Notes on the Real Business Cycle Model

Pengfei Wang

Hong Kong University of Science and Technology

pfwang (Institute) Notes on the Real Business Cycle Model 03/09 1 / 46

Introduction: Basic Facts about Business Cycle

Introduction: Basic Facts about Business Cycle 2

Introduction: The central questions

What cuases business cycles?

Multiplier-accelerator theoryClean-up theorySunspots theory

Need to distinguish

Source of ShocksPropagation Mechanisms

How Should Government Policy Respond to Business Cycles?

Multiplier-accelerator theory

Clean-up theorySunspots theory

Need to distinguish

Multiplier-accelerator theoryClean-up theory

Sunspots theory

Need to distinguish

Source of Shocks

Propagation Mechanisms

Need to distinguish

Two Major Schools of Thoughts:

The Classical School�>Doctrine: Supply determines demand

supply shocks are the major source of business cycles.

Supply shocks include shocks to technology, to endowment, to costsof production, investment e¢ ciency,etc.

Propagation mechanisms include adjustment costs, capitalaccumulation, time to build,etc.

Policy recommendation: Do nothing.

Features: Very rigorous but against common sense.

The Keynesian School�> Doctrine: Demand determines supply

demand shocks are the major source of business cycles.

Demand shocks include shocks to consumption, to investment, togovernment spending, to exports, and to money supply;

Shocks to consumption or investment also include extrinsicuncertainty (expectations, animal spirits, sunspots).

Propagation mechanisms include interest rates, portfolio allocations,asset bubbles (including the stock market), banking and �nancialcrises, and international trade and currency crises.

Policy recommendation: Intervene

Features: Appealing to common sense but very vague and imprecise.

Key Assumptions:

prices adjust instantaneously to clear markets

rational expectations

perfect competition

perfect risk sharing

no asymmetric information

no externalities

A Benchmark Model:

A social planner (or representative agent) chooses paths of consumption,investment and employment (hours) to solve

objective function

maxfct ,kt+1,ntg∞

∑t=0

βt [log ct + γ log (1� nt )] (1)

Resource Constraint

ct + kt+1 � (1� δ) kt = Atkαt n1�αt , (2)

Shock ProcesslogAt = ρ logAt�1 + εt . (3)

A Benchmark Model:

objective function

maxfct ,kt+1,ntg∞

∑t=0

Resource Constraint

A Benchmark Model:

objective function

maxfct ,kt+1,ntg∞

∑t=0

Resource Constraint

The Lagrangian:

The Lagrangian is given by

L = E0

�∑∞t=0 βt [log ct + γ log (1� nt )]

+λt�Atkα

t n1�αt � ct � kt+1 + (1� δ) kt

� � . (4)

state variables in period tkt ;At (5)

choice variables in period t

ct , nt ,λt , kt+1 (6)

The Lagrangian:

L = E0

+λt�Atkα

t n1�αt � ct � kt+1 + (1� δ) kt

� � . (4)

ct , nt ,λt , kt+1 (6)

The Lagrangian:

L = E0

+λt�Atkα

t n1�αt � ct � kt+1 + (1� δ) kt

� � . (4)

ct , nt ,λt , kt+1 (6)

The First Order Conditions:

with respect to consumption :

1ct= λt (7)

with respect to labor:

1� nt= λt

�(1� α)Atkα

t n�αt

�(8)

with respect to kt+1

λt = βEt�λt+1

�αAt+1kα�1

t+1 n1�αt+1 + 1� δ

��(9)

1ct= λt (7)

1� nt= λt

�(1� α)Atkα

t n�αt

�(8)

λt = βEt�λt+1

�αAt+1kα�1

t+1 n1�αt+1 + 1� δ

��(9)

1ct= λt (7)

1� nt= λt

�(1� α)Atkα

t n�αt

�(8)

λt = βEt�λt+1

�αAt+1kα�1

t+1 n1�αt+1 + 1� δ

��(9)

The First Order Conditions (continued):

with respect to λt :

plus a standard transversality condition:

limT!∞

E0βT λT kT+1 = 0, (11)

and the law of motion for technology

logAt = ρ logAt�1 + εt . (12)

limT!∞

E0βT λT kT+1 = 0, (11)

limT!∞

E0βT λT kT+1 = 0, (11)

Equilibrium:

An equilibrium is a set of decision rules:

xt = x (kt ,At )

for x = fct , kt+1, nt ,λtg such that equations (7)-(12) are satis�ed.

Steady State :

In a steady state, (9) implies

1 = βhαyk+ 1� δ

1� β (1� δ). (14)

and (10) impliescy= 1� δ

ky. (15)

Steady State :

1� β (1� δ). (14)

ky. (15)

Steady State :

1� β (1� δ). (14)

ky. (15)

Steady State (continued) :

Hence, the great ratios are given by

=δαβ

1� β (1� δ)(16)

= 1� δαβ

1� β (1� δ)(17)

1� β (1� δ). (18)

Note that the steady-state rate of saving is given by

s� =iy=

δαβ

1� β (1� δ). (19)

Hence, the great ratios are given by

=δαβ

1� β (1� δ)(16)

= 1� δαβ

1� β (1� δ)(17)

1� β (1� δ). (18)

Note that the steady-state rate of saving is given by

s� =iy=

δαβ

1� β (1� δ). (19)

To solve for the steady-state levels, we note that (7) and (8) imply

γn1� n = (1� α)

n� =(1� α) yc

γ+ (1� α) yc< 1. (21)

and by yk = A

� nk

�1�α, we have

k� =

�A�ky

�� 11�α

n� (22)

y � = A (k�)α (n�)1�α ; i� = s�y �; c� = (1� s�) y � (23)

γn1� n = (1� α)

n� =(1� α) yc

γ+ (1� α) yc< 1. (21)

and by yk = A

� nk

�1�α, we have

k� =

�A�ky

�� 11�α

n� (22)

y � = A (k�)α (n�)1�α ; i� = s�y �; c� = (1� s�) y � (23)

γn1� n = (1� α)

n� =(1� α) yc

γ+ (1� α) yc< 1. (21)

and by yk = A

� nk

�1�α, we have

k� =

�A�ky

�� 11�α

n� (22)

y � = A (k�)α (n�)1�α ; i� = s�y �; c� = (1� s�) y � (23)

A Decentralized Version-The �rms :

Aggregate Production Technology:

Yt = AtK αt (ntLt )

1�α , (24)

where n is hours per worker and L is the labor force (without loss ofgenerality, assuming its growth rate be zero).

The per-worker production function is given by

yt = Atkαt n1�αt . (25)

Pro�t maximization:

max [f (kt , nt )� wtnt � (rt + δ)kt ] (26)

which implies the following factor demand functions:

rt + δ = αAtkα�1t n1�α

t (27)

wt = (1� α)Atkαt n�αt . (28)

1�α , (24)

where n is hours per worker and L is the labor force (without loss ofgenerality, assuming its growth rate be zero).The per-worker production function is given by

t (27)

wt = (1� α)Atkαt n�αt . (28)

1�α , (24)

t (27)

wt = (1� α)Atkαt n�αt . (28)

1�α , (24)

t (27)

wt = (1� α)Atkαt n�αt . (28)

A Decentralized Version-The households :

A representative worker�s problem:

maxfct ,st+1,nst g∞

∑t=0

βt [log ct + γ log (1� nst )] (29)

subject toct + st+1 = (1+ rt ) st + wtnst . (30)

First order conditions:

= λt (31)

1� nst= λtwt

λt = βEtλt+1 (1+ rt+1) .

∑t=0

= λt (31)

1� nst= λtwt

λt = βEtλt+1 (1+ rt+1) .

∑t=0

= λt (31)

1� nst= λtwt

λt = βEtλt+1 (1+ rt+1) .

A Decentralized Version-Equilibrium :

Equilibrium: In equilibrium, prices clear the markets and supply meetsdemand:

st+1 = kt+1 (32)

nst = nt . (33)

Hence we have:

= λt (34)

1� nt= λt (1� α)Atkα

t n�αt (35)

λt = βEtλt+1�αAt+1kα�1

t+1 n1�αt+1 + 1� δ

�(36)

ct + kt+1 = (1� δ) kt + Atkαt n1�αt , (37)

A Decentralized Version-Equilibrium :

Equilibrium: In equilibrium, prices clear the markets and supply meetsdemand:

st+1 = kt+1 (32)

nst = nt . (33)

Hence we have:

= λt (34)

t n�αt (35)

λt = βEtλt+1�αAt+1kα�1

t+1 n1�αt+1 + 1� δ

�(36)

ct + kt+1 = (1� δ) kt + Atkαt n1�αt , (37)

A special case :

Equations (3) and (4) become

1ct= βEt

�1ct+1

αyt+1kt+1

�(38)

ct + kt+1 = Atkαt n1�αt . (39)

Guess the decision rule:ct = ξyt . (40)

Then we have,

1yt= βEt

�1yt+1

αyt+1kt+1

�= βαEt

, (41)

which implieskt+1 = βαyt . (42)

A special case :

1ct= βEt

�1ct+1

αyt+1kt+1

�(38)

Then we have,

1yt= βEt

�1yt+1

αyt+1kt+1

�= βαEt

, (41)

A special case :

1ct= βEt

�1ct+1

αyt+1kt+1

�(38)

Then we have,

1yt= βEt

�1yt+1

αyt+1kt+1

�= βαEt

, (41)

A special case continued:

Substituting this into (39) gives

ct = (1� βα) yt . (43)

(8) then implies

1� nt=

1(1� βα) yt

�(1� α)

1� α

1� βα

1nt, (44)

which implies

nt =1�α1�βα

γ+ 1�α1�βα

2 (0, 1). (45)

ct = (1� βα) yt . (43)

(8) then implies

1� nt=

1(1� βα) yt

�(1� α)

1� α

1� βα

1nt, (44)

which implies

nt =1�α1�βα

γ+ 1�α1�βα

2 (0, 1). (45)

ct = (1� βα) yt . (43)

(8) then implies

1� nt=

1(1� βα) yt

�(1� α)

1� α

1� βα

1nt, (44)

which implies

nt =1�α1�βα

γ+ 1�α1�βα

2 (0, 1). (45)

we have already obtained

nt =1�α1�βα

γ+ 1�α1�βα

2 (0, 1). (46)

Hence the decision rules are given by

yt = Atkαt n1�α (47)

ct = (1� βα)Atkαt n1�α (48)

kt+1 = βαAtkαt n1�α. (49)

we have already obtained

nt =1�α1�βα

γ+ 1�α1�βα

2 (0, 1). (46)

Hence the decision rules are given by

yt = Atkαt n1�α (47)

ct = (1� βα)Atkαt n1�α (48)

kt+1 = βαAtkαt n1�α. (49)

A special case -Log linearization:

take log we have

yt = At + αkt (50)

ct = At + αkt (51)

kt+1 = At + αkt . (52)

At = ρAt�1 + εt

1� ρLεt =

∑j=0

ρj εt�j , (53)

where L is a lag operator LX (t) = X (t � 1), LjX (t) = X (t � j)the decision rule for capital can be rewritten as a moving-averageprocess:

kt+1 =1

1� αLAt =

1(1� αL) (1� ρL)

take log we have

yt = At + αkt (50)

ct = At + αkt (51)

kt+1 = At + αkt . (52)

At = ρAt�1 + εt

1� ρLεt =

∑j=0

ρj εt�j , (53)

where L is a lag operator LX (t) = X (t � 1), LjX (t) = X (t � j)

the decision rule for capital can be rewritten as a moving-averageprocess:

kt+1 =1

1� αLAt =

1(1� αL) (1� ρL)

take log we have

yt = At + αkt (50)

ct = At + αkt (51)

kt+1 = At + αkt . (52)

At = ρAt�1 + εt

1� ρLεt =

∑j=0

ρj εt�j , (53)

where L is a lag operator LX (t) = X (t � 1), LjX (t) = X (t � j)the decision rule for capital can be rewritten as a moving-averageprocess:

kt+1 =1

1� αLAt =

1(1� αL) (1� ρL)

A special case -Log linearization(continued):

or as an AR(2) process:

(1� αL) (1� ρL) kt+1 = εt (54)�1� (α+ ρ) L+ αρL2

�kt+1 = εt (55)

orkt+1 = (α+ ρ) kt � αρkt�1 + εt . (56)

Utilizing (50), we can also express consumption and output as ARMA(p, q) processes (p = 2, q = 0):

1� ρLεt +

(1� αL) (1� ρL)εt�1 (57)

(1� αL) (1� ρL) xt = (1� αL) εt + αεt�1

= εt . (58)

(1� αL) (1� ρL) kt+1 = εt (54)�1� (α+ ρ) L+ αρL2

�kt+1 = εt (55)

orkt+1 = (α+ ρ) kt � αρkt�1 + εt . (56)

1� ρLεt +

(1� αL) (1� ρL)εt�1 (57)

(1� αL) (1� ρL) xt = (1� αL) εt + αεt�1

= εt . (58)

(1� αL) (1� ρL) kt+1 = εt (54)�1� (α+ ρ) L+ αρL2

�kt+1 = εt (55)

orkt+1 = (α+ ρ) kt � αρkt�1 + εt . (56)

1� ρLεt +

(1� αL) (1� ρL)εt�1 (57)

(1� αL) (1� ρL) xt = (1� αL) εt + αεt�1

= εt . (58)

(1� αL) (1� ρL) kt+1 = εt (54)�1� (α+ ρ) L+ αρL2

�kt+1 = εt (55)

orkt+1 = (α+ ρ) kt � αρkt�1 + εt . (56)

1� ρLεt +

(1� αL) (1� ρL)εt�1 (57)

(1� αL) (1� ρL) xt = (1� αL) εt + αεt�1

= εt . (58)

A special case -impulse responses:

Consider the impulse responses of capital to a one unit increase intechnology at time t. Since

kt+1 = (α+ ρ) kt � αρkt�1 + εt , (59)

or using the state-space representation, we have�kt+1kt

�α+ ρ �αρ1 0

��ktkt�1

�1 00 0

��εtεt�1

�, (60)

A special case -impulse responses:

Consider the impulse responses of capital to a one unit increase intechnology at time t. Since

kt+1 = (α+ ρ) kt � αρkt�1 + εt , (59)

or using the state-space representation, we have�kt+1kt

�α+ ρ �αρ1 0

��ktkt�1

�1 00 0

��εtεt�1

�, (60)

Log linearization:

First-order conditions:1ct= λt (61)

1� nt= λt

�(1� α)Atkα

t n�αt

�(62)

λt = βEt�λt+1

�αAt+1kα�1

t+1 n1�αt+1 + 1� δ

��(63)

ct + kt+1 � (1� δ) kt = Atkαt n1�αt . (64)

Log linearization (continued ):

Denote xt = xt�xx ' log(xt )� log(x) as the percentage change from

steady-state.

log-linearizing equationyt = xα

t (65)

yieldsyt = αxt (66)

1txβ2t (67)

yiedsyt = αx1t + βx2t (68)

steady-state.

t (65)

1txβ2t (67)

steady-state.

t (65)

1txβ2t (67)

log-linearizing equationyt = x1t + x2t (69)

yieldsyt =

x1x1 + x2

x1t +x2

x1 + x2x2t (70)

log-linearizing equation

∑j=1xjt (71)

∑j=1

�xjy

�xjt (72)

log-linearizing equationyt = x1t + x2t (69)

yieldsyt =

x1x1 + x2

x1t +x2

x1 + x2x2t (70)

∑j=1xjt (71)

∑j=1

�xjy

�xjt (72)

yt = Et [x1t+1 + x2t+1] (73)

yieldsyt =

x1x1 + x2

Et x1t+1 +x2

x1 + x2Et x2t+1 (74)

log-linearizing equationyt = Etxα

t+1 (75)

yiedsyt = αEt xt+1 (76)

yt = Et [x1t+1 + x2t+1] (73)

yieldsyt =

x1x1 + x2

Et x1t+1 +x2

x1 + x2Et x2t+1 (74)

log-linearizing equationyt = Etxα

t+1 (75)

yiedsyt = αEt xt+1 (76)

Log linearizing the f.o.cs:

Log linearization the equation

1ct= λt (77)

gives�ct = λt (78)

t n�αt (79)

yiedsn

1� n nt = λt + At + αkt � αnt (80)

where sc = cy , si =

Log linearizing the f.o.cs:

1ct= λt (77)

gives�ct = λt (78)

t n�αt (79)

yiedsn

1� n nt = λt + At + αkt � αnt (80)

where sc = cy , si =

Log linearizing the f.o.cs(continued):

λt = βEt�λt+1

�αAt+1kα�1

t+1 n1�αt+1 + 1� δ

��(81)

yields

λt = Et�

λt+1 + (1� β (1� δ))�(α� 1) kt+1 + (1� α) nt+1 + At

�(82)

and log-linearization the equation

yields

sc ct + si

�1δkt+1 �

(1� δ)

�= At + αkt + (1� α) nt (84)

and the technology follows

At = ρAt�1 + εt (85)

λt = βEt�λt+1

�αAt+1kα�1

t+1 n1�αt+1 + 1� δ

��(81)

yields

λt = Et�

λt+1 + (1� β (1� δ))�(α� 1) kt+1 + (1� α) nt+1 + At

�(82)

yields

sc ct + si

�1δkt+1 �

(1� δ)

�= At + αkt + (1� α) nt (84)

At = ρAt�1 + εt (85)

λt = βEt�λt+1

�αAt+1kα�1

t+1 n1�αt+1 + 1� δ

��(81)

yields

λt = Et�

λt+1 + (1� β (1� δ))�(α� 1) kt+1 + (1� α) nt+1 + At

�(82)

yields

sc ct + si

�1δkt+1 �

(1� δ)

�= At + αkt + (1� α) nt (84)

At = ρAt�1 + εt (85)

Reduced forms:

we can write consumption and labor as�ctnt

�= A1

�ktλt

�+ A2At (86)

and equation (82) and (84) become

�kt+1λt+1

�= B

�ktλt

�+ R1Et At+1 + R2At . (87)

A unique equilibrium exists if one of the eigenvalues of B lies outsidethe unit circle and the other lies inside the unit circle.

Reduced forms:

�= A1

�ktλt

�+ A2At (86)

�kt+1λt+1

�= B

�ktλt

�+ R1Et At+1 + R2At . (87)

Reduced forms:

�= A1

�ktλt

�+ A2At (86)

�kt+1λt+1

�= B

�ktλt

�+ R1Et At+1 + R2At . (87)

Solution Method 1:

(Remember to put back Et operator later)�kt+1λt+1

�= PΛP�1

�ktλt

�+ R1At+1 + R2At , (88)

which implies

P�1�kt+1λt+1

�= ΛP�1

�ktλt

�+ P�1R1At+1 + P�1R2At , (89)

or �x1t+1x2t+1

�λ1 00 λ2

��x1tx2t

�+ R1At+1 + R2At . (90)

Solution Method 1:

�= PΛP�1

�ktλt

�+ R1At+1 + R2At , (88)

which implies

P�1�kt+1λt+1

�= ΛP�1

�ktλt

�+ P�1R1At+1 + P�1R2At , (89)

or �x1t+1x2t+1

�λ1 00 λ2

��x1tx2t

�+ R1At+1 + R2At . (90)

Solution Method 1:

�= PΛP�1

�ktλt

�+ R1At+1 + R2At , (88)

which implies

P�1�kt+1λt+1

�= ΛP�1

�ktλt

�+ P�1R1At+1 + P�1R2At , (89)

or �x1t+1x2t+1

�λ1 00 λ2

��x1tx2t

�+ R1At+1 + R2At . (90)

Solution Method 1 (continued):

Suppose jλ2j > 1, then

Etx2t+1 = λ2xt + r1Et At+1 + r2At , (91)

orx2t =

1λ2Etx2t+1 �

�r1Et At+1 + r2At

Denote the foward operator as F ,

Fxt = Etxt+1,FFxt = FEtxt+1 = Et [Et+1xt+2] = Etxt+2 (92)

Iterating this forward and applying the law of iterated expectations, astationary solution for this di¤erence equation is given by

x2t =� 1

1� 1λ2F

�r1Et At+1 + r2At

= � r1

∑j=0

�jEt At+1+j �

∑j=0

�jEt At+j

= � r1λ2

1� ρλ2

At �r2λ2

11� ρ

= �ρr1 + r2λ2 � ρ

At . (93)

Denote the foward operator as F ,

Fxt = Etxt+1,FFxt = FEtxt+1 = Et [Et+1xt+2] = Etxt+2 (92)

Iterating this forward and applying the law of iterated expectations, astationary solution for this di¤erence equation is given by

x2t =� 1

1� 1λ2F

�r1Et At+1 + r2At

= � r1

∑j=0

�jEt At+1+j �

∑j=0

�jEt At+j

= � r1λ2

1� ρλ2

At �r2λ2

11� ρ

= �ρr1 + r2λ2 � ρ

At . (93)

since �x1tx2t

�� P�1

�ktλt

�, (94)

we havex2t = p21kt + p22λt .

Thus, the equilibrium path for λt is given by

λt = �p21p22kt �

ρr1 + r2λ2 � ρ

At . (95)

since �x1tx2t

�� P�1

�ktλt

�, (94)

we havex2t = p21kt + p22λt .

Thus, the equilibrium path for λt is given by

λt = �p21p22kt �

ρr1 + r2λ2 � ρ

At . (95)

Once we obtain λt , we have�ctnt

�= Π

�ktAt

�(96)

kt+1 = π

�ktAt

�. (97)

Once we obtain λt , we have�ctnt

�= Π

�ktAt

�(96)

kt+1 = π

�ktAt

�. (97)

Solution Method 2

We have �ctnt

�= A2�3

0@ ktAtλt

1A (98)

0@ kt+1At+1λt+1

1A = B3�3

0@ ktAtλt

1A = PΛP�1

0@ ktAtλt

1A . (99)

Solution Method 2

We have �ctnt

�= A2�3

0@ ktAtλt

1A (98)

0@ kt+1At+1λt+1

1A = B3�3

0@ ktAtλt

1A = PΛP�1

0@ ktAtλt

1A . (99)

Solution Method 2 (continued)

De�ne

xt = P�1

0@ ktAtλt

1A , (100)

and suppose jλ3j > 1. Then a stationary solution is given by

x3t = 0, (101)

orp31kt + p32At + p33λt = 0, (102)

orλt = �

p31p33kt �

p32p33At . (103)

De�ne

xt = P�1

0@ ktAtλt

1A , (100)

x3t = 0, (101)

orp31kt + p32At + p33λt = 0, (102)

orλt = �

p31p33kt �

p32p33At . (103)

De�ne

xt = P�1

0@ ktAtλt

1A , (100)

x3t = 0, (101)

orp31kt + p32At + p33λt = 0, (102)

orλt = �

p31p33kt �

p32p33At . (103)

De�ne

xt = P�1

0@ ktAtλt

1A , (100)

x3t = 0, (101)

orp31kt + p32At + p33λt = 0, (102)

orλt = �

p31p33kt �

p32p33At . (103)

An example

Suppose we have �ctkt

�= QEt

�ct+1kt+1

�(104)

Where ct is the endogenous variable and kt is the state variable. Thematrix Q is:

Q =��2.5 �91.5 5

�(105)

Solve ct in term of kt .

An example

Suppose we have �ctkt

�= QEt

�ct+1kt+1

�(104)

Where ct is the endogenous variable and kt is the state variable. Thematrix Q is:

Q =��2.5 �91.5 5

�(105)

Solve ct in term of kt .

An example

Follow the procedure in class, The system can be written as :�ctkt

��3 21 �1

��0.5 00 2

��1 �2�1 �3

� �ct+1kt+1

�(106)

Therefore we havect + 2kt = 0 (107)

orct = �2kt (108)

An example

Follow the procedure in class, The system can be written as :�ctkt

��3 21 �1

��0.5 00 2

��1 �2�1 �3

� �ct+1kt+1

�(106)

Therefore we havect + 2kt = 0 (107)

orct = �2kt (108)

Calibration

There are 5 structural parameters in the model, fα, β, δ,γ, ρg ,so weneed �ve conditions

we can use the steady-state relationships to back solve the parametervalues so that the implied steady-state values of model are consistentwith empirical data.

In the steady-state

1 = β (1+ r �)

α =(r � + δ)k�

y �; 1� α =

y �=

1� β (1� δ).

If r = 4% a year (i.e., 1% a quarter), then in a quarterly model wehave β = 1

1.01 = 0.99.

Calibration

In the steady-state

1 = β (1+ r �)

α =(r � + δ)k�

y �; 1� α =

y �=

1� β (1� δ).

1.01 = 0.99.

Calibration

In the steady-state

1 = β (1+ r �)

α =(r � + δ)k�

y �; 1� α =

y �=

1� β (1� δ).

1.01 = 0.99.

Calibration

In the steady-state

1 = β (1+ r �)

α =(r � + δ)k�

y �; 1� α =

y �=

1� β (1� δ).

1.01 = 0.99.

Calibration (continued)

wn�y � = 1� α; and wn�

y � = 0.65 in data, so we have α = 0.35.

k �y � = 10 in data, by α yk = (r + δ), we have

δ =α

k/y� r = 0.035� 0.01 = 0.025. (109)

implies a steady-state rate of saving equal to

s� = δky= 0.25 = 25%, (110)

and a steady-state cy = 1� s = 0.75.

y � = 0.65 in data, so we have α = 0.35.k �y � = 10 in data, by α yk = (r + δ), we have

δ =α

k/y� r = 0.035� 0.01 = 0.025. (109)

s� = δky= 0.25 = 25%, (110)

y � = 0.65 in data, so we have α = 0.35.k �y � = 10 in data, by α yk = (r + δ), we have

δ =α

k/y� r = 0.035� 0.01 = 0.025. (109)

s� = δky= 0.25 = 25%, (110)

Also, we know that the fraction of hours worked each week is aboutn� = 40

24�7 � . 24 (which implies that the fraction of hours worked ina quarter is also 0.24). Given the steady-state relationship,

γn1� n = (1� α)

yc=0.760.24

� 0.650.75

= 2.74, (111)

Finally, to calibrate ρ, we can estimate the Solow residual using

At = yt � αkt � (1� α) nt ,

and then estimate ρ by

At = ρAt�1 + et . (112)

Also, we know that the fraction of hours worked each week is aboutn� = 40

24�7 � . 24 (which implies that the fraction of hours worked ina quarter is also 0.24). Given the steady-state relationship,

γn1� n = (1� α)

yc=0.760.24

� 0.650.75

= 2.74, (111)

Finally, to calibrate ρ, we can estimate the Solow residual using

At = yt � αkt � (1� α) nt ,

and then estimate ρ by

At = ρAt�1 + et . (112)

Simulation

The decisions rules can be arranged into:0BBB@ctntyt...

1CCCA = Π�ktAt

�(113)

�ktAt

�= M

�kt�1At�1

�εt . (114)

Simulation (continued)

Starting from the steady state at t = 0, we have k0�j = 0 andA0�j = 0 for j > 0. Given the sequence, fεtgTt=0 (drawn from arandom generator), (8) implies�

�ε0 (115)

�k1A1

�= M

�k0A0

�ε1 (116)�

kt+1At+1

�= M

�kt�1At�1

�εt+1. (117)

Substituting the generated sequences,�kt , At

Tt=0 , into (113)

produces the sequences fct , nt , yt , ...gTt=0.

Notes on the Real Business Cycle Model

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Transcript of Notes on the Real Business Cycle Model