Macroeconomics I Problem Set No. 1: Solution. Alessandro Piergallini University of Rome Tor Vergata

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1 Macroeconomics I Problem Set No. 1: Solution Alessandro Piergallini University of Rome Tor Vergata [email protected] November 2007

2 E1 Consider a growth model with a linear production function: Y = F (K ; L ) = AK + BL (1.1) where A > 0 and B > 0 are technological coe cients, Accumulation of physical capital is given by dk dt = K = sy K ; (1.2) where s : saving rate, : rate of physical depreciation, and the labour force (L) grows at a constant rate equal to n 1. Verify which conditions of the neoclassical production function hold and which are violated in Derive the intensive form for the production function (1.1) and the law of motion of capital (1.2) 3. Under which condition there exists a steady state equilibrium for the per capita stock of capital ()? 4. Show graphically the case in which the model leads to a steady state equilibrium and the case in which there is endogenous growth. Solution: 1. The conditions of the neoclassical production function are: constant returns to scale: this property is satis ed; Y = AK + BL A(K) + B (L) = Y (X) positive and diminishing returns to private = A (X); = 0 = B (X); = 0 2 this function doesn t exhibit diminishing returns of capital and labour as the second derivatives are not negative; 1 () ()

3 Inada conditions are violated: lim K!1 () = A 6= 0 (); lim = A 6= +1 K!0@K = B 6= 0 (); lim = B 6= +1 Essentiality is not satis ed as if K = 0 ) Y = BL 6= 0 () if L = 0 ) Y = AK 6= 0 () 2. To derive the intensive form for the production function we divide both sides of 1.1 by L de ning y Y L Y L and K L = AK L + B y = f () = A + B (1.3) deviding both sides of 1.2 by K ; we can derive the rate of growth of capital: K K = s Y K (1.4) to re-write 1.4 in per capita terms, remind that: = K K : L L = K K n (1.5) substituting 1.5 in 1.4 and deviding the numerator and the denominator of Y =K by L we obtain: = sy ( + n) (1.6) 3. A steady state equilibrium for the per capita stock of capital implyes thet the system (1.3 and 1.6) leads to a value of for wich its rate of growth is equal to zero: = 0 2

4 it means that f () s sa + B = ( + n) condition 1.7 can be veri ed only if sa < ( + n) = ( + n) (1.7) 4. The per capita stock of capital converge to a steady state level only if sa < ( + n), graphically: 3

5 When sa > ( + n) savings will be always grater than capital depreciation and will keep growing (it is a sort of endogenous growth), graphically: 1 E2 Consider the following neoclassical model of optimal growth. The lifetime utility function of the representative consumer is of the following form: U = Z 1 where c : per capita consumption at time t, : rate of time preference, and u [] is such that u c [] > 0 and u cc [] < 0. Accumulation of physical capital is given by d dt 0 u [c ] e t dt; (3.1) = = f [] c (n + ) ; (3.2) where : per capita physical capital, n : growth rate of population, : rate of physical depreciation, f []: production function satisfying the inada conditions. 1. Solve the intertemporal optimization problem and interpret the optimality conditions. 4

6 2. Which is the modi ed golden rule? Compare it with the golden rule of the Solow growth model. 3. Derive the phase diagram to describe the dynamic properties of the system. 4. Using the phase diagram, show what happens in the case of a permanent reduction in the rate of time preference and brie y explain your results. Solution 1. The objective function for the social planner (or the representative family) is given by W = Z 1 0 u [c ] N e t dt; (1) where N denotes total population in period t. Because the rate of population growth is n > 0, we have N = N (0) e nt : (2) Setting N (0) = 1, for analytical simplicity, and substituting (2) into (1), the welfare function can be re-written as W = Z 1 0 u [c ] e 0t dt; (3) where 0 = (with 0 > 0). n. We shall impose > n to have a well-de ned optimizing problem To solve the maximizing problem of the representative family we set up the Hamiltonian function H = u [c ] + [f [] c (n + ) ] : (4) The rst-order conditions for optimality are = 0 Rearranging the costate equation (6) yields u 0 [c ] = = 0 f 0 [] + (n + ) ; (6) lim e 0t = 0: (7) t!1 = f(0 + n + ) f 0 []g : (8) Di erentiating both sides of (5) with respect to t, one obtains u 00 [c ] c = (9) 5

7 Dividing both sides by u 0 [c ] (and recalling (5)) we can write (9) as that is where [c ] = equation u 00 [c ] c u 0 [c ] c c = c c = ; (10) [c ] ; (11) u 0 [c]. Substituting (8) into (11) we obtain the Keynes-Ramsey u 00 [c]c c = [c ] ff 0 [] ( 0 + n + )g c: (12) 3. The dynamics of the system are described by (3.2) and (12). The demarcation lines are K = 0! D 1 : c = f [] (n + ) ; (13) c = 0! D 2 : f 0 [] = 0 + n + = + : (14) D 1 is a concave function which has a maximum in correspondence of the golden rule capital stock, (implicitly de ned by the condition f 0 [] = (n + )), while D 2 is a vertical line in correspondence of the modi ed golden rule capital stock, mg (implicitly de ned by the condition f 0 [] = ( + )). Since mg < (this follows from the condition > n), this vertical line de ning the locus c = 0 must lie to the left of. To de ne the stream lines, notice = 1 < 0; = [c ] c f 00 [] < 0: (16) The dynamics of the system are of the saddle path-type. 4. A permanent reduction in the rate of time preference determines a translation of the vertical line D 2 to the right, mg = 1 < 0. f 00 [k] c=0 steady state values of c and k are both higher. For the transversality condition to be satis ed, consumption on impact jumps downwards to the new saddle path. 5. The e ect on of an increase in capital share is straightforward. If in a Cobb-Douglas production function the share of capital increase " also the marginal productivity of capital grows. Looking at the equilibrium condition of the Ramsey model f 0 (k t ) = + a greater marginal productivity implies that the equilibrium level of K t rise as the share of capital does. 6

8 Formally: k = + t If we log-linearize the above equation: ln + ( 1)lnk ln( + ) = 0 Then by total di erentiate it is possible to look at the e ect on k t of a marginal change of 1 + lnkdk + ( 1) 1 k dk = 0 Of course we ignore the term ln( + that is constant. Rearranging the expression 1 dk d = + + lnk (1 ) 1 > 0 k The e ect is positive. 7