DSGE (3): BASIC NEOCLASSICAL MODELS (2): APPROACHES OF SOLUTIONS

DSGE (3): Stochastic neoclassical growth models (2): how to solve? 27 DSGE (3): BASIC NEOCLASSICAL MODELS (2): APPROACHES OF SOLUTIONS. Recap of basic neoclassical growth models and log-linearisation 2. Solutions by hands Uhlig style 3. Solutions by matrix/vectors Blanchard-Kahn techniques Further Reading: Uhlig (999) toolkit is a very good starting point to read (the first part on stochastic neoclassical growth models) Uhlig, H. (999), "A toolkit for analysing nonlinear dynamic stochastic models easily" in Marimon R. and A. Scott, eds, Computational Methods for the Study of Dynamic Economies, Oxford University Press. Earlier versions of the above chapter can be downloaded from https://www.wiwi.hu-berlin.de/de/professuren/vwl/wipo/research/matlab_toolkit The main industry method Blanchard-Kahn techniques is still worth a read. Blanchard, O. J. and C. M. Kahn (98), The solution of linear difference models under rational expectations, Econometrica, 48, 35-32. The Blanchard-Kahn techniques are discussed in Chapter 6 of ABCs of RBCs. However, the way it presents might not be so friendly to those who first see it. This lecture note is a simpler version of it and by no means exhausting.. Recap of basic neoclassical growth models and loglinearisation As we discussed before, basic RBC models are equivalent to neoclassical growth models with fluctuations + decisions for leisure/ labour supply. In the last lecture note, we set up basic stochastic neoclassical growth models with constant labour and stationary TFP with shocks and the social planner is optimising the following intertemporal objective function, Subject to VV = max EE tt ββ ss tt uu(cc ss ), cc ss,kk ss+ ss=tt kk ss+ = ( δδ)kk ss + ZZ ss kk ss αα cc ss, ss = tt, tt +, And ln ZZ tt+ = ρρ ln ZZ tt + εε tt+, with mean value of ZZ =. where kk ss is per capita capitals, cc ss is per capita consumption, and ZZ ss is TFP at period ss. The optimisation gives the following market-clearing results with (cc ss ) = cc ss θθ Euler equation: cc θθ tt = ββee tt RR tt+ cc θθ tt+ θθ,

DSGE (3): Stochastic neoclassical growth models (2): how to solve? 27 Production function: yy tt = ZZ tt kk tt αα Budget constraint: kk tt+ = ( δδ)kk tt + yy tt cc tt Real returns: RR tt = ( δδ) + ααzz tt kk tt αα TFP shocks: ln ZZ tt+ = ρρ ln ZZ tt + εε tt+, with mean value of ZZ = With log-linearisation, we have following the first-order linear results for the above nonlinear equations, from the discussions of the last lecture note, Basic neoclassical growth models Nonlinear equations Log-linearised equations around equilibrium Euler equation cc θθ tt = ββee tt RR tt+ cc tt+ = EE tt (cc tt cc tt+) + θθ RR tt+ Production function αα yy tt = ZZ tt kk tt yy tt = zz tt + ααkk tt Feasibility/Budget constraint kk tt+ = ( δδ)kk tt + yy tt cc tt kkkk tt+ = ( δδ)kkkk tt + yyyy tt cccc tt Real returns/rental rate αα RR tt = ( δδ) + ααzz tt kk tt RRRR tt = ααkk αα zz tt + (αα )kk tt TFP shocks ln ZZ tt+ = ρρ ln ZZ tt + εε tt+ zz tt+ = ρρzz tt + εε tt+ This is probably the most troublesome part of the whole short course of DSGE models in terms of the tediousness of linear algebra (OK not really, but anything more tedious than this, we go for numerical solutions by Dynare instead.) As we explained before, there are no closed-form solutions, for most of complicated RBCs/DSGEs models. So we usually solve for steady state and study dynamics around steady states due to shocks/fluctuations from TFP/technology. The above 5 equations does not sound good at the first look. We need some tidying up so that there are only 3 equations for 3 variables for consumption, capitals, a TFP shocks. Steady-state equilibrium As we have discussed before, the steady state have the following relationships, = RR, yy = ββ ZZkkαα, = δδδδ + yy cc, RR = ( δδ) + ααααkk αα, ZZ = Obviously we have = ( δδ) + ααααkkαα ββ to give steady state capital value, output, and consumption kk = ββ +δδ αααα αα ; yy = ZZ ββ +δδ αααα And you can get consumption too from = δδδδ + yy cc. If say ZZ increases, then kk and cc also increase. If we have lower discount rate (higher ββ), then kk and cc both go up the usual story. For most of RBCs and DSGEs models, we would not be so lucky and we need computer codes to obtain the steady-state values for consumption, output, and capitals. However, it is relatively easy to do as they can be obtained/solved by nonlinear systematic equations numerically. Note that as we mentioned before that the neoclassical growth models are usually solved by recursive dynamic programming methods. Here we are solving it along the αα αα 2

DSGE (3): Stochastic neoclassical growth models (2): how to solve? 27 approached used by RBCs and DSGEs methods we have added stochastic part to neoclassical growth models. And by add leisure/labour choices to stochastic neoclassical growth models, then we are golden to have RBCs models. Tidying up the dynamic system Combine Euler equation = EE tt (cc tt cc tt+) + RR θθ tt+ and real returns RRRR tt+ = ααkk αα zz tt+ + (αα )kk tt+, we have = EE tt (cc tt cc tt+) + θθ ββββkkαα zz tt+ + (αα )kk tt+ cc tt = EE tt θθ ββββkkαα (αα )kk tt+ + cc tt+ + θθ ββββkkαα zz tt+ Combine feasibility/budget constraint and production function: kkkk tt+ = ( δδ)kkkk tt + yyyy tt cccc tt and yy tt = zz tt + ααkk tt, kkkk tt+ = ( δδ)kkkk tt + yyzz tt + ααkk tt cccc tt kk tt+ = ( δδ)kk tt + yy kk zz tt + ααkk tt cc kk cc tt kk tt+ = αα yy kk + δδ kk tt + cc kk cc tt + yy kk zz tt To simplify the notations, let s have cc tt = EE tt aa kk tt+ + cc tt+ + aa 2 zz tt+ kk tt+ = aa 3 kk tt + aa 4 cc tt + aa 5 zz tt aa = θθ ββββkkαα (αα ); aa 2 = θθ ββββkkαα ; aa 3 = αα yy + δδ; aa kk 4 = cc ; aa kk 5 = yy. kk And of course we also have zz tt+ = ρρzz tt + εε tt+. 2. Solutions by hands Uhlig style Basic neoclassical growth models Euler equation Production function Feasibility/Budget constraint Real returns/rental rate TFP shocks Log-linearised equations around equilibrium = EE tt (cc tt cc tt+) + θθ RR tt+ yy tt = zz tt + ααkk tt kkkk tt+ = ( δδ)kkkk tt + yyyy tt cccc tt RRRR tt = ααkk αα zz tt + (αα )kk tt zz tt+ = ρρzz tt + εε tt+ cc tt = EE tt aa kk tt+ + cc tt+ + aa 2 zz tt+ kk tt+ = aa 3 kk tt + aa 4 cc tt + aa 5 zz tt Now we guess the solutions have the following forms, kk tt+ = ηη kkkk kk tt + ηη kkkk zz tt cc tt = ηη cccc kk tt + ηη cccc zz tt (Why the solutions have the above forms? Think feasibility constraint and the meaning of Bellman equation/dynamic programming.) And substitute back to kk tt+ = aa 3 kk tt + aa 4 cc tt + αα 5 zz tt, we then have ηη kkkk kk tt + ηη kkkk zz tt = aa 3 kk tt + aa 4 ηη cccc kk tt + ηη cccc zz tt+ + aa 5 zz tt 3

DSGE (3): Stochastic neoclassical growth models (2): how to solve? 27 ηη kkkk kk tt + ηη kkkk zz tt = (aa 3 + aa 4 ηη cccc )kk tt + (aa 5 + aa 4 ηη cccc )zz tt. The equations hold if ηη kkkk = aa 3 + aa 4 ηη cccc ηη kkkk = aa 5 + aa 4 ηη cccc. And substitute back to cc tt = EE tt aa kk tt+ + cc tt+ + aa 2 zz tt+, we then have ηη cccc kk tt + ηη cccc zz tt = EE tt aa ηη kkkk kk tt + ηη kkkk zz tt + ηη cccc kk tt+ + ηη cccc zz tt+ + aa 2 zz tt+ ηη cccc kk tt + ηη cccc zz tt = EE tt aa ηη kkkk kk tt + ηη kkkk zz tt + ηη cccc ηη kkkk kk tt + ηη kkkk zz tt + (ηη cccc + aa 2 )zz tt+ Now using zz tt+ = ρρzz tt + εε tt+, ηη cccc kk tt + ηη cccc zz tt = EE tt aa ηη kkkk kk tt + ηη kkkk zz tt + ηη cccc ηη kkkk kk tt + ηη kkkk zz tt + (ηη cccc + aa 2 )(ρρzz tt + εε tt+ ) And taking (rational) expectations ηη cccc kk tt + ηη cccc zz tt = aa ηη kkkk kk tt + ηη kkkk zz tt + ηη cccc ηη kkkk kk tt + ηη kkkk zz tt + ρρ(ηη cccc + aa 2 )zz tt ηη cccc kk tt + ηη cccc zz tt = (aa ηη kkkk + ηη cccc ηη kkkk )kk tt + aa ηη kkkk + ηη cccc ηη kkkk + ρρ(ηη cccc + aa 2 )zz tt (Dangerous here as it is linear system from first-order approximations and leave no room of direct impact of changes in uncertainty on variables.) The equations hold if we have ηη cccc = (aa + ηη cccc )ηη kkkk ηη cccc = ρρ(ηη cccc + aa 2 ) + (aa + ηη cccc )ηη kkkk. Now we have 4 equations for 4 unknown parameters.. ηη kkkk = aa 3 + aa 4 ηη cccc ; ηη kkkk = aa 5 + aa 4 ηη cccc ; ηη cccc = (aa + ηη cccc )ηη kkkk ; ηη cccc = ρρ(ηη cccc + aa 2 ) + (aa + ηη cccc )ηη kkkk. After further detailed tediousness, we have the solutions kk tt+ = ηη kkkk kk tt + ηη kkkk zz tt ; cc tt = ηη cccc kk tt + ηη cccc zz tt ηη kkkk = aa aa 4 + aa 3 2 aa 2 aa 4 + aa 3 aa 2 3 ηη cccc = ηη kkkk aa 4 aa 3 aa 4 ηη cccc = ρρaa 2 + ηη cccc aa 5 + aa 5 aa ρρ ηη cccc aa 4 aa aa 4 ηη kkkk = aa 5 + aa 4 ηη cccc With aa = θθ ββββkkαα (αα ); aa 2 = θθ ββββkkαα ; aa 3 = αα yy kk + δδ; aa 4 = cc kk ; aa 5 = yy kk. Note that we don t take the root of ηη kkkk > as it will lead to explosive paths (Why? Violating transversality conditions and/or saddle paths of consumption?) With the same input as Uhlig s paper: ββ =.99; δδ =.25; θθ =.; αα =.36, ZZ =, ρρ =.95. Of course even with this approximate first-order solutions, we would need some help from computer for numerical simulations. And aa =.222; aa 2 =.348; aa 3 =.; aa 4 =.725; aa 5 =.975; ηη kkkk =.9653; ηη cccc =.682; ηη cccc =.352; ηη kkkk =.754. And more importantly kk tt+ =.9653kk tt +.754zz tt cc tt =.682kk tt +.352zz tt I tried my best to follow Uhlig s way for the current setting anyway, what do we learn here? Not very practical hand-on solutions as computers would do a better job to obtain the numerical answer. 4

DSGE (3): Stochastic neoclassical growth models (2): how to solve? 27 We can now test the impulse response analysis. Let s say there is one-off positive shocks to TFP for increasing by %, we then have capital % changes as follows, kk tt+ =.9653kk tt +.754zz tt = +.754. =.754% and consumption changes by cc tt =.682kk tt +.352zz tt =.682.754 +.352. =.352% and output and real returns can be computed accordingly yy tt = zz tt + ααkk tt = ; RR tt = ααkkαα zz tt + (αα )kk tt =? RR Figure from Uhlig (999) Note that kk tt, cc tt are all in terms of rates of changes, so is zz tt as ZZ =. After getting kk tt+ and cc tt, we can get go on getting kk tt+2 and cc tt+, and for output and real returns. Then we have usual explanations for the pathways of all variables from basic neoclassical models. It seems nice, but what are TFP shocks? Or more precisely, what are negative TFP (technology) shocks? Uhlig starts use capital endowment notion as kk tt. 3. Solutions by matrix/vectors Blanchard-Kahn techniques The above by-hand solutions are hard to come by actually you can also get such similar solutions for Hansen (985) RBC model in lecture note 5... As the loglinearisation creates a linear system to solve N equations for N unknown variables. We can use matrix/vector to solve and obtain the answers. There are many methods to solve for such linear systems by matrix. Chapter 6 of ABCs of RBCs introduce some of them, and we are not going through all of them. Here we go through Blanchard-Kahn (98) method to solve the above stochastic neoclassical growth model, as it is the industry-standard and easy to understand. So what do we have for the log-linearisation system of stochastic neoclassical growth model? cc tt = EE tt aa kk tt+ + cc tt+ + aa 2 zz tt+ kk tt+ = aa 3 kk tt + aa 4 cc tt + aa 5 zz tt zz tt+ = ρρzz tt + εε tt+ 5

DSGE (3): Stochastic neoclassical growth models (2): how to solve? 27 aa = θθ ββββkkαα (αα ); aa 2 = θθ ββββkkαα ; aa 3 = αα yy kk + δδ; aa 4 = cc kk ; aa 5 = yy kk. The TFP shocks equation is stable as long as < ρρ <. However, there are no guarantee aa. When we obtain the final solutions, there are no guarantees that the system is stable and some economics constraints would need to be imposed. Anyway, we can write the above three equations in matrix/vector form, aa 2 aa zz tt+ zz tt EE tt kk tt+ = aa 5 aa 3 aa 4 kk tt cc tt+ ρρ Or we can write in following form, cc tt EE tt AAxx tt+ = BBxx tt, aa 2 aa zz tt+ zz tt where AA =, BB = aa 5 aa 3 aa 4, xx tt+ = kk tt+, xx tt = kk tt. ρρ cc tt+ cc tt For illustration purpose, let s just use last section s benchmark value, and we have.348.222 zz tt+ zz tt EE tt kk tt+ =.975..725 kk tt, cc tt+.95 cc tt as we know from the last section that aa =.222; aa 2 =.348; aa 3 =.; aa 4 =.725; aa 5 =.975 Obviously there two ways to do it: either move AA to the right-hand side or move BB to the left-hand side. Both ways would give you the same results in the end. So we just pick one, say. EE tt BB AAxx tt+ = EE tt CCxx tt+ = xx tt, and.348.222 CC =.975..725 ρρ.526.348.222 =.78.99.6.526 =.4.996.78.348.222 Now the standard linear algebra method to solve for the above system is via eigenvalues/eigenvectors/eigenmatrix. Say you are not too rusty on them: CCCC = MMΛ CC = MMΛMM and we can write EE tt CCxx tt+ = xx tt as follows EE tt MMΛMM xx tt+ = xx tt, λλ where Λ is the eigenmatrix λλ 2 for CC and MM is corresponding eigenmatrix. λλ 3 We can rewrite EE tt MMΛMM xx tt+ = xx tt as follows, EE tt ΛMM xx tt+ = MM xx tt EE tt Λww tt+ = ww tt, where ww tt+ = MM xx tt+ and ww tt = MM xx tt. For eigenvalue matrix and eigenmatrix are as follows for our input.526 ΛΛ =.36.744.9556 MM =.865.856.894.4787.5259.448 6

DSGE (3): Stochastic neoclassical growth models (2): how to solve? 27 5.734 MM = 5.48.5264.53..35.678.9992 Thus, we have for EE tt Λww tt+ = ww tt as follows,.526 ww tt+ ww tt.36 ww2 tt+ = ww2 tt.9556 ww3 tt+ ww3 tt Discussions: λλ =.526 and λλ 2 =.36 are two stable eigenvalues for the ww and ww2 processes (Why?) and λλ 3 =.9556 is unstable one for ww3. We usually sort the order by stabilities. the implication of xx tt : In this case, unstable one for ww3 and stables one for ww and ww2. By the Blanchard-Kahn condition, the number of unstable eigenvalues is the same as the number of control/choice variables. We have control/choice variable here for the stochastic neoclassical growth model (why only one control/choice variable here?) This means the Blanchard-Kahn condition is satisfied. If studying neoclassical growth models before with phase diagram, we would know for a saddle path, we need one stable root and one unstable root in the phase diagram. If the Blanchard-Kahn condition is not satisfied here, we have problems with transversality conditions and we need more advanced techniques, which are not discussed here What we have here for EE tt Λww tt+ = ww tt :.526 ww tt+ ww tt.36 ww2 tt+ = ww2 tt.9556 ww3 tt+ ww3 tt The unstable/explosive one implies that transversality conditions need to be satisfied say so that ww3 tt+tt = (λλ TT 3 )ww3 tt implies ww3 tt = for every tt. So we have the answers as follows ww tt+ =.526 ww tt =.95ww tt and ww2 tt+ =.36 ww2 tt =.9653ww2 tt Remember that This means that we have ww tt = MM xx tt ww tt 5.734 zz tt ww2 tt = 5.48.5264.53 kk tt.35.678.9992 cc tt This means that.352zz tt +.678kk tt.9992cc tt = and cc tt =.682kk tt +.352zz tt, which is the same as results by Uhlig s method in the previous section We need to find out the process for kk tt+ and zz tt+. 5.734 zz tt+ ww tt+ 5.48.5264.53 kk tt+ = ww2 tt+.35.678.9992 cc tt+.95ww tt =.9653ww2 tt ww tt 5.734 zz tt ww2 tt = 5.48.5264.53 kk tt.35.678.9992 cc tt 7

DSGE (3): Stochastic neoclassical growth models (2): how to solve? 27 And after linear algebraic tidying up, we have kk tt+ =.9653kk tt +.754zz tt zz tt+ =.95zz tt + εε tt+. Details: zz tt+ =.95zz tt + εε tt+ is easy to get. 5.734 zz tt+ ww tt+.95ww tt From 5.48.5264.53 kk tt+ = ww2 tt+ =.9653ww2 tt.35.678.9992 cc tt+ 5.48zz tt+ +.5264kk tt+ +.53cc tt+ =.9653ww2 tt 5.48.9653 zz tt+ +.5264.9653 kk tt+ +.53.9653 cc tt+ = ww2 tt 5.677zz tt+ +.5453kk tt+ +.88cc tt+ = ww2 tt ww tt 5.734 zz tt Also we have the following from ww2 tt = 5.48.5264.53 kk tt,.35.678.9992 cc tt 5.48zz tt +.5264kk tt +.53cc tt = ww2 tt cc tt+ =.682kk tt+ +.352zz tt+ 5.677.95zz tt +.5453kk tt+ +.88cc tt+ = 5.48zz tt +.5264kk tt +.53cc tt.5453kk tt+ +.88cc tt+ =.869zz tt +.5264kk tt +.53cc tt Substituting cc tt+ =.682kk tt+ +.352zz tt+ and cc tt =.682kk tt +.352zz tt 5.677.95zz tt +.5453kk tt+ +.88.682kk tt+ +.352.95zz tt = 5.48zz tt +.5264kk tt +.53.682kk tt +.352zz tt kk tt+ = (.5264 +.53.682)/(.5453 +.88.682)kk tt +.754zz tt kk tt+ =.9653kk tt +.754zz tt which is the same as Uhlig s analytic solutions for kk tt+ and the process for zz tt+ is just as given initially. Formal presentation of the above procedure Of course most of textbooks would do a formal approach, by separating stable/explosive variables, say that EE tt CCxx tt+ = xx tt EE tt CC ssssssssssss tt+ = ssssssssssss tt eeeeeeeeeeeeeeeeee tt+ eeeeeeeeeeeeeeeeee tt And CC = MMΛMM, where ΛΛ = ΛΛ and let NN = MM ΛΛ = NN NN 2. Then we 2 NN 2 NN 22 have, EE tt ΛΛ NN NN 2 ssssssssssss tt+ = NN NN 2 ssssssssssss tt ΛΛ 2 NN 2 NN 22 eeeeeeeeeeeeeeeeee tt+ NN 2 NN 22 eeeeeeeeeeeeeeeeee tt EE tt ΛΛ ssssssssssss tt+ = NN NN 2 ssssssssssss tt, ΛΛ 2 eeeeeeeeeeeeeeeeee tt+ NN 2 NN 22 eexxpppppppppppppp tt where ssssssssssss tt+ eexxxxxxxxxxxxxxxx = NN NN 2 ssssssssssss tt+ and ssssssssssss tt NN tt+ 2 NN 22 eeeeeeeeeeeeeeeeee tt+ eexxxxxxxxxxxxxxxx = tt NN NN 2 ssssssssssss tt. NN 2 NN 22 eeeeeeeeeeeeeeeeee tt by the Blanchard-Kahn condition, after tidying up, the explosive variables gives = NN 2 ssssssssssss tt + NN 22 eeeeeeeeeeeeeeeeee tt eeeeeeeeeeeeeeeeee tt = NN 22 NN 2 ssssssssssss tt. 8

DSGE (3): Stochastic neoclassical growth models (2): how to solve? 27 And for stable variables, ssssssssssss tt = NN ssssssssssss tt + NN 2 eeeeeeeeeeeeeeeeee tt = (NN NN 22 NN 2 )ssssssssssss tt EE tt ΛΛ ssssssssssss tt+ = NN ssssssssssss tt + NN 2 eeeeeeeeeeeeeeeeee tt = (NN NN 22 NN 2 )ssssssssssss tt EE tt ssssssssssss tt+ = Λ (NN ssssssssssss tt + NN 2 eeeeeeeeeeeeeeeeee tt ) = Λ (NN NN 22 NN 2 )ssssssssssss tt And EE tt ssssssssssss tt+ = NN ssssssssssss tt+ + NN 2 eeeeeeeeeeeeeeeeee tt+ = (NN NN 22 NN 2 )ssssssssssss tt+. Combining the above equations gives EE tt ssssssssssss tt+ = (NN NN 22 MM 2 ) Λ (NN NN 22 NN 2 )ssssssssssss tt. Discussions: It is very easy to write Matlab codes or C++ codes to compute above matrix/vector multiplication. You can use Uhlig (999) toolkit or Dynare to solve it without going through the above procedure we will discuss how to use Dynare in the next lecture note. Note that in ABCs of RBCs, it moves matrix AA of EE tt AAxx tt+ = BBxx tt to the right-hand side (pages 28-34, though it is used to RBCs principles are the same) so that EE tt AAxx tt+ = AA BBxx tt = CCxx tt EE tt xx tt+ = CCCCMM xx tt EE tt MM xx tt+ = MM CCCCMM xx tt = ΛMM xx tt And you can follow the similar procedure to obtain the same results. The only differences are that λλ < means stable variables and λλ > implies explosive variables here. If you write a program to do the computations, you can check if you have the following results DIG (for eigenmatrix)=.95.9653.464 M (matrix) =.744 -.865 -.856 -.894 -.4787 -.5259.448 N (inverse of M) = 5.734 -. -. -5.48 -.5264 -.53 -.35 -.678.9992 unstable_for_consumption =.352.682 9

DSGE (3): Stochastic neoclassical growth models (2): how to solve? 27 stable_for_shocks_and_capitals =.95 -..754.9653 As you can see that the process of exogenous TPF shocks reveal itself in the end. Thus, many textbooks/papers don t put it into the matrix/vector. Alternatively, you could just put ALL equations into the matrix/vector without tidying up first and you will still have the same answers numerically, e.g. the approach set up by chapter 6 of ABCs of RBCs. Instead of showing M-file in Matlab or using Uhlig (999) toolkit, we use the easy way out by Dynare in lecture note 4