Final Exam - Solutions

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1 EC 70 - Math for Economsts Samson Alva Department of Economcs, Boston College December 13, 011 Fnal Exam - Solutons 1. Job Search (a) The agent s utlty functon has no explct dependence on tme, except through the exponental dscount factor, and nether the wage offer process nor the annuty process has explct tme dependence. Thus, gven that there s no means to save and that annuty s constant every perod, the only thng that vares from one perod to the next s the most recent job offer held by the agent. If n two dfferent perods the agent held an offer for some amount w, then from the agent s perspectve, the futures startng from each of those perods are equvalent to the extent that the probablty of any future path of all varables s the same f the same contngent plan s used. Ths statonarty s the reason why an agent that has chosen to work at some job wll never fnd t optmal to qut the job and search for a new one at some later date. Consder the agent s decson when the job offer was frst receved. Gven that the choce was optmal n ths perod, then at some later date, when the agent s consderng whether to contnue to work at the job or search for another job, the agent realzes that nothng has changed about her nformaton about wage offers. Her current offer s the same as when she frst began her job, and the wage offer dstrbuton s the same as well. Moreover, because she cannot save, and because the annuty s constant, nothng about her decson problem has changed, and so, gven that t was optmal to start workng at her current job, t s optmal for her to contnue to work at her current job. (b) The above argument about never quttng a job once startng t mples that the agent has a threshold wage offer, above whch she would accept the offer and stay at the job forever, effectvely stoppng the job search process, and below whch she rejects the offer and contnues to search. Let ths threshold wage be denoted by. Denotng by ŵ the most recent wage offer, the Bellman for the decson problem s: V (ŵ t ) = max e t (y 0 + e t ŵ + β E[V (ŵ t+1 )]) where ŵ t+1 = e t ŵ t + (1 e t )w t. Keep n mnd that w t s only observed f e t = 0, and so s a random varable at tme t, when the agent s decdng whether to work or search. Also, f the agent searches and receves offer w t, the agent cannot work at ths job untl the next perod, and so receves no wage n the perod she s searchng. 1

2 Gven the prevous dscusson that the agent wll choose to work f the most recent wage offer s greater than, and choose to search f the most recent wage offer s below, we know f the most recent wage offer s equal to, the agent s ndfferent to searchng and to workng. If the agent choose to work, the agent s lfetme utlty s just the present-dscounted value of the stream of ncome y 0 +. Gven the dscount factor of β, ths lfetme utlty level s y 0+. If the agent 1 β chooses to search, the agent receves y 0 n utlty n the perod of search, and the dscounted expected value of the offer receved from today s search. Thus, the value of search f the current wage offer s s y 0 + β E[V (w t )], where w t s drawn..d from a dstrbuton wth c.d.f. H. So, when the current wage offer s, the agent must be ndfferent, mplyng: y β = y 0 + βe[v (w t )], (1) where w t, the newly-receved wage offer, s dstrbuted accordng to H. Now, usng the dea that the agent wll accept any offer w t that s greater than, whch occurs wth probablty 1 H( ), we know that V (w t ) = y 0+w t for any w 1 β t, snce the agent wll receve ncome stream y 0 + w t forever. On the other hand, f the agent receves a wage offer w t <, the agent wll reject the offer and search agan. Thus, f w t <, whch occurs wth probablty H( ), V (w t ) = y 0 + β E[V (w t+1 )], where w t+1 s the next wage offer. However, E[V (w t+1 )] = E[V (w t )], gven our assumptons about the wage offer process. Thus, assumng the p.d.f. assocated wth H exsts, E[V (w)] = 0 V (w)h(w)dw = H( )(y 0 +β E[V (w)])+ Rearrangng equaton (), we obtan (1 βh( )) E[V (w)] = H( )y 0 + Rearrangng equaton (1), we obtan (y 0 + w) 1 β h(w)dw. β(1 β) E[V (w)] = (y 0 + ) (1 β)y 0. Puttng these results together to elmnate E[V ( )], we get (1 βh( ))(y 0 + (1 β)y 0 ) = β(1 β)h( )y 0 + β (y 0 + w) h(w)dw () 1 β (y 0 + w)h(w)dw. The frst term n the ntegral evaluates to βy 0 (1 H( )), and so the above equaton becomes (1 βh( )) +β(1 βh( ))y 0 = β(1 β)h( )y 0 +β(1 H( ))y 0 +β wh(w)dw.

3 All the terms nvolvng y 0 cancel out, leavng the equaton Ths smplfes to (1 βh( )) =β =β =β (1 β) = β wh(w)dw (w )h(w)dw + β h(w)dw (w )h(w)dw + β(1 H( )). (w )h(w)dw. Notce that the left- and rght-hand sdes of the equaton are functons of. The left-hand sde s an ncreasng functon, startng at a value of 0 when = 0. The rght-hand sde s a decreasng functon of snce the lower lmt of the ntegral s, and snce the term n parentheses s always postve but decreasng n for any value of w wthn the ntegraton lmts. Moreover, the rght-hand sde has value 0 when = w. Thus, by the ntermedate value theorem, there must exst some that solves the equaton f the cumulatve dstrbuton functon H s absolutely contnuous (and so the ntegral s contnuous). Ths equaton mplctly determnes the threshold wage that the agent uses to decde whether to accept or reject a wage offer. If we assume that wage offers are unformly dstrbuted on [0, w], then h(w) = 1 w, and the above equaton smplfes to whch yelds a quadratc equaton n, The two roots of ths quadratc are (1 β) = β w (w ), β w + βw = 0. = w β (1 ± 1 β ), of whch only the negatve root makes sense (.e. the postve root yelds > w). Thus, the threshold wage = w β (1 1 β ). (c) Clearly, the threshold wage s ndependent of y 0.e. y 0 = 0. The ntuton s qute smple. Our agent receves the annuty no matter what she does. Snce she has no way to earn nterest ncome va a savngs opton, she has to consume the annuty n the perod t s receved. Moreover, she has lnear utlty and so her decson about whether to begn workng or not, gven her level of consumpton guaranteed by her annuty ncome, s at the margn the same no matter the level 3

4 of y 0. Ths would not be true f she had, say, a concave utlty functon. Then, as her annuty amount ncreases, the margnal value of workng and earnng wage ncome ths perod decreases. So, the opportunty cost of searchng decreases. Moreover, at hgher consumpton levels, (most) concave utlty functons appear more lnear for a gven spread n consumpton, and so the dsutlty from the rsk assocated wth searchng decreases. Ths reasonng would suggest that hgher levels of annuty should nduce our agent to ncrease the threshold wage for a gven utlty functon. On the other hand, a smlar lne of reasonng suggests that ncreasng the concavty of the utlty functon for a gven level of annuty should reduce the threshold wage. For a general utlty functon, the threshold wage can be shown to be the soluton to the followng equaton, usng an analyss smlar to the one conducted above: (1 β) [u(y 0 + ) u(y 0 )] = β [u(y 0 + w) u(y 0 + )] h(w)dw. In ths case, after applyng the Implct Functon Theorem and smplfyng, one can obtan = 1 + (1 β)u (y 0 ) + β w u (y 0 + w)h(w)dw y 0 (1 β)u (y 0 + ) + β u (y 0 + )h(w)dw. The fractonal term s postve, but the magntude s not apparent.. Managng a Hydroelectrc Power Plant (a) The current-value Hamltonan s H(L, z, µ) ALz + µ(r z) + λ 1 z λ (z Z), whch takes nto account the constrants on the control. Accordng to the maxmum prncple, at the optmum the paths of the control, the state and the costate satsfy: H z = AL(t) µ(t) + λ 1 (t) λ (t) = 0 λ 1 (t) 0, z(t) 0, λ 1 z(t) = 0 λ (t) 0, z(t) Z, λ (Z z(t)) = 0 H L = Az(t) = ρµ(t) µ(t) H µ = R(t) z(t) = L(t) lm t e ρt µ(t) = 0, where the last condton s a transversalty condton ensurng the shadow value of the state/stock varable L (n present-value terms) approaches 0. Ths condton makes sense, snce havng a postve shadow value mples that the rate of outflow could have been ncreased, whch would mprove the lfe-tme payoff. Ths form of the transversalty s present, rather than one nvolvng L drectly, snce we don t have to be concerned wth a non-negatvty condton on L. That L appears n the objectve mples that L wll never be negatve at an optmum. 4

5 (b) Snce R(t) = R, a constant, the requrement for steady-state s that L(t) = 0 and µ(t) = 0 for all t. These two condtons mply that z(t) = R (assume R (0, Z), otherwse the maxmum rate of outflow s less the the rate of nflow, and the lake wll grow wthout bound), and ρµ(t) = Az(t) = AR. Fnally, gven that R (0, Z), z(t) avods the boundares of ts constrant set, and so λ 1 (t) = λ (t) = 0, mplyng µ(t) = AL(t). Thus, L(t) = R. In summary, n the ρ steady-state, the control, state, and costate have values z = R, L = R ρ, µ = AR ρ. (c) Lnearty of the control n the Hamltonan mples that the optmal control s dscontnuous, takng values at the boundares when not n steady-state. Ths type of soluton s called a bang-bang soluton, because the control does not take on values n the nteror of the admssble set, except perhaps n steady-state. If the steady-state path s ncluded n the optmal soluton, then the value of z(0) depends on whether the level of water L(0) s above or below the steady-state. If t s below, then z(0) equal 0 untl the level of water reaches the steady-state, whch occurs at some tme τ, after whch tme z(t) = R. Smlarly, f L(0) s greater than the steady-state level, then z(0) = Z untl some tme τ, causng the level of water to fall untl t reaches the steady-state level, at whch pont z(t) = R. Unlke n, say, the optmal growth problem, steady-state can be reached n a fnte tme, and the control s pecewse contnuous, actually pecewse constant. Thus, the only queston remanng s whether ths soluton of gettng to the steadystate and then stayng there s n fact the optmal soluton. The only other canddate soluton s one where z keeps swtchng between the two extremes of 0 and Z. 3. Durable Good Monopolst wth Myopc Consumers (a) A consumer buys the product f and only f hs valuaton v p. Each consumer draws hs valuaton randomly from the unform dstrbuton, and so the probablty that v p s 1 p. Snce each consumer s valuaton s drawn ndependently and dentcally from the unform dstrbuton, and we have a contnuum of consumers, the dstrbuton of valuatons n the market place ends beng exactly the dstrbuton from whch the valuatons are drawn. Thus, 1 p s also the market demand.e. q 1 (p) = 1 p. (b) Frst perod proft s p 1 (1 p 1 ) and so the proft-maxmzng prce s p 1 = 1, and the proft level s 1 4. (c) All consumers wth valuatons above p 1 wll buy the good n the frst perod and leave, and so the hghest valuaton consumer left n the market has valuaton p 1. Also, 1 p 1 consumers buy n the frst perod, and so the resdual demand n perod s q (p ) = p 1 p. Takng p 1 as gven, the proft-maxmzng prce s p 1. The perod proft s p

6 (d) Thus, the present dscounted value of proft n perod 1, π = p 1 (1 p 1 ) + β p 1 4. The frst-order condton yelds 1 p 1 + β p 1 = 0, whch yelds p 1 = 4 β. The dscounted proft level s 1 4 β. (e) Settng a prce n some perod that s hgher than the prce n some prevous perod wll result n zero sales, because every consumer who mght have purchased the product ths perod wll have already exted the market. Thus, prces must (weakly) decrease n order to have any chance to makng any sales. (f) At tme t, every consumer whose valuaton s above the lowest prce to date wll have exted the market.e. the only consumers remanng n the market are those whose valuatons are less than or equal to mn t {p 1,..., p t 1 }. Snce we know prces must decrease, we can conclude that mn t {p 1,..., p t 1 } = p t 1. Thus, at tme t, every consumer wth valuaton below p t 1 remans. Thus, at tme t, demand q t = p t 1 p t. (g) The monopolst s problem s max {p t} T t=1 T β t 1 p t (p t 1 p t ), t=1 where p 0 s defned to be 1. It should be apparent that boundary solutons wll not occur, so ths s just an unconstraned maxmzaton problem wth T choce varables {p 1,..., p T }. The frst order condton for the prce at a generc tme perod t < T s and the frst order condton for p T s These smplfy to β t 1 ((p t 1 p t ) p t ) + β t p t+1 = 0 β T 1 ((p T 1 p T ) p T ) = 0. p t 1 p t + βp t+1 = 0, t {1,..., T 1} p T 1 p T = 0. Ths second-order dfference equaton determnes the optmal prcng polcy of the monopolst. (h) Start wth the last perod. We see that p T = p T 1. Next, we have that p T p T 1 + βp T = 0, whch mples p T 1 = p T +βp T. Substtutng our soluton for 6

7 p T and rearrangng yelds p T 1 = 4 p T 4 β = γ 1 p T, where γ β Next, notce that we have p t = p t 1+βp t+1 for any t < T. Let s prove that the p dfference equatons yeld p t = γ t 1 T t for any t > 0, where γ T t s some constant that depends on t. Notce that we have p T = p T 1, so γ 0 = 1. We also know that γ 1 = 4. Now, suppose our conjecture holds for every tme perod 4 β t p {t + 1,..., T }. Then, gven that p t+1 = γ t T (t+1), we can substtute for p t+1 n the dfference equaton p t = p t 1+βp t+1 to get p t = p βγ T (t+1) t 1+ p t, whch mples that p t = 4 4 βγ T (t+1) p t 1. Thus, we have that γ T t = 4 4 βγ T (t+1), whch s a dfference equaton that defnes the sequence {γ s } T s=0 1 p. Now, p 1 = γ 0 T 1 More generally, p t = 1 t = γ T 1 t γ s. s=1 and p = γt p 1 = γ T 1 γ T. Startng wth γ 0 = 1, one can show that γ s+1 > γ s for all s 0. Moreover, γs < 1 for all s 0. Thus, we see that prces do n fact decrease over tme. Also, notce that γ s+1 β = 4γ s (4 βγ s ) = γ s 4 γ s+1 > 0 for all s 0. Thus, each term n the product comprsng p t s ncreasng n β, and so gven a tme perod t, p t s ncreasng n β. Thus, a more patent monopolst charges a hgher prce that a less patent monopolst n every perod. Also, as the number of tme perods ncreases, the prce n any partcular tme perod ncreases; to see ths, notce that p 1 = γ T 1 s ncreasng n T snce γ T 1 s ncreasng n T, as dscussed earler. A smlar argument works for every subsequent prce. Thus, as the decson horzon s extended, the monopolst chooses a sequence of prces that decreases more slowly. In fact, notce that the recursve equaton for γ s must converge so that lm s (γ s+1 γ s ) = 0. To fnd the lmtng γ, solve γ = 4 4 βγ. We get a quadratc equaton n γ, βγ 4γ + 4, whch yelds the soluton γ = β (1 ± 1 β), of whch only the negatve root makes sense, snce we need γ < 1 for the prce equaton to yeld a prce between 0 and 1. Thus, as the tme horzon T extends, the rato p t+1 p t s ncreasngly well approxmated by γ, wth the approxmaton best for the early perods and gettng worse as the termnal perod approaches. We 7

8 shall see that ths rato s the exact result for the nfnte horzon verson of the problem. The followng Excel fle models ths problem, producng the optmal prce sequence, the dscounted and undscounted sum of profts, and the sequence {γ s }, whle allowng for any value of β [0, 1] and T {1,..., 95}: bc.edu/samson-alva/ec70f11/durablemonopoly.xls. () In general, we would need to keep track of the entre hstory of prces, but snce we know that today s demand s determned by the lowest prce set n the past, and that ths prce s the most recent prce set, as argued above, the only thng that needs to be kept track of s the last prce. Thus, the Bellman s V (p t 1 ) = max p t (p t (p t 1 p t ) + βv (p t )) (j) We have that the optmal prce p t = γp t 1 for any t > 0, and also that V (p t ) = A(p t ), where γ and A are coeffcents to be determned. The frst-order condton for optmalty s p t 1 p t + βv (p t ) = 0. But wth V (p t ) = A(p t ), so V (p t ) = Ap t. Usng ths, and the relatonshp p t = γp t 1, we obtan p t 1 = (1 βa)p t = 1 = (1 βa)γ. Next, the envelope theorem appled to the Bellman equaton mples V (p t 1 ) = p t. Usng our conjectures about both V and p t, we see that Ap t 1 = γp t 1, mplyng γ = A, = 4A(1 βa) = 1. Thus, combnng the two equatons nvolvng γ and A, we get γ( βγ) = 1 = βγ γ + 1 = 0. Solvng ths quadratc for γ gves us two roots, γ = 1 β (1 ± 1 β) of whch only the negatve root yelds a value between 0 and 1. Thus, we have that γ = 1 (1 1 β) and A = γ = 1 (1 1 β). Thus, p β β t = 1 (1 1 β)p β t 1, whch s the same prcng polcy obtaned as T approached nfnty n the fntehorzon verson of the problem. The dscounted sum of profts s just the value functon at tme 1, wth state varable havng value 1, and so s equal to 1 (1 1 β). Next, note that β 8

9 p t = γ t p 0 = γ t. So, the undscounted sum of profts s p t (p t 1 p t ) = t=1 γ t (γ t 1 γ t ) = (1 γ) t=1 t=1 γ t 1 = 1 γ γ (γ + γ 4 + γ ) = 1 γ γ γ 1 γ = γ 1 + γ = 1 1 β β β = 1 β 3 + β (k) We have to use L Hôptal s Rule to evaluaton the lmt of γ as β approaches 0. So, β 1 β lm = lm = 1 β 0 β β 0 1. Usng the expressons above for dscounted proft and undscounted proft, we see that wth β = 0, they are 1 and 1, respectvely. Essentally, γ = 1 s obtaned 4 3 because the monopolst doesn t care at all about the future, and so s really just solvng a sequence of statc optmzaton problems. Gven the lnear demand wth slope 1, the statcally optmal thng to do s to charge half of the maxmum valuaton n the market, and that s exactly what the monopolst does n every perod. The dscounted value of profts s just the proft n the current perod, snce the future doesn t matter at all. But the undscounted sum of profts s greater, though not as great as the total market surplus, whch equals the total area under the demand curve.e. 1. (l) The lmt of γ as β approaches 1 s 1. Thus, the optmal prcng rule s p t = p t 1. But ths doesn t make any sense, snce t mples that prce stays at 1. Nevertheless, the dscounted and undscounted profts (whch are the same n the lmtng case of β = 1) converge to 1, whch mples that the monopolst extracts all consumer surplus. So, magne a value of β very close but not equal to 1. The prce falls extremely slowly, and sales n each perod s very low. But, the consumers who buy n a partcular perod pay very close to ther valuaton, and so very lttle consumer surplus s generated. Instead, the monopolst succeeds n extractng almost all the surplus as profts. As the monopolst becomes more patent, t s more wllng to wat to earn these extra profts by very gradually decreasng prce. Notce that no matter what the dscount factor of the monopolst s, eventually every consumer wth non-zero valuaton wll get to buy the product, because the prce wll eventually fall below any postve number as t heads for 0. Thus, there s no deadweght loss n ths sense. However, f consumers dscount future utlty, clearly the delay nduces a cost, and so ths smplstc welfare analyss does not gve us the entre pcture. The model however does a good job of explanng the prcng polcy of technol- 9

10 ogy companes lke Apple wth Phones or Mcrosoft wth the Xbox, who greatly restrct the quantty of a new durable good ntally n order to rase the prce and extract the surplus from hgh-value consumers, and subsequently release more and more goods at lower prces. The rate of decrease of the prce depends on how mpatent the monopolst s. The lmtaton of ths model s that consumers are assumed to be myopc. In fact, people strategcally tme when to buy the durable, antcpatng that prces wll fall, and so a more robust analyss wll model ths stuaton as a game. 4. Optmal Allocaton wth Envous Consumers There s a fxed total Y > 0 of goods at the dsposal of socety. There are two consumers who envy each other. If consumer 1 gets y and consumer gets z, ther utltes are U 1 = y kz and U = z ky, respectvely, where k > 0 s a constant. The allocaton must satsfy y + z Y, and must maxmze U 1 + U. Also, all allocatons must be non-negatve. (a) The planner s problem s max y,z (y kz + z ky ) subject to y + z Y, y 0 and z 0. Defne the Lagrangan L (y kz + z ky ) λ(y + z Y ) = (1 λ)(y + z) k(y + z ) + λy. The Karush-Kuhn-Tucker condtons are L y = 1 λ ky 0, y 0, yl y = 0 L z = 1 λ kz 0, z 0, zl z = 0 L λ = Y y z 0, λ 0, λl λ = 0 Suppose λ > 0. Then y + z = Y. Snce Y > 0, at least one of y or z s postve. Wthout loss of generalty, suppose y > 0. Then, 1 λ ky = 0, whch mples λ = 1 ky. Then, 1 (1 ky) kz 0, whch mples k(y z) 0. But then y z 0, mplyng z y > 0, and thus we conclude that z > 0, and that L z = 0. So, we obtan λ = 1 kz, and thus y = z = Y. Also, λ = 1 ky. Suppose λ = 0. Then, y 1 > 0 and x 1 > 0. But complementary slackness k k condtons yl y = 0 and zl z = 0 then mply that L y = L z = 0, and so y = z = 1. k Thus, y + z = 1 Y. k (b) Suppose Y > 1. Now, let s suppose that the resource constrant bnds at the k optmum. Then t must be the case that λ > 0; n fact, we showed above that λ = 1 ky. But wth Y > 1 we get λ < 0, whch volates the KKT condtons. k Thus, the resource constrant cannot bnd the optmum n ths case. (c) The planner treats the two agents symmetrcally, but there are consumpton externaltes due to envy. The planner eventually encounters the stuaton that the margnal ncrease n envy n the socety exceeds the margnal drect beneft to the consumers of more goods. Thus, these goods have no value to agents or to the planner. 10

11 (d) The planner s problem s max {x } x k(x + x 3) subject to x Y, x 0 for all {1,, 3}. Defne the Lagrangan L x k(x + x 3) λ( x Y ) = (1 λ) x k(x + x 3) + λy. The Karush-Kuhn-Tucker condtons are {, 3}, L = 1 λ kx 0, x 0, x L = 0 L 1 = 1 λ 0, x 1 0, x 1 L 1 = 0 L λ = Y x 0, λ 0, λl λ = 0 Suppose x 1 > 0. Then λ = 1, whch mples that x = x 3 = 0, and snce the constrant s bndng, we get that x 1 = Y. Thus, (x 1, x, x 3, λ) = (Y, 0, 0, 1) solves the frst-order condtons. Next, suppose x 1 = 0. Then λ 1, whch mples that x 0, for {, 3}. But ths mples that x = 0, and so x =. Thus, the constrant s slack, whch mples that λ = 0. But ths yelds a contradcton wth the condton that λ 1. Thus, the only soluton s that the planner allocated all the resources to agent 1. The ntuton s that the consumpton externalty created by the envousness of agents nduces the planner to reduce the amount of envy created. But snce utlty s lnear for all agents, other than the envy component, the planner faces no dffculty n reducng the allocaton to agents that are enved and reallocatng the good to the unenved agent. (e) The ntuton from the prevous part contnues to hold. Agent 1 s the only agent that s not enved, and reducng the allocaton of some enved agent, say k, and correspondngly ncreasng the allocaton to agent 1 elmnates some envy, but the ncrease n agent 1 s utlty equal exactly the decrease n agent k s utlty, and so the planner acheves a hgher value of total utlty. Formally, the planner s problem s max {x } x k n = x subject to x Y, 11

12 x 0 for all {1,..., n}. Defne the Lagrangan L x k n x λ( = x Y ) = (1 λ) x k n x + λy. = The Karush-Kuhn-Tucker condtons are {,..., n}, L = 1 λ kx 0, x 0, x L = 0 L 1 = 1 λ 0, x 1 0, x 1 L 1 = 0 L λ = Y x 0, λ 0, λl λ = 0 Suppose x 1 > 0. Then λ = 1, whch mples that x = 0 for all {,..., n}, and snce the constrant s bndng, we get that x 1 = Y. Next, suppose x 1 = 0. Then λ 1, whch mples that x 0, for {,..., n}. But ths mples that x = 0, and so x =. Thus, the constrant s slack, whch mples that λ = 0. But ths yelds a contradcton wth the condton that λ 1. Thus, the only soluton s for the planner to allocate all of Y to agent 1. (f) The planner s problem s subject to max {x } (x kx ) x Y, x 0 for all {1,, 3}. Defne the Lagrangan L (x kx ) λ( x Y ) = (1 λ) x k x + λy. The Karush-Kuhn-Tucker condtons are L = 1 λ kx 0, x 0, x L = 0 L λ = Y x 0, λ 0, λl λ = 0 Suppose λ > 0. Then x = Y. Snce Y > 0, x > 0 for some ; w.l.o.g. suppose x 1 > 0. Then, λ = 1 kx 1, and so 1 (1 kx 1 ) kx 0, mplyng k(x 1 x ) 0. But then x 1 x 0 and so x x 1 > 0, so we conclude that x > 0 for all, and thus λ = 1 kx. Summng over all agents, 3λ = 3 ky, and so λ = 1 ky. Also, λ = 1 kx 3 = 1 kx j, so x = x j, and thus x = 1Y 3 for all. Suppose λ = 0. Then x 1 > 0 for all. But complementary slackness k condtons x L = 0 mples that L = 0 and thus x = 1 for all. Thus, ths k soluton obtans when x = 3 < Y. k The ntuton s smlar to that of part a). Snce all agents are enved (and symmetrcally), the planner wll dstrbute resources evenly. However, whle ntrnsc 1

13 utlty from the good s lnear, the dsutlty generated va envy s quadratc, and so eventually the planner wll stop dstrbutng the good because envousness ncreases at the margn more than the utlty of the recpent of the margnal unt of the good. Thus, t s possble that the resource constrant does not bnd. (g) Followng the argument of the prevous part, t s straghtforward to obtan the followng: f Y n then x k = Y for all ; f Y > n, then x n k = 1 for all. k Okay, now let s push ths generalzaton further. Suppose that you have an economy wth n agents, and agent has utlty functon U = x k n j=1 A jx j, where A j [0, 1] represents the degree of envy agent has for agent j s allocaton, where k > 0 captures the maxmum possble degree of envy. Assume that for all, A = 0; an agent doesn t envy hmself. Assume the planner wshes to maxmze the sum of utltes of all agents, n =1 U by allocaton the quantty of goods Y, subject to the constrants n =1 x Y, and x 0. (h) The Lagrangan s L = x k A j x j λ( j x Y ). Defne α j = A j; ths s a metrc of the degree to whch agent j s enved by other agents. The Lagrangan can be rewrtten as L = (1 λ) x k α x + λy. The KKT condtons are L = 1 λ kα x 0, x 0, x L = 0 L λ = ( x Y ) 0, λ 0, λl λ = 0 Note that α x 1 λ, and that equalty obtans f x k > 0, from the complementary slackness condton x L = 0. Let I be the set of agents that receve a strctly postve amount of the good at the optmum, and J be the set of agents that receve a zero allocaton at the optmum. For any I, α x = 1 λ, and α k x = α x for all, I. Notce that f α = 0 for some I, then α x = 0 for all I, and so α = 0. Thus, ether α > 0 for all I or α = 0 for all I. In words, we have the result that f any unenved agent (an agent wth α = 0) receves a postve allocaton, then all agents that receve a postve allocaton must be unenved. Also, we have the result that f any enved agent (an agent wth α > 0) receves a postve allocaton, then any agent that receves a postve allocaton must be enved. Now, consder an agent j J, assumng ths set s nonempty. Suppose there exsts some agent I. Snce α j x j 1 λ = α k x, we have that α j x j α x. But 13

14 snce j J, x j = 0, so α x 0, whch mples that α = 0, snce x > 0. In words, we have the result that f any agent receves a zero allocaton then only unenved agents can receve a strctly postve allocaton. Fnally, notce that f there exsts a j J and an I such that α j > 0 and α > 0, then we have that α x = 1 λ α k j x j, whch mples that x α j α x j. But j J mples that x j = 0, whch mples that x 0, whch contradcts our assumpton that I. Thus, n words, f any enved agent receves a strctly postve allocaton, then all enved agents receve a strctly postve allocaton. Puttng all the results together, we have the followng pcture. If there s at least one unenved agent, then every enved agent receves a zero allocaton, and at least one of the unenved agents receves a strctly postve allocaton. If there are no unenved agents, then every (enved) agent receves a strctly postve allocaton, snce Y > 0. Frst, let s suppose there are no unenved agents.e. α > 0 for all. Defne α H 1 α, the harmonc mean of the set {α }. Then, we know that x > 0 for all, and so α x = 1 λ. Consder two cases: λ > 0 mples k x = Y, so n(1 λ) = Y, so x kα H = αh Y. Next, f λ = 0, then x α n = 1 1 and so α k x = 1 n. α H k Now, let s suppose there s a set of unenved agents, denoted I. We know that only unenved agents can receve a strctly postve allocaton. Take some such agent. Then, α x = 1 λ. But snce α k = 0, ths mples that λ = 1, and so I x = Y. Thus, for any agent j, α j x j 0, and so we have a whole class of solutons defned as follows: every enved agent receves none of the good (x j = 0 for all j I), and any allocaton of the good that satsfed I x = Y and x 0 for all I s an allocaton. Ths set of solutons resembles an I -dmensonal smplex. () Suppose you are agent and the agent whom you envy more s agent j. Suppose agent j was unenved. If there s some other agent that s also unenved (possbly you), then your ncreased envy of agent j means hs new allocaton s zero, snce the presence of an unenved agent means that enved agents receve nothng, and so agent j s allocaton decreases (possbly strctly, snce an unenved agent can receve zero or a postve amount of the good). If you are an unenved agent, then there exst optmal solutons that result n you recevng a larger allocaton. If you are an enved agent, then your allocaton remans at zero. Lastly, suppose that agent j was the only unenved agent before your envy for hm ncreased. Then, he used to receve the entre amount Y, but now that he s enved, he wll receve strctly less than Y, and so your envy has decreased hs allocaton. Moreover, your allocaton strctly ncreases snce you used to receve zero, and now you ll receve some postve amount. Suppose agent j was enved. If there s some other agent that s unenved (possbly you), then your ncreased envy of j does nothng to hs allocaton, snce t s already zero. If you are also enved, then your allocaton also remans at zero. However, f you were unenved, there exst optmal allocatons where you wll receve strctly more now that your envy of j has ncreased. Lastly, suppose 14

15 there are no unenved agents. Then your ncreased envy of j mples that α j strctly ncreases, whch mples that x j strctly decreases. To see ths, notce that αh n α = 1+ α, whch clearly decreases when α j ncreases. There are two j α j cases to consder when examnng the effect of your ncreased envy of j on your own allocaton. Frst, suppose x > Y. Snce α H ncreases when α j ncreases, x = αh Y must strctly ncrease. Next, suppose that α n x Y. Now, the constrant s almost slack, and so x = 1 α, whch s unchanged. k To summarze, f agent s envy for agent j ncreases, then the planner s allocaton to agent j s weakly decreased and the allocaton to agent s weakly ncreased, and there exst stuatons where ths weak changes are strct. 15