Fixed Term Employment Contracts. in an Equilibrium Search Model

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1 Supplemental material for: Fixed Term Employment Contracts in an Equilibrium Search Model Fernando Alvarez University of Chicago and NBER Marcelo Veracierto Federal Reserve Bank of Chicago This document contains 4 appendices: Appendix A: Analysis of the Island Planning Problem. Appendix B: Analysis of the Simplied Island Planning Problem. Appendix C: Proofs. Appendix D: Denition of Auxiliary Competitive Equilibrium ( ACE ). i

2 Appendix A: Analysis of the Island Planning Problem Consider the problem of the planner of an island that receives workers per period and that starts with workers ( ) where is the number of workers with tenure =1 2 Dene as the set of possible workers tenure proles, =[0] 1 + The planners value function : solves à X! X X [ ]( )=max {} =0 + [ ] [ ] (20) subject to =0 =0 =1 ¾ + ( ) ( 0 ) for =1 2 Proposition 5 maps concave functions into concave ones. We use the following notation for subgradients. Let : be a concave function. We use () to denote its subgradient at (if it is clear the value of from the context we simply use ). In our case we use () for =12 (and when it is clear) to denote the projection of () into the subspace of the 0. Abusing notation, we use () (and when it is clear) to denote a generic element of () so that () () The next proposition gives a useful result, ordering the subgradients of Proposition 6 Consider a function satisfying (21) (22) for all and 0 where () Then, [ ] 1 [ ] 2 [ ] 1 [ ] (23) [ ] 1 [ ] + 1 (24) for all and 0 where ³ [ ] 1 [ ] 2 [ ] 1 [ ] [ ]() i

3 Intuitively it follows from the assumption that workers are perfect substitutes and from the fact that the s are increasing in. The next set of results establish that the xed point = [ ] is dierentiable and its derivatives are indeed given by in equation (4). The results in the next three lemmas and two propositions are analogous to standard manipulations of rst order conditions, except for the fact that may not be dierentiable. Let dene the function ˆ () as follows: ˆ : +1 + Ã X! X X ˆ () = + + ( ) ( 0 ) =0 =0 The rst lemma shows a standard saddle-type result for the problem dening [ ] =1 Lemma 7 Let be concave. Fix and let n o [ ]() = max ˆ ()+ ˆ :0 () = argmax n ˆ () :0 o (25) Then ˆ + =( [ ] 0 [ ] 1 [ ] ) [ ]() if and only if is a Lagrange multiplier, i.e. ˆ ( )+ ( ) ˆ ( )+ ( ) (26) ˆ ()+ ( ) for all non-negative, where ˆ =( ) = (), 0 =0and = 0 Notice that since ˆ is concave and the restrictions are linear, () solves problem (25) if and only if there is a saddle ( ) as in equation (26) -see, for example, Analytical Method in Economics, Takayama, Theorem The next lemma shows the Kuhn-Tucker conditions for this problem. Lemma 8 Let be concave. A necessary and sucient condition for = { } =0 to solve arg max ˆ () s.t. 0 n o given,isthatthereexistsa ˆ ˆ ( ) such that ( ) is a saddle where =0 = ˆ (27) Given our previous results we can now write the analogous to the Euler equations. ii

4 Proposition 9 Let be concave. Fix Then, 0 is an optimal choice given if and only if for all { [ ] ()} =0 [ ]() there is a n ˆ o =0 ˆ ( ) such that for every =0 [ ] () = ˆ ( )+ Ã X! ˆ ( ) ( )+ =0 with = if 0 min(+1) ( 0 ) ˆ ( ) 0 0 = ( [ ] () ( )) ( ), where we let = 0 and 0 =0. The next lemma shows that employment is bounded below, and hence marginal productivity is bounded above. Lemma 10 There is an 0 such that for all By this lemma, the solution for bounded. X () 0 =0 in equation (4) is well dened because ³ P =0 is uniformly Proposition 11 Let be the xed point of Assume that 0 Then is dierentiable with respect to when 0 The following proposition and corollaries are important to characterize the solution of the problem and reduce its dimensionality when the separation costs satisfy equations (14)-(15). Proposition 12 Suppose that the separation costs satisfy equations (14)-(15). Let satisfy (21). Then the policies for [ ] satisfy the following. Let =( ) [0] + be feasible given.consider an alternative ³ = such that: i) it is feasible for,ii) 1 X = =0 1 X =0 and = and iii) there is a 0 such that for all 0 1 and that =0for all 0 1 Then is weakly preferred to Corollary 13 The optimal policy can be chosen with the following property: (*) If for some 1 1, then 0 =0for all 0 : 0 1 The following corollary states that when the separation costs satisfy equations (14)-(15).the ergodic set is a subset of E, whichisgivenbyequation(16). iii

5 Corollary 14 If E and 0 is given by the optimal policy 0 =( )=( ) then 0 E The following proposition also applies to the case when the separation costs satisfy equations(14)-(15). It shows that if some permanent workers are red then all temporary workers must have been red. Proposition 15 Suppose that the separation costs satisfy equations (14)-(15). Then, every 6=. =0for Appendix B: Analysis of the Simplied Island Planning Problem Throughout this appendix separation costs are assumed to satisfy equations (14)-(15). The planner s value function :[0 ] + has to satisfy the functional equation : ½ ¾ []( ) =max ( + )+ [ ]+( )[ ]+ ( ) ( 0 ) (28) subject to 0 and 0 where the law of motion is given by 0 =min{ + } and 0 = +max{ + 0} Proposition 16 Consider and such that ( )= ( ) (29) for all ( ) E Then []( )= [ ]( ) (30) for all ( ) E Lemma 17 Assume that satises (21). Consider and ˆ and such that = ˆ 1 + ˆ ˆ 1 and = ˆ (31) for any ˆ E and then ³ [ ]() [ ] ˆ Proposition 18 Let be the function corresponding to as in (29) dened for E Assume that ( ) is concave, and satises (21). Then []( ) is concave in iv

6 Remark 19 The previous proposition is not obvious since the feasible set of the problem dened by the right hand side of [] in (28) is not convex. We now introduce the which is the objective function being maximized in [] The derivatives of are used to dene the functions ˆ and ˆ Denition 20 Given dene ( ) as + ( )= ( + ) ( ) ( +min{ ( 1) } + min { ( 1) } 0 ) ( 0 ) Consider an island planner with no temporary workers ( =0)andagiven The quantity ˆ () is the number of permanent workers that leaves the island s planner indierent between ring one permanent worker and keeping all ˆ () of them. Denition 21 Let be dened as in denition (20). For each dene ˆ () such that 0 (0 ˆ () ) Consider an island planner with 0 ˆ () so it does it not want to re any permanent worker for that The quantity ˆ ( ) is the number of temporary workers that leaves the island s planner indierent between ring one" transitory worker and keeping all ˆ ( ) of them Formally: Denition 22 Let be dened as in denition (20). For each dene ˆ ( ) as follows: (i) if 0 for all ( ), thenˆ ( ) = (ii) if 0 for all (0) then ˆ ( ) =0 (iii) otherwise ˆ ( ) solves 0 ˆ ( ) The remaining of this appendix shows that ˆ ˆ exist, they are unique, and ˆ is decreasing in The proofs are complicated by the fact that is not dierentiable. Proposition 23 Let be given by as in (29). Assume that is concave and satises (21). The function ( ) is strictly concave. Remark 24 It is standard to show that [] is increasing in and if has these properties. Remark 25 Assume that satises (21) and (22). Let be dened as in (29). Denote by [] the subgradient of []( ) when is considered as a function of and AcorollaryofProposition(16)andProposition(6) is that [] [] [] + for all ³ [] [] []( ) v

7 Proposition 26 Fix. Assume that satises (21), (22), and is concave. Dene as in (29). Let ³ [] [] []( ). Then [] Moreover, there exists a () such that for all () and, [] = for any [] [] ( ) Given dene ( ) ( +min{ ( 1) } + min { ( 1) } 0 ) ( 0 ) as a function of and and Let be its subgradient with respect to ( ) Lemma 27 Dene as in (29). Assume that is concave and satises + for all Fix any Let ( ) Then, + Let ( ) be the subgradient of when considered as a function of ( ). Lemma 28 Assume that is concave and satises + for all Fix any. For all ( ): + (1 ) Corollary 29 Let be the optimal choice of employment for Problem (28). If and 0 then =0 If this were not true, i.e. if and 0, then = =0,whichcontradictsLemma28. Lemma 30 Let be given by as in (29), assume that is concave and satises (21). Let be dened as in denition (20). For each there is a unique ˆ satisfying (21). Moreover, 0 ˆ () () + Using the concavity of and strict concavity of we dene ˆ as follows. Lemma 31 Let be given by as in (29), assume that is concave and satises (21). Let be dened as in denition (20). Then for each ( ) 0 ˆ () there exists a unique ˆ that satises (22). Proposition 32 Assume that is concave and satises (21) and (22). Let be given by as in (29). Assume, without loss of generality that is concave in ( ). Then, i) The optimal decision rules of [] are described by the set of Inaction for as ( ) =min ˆ ( ) ª and ( ) =min{ ˆ ()} for all ii) [ ] is concave, and satises (21) and (22). iii) [] and [ ] satisfy (29), and [] is concave. vi

8 Lemma 33 Let be concave, and satisfy (21) and (22). Let be dened as in (29). Let ˆ, ˆ and be dened as in (18), (19), and equation (18), respectively. Then, the subgradients of [] are as follows: If 6= for =1 2 1 then []( ) is dierentiable with respect to If ( ) ( ()): If ( ) ³ () : If ( ): = ˆ ( ) : [] ( ) = ( + )+ [] ( ) =( + )+ ( 0 ) ( 0 ) ( 0 ) ( 0 ) [] ( ) [] ( ) = ( + )+ ( ) Denition 34 We say that ( ) is decreasing in if it satises the following property. If 0 dene h i 0 0 and satisfying 0 0 = ( 0 ) and = ( ) Then, 0 and 0 Notice that if is dierentiable at ( ), thispropertysimplysaysthat ( ) is decreasing in Lemma 35.Let be concave, and satisfy (21) and (22). Let be dened as in (29). Assume that the subgradient of is decreasing in, i.e. itsatises denition (34). Let ˆ ( ) be dened as in (19) for the optimal rule that attains the right hand side of [] Then, the subgradient of [] is decreasing in too, i.e. it satises denition (34) and ˆ ( ) is weakly decreasing in Proposition 36 Let be the xed point of Let ˆ be dened as in denition (18). Then ˆ ( ) is decreasing in. Moreover,ifˆ is not a multiple of then ˆ is strictly decreasing in vii

9 Appendix C: Proofs Proof of Proposition 5 The proof is standard, so we omit it. Proof of Proposition 6 We rst show that (24) holds. Consider two states 0and 0 0 where 0 is obtained from by increasing the number of workers with tenure by and by decreasing the number of workers with tenure 1 by : 0 = for = = 1 and 0 = + It suces to show that there is a feasible policy for 0 that produces a reduction in total payo at most by 1 and thus [ ( 0 )] [ ()] ( 1 ) To establish this, consider two cases, depending on whether in the original plan more than workers with tenure 1 were red or not. Let be a positive number smaller than 1 2 In the case where more than workers with tenure 1 were red, reduce the ring of workers with tenure 1 by and increase the ring of workers with tenure by Then there is a reduction in current payo of ( 1 ),andnochangeinthefuturestate.inthesecond case, let 1 ( [ ( 0 )] [ ()]) 1 ³ h i 1 [ ( ) ] where = for =2 1 1 = 1 = + which is feasible given the stated assumptions. Thus using the properties of directional derivatives and subgradients of concave functions 1 ³ lim ( ) = min ( 1 ) {( 1 )( )} = min ( 1 ) = lim 0 ½ lim 0 ( 1 )( ) min {( 1 )( )} ( 1 ) ( 1 ) ¾ viii

10 where we use theorem 24.4, page 233, of Rockafellar (1997) which shows that the graph of is closed for a concave function on,thehypothesisthat(22)holdsforallsubgradientswith0, andwherewedenote ( 1 )( ) ( ) 1 ( ) Finally since for all subgradients 1 [ ] () [ ] 1 () lim 0 ( [ ( 0 )] [ ()]) then [ ] () [ ] 1 () ( 1 ) The argument to show that (23) holds follows a similar argument, where we let 0 = + and 0 +1 = +1 for =1 1 Proof of Lemma 7. Let be a Lagrange multiplier, then ( )=0 Consider 0,and 0 = ( 0 ), then [ ]() ˆ = ˆ ( ( ) ) ˆ ( ( 0 ) )+ ( ( ) ( 0 )) ˆ ( ( 0 ) )+ ( 0 ) = [ ]( 0 ) ˆ 0 + ( 0 ) thus ˆ + is a subgradient of [ ] Let ˆ + be a subgradient of [ ](). Sinceworkerscanalwaysbesent back and get ˆ,then 0 Also, [ ]()= [ ]( )+ ˆ [ ] for = (). Then,bydenition of subgradient ˆ [ ]= [ ]() [ ]( ) ³ ˆ + ( ) or 0= ˆ () ˆ ( ) ( ) but so ( )=0.Thisequality,togetherwiththedenition of a subgradient, implies that is a Lagrange multiplier. ix

11 Proof of Lemma 8. Let ( ) be a saddle satisfying (27). Then, by Theorem 2.9 in Takayama, is optimal. Let be optimal. Then, by Theorem 2.9 in Takayama there are 0 such that ( ) is a saddle. It remains to show that = ˆ for some subgradient. From the denition of a saddle, ˆ ( )+ ( ) ˆ ()+ ( ) or ˆ ( ) ˆ ()+ ( ) which is the denition of a subgradient. Proof of Proposition 9. Let be optimal. Take any { [ ] ()} =0 alagrangemultiplier,where { [ ] ()} =0 = + ˆ [ ]() By Lemma 7 is By Lemma (8), = ˆ for some subgradient. Then ˆ 0 and Let { [ ] ()} =0 0= ˆ ( )= that the above conditions are satised. Dene as ³ [ ] () ˆ ( ) [ ](), andlet be a subgradient of ˆ evaluated at some 0 such = { [ ] ()} =0 ˆ = { } =0 where the last equality follows from the assumed properties. We will show that ( ) is a saddle. From the above conditions, 0= ( ) Hence, ˆ ( )+ ( ) ˆ ( )+ ( ),forevery 0 Since, by the above conditions, is a subgradient of ˆ evaluated at 0,itfollowsthat ˆ () ˆ ( )+ ( ),forevery Hence, ˆ ( )+ ( ) ˆ ()+ ( ),forevery so that is optimal. If 0 Ã X! ˆ ( ) = + =0 + min{+1} ( 0 ) x

12 follows since ( + )() = ()+() (see Rockafeller, Theorem 23.8), and is dierentiable with derivative. When =0 the subgradient of is any number greater than and hence the previous expression holds with inequality Proof of Lemma 10. By contradiction, for all 0 there is a such that X () Take such that ( ) where = { : } Since 0 = 0 0 () 0 From 9 0=[ [ ] 0 () ][ 0 0 ] =0 thus but so ˆ 0 ( )=0 Since [ ] 0 ()= [ ] 0 ()= ˆ 0 ( )+ à X! 0 = ˆ 0 ( ) = ( 0 ) ( 0 ) 0 Proof of Proposition 11. Let be such that 0 Assume that ª is optimal. Take a subgradient ()= [ ] () First consider the case where =0.Then h [ ] () ˆ ()i =0 and its unique solution is [ ] ()= ˆ () provided that 0 Thus, as an special case, if 0 =0 ˆ (0) is the derivative of Now consider the case where 0 We use the formula in Proposition 9 and replace its value repeatedly, solving it forward until = the rst time that for this cohort, employment is smaller than the number of workers present at the location. Since 0 at each iteration à X! [ ] () = = ( 0 ) xi

13 Notice that in this case, we argue above that = ˆ () Thus, we nd that the unique solution for [ ] () is (). Hencethesubgradientisunique,andthus () is dierentiable. Proof of Proposition 12 Replacing any policy by one with these properties can not decrease output but can decrease the separation cost Proof of Proposition 15. From Proposition 9 and we have that à X! [ ] ()= = ( 0 )= Suppose that 0. Then, à X! [ ] ()= = ( 0 ) Therefore, +1 [ ] () [ ] () = ( 0 ) If = 1 it follows that [ ] ()= [ ] ()=. ButfromLemma7, [ ] (). If 1 it follows that +1 [ ] () + = ( 0 ) where the inequality follows from equations (23) and (24). Therefore, [ ] () (1 ) But from Lemma 7, [ ] (). Proof of Theorem 1. To show this theorem we characterize the competitive equilibrium of a particular decentralization of the economy. Since the rst welfare theorem holds, characterizing this equilibrium gives us a characterization of the ecient allocations. We call this equilibrium auxiliary competitive equilibrium or ACE for short. See Appendix G below for a denition of the ACE. The characterization of a stationary ACE coincides with conditions i) to v) of Theorem 1. We start by providing some of the necessary conditions that an ACE must satisfy. Lemma 37 Let { ( ) ( ) ( ) ( ) ; all } be an ACE. Then, there is sequence { } where is the value of search at for which: xii

14 i) without loss of generality, ( )= ii) X X = = max{ } 0 = [ ] +1 ( 0 ) and iii) for all 0 = 0 0 The proof of this Lemma follows directly from the linearity of the problem of rms type II. This Lemma shows, among other things, that the value to a rm type I of reallocating (ring) a worker does not depend on the characteristic of the island, so that does not depend on ( ) and the value of search is related to the value of selling (assigning) a worker to a dierent island randomly, i.e. in proportion to the number of islands of each type. We will show that the ACE allocation can be obtained by solving a particular dynamic programing problem given two numbers ( ) and by checking two appropriate equilibrium conditions. We develop this characterization in asequenceofresults. The solution of the dynamic programing problem will give the equilibrium quantities chosen by rms type I and the equilibrium prices ( ). Thisproblemhastheinterpretationofthemaximizationproblemsolvedfora coalition of rms type I that are endowed with a ow U = { } =0 of newly arrived workers. These rms also take as given a ow C = { } =0 of consumption, which they use to discount future payos. We refer to this problem as the island planning problem, i.e. the problem of a planner inchargeoftheislandemployment decisionbytenure. The planner chooses how many workers to employ at each tenure level and how many to send back, obtaining for each of them, net of the cost Denition 38 Let : ( 1 ; U C) X = max { =0 + =0 X [ ] =0 0 ( +1 ) 0 ( ) X +1 X [ ] =1 +1 ( ; +1 U C) } ( +1 ) } xiii

15 subject to 0 = =0 1 where U = { ; all 0} + and C = { ; all 0} +. The next Lemma links the island planning problem with the equilibrium quantities chosen by rms type I and the prices { } Lemma 39 Let ( ) ( ) ( ) ( ) ; all ª be an auxiliary competitive equilibrium given initial conditions 1 ( 0 ).Dene for { }, andlet ˆ () be its optimal policy. Then, ( ) ª solves for all the initial conditions i.e. ˆ = for all and = ; U for all where ( ; U ) is and element of the subgradient of ; ª with respect to The proof of this Lemma follows from comparing the island planning problem with the problem of rms type I in a competitive equilibrium, and from the denition of a subgradient. The next Lemma gives the characterization of an ACE. Lemma 40.Letsomearbitraryinitialdistribution ( 0 ) be given. Let also some arbitrary sequence { :all } be given. Let ˆ () be the optimal policy of the island planning problem in condition (38) dened for { } Dene ª as = ˆ n o where has been generated by ˆ and the initial condition i.e. 0 = = ˆ for =1 1 = ˆ ˆ and let ( ; ) be an element of the subgradient of ; ª with respect to i) Dene +1ª as = ; ii) Dene the value of search { } as = 0 +1 X X 0 ( ( 0 ) for all ) +1 xiv

16 iii) Dene { } as = X X X ( 0 ) 0 for all iv) Dene as v) Dene as vi) Suppose that = X X X =0 = 0 ( ) = 1 0 ( ) X ( 0 ) =1 vii) Suppose that the following optimal labor force participation conditions are satised ( ) 0 =max 0 ( + +1 ) 0 ( ) +1 " 0 ( ) # ( ) +1 =0 for all. Then ( ) ( ) ( ) ( ) ; all ª is an auxiliary competitive equilibrium (ACE) given the initial conditions 1 and The proof of this Lemma follows by construction, the denition of competitive equilibrium, and the properties of the island planning problem dened in (38). Since the rst welfare theorem holds for this economy, the characterization of the allocation for an ACE in the previous Lemma applies to the ecient allocations. Now we dene a stationary ACE in terms of the objects used in our previous characterization of the ACE. Denition 41 We say that the ACE { },foraninitialmeasure, isastationary equilibrium if there are constants, and functions, : +1 + =0, : +1 + for which = all = all = all = all = all = all xv

17 and where dening 0 : +1 + as 0 0 () = 0 () = 1 () for =1 1 0 () = ()+ 1 () and letting be an invariant distribution of the joint process () with transition given by ( 0 ), wehave ( ) () = () where () is the invariant distribution of. Finally, since a stationary ACE is a particular type of ACE, then by the previous application of the rst welfare theorem, the stationary version of conditions i) to iv) in Lemma 40 characterizes a stationary ecient allocation. Since the stationary version of conditions i) to vii) in Lemma 40 coincide with conditions i) to v) of Theorem 1, we have nished its proof. Proof of Theorem 2 To prove the theorem it is convenient to dene a sequential economy that corresponds to the island planning problem taking as given. Thiseconomyhasarm whose problem corresponds to the one of the rm with value function in the recursive competitive equilibrium (RCE) and afamilywhoseproblemhassolutionthatgivesthe workers value function in the RCE. I) We dene this economy in a standard Arrow-Debreu sequential way. This denition allows to use the rst and second welfare theorems to link the allocation that solves the sequential island planning problem with an allocation that solves the rms and workers problem in the island economy as well as with the equilibrium wages The commodities for the sequential island economy with initial state ( 0 ) is given by processes for employment by tenure and consumption () = : =0 ª We use to denote the labor choice of the rms in a sequential island problem. We use the and for hiring and ring of workers of tenure. Thenetoutputofrms is to produce the following date history amount of consumption good X =0 X (32) The choices of for the rms are subject to the restrictions that 1 ( 1 )= for =1 the law of motion for workers of dierent tenure 0 =1 = (33) = (34) xvi

18 and the non-negativity constraints for hiring and ring 0, 0 0 for all =1 0 We use to denote the labor choice and the consumption choice of the household in the sequential island problem. This household owns as an endowment a stream of unemployed workers per period, that arrives to the island every period. The household is risk neutral in terms of consumption ( ) Its decision is to assign a worker to work on the island or permanently work outside the island, which gives value per worker, in units of the nal good. The utility function of the household is X X 0 = X =1 The household is subject to the following restrictions: 1 ( 1 )= for =0 and for all non-negative is subject to: 0 (35) 11 1 for = Market clearing for the sequential economy is given by = Ã X =! X =0 =1 for all =0 and all Prices in this sequential island economy are given by intertemporal consumption prices, ( ) and wages by tenure ( ) in terms of date, history consumption goods. Given the household preferences for consumption we impose = 0 With these prices the problem of the rm is to maximize prots, i.e. 0 ( 0 ) = max {} X X =0 ³X X =0 X 0 =1 xvii

19 subject to 1 = and the laws of motion for and Let ( ) ( 0 ) be the multipliers of restrictions (33)-(34). The rst order conditions for the rm s problem are: Ã X! + X ˆ (+1 ) 0 +1 =0 for =0,withequalityif ( ) 0. For =1 Ã X! + X ˆ min{+1}+1 +1 (+1 ) ˆ (36) +1 =0 with equality if ( ) 0 The rst order conditions for ( ) is ˆ 0 with equality if ( ) 0 The rst order conditions for ( ) is ˆ 0 with equality if ( ) 0 Equation (33) thus gives: 11 1 then ˆ = 11 1 then ˆ =0 and equation (34) gives then ˆ = then ˆ =0 Now we turn to the household problem in a sequential island economy. Letting ( 0 )ˆ ( ) be the Lagrange multiplier for (35), the rst order conditions for the household problem are equivalent to =max + X (+1 ) +1 for =0 1 and =max + X (+1 ) +1 where with slackness =ˆ then = xviii

20 and ( ) 0 = + X (+1 ) for =0 1 and analogously for = To see why this is the case, write the Lagrangian of the household problem as X X 0 =0 X + =0 0 1 X =0 +ˆ X + ˆ ˆ =1 The rst order conditions for this problem are as follows. For ( ): +ˆ + X +1 +ˆ (+1 ) 0 with = if ( ) 0 for =0 1 1 and +ˆ + X +1 +ˆ+1 +1 (+1 ) 0 with = if ( ) 0 The slackness conditions are: if ( ) 11 1 then ˆ ( )=0for =0 1, andif ( ) then ˆ ( )=0 To compare a competitive equilibrium with the planning problems it helps dening a sequential island planning problem. In this problem the planner maximizes the expected discounted value of net output (32) plus the value of sending workers outside the island (which gives per worker), subject to the feasibility constraints (33), (34) and (35). This is the sequential version of the recursive island planning problem. Let 0 ( 0 ) be the value attained by this planning problem. Let ( ) ( 0 ) be the multipliers of constraints (33)-(34) and ( ) ( 0 ) the multiplier of the constraints (35). The rst order conditions for the sequential island planning problem are equivalent to the following. For =0: + ³X = max + X (+1 )+ X (+1 ) xix

21 with ( )=0if ( ) and + ³X = + X (+1 )+ X (+1 ) if ( ) 0 For =1 1 we have + + ³X = max + X (+1 ) +1 with ( )=0if ( ) 11 1 and + + ³X = + X (+1 ) +1 if ( ) 0 For = we have + + ³X = max + X (+1 ) +1 with ( )=0if ( ) and + + ³X = + X (+1 ) +1 if ( ) 0 To see why this is the case, write the Lagrangian for the planning problem as: X =0 0 ( 0 )=max {} X =1 X X 0 = X = X = X = ª xx

22 The rst order conditions are the following. For =0: X +1 ³X (+1 )+ X (+1 ) 0 with equality if ( ) 0 For =1 1: X +1 ³X (+1 )+ X (+1 ) 0 with equality if ( ) 0For = : X +1 ³X (+1 )+ X (+1 ) 0 with equality if ( ) 0 The rst order condition for ( ) is 0 with equality if ( )=0 The rst order condition for ( ) is 0 with equality if ( ) 0 (II) We now show i), the rst welfare theorem, and iii). We start with an island RCE { }. Pick an arbitrary state ()=( 0 ) in the support of We construct the sequential competitive equilibrium with ( 0 ) as initial condition as follows. Let wages be: = 1 1 and let multipliers and employment be +ˆ = 1 1 = 1 1 where = ( ) for (37) = and 1 ( 0 ) = xxi

23 It is immediate to verify that { ˆ} solves the rst order conditions for the household problem in a sequential island equilibrium, and hence it solves the household problem. For future reference we dene 0 ( 0 )=ˆ 0 ( 0 )+ Dene the Lagrange multiplier and employment for the rms problem as: It is immediate to verify that ˆ = = 1 1 n ˆ o solves the rst order conditions for the rm s problem in a sequential island competitive equilibrium, and hence it solves the rm s problem. Let 0 be the value of the rm in the sequential island CE. For future reference, from the envelope theorem, we have 0 ( 0 ) = ˆ 0 ( 0 ) By the rst welfare theorem applied to the sequential island economy, {} = {} is a Pareto optimal allocation, and hence solves the sequential island planning problem. By inspection, the Lagrangemultipliers ª that satisfy the rst order conditions for the sequential planning problem are identical to the Lagrange multipliers for the n o rm s problem ˆ and the Lagrange multipliers {ˆ } for the households problem in the sequential competitive equilibrium. From these rst order conditions: 0 ( 0 )=ˆ 0 ( 0 )+ = 0 ( 0 )+ = 0 ( 0 ) for =0and 0 ( 0 )+ 0 ( 0 ) = ˆ 0 ( 0 )+ + ˆ 0 ( 0 ) = 0 ( 0 )+ + 0 ( 0 )= 0 ( 0 ) for =1 The allocation described by is, by hypothesis, recursive, so it solves the recursive island planning problem with initial condition ( 0 ) Repeating this argument for each initial condition ( 0 ) we show that 0 () = () 0 () = () 0 () = () for all () Hence we have shown the rst welfare theorem for the recursive representation of the island problem, and that (7), condition iii) of the theorem, holds. (III). We now show ii), the second welfare theorem, and condition iii) of Theorem 2. We start with a solution of the recursive planning problem, () and () which, by the envelope theorem, satisfy () = + () xxii

24 for =0 and () = + ()+ () for =1. If there were more than one pair for a given select the one that only depends on (). From the principle of optimality, the solution to the recursive island problem is the same as the value function for the sequential island problem 0 so that ()= 0 (). Choose any initial state ( 0 ) to be used as initial condition to the sequential island problem. Dene = 1 1 = 1 1 = 1 1 where 1 is dened as in (37) using the optimal decision rule from the recursive planning problem. By comparing the rst order conditions for the recursive island planing problem with the rst order conditions for the sequential island planning problem, it can be seen that ª so dened solve the rst order conditions for the sequenctial island planning problem. Next dene wages as follows: X = + X (+1 ); (38) +1 =0 for =0;for =1 1 let X = + X (+1 ); (39) +1 =0 nally, for = X = + X (+1 ) (40) +1 =0 The function 0 ( 0 ) is dened as the solution to the rm s problem for wages in the sequential island equilibrium. Given wages the functions are dened as: for =0 = + (41) Dene the candidate multipliers for the sequential rm problem as ˆ = Given wages and multipliers ˆ it is easy to verify that the allocation = and its implied { } solve the rst order conditions for the rms in the sequential island economy. To verify this, one uses the rst order conditions for the island planning problem in the island sequential economy. From the envelope condition it is immediate that 0 ( 0 ) = ˆ 0 ( 0 ) Dene the candidate multipliers for the sequential household problem ˆ =.Givenwages and multipliers ˆ,itiseasytoverifythattheallocation = solves the rst order conditions for the household s problem in xxiii

25 the sequential island economy. To verify this, use the rst order conditions for the island planner problem in the sequential island economy. Thus we have established that the sequential allocation constructed out of the solution of the recursive island planning problem from an initial state ( 0 ) can be decentralized as a sequential island competitive equilibrium. Finally, we dene the elements of the recursive island competitive equilibrium as follows: ( 0 ) = 0 ( 0 ) ( 0 ) = 0 ( 0 ) ( 0 ) = 0 ( 0 ) By repeating this construction for all ( 0 ) in the support of we construct the functions and These functions constitute a RCE since they are constructed from the sequential island competitive equilibrium. From the previous arguments we have: for =0 and () = + ()= + 0 ( ) (42) = +ˆ 0 () = 0 () () = + ()+ ()= + 0 ()+ () (43) for =1,andthusconditioniii)issatised. = +ˆ 0 ()+ ˆ 0 () = 0 ()+ 0 () Proof of Proposition 4 When separation costs satisfy equations (14)-(15), the wages given by equations 38-40) in the proof of Proposition 2 become the following. For =012 2 X = ; =0 for = 1 1 X = + X (+1 ); +1 =0 nally, for = X = + X (+1 ) +1 =0 The multiplier [0] and ( )= if ( ) 0 i.e. if permanent workers are being red. Thus, conditions (a) and (b)the inequalities in (b) follow from these denitions and the properties of We establish condition (c). Since Proposition 2 has shown the rst and second welfare theorems, we can, without loss of generality, start with an ecient allocation and examine the equilibrium value function for workers that xxiv

26 we constructed in equation (41) (in the proof of Proposition 2). Using equations (42), (43) (also in the proof of Proposition 2) we have that () = () for =0 1 () = ()+ () In Proposition (6) we have shown that Thus Finally since is part of an equilibrium, it satises 1 () = 1 ()+ [ ( () 0 ) ] () = ()+ [ ( () 0 ) ] and since we have already established (b), 1 and thus we have 1 Proof of Proposition 16 By Proposition 12 and its corollaries, [] = [ ] in E Proof of Lemma 17 If follows directly from the denition of E and the assumed property (21). Proof of Proposition 18. Take ( 1 1 ) and ( 2 2 ) and consider their convex combination ( )=( 1 + (1 ) 2 1 +(1 ) 2 ) Let the unique corresponding elements in E for ( 1 1 ) and ( 2 2 ) be 1 and 2 Consider = 1 +(1 ) 2,whichisnotnecessarilyonE Let ˆ be the unique element in E that corresponds to Note that ( ) satises Then, = 1 X =1 ˆ and = ˆ []( 1 1 )+(1 ) []( 2 2 ) = [ ]( 1 )+(1 ) [ ]( 2 ) [ ]( ) ³ [ ] ˆ = []( ) where the rst equality follows from Proposition 16, the rst inequality follows from concavity of and Proposition 5,thesecondinequalityfollowsfromLemma17,andthelastequalityfollowsfromProposition16. xxv

27 Proof of Proposition 23. First dene X 1 X ˆ () = ( ) =0 =0 + ( ) ( 0 ) Since and are concave, then ˆ is concave. Now take for =1 2 and consider =( 1 + (1 ) 2 1 +(1 ) 2 ) Let the unique corresponding elements to in [0] + that satises property (*) in Corollary 13 be denoted by for =1 2 Dene = 1 +(1 ) 2 Note that 1 X = =0 and = Dene as the unique element in [0] + that satises property (*) and such that 1 X = =0 1 X =0 and = Then 1 1 +(1) 2 2 = ˆ ³ 1 +(1) ˆ ³ 2 ˆ ˆ ³ = where the rst equality follows from construction of and since by assumption and satisfy (29), the rst inequality follows from the concavity of ˆ the second inequality follows by assumption (21) and Proposition 12 and its corollaries, and the last equality follows from the same argument as the one in Proposition 16. Proof of Proposition 26 Dene the operator as []( ) = max { ( + )+ [ ]+( )[ ]+ 0 0 ¾ + ( + ( ) + ( ) 0 ) ( 0 ) Comparing this problem with (28) the constraints and were removed, hence []( ) []( ) The optimal policies do not depend on and, thusthefunction [] is linear with derivatives [] ( ) = [] ( ) = xxvi

28 for all By concavity of [] []( 0) []( )+ [] (0 ) or []( ) [] + []( 0) where ³ [] [] []( ) Rearranging and using the linearity of [] : [] + []( 0) []( ) []( ) = []( 0)+[ ] for all Thus by monotonicity of [] and [] on : [] ( ) + [](0 0) [](( 1) 0)+[ ] Hence sup [] ( ) + [](00) [](( 1) 0)+[ ] [0(1)] or lim inf [](00) []( ( 1) 0) Ã = 0 lim inf lim sup " [ ] sup [0(1)] sup [] ( ) [0(1)] # [] ( ) On the other hand, for the original problem (28), for ( 0 ), afeasiblepolicyfor 0 is to set 0 = ( 0 ) in which case each additional unit of yields. Hencetherightderivativeof []( 0 ) is greater or equal than Since [] is concave, then [] ( 0 ) for all ( 0 ). Combining the two inequalities, for large enough [] ( ) = for all! Proof of Lemma 27. Consider two cases. First ( 1). Inthiscase ( )= which implies that ( )= ( + 0 ) ( 0 ) thus = = where ( ) ( + 0 ) for the corresponding elements. Second, if ( 1) ( )= ( + ( 1) 0 ) ( 0 ) xxvii

29 thus = = Since, by assumption, + we have shown the required result, except for the case where =( 1) This case follows by continuity, since the graph of the subgradient of a concave function is closed (Rockafellar, 1997, Theorem 24.4, page 233). Proof of Lemma 28. By denition of : = ( ) ( )+ = ( ) + where ( ) Then = + (1 ) where the inequality follows from the previous lemma. Proof of Lemma 30. The existence of ˆ follows by the concavity of with respect to, theinadaconditions on,andfromproposition26,whichshowsthat = for large. Theuniquenessof follows by the strict concavity of.thatˆ follows from concavity of with respect of and Lemma 28. Proof of Lemma 31 The existence and uniqueness of ˆ follows from the strict concavity of Proof of Proposition 32 It follows from the denition of ˆ ˆ and various of the previous results. Proof of Lemma 33. The rst statement follows by considering the case where E so that there is a {1 2 1} and such that 1 =( 00 ) for (0) Thus ( 0 )= (( 1) + ) for all (0) ( 0 )= (( 1) + ) for =1 2 1 The second and third claim follows from the form of the optimal decision rules, i.e. the denition of the range of inaction and the strict concavity of. The third follows since, for ˆ ( ) it is feasible to re any extra temporary workers, so that we know the right derivative of [] with respect to xxviii

30 Proof of Lemma 35. First we establish that ˆ ( ) is decreasing in Then we use this result, to show that [] is decreasing in By denition of ˆ 0 ˆ ( ) for the case where 0 ˆ The main idea is to show that ( ) is decreasing in and then use that, by concavity, ( ) is decreasing in The subgradient is given by ( ) = ( + ) + ( ) where ( ) is given by ( ) = ( +min{ ( 1)} + min { ( 1)} 0 ) ( 0 ) We can then write by cases as ( ) = ( ) = ( + 0 ) ( 0 ) if ( 1) (+ ( 1) 0 ) ( 0 ) for ( 1) and hence its subgradients are ( ) = ( + 0 ) ( 0 ) if ( 1) ( ) = ( 0 ) ( 0 ) ( 0 ) ( 0 ) if = ( 1) ( ) = (+ ( 1) 0 ) ( 0 ) for ( 1) Now we are ready to show that ( ) is strictly decreasing in Consider rst the case where ( 1) In this case the result follows from the hypothesis that is decreasing in and strictly concave. Consider the case where ( 1) In this case the result follows from the concavity of so that is decreasing, and the strict concavity of Finally, for the case where =( 1) we combine the previous two arguments for the right and left derivatives. Having established that ( ) is strictly decreasing in then it follows that ˆ is decreasing in since is decreasing in by concavity of case, The cases where ˆ = or ˆ =0are similar. Now we turn to show that [] is decreasing in We consider three cases. First, let ( ) ( ()) In this [] ( ) = ( + )+ ( ) = ( )+ and thus [] is decreasing in since, as shown above, ( ) is decreasing in In the case where ( ) ³ () then [] =, andhence [] is dierentiable, and its derivative constant, so that it is weakly xxix

31 decreasing in Finally, consider the case where ( ) is such that = ˆ ( ) As shown above, ˆ is weakly decreasing in, thusfor 0 ˆ ( 0 ) Also, the subgradient of [] is [] ( ) [] ( ) = ( + )+ ( ) If ˆ ( 0 )=ˆ ( ) then, using the expression for the left derivative of [] it follows since is concave and, as shown above, is decreasing in If ˆ ( 0 ) ˆ ( ) then, it must be that the point ˆ ( ) 0 is in the interior of the complement of the range of inaction, and thus [] ˆ ( ) 0 = Thus [] has decreased in this case too, since the subgradient has collapsed to its right derivative. Proof of Proposition 36. That ˆ is decreasing in follows from using Lemma 35. Notice that starting with 0 =0and 0 =0satises all the hypothesis of this lemma. Since all these properties are preserved in the limit, they hold for the xed point. To see that ˆ is strictly decreasing, consider rst the case where (1) In this case it follows from the fact that in a neighborhood of those points ( + 0 ) is dierentiable with respect to, satises = ˆ ( )+ + ˆ ( )+ 0 ( 0 ) is decreasing in and is strictly decreasing. A similar argument holds when ( 1) where = ˆ ( )+ + + ˆ ( ) 0 ( 0 ) xxx

32 Appendix D: Denition of Auxiliary Competitive Equilibrium ( ACE ) This appendix denes the auxiliary competitive equilibrium (ACE) used in the proof of Theorem 1. There are two types of rms, type I and II, and families. There are as many markets to buy and sell workers as islands of type ( ) Preferences of the family. The families own all rms of both types. Their preferences are given by X [ ( )+ ] where is consumption of the market good and is the consumption of home goods. Their budget constraint is the following: X [ + ] X 0 ( 0 ) ( 0 ) ( 0 ) 0 where is the Arrow-Debreu price of the consumption good, is the price of home goods, and 0 ( 0 ) are the total prots of rms type I on and island of type ( 0 ). Observethat,atequilibrium, =1 0 ( ). Labor is allocated initially to the two types of rms. Problem of rms type I. There is a continuum of rms type I on each island of type ( 0 ) that buys workers from a central location at price ( ) They start at period =0with a prole of workers given by their type Workers that are bought in this period are given tenure =0 They operate the technology.ifthey sell workerstothecentral location, they obtain a price ( ) If they sell workers with tenure 0at the island level, they loose per worker. The sequence problem for rms type I who buy workers at price,sellthematprice,andpaythe separation costs is the following. For each ( 0 ),theymaximize: 0 ( 0 ) (44) = X X X 0 =0 =1 + X X X X =0 =0 by choice of { } 0 subject to to the technological constraints in hiring and ring: =1 for =0 1 = 11 1 for =1 2 1 = given initial conditions 0 0 = for =1 2 xxxi

33 Problem of rms type II. There is a continuum of rms type II that sells workers to each island, subject to the undirected search technology, and buys them back from islands. Firms also operate the home production technology. Purchases are denoted by ( ) with price ( ) and sales are denoted by ( ) with price ( ). The sequence problem for rms type II that produce home goods and reallocate workers is the following: X + X X X ( 0 ) (45) X X X X ( 0 ) =0 by choice of { } 0 subject to the the undirected search technology, so that they cannot sell dierent quantities to dierent islands, which is written as 1 = for all and the ow constraint stating that workers bought can be allocated to either increase the stock producing the home good or to search: + 1 X X X ( 0 ) for all =0 where 1 and 1 are given. Market clearing. For nal goods: X X X ( 0 )+ = X =1 X X ( 0 ); =1 for the market of new (tenure =0)workers: 1 = 0 for all ; for the market of incumbent (tenure 0) workers: = for all for the home production good: = xxxii