Solutions to Problem Set 1

Transcription

1 Eon602: Maro Theory Eonomis, HKU Instrutor: Dr. Yulei Luo September 208 Solutions to Problem Set. [0 points] Consider the following lifetime optimal onsumption-saving problem: v (a 0 ) max f;a t+ g t t () subjet to a t+ R (a t ) ; t 0; ; (2) a 0 a (0) ; a T + 0; (3) where is the onsumer s rate of time preferene ( ), is the inverse of the elastiity of intertemporal substitution, R + r is the gross interest rate, and given initial level of asset holdings (a 0 ). (a) Use the ow budget onstraints, (2), and the no Ponzi game ondition ( npg), lim t a t+ R t+ 0, to derive the intertemporal (lifetime) budget onstraint. Solution: (2) implies that a R (a 0 0 ) ; a 2 R (a ) ; a T + R (a T T ) : Combining all equations together and eliminating a ; a 2 ; ; a T gives a T + R T + + T R T + + R + 0 a 0 ) TX R t a 0 (4) where we use the fat that lim T a T + R T + 0: (b) Use optimal ontrol (the Lagrange multiplier method) to derive the onsumption Euler equation that links onsumption in two onseutive periods, t and t + ; and then ombine it with the intertemporal budget onstraint to nd optimal onsumption ( ) as a funtion of asset holding a t and model parameters (R; ; ). What is the onsumption funtion when R?

2 Solution: The Lagrangian is L X t u( ) + a 0 R t where is the onstant Lagrangian multiplier for the lifetime budget onstraint (4). The FOCs for an optimum are then t u 0 ( ) ; where t 0; ; : (5) Rt Sine is a onstant, the above FOCs implies that the Euler equations are t Sine u( ), we have whih means that u 0 ( ) Ru 0 (+ ); where t 0; ; : R ; (6) + + (R) : Combining it with the lifetime budget onstraint (4) gives (R) t R t 0 a 0 ) where we have to impose the restrition that, (R) R t an be reovered as follows (R) t 0 a 0 ) R 0 ( ) a 0 ; (7) (R) t 0 Similarly, we an determine the optimal time path of asset holdings: 2 (0; ), and onsumption in the periods (R) t ( ) a0 ; 8t : (8) a t+ Ra t (R) t ( ) a0 : (9) Note that sine a R (a 0 0 ) and 0 ( ) a 0, a 0 an be rewritten as a funtion of a and thus an be expressed as a funtion of a. Following the same logi, we an always write as a funtion of a t whih is just the onsumption funtion: ( ) a t : (0) In the speial ase in whih R, we obtain ( ) a t ; () a t+ Ra t ( ) a 0 : (2) 2

3 () Given the same restrition R, use dynami programming (the Bellman equation) to solve for the onsumption funtion for the same optimization problem, (). Solution: The Bellman equation for this speial ase ( J (a t ) max t + J (a t+ ) ) ; (3) where a t+ R (a t ). Substituting the onstraint into (3) gives ( ) J (a t ) max t + J (R (a t )) : (4) The FOC is thus t RJ 0 (R (a t )) 0; (5) whih gives optimal onsumption as (a). Substituting it into (4): Taking di erentiation w.r.t. a gives J (a) (a) + J (R (a (a))) : (6) J 0 (a) (a) 0 (a) + RJ 0 (R (a (a))) 0 (a) ; (a) 0 (a) + J 0 (R (a (a))) 0 (a) if R ; (7) whih means that J 0 (a) u 0 () (a) : (8) Combining (7) with (8), we have u 0 ( ) u 0 (+ ) ) + : (9) Substituting + into the intertemporal budget onstraint, P t R t R t 0 R t a 0 ) a 0, we have 0 R R a 0. (20) Given that 0 ; 8t, we have R R a 0: (2) Similarly, the optimal time path of asset holdings: R a t+ R a t R a 0 ; 8t 0; (22) 3

4 whih means that a t a 0 ; 8t : (23) Therefore, the onsumption funtion at time t is R R a t: (24) Note that we an also solve for the onsumption funtion by guessing and verifying the value funtion, J (a). 2. [8 points] Consider the following onsumption-saving problem with a nonlinear saving tehnology disussed in the leture: max U 0 f;k t+ g t ln ( ) subjet to k t+ k t ; 0; k t 0 given the initial asset holdings k 0 k (0). (a) The Bellman equation for this problem an be written as: n o n V (k) max ln () + V ek ln ; e k or V (k) max e k o k e k + V ek ; where k and e k are urrent and future levels of apital stok, respetively. Guess that V (k) E ln (k) + F and use the guess-and-verify method to solve this Bellman equation. Solution: First we guess that V (k) E + F ln k. The FOC for e k is: Substituting it into the Bellman equation: E + F ln k ln Mathing the oe ients: F k e k + F 0 e e k F k + F k : F + F k + E + F ln F + F k : ln ( ( )) + ln () and E e k k : (b) Assume that V 0 (k) 0 for any k. Here V 0 (k) is the guessed initial input in the iteration proess: n o V j+ (k) max ln k e k + V j ek ; 8k; (25) e k where j 0. Use the value funtion iteration method to analytially iterate the value funtion and solve for the value funtion and the onsumption funtion. 4

5 . Solution: Starting from V 0 (k) 0, the rst iteration is de ned by maximizing the funtional funtion: n o V (k) max ln () + V 0 ek ; e k s:t: k e k: In this ase, given that for all k the value of V 0 (k) is 0, it is optimal to onsume all apital, i.e., k. Substituting this optimal solution into the above Bellman equation yields: V (k) ln (k ) + 0 ln + ln k: Given that V (k) 6 V 0 (k), we know that our proposed V 0 is not the solution to the Bellman equation. Having the result from the rst iteration, we an now iterate again: n o V 2 (k) max ln () + V ek ; e k s:t: k e k: Maximizing this problem gives us the FOC: whih implies e k k e k e k ; + k and + k : Substituting these optimal solutions into the Bellman equation yields V 2 (k) ln + k + ln + ln + k where F + E ln k; E ( + ) ; F ln + + ln + + : It is straightforward to show that V (k) 6 V 2 (k). Similarly, we an iterate again to ompute V 3 (k): The FOC for e k implies that e k n o V 3 (k) max ln k e k + V 2 ek : e k + ()2 + + () 2 k and () 2 k :

6 Substituting these solutions into the Bellman equation yields: V 2 (k) ln k e k + V 2 ek F 2 + E 2 ln k; where E + + () 2 ; F 2 ln + + () () 2 ln + () () 2 + F : Here again we an hek that V 3 (k) 6 V 2 (k). However, we do start seeing a patterb emerging. As we allow the round of iteration, j, to inrease to in nity, we have + + () () j lim E j j : And F j ln + + () () j + ln + + () () j j ln () () j 2 + () () j 2 ln + + () () j 2 + () + + () () j 3 ln + + j 2 () ln + + () () j () () j 2 Note that the rst line in the above equation an be simpli ed as: lim ln j + + () () j + ln + + () () j 2 Xj lim t ln j + + () () j t The remaining part of the equation an be simpli ed as: Xj 2 lim j Xj 2 lim j t t ln ( ( )) : + + () () j 2 + () () j 2 ln + + () () j 2 ln () ln () ( ) ( ) + + j ln 6

7 3. [2 points] Consider the following Ramsey-Cass-Koopmans optimal growth model. First, we assume that the growth rate of population in the model eonomy is exogenously given and is n > 0, i.e., L t+ + n; (26) L t where L 0 is given,. Seond, we assume that the resoure onstraint in the eonomy is K t+ ( ) K t + K t (AL t ) C t ; (27) where K 0 is given, A > 0; 2 (0; ) ; and 2 (0; ) is the depreiation rate. Third, we assume that the benevolent planner hooses sequenes of onsumption and apital in per apita terms to solves the following dynami optimization problem: max t t ; (28) f;k t+ g subjet to the resoure onstraint in per apita terms, where C t L t and k t+ K t+ L t+, where > 0 and 2 (0; ). (a) 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 erene equations system to show that the model eonomy is saddle-point stable in the neighborhood of the steady state. Solution: The resoure onstraint in per apital terms: ( + n) k t+ ( ) k t + A k t : (29) we an set up the Lagrangian funtion as follows ( ) L t t + t [( ) k t + Akt ( + n) k t+ ] The FOCs w.r.t. and k t+ are where e +n onsumption Euler equation t t ; (30) t e + Akt+ t+ ; (3) < is the e etive disount fator. Combining the two equations give the t e + Ak t+ t+ ) + h i e + Ak t+ : (32) In the steady state, + and k t+ k t, and the steady state, k;, satis es e + A k ; (33) k + A k 0: (34) 7

8 From whih, we obtain that k A k + A k > 0; ( ) A ( ) e + > 0; + where e +. Linearizing the dynami system, (32) and (29), around k; gives (+ ) e + A k ( ) + e ( ) A k 2 k t+ k ; k t+ k ( ) + + A k k t k ; (35) whih an be redued to + ( ) + e ( ) A k 2 k t+ k ; k t+ k ( ) + + A k k t k ; (36) by using the fat that e +A k and whih an be written (denote ex t+ x t+ x; x k; ) e ( ) A k 2 e+ 0 e 0 + : 0 e kt+ e kt 0 {z } {z } {z } {z } {z } u v d J Multiplying J on both sides gives 0 e+ + J 0 e 0 e 0 kt+ e kt 0 {z } {z } {z } {z } {z } I u M v d Iu Kv 0; (37) where K J M e ( ) A k e ( ) A k 2 ( ) A k 2 ; M To determine the stability properties of the model, we need to know the eigenvalues (b ; b 2 ) of the 2-by-2 oe ient matrix K. Two important results in linear algebra are that the trae and determinant of K are the sum and produt of the eigenvalues (b ; b 2 ), respetively. So T rae (K) b + b 2 Det (K) b b 2 e > : e ( ) A k 2 + e > 2; 8

9 Hene, the disriminant should be positive beause T rae (K) 2 4Det (K) > e ( ) A k 2 e ( ) A k e e 2 > 0; e beause e ( ) A k 2 > 0. Therefore, both roots are real. Also, beause Det > and T rae > 2; the two roots must individually be positive. We an also judge the magnitudes of the two roots as follows: jbi Kj 0 () p (b) (b b ) (b b 2 ) 0 ) p () ( b ) ( b 2 ) T rae + Det e ( ) A k 2 < 0 This an only be true if one root (say b ) is less than and the other root is greater than : We an then onlude (and on rm the preditions of the PD) that the equilibrium is saddle-point. After obtaining 0 < b < and b 2 > ; we an have the general solution for this system: e k A b t + k 2A 2 b t 2 e kt A b t + A 2b t : (38) 2 where k and k 2 are two onstants determined by the roots (here we ignore their detailed values). Given k 0 ; k 0 k A + A 2 e 0 0 k A + k 2 A 2 Note that b t 2 as t beause b 2 > : To guarantee a onvergent time path to the I.E. (i.e., to kill the explosive path), the endogenous e 0 need to be set in the right way and make A 2 be 0: 0 k k 0 k : (39) Hene, given any k 0 ; we an nd 0 s.t. the eonomy jump to the pair of stable branhes and then move to the saddle point equilibrium. [Optional] The following are about the details about the transitional dynamis of e ; e k t obtained above. Given that the eonomy is haraterized by the following rst-order di erene equation system: e+ e kt+ K e e kt we ; (40) 9

10 where e ( ) A k 2 K ( ) A k 2 ; e and as shown above 0 < b < and b 2 > are two orresponding eigenvalues. Note that given the model parameters and steady state values, we an deompose the 2-by-2 matrix K as follows K QQ ; (4) where Q has the eigenvetors of K as its olumns, and has eigenvetors of K down its diagonal: b 2 0 : (42) 0 b The solution to system (40) an be written e+ b t 2 0 Q e kt+ 0 b t Q e 0 ; (43) where e k 0 is given. Sine b 2 >, we have one explosive eigenvalue and the system will blow-up as t unless the endogenous variable e 0 jumps in the right way so as to kill the explosive dynamis. To see how this works, we an write the system as follows e+ q q 2 b t 2 0 q 22 q 2 e 0 e kt+ det (Q) q 2 q 22 0 b t ; q 2 q q b t 2 q 2 b t q 22 e 0 q 2 det (Q) q 2 b t 2 q 22 b t ; q 2 e 0 + q q b t 2 q 22 e 0 q 2 + q 2 b t q 2 e 0 + q 5 ; det (Q) q 2 b t 2 q 22 e 0 q 2 + q 22 b t q 2 e 0 + q q q 2 where we assume that Q. Sine b t 2 as t, we have to shut these explosive q 2 q 22 paths down by setting e 0 just right (this is our one degree of freedom, it is the only endogenous variable not pinned down yet). In other words, we hoose onsumption so that we are not on the explosive path that violates the transversality ondition (TVC). Apparently, we should set e 0 q 2 q 22 (44) to rule out the unstable dynamis. Hene our omplete solution is 2 3 e+ 4 q 2b t q q 2 2 e q k q 5 ; e kt+ det (Q) q 22 b t q q 2 2 e q k q q q 2 q 2 2 q 22 + q 5 b t det (Q) e q q 22 q 2 k 0 : (45) 2 q 22 + q 0

11 In pratie we an rst ompute the matrix K, then ompute its eigenvalues and eigenvetors to get the matrix Q and pik the stable eigenvalue so that we have a bounded solution. (b) Use the phase diagram to show that the steady state is saddle-point stable. Solution: For the phase diagram analysis, refer to leture notes.