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Hybrid New Keynesian Model

조회 수: 5 (최근 30일)
Farah Shahpoor
Farah Shahpoor 2018년 12월 14일
Hello, I have a problem with the timepath of the variables, when I program the Hybrid NKM under commitment. Iam using the Schur decomposition as usual in the hybrid model.
The code I used :
%Parameters
gam_f = 0.5
gam_b =0.5
beta= 1
gam_x = 0.2
lambda = 0.5
AR_par = 0.8
% System is: (w;v)(+1) = A*[w;v] + [ 1;0;0;0;0]*eps
A11= [AR_par 0 0;0 0 0; 0 0 0];
A12= [ 0 0; 1 0; 0 1];
A21 =[ 0 -gam_f/beta^2*gam_b 0; -1/beta*gam_f 0 -gam_b/gam_f];
A22 =[ 1/beta^2*gam_f gam_x/beta^2*lambda*gam_b; -gam_x/beta*gam_f 1/beta*gam_f];
A = [ [A11 A12] ; [A21 A22] ]
% Solve the System
disp('Schur decomposition')
[Z, T] = schur(A, 'complex')
disp ('reorder eigenvalues in increasing order along the principal diagonal')
[Z, T] = ordschur(Z,T, 1:5)
if abs(sum(sum(Z*T*Z'-A))) > 0.0001 && sum(sum(Z'*Z-eye(lenght(Z)))) > 0.0001
disp('Error in Schur decomposition')
end
disp('check Blanchard-Kahn')
abs_diag_T = abs(diag(T))'
%%Calculating the solution time path
% z(+1) = E[z(+1)] + Z_11^-1 * [1;0;0] * eps
T_11 = T(1:1,1:1);
Z_11 = Z(1:1,1:1);
Z_21 = Z(4:5,1:1);
T=30;
z_solution= NaN(3,T);
w_solution= NaN(3,T);
w_solution(:,1)=[ 1; 0; 0];
z_solution(:,1)=inv(Z_11)* w_solution(: ,1);
v_solution= NaN(2,T);
for t=2:T
z_solution(:,t)= T_11* z_solution(: ,t-1 );
w_solution(:,t)= Z_11 * z_solution(:,t);
v_solution(:,t)= Z_21 *inv(Z_11)* w_solution(:,t);
end
Funktion ohne Link?
Specially the initial jump of output and inflation looks completly different as it should be. Does anybody have a hint?

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