Lecture Notes 4: Consumption 1

Transcription

1 Leure Noes 4: Consumpon Zhwe Xu hs noe dsusses households onsumpon hoe. In he nex leure, we wll dsuss rm s nvesmen deson. I s safe o say ha any propagaon mehansm of maroeonom model s losely relaed o he ndvduals onsumpon and nvesmen desons. hus, before we sudy he real busness yle heory, we rs dsuss he onsumpon and he nvesmen hrough some sylzed dynam models. Permanen-Inome Hypohess We sar wh a smple dynam onsumpon-savng model under erany. Consder an ndvdual who lves for perods whose opmzaon problem s max f ;s g X u ( ) = subje o budge onsrans and ermnal onsran + s = y + s ; () s 0: (2) Noe ha he ermnal onsran s ndeed he "no-ponz-game" ondon. he nome ow fy g s deermns and exogenously gven. As he perod s ne, here we do no requre a dsounng rae o avod dvergen lfe-yle uly. o solve he model, we replae onsumpons f g n he uly by fs g and fy g : he opmzaon problem hen an be expressed as max fs g X = u (y + s s ) = max fs g u (y + s 0 s ) + u (y 2 + s s 2 ) + ::: + u (y + s 2 s ) + u (y + s s ) + s ; where s he Lagrangan mulpler for he ermnal ondon (2). Frs order ondons w.r.. s are gven by u 0 ( ) = u 0 ( + ) ; (3) hs noe s parly borrowed from Prof. Y Wen s leure noes a snghua Unversy. I hank hm for kndly sharng hs maerals.

2 2 for = ; ::; : u 0 ( ) = : (4) And he Kuhn-uker ondon gves us s = 0: (5) Noe ha hs ondon s analog o he ransversaly ondon n he n ne horzon model. (3) mples ha = ; for all : (6) As u 0 () > 0; (4) and (5) mply ha > 0 and s = 0: Sne he opmal onsumpon s onsan over he me, from he budge onsran, we have = s 0 + X y, for all : (7) = Inuvely, due o he onave uly funon, he onsumer wans o smooh hs/her onsumpon ow. Sne here s no unerany, he onsumer an fully smooh hs/her onsumpon for eah perod by onsumng he average permanen nome. he opmal onsumpon (7) has several mporan mplaons. Frs, mples he onsumpon manly depends on he average level of he lfe-me nome, hs s beause he average nome s muh smooher han he nome n parular perod. Hene, ransory hanges (or urren hange) n nome do no have sgn an e es on he onsumpon. Seond, savng plays he role of bu er-sok. ha s, any exra nome ha s hgher han he average nome wll be sored o ompensae he perods wh low nome. o see hs, we have Dsussons s s = y = y s 0 +! X y : (8) =. When nome s onsan over me: y = y, we have = s 0 + y; (9) s s = s 0; (0) s = s 0 : () In hs ase, nome per-perod re es he permanen nome and here s no emporary nome, herefore, n eah perod, onsumer uses all of he nome o onsume.

3 3 2. Change of nome. Consder wo ases: a. Expeed hange of nome. Suppose onsumer expes ha he nome n perod wll nrease an amoun of ;.e., ~y = y + ; hen he onsumpon s = s 0 + X y j + ; for all : (2) j= b. Unexpeed hange of nome. Suppose n perod ; he nome suddenly nreases ; hen he onsumpon pah s 8 >< = >: s 0 + s 0 + X y j, for < j= X j= y j + + ; for : (3) 3. Rardan Equvalene. PIH says ha he onsumpon only depends on he permanen nome, hus any poly ha does no hange he nome permanenly would no a e he onsumpon. For example, f he governmen redues ax oday bu wll nrease he ax n he fuure o balane he governmen spendng, and f he onsumers expe hs, hen he onsumpon wll no hange orrespondng o he urren ax reduon. 2 Consumpon under Unerany Suppose ha he nome fy g follows a sohas proess, and he onsumer s opmzaon problem s gve by max E 0 subje o budge onsran: X u ( ) + s = y + s ; = 0; ; 2; :::; ; (4) and ermnal ondon s 0: (5) Pung he budge onsran no he uly funon yelds " # X max E 0 u (y + s s ) + s : he F.O.Cs are u 0 (y + s s ) = E u 0 (y + + s s + ) ; (6)

4 4 for = 0; :::; : and Equaon (6) mples or u 0 (y + s s ) = ; (7) s = 0: (8) u 0 ( ) = E u 0 ( + ) ; for all ; (9) u 0 ( ) = u 0 ( ) + " ; (20) where " s..d. whe nose sas es E (" ) = 0. he above proess s alled random walk. Example: suppose he uly funon akes he form hen he opmal onsumpon sas es u () = 2 2 : (2) E + = ; (22) or equvalenly = + " : (23) o deermne wha he " s, pu las equaon no budge onsran (4) and ake expeaon on boh sdes, we have and = + s + + E X y + ; (24) s = y + s = y + + s for all : Smlarly, for f ; s g we have + E X y + ; (25) = + 2 s E X+ y + : (26) s = y s E X+ y + (27)

5 5 Pluggng las equaon no (24), we ge = = + s s 2 + E X " + y E X+ # X y + E y + y : (28) Fnally, from (28) and (26), we an show ha = + E! X X y + E y + ; (29) Las equaon ndaes ha he " s ndeed he predon error for he permanen nome. here s a very mporan mplaon n hs example. Reall ha he opmal onsumpon s gven by (24), whh mples ha when makng onsumpon hoe, he onsumer only ares abou he expeaon of he nome ow. he rsk or he volaly (seond-order momen of y ) does no a e onsumer s deson. We all hs ype of deson as "Cerany Equvalene". Noe ha, he erany equvalene only appears when he margnal uly u 0 () s lnear n ; or u 000 = 0: o see hs, le us look a F.O.C (9). If u 0 () s lnear, hen we have u 0 ( ) = E hu 0 ( ) = u 0 (E [ ]) ; (30) ha s = E [ + ] ; for all. (3) However, f u 0 () s no lnear, or u 000 () > 0; hen we have u 0 ( ) = E hu 0 ( + ) > u 0 (E [ + ]) ; (32) or < E [ + ] : (33) In hs ase, he onsumpon s less han ha n "Cerany Equvalene" ase, or n oher words, he savng n hs ase s relavely large. We all he "over-savng" behavor n hs ase as "preauonary savng". 2. Preauonary Savng n he In ne Horzon Model Le us onsder an n ne horzon model. Consumer s opmzaon problem s X max E 0 ; > 0;

6 6 subje o budge onsran F.O.C s gven by + s = ( + r) s + y : (34) h = ( + r) E + : (35) For smply, we assume ha he onsumpon f g follows log-normal dsrbuon,.e. log + ~ N E (log + ) ; E 2 ; (36) hen we have E ( + ) = e E(log +)+ 2 E2 : (37) Sne = e log ; we have Pung (38) no (35) and akng log on boh sdes, we have E + = e E(log +)+ 2 2 E2 : (38) log = log [ ( + r)] E (log + ) E 2 ; (39) or E log + + ' E = log [ ( + r)] + 2 E 2 : (40) Preauonary savng mples ha he expeed onsumpon growh s a eed by he unerany. 2.2 Preauonary Savng under Cred Consran Indeed, he preauonary savng behavor does no neessarly requre onvex margnal uly, as long as he eonomy su ers red onsran. o see hs, le us onsder a smple hree-perod model. he onsumer s opmzaon problem s max E [u ( ) + u ( 2 ) + u ( 3 )] subje o budge onsrans + s = y + s 0 ; 2 + s 2 = y 2 + s ; 3 + s 3 = y 3 + s 2 ; and ermnal ondon s 3 0; (4)

7 7 Besdes, we nrodue red onsran for perod 2 s 2 0: (42) As we already know ha whou red onsran, he opmal onsumpon n perod s = 3 s E (y + y 2 + y 3 ) : (43) Now f we onsder he red onsran, hen he problem beomes 2 max E 6 4 FOCs are 3 u ( ) + u ( 2 ) + u ( 3 ) + (y + s 0 s ) + 2 (y 2 + s s 2 2 ) : (y 3 + s 2 s 3 3 ) 5 + s s 3 u 0 ( ) = ; u 0 ( 2 ) = 2 ; u 0 ( 3 ) = 3 ; = E 2 ; 2 = + E 2 3 ; 3 = 2 : I an be shown ha u 0 ( ) = E + E u 0 ( 3 ) : (44) Sne he Lagragan mulpler 0; we have u 0 ( ) = E u 0 ( 2 ) E u 0 ( 3 ) : Suppose u 0 () s lnear, hen we have = E ( 2 ) : E ( 3 ) : Furhermore, from he budge onsran, we have + E 2 + E 3 = s 0 + E (y + y 2 + y 3 ) ; (45) or 3 s E (y + y 2 + y 3 ) : (46)

8 8 3 Consumpon and Rsky Asses So far, we only onsder he rsk-free asse, ha s, he neres rae r + s deermns n perod : We now onsder a bunh of asses (ndexed by ) wh sohas raes of reurn. F.O.C for asse s gven by = E h + r+ + g + : (47) For he erm + r + + g + ; afer seond-order aylor expanson around r + = g + = 0; we have + r+ + g + = + r + g+ r+g + ( + ) + 2 And hus for any ; we have = E r+ ( + ) + 2 E g+ E r+ E g+ ov r+; g+ g + 2 : (48) E g+ 2 ( + ) + var g 2 + : (49) If we gnore erms E r + E g + and E g + 2 whh are very small, he d erene of raes of reurn beween asse and j s E r+ E r j + = h ov r+; g+ ov r j + ; g + = ov r+ r j + ; g + = orr r+ r j + ; g + sd r+ r j + sd g+ : (50) If we ake j as rskless asse, hen he LHS s he rsk premum. In he U.S. daa ( ), he rsk premum s abou 7%, and he orrelaon beween r+ r + and g+ s abou 0.27, he sandard devaon of exess reurn s 4.4%, and he sandard devaon of onsumpon growh rae s.%. herefore, he mpled oe en of relave rsk averson s = 0:07 = 63: (5) 0:27 0:0 0:44 In order o explan he rsk premum observed n he daa, he o en mus be unreasonably large, hs s so-alled rsk premum puzzle.