Solutions to selected problems from homework 1.

Transcription

1 Jan Hagemejer 1 Soltons to selected problems from homeork 1. Qeston 1 Let be a tlty fncton hch generates demand fncton xp, ) and ndrect tlty fncton vp, ). Let F : R R be a strctly ncreasng fncton. If the tlty fncton s defned by x) F x)) hat are the demand fnctons generated by anser n terms of xp, ) and vp, )). In fact, by mstake I ommted the part of the qeston that I ntended to nclde and ths has mslead many of yo. The mssng sentence shold be Ho are the ndrect tlty fnctons related. Of corse everybody ho handed n the homeork gets fll credt for the qeston 1. So here s the solton: Ste. Yo can sho that the tlty maxmzaton of leads to the same demand xp, ) maxmzaton of. Yo do not have to do that snce e sad that n the lectre, bt here t s: L 1 x) λ p x ) and L 2 F x)) λ p x ) f e take the FOC s th respect to x and x j for L 1 e get: x Ux)/ x λp 0 x j Ux)/ x j λ 0 these lead to: for any par of and j. Ux)/ x Ux)/ x j p 1) For L 2 e get: x F )/ Ux)) Ux)/ x ) λp 0 x j F )/ Ux)) Ux)/ x j λ 0 And these lead to the same condton as n 1). Snce the bdget constrant s the same for both problems, xp, ) x p, ). Ste. What abot the V p, )? We kno that By sbstttng xp, ) nto e kno that: F ) v p, ) F xp, )) F vp, )). 1 I have sed some of the LATEX code prepared by the grop by Marana Abbe, Los Hales, Domnka Lczak, Matthe Maycock and Assh Sbed. I personally se L Y X hch s the What-Yo-See-Is-What-Yo-Get ell, sort of) overlay to LATEX, hch may be less scary to those of yo that ant to start sng t. 1

2 Qeston 2 Let be a tlty fncton hch generates Hcksan demand fncton hp, ) and expendtre fncton ep, ). Let F : R R be a strctly ncreasng fncton. If the tlty fncton s defned by x) F x)) hat are the Hcksan demand fnctons generated by anser n terms of hp, ) and ep, )). Ho are the expendtre fnctons related? 1. We kno that x) F x)). So e kno th certanty that h p, ) hp, F 1 )) hp, ), snce F 1 ). 2. To calclate expendtre e take e p, ) l h l p, ) l h l p, F 1 )) l h l p, ) ep, ). Qestons 3 and 4 Note that these qestons are related. Yo can do t 2 ays. Frst of all, please note that the fncton s constrcted n sch a ay that for any postve level of tlty, the zero consmpton s mpossble. Therefore e can rle ot corner soltons x 0), One ay: Qeston 3. Rederve Cobb-Doglas Walrasan and Hcksan demands for the general case: th n α 1. Ux) We frst transform the tlty fncton from n xα to n α ln x, and no e try to maxmze ths fncton th regards to the constrant n p x. To do ths, e se the method of Lagrange mltplers, and have or Lagrange eqaton as: [ n ] ) Lx, λ) α ln x λ p x. The method of Lagrange mltplers then reqres that e dfferentate ths eqaton th respect to each x and λ and that these partal dervatves shold be eqal to zero. Ths yelds the eqatons: α λp 0 x x n λ p x 0 hch yelds that α λp x, and so e can se the fact that α s sm p to 1: 1 α λp x λ p x λ. mplyng that λ 1/ and so α p x /, meanng that x p, ) α /p. Otherse e cold get the same thng I do t becase ths s sally the ay to go hen e have large L) by takng the rato of the FOC s for l and. And have: Sbsttte to the last FOC: α x l x p x x l α p p x p x l α p α x l x l α x l Whch yelds the same: x l. To calclate the Hcksan demand e solve the expendtre mnmzaton problem. Assmng constant tlty e solve the Lagrangan: 2

3 Lx, ) n p x λ Agan e dfferentate L th respect to each x and λ and ask that these formla are eqal to zero. Ths: ) n p λα xα 0 x x x p λα 0 x x j λα j 0 x j λ 0 Takng the frst to FOC s for any and j e have by dvdng correspondng eqatons: p α α j x j x x j p α j x α Usng the constrant from dfferentatng th respect to λ and sbstttng x j from above, e have: j1 x αj j ) αj p α j x α ) αj x αj p αj x p α x x p α x p α ) αj α ) n k1 α k n ) αj αj j1 j1 ) αj αj ) αj j1 α ) αj p j1 Qeston 4. fncton. For the general Cobb-Doglas tlty fncton, derve the ndrect tlty fncton and the expendtre We have that the demand fncton for the good x s x p, ) α /p. The ndrect tlty fncton s vp, ) xp, )), hch th the above gves: Ux) V p, ) x p, ) α ) α α p To derve the expendtre fncton ep, ) e se the Hcksan demand. For a gven set of prces and tlty the Hcksan demand tells s ho mch of each good to get, and so e mltply the demand for each good by ts prce, and ths s the expendtre: 3

4 ep, ) e p, ) p h p, ) α ) αj j1 ) αj j1 ) αj j1 α Alternatve solton: once e have the Walrasan demand, e can have the ndrect tlty fncton. From ndrect tlty fncton e can get the expendtre fncton. From expendtre fncton, e can get Hcksan demand shorter path): From the frst half of qeston 3, e have Walrasan demand x p, ) α /p. Sbsttte t to U ) to fnd V p, ): No let s se dalty V p, ) and ep, ) : same reslt as before). No the Hcksan demand s: V p, ) ) α α p ) α ep, ) α p ) α ep, ) α ep, ) n α p ep, ) / h l p, ) α p ep, ) α p ) α ) α ep, ) ) α p α ) α p α ) α α Agan, same as before. The lesson from ths exercse s that f e do the tlty maxmzaton, there s no pont dong the cost mnmzaton, as e can get both the expendtre fncton and the Hcksan demand throgh dalty. p Qeston 5 For the tlty fncton x) L l1 lnx l γ l ), here N l1 1 and γ l < 0 fnd the demand fncton and ndrect tlty fncton for the case l 2 look for corner soltons). Note that ths fncton s smlar to a Cobb-Doglas fncton, f yo take expux)) L l1 x l γ l ). Try to dra the ndfference crves, they ll look smlarly to those of Cobb-Doglas bt the asymptotes ll no be n x 1 γ 1, and x 2 γ 2. The ndfference crosses the x 1 and x 2 axs, therefore e defntely have corner soltons. Or Lagrange problem ll have to neqalty constrants: x 1 0 and x 2 0 th Lagrange mltplers θ 1 and θ 2. The mltpler on the bdget constrant ll be λ. The fll Lagrange fncton s the constrants are th a mns n front and the constrants th a pls): Lx, λ, γ 1, γ 2 ) lnx 1 γ 1 ) + α 2 lnx 2 γ 2 ) λ x 1 + x 2 ) + θ 1 x 1 + θ 2 x 2 4

5 The frst order condtons: λ + θ 1 0 x 1 x 1 γ 1 α 2 λ + θ 2 0 x 2 x 2 γ 2 λ x 1 x 2 0 And the remanng to: θ 1 x 1 0 and θ 1 x 1 0 θ 2 x 2 0 and θ 2 x 2 0 Case 1, nteror solton th: λ > 0, θ 1 0, θ 2 0. The rle s: f the mltpler s > 0 then the correspondng condton s satsfed th eqalty - bndng constrant, f t s 0 then th neqalty - not bndng). So n ths case x 1 > 0 and x 2 > 0. Take frst to and dvde one th another: α 2 / / x 1 γ 1 x 2 γ 2 Solve for x 2 : Sbsttte to bdget constrant: x 2 γ 2 x 1 γ 1 α 2 x 2 α 2 x 1 γ 1 ) + γ 2 x 1 + x 2 x 1 + α 2 x 1 γ 1 ) + γ 2 ) x 1 + α 2 x 1 γ 1 ) + γ 2 ) x ) x 1 γ 1 ) + γ 2 ) x )x 1 γ 1 1 ) + γ 2 x 1 γ 1 1 ) + γ 2 x 1 + γ 1 γ 1 γ 2 and by symmetry: We shold also check hen ths s 0 : x 1 γ 1 γ 2 ) + γ 1 x 2 α 2 γ 1 γ 2 ) + γ 2 5

6 γ 1 γ 2 ) + γ 1 > 0 and: γ 1 + γ 2 γ 1 )/ α 2 γ 1 + α 2 γ 2 γ 2 )/α 2 Case 2, corner solton th: λ > 0, θ 1 > 0, θ 2 0. Therefore x 1 0. The FOC s become: λ θ 1 0 x 1 x 1 γ 1 α 2 λ 0 x 2 x 2 γ 2 λ x 2 0 From the thrd FOC e have that x 2 and e kno that x 1 0. When does that happen? We have to take nto accont the remanng condtons: α 2 α λ < 0 snce θ 1 > 0, the second FOC becomes: λ, and λ 2 γ 1 / γ γ 2/, so: 2 α 2 < γ 1 γ 2 / γ 2 / ) < γα 2 γ 2 / ) < γ 1 α 2 < γ 1 α 2 + γ 2 / < γ 1 + γ 1 + γ 2 / < γ 1 + γ 2 γ 1 )/ Case 3, by symmetry: corner solton th: λ > 0, θ 1 0, θ 2 > 0. Therefore x 2 0. The solton: And ths happens hen: x 1 The fll solton: xp, ) < α 2 γ 1 + α 2 γ 2 γ 2 )/α 2 0,, < γ 1 + γ 2 γ 1 )/ γ 1 γ 2) α + γ 1, 2 γ 1 γ 2) + γ 2, max{ γ 1 + γ 2 γ 1 )/, α 2 γ 1 + α 2 γ 2 γ 2 )/α 2 }, 0, < α 2 γ 1 + α 2 γ 2 γ 2 )/α 2 We can also compte the ndrect tlty fncton by sbstttng the reslts nto x) : ln γ 1 ) + α 2 ln γ 2 ). < γ 1 + γ 2 γ 1 )/ ln α1 γ1p1 p2γ2) p V p, ) 1 ) + α 2 ln α2 γ1p1 p2γ2) ), max{ γ 1 + γ 2 γ 1 )/, α 2 γ 1 + α 2 γ 2 γ 2 )/α 2 } ln γ 1 ) + α 2 ln γ 2 ), < α 2 γ 1 + α 2 γ 2 γ 2 )/α 2 6

7 Qeston 6 For the Leonteff tlty fncton x) mn{x 1 /a 1, x 2 /a 2,..., x l /a l } here a l > 0, l, fnd the demand fncton and the ndrect tlty fncton. What s the expendtre fncton? Ste. Note that th postve prces, e ll have demands sch that: x 1 a 1 x 2 a 2... x L a L Therefore, x 1 a 1 x a,. And: x a x 1 a 1 Sbstttte t to the bdget constrant: Therefore: And by symmetry: So the ndrect tlty fncton s: p x x 1 x j x 1 p a x 1 p a a 1 a 1 a 1 p a a j p a And the expendtre fncton by dalty): vp, ) mn{ a j p /a j } mn{ a p } a p a ep, ) p a ep, ) p a 7