8. Linear Contracts under Risk Neutrality

Transcription

1 8. Lnr Contrcts undr Rsk Nutrlty Lnr contrcts r th smplst form of contrcts nd thy r vry populr n pplctons. Thy offr smpl ncntv mchnsm. Exmpls of lnr contrcts r mny: contrctul jont vnturs, quty jont vnturs, crop-shrng contrcts, nd fxd-prc contrcts, tc. Km nd Wng (998), nd Wng nd Zhu (005) provd som of th bst rfrncs. A ky thortcl quston s: undr wht condtons s lnr contrct optml? Whn both prts n contrctul rltonshp r rsk nutrl, w sy thr s doubl rsk nutrlty; smlrly, whn both prts n contrctul rltonshp hv morl hzrd, w sy thr s doubl morl hzrd. Ths scton shows tht, wth morl hzrd nd uncrtnty, lnr contrct cn b optml only undr doubl rsk nutrlty, nd t cn b th frst bst only undr sngl morl hzrd. From (.), w know tht th optml contrct s gnrlly nonlnr f on of th prts s rsk vrs. For xmpl, f th prncpl s rsk nutrl wth utlty functon uz ( ) = z for ll z, thn (.) mpls é - - æ ( ) f ( x; ) ö ù s ( x) = u l m ç +. çè f( x; ) ø êë úû For som populr output dstrbutons, f ( x; )/ f( x; ) s lnr n x. For xmpl, f output follows th norml dstrbuton Ns (, ), thn f ( x; )/ f( x; ) = ( x- )/ s. If tht s th cs, th optml contrct s nonlnr for rsk-vrs gnt. Hnc, to fnd optml lnr contrcts, w ssum doubl rsk nutrlty n ths scton nd spcfy utlty functons uz ( ) = z nd vz ( ) = z, for ll z Î, for th gnt nd prncpl, rspctvly. 8.. Sngl Morl Hzrd Lt b th gnt s ffort nd f( x; ) b th dnsty functon of output x. Lt R( ) b th xpctd rvnu R( ) = xf( x; ) dx. Th Frst Bst Wth vrfbl ffort, th prncpl offrs contrct of th form ( s, ()). Hnc, th prncpl s problm s p = mx [ x-s( x)] f( x; ) dx st.. s( x) f( x; ) dx³ c( ) Pg 8 of 66

2 Snc th IR condton must b bndng, ths problm bcoms problm of socl wlfr mxmzton: p = mx R ( ) -c ( ) -u st.. s( x) f( x; ) dx³ c( ) Snc th prncpl s objctv functon s ndpndnt of th contrct s( x ), ths problm cn b solvd n two stps. Frst, by mxmzng R( ) -c( ) - u, th frst-bst ffort s dtrmnd by R ( ) = c ( ). Scond, gvn th optml ffort, n optml contrct wll hv to stsfy th IR condton only. To fnd n optml contrct, consdr fxd contrct sx ( ) = g. By th bndng IR con- dton, s ( x) = c( ) + u s such contrct. Ths s frst-bst contrct snc t supports th frst-bst ffort. Th Scond Bst Wth unvrfbl ffort, th prncpl wll stll offr contrct of th form ( s, ()). But th prncpl hs to provd ncntvs for th gnt to ccpt ths. For ths purpos, n IC s ntroducd. Hnc, th prncpl s problm s p = mx [ x-s( x)] f( x; ) dx st.. IC : s( x) f ( x; ) dx = c ( ) IR: s( x) f( x; ) dx³ c( ) Snc th IR condton must b bndng, ths problm bcoms problm of socl wlfr mxmzton: p = mx R ( ) -c ( ) -u st.. IC : s( x) f ( x; ) dx = c ( ) IR: s( x) f( x; ) dx= c( ) (.4) Snc th prncpl s objctv functon s ndpndnt of s( x ), ths problm cn b solvd n two stps. Frst, by mxmzng R( ) -c( ) - u, th scond-bst ffort s dtrmnd by R ( ) = c ( ), Pg 9 of 66

3 mplyng =. Scond, gvn th optml ffort, n optml contrct wll hv to stsfy th IC nd IR condtons only. W cn fnd mny contrcts tht cn stsfy ths condtons for gvn. For lnr contrct sx ( ) = bx+ g, th IC nd IR condtons bcom whch mply Ths sx ( ) b xf ( x; ) dx = c ( ), b xf ( x; ) dx + g = c( ) + u, c ( ) b = =, g = u + c( )- br( ) =- p. R ( ) = bx+ g s scond-bst contrct snc t supports th scond-bst ffort. Proposton.4. Wth sngl morl hzrd nd doubl rsk nutrlty, th lnr contrct s ( x) = x- p s optml, mplyng th frst-bst ffort. Intrstngly, ths soluton cn b mplmntd by chng of ownrshp: th prncpl cn smply sll th frm to th gnt for pymnt p. Whn th gnt s th ownr, th ncntv problm dspprs. Ths soluton lds to n mportnt d: ncntv problms my b solvd through n orgnztonl pproch. Cors (960) contrbuts prcsly by proposng ths d. Howvr, n ths chptr, s w mntond t th bgnnng, ll contrcts r trtd s complt contrcts. Wth complt contrcts, n ownrshp trnsfr s not llowd. W wll llow ownrshp trnsfrs from th nxt chptr onwrds whn w us ncomplt contrcts. 8.. Doubl Morl Hzrd W hv so fr llowd th gnt to nvst only. In ths subscton, w llow both prts n jont vntur to nvst. Consdr two gnts, M nd M, nggd n jont projct. Efforts (nvstmnts) nd rspctvly from M nd M r prvt nformton. Lt b th ffort spc. Lt c ( ) nd c( ) b th prvt costs of ffort. Lt h = h(, ) b th jont ffort, nd x = X( w, h) b th x post rvnu dpndng on th stt w nd jont ffort h. Lt x follow dnsty functon f( x, h) x nt. Thus, th xpctd rvnu s R(, ) º xf[ x; h(, )] dx. Assumpton.7. h (, ) s strctly ncrsng n nd. Assumpton.8. c( ) nd c( ) r convx nd strctly ncrsng. Assumpton.9. R(, ) s concv nd strctly ncrsng n nd. Pg 30 of 66

4 Gvn contrct [ s( x), s( x)] tht spcfs pymnts s ( x ) nd s ( ) x to gnts nd, rspctvly, th two gnts ply Nsh gm to dtrmn thr fforts. In othr words, gvn, gnt chooss hs ffort by mxmzng hs own xpctd utlty: mx s ( x) f[ x; h(, )] dx- c ( ), ÎE whch mpls th FOC nd SOC: h s ( x) f [ x; h(, )] dx = c ( ), h h s ( x) f [ x; h(, )] dx+ ( h ) s ( x) f [ x; h(, )] dx< c h hh Smlrly, gvn, gnt chooss hs ffort by mxmzng hs own utlty. Assum tht contrctng ngotton lds to socl wlfr mxmum. Ths ssumpton s stndrd n th ltrtur nd s mposd on ny ngotton outcom. Thn th problm of mxmzng socl wlfr cn b wrttn s V = mx R(, ) -c ( ) -c ( ) s Î S,, Î st.. IC : h s ( x) f [ x; h(, )] dx= c ( ), h IC : h s( x) fh[ x; h(, )] dx= c ( ), SOC:for, SOC :for, RC : s ( x) + s ( x) = x, for ll x Î, + (.4) whr th lst constrnt s th rsourc constrnt (RC). Snc w llow fxd trnsfr n th contrct, IR condtons r unncssry. Th frst-bst problm s mx R (, ) -c( ) -c( ), Î s.t. RC : s ( x) + s ( x) = x, for ll x Î, + whr th contrct nds to stsfy RC only. Proposton.5. Undr Assumptons.7.9, wth doubl morl hzrd nd doubl rsk nutrlty, thr xsts lnr output-shrng rul s ( x) = x, Pg 3 of 66

5 whr c ( ) =, R ( ), tht nducs th scond-bst fforts > 0 dtrmnd by mx R (, ) -c( ) -c( ), Î h (, ) st.. R (, ) = c ( ) + c h (, ) In ddton, 0< <, nd th frst-bst outcom s not obtnbl. Proof. Condtons IC nd IC mply 6 So th problm s quvlnt to h (, ) R (, ) = c ( ) + c sî S,, ÎE h (, ) mx R (, ) -c( ) -c( ) h (, ) st.. R (, ) = c ( ) + c ( ), h (, ) IC, SOC, SOC, RC. (.43) (.44) Ths problm cn b solvd n two stps. Frst, w fnd soluton (, ) from th followng problm: mx R (, ) -c( ) -c( ), ÎE h (, ) st.. R (, ) = c ( ) + c, h( ) (.45) Ths problm s not rltd to contrct. Scond, gvn (, ), w look for contrct s ( x ) tht stsfs IC, SOC, SOC nd RC. Thr r mny such contrcts. Consdr smpl shrng contrct of th form s( x) = x for =, wth c ( ) =. R (, ) It s sy to vrfy tht ths contrct stsfs IC. For RC, by (.43), w hv c c R + =. h h h Thus, 6 Also: h (, ) R (, ) = c ( ) + c h (, ) Pg 3 of 66

6 c c c c R R + = + = + = = =. R R R h xfhdx h xfhdx h xfhdx Condtons IC nd RC hv now bn vrfd. Snc > 0 for both = nd, w must hv 0< <. Furthr, for contrct s( x) = x, w hv ( ) [ ; (, )] = [ ; (, )] = (, ) 0. s x f x h dx xf x h dx R Hnc, by Assumpton.8 nd Assumpton.9, condton SOC s stsfd. Condton SOC cn lso b vrfd smlrly. Fnlly, snc th frst-bst soluton (, ) stsfs R (, ) = c ( ) = c ( ), by condton (.43), th soluton (, ) of problm (.45) cnnot b th frst bst. Th proposton s thus provn. Q.E.D. Th modl n (.4) s modl of two qul prtnrs. An ltrntv stup s prncplgnt modl. If on of th gnts, sy M, s th prncpl nd th othr s th gnt, w nd to dd n IR condton of th form s ( x) f[ x; h(, )] dx- c ( ) ³ u for th gnt nto problm (.4). In ths cs, Proposton.5 stll holds xcpt tht th optml contrct s lnr contrct wth th form s ( x) = x+ b, whr b s dtrmnd by th IR condton. In summry, n optml lnr contrct xsts undr doubl rsk nutrlty; t s th frst bst undr sngl morl hzrd nd t s th scond bst undr doubl morl hzrd. Exmpl.3. Consdr th followng prmtrc cs: h, = +, X h = Ah, c =, ( ) m m ( ) ( ) / whr m, m > 0, A s rndom vrbl wth A > 0 nd E( A ) =. Th scond-bst soluton s Th frst-bst soluton s Q.E.D. 3 3 m + m m + m m + m m m m =, =, =. = m, = m. Pg 33 of 66

7 Thr s no problm of rsk shrng n modl undr doubl rsk nutrlty snc both prts cr bout xpctd ncoms only. Hnc, undr sngl morl hzrd, whn sngl mchnsm ( contrct) s suffcnt to hndl th ncntv problm, th optml soluton chvs th frst bst. Howvr, undr doubl morl hzrd, whn sngl mchnsm s not suffcnt to hndl th two ncntv problms, th optml soluton cnnot chv th frst bst. For th optml lnr soluton n Proposton.5, th vlu of rflcts th rltv mportnc of plyr n th projct. Exmpl.3 ndd shows ths, whr ndvdul s mrgnl contrbuton to th projct s rprsntd by th prmtr m nd th output shr ndd rflcts hs mportnc. Bhttchryy nd Lfontn (995) r th frst to provd such rsult n Proposton.5. Thr rsult s for spcl output procss of th form x = h+ wth spcl dstrbuton functon F( xh ; ) = F ( x-h), whr s rndom shock wth dstrbuton functon F. Km nd Wng (998) r th frst to provd ths gnrl thory on optml lnr contrcts, wth Wng nd Zhu (005) provdng th proof for t. Th optmlty of lnr contrcts n ths scton s bsd on rsk nutrlty. Wthout doubl rsk nutrlty, lnr contrcts r gnrlly not optml. Howvr, lnr contrcts r vry populr n rlty nd most prts nvolvd r lkly to b rsk vrs. Ths s puzzlng wthn th frmwork of complt contrcts. Our thory of ncomplt contrcts n Chptr wll provd rsoluton to ths puzzl. Lnr contrcts must b optml ncomplt contrcts. Pg 34 of 66