# Graduate Macro Theory II: Notes on Neoclassical Growth Model

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1 Graduae Macro Theory II: Noes on Neoclassical Growh Model Eric Sims Universiy of Nore Dame Spring Basic Neoclassical Growh Model The economy is populaed by a large number of infiniely lived agens. These agens consume, save in physical capial, and supply one uni of labor each period inelasically. Time runs from = 0,...,. The populaion a each poin in ime is N = 1 + n) N 0, wih N 0 given. Households ge flow uiliy from consumpion, C, by an increasing and concave funcion, uc ). They discoun he fuure by b < 1. Firms produce oupu using capial and labor, where hese inpus are urned ino oupus hrough a consan reurns o scale producion funcion. The firms are owned by he households. There are wo oher exogenous inpus o producion: Z, which is called labor augmening echnology; and A, which is neural echnological progress. Assume ha Z = 1 + z) Z 0, Z 0 given. Suppose ha A follows some saionary sochasic process. The producion funcion is: Y = A fk, Z N ) 1) Oupu mus be eiher consumed or used as invesmen in new capial goods. aggregae accouning ideniy is: Hence he Y = C + I 2) Capial accumulaes according o he following equaion, wih 0 < δ < 1 he depreciaion rae on capial. There is an assumed one period delay beween when new capial is accumulaed and when i is producive: Equaions 1)-3) can be combined: K +1 = I + 1 δ)k 3) K +1 = A F K, Z N ) C + 1 δ)k 4) 1

2 The equilibrium of he economy can be expressed as he soluion o a social planner s problem: max C,K +1 b E uc ) =0 s.. K +1 = A F K, Z N ) C + 1 δ)k There are wo sources of non-saionariy in he model populaion growh, n; and echnological progress, z. We need o make he variables in he model saionary. To do his, define lower-case variables as heir large counerpars divided by Z N. Hence: y C Z N Y Z N K Z N Divide boh sides of he accouning ideniy by Z N : K +1 = A F K, Z N ) C + 1 δ) K Z N Z N Z N Z N Because F K, Z N ) has consan reurns o scale, we can wrie Define f ) = F, 1). Then we can wrie he consrain as: AF K,ZN) Z N = A F K Z N, ZN Z N ). K +1 Z N = A f ) + 1 δ) To ge he lef hand side in erms of he derended variables, we need o muliply and divide by Z +1 N +1 : K +1 Z +1 N +1 Z +1 N +1 Z N = A f ) + 1 δ) Since Z +1 = 1 + z)z and N +1 = 1 + n)n, his can simplify o: 2

3 γ+1 = A f ) + 1 δ) γ = 1 + z)1 + n) Assume ha wihin-period uiliy akes he following funcional form: uc ) = c1 σ 1 σ, wih σ > 0. Normalizing Z 0 = N 0 = 1, we can wrie: uc ) = 1 + z)1 + n) 1 σ) c 1 σ 1 σ = γ1 σ) c 1 σ 1 σ. Hence, we can re-wrie he planner s problem in erms of derended variables as, defining β = bγ 1 σ : max,+1 =0 β c 1 σ E 1 σ s.. γ+1 = A f ) + 1 δ) 2 Firs Order Condiions We can find he firs order condiions necessary for an inerior soluion a couple of differen ways eiher he mehod of Lagrange mulipliers or by expressing he problem as a dynamic program. I ll begin wih he Lagrangian formulaion. I wrie Lagrangians as curren value Lagrangians, which means ha he discoun facor muliplies he consrains. An alernaive formulaion is o use a presen value Lagrangian, in which case he discoun facor does no muliply he consrains. The Lagrangian is: L = =0 The firs order condiions are: c β 1 σ ) E 0 1 σ + λ A f ) + 1 δ) γ+1 ) L = 0 c σ = λ 5) L = 0 γλ = βe λ +1 A+1 f +1 ) + 1 δ) ) +1 6) L = 0 γ+1 = A f ) + 1 δ) λ 7) This Lagrangian formulaion is called curren value because he Lagrange muliplier is he marginal uiliy of consumpion a ime. A presen value Lagrangian would have he muliplier equal o he marginal uiliy of consumpion a ime discouned baco he beginning of ime. 3

4 FOC 5) and 6) can be combined o yield: γc σ = βe c σ +1 A+1 f +1 ) + 1 δ) ) 8) The ransversaliy condiion says ha he presen discouned value of he capial sock a he end of ime is zero: lim E 0β λ +1 = 0 9) An alernaive way o find he firs order condiions is by seing he problem up as a dynamic programming problem. The sae variable here is A is also a sae, bu I m going o ignore i for now, which is fine since i is exogenous). The choice is. We can express he problem recursively as: V ) = max c 1 σ 1 σ + βv +1) s.. γ+1 = A f ) + 1 δ) We can simplify his by imposing ha he consrain holds and eliminaing : V ) = max +1 The firs order condiion is: A f ) + 1 δ) γ+1 ) 1 σ 1 σ + βv +1 ) γc σ = βv +1 ) 10) Wha is V +1 )? Assuming ha +1 has been chosen opimally, he dynamic programming problem can be wrien: V ) = A f ) + 1 δ) γ+1 ) 1 σ 1 σ + βv +1 ) The derivaive wih respec o he argumen,, is: 4

5 V ) = c σ A f ) + 1 δ) ) To ge his we ignore erms involving d+1. This is an applicaion of he envelope heorem, which says ha, if +1 is chosen opimally, he opimal choice isn going o change for a small change in. This resuls from he fac ha funcions are fla near he opimum. Evaluaing his derivaive a +1, we ge: V +1 ) = c σ +1 A+1 f +1 ) + 1 δ) ) Plugging his back ino he firs order condiion, 10), we ge: γc σ = βe c σ +1 A+1 f +1 ) + 1 δ) ) 11) This is idenical o 8). 3 The Seady Sae The soluion of he model is a policy funcion given he saes, wha is he opimal choice of he conrol. Excep for very special cases in paricular δ = 1), here is analyical soluion for his model. We can analyically characerize he soluion for a special case in which he variables of he model are consan, however. This is called he seady sae. Assume ha he producion funcion akes he following form: Y = A K α Z N ) 1 α 0 < α < 1 12) I follows ha he per effecive worker producion funcion is hen: y = A k α 13) The non-sochasic seady sae is defined as a siuaion in which all variables are consan and where he only source of uncerainy which in his case is A ) is held consan a is uncondiional mean. In paricular, his requires ha +1 = and +1 =. Denoe hese values k and y, respecively. Le A denoe he seady sae value of A, which is equal o is uncondiional mean. We can analyically solve for he seady sae capial sock from equaion 8): 5

6 γc σ = βc σ αa k α δ) ) γ β 1 δ) = αa k α 1 k = αa γ/β 1 δ) ) 1 1 α Assume ha γ/β > 1 δ) for his o have a well-defined seady sae. From he accumulaion equaion, we know ha: 14) A k α = c + δk Hence we can solve for seady sae consumpion as: c = A αa γ/β 1 δ) ) α 1 α δ αa γ/β 1 δ) ) 1 1 α We can consider a couple of simple comparaive saics. If A increases, hen k and c also increase. If b goes up households become more paien), hen k and c boh go up as well. 15) 4 The Phase Diagram We can qualiaively characerize he full soluion o he model hrough a phase diagram. In a wo dimensional world, he phase diagram ypically pus he endogenous sae variable on he horizonal axis in his case ) and he jump variable on he verical axis in his case ). Phase diagrams are more naural in coninuous ime; we will proceed in discree ime wih one sligh abuse of noaion. We wan o find he ses of poins where he sae capial) and jump variables are no changing in, ) space. Call hese wo ses of poins he +1 = 1 isocline and he +1 = 1 isocline. There is only one value of +1 consisen wih +1 = 1, which is he seady sae capial sock if A is a is mean. We can see his from he firs order condiion. +1 = 1 : +1 = αa +1 γ/β 1 δ) ) 1 1 α 16) The +1 = 1 isocline is found by looking a he capial accumulaion equaion: +1 = A k α + 1 δ) +1 = = A k α δ 6

7 Hence he isocline is defined by: +1 = 1 : = A k α δ 17) A complicaion arises because shows up in he +1 = 1 isocline, whereas he +1 = 1 isocline depends on +1. This problem would no be presen in coninuous ime, which is why coninuous ime is more naural for phase diagram. I m going o simply circumven his issue by assuming ha +1, and will rea he +1 in he +1 = 1 isocline as. Given his simplifying assumpion, we can graph each of hese lines in a plane wih on he verical axis and on he horizonal axis. The +1 assuming ha A is a is seady sae). The +1 see ha is slope is dc d = αa k α 1 is posiive because αa k α 1 will be negaive and will approach δ because αa k α 1 ) 1 = 1 isocline is a verical line a = k = 1 isocline is a bi more complicaed. We can δ. When is small i.e. near he origin), hen his slope will be large. When is large far away from he origin), he slope will go o zero). The peak occurs where = αa 1 α δ, which is greaer han k when evaluaed a he seady sae value of A. The acual seady sae is where he wo isoclines cross. The above picure shows he isoclines and he seady sae. I also shows he i) saddle pah and ii) some unsable dynamic lines. These dynamics can be derived as follows. Below he +1 = 1 isocline, is oo small, and hence +1 >, and we draw arrows poining he righ, denoing he direcion in which will be expeced o ravel. Above he +1 = 1 isocline, is oo big, and +1 <, and hence we draw arrows poining lef. To ge he dynamic arrows relaive o he +1 = 1 isocline, we have o engage in a sligh abuse of erminology. Technically wha governs he evoluion of is where +1 is relaive o k, bu wha shows up in he diagram is. Le s ignore his disincion and rea +1. To he righ of he +1 means ha αa +1 k α 1 +1 he righ of he +1 = 1 isocline, is oo big ; his will be small, and consumpion will be expeced o decline. Hence, o = 1 isocline we draw arrows poining down, showing he expeced direcion of 7

9 consumpion iniially jumping up, which is wha will happen under plausible parameerizaions of σ. Thereafer he sysem mus ride he new dynamics and approach he new seady sae. Since +1 >, invesmen increases on impac, so ha consumpion does no increase as much as oupu. Below he phase diagram I show he impulse responses, which race ou he dynamic responses of consumpion and he capial soco he shock. Again, noe ha consumpion jumps on impac, whereas he capial sock does no, and from hereafer hey ride he dynamics o a seady sae. Now consider an unexpeced bu emporary increase in A. Here he new isoclines shif in he same way hey do in he case of a permanen shif in A. I show he new isoclines as dashed lines, and also show he new saddle pah as a dashed line. Here, however, consumpion canno jump all he way o he new saddle pah as i did in he case of he permanen change in A. Consumpion mus jump and ride he unsable new sysem dynamics unil ime T, when A goes baco is original value, a which poin in ime i mus be back on he original saddle pah. From here on i mus follow he original saddle pah back ino he original seady sae. 9

10 The figure above shows he ime pah of c wih a scrachy line. I jumps up, rides he unsable dynamics of he new sysem, and his he original saddle pah a exacly ime T. Imporanly, we see from he picure ha i jumps less han i would if he change in A were permanen if i were permanen i would jump all he way o he new saddle pah). Since c jumps by less on impac bu he change in oupu is he same as if he shock were permanen, invesmen mus jump by more. We can back ou he ime pah of invesmen from he impulse response for he capial sock, shown in he figure below he phase diagram. These resuls are consisen wih he inuiion from he permanen income hypohesis. This exercise urns ou o provide imporan insigh ino more complicaed problems. In hese dynamic sysems you can almos never find he analyical soluion for a general case, hough you can do so for he i) seady sae and ii) wha he soluion would look like if he jump variable didn jump a all. For he case of emporary bu persisen exogenous shocks, he full soluion is somewhere in beween hose wo exreme cases beween i) and ii)). If he shock is no very persisen, a good approximaion o he soluion would have he jump variable no changing a all. 10

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