Notes on Kehoe Perri, Econometrica 2002

Transcription

1 Notes on Kehoe Perr, Econometrca 2002 Jonathan Heathcote November 2nd 2005 There s nothng n these notes that s not n Kehoe Perr NBER Workng Paper 7820 or Kehoe and Perr Econometrca However, I have gathered all the equatons together, and added a few more steps n the computatons n places. 1 Model Planner s objectve " max λ 1 β t π(s c1 (s t,l 1 (s t + λ 2 c2 (s t,l 2 (s t # t=0 s t t=0 s t Resource constrants c (s t +k (s t = F (k (s t 1,A (s t l (s t + (1 δk (s t 1 t, s t Enforcement constrants r=t s r β r t π(s r s t U (c (s r,l (s r V (k (s t 1,s t, t, s t Autarky problem V (k (s t 1,s t =max β r t π(s r s t U(c (s r,l (s r r=t s r subject to c (s r +k (s r F (k (s r 1,A (s r l (s r + (1 δk (s r 1 r, s r 1

2 Lagrangan for planner s problem max λ 1 β t π(s t U c1 (s t,l 1 (s t + λ 2 β t π(s t U c2 (s t,l 2 (s t {c (s t,l (s t,k (s t } t=0 s t t=0 s " t # + β t π(s t μ (s t β r t π(s r s t U (c (s r,l (s r V (k (s t 1,s t t=0 s t r=t s r + β t π(s t γ(s t F (k (s t 1,A (s t l (s t + (1 δk (s t 1 c (s t +k (s t t=0 s t Now note that π(s r =π(s r s t π(s t. Note also that there s a partal summaton formula of Abel whch ponts out that P β t μ t β r t P u(c r = β t M t u(c t t=0 r=t t=0 where M t = M t 1 + μ t,m 1 =0 Lets apply ths to the Lagrangan max + {c (s t,l (s t,k (s t } t=0 s t + resource constrants β t π(s t M (s t U c (s t,l (s t μ (s t V (k (s t 1,s t where M (s t =M (s t 1 +μ (s t M (s 1 =λ KP nstead choose to wrte max β t π(s t M (s t 1 U c (s t,l (s t + μ (s t U c (s t,l (s t V (k (s t 1,s t {c (s t,l (s t,k (s t } t=0 s t + β t π(s t γ(s t F (k (s t 1,A (s t l (s t + (1 δk (s t 1 c (s t +k (s t t=0 s t (we wll use ths to compute frst order condtons 2 Frst order condtons 1. c (s t β t π(s t M (s t 1 U c (s t +β t π(s t μ (s t U c (s t β t π(s t γ(s t =0 2

3 Combnng the FOCs for the two countres, we get U 1c (s t M 1 (s t 1 +μ 1 (s t = U 2c (s t M 2 (s t 1 +μ 2 (s t or U 1c (s t U 2c (s t = M 2(s t 1 +μ 2 (s t M 1 (s t 1 +μ 1 (s t (note ths mples rato of sum of past multplers defnes relatve margnal values - mplcatons wthout partcpaton constrants 2. l (s t β t π(s t M (s t U l (s t +β t π(s t μ (s t U l (s t +β t π(s t γ(s t F l (s t =0 Combnng ths wth the FOC for c (s t gves U l(s t (M (s t +μ (s t F l (s t or U l(s t F l (s t = U c (s t M (s t +μ (s t = U c (s t 3. k (s t β t π(s t γ(s t +β t+1 s t+1 π(s t,s t+1 γ(s t,s t+1 r (s t,s t+1 μ (s t,s t+1 V k (k (s t, s t,s t+1 =0 U c (s t M (s t 1 +μ (s t = β s t+1 π(s t,s t+1 π(s t Uc (s t+1 M (s t +μ (s t+1 r (s t+1 μ (s t,s t+1 V k (k (s t, s t,s t+1 U c (s t = β π(s t+1 s t U c (s t+1 M (s t+1 s t+1 V k (k (s t, (s t,s t+1 M (s t r (s t,s t+1 μ (s t,s t+1 M (s t = β π(s t+1 s t U c (s t+1 r (s t+1 + μ (s t+1 Uc M s (s t (s t+1 r (s t+1 V k (k (s t, (s t,s t+1 t+1 where r (s t+1 =F k (s t+1 +(1 δ From the autarky problem, V k (k (s t,s t+1 =Uc Aut (s t+1 r Aut (s t+1 Comment: If no constrants are bndng, the captal accumulaton FOC s standard. Otherwse thngs look dfferent. If you thnk agent 1 s constrant 3

4 mght bnd tomorrow (μ 1 (s t+1 > 0 then the reason must be that you thnk autarky consumpton wll be hgh, so V 1k (k 1 (s t,s t+1 wll be low - lower than U c (s t+1 r (s t+1 - so the RHS of the nter-temporal FOC s larger that n the absence of the enforcement constrant. One mght thnk that ths would suggest the planner should ncrease k 1 (s t, to reduce r (s t+1. However, ncreasng k 1 (s t drves V 1k (k 1 (s t,s t+1 further down, makng the RHS even larger. Thus the planner actually reduces k 1 (s t. Ths s (roughly the ntuton for why n ths economy, net exports are pro-cyclal: when country 1 has a postve persstent productvty shock - and s therefore tempted to default - the planner reduces k 1 (s t and ncreases k 2 (s t. The reducton s k 1 (s t reduces the value of default n the next perod, and the ncreased k 2 (s t makes potental future transfers from country 2 larger, ncreasng value nsde the contract. 3 New State Varables As n the smpler Kocherlakota economy, Kehoe and Perr then defne some new varables v (s t = μ (s t M (s t 1 v (s t = M (s t μ (s t M (s t = M (s t 1 M (s t z(s t = M 2(s t M 1 (s t Then M (s t = M (s t 1 1 v (s t and thus z(s t = 1 v 1(s t 1 v 2 (s t z(st 1 Now the frst order condtons for consumpton and captal can be rewrtten as and U 1c (s t U 2c (s t = M 2(s t +μ 2 (s t M 1 (s t 1 +μ 1 (s t = z(st = 1 v 1(s t 1 v 2 (s t z(st 1 U c (s t =β s t+1 π(s t+1 s t Uc (s t+1 r (s t+1 1 v (s t+1 v (s t+1 1 v (s t+1 V k(k (s t,s t+1 Kehoe and Perr focus on Markov shocks, so that π(s t s t 1 =π(s t s t 1. Then look for decson rules c (x t,l (x t,k (x t together wth polcy rules for z(x t and v (x t where the state x t = z(s t 1,k 1 (s t 1,k 2 (s t 1,s t. Thus ths s the same state space we used to solve the Kocherlakota model, except that k 1 (s t 1 and k 2 (s t 1 have been added. 4

5 4 Computaton procedure Let x =(z,k 1,k 2,s be the state (note that the notaton s a bt confusng, snce the understandng s that z, k 1 and k 2 are nherted from the prevous perod, and do not depend on the current state s. Defne also value functons W (x =U(c (x,l (x + β s 0 π(s 0 sw (x 0 Defne a grd on the state space (.e a three dmensonal grd over the three contnuous varables z, k 1 and k 2 for each possble (dscrete value for s. Guess c 0 (x,l0 (x,k00 (x,z00 (x,v 0(x,W0 (xª x. One good ntal guess s the soluton to the plannng problem wthout enforcement constrants. Note that these allocatons correspond to the allocatons that would emerge n a decentralzed economy wth a complete set of asset markets and no enforcement problems. Now procede to update the guess. Suppose, we are on the n th teraton n updatng the vector of unknown functons, and suppose we are at pont q n the grd. Frst assume nether enforcement constrant bnds. Thus mmedately v (q = 0 and z 0 (q =z. Compute (c (q,l (q,k 0 (q (6 numbers that satsfy 1. The consumpton lesure FOC for both agents (2 equatons U l (q F l (q = U c(q 2. The expresson for z (1 equaton U 1c (q U 2c (q = z0 (q 3. The world resource constrant (1 equaton [c (q+k(q] 0 = [F (k (q,a (ql (q + (1 δk (q] 4. The FOCs for captal (2 equatons U c (c (q,l (q = β π(s 0 s s0 U c (c n 1 (x 0,l n 1 (x 0 F k(k 0(q,A(s0 l n 1 (x δ (x 0 Note that, to evaluate c,l and v n the next perod, we use the old guesses for the decson rules polcy functons (superscrpt n 1. The unknowns n ths equaton are c (q, l (q and k 0 (q. Note, however, that although we use the old functons for next perod varables, we do evaluate them at the correct pont: x 0 =(z 0 (q,k1 0 (q,k0 2 (q,s0. vn 1 (x 0 (x 0 V k(k 0 (q,s 0 5

6 Thus at ths pont we solve a system of 6 non-lnear equatons n 6 unknowns. Then we check whether ether of the two enforcement constrants s volated. If nether s, we are done wth ths grd pont. Alternatvely suppose one of the enforcement constrants s volated (say the one for country 1. Then we solve a dfferent system of equatons. Now we look for (c (q,l (q,k 0(q,v 1(q,z 0 (q (8 numbers that satsfy 1. The consumpton lesure FOC for both agents (2 equatons U l (q F l (q = U c(q 2. The expresson for z (1 equaton U 1c (q U 2c (q = z0 (q 3. The law of moton for z (1 equaton z0(q = 1 v 1(q z 1 4. The world resource constrant (1 equaton [c (q+k(q] 0 = [F (k (q,a (ql (q + (1 δk (q] 5. The FOCs for captal (2 equatons U c (c (q,l (q = β π(s0 s U c (c n 1 (x 0,l n 1 s0 (x 0 F k(k 0(q,A(s0 l n 1 (x δ (x 0 6. The enforcment constrant for country 1 s an equalty (1 equaton vn 1 (x 0 (x 0 V k(k 0 (q,s 0 U c (c (q,l (q + β s 0 π(s 0 sw n 1 (x 0 =V (k,s Note that n equlbrum, f the planner s dong ts job, t wll not be the case the both enforcement constrants bnd smultaneously. Once decson rules have been updated at one pont n the grd, we move to the next pont, and contnue untl decson rules have been updated. Now (and only now we can update the value functons, for each x. For example, for x = q W n (q =U(c n (q,l n (q + β s 0 π(s 0 sw n 1 (x 0 (q If the new value functons are cloes to the old for every x, we are done. Otherwse, we need to repeat the entre process. It s mportant the the ntal guess for the value functons s unformly greater than or equal to the true value of the value functon. 6