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1、Advanced MacroeconomicsReal-Business-Cycle Theory, Part IINovember 17, 2014A Special CaseSimplifying AssumptionsI = 1IGt= 0 s = enl =1 1 +(1 s)bConstant saving rate and labor supply per personA Special CaseThis simple model provides an example of an economy where real shocks drive output movements.T
2、he output movements are the optimal responses to the shocks.In this model, fluctuations do not reflect any market failures, and government interventions to mitigate them can only reduce welfare.Observed aggregate output movements represent the time-varying Pareto optimum.A Special CaseOutput fluctua
3、tions are determined by the two state variables: A and KlnYt= lnKt+(1)(lnAt+lnLt)Given that Kt= sYt1, Lt=lNt, ln(At) =A+gt +At, ln(Nt) =N +ntlnYt=ln s +lnYt1+(1)(A+gt +At+lnl +N +nt)A Special CaseLetYtbe the difference between lnYtand the value it would take if ln(At) =A+gtIYt= lnYtlnYtlnYt=ln s +ln
4、Yt1(1)+(1)(A+gt +lnl +N +nt)lnYt=ln s +lnYt1(2)+(1)(A+gt +At+lnl +N +nt)(2) - (1):eYt= eYt1+(1)AtA Special CaseeYt= eYt1+(1)AtAt=1 1(eYteYt1)Given thatAt= AAt1+A,teYt=eYt1+(1)(AAt1+A,t)=eYt1+(1)(A 1(eYt1eYt2)+A,t)=( +A)eYt1AeYt2+(1)A,tA Special CaseeYt= ( +A)eYt1AeYt2+(1)A,tOutput can have a “hump-s
5、haped” response to disturbances.Investment ( sYt) and consumption (1 s)Yt) are equally volatileIEmpirical: investment varies much more than consumptionlabor input (l) does not varyIEmpirical: employment and hours are strongly procyclicalReal wage (wt= (1)Yt/Lt) is highly procyclical: Ltdoes not resp
6、ond to technology shocks wtrises one-for-one with YtIEmpirical: wtis only moderately procyclicalLog-LinearizationLet Xtbe a strictly positive variable,Xtits balanced-growth-path valueXt lnXtlnXtSuppose that we have an equation of the following form:f (Xt,Yt) = g(Zt)This equation is clearly also vali
7、d at its balanced-growth-path valuesf (Xt,Yt) = g(Zt)Log-LinearizationRewrite the variables using the identity (Xt= exp(ln(Xt) and then take logs on both sidesln? f (eln(Xt),eln(Yt)? = ln? g(eln(Zt)?Take a first order Taylor approximation around their balanced-growth-path values (ln(Xt),ln(Yt),ln(Zt
8、)LHS = ln(f (Xt,Yt)+1 f (Xt,Yt)?f1(Xt,Yt)Xt(lnXtlnXt) +f2(Xt,Yt)Yt(lnYtlnYt)?RHS = ln(g(Zt)+1 g(Zt)g0(Zt)Zt(lnZtlnZt)By equating LHS and RHS,f1(Xt,Yt)XtXt+f2(Xt,Yt)YtYt= g0(Zt)ZtZtLog-LinearizationA simpler methodXt=XteXtXte0+Xte0(Xt0)Xt(1+Xt)f (Xt)f (Xt)+f0(Xt)(XtXt)f (Xt)1+f0(Xt) f (Xt)Xt(XtXt Xt)
9、!f (Xt)1+f0(Xt) f (Xt)XtXt!Log-LinearizationA simpler methodf (Xt,Yt)f (Xt,Yt)+f1(Xt,Yt)(XtXt)+f2(Xt,Yt)(YtYt)f (Xt,Yt)1+f1(Xt,Yt) f (Xt,Yt)Xt(XtXt Xt)+f1(Xt,Yt) f (Xt,Yt)Yt(YtYt Yt)f (Xt,Yt)? 1+f1(Xt,Yt) f (Xt,Yt)XtXt+f2(Xt,Yt) f (Xt,Yt)YtYt?Note:XtYt 0Log-LinearizationA simpler methodXt+YtXt(1+Xt)
10、+Yt(1+Yt)XtYtXt(1+Xt)Yt(1+Yt) XtYt(1+Xt+Yt)Xt YtXt Yt(1+XtYt)(Xt) (Xt)(1+Xt)Log-LinearizationA simpler methodZt= Xt+YtZt1Xt+Yt(XtXt+YtYt)Zt= XtYtZtXt+YtZt=Xt YtZtXtYtZt= (Xt)Zt XtGeneral CaseV(Kt,At,Gt) = max lt,Kt+1?u(ct,1lt)NtH+ eEtV(Kt+1,At+1,Gt+1)?s.t.Kt+1 H=Kt H+wtltNt H+rtKt HGt HctNt HGeneral
11、 CaseFOC ct 1lt=wt b1 ct= eEt?1 ct+1(1+rt+1)?General CaseFOC1ct 1lt=(1)KtA1tLt bFOC2 1 ct= eEt?1 ct+1(1+K1t+1(At+1Lt+1)1)?General Case: FOC1ct 1lt=(1)KtA1tLt bLHS:LHS ct 1lt(1+1 1lt ct 1lt ct ct+ ct (1lt)2 ct 1ltltlt) ct 1lt(1+ ct+lt 1ltlt)RHSRHS(1)KtA1tLt b(1+Kt+(1)AtLt)General Case: FOC1Ct= ct,Lt=
12、ltIe.g., lnCt= lnct+lnNt lnCtlnCt= lnctln ct+(lnNtlnNt)By equating the LHS and RHSCt+(lt 1lt+)Lt= Kt+(1)AtNote: Along the balanced growth path,lt= lGeneral Case: FOC1Method of undetermined coefficients: assume consumption and employment take the following formCtCKKt+CAAt+CGGt LtLKKt+LAAt+LGGtThen, t
13、he log-linearized version of FOC1 can be modified as follows:CKKt+CAAt+CGGt+ (lt 1lt+)(LKKt+LAAt+LGGt)= Kt+(1)AtGeneral Case: FOC1Unknown coefficients must satisfyCK+(lt 1lt+)LK=CA+(lt 1lt+)LA=1CG+(lt 1lt+)LG=0General Case: FOC21 ct= eEth 1 ct+1(1+rt+1)iLHS:LHS1 ct(1+1 c2t ct1 ct ct)1 ct(1 ct)RHSRHS
14、eEt1+ rt+1 ct+1(1+1 ct+1 rt+1 1+ rt+1 ct+1 rt+1+1+ rt+1( ct+1)2 ct+1 1+ rt+1 ct+1 ct+1)eEt?1+ r t+1 ct+1(1+ rt+1 1+ rt+1 rt+1 ct+1)?General Case: FOC2By equating the LHS and RHS1 ct(1 ct) = eEt?1+ r t+1 ct+1(1+ rt+1 1+ rt+1 rt+1 ct+1)? ct= Et? rt+1 1+ rt+1 rt+1 ct+1?Ct= Et? rt+1 1+ rt+1 rt+1Ct+1?Note: Along the balanced growth path, rt+1= rGeneral Case: FOC2We derive rt+1in two stepsStep OneKt+1= (1)Kt+Kt(AtLt)1CtGtKt+1(1+Kt+1) =?(1)Kt(1+Kt)Ct(1+Ct)Gt(1+Gt) +Kt(AtLt)1(1+Kt+(1)At+(1)Lt)?Kt+1(Kt
