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Economics 202ALecture Outline #3 (version 1.0)Maurice ObstfeldSteady State of the Ramsey-Cass-Koopmans ModelIn the last few lectures we have seen how to set up the Ramsey-Cass-Koopmans Model in discrete time, and with an in…nite horizon. We havereviewed the solution method for (bivariate) di¤erence equation systems andderived an exact solution to the model in a special case. I have not yetdeveloped the main qualitative implications of the model however. Thesecan perhaps be drawn out most conveniently by de…ning the model’s steadystate (or balanced growth allocation) and linearizing the model around thatlong-run destination. As we saw, when f00(k) = 0 as in the last lecture,the model has no steady state in c and k, but in the customary case withf00(k) < 0 a well-de…ned steady state exists. It is instructive to examine itsproperties.Let c andk denote the steady-state values. Then they must satisfy theintertemporal Euler equationu0(c) = 1 + f0k  u0(c);which is equivalent tof0k=1  + : (1)Intuitively, this states that the net marginal product of steady-state capital,f0k , equals the rate of pure time preference.1Steady state values must also ensure thatk =fk+ (1  )k  c1 + n;or, solving for c, thatc = fk (n + )k: (2)1If  = 1=(1 + ); then we call  the rate of pu re time preference. With this notationeq. (1) becomesf0k=  + :1Di¤erentiation of eq. (2) shows that per capita consumption is maximizedwhen the constant capital stock equals k ; wheref0(k ) = n + : (3)This is the “golden rule” point where the marginal product of capital justequals replacement needs. Because the model assumed (1+n) < 1, however,it follows thatn <1  :Because f00(k) < 0, a comparison of (3) with (1) shows thatk < k :Given the consumer optimization assumed in this model, there is no possi-bility of a dynamically ine¢ cient steady state with capital over-accumulation.This is one di¤erence compared to the Solow model.An excellent exercise is to linearize this model in the neighborhood ofc;kand investigate its dynamic properties, in particular showing that thetwo chacteristics roots are, respectively, greater than and less than 1. The…rst question on Problem Set 2 involves an example like that one, so I willnot pursue the linearization here. Instead, I will look in greater detail atthe continuous-time version of this model, using that as a springboard to adiscussion of optimal control theory.The Ramsey-Cass-Koopmans Model in Continuous TimeThe …rst tool we need is compound interest. Please bear with me if thisis familiar ground.Suppose you invest $1 for a year at 3% interest, compounded annually.After a year you will have1 +:0311= $1:03:What if, instead, interest is compounded every six months? In other words,the interest plus principal accrued after six months is reinvested for anothersix months at a 3% annualized rate. In that case, after a year you will have(approximately)1 +:0322 $1:030225:2Clearly you get more money if interest is compounded periodically. and themore frequently it is compounded, the more you earn. For example, withquarterly compounding you end the year holding1 +:0344 $1:03033:Weekly compounding gets you yet more, about $1.03044.The numbers are approaching a limit as the time intervals become pro-gressively …ner. That limit is the (hypothetical) case of continuous com-pounding, which yieldslimh!0(1 + :03h)1=h e:03 $1:03045:With this background, let’s imagine time is measured in intervals of lengthh, so that t = 0; h; 2h; 3h; 4h; etc. The intertemporal objective to be maxi-mized is now1Xt=0(1 + nh)t=h(1 + h)t=hhu(ct);where we multiply u(ct) by h on the theory that the ‡ow of utility fromconsumption at rate ctover a period of length h is proportional to the lengthof the period. In the limit as h ! 0, this objective takes the form of anintegral,1Z0e(n)tu [c(t)] dtMaximization is carried out subject to the constraintkt+h=hf(kt) + kt hkt hct1 + nh:Above, f(kt) is interpreted as the rate of output ‡ow and  as the rate ofdepreciation (per unit time).The Lagrangian for the optimization problem isL =1Xt=01 + nh1 + ht=hfhu(ct) + t[hf(kt) + kt hkt hct (1 + nh)kt+h]g :3First-order conditions are@L@ct=1 + nh1 + ht=hh [u0(ct)  t] = 0;@L@kt+h= 1 + nh1 + ht=h(1+nh)t+1 + nh1 + h(t+h)=ht+h[hf0(kt+h) + 1  h] = 0:These can be simpli…ed to readu0(ct) = t(4)(a familiar condition) andt+h th=t[ +   f0(kt+h)]1 + hf0(kt+h)  h:Going to the limit of continuous time (and assuming that (t) has aright-hand derivative) gives uslimh!0t+h th=_(t) = (t) f +   f0[k(t)]g : (5)Because the accumulation equation can be rewritten askt+h kth=f(kt)  (n + ) kt ct1 + nh;its continuous-time limit is_k(t) = f [k(t)]  c(t)  (n + )k(t): (6)Also necessary is the transversality conditionlimt!1e(n)tu0[c(t)] k(t) = 0:To reduce this all to a di¤erential equation system in c and k, note thatbecause u0(ct) = t,_(t) = u00[c(t)] _c(t):By condition (5), u00[c(t)] _c(t) = u0[c(t)] f +   f0[k(t)]g ; or4_c(t)c(t)= u0[c(t)]c(t)u00[c(t)]ff0[k(t)]  ( + )g :This equation and eq. (6) constitute the desired system in c and k. Tosimplify the notation, I assume that u(c) has the isoelastic form u(c) =c111 1;in which case the last equation becomes_c(t)c(t)=  ff0[k(t)]  ( + )g : (7)In this equation (which is the continuous-time analog of the dynamic Eulerequation), ; as noted earlier in this course, is the intertemporal elasticity ofsubstitution. Together with eq. (6), eq. (7) de…nes the system’s p otentialdynamic paths.The parameter  helps determine the response of consumption to changesin the interest rate (or marginal product of capital). When  is high, marginalutility declines more gently when consumption rises. Thus, for example, iff0[k(t)]   rises above , c(t) will fall sharply, making_c(t)c(t)strongly positive.We will look more closely at the relationship among , saving, and interestrates later in the course.The system consisting of (7) and (6) has a useful phase diagram rep-resentation; I borrow the diagrams from David Romer’s book. The steadystate point in this


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