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Notes on Kehoe Perri

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<str<strong>on</strong>g>Notes</str<strong>on</strong>g> <strong>on</strong> <strong>Kehoe</strong> <strong>Perri</strong>J<strong>on</strong>athan HeathcoteOctober 10, 2002There is nothing in these notes that is not in <strong>Kehoe</strong> <strong>Perri</strong> NBER WorkingPaper 7820 or <strong>Kehoe</strong> and <strong>Perri</strong> Ec<strong>on</strong>ometrica 2002. However, I have gatheredall the equati<strong>on</strong>s together, and added a few more steps in the computati<strong>on</strong>s inplaces.1 ModelPlanner’s objective"X∞ Xmax λ 1 β t π(s t )U ¡ c 1 (s t ),l 1 (s t ) ¢ X∞ X+ λ 2 β t π(s t )U ¡ c 2 (s t ),l 2 (s t ) ¢#t=0 s t t=0 s tResource c<strong>on</strong>straintsX £ci (s t )+k i (s t ) ¤ = X£F (ki (s t−1 ),A i (s t )l i (s t )) + (1 − δ)k i (s t−1 ) ¤ ∀t, s ti=1,2i=1,2Enforcement c<strong>on</strong>straints∞X Xr=ts r β r−t π(s r |s t )U ¡ c i (s r ),l i (s t ) ¢ ≥ V i (k i (s t−1 ),s t ) ∀i, t, s tAutarky problem∞X XV i (k i (s t−1 ),s t )=max β r−t π(s r |s t )U(c i (s r ),l i (s r ))r=t s rsubject toc i (s r )+k i (s r ) ≤ F (k i (s r−1 ),A i (s r )l i (s r )) + (1 − δ)k i (s r−1 )∀r, s r1


Lagrangian for planner’s problem∞Xmax λ X1 β t π(s t )U ¡ c 1 (s t ),l 1 (s t ) ¢ X∞ X+ λ 2 β t π(s t )U ¡ c 2 (s t ),l 2 (s t ) ¢{c i (s t ),l i (s t ),k i (s t )}t=0 s t t=0 s t∞X X+ β t π(s t ) X "X ∞Xµ i (s t ) β r−t π(s r |s t )U ¡ c i (s r ),l i (s t ) ¢ #− V i (k i (s t−1 ),s t )t=0 s ti=1,2 r=t s⎡r ⎤∞X X+ β t π(s t )γ(s t ) ⎣ X £F (ki (s t−1 ),A i (s t )l i (s t )) + (1 − δ)k i (s t−1 ) ¤ − X £ci (s t )+k i (s t ) ¤ ⎦t=0 s t i=1,2i=1,2Now note that π(s r )=π(s r |s t )π(s t ).Note also that there is a ’partial summati<strong>on</strong> formula of Abel’ which pointsout that∞P X∞β t µ t β r−t Pu(c r )= ∞ β t M t u(c t )t=0r=tt=0whereM t = M t−1 + µ t ,M −1 =0Lets apply this to the Lagrangianmax + X∞ X{c i (s t ),l i (s t ),k i (s t )}t=0 s t+ resource c<strong>on</strong>straintsXβ t π(s t ) £ M i (s t )U ¡ c i (s t ),l i (s t ) ¢ − µ i (s t )V i (k i (s t−1 ),s t ) ¤i=1,2whereM i (s t )=M i (s t−1 )+µ i (s t )M i (s −1 )=λ iFor some reas<strong>on</strong> KP instead choose to write choose instead to write∞X X Xmaxβ t π(s t ) £ M i (s t−1 )U ¡ c i (s t ),l i (s t ) ¢ + µ i (s t ) £ U ¡ c i (s t ),l i (s t ) ¢ − V i (k i (s t−1 ),s t ) ¤¤{c i (s t ),l i (s t ),k i (s t )}t=0 s t i=1,2+ resource c<strong>on</strong>straints2 First order c<strong>on</strong>diti<strong>on</strong>s1. c i (s t )β t π(s t )M i (s t−1 )U ic (s t )+β t π(s t )µ i (s t )U ic (s t ) − β t π(s t )γ(s t )=0Combining the FOCs for the two countries, we getU 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 ) ¢2


orU 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 )2. l i (s t )β t π(s t )M i (s t )U il (s t )+β t π(s t )µ i (s t )U il (s t ) − β t π(s t )γ(s t )F ilCombining this with the FOC for c i (s t ) gives3. k i (s t )U il (s t )(M i (s t )+µ i (s t ))F il= U ic (s t ) ¡ M i (s t )+µ i (s t ) ¢or U il(s t )F il= U ic (s t )−β t π(s t )γ(s t )= β X t+1 π(s t+1 |s t )γ(s t+1 )[F ik +(1− δ)] − β X t+1 π(s t+1 |s t )µ i (s t+1 )V ik (k i (s t ),s t+1 )s t+1 s t+1U ic (s t ) ¡ M i (s t−1 )+µ i (s t ) ¢= β X π(s t+1 |s t ) £ U ic (s t ) ¡ M i (s t )+µ i (s t+1 ) ¢ [F ik +(1− δ)] − µ i (s t+1 )V ik (k i (s t ),s t+1 ) ¤s t+1U ic (s t )=β X ·π(s t+1 |s t ) U ic (s t ) M i(s t+1 )M i (s t [F ik +(1− δ)] − µ i(s t+1 ¸))M i (s t ) V ik(k i (s t ),s t+1 )s t+1As in the simpler Kocherlakota ec<strong>on</strong>omy, <strong>Kehoe</strong> and <strong>Perri</strong> then define somenew variablesv i (s t )= µ i(s t )M i (s t )z(s t )= M 2(s t )M 1 (s t )ThenM i (s t )= M i(s t−1 )1 − v i (s t )and thusz(s t )= 1 − v 1(s t )1 − v 2 (s t ) z(st−1 )Now the first order c<strong>on</strong>diti<strong>on</strong>s for c<strong>on</strong>sumpti<strong>on</strong> and capital can be rewrittenasU 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 )3


andU ic (s t )=β X ·π(s t+1 |s t ) U ic (s t 1)1 − v i (s t+1 ) [F ik +(1− δ)] − v i(s t+1 ¸)1 − v i (s t+1 ) V ik(k i (s t ),s t+1 )s t+1<strong>Kehoe</strong> and <strong>Perri</strong> focus <strong>on</strong> Markov shocks, so that π(s t |s t−1 )=π(s t |s t−1 ).Then look for decisi<strong>on</strong> rules c i (x t ),l i (x t ),k i (x t ) together with policy rulesfor z(x t ) and v i (x t ) where the state x t = ¡ z(s t−1 ),k 1 (s t−1 ),k 2 (s t−1 ),s t¢. Thusthis is the same state space we used to solve the Kocherlakota model, exceptthat k 1 (s t−1 ) and k 2 (s t−1 ) have been added.3 Computati<strong>on</strong> procedureLet x =(z,k 1 ,k 2 ,s) be the state (note that the notati<strong>on</strong> is a bit c<strong>on</strong>fusing, sincethe understanding is that z, k 1 and k 2 are inherited from the previous period,and do not depend <strong>on</strong> the current state s).Define also value functi<strong>on</strong>sW i (x) =U(c i (x),l i (x)) + β X s 0π(s 0 |s)W i (x)Define a grid X <strong>on</strong> the state space (i.e a three dimensi<strong>on</strong>al grid over thethree c<strong>on</strong>tinuous variables z, k 1 and k 2 for each possible (discrete) value for s.Guess © c 0 i (x),l0 i (x),k00 i (x),z00 (x),vi 0(x),W0 i (x)ª ∀x ∈ X. One good initialguess is the soluti<strong>on</strong> to the planning problem without enforcement c<strong>on</strong>straints.Note that these allocati<strong>on</strong>s corresp<strong>on</strong>d to the allocati<strong>on</strong>s that would emerge in adecentralized ec<strong>on</strong>omy with a complete set of asset markets and no enforcementproblems.Now procede to update the guess. Suppose, we are <strong>on</strong> the n th iterati<strong>on</strong> inupdating the vector of unknown functi<strong>on</strong>s, and suppose we are at point q in thegrid.First assume neither enforcement c<strong>on</strong>straint binds. Thus immediately v i (q) =0 and z 0 (q) =z.Compute (c i (q),l i (q),ki 0 (q)) (6 numbers) that satisfy1. The c<strong>on</strong>sumpti<strong>on</strong> leisure FOC for both agents (2 equati<strong>on</strong>s)U il (q)F il (q) = U ic(q)2. The expressi<strong>on</strong> for z (1 equati<strong>on</strong>)U 1c (q)U 2c (q) = z0 (q)3. The world resource c<strong>on</strong>straint (1 equati<strong>on</strong>)X[c i (q)+ki(q)] 0 = X[F (k i (q),A i (q)l i (q)) + (1 − δ)k i (q)]i=1,2i=1,24


4. The FOCs for capital (2 equati<strong>on</strong>s)U c (c i (q),l i (q))= β X ·π(s0|s) U c (c n−1i (x 0 ),lin−1s0(x 0 )) F k(ki 0(q),A(s0 )l n−1i(x 0 )) + (1 − δ)1 − v n−1−i(x 0 )Note that, to evaluate c i ,l i and v i in the next period, we use the oldguesses for the decisi<strong>on</strong> rules policy functi<strong>on</strong>s (superscript n − 1). The’unknowns’ in this equati<strong>on</strong> are c i (q), l i (q) and ki 0 (q). Note, however, thatalthough we use the old functi<strong>on</strong>s for next period variables, we do evaluatethem at the correct point: x 0 =(z 0 (q),k1(q),k 0 2(q),s 0 0 ) .Thus at this point we solve a system of 6 n<strong>on</strong>-linear equati<strong>on</strong>s in 6 unknowns.Then we check whether either of the two enforcement c<strong>on</strong>straints is violated.If neither is, we are d<strong>on</strong>e with this grid point.Alternatively suppose <strong>on</strong>e of the enforcement c<strong>on</strong>straints is violated (say the<strong>on</strong>e for country 1). Then we solve a different system of equati<strong>on</strong>s. Now we lookfor (c i (q),l i (q),k 0 i (q),v 1(q),z 0 (q)) (8 numbers) that satisfy1. The c<strong>on</strong>sumpti<strong>on</strong> leisure FOC for both agents (2 equati<strong>on</strong>s)2. The expressi<strong>on</strong> for z (1 equati<strong>on</strong>)U il (q)F il (q) = U ic(q)U 1c (q)U 2c (q) = z0 (q)3. The law of moti<strong>on</strong> for z (1 equati<strong>on</strong>)z0(q) = 1 − v 1(q)z14. The world resource c<strong>on</strong>straint (1 equati<strong>on</strong>)X[c i (q)+ki(q)] 0 = X[F (k i (q),A i (q)l i (q)) + (1 − δ)k i (q)]i=1,2i=1,25. The FOCs for capital (2 equati<strong>on</strong>s)U c (c i (q),l i (q))= β X ·π(s0|s) U c (c n−1i (x 0 ),lin−1s0(x 0 )) F k(ki 0(q),A(s0 )li n−1 (x 0 )) + (1 − δ)1 − vi n−1−(x 0 )6. The enforcment c<strong>on</strong>straint for country 1 is an equality (1 equati<strong>on</strong>)vn−1 i(x 0 ¸)1 − v n−1i(x 0 ) V ik(k0(q),s 0 )vn−1 i (x 0 ¸)1 − vi n−1 (x 0 ) V ik(k0(q),s 0 )U c (c i (q),l i (q)) + β X s 0π(s 0 |s)W n−1i (x 0 )=V i (k i ,s)5


Note that in equilibrium, if the planner is doing its job, it will not be thecase the both enforcement c<strong>on</strong>straints bind simultaneously.Once decisi<strong>on</strong> rules have been updated at <strong>on</strong>e point in the grid, we move tothe next point, and c<strong>on</strong>tinue until decisi<strong>on</strong> rules have been updated. Now (and<strong>on</strong>ly now) we can update the value functi<strong>on</strong>s, for each x ∈ X. For example, forx = qW ni (q) =U(c n i (q),l n i (q)) + β X s 0π(s 0 |s)Win−1 (x 0 (q))If the new value functi<strong>on</strong>s are cloes to the old for every x, we are d<strong>on</strong>e.Otherwise, we need to repeat the entire process.It is important the the initial guess for the value functi<strong>on</strong>s is uniformlygreater than or equal to the true value of the value functi<strong>on</strong>.6

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