt+1[ln ct+1 + θ ln(1− ht+1) + λt+1(ezt+1 kαt+1h1−α
t+1 + (1− δ)kt+1 − ct+1 − kt+2)]
+ ...)
Take derivatives wrt ct, ht, kt+1
Business Cycles Problem Set 4, Exercise 2
Question 1
Derive the first order condition for ct, ht, kt+1
Write the Lagrangian
E0
∞∑t=0
βt[ln ct + θ ln(1− ht) + λt(ezt kαt h1−α
t + (1− δ)kt − ct − kt+1)]=
E0([ln c0 + θ ln(1− h0) + λt(ez0 kα0 h1−α
0 + (1− δ)k0 − c0 − k1)]+ ...
βt[ln ct + θ ln(1− ht) + λt(ezt kαt h1−α
t + (1− δ)kt − ct − kt+1)]+
βt+1[ln ct+1 + θ ln(1− ht+1) + λt+1(ezt+1 kαt+1h1−α
t+1 + (1− δ)kt+1 − ct+1 − kt+2)]
+ ...)
Take derivatives wrt ct, ht, kt+1
Business Cycles Problem Set 4, Exercise 2
Question 1
Derive the first order condition for ct, ht, kt+1
Write the Lagrangian
E0
∞∑t=0
βt[ln ct + θ ln(1− ht) + λt(ezt kαt h1−α
t + (1− δ)kt − ct − kt+1)]=
E0([ln c0 + θ ln(1− h0) + λt(ez0 kα0 h1−α
0 + (1− δ)k0 − c0 − k1)]+ ...
βt[ln ct + θ ln(1− ht) + λt(ezt kαt h1−α
t + (1− δ)kt − ct − kt+1)]+
βt+1[ln ct+1 + θ ln(1− ht+1) + λt+1(ezt+1 kαt+1h1−α
t+1 + (1− δ)kt+1 − ct+1 − kt+2)]
+ ...)
Take derivatives wrt ct, ht, kt+1
Business Cycles Problem Set 4, Exercise 2
Question 1
wrt ct
1/ct − λt = 0
wrt ht
− θ
1− ht+ (1− α)λtezt kαt h−αt = 0
wrt kt+1
−λt + βEtλt+1
(αezt+1 kα−1
t+1 h1−αt+1 + (1− δ)
)= 0
Business Cycles Problem Set 4, Exercise 2
Question 1
wrt ct
1/ct − λt = 0
wrt ht
− θ
1− ht+ (1− α)λtezt kαt h−αt = 0
wrt kt+1
−λt + βEtλt+1
(αezt+1 kα−1
t+1 h1−αt+1 + (1− δ)
)= 0
Business Cycles Problem Set 4, Exercise 2
Question 1
wrt ct
1/ct − λt = 0
wrt ht
− θ
1− ht+ (1− α)λtezt kαt h−αt = 0
wrt kt+1
−λt + βEtλt+1
(αezt+1 kα−1
t+1 h1−αt+1 + (1− δ)
)= 0
Business Cycles Problem Set 4, Exercise 2
Substitute out the lagrangian multiplier and we get
θct
1− ht= (1− α)ezt kαt h−αt
1 = βEtct
ct+1
(αezt+1 kαt+1h−αt+1 + (1− δ)
)
Introduce the following variables
yt = ezt kαt h1−αt
it = kt+1 − (1− δ)kt
rt = αyt
kt
wt = (1− α) yt
ht
Notice also that
ezt kαt h−αt =yt
ht
ezt kα−1t h1−α
t =yt
kt
Business Cycles Problem Set 4, Exercise 2
Substitute out the lagrangian multiplier and we get
θct
1− ht= (1− α)ezt kαt h−αt
1 = βEtct
ct+1
(αezt+1 kαt+1h−αt+1 + (1− δ)
)Introduce the following variables
yt = ezt kαt h1−αt
it = kt+1 − (1− δ)kt
rt = αyt
kt
wt = (1− α) yt
ht
Notice also that
ezt kαt h−αt =yt
ht
ezt kα−1t h1−α
t =yt
kt
Business Cycles Problem Set 4, Exercise 2
Substitute out the lagrangian multiplier and we get
θct
1− ht= (1− α)ezt kαt h−αt
1 = βEtct
ct+1
(αezt+1 kαt+1h−αt+1 + (1− δ)
)Introduce the following variables
yt = ezt kαt h1−αt
it = kt+1 − (1− δ)kt
rt = αyt
kt
wt = (1− α) yt
ht
Notice also that
ezt kαt h−αt =yt
ht
ezt kα−1t h1−α
t =yt
kt
Business Cycles Problem Set 4, Exercise 2
Question 2
Equilibrium Condition
wt =θct
1− ht(1)
1 = βEtct
ct+1(rt+1 + (1− δ)) (2)
yt = ezt kαt h1−αt (3)
it = kt+1 − (1− δ)kt (4)
rt = αyt
kt(5)
wt = (1− α) yt
ht(6)
yt = ct + it (7)
7 endogenous variables, ct, ht, kt+1, yt, rt,wt, 7 equations. 1 exogenous.
Compute the non stochastic steady state, i.e. εt = 0 for all t. Thus z = 0
Business Cycles Problem Set 4, Exercise 2
Question 2
Equilibrium Condition
wt =θct
1− ht(1)
1 = βEtct
ct+1(rt+1 + (1− δ)) (2)
yt = ezt kαt h1−αt (3)
it = kt+1 − (1− δ)kt (4)
rt = αyt
kt(5)
wt = (1− α) yt
ht(6)
yt = ct + it (7)
7 endogenous variables, ct, ht, kt+1, yt, rt,wt, 7 equations. 1 exogenous.
Compute the non stochastic steady state, i.e. εt = 0 for all t. Thus z = 0
Business Cycles Problem Set 4, Exercise 2
Question 2
Equilibrium Condition
wt =θct
1− ht(1)
1 = βEtct
ct+1(rt+1 + (1− δ)) (2)
yt = ezt kαt h1−αt (3)
it = kt+1 − (1− δ)kt (4)
rt = αyt
kt(5)
wt = (1− α) yt
ht(6)
yt = ct + it (7)
7 endogenous variables, ct, ht, kt+1, yt, rt,wt, 7 equations. 1 exogenous.
Compute the non stochastic steady state, i.e. εt = 0 for all t. Thus z = 0
Business Cycles Problem Set 4, Exercise 2
Question 2
from (2)
1 = β (r + (1− δ))r = 1/β + δ − 1
from (5)
y/k =1/β + δ − 1
α
from (3)
y/h = (y/k)α
α−1 =
(1/β + δ − 1
α
) αα−1
from (6)
w = (1− α)y/h = (1− α)(
1/β + δ − 1α
) αα−1
from (4)
i/y = δk/y =δα
1/β + δ − 1
Business Cycles Problem Set 4, Exercise 2
Question 2
from (2)
1 = β (r + (1− δ))r = 1/β + δ − 1
from (5)
y/k =1/β + δ − 1
α
from (3)
y/h = (y/k)α
α−1 =
(1/β + δ − 1
α
) αα−1
from (6)
w = (1− α)y/h = (1− α)(
1/β + δ − 1α
) αα−1
from (4)
i/y = δk/y =δα
1/β + δ − 1
Business Cycles Problem Set 4, Exercise 2
Question 2
from (2)
1 = β (r + (1− δ))r = 1/β + δ − 1
from (5)
y/k =1/β + δ − 1
α
from (3)
y/h = (y/k)α
α−1 =
(1/β + δ − 1
α
) αα−1
from (6)
w = (1− α)y/h = (1− α)(
1/β + δ − 1α
) αα−1
from (4)
i/y = δk/y =δα
1/β + δ − 1
Business Cycles Problem Set 4, Exercise 2
Question 2
from (2)
1 = β (r + (1− δ))r = 1/β + δ − 1
from (5)
y/k =1/β + δ − 1
α
from (3)
y/h = (y/k)α
α−1 =
(1/β + δ − 1
α
) αα−1
from (6)
w = (1− α)y/h = (1− α)(
1/β + δ − 1α
) αα−1
from (4)
i/y = δk/y =δα
1/β + δ − 1
Business Cycles Problem Set 4, Exercise 2
Question 2
from (2)
1 = β (r + (1− δ))r = 1/β + δ − 1
from (5)
y/k =1/β + δ − 1
α
from (3)
y/h = (y/k)α
α−1 =
(1/β + δ − 1
α
) αα−1
from (6)
w = (1− α)y/h = (1− α)(
1/β + δ − 1α
) αα−1
from (4)
i/y = δk/y =δα
1/β + δ − 1
Business Cycles Problem Set 4, Exercise 2
Question 2
from (7)
c/y = 1− i/y = 1− δα
1/β + δ − 1
from (1)
(1− α)y/h =θc
1− h1− αθ
1− hh
= c/y = 1− δα
1/β + δ − 1
h =1
1 + θ1−α c/y
Business Cycles Problem Set 4, Exercise 2
Question 2
from (7)
c/y = 1− i/y = 1− δα
1/β + δ − 1
from (1)
(1− α)y/h =θc
1− h1− αθ
1− hh
= c/y = 1− δα
1/β + δ − 1
h =1
1 + θ1−α c/y
Business Cycles Problem Set 4, Exercise 2
Question 3, Log linearization
Consider the following function
F(s1,t, s2,t, s3,t) = 0
The log linear approximation is
F1 s1 s1,t + F2 s2 s2,t + F3 s3 s3,t = 0
where sj,t = ln sj,t − ln sj for j = 1, 2, 3 and Fj is the derivative of sj
variables evaluated at the steady state. See Uhlig(1998).
Notice that in a neighborhood of the steady state
sj,t = ln sj,t − ln sj 'sj,t − sj
sj
Then the latter represent the percentage deviation of a variable from thesteady state.
Business Cycles Problem Set 4, Exercise 2
Question 3, Log linearization
Consider the following function
F(s1,t, s2,t, s3,t) = 0
The log linear approximation is
F1 s1 s1,t + F2 s2 s2,t + F3 s3 s3,t = 0
where sj,t = ln sj,t − ln sj for j = 1, 2, 3 and Fj is the derivative of sj
variables evaluated at the steady state. See Uhlig(1998).
Notice that in a neighborhood of the steady state
sj,t = ln sj,t − ln sj 'sj,t − sj
sj
Then the latter represent the percentage deviation of a variable from thesteady state.
Business Cycles Problem Set 4, Exercise 2
Question 3, Log linearization
Consider the following function
F(s1,t, s2,t, s3,t) = 0
The log linear approximation is
F1 s1 s1,t + F2 s2 s2,t + F3 s3 s3,t = 0
where sj,t = ln sj,t − ln sj for j = 1, 2, 3 and Fj is the derivative of sj
variables evaluated at the steady state. See Uhlig(1998).
Notice that in a neighborhood of the steady state
sj,t = ln sj,t − ln sj 'sj,t − sj
sj
Then the latter represent the percentage deviation of a variable from thesteady state.
Business Cycles Problem Set 4, Exercise 2
Question 3, Log linearization
Eq (1)
(1− α)yt/ht =θct
1− ht
(1− α)1/hyyt − (1− α) yh2 hht = θ
c1− h
ct + θc
(1− h)2 hht
since (1− α)y/h = θc1−h , we get
yt − ht = ct +h
1− hht
yt = ct +1
1− hht
Eq (2)
1 = βEtct
ct+1(rt+1 + (1− δ))
0 = Et (β(r + 1− δ)(ct − ct+1) + βrrt+1)
0 = Et (ct − ct+1 + βrrt+1)
since at the steady state β(r + 1− δ) = 1
Business Cycles Problem Set 4, Exercise 2
Question 3, Log linearization
Eq (1)
(1− α)yt/ht =θct
1− ht
(1− α)1/hyyt − (1− α) yh2 hht = θ
c1− h
ct + θc
(1− h)2 hht
since (1− α)y/h = θc1−h , we get
yt − ht = ct +h
1− hht
yt = ct +1
1− hht
Eq (2)
1 = βEtct
ct+1(rt+1 + (1− δ))
0 = Et (β(r + 1− δ)(ct − ct+1) + βrrt+1)
0 = Et (ct − ct+1 + βrrt+1)
since at the steady state β(r + 1− δ) = 1
Business Cycles Problem Set 4, Exercise 2
Question 3, Log linearization
Eq (1)
(1− α)yt/ht =θct
1− ht
(1− α)1/hyyt − (1− α) yh2 hht = θ
c1− h
ct + θc
(1− h)2 hht
since (1− α)y/h = θc1−h , we get
yt − ht = ct +h
1− hht
yt = ct +1
1− hht
Eq (2)
1 = βEtct
ct+1(rt+1 + (1− δ))
0 = Et (β(r + 1− δ)(ct − ct+1) + βrrt+1)
0 = Et (ct − ct+1 + βrrt+1)
since at the steady state β(r + 1− δ) = 1
Business Cycles Problem Set 4, Exercise 2
Question 3, Log linearization
Eq (3)
yt = zt + αkt + (1− α)ht
Eq (5)
rt = yt − kt
Eq (6)
wt = yt − ht
Eq (4)
i it = kkt+1 − (1− δ)kkt
i/yit = k/ykt+1 − (1− δ)k/ykt
Eq (7)
yyt = cct + iit
yt = c/yct + i/yit
Business Cycles Problem Set 4, Exercise 2
Question 3, Log linearization
Eq (3)
yt = zt + αkt + (1− α)ht
Eq (5)
rt = yt − kt
Eq (6)
wt = yt − ht
Eq (4)
i it = kkt+1 − (1− δ)kkt
i/yit = k/ykt+1 − (1− δ)k/ykt
Eq (7)
yyt = cct + iit
yt = c/yct + i/yit
Business Cycles Problem Set 4, Exercise 2
Question 3, Log linearization
Eq (3)
yt = zt + αkt + (1− α)ht
Eq (5)
rt = yt − kt
Eq (6)
wt = yt − ht
Eq (4)
i it = kkt+1 − (1− δ)kkt
i/yit = k/ykt+1 − (1− δ)k/ykt
Eq (7)
yyt = cct + iit
yt = c/yct + i/yit
Business Cycles Problem Set 4, Exercise 2
Question 3, Log linearization
Eq (3)
yt = zt + αkt + (1− α)ht
Eq (5)
rt = yt − kt
Eq (6)
wt = yt − ht
Eq (4)
i it = kkt+1 − (1− δ)kkt
i/yit = k/ykt+1 − (1− δ)k/ykt
Eq (7)
yyt = cct + iit
yt = c/yct + i/yit
Business Cycles Problem Set 4, Exercise 2
Question 3, Log linearization
Eq (3)
yt = zt + αkt + (1− α)ht
Eq (5)
rt = yt − kt
Eq (6)
wt = yt − ht
Eq (4)
i it = kkt+1 − (1− δ)kkt
i/yit = k/ykt+1 − (1− δ)k/ykt
Eq (7)
yyt = cct + iit
yt = c/yct + i/yit
Business Cycles Problem Set 4, Exercise 2
Question 3, Sum up
yt = ct +1
1− hht
0 = Et (ct − ct+1 + βrrt+1)
yt = zt + αkt + (1− α)ht
rt = yt − kt
wt = yt − ht
i/yit = k/ykt+1 − (1− δ)k/ykt
yt = c/yct + i/yit
and
zt = ρzt−1 + εt εt ∼ N(0, σ)
Business Cycles Problem Set 4, Exercise 2
Question 3, Sum up
yt = ct +1
1− hht
0 = Et (ct − ct+1 + βrrt+1)
yt = zt + αkt + (1− α)ht
rt = yt − kt
wt = yt − ht
i/yit = k/ykt+1 − (1− δ)k/ykt
yt = c/yct + i/yit
and
zt = ρzt−1 + εt εt ∼ N(0, σ)
Business Cycles Problem Set 4, Exercise 2
Question 4
Let xt = [ct, ht, kt+1, yt, it, rt,wt], show that the system satisfies
0 = Et [Fxt+1 + Gxt + Hxt−1 + Lzt+1 + Mzt]
zt+1 = Nzt + εt+1
8 LINEAR equations, 7 endogenous variables and 1 exogenous. Beinglinear, the latter system can be written in matrix format.
Give an expression for F,G,H,L,M
Business Cycles Problem Set 4, Exercise 2
Question 4
Let xt = [ct, ht, kt+1, yt, it, rt,wt], show that the system satisfies
0 = Et [Fxt+1 + Gxt + Hxt−1 + Lzt+1 + Mzt]
zt+1 = Nzt + εt+1
8 LINEAR equations, 7 endogenous variables and 1 exogenous. Beinglinear, the latter system can be written in matrix format.
Give an expression for F,G,H,L,M
Business Cycles Problem Set 4, Exercise 2
Question 4
Let xt = [ct, ht, kt+1, yt, it, rt,wt], show that the system satisfies
0 = Et [Fxt+1 + Gxt + Hxt−1 + Lzt+1 + Mzt]
zt+1 = Nzt + εt+1
8 LINEAR equations, 7 endogenous variables and 1 exogenous. Beinglinear, the latter system can be written in matrix format.
Give an expression for F,G,H,L,M
Business Cycles Problem Set 4, Exercise 2
Question 4
xt+1 = [ct+1, ht+1, kt+2, yt+1, it+1, rt+1,wt+1]
F =
0 0 0 0 0 0 0−1 0 0 0 0 βr 00 0 0 0 0 0 00 0 0 0 0 0 00 0 0 0 0 0 00 0 0 0 0 0 00 0 0 0 0 0 0
xt = [ct, ht, kt+1, yt, it, rt,wt]
G =
1 1/(1− h) 0 −1 0 0 01 0 0 0 0 0 00 1− α 0 −1 0 0 00 0 0 1 0 −1 00 −1 0 1 0 0 −10 0 k/y 0 i/y 0 0
c/y 0 0 −1 i/y 0 0
Business Cycles Problem Set 4, Exercise 2
Question 4
xt+1 = [ct+1, ht+1, kt+2, yt+1, it+1, rt+1,wt+1]
F =
0 0 0 0 0 0 0−1 0 0 0 0 βr 00 0 0 0 0 0 00 0 0 0 0 0 00 0 0 0 0 0 00 0 0 0 0 0 00 0 0 0 0 0 0
xt = [ct, ht, kt+1, yt, it, rt,wt]
G =
1 1/(1− h) 0 −1 0 0 01 0 0 0 0 0 00 1− α 0 −1 0 0 00 0 0 1 0 −1 00 −1 0 1 0 0 −10 0 k/y 0 i/y 0 0
c/y 0 0 −1 i/y 0 0
Business Cycles Problem Set 4, Exercise 2
Question 4
xt−1 = [ct−1, ht−1, kt, yt−1, it−1, rt−1,wt−1]
H =
0 0 0 0 0 0 00 0 0 0 0 0 00 0 α 0 0 0 00 0 −1 0 0 0 00 0 0 0 0 0 00 0 (1− δ)k/y 0 0 0 00 0 0 0 0 0 0
L =(
0 0 0 0 0 0 0)′
M =(
0 1 0 0 0 0 0)′
Business Cycles Problem Set 4, Exercise 2
Question 4
xt−1 = [ct−1, ht−1, kt, yt−1, it−1, rt−1,wt−1]
H =
0 0 0 0 0 0 00 0 0 0 0 0 00 0 α 0 0 0 00 0 −1 0 0 0 00 0 0 0 0 0 00 0 (1− δ)k/y 0 0 0 00 0 0 0 0 0 0
L =
(0 0 0 0 0 0 0
)′
M =(
0 1 0 0 0 0 0)′
Business Cycles Problem Set 4, Exercise 2
Question 4
xt−1 = [ct−1, ht−1, kt, yt−1, it−1, rt−1,wt−1]
H =
0 0 0 0 0 0 00 0 0 0 0 0 00 0 α 0 0 0 00 0 −1 0 0 0 00 0 0 0 0 0 00 0 (1− δ)k/y 0 0 0 00 0 0 0 0 0 0
L =
(0 0 0 0 0 0 0
)′M =
(0 1 0 0 0 0 0
)′
Business Cycles Problem Set 4, Exercise 2
Question 5, Guess the solution
Guess that the solution takes the following form
xt = Pxt−1 + Qzt
and that P solves a quadratic equation
Plug the latter in our system and we get
0 = Et [Fxt+1 + Gxt + Hxt−1 + Lzt+1 + Mzt] =
= Et
[F(P2xt−1 + PQzt + Qzt+1) + G(Pxt−1 + Qzt) + Hxt−1 + Lzt+1 + Mzt
]=
= Et[F(P2xt−1 + PQzt + QNzt + Qεt+1) + G(Pxt−1 + Qzt)
+ Hxt−1 + LNzt + Lεt+1 + Mzt] =
= xt−1(FP2 + GP + H) + zt(FPQ + FQN + GQ + LN + M)
the solution must hold for every xt−1 and zt.
first solve P2 + GP + H = 0 and get P
given P solve for Q using (FPQ + FQN + GQ + LN + M) = 0
Business Cycles Problem Set 4, Exercise 2
Question 5, Guess the solution
Guess that the solution takes the following form
xt = Pxt−1 + Qzt
and that P solves a quadratic equation
Plug the latter in our system and we get
0 = Et [Fxt+1 + Gxt + Hxt−1 + Lzt+1 + Mzt] =
= Et
[F(P2xt−1 + PQzt + Qzt+1) + G(Pxt−1 + Qzt) + Hxt−1 + Lzt+1 + Mzt
]=
= Et[F(P2xt−1 + PQzt + QNzt + Qεt+1) + G(Pxt−1 + Qzt)
+ Hxt−1 + LNzt + Lεt+1 + Mzt] =
= xt−1(FP2 + GP + H) + zt(FPQ + FQN + GQ + LN + M)
the solution must hold for every xt−1 and zt.
first solve P2 + GP + H = 0 and get P
given P solve for Q using (FPQ + FQN + GQ + LN + M) = 0
Business Cycles Problem Set 4, Exercise 2
Question 5, Guess the solution
Guess that the solution takes the following form
xt = Pxt−1 + Qzt
and that P solves a quadratic equation
Plug the latter in our system and we get
0 = Et [Fxt+1 + Gxt + Hxt−1 + Lzt+1 + Mzt] =
= Et
[F(P2xt−1 + PQzt + Qzt+1) + G(Pxt−1 + Qzt) + Hxt−1 + Lzt+1 + Mzt
]=
= Et[F(P2xt−1 + PQzt + QNzt + Qεt+1) + G(Pxt−1 + Qzt)
+ Hxt−1 + LNzt + Lεt+1 + Mzt] =
= xt−1(FP2 + GP + H) + zt(FPQ + FQN + GQ + LN + M)
the solution must hold for every xt−1 and zt.
first solve P2 + GP + H = 0 and get P
given P solve for Q using (FPQ + FQN + GQ + LN + M) = 0
Business Cycles Problem Set 4, Exercise 2
Question 5, Guess the solution
Guess that the solution takes the following form
xt = Pxt−1 + Qzt
and that P solves a quadratic equation
Plug the latter in our system and we get
0 = Et [Fxt+1 + Gxt + Hxt−1 + Lzt+1 + Mzt] =
= Et
[F(P2xt−1 + PQzt + Qzt+1) + G(Pxt−1 + Qzt) + Hxt−1 + Lzt+1 + Mzt
]=
= Et[F(P2xt−1 + PQzt + QNzt + Qεt+1) + G(Pxt−1 + Qzt)
+ Hxt−1 + LNzt + Lεt+1 + Mzt] =
= xt−1(FP2 + GP + H) + zt(FPQ + FQN + GQ + LN + M)
the solution must hold for every xt−1 and zt.
first solve P2 + GP + H = 0 and get P
given P solve for Q using (FPQ + FQN + GQ + LN + M) = 0
Business Cycles Problem Set 4, Exercise 2
Question 5, Guess the solution
Guess that the solution takes the following form
xt = Pxt−1 + Qzt
and that P solves a quadratic equation
Plug the latter in our system and we get
0 = Et [Fxt+1 + Gxt + Hxt−1 + Lzt+1 + Mzt] =
= Et
[F(P2xt−1 + PQzt + Qzt+1) + G(Pxt−1 + Qzt) + Hxt−1 + Lzt+1 + Mzt
]=
= Et[F(P2xt−1 + PQzt + QNzt + Qεt+1) + G(Pxt−1 + Qzt)
+ Hxt−1 + LNzt + Lεt+1 + Mzt] =
= xt−1(FP2 + GP + H) + zt(FPQ + FQN + GQ + LN + M)
the solution must hold for every xt−1 and zt.
first solve P2 + GP + H = 0 and get P
given P solve for Q using (FPQ + FQN + GQ + LN + M) = 0
Business Cycles Problem Set 4, Exercise 2
Question 6, Dynare
Declare the variablesvar y c k h i r w z;
Declare the exogenousvarexo ez;
declare the parametersparameters cy ky alpha beta delta rho hbar theta iy
r0;
Assign numerical values to the parameters
Write down the equilibrium conditionsmodel; ... end;
set the standard deviation of the shocksshocks; var ez; stderr 0.0072; end;
simulate data and compute impulse responsestoch simul(order=0,periods=115,irf=15);
save all in a rbc.mod file, type in matlab dynare rbc
Business Cycles Problem Set 4, Exercise 2
Question 6, Dynare
Declare the variablesvar y c k h i r w z;
Declare the exogenousvarexo ez;
declare the parametersparameters cy ky alpha beta delta rho hbar theta iy
r0;
Assign numerical values to the parameters
Write down the equilibrium conditionsmodel; ... end;
set the standard deviation of the shocksshocks; var ez; stderr 0.0072; end;
simulate data and compute impulse responsestoch simul(order=0,periods=115,irf=15);
save all in a rbc.mod file, type in matlab dynare rbc
Business Cycles Problem Set 4, Exercise 2
Question 6, Dynare
Declare the variablesvar y c k h i r w z;
Declare the exogenousvarexo ez;
declare the parametersparameters cy ky alpha beta delta rho hbar theta iy
r0;
Assign numerical values to the parameters
Write down the equilibrium conditionsmodel; ... end;
set the standard deviation of the shocksshocks; var ez; stderr 0.0072; end;
simulate data and compute impulse responsestoch simul(order=0,periods=115,irf=15);
save all in a rbc.mod file, type in matlab dynare rbc
Business Cycles Problem Set 4, Exercise 2
Question 6, Dynare
Declare the variablesvar y c k h i r w z;
Declare the exogenousvarexo ez;
declare the parametersparameters cy ky alpha beta delta rho hbar theta iy
r0;
Assign numerical values to the parameters
Write down the equilibrium conditionsmodel; ... end;
set the standard deviation of the shocksshocks; var ez; stderr 0.0072; end;
simulate data and compute impulse responsestoch simul(order=0,periods=115,irf=15);
save all in a rbc.mod file, type in matlab dynare rbc
Business Cycles Problem Set 4, Exercise 2
Question 6, Dynare
Declare the variablesvar y c k h i r w z;
Declare the exogenousvarexo ez;
declare the parametersparameters cy ky alpha beta delta rho hbar theta iy
r0;
Assign numerical values to the parameters
Write down the equilibrium conditionsmodel; ... end;
set the standard deviation of the shocksshocks; var ez; stderr 0.0072; end;
simulate data and compute impulse responsestoch simul(order=0,periods=115,irf=15);
save all in a rbc.mod file, type in matlab dynare rbc
Business Cycles Problem Set 4, Exercise 2
Question 6, Dynare
Declare the variablesvar y c k h i r w z;
Declare the exogenousvarexo ez;
declare the parametersparameters cy ky alpha beta delta rho hbar theta iy
r0;
Assign numerical values to the parameters
Write down the equilibrium conditionsmodel; ... end;
set the standard deviation of the shocksshocks; var ez; stderr 0.0072; end;
simulate data and compute impulse responsestoch simul(order=0,periods=115,irf=15);
save all in a rbc.mod file, type in matlab dynare rbc
Business Cycles Problem Set 4, Exercise 2
Question 6, Dynare
Declare the variablesvar y c k h i r w z;
Declare the exogenousvarexo ez;
declare the parametersparameters cy ky alpha beta delta rho hbar theta iy
r0;
Assign numerical values to the parameters
Write down the equilibrium conditionsmodel; ... end;
set the standard deviation of the shocksshocks; var ez; stderr 0.0072; end;
simulate data and compute impulse responsestoch simul(order=0,periods=115,irf=15);
save all in a rbc.mod file, type in matlab dynare rbc
Business Cycles Problem Set 4, Exercise 2
Question 6, Dynare
Declare the variablesvar y c k h i r w z;
Declare the exogenousvarexo ez;
declare the parametersparameters cy ky alpha beta delta rho hbar theta iy
r0;
Assign numerical values to the parameters
Write down the equilibrium conditionsmodel; ... end;
set the standard deviation of the shocksshocks; var ez; stderr 0.0072; end;
simulate data and compute impulse responsestoch simul(order=0,periods=115,irf=15);
save all in a rbc.mod file, type in matlab dynare rbc