Econ6012: Macro Theory Instructor: Dr. Yulei Luo SEF, HKU October 2014 Suggested Solutions to Problem Set 1 1. [10 points] Consider the following lifetime optimal consumption-saving problem: T 1 X 1 t ct v (a0 ) = max (1) 1 fct ;at+1 g t=0 subject to at+1 = R (at ct ) ; t = 0; a0 = a (0) ; aT +1 ;T (2) 0; (3) where is the consumer’s rate of time preference ( 1), is the inverse of the elasticity of intertemporal substitution, R = 1 + r is the gross interest rate, and given initial level of asset holdings (a0 ). (a) Use the ‡ow budget constraints, (2), and the terminal condition to derive the intertemporal (lifetime) budget constraint. Solution: (2) implies that a1 = R (a0 c0 ) ; a2 = R (a1 c1 ) ; aT +1 = R (aT cT ) : Combining all equations together and eliminating a1 ; a2 ; aT +1 cT + + T +1 R RT + c1 + c0 R T X ct t=0 Rt ; aT gives = a0 =) = a0 where we use the fact that aT +1 = 0 (and then limT !1 (4) aT +1 RT +1 = 0): (b) Use optimal control (the Lagrange multiplier method) to derive the consumption Euler equation that links consumption in two consecutive periods, t and t + 1; and then combine it with the intertemporal budget constraint to …nd optimal consumption (ct ) as a function of asset holding at and model parameters (R; ; T ). What is the consumption function when R = 1? 1 Solution: The Lagrangian is L= T X t u(ct ) + T X ct Rt a0 t=0 t=0 ! where is the constant Lagrangian multiplier for the lifetime budget constraint (4). The FOCs for an optimum are then 1 ; where t = 0; Rt t 0 u (ct ) = Since ; T: (5) is a constant, the above FOCs implies that the Euler equations are u0 (ct ) = Ru0 (ct+1 ); where t = 0; Since u(ct ) = c1t 1 1 ;T 1: , we have 1 ct = R 1 ; where t = 0; ct+1 ;T 1; (6) which means that ct+1 = ( R)1= ct ; where t = 0; ;T 1 Combining it with the lifetime budget constraint (4) gives T ( R)1= X Rt t=0 t c0 = a0 =) t=0 c0 = where , follows ( R)1= R T X 1 !t ! (7) , and consumption in the periods t ct = ( R)1= t c0 = ( R)1= c0 = a0 =) a0 ; T +1 1 ( R)1= R 1 t T +1 1 1 can be recovered as a0 ; 8t 1: (8) Similarly, we can determine the optimal time path of asset holdings: at+1 = Rat ( R)1= 1 t 1 T +1 a0 : (9) Note that since a1 = R (a0 c0 ) and c0 = 1 1 T +1 a0 , a0 can be rewritten as a function of a1 and thus c1 can be expressed as a function of a1 . Following 2 the same logic, we can always write ct as a function of at which is just the consumption function: ! 1 T t X 1 ct = a = at : (10) T t+1 t 1 =0 In the special case in which R = 1, we obtain ct = 1 T +1 1 1 at+1 = Rat ct = a0 ; 8t 1 T t X R =0 (11) a0 ; T +1 ! 1; (12) 1 at : (13) (c) Assume that T = 1. Given the same restriction R = 1, use dynamic programming (the Bellman equation) to solve for the consumption function for the same optimization problem, (1). Solution: The Bellman equation for this special case ( ) c1t 1 + J (at+1 ) ; J (at ) = max ct 1 ct ). Substituting the constraint into (14) gives ( ) c1t 1 J (at ) = max + J (R (at ct )) : ct 1 (14) where at+1 = R (at (15) The FOC is thus RJ 0 (R (at ct ct )) = 0; (16) which gives optimal consumption as c = c (a). Substituting it into (15): J (a) = c (a)1 1 1 + J (R (a c (a))) : (17) Taking di¤erentiation w.r.t. a gives J 0 (a) = c (a) = c (a) c0 (a) + RJ 0 (R (a c0 (a) + J 0 (R (a c (a))) 1 c (a))) 1 c0 (a) ; c0 (a) if R = 1; (18) which means that J 0 (a) = u0 (c) = c (a) 3 : (19) Combining (18) with (19), we have u0 (ct ) = u0 (ct+1 ) =) ct = ct+1 : Substituting ct = ct+1 into the intertemporal budget constraint, we have ! 1 1 X X 1 ct = c0 = a0 =) Rt Rt P1 ct t=0 Rt = a0 , t=0 t=0 c0 = Given that ct = c0 ; 8t (20) R 1 R a0 . (21) 1, we have ct = R 1 R a0 : (22) Similarly, the optimal time path of asset holdings: R at+1 = R at 1 R a0 ; 8t 0; (23) which means that at = a0 ; 8t 1: (24) Therefore, the consumption function at time t is ct = R 1 R at : (25) Note that we can also solve for the consumption function by guessing and verifying the value function, J (a). 2. [8 points] Consider the following continuous-time optimal consumption-saving problem: ! Z T ct1 1 dt; (26) max exp ( t) ct 1 0 subject to a0t = r (at ct ) ; (27) where is the discount rate, r is the instantaneous interest rate, given a0 , and aT 0. (a) Find the second-order di¤erential equation (SODE) governing the evolution of at . Solution: The Hamiltonian for this problem is: 4 ct1 1 H(t; ct ; at ; t ) = exp ( t) + t r(at ct ); 1 where t is the co-state variable. The FOCs w.r.t. ct , at , and t are: Hc = 0 =) exp ( 0 t Ha + H = 0 =) r t t) ct + = 0 =) r(at 0 t tr = 0; (28) = 0; (29) ct ) = a0t ; (30) respectively. The TVC are: T = 0; aT = 0; T aT = 0. Note that T must be positive; otherwise, the Inada condition is not satis…ed because cT goes to 1 when T = 0. Following the procedure discussed in the lecture, we can easily derive the consumption Euler equation: r c0t = : (31) ct Taking derivatives w.r.t. t on both sides of the budget constraint yields a00t = r a0t c0t ; (32) and combining it with the Euler equation and the budget constraint leads to the second-order di¤erential equation in terms of at as follows: a00t + r r( r a0t r) at = 0: (33) The characteristic equation corresponding to this equation is simply: 2 + r r( r and the two eigenvalues are are 1 = of this SODE is at = A exp ( r 1 t) and r) 2 = 0; = r. Hence, the general solution + B exp ( 2 t) ; (34) where the two undetermined coe¢ cients, A and B, can be pinned down using two boundary conditions a(0) = a0 and aT = 0: A + B = a0 , A exp ( 1T ) + B exp ( 2T ) = 0. Solving them yields A= exp ( 2 T ) exp ( 1 T ) exp ( 2T ) a0 ; B = 5 exp ( 1 T ) exp ( 1 T ) exp ( 2T ) a0 : 1. (b) Find optimal consumption as a function of t and model parameters (T , r, ). Solution: Using the expression for at , (34), and the budget constraint, ct = rat a0t , we can easily solve for the consumption function: r ct = where A, B, a0 . 1, 1 1 and r 2 A exp ( 1 t) + 1 2 r B exp ( 2 t) : (35) are determined by model parameters: , , r, T , and 2. [12 points] Consider the following Ramsey-Cass-Koopmans optimal growth model. First, we assume that the growth rate of population in the model economy is exogenously given and is n > 0, i.e., Lt+1 = 1 + n; Lt (36) where L0 is given,. Second, we assume that the resource constraint in the economy is Kt+1 = (1 ) Kt + Kt (ALt )1 Ct ; (37) where K0 is given, A > 0; 2 (0; 1) ; and 2 (0; 1) is the depreciation rate. Third, we assume that the benevolent planner chooses sequences of consumption and capital in per capita terms to solves the following dynamic optimization problem: ! 1 1 X c t t ; (38) max 1 fct ;kt+1 g t=0 subject to the resource constraint in per capita terms, where ct = Ct =Lt and kt+1 = Kt+1 =Lt+1 , where > 0 and 2 (0; 1). (a) Derive the resource constraint in per capita terms. Set up the Lagrangian function and …nd the consumption Euler equation for this model. Solution: The resource constraint in per capital terms: (1 + n) kt+1 = (1 ) k t + A1 kt ct : (39) (b) Find the intertemporal equilibrium (i.e., the steady state) for this model, linearize the model around the steady state, and use the resulting two di¤erence equations system to show that the model economy is saddle-point stable in the neighborhood of the steady state. 6 ct+1 ct Solution: In the steady state, k; c , satis…es e 1 kt+1 kt = 1 and + A1 k k + A1 k 1 = 1, and the steady state, = 1; (40) c = 0: (41) From which, we obtain that where gives c 1 k = 1 A1 c = k + A1 1 e 1=( 1 + e 1) 1 1=(1 A1 + = ) > 0; k > 0; = 1+ . Linearizing the dynamic system, (??) and (??), around k; c c) = e 1 (ct+1 kt+1 = k + A1 (ct 1 k 1 c + A1 c) + 1 c) + c e ( (ct 1 k kt k ; kt+1 k ; 1) A1 ct+1 c = (ct kt+1 = k c) + (ct by using the fact that e1 = 1 x et+1 = xt+1 x; x = k; c) " 1e 1 ( 1) A1 k | 0 {z 2 c 1 k u Iu K = = J " 1 M= 1e ( 1 0 (43) 1e 1) A1 v M M " k ; kt # " # " # #" 1 0 e ct+1 e ct 0 + = : 1 e e 1 kt+1 kt 0 {z }| {z } | {z } }| {z } | u 1 1 k and which can be written (denote Multiplying J 1 on both sides gives " #" " #" # 1 0 1 0 e ct+1 +J 1 1 e 1 0 1 kt+1 | {z }| {z } | {z }| where c #" J I 2 k + A1 c) + 1 +A1 1 1) A1 1) A1 ( k d Kv = 0; 2 c 1 k 2 c 1 ( 7 # 1) A1 1 1 d # " # e ct 0 = e kt 0 {z } | {z } v 1" k 1 1 2 c (44) 0 1 # 2 kt+1 k ; (42) which can be reduced to e ( k ; # To determine the stability properties of the model, we need to know the eigenvalues (b1 ; b2 ) of the 2-by-2 coe¢ cient matrix K. Two important results in linear algebra are that the trace and determinant of K are the sum and product of the eigenvalues (b1 ; b2 ), respectively. So T race (K) = b1 + b2 = 1 Det (K) = b1 b2 = 1 > 1: e 1e 1) A1 ( k 2 c+ 1 > 2; e Hence, the discriminant should be positive because = T race (K)2 > 1 1e ( 4Det (K) = 1 1) A1 k 2 c 1e 1 e 1) A1 ( k 2 c+ 2 > 0; 1 e 2 4 1 e 2 c > 0. Therefore, both roots are real. Also, 1) A1 k because 1 e ( 1 > 1 and T race > 2; the two roots must individually be because Det = positive. We can also judge the magnitudes of the two roots as follows: jbI Kj = 0 () p (b) = (b b1 ) (b p (1) = (1 b1 ) (1 b2 ) = 1 1e ( 1) A1 k = b2 ) = 0 =) T race + Det 2 c<0 This can only be true if one root (say b1 ) is less than 1 and the other root is greater than 1: We can then conclude (and con…rm the predictions of the PD) that the equilibrium is saddle-point. After obtaining 0 < b1 < 1 and b2 > 1; we can have the general solution for this system: " # " # e ct k1 A1 bt1 + k2 A2 bt2 = : (45) e kt A1 bt1 + A2 bt2 where k1 and k2 are two constants determined by the roots (here we ignore their detailed values). Given k0 ; e k0 = k0 k = A1 + A2 e c0 = c0 c = k 1 A1 + k 2 A2 Note that bt2 ! 1 as t ! 1 because b2 > 1: To guarantee a convergent time path to the I.E. (i.e., to kill the explosive path), the endogenous e c0 need to be 8 set in the right way and make A2 be 0: c0 c = k1 k0 k : (46) Hence, given any k0 ; we can …nd c0 s.t. the economy “jump” to the pair of stable branches and then move to the saddle point equilibrium. [Optional ] The following are about the details about the transitional dynamics of e ct ; e kt we obtained above. Given that the economy is characterized by the following …rst-order di¤erence equation system: # # " " e ct e ct+1 ; (47) =K e e kt kt+1 where K= " 1 1e ( 1) A1 k 2 1 c 1) A1 ( 1 e 1 k 2 c # ; and as shown above 0 < b1 < 1 and b2 > 1 are two corresponding eigenvalues. Note that given the model parameters and steady state values, we can decompose the 2-by-2 matrix K as follows K=Q Q 1 ; (48) where Q has the eigenvectors of K as its columns, and K down its diagonal: " # b2 0 = : 0 b1 The solution to system (47) can be written " # " # e ct+1 bt2 0 =Q Q e kt+1 0 bt1 1 " e c0 e k0 # has eigenvectors of (49) ; (50) where e k0 is given. Since b2 > 1, we have one explosive eigenvalue and the system will blow-up as t ! 1 unless the endogenous variable e c0 “jumps” in the right way so as to kill the explosive dynamics. To see how this works, we can write the system as follows " #" #" #" # " # q11 q12 bt2 0 q22 q12 e c0 e ct+1 1 ; = e e det (Q) q21 q22 0 bt1 q21 q11 k0 kt+1 " #" # q11 bt2 q12 bt1 q22 e c0 q12 e k0 1 = ; det (Q) q21 bt2 q22 bt1 q21 e c0 + q11 e k0 2 3 t q e t e e q b c q k + q b q e c + q k 11 2 22 0 12 0 12 1 21 0 11 0 1 4 5; = t t det (Q) q21 b q22 e c0 q12 e k0 + q22 b q21 e c0 + q11 e k0 2 1 9 # q11 q12 . Since bt2 ! 1 as t ! 1, we have to where we assume that Q = q21 q22 shut these explosive paths down by setting e c0 just right (this is our one degree of freedom, it is the only endogenous variable not pinned down yet). In other words, we choose consumption so that we are not on the explosive path that violates the transversality condition (TVC). Apparently, we should set " e c0 = q12 e k0 q22 to rule out the unstable dynamics. Hence our complete solution is 2 3 # " q12 e t e q b q k + q k 12 1 21 q22 0 11 0 e ct+1 1 4 5; = q e t 12 det (Q) q22 b1 kt+1 q21 q22 e k0 + q11 e k0 2 3 q12 + q q q 11 12 21 q22 1 4 5 bt1 e = k0 : q12 det (Q) q22 q21 q22 + q11 (51) (52) In practice we can …rst compute the matrix K, then compute its eigenvalues and eigenvectors to get the matrix Q and pick the stable eigenvalue so that we have a bounded solution. (c) Use the phase diagram to show that the steady state is saddle-point stable. Solution: For the phase diagram analysis, refer to lecture notes. 10

© Copyright 2024