( ) 2 ( ) ( ) Problem Set 4 Suggested Solutions. Problem 1

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1 Problem Set 4 Suggested Solutons Problem (A) The market demand functon s the soluton to the followng utlty-maxmzaton roblem (UMP): The Lagrangean: ( x, x, x ) = + max U x, x, x x x x st.. x + x + x y x, x, x 0 (,, ; λµ,, µ, µ ) = + + λ[ ] L x x x x x x y x x x ( x 0) ( x 0) ( x 0) + µ + µ + µ The frst-order condtons: Set (I) ( λµ µ µ ) U( x) L x;,,, = 0 = λ µ λ µ = x x x x x ( λµ µ µ ) U( x) L x;,,, = 0 = λ µ λ µ = x x x x x ( ; λµ,, µ, µ ) U( x) L x x = 0 = λ µ λ µ = x (.) (.) (.) Set (II) x+ x+ x y (.) λ 0 (.) λ y x x x = 0 (.) x x = 0 (.)

2 x x = 0 (.) x x = 0 (.) We have the followng ossble cases: (.),(.),(.) (Interor Soluton) x, x, x > 0 µ = 0 By (.), snce we are gven > 0 =,,, we get λ = > 0. Hence, by (.): x+ x+ x= y (.) x = (.) x From (.): x x = = = x x x x = 7 Thus, x = 7 Fnally: y x y + + = = = 7 x x x y x Hence, a canddate soluton s: x (,,, y) 7 7 = 7 y x x x = = 0, > 0 Thus, µ = 0 and by (.): λ = > 0

3 y Hence, by (.): x = y x= From (.): µ = > 0 Smlarly, from (.): µ = > 0 0 Hence, a canddate soluton for the UMP s: x (,,, y) = 0 y x, x > 0, x = 0 Thus, µ = 0 and by (.) or (.): λ > 0 Hence, by (.): x+ x= y (.) x = (.) x y y Hence, x = and x = But from (.): µ = > 0 snce the rce vector s strctly ostve. Ths result clearly volates our earler remse that µ = 0. Therefore, ths case cannot yeld a canddate soluton. Note that we don t need to check the sub-cases of each of the cases (a) x = x = 0, x > 0 and (b) x, x > 0, x = 0 where one of x, x s strctly ostve whle the other beng zero because, by the functonal form of the gven utlty functon, the zero element drves the entre frst term n the utlty functon to zero rresectve of the fact that the other element s non-zero. Hence, f for examle x = 0, there s no ont s wastng money on x. In other words, () otmal. x = 0, x > 0 and, smlarly, () x = 0, x > 0 could never be It s easy to verfy that, from the two canddate solutons that we found above, the frst one yelds the hghest level of utlty. Secfcally, y y U( x (,,, y) ) = + > = U( x (,,, y) ) 7 One would reach the same concluson by notcng nstead that from (.): µ = > 0.

4 Fnally, the Marshallan demand s gven: 7 (,,, ) x y 7 = y 7 The ndrect utlty functon was already calculated: y V (,,, y) U( x(,,, y) ) 7 (V) The Hcksan demand functon s the soluton to the followng exendture-mnmzaton roblem (EMP): mn ( x, x, x ) st.. whch s, of course, equvalent to: (,, ) U x x x = x x + x u x, x, x 0 max ( x, x, x ) ( x 0) ( x 0) ( x 0) x + x + x ( x x x ) + + st.. U( x ), x, x = xx + x u x, x, x 0 The Lagrangean: L( x, x, x ; λµ,, µ, µ ) ( x x x ) λ u ( x x ) x + µ + µ + µ =

5 The frst-order condtons: Set (I) ( ;,,, ) U( x) L x λµ µ µ λ = 0 λ = µ µ = x x x x x ( ;,,, ) U( x) L x λµ µ µ λ = 0 = µ µ = x x x x x ( ; λµ,, µ, µ ) U( x) L x x = 0 = µ µ = λ x (.) (.) (.) Set (II) x x + x u (.) λ 0 (.) λ u xx + x = 0 (.) x x = 0 (.) x x = 0 (.) x x = 0 (.) We have the followng ossble cases: (Interor Soluton) (.),(.),(.) x, x, x > 0 µ = 0 By (.), snce we are gven > 0 =,,, we get λ = > 0 Hence, by (.): (.) (.) x = x From (.): u = x x + x x = x = 7 x 5

6 Thus, x = 7 u = xx + x x = u 9 Fnally: 7 Hence, a canddate soluton s: h (,,, u) 7 = u 9 x x x = = 0, > 0 Thus, µ = 0 and by (.): λ = > 0 Hence, by (.): x = u From (.): µ = > 0 Smlarly, from (.): µ = > 0 0,,, = 0 u Hence, a canddate soluton for the EMP s: h ( u) x x x, > 0, = 0 Thus, µ = 0 and by (.) or (.): λ > 0 Hence, by (.): u = x x (.) (.) Hence, x = x u x = and x = u u u u u But from (.): µ = = > 0. Ths result clearly volates our earler remse that µ = 0. Therefore, ths case cannot yeld a canddate soluton. One would reach the same concluson by solvng nstead for notcng nstead for µ from (.). 6

7 Note that we don t need to check the sub-cases of each of the cases (a) x = x = 0, x > 0 and (b) x, x > 0, x = 0 where one of x, x s strctly ostve whle the other beng zero for exactly the same reason as n the UMP. It s easy to verfy that, from the two canddate solutons that we found above, the frst one yelds the lowest level of exendture. Secfcally, m h,,, u ( ( )) = h + h + h = u 7 < u = h + h + h ( (,,, )) = m h u Fnally, the Hcksan demand s gven: 7 (,,, ) h u 7 = u 9 The exendture functon was already calculated: m,,, u = m h,,, u ( ) ( ) h (,,, u) h (,,, u) h (,,, u) = + + = u 7 (E) 7

8 At the soluton onts for the UMP and EMP, we can nterchange usng the followng denttes: x y, = h u, = h V, y,. ( ) To see ths, consder the EMP when the agent s requred to acheve at least the utlty level corresondng to hs ndrect utlty level from (V) when rces and ncome are y., gven by We get from (V) for u = V (,,, y) y u = + 7 For ths requred reservaton level of utlty, the soluton to the EMP wll be gven: 7 7 h(,,, u) = = 7 7 y u = = x y 7 y 7 (,,, ). x( m, ( u, )) = h( u, ) To see ths, consder the UMP when the agent s ncome s the mnmum amount he needs n hs EMP n order to acheve at least a utlty level u. We get from (E) that the mnmum amount of money requred to acheve a reservaton level of utlty equal to u s gven: m(, u) = u 7 8

9 When ths s the amount of ncome avalable for the UMP, the soluton to the UMP wll be gven: 7 7 x(,,, m(, u) ) = = 7 7 e(, u ) u = = h (,, 7, u ) u 9. (, (, )) m V y = y To see ths, consder the EMP when the agent s requred to acheve at least the utlty level corresondng to hs ndrect utlty level from (V) when rces and ncome are gven by ( y., ) y We get from (V) for u = V (,,, y) : u = + 7 Now consder the EMP when the agent s requred to acheve ths as the reservaton level of hs utlty. From (E), the mnmum amount of money requred to acheve ths reservaton level of utlty s gven: y m(, u) = + = y (, (, )) V m u = u Consder the UMP when the agent s ncome s the mnmum amount he needs n hs EMP n order to acheve at least a utlty level u. We get from (E) that the mnmum amount of money requred to acheve a reservaton level of utlty equal to u s gven: m(, u) = u 7 9

10 When ths s the amount of ncome avalable for the UMP, the soluton to the UMP wll yeld a utlty level gven by (V): m(, u) V (, m(, u) ) = + = u + = u (B) Ths nvolves only algebra and I wll leave t to you to verfy. (C) Ths nvolves only algebra and I wll leave t to you to verfy. (D) Consder the followng (unconstraned) mnmzaton roblem:,, The frst order condtons: = = + = 7 mn V,,, x x V,,, x = = = = x 0 0 x 7 7 (D.) V,,, x = = = = x 0 0 x 7 7 (D.),,, = = = 0 = 0 x + x = 7 7 V x x x (whch holds whenever (D.) and (D.) do) 0

11 The mnmzed exendture (.e. the value functon) of the roblem s found by luggng the soluton nto the objectve. mn V,,,,, x = x = = + 7 = x x x 7 = x = x = x + 9 = x = x + = x + ( x x ) (,, x ) = U x x Consder now the followng (constraned) mnmzaton roblem: mn u (,, ) st.. e u = u x + x + x (, ),, 0 7

12 whch s, of course, equvalent to: max u (,, ) st.. u x x x,, 0 7 Note frst that we are lookng for a soluton to ths roblem, whch ought to be vald for all,, 0. In ths resect, ths roblem dffers substantally from the constraned otmzaton roblems that we have seen thus far n that we are no longer allowed to consder cases for our choce varables,,. In other words, comng u wth a canddate soluton that requres, for examle,, > 0 but s not vald when, > 0, = 0 wll not do here. We need to fnd a soluton that wll oerate as an dentty n the sense that wll be vald across all nteror and corner cases as long as,, 0. The Lagrangean: L(,, ; λµ,, µ, µ ) = u+ λ x x x + u 7 + µ + µ + µ ( 0) ( 0) ( 0) The frst-order condtons: Set (I) ( ; λµ,, µ, µ ) λ L = 0 = λ x µ 7 ( ; λµ,, µ, µ ) λ L 7 = 0 = λ x µ ( ; λµ,, µ, µ ) λ L = 0 λu = λ x µ 7 (D.I) (D.II) (D.III)

13 Set (II) u x+ x + x 7 (D..) λ 0 (D..) λ x + x + x u = 0 7 (D..) 0 µ = 0 (D..) 0 µ = 0 (D..) 0 µ = 0 (D..) Agan, we need a soluton, whch s to work for any,, 0. Therefore, t ought to work also when we have an nteror soluton.e.,, > 0 But ( D..),( D..),( D..),, > 0 µ = 0. Therefore, n our soluton, we requre: µ = 0 Smlarly, we must have λ > 0 n our soluton. For f we don t, we leave,, comletely free relatve to x, x and ths wll create roblems n the case where, > 0 = 0. Gven that µ = 0, observe that λ = 0 has all equatons but (D..) satsfed trvally rresectvely of the values of,, and x, x, x. To satsfy all condtons, we only need to worry about (D..). For, > 0 = 0, ths gves: x + x 0 whch cannot be satsfed for any, 0 > unless x = x = 0. But recall that the exendture functon m(, u ) was derved as the value functon of the EMP whch does admtted only a strctly nteror soluton where x, x, x > 0. Hence, we cannot have x = x = and consequently we cannot have λ = 0. 0

14 Take therefore µ = 0 and λ > 0 DI. x = DII. x From (D..) we get: u = x + x + x 7 u = x+ x + x + 7 Proceed now as n the dervaton of the result for the revous mnmzaton roblem (see. ) to derve the functonal form of u: u = x x + x (D) Ths nvolves only algebra (albet qute tedous n showng concavty of m(, u ) and convexty of V (, y) snce one needs to establsh the defnteness of the 4x4 Hessan matrces) and I wll leave t to you to verfy. (E) Ths nvolves only algebra and I wll leave t to you to verfy. (F) Ths nvolves only algebra and I wll leave t to you to verfy. Note, however, that there are some tyos n the text of ths art of the roblem set. The gven relatons should read: (. ) x( y, ) h( u, ) h( u, ) V y, = + u (, ) (, ) j j u= V y u= V y j (, ) x y, h u, V y, = u y j u= V y 4

15 (.) (, y) x x( y, ) h( u, ) y V y, = + (, j V y j ) u= V(, y) j y (, ) x y, h u, x y, = x (, y) y j j u= V y (, ) x y, m u, x y, = x (, y) y j j u= V y Problem (A) Consder the roft-maxmzaton roblem (PMP) : max ( z, z, z ) st.. z = = z 5 z z z z, z 0 The Lagrangean: L z z z z z z z z z = 5,, ; λµ,, µ = + λ + µ ( 0 ) + µ ( 0 ) The frst-order condtons: Set (I) ( ; λµ,, µ ) L z z = 0 λ = (.) ( ;,, ) L z λµ µ λ 4 = 0 = + z z 5 z z 5 L z; λµ,, µ λ 4 = 0 + z z 5 z z z 5 (.) (.) Note that only the two nut varables z, z are constraned (to be non-ostve) n ths roblem. 5

16 Set (II) z z z 5 (.) λ z λ 0 (.) z z 5 = (.) 0 z 0 µ z = 0 (.) z 0 µ z = 0 (.) From (.), t s clear that 0 λ >. Therefore, from (.): 5 z = z z Note that an obvous ont that satsfes all of our frst order condtons s the ont (,, ) ( 0,0,0) z z z z = = wth µ = > 0, µ = > 0. However, not roducng nothng at all results n zero roft. Consequently, any other vector z that generates a ostve amount of roft would be referred to the zero-ont and, thus, ths ont cannot be otmal. For exactly the same reason, any vector z nvolvng z = 0 for ether of =, cannot be otmal as t gves zero outut (.e. zero revenues from sales and non-ostve roft). We have only the followng case to consder: (.) (Interor Soluton) z =. z From (.) and (.): (.),(.) z, z < 0 µ = = ( zz ) z= z z z= = And z 5 = = = = z 6

17 Therefore z(,, ) = ( z, z, z ) =,, The roft functon s gven: π = z = = = π I wll leave to you to verfy that z = {,, } convex n (the latter exercse wll be rather tedous as you need to determne the defnteness of a x Hessan matrx). as well as that π s (B) The free dsosal roerty can be wrtten as follows: z = z, z, z T z% R : z% z z% z z% z T { } { } : 0 T = z R z >> We are gven that the roducton ossblty set (PPS): π s closed. { } : 0 Ths means that t ought to nclude ts boundary: π T = z R z = >> Consder ths boundary set. It conssts of those onts z that satsfy the relaton: π 0 (I). z = z = >> = 7

18 Note, however, that snce ths relaton s to hold for all strctly ostve rce vectors, t s really an dentty relaton wth resect to. Therefore, f we consder the gradent of each sde of (I) wth resect to, the two sdes should agree 4..e. = z = z = π >> 0 z = π >> 0 z π ( = ) = >> 0 = = z = π = {,,} >> 0 π In other words, we get: z = {,, } and ths ought to be a necessary condton, f (I) s to hold as an dentty for all rce vectors >> 0. Regardng the gven roft functon, we have: ( ) 4 + z = z = z = (I.) Note that (I.) s requred to hold as a system of equatons, for all >> 0. Therefore, t defnes the followng nter-relaton between the elements of the suly vector z: z z z, z 0 z > = + 8z = ( z + z) Essentally, I am usng the followng argument. Let f ( x) = g ( x) x domf domg (I) Then, clearly f g and, consequently, domf domg domf domg mles f x = g x x domf domg. f g domf domg domf domg. In other words, (I) 8

19 We are now n a oston of beng able to wrte the boundary set for the PPS under examnaton. { : π 0} T = z R z = >> { z R : z 0, z 0, 8z ( z z )} * z, z, ( z z ) : z, z R + = > > = + { } = + Recall now that the PPS has to satsfy the free dsosal roerty. Hence, t wll consst of the boundary set T and all vectors that le to the south-west of any gven ont on ths boundary..e. T = z, z, z R : z > 0, z > 0, 8z z + z or { } {,, : 0, 0, 8 } T = z z z R z > z > z + z z Notes: To comlete the dervaton of the roducton ossblty set T, you should verfy that the set gven here s convex. Ths s qute straght forward as t suffces to show that the boundary set s concave. ( + ) The roblem also asks for us to verfy that π = s ndeed the roft functon assocated wth ths roducton ossblty set. Ths calls merely for you to reeat art (A) of the roblem usng T = {( z ), z, z R : z > 0, z > 0, z + z 8z} as your roducton ossblty set. I wll, thus, leave t to you to show that ths actually works 5. 5 It would be agan clear from the FOC s on the Lagrangean that λ > 0. Thus: z + z = 8z z = z, z, z = 0,0,0 wth An obvous ont, agan, that satsfes all of the FOC s s the ont µ = > 0, µ = > 0. It s ruled out, though, for exactly the same reasonng as n art (A). However, here we cannot rule out a ror any vector z nvolvng z = 0 for ether of =, as long as t s not for both, snce such a vector does not necessarly gve zero total outut (as was the case n art (A)). 4 For z = 0, z > 0, the canddate soluton s z = ( z, z, z) = 0,,. Smlarly, z > 0, z = 0 gves z =,0,. The former gves a roft of whle the latter gves +. Nether beats the nteror soluton s roft. 9