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Lecture Note: Business Cycles ModelVimut VanitcharearnthumUniversity of the Thai Chamber of Commerce

March 5, 2009

1 A Prototype ModelA representative agent maximizes

Eo L0 (3tu (Ct, 1 - nt) . (1)t=O

where u(Ct, 1 - nt) = a In(ct) + (1 - a) In(1 - nt) . The budget constraint isCt + kt+1 = eZtkfni-O+ (1 - 6)kt . (2)

The technology shock is assumed to follow the AR(I) process. That is,Zt+l = pZ t + Et+l . (3)

The first-order conditions are1a-

Ct1(l-a)--.f 1- nt

At1.1 Steady State

I-aI-n1

{3c

(4)A (1 - B)eZt kOn -0t t , (5)

(6)

(7)(8)(9)


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Variables Steady State Valuesr 0.0454k 10.52y 1.19c 0.93

Table 2: Computed Steady State Values1.3 Log-linearized

C f N l ~ o \ '-I.- - ~ - " ~ wl(Q..,- ~ ~ t v t3 = At, IS ( (10)n A A

--1-nt At + ekt - ent + Zt , (11)- n '\t+l + 3A(e - 1)rkt+1+ /3(1 - e)rn:+1 + /3rZt+l ,(12) (~ + ~ ~ ~ t ,+ (1 - e)1/&)+ ( ~ _ ~ 8)kkt . \ (13) f

L.- ~ J . P J s - t " ' e o ~ ~ - t t - v. r--' e } C o ~ l I ' e o v t J ~ \ - - t i l \ " L2 Constructing A State-space System

(11) can be rewritten as( e - _n_) nt = ,\t + ekt + Zt , (14)1 -n

~ k ~ ( l t " , > r or, tot~ Ao { ~ G \ l ( t , nt =


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--

where r k / ' 0 n ;C]< \:I __1Ao = t,6(l- 1 I ) ~ [ 4 > o -l j)2+ { 3 r ( ~ -= 1I)4>J ~ { 3 ~ ( 1 ~ ~ ~ l 2 y[O + (1 - .8)0] + (1 - 6)k C -1+ (1 - 8)Y1 y(1 + (1 - 8)1 ]

Al = 0 1 0[ o . 0 pThe above system can also be rewritten as

+ B Et+l , (17)

where A = Ail l A l and B = Ail l [ ] . Using these numerical values, weare able to construct the matrices A and B below

[ ~ : : ~ ]. [ ! O ~ ; 4 ( ~ . ; } --1l!Li ] .[ ~ ]+ [ - O 77 ] Et+ l ' (18)Zt+l 0 0 0.8 Zt 1A BTo decouple the dynamics of this system, we need to find the eigenvaluesand the eigenvectors of the matrix A. Matlab can help us find them easily.

Just use the command "eig". For example, simply type[V,D] = eig(A)will give us the matrix of eigenvectors (V), and the diagopal matrix of. 1 D A I " n ~ h t 9 . U IM ' IVelgenva ues, . ( ~ - \ ' t b {OjVlloll\.tAHV , f-In our model, the diagonal matrix i S ~ i S tv h Z : ~ : J \i(\.tvv \'- (Iv... 0 A.\ /J-tJ.. ,} J.I'.t-' toll/", v ~ ~ ~ '--- 't _ [1.978 0 0] b(vJ Kt s.D - 0 0.399 0 .o 0 0.8Notice that the eigenvalue in the first row has a value outside a unitcircle. This is the unstable root that we can use to eliminate >. from our

dynamical system. That is, we are able to write>' as a linear combinationof it, and z, the endogenous and exogenous state variables of our model.

It-.q =- PZt + ~ f , - t , 4~ ' t + \ "::; 1 . l . f ~ kt- + to, 'l-') At


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v.J == /yw(v)The above state-space ystem can be rewritten as - W(1 11)/W(1/2-)

kt+ I ] = [ k t] + -0.77 ]t + I DV- I At [ E t+ I , (19)[ ~ + 1or,

co- -11tk V -I J ~ = ZR[ t ]+ [ - o 77 ] Et+I , (20)5 ~ ~ t + I 11 Zt 1d ,,,,,e", , , . , , . j ~ N \ 0et rid of ).++1 T h ~ F s t - e q u a ' b ~ W ~ i ~ t g i s system is/ ~ ( J i ) VI,lkt+l + VI,2.xt+1 + VI,azt+l)= 1.98(VI,lkt + VI,2.x t + VI,aZt) ,I -where v(l,j) is the first row of matrix V-I.

Since this is the equation associating with the unstable root (1.98 > 1),th e linear combination in th e parenthesis has to be zero. Therefore, we areable to write vtt",'() A I rT\_ . kt-,--t-, -::. S , 4 ? Kt' -1: 0.'Z)\(-.!J

A AI V I3 .At = ' ~ - -'Zt . -+- " l- I. { CVI 2 VI 2 tr'~ L ~ ~ ~ , i ) ' ..",.,se this relationship in ~ ~ a ~ - - t o 6 P t - a i R _k+I = 0.399kt + 0.239zt . "" -""-"",' ...

Finally, the law of motion ofthestate variables in this model is""---------

k t+I ] = /'["/0'399. ,,,,0, .23;/1 [ kt ] + [ ]Et+I (21)[ Zt+I ,0;8 J Zt 1v..DD3 Impulse Response Function /Suppose we are interested in the dynamic impact of a one-time increase in /the innovation to technological change, Et . /The impact of Et on kt+I an d Zt+ I is

ltft,u of (\I\o-tllm u f [ kt+I ] = [0.399 0.239] [ ] 0.239 ]L r.- Zt+I ' 0.8 1 [ 0.8 /l C l . \ ~ (l r M/) tiOVl Or 1; /1\

-:;:! -Ct 5

ACt-

IJvAV1t :::.


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The impact of Et on k+2 and Zt+2 isk+2 ] = [0.399 0.239] [0.399 0.239"] [ 0 ]

"[ Zt+2 0 0.8 0 0.8 1

= [ O O ~ 6 3 4 4 ] .The impact of Et on kt+i and Zt+i is

kt+i "] = [0.399 0.239] i [ 0 ][ Zt+i 0 0.8 1Here is the Matlab code for computing the impulse response function,M = zeros(2,200);For i = 1:200 1;1:':,; A\i*[O;1 ; '(\ ('1-\/ 'C . / )V0~ I ~ - , ~ ( ) 9" l""-'\ J

p r .) -::; l\> . ~ o / \ } ')t!'Co')v \)0'"v " rN .0v' l (V]

-1V Vt-t-rL--v-_...j\ Ii\ 0. 'J e" V\ ~ n V l ~ fj (IV\

6


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.I1O ..000 Mauro! Date ? .I t?1.. .

I' 1 hMere.. Xok =- lA\, 0 -- fA1 0 0 . - ,I 'A4; \ 0 }A2. ()li j 0 }A 1..

() ,4\

-z..,.....- 0 xoc D1 - -X

AAC1 71'") Kt + AC172-) . . 'ri(1,3) l . ..... , 4 (13 )2

1\'(1 ~ - A(17-)


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MFile(MATLAB)

%ThisisanMfileformyfirstRBCModel.

%Settingparametervalues.

beta=0.98;

theta=0.4;

alpha=0.33;

delta=0.025;

n=0.28;

rho=0.8;

phi1=1/(theta(n/(1n)));

phi0=phi1*theta;

%Steadystatevalue

r=1/beta (1delta);

yk=r/theta;

kn=yk^(1/(theta1));

k=kn*n;

y=yk*k;

c=ydelta*k;

%ConstructMatrices

a4=beta*(1theta)*r*(phi01);

a5=2+beta*r*(1

theta)*phi1;

a6=beta*r*(1theta)*phi1+1;

A0=[k00;a4a5a6;001];

a1=y*(theta+(1theta)*phi0)+(1delta)*k;

a2=(1/c)+(1theta)*y*phi1;

a3=y*(1+(1theta)*phi1);

A1=[a1a2a3;010;00rho];


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%CalculatingMatrix

A=inv(A0)*A1;

B=inv(A0)*[0;0;1];

[V,D]=eig(A);

W=inv(V);

%The1steigenvaluesisoutsidetheunitcircle.Therefore,weimposethat

%thetransformedvariableinthe1strowofmatrixV*Xequalszero

DD=[A(1,1)(A(1,2)*(W(1,1)/W(1,2)))A(1,3)(A(1,2)*(W(1,3)/W(1,2)));0rho];

%ComputingImpulseResponseFunction

%onetimeincreaseintechnologyshock

M=zeros(2,200);

fori=1:200

M(:,i)=DD^i*[0;1];

end;

Y=M';

plot(Y)

%ImpulseResponseFunctionforConsumption,IncomeandLabor

c_hat=zeros(200,1);

y_hat=zeros(200,1);

n_hat=zeros(200,1);

fori=1:200

c_hat(i)=[W(1,1)/W(1,2)W(1,3)/W(1,2)]*M(:,i);

n_hat(i)=[phi1*(theta(W(1,1)/W(1,2)))phi1*(1(W(1,3)/W(1,2)))]*M(:,i);

y_hat(i)=(r/theta)*M(1,i)+M(2,i);

end;

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