弹簧一样的脉冲响应图,图片在附件,求大神帮忙看看,代码如下:var c q b y lab d r z j pai q_y h;
varexo e eps;
parameters beta delta yita alpha ds rhor rhopai rhoy rhoq rhod rhob rhoz rhoj;
beta=0.987;
delta=0.012;
yita=6.16;
alpha=0.4;
ds=0.7;
rhor=0.73;
rhopai=0.36;
rhoy=0.36;
rhoq=0.13;
rhod=0.3;
rhob=0.5;
rhoz=0.8;
rhoj=0.8;
model;
#A=((beta*ds-delta-ds)*beta/(1-beta+beta*delta));
#labs=((1-A)*(1-alpha)/(1+A*alpha))^(1/(1+yita));
#ys=labs^(1-alpha);
#qs=beta*(1-alpha+alpha*labs^(1+yita))/((1-beta+delta*beta)*labs^(alpha+yita));
#rs=1/beta;
#bs=ds*beta*qs;
beta*(exp(j)/h+(1-delta)*q/c+alpha*y/(h*c))=q(-1)/c(-1);
c*lab^(1+yita)=(1-alpha)*y;
r/pai=c(+1)/(beta*c);
b(-1)*r(-1)/pai+c+q*(h(+1)-(1-delta)*h)=y+b;
r*b=d*q(+1)*h(+1)*pai(+1);
r=r(-1)^rhor*(pai(-1)^(1+rhopai)*(y(-1)/ys)^rhoy*(q(-1)/qs)^rhoq*rs)^(1-rhor);
z=rhoz*z(-1)+e;
j= rhoj*j(-1)+eps;
y=exp(z)*(h^alpha)*(lab^(1-alpha));
h=1;
d/ds=(d(-1)/ds)^rhod*(b(-1)/bs)^(-rhob);
q_y=q/y;
end;
initval;
c=0.3773;
q=31.5680;
b=21.8103;
y=1.0434;
lab=1.0733;
d=ds;
r=1.0132;
z=0;
j=0;
pai=1;
q_y=30.25638;
h=1;
end;
steady;
check;
resid;
shocks;
var e=0.0001;
var eps=0.0001;
end;
stoch_simul(periods=900,order=1,irf=800,drop=400);