DEPOCEN Workng Paper Seres No. 28/2 No arbtrage condton and exstence of equlbrum n nfnte or fnte dmenson wth expected rsk averse utltes Tha Ha Huy * Cuong Le Van ** Manh Hung Nguyen *** * Unversté Pars 1, CNRS CES, 16 Bd de l Hôptal, 7513 Pars, France. ** PSE, Unversty Pars1, CNRS CES, 16 Bd de l Hôptal, 7513 Pars, France. *** Toulouse School of Economcs, LERNA-INRA. The DEPOCEN WORKING PAPER SERIES dssemnates research fndngs and promotes scholar exchanges n all branches of economc studes, wth a specal emphass on Vetnam. The vews and nterpretatons expressed n the paper are those of the author(s) and do not necessarly represent the vews and polces of the DEPOCEN or ts Management Board. The DEPOCEN does not guarantee the accuracy of fndngs, nterpretatons, and data assocated wth the paper, and accepts no responsblty whatsoever for any consequences of ther use. The author(s) remans the copyrght owner. DEPOCEN WORKING PAPERS are avalable onlne at http://www.depocenwp.org
No arbtrage condton and exstence of equlbrum n nfnte or fnte dmenson wth expected rsk averse utltes Tha Ha Huy a, Cuong Le Van b, Manh Hung Nguyen c a Unversté Pars 1, CNRS CES, 16 Bd de l Hôptal, 7513 Pars, France. b PSE, Unversty Pars1, CNRS CES, 16 Bd de l Hôptal, 7513 Pars, France. c Toulouse School of Economcs, LERNA-INRA. July 5, 28 Abstract We consder a general equlbrum model n asset markets wth a countable set of states and expected rsk averse utltes. The agents do not have the same belefs. We use the methods n Le Van - Truong Xuan (JME, 21) but one of ther assumpton whch s crucal for obtanng ther result cannot be accepted n our model when the number of states s countable. We gve a proof of exstence of equlbrum when the number of states s nfnte or fnte. Keywords: No-arbtrage Condtons, the two-perod wealth model, No Unbouded Arbtrage, Weak No Market Arbtrage. JEL Classfcaton: C62, D5, D81,D84,G1. E-mals: tha.ha-huy@m4x.org, levan@unv-pars1.fr, nguyen@toulouse.nra.fr 1
1 Introducton Expected utlty wth addtve probablty theores, e.g, Savage s (1954) and Anscombe and Aumann s (1963) are known as standard formulaton of decson under uncertanty. Snce the semnal paper of Hart (1974), the queston of exstence of equlbrum n the unbounded securtes exchange model has been a subject of much development. In fnte dmenson economes, one of a crucal assumpton nterpreted as a no-arbtrage-condton be used to prove the compactness of the ndvdually ratonal utlty (see,e.g, Werner 1987, Nelsen 1989, Page and Wooders 1996, Allouch et al. 22). Ths assumpton together wth other standard assumptons are suffcent condton for the exstence of equlbrum. However, n nfnte dmenson economes, the no-arbtrage condton are not suffcent to ensure the compactness of the utlty set. Therefore, to fnd the condtons for whch the compactness of utlty set holds s nterested by many authors. (e.g,cheng 1991, Dana et al, (1999), Dana and LeVan 2). Recently, Le Van and Truong Xuan (21) have proved the compactness of utlty set ( and hence the exstence of equlbrum followed), n asset market wth consumpton set equal to L p, separable utltes and the contnuum states whch belong to [,1]. Followng ths drecton, we consder a general equlbrum model n asset markets wth a countable set of states and expected rsk-averse utltes. The agents do not have the same belefs. We use the methods n Le Van - Truong Xuan (21) but one of ther assumpton whch s crucal for obtanng ther result cannot be accepted n our model when the number of states s countable. Moreover, by assumng the exstence of a common margnal utlty prce, the proof we gve s more natural and smple than the one gven n Le Van and Truong Xuan (21). The exstence of a quas-equlbrum n L 1 can be also derved. The paper s organzed as follows. In Secton 2, we gve a proof of exstence of equlbrum n a model wth expected rsk-averse utltes the number of states s nfntely countable. Secton 3 we consder the case of contnuum states as n Le Van and Truong Xuan (21) but we relax one of ther crucal assumpton. Secton 4 prove the exstence of equlbrum n the case of fnte number of states by explotng the smlarty of NUBA and WNMA. 2 The model wth nfntely countable states Frst, we consder the case where the set of states possble s countable. There are m agents ndexed by 1,..., m. Each agent has a probablty (πs) n the set := {π R : + π s = 1}. Let us denote the probablty π = 1 m m =1 π, a consumpton set X = L p (π) wth 1 p and an endowment e L p (π). We assume that for each agent, there exsts a concave, strctly 2
ncreasng functon u from R to R and consumer choose a portfolo x = (x s) Lp (π) to solve the problem max U (x ) = max πsu (x s) We recall the noton set of ndvdually ratonal attanable allocatons A s defned by A = {(x ) (L p ) m m x = =1 m =1 e The ndvdually ratonal utlty set U s defned by and U (x ) U (e ) for all.} U = {(v 1, v 2,..., v m ) R m x A s.t U (e ) v U (x ) for all.} Let us denote, for each agent, a := nf u (z), b := sup u (z). Assumpton 1 p (L p ), (λ ) R m ++, (x 1,..., x m ) A such that:, s, p s = λ π su (x s) and nf s u (x s) = m > a sup u (x s) = M < b s Remark From the assumpton 1, we know that all the probabltes π are equvalences and hence equvalences wth π. Assumpton 2 For all = 1, 2,..., m, b = +. Proposton 1 Wth the assumpton 1 there exsts C > such that for all (x 1,..., x m ) A, we have: for all. + p s x s < C Proof: From the condton a < m = nf s u (x s) sup s u (x s) = M < b, there exst η > such that for all. Then we defne the prce q such that,, j, a < u (x s)(1 + η) < b (1) q s = λ π su (x s)(1 + η) = λ j π j su j (x j s)(1 + η). 3
It follows from ( 1) that, for each, there exst z L such that s, q s = λ u (z s). Note that s, p s < q s. Denote { x + : = { x : = x f x > f x x f x < f x From the concavty of the utlty functon u we have λ πsu (x s) λ πsu (x + s ) λ Therefore, λ πsu (z s) λ πsu ( x s ) λ π su (x s)(x s x + s ), π su (z s)(z s + x s ) λ πsu (z s)x λ πs[u (z s) + u (x s) u (x + s ) u (x s )] whch mples λ πsu (z s)z s + λ πsu (x s)x + s. q s x s λ [U (z ) + U (x ) U (x )] λ [U (z ) + U (x ) U (e )] q s z s + + q s z s + p s x + s p s x + s. Hence,, where + (q s p s )x s C + + p s x s C = λ [U (z ) + U (x ) U (e )]. Thus we have m =1 (q s p s )x s = m m C + p s x s =1 =1 m C + p s e s = C 1. =1 4
Snce x s,, s, we get =1 for all. We also have m (q s p q )(x + s whch mples m (q s p s )x + s = =1 (q s p s )x s C 1 x s ) = (q s p s )e s (q s p s )e s C 2 m =1 (q s p s )x s Thus whch mples (q s p s ) x s C 1 + C 2 =: C η p s x s C. Remark 1. The condton (x ) L s not suffcent for the exstence of Assumpton 1, because the utlty functon can be lnear by peces. 2. In the proof of boundedness above, we used the property U (x ) U (e ). However, We can use a weaker assumpton that there exsts a constant a such that U (x ) a for all. Note that p s = λ πsu (x s) λ πsm, so there exst the constant C > such that: πs x s C wth all (x 1, x 2,..., x m ) A. From ths property and by usng Jensen s nequalty, we have the followng Lemma Lemma 1 There exsts C > such that for all (x 1, x 2,..., x m ) A, U (x ) < C Thus we get the followng Theorem whch wll be used later. Theorem 1 Wth the assumptons 1 and 2, n the case X = L (π), for all ɛ >, there exsts N > such that πs x s < ɛ s=n for all (x 1, x 2,..., x m ) A, for all. 5
Proof: Assume the contrary, there exsts a sequence (x 1 (n), x 2 (n),..., x m (n)) A and a constant c > such that x s(n) > c s=n Wthout loosng of the generalty, we can suppose that s=n x s(n) c >. We can assume that there exsts lm n s=n π s x s = c > lm n s=n π sx + s (n) lm n s=n π sx s (n) = c. For every s, j such that x j s(n) < x s e s m 1. There s an fnte j, then for the smplcty, we can assume that there exsts j fxed such that and j satsfes the propertes: 1. En N {s n}, x s > for all s and lm n s En πsx s = c > 2. For all s E n x j s(n) x s(n) e s Wth each M >, defne the set S n E n the states s satsfes: x s(n) e s > M and for all s S n we have: x j s(n) e s x s(n) < M We can see that lm n + π sx s = c. The two probabltes π and π j are equvalent, then we have lm n πj sx s = c j >. Now consder the sequence (y 1 (n), y 2 (n),..., y m (n)) defned by: y s(n) := x s(n) x s e s y j s(n) := x j s(n) + x s e s + M wth s S n M wth s S n y k s = x k s wth every k, j or s / S n. Remarks that y s(n) x s(n) and y j s(n) x j s(n) for all s. We wll prove that (U l (y l (n))) l=1,m s bounded below, but s not bounded above, that leads us to 6
a contradcton wth the Lemma 1. U (y (n)) U (x (n)) = πs(u (ys(n)) u (x s(n))) u (M) π su (x s(n) x s e s + M)( x s e s + M) π su (M)( x s(n) ) + u (M)( e s + M) π sx s(n) + u (M)( e s + M) π s π s Let n + we have: lm nf U (y (n)) v u (M)c n + so for great n, U (y (n)) s bounded below. Now we wll see U j (y j (n)). U j (y j (n)) U j (x j (n)) = πs(u j j (ys(n)) j u j (x j s(n))) π j su j (x j s(n) + x s(n) e s M)( x s(n) e s M) U j (y j (n)) U j (x j (n)) So we have the lmt π j sb j ( x s(n) e s uj ( M) M) πsx j s(n) M lm nf U j (y j (n)) v j + uj ( M)c j n + s S,j N So f b = +, we can choose M very large, and the lmt of U j (y j (n)) s unbounded above: a contradcton. The next Lemma show that the sum s bounded unformly. Lemma 2 If p =, for all ɛ >, there exst N > such that for all (x 1, x 2,..., x m ) A, for all, πsu (x s) < ɛ s=n π s 7
Proof: Fxe a R arbtrarly, we have u (a) u (x s) u (a)(a x s), so wth every N >, πsu (x s) [u (a) u (a)a] s=n πs + u (a) s=n πsx s From the Lemma above, n choosng N large suffcently, we have sn π su (x s) < ɛ wth every (x 1, x 2,..., x m ) A. We know that U s bounded. Suppose that there exsts a sequence n U (v 1 (n), v 2 (n),..., v m (n)) (v 1, v 2,..., v n ). We have to fnd that f (v 1, v 2,..., v m ) A. Denote the sequence (x 1 (n), x 2 (n),..., x m (n)) A such that U (x (n)) = v (n) for all. s=n Theorem 2 Under the Assumptons 1 and 2, U s closed for every p. Proof: Note that L L p for every 1 p. We have two cases: I There exsts M >,N > such that for all n > N, for all, s: x s(n) < M. II For all M >, there exsts n,, s such that x s(n) > M. Consder the frst case. From the Theorem 1, we know that A s the subset of a compact set of the product topology, then we can assume that x (n) y n ths topology for all. For all s, lm n x s(n) = y s. y s M for all, s, then y L for all. For all ɛ >, choose N > n the theorem 1 and the lemma 2, such that the sum s=n u (x s(n)) < ɛ, we have: u (x s(n)) = N πsu (x s(n)) + πsu (x s(n)) N πsu (x s(n)) + ɛ s=n+1 lm n πsu (x s(n)) for all N suffcently large, we have: lm n πsu (x s(n)) N u (ys) + ɛ πsu (ys) for all, U (y ) v (v 1, v 2,..., v m ) U. Then we consder the second case. Suppose that for every M > there exsts and an nfnte n such that x s(n) > M wth an s, wthout losng the 8
generalty, we can assume that s true for all n and for an fxed. Choose M suffcently large such that for all, denote T n = {s : x s < M 1}, we have s T n π s > 1 2 Wth each M >, denote the sets E n, S n as above. Choose M suffcently large such that M > M and S n T n =. π and π j are equvalents, so there exst h > such that: We can choose M such that: hπ s π j s 1 h π s u j ( M) > uj ( M) h 2 We consder two cases: 1. t Sn for an nfnte n. 2. lm n mn Sn =. Consder the frst case. Then lm nf s Sn π s >. Wthout lost of generalty, we can suppose that there exst s S a = lm π n sx s n () s Tn π s Remarks that < a < 1 2. Defne the sequence (y (n)): ys(n) = x s(n) x s(n) e s + M f s Sn ys(n) = x s(n) + a f s Tn y s(n) = x s(n) others cases ys(n) j = x j s(n) + xj s(n) e s M f s Sn ys(n) j = x j s(n) a f s Tn y j s(n) = x j s(n) others cases And y k (n) = x k (n) for all k, j. Easly, we see that m k=1 yk (n) = e. We are estmatng U (y (n)) and U j (x j (n)). πs[u (ys(n)) u (x s(n))] + U (y (n)) U (x (n)) = s T n au (M) s T n π s[u (y s(n)) u (x s(n))] πsu (x s(n) + a) πsu (x s(n) x s(n) e s + M)[ x s(n) e s M] s Tn s Sn πs u (M) πs[ x s(n) e s M] au (M) s T n π s u (M) s Sn π x s s + u (M) s Sn s Sn 9 πs[ e s + M] >
wth n suffcently great. So we have the result that lm nf n U (y (n) > lm nf n U (n) = v. U j (y j (n)) U j (x j (n)) = s Tn a πsu j j (x j s(n) a) + s Tn s Sn πs j + u j ( M) au j ( M) s T n au j ( M) s T n a h uj ( M) s T n πs[u j j (ys(n)) j u j (x j s(n))] + πs[u j j (ys(n)) j u j (x j s(n))] π su j (x j s(n) + x s(n) e s π j s + u j ( M) π s + hu j ( M) π j s π j s x s x s π s x s M)[ x s(n) e s M] We know that u j ( M) > u j ( M)/h 2. Then lm nf n U j (y j (n)) > lm nf n U j (x j (n)) = v j too. Now we wll show that we can construct a sequence (z k (n)) such that lm n U k (z k (n)) > v k. Choose k, j above. Choose ɛ > very small such that lm nf n U (y (n)) ɛu ( M) > v. Choose t T n arbtrarly, we defne the new sequence (z l (n)) as: z t(n) = y t(n) ɛ z k t (n) = y k t (n) + ɛ z l s(n) = y l s(n) n others cases U (z (n)) U (y (n)) = π t[u (y t(n)) u (z t(n))] and then lm nf n u (z (n)) > v. π tu (y t(n) ɛ)ɛ > u ( M)ɛ U k (z k (n)) U k (y k (n)) = π k t [u k (y k t (n)) u k (z k t (n))] π k t u k (y k t (n) + ɛ)ɛ > π k t u k (M)ɛ then lm nf n U k (z k (n)) > v k. By the nducton, we can construct the sequence (z (n)) such that m =1 z (n) = e and lm nf n U (z (n)) > v for all. Then there exst n such that U (z (n)) > v for all = 1, m (v 1, v 2,..., v m ) U. Now we return to the case lm n nf S n = +. In ths case, we wll construct 1
a sequence satsfy the propertes: lm nf U (y (n)) = v and sup n sup s ys(n) < +. If those propertes are true for a sequence (x (n)), we have nothng to do, n the converse case, there exst such that for all M, there exst an nfnte n and s s.t x s(n) > M. Defne, M, Sn, as above, remarks that j xj s(n) = e s x s(n) <. Then we have j xj+ s (n) < j xj s (n). Then there exsts a sequence zs(n) x j s (n) such that j zj s(n) = j xj+ s (n). We defne the sequence (y (n)): y s(n) = e s f s S n y s(n) = x s(n) f s / S n y j s(n) = x j s(n) + z j s(n) f s S n y j s(n) = x j s(n) f s / S n We can check that m k=1 yk (n) = e. We have nf Sn +, so from the Lemma 2 U (y (n)) U (x (n)) πs u (y (n)) u (x (n)) snf S n and U j (y j (n)) U j (x j (n)) = πs[u j (ys(n)) j u j (x j s(n))] πsu j j (xs(n) j + zs(n))z j s(n) j u j () πsz j s(n) j So lm n U (y (n)) = v and for n great enough, for all s, we have y s(n) M() e s. By nducton, n applyng the same method, we can construct our sequence wth the propertes desred. We have the sequence (y (n)) A satsfy: lm U (y (n)) = v n M > such that y (n) < M From Proposton 1, we can suppose that lm y (n) = y n the L 1. y (n) < M y L for all. Then we have (y 1, y 2,..., y m ) A wth U (y ) v, then (v 1, v 2,..., v m ) U. U s closed and bounded n L p, so U s compact. Now we wll drop the condton of Assumpton 2, b = + for all, we wll prove that wth only Assumpton 1, there s an quas-equlbrum n L 1. Theorem 3 Wth Assumpton 1, there s an quas-equlbrum n L 1. 11
Proof: We construct the sequence of utltes concave functons u N : R R such that u N (x) = u (x) wth x [ N, + ), for all N, u N ( ) = and u N u N+1. Remark that x, lm N u N (x) = u (x). From the Theorem 2 and [2] we know that for N suffcently large such that x s, e s [ N, + ), s, there exsts an equlbrum general (p (N), x (N)). From the Theorem 1, we know that (x (N)) s n a compact set of the topology L 1, and for all ɛ >, there exst N > such that for all N we have: πs x s (N) < ɛ sn n L 1, the sequence (x (N)) converge to (x ). And the prce sequence p (N) converge to p. Suppose that there exst x L 1 such that U (x ) > U (x ). Choose < ɛ < U (x ) U (x ). There exst M > such that lm M πsu (x s) > U (x ) + ɛ N so for N suffcently large we have M πsu N(x s) > U (x ) + ɛ M πsu N(x s) > U (x ) + ɛ We can choose M very large such that s>m π s x s (N) < ɛ for every N. Construct the sequence (x (N)) satsfy: x s(n) = x s for s M, x s(n) = x s (N) f s > M, then we have U (x (N)) > U (x ) p (N).x (N) > p (N).x for every N suffcently large. Let M and N tend to nfnty, we have p.x p.x. 3 The model wth contnuum states In ths secton, we wll gve a proof wth a smlar result as the secton above. In usng a utlty functon less general than [4], we can have the result wthout the assumpton H4 n ther paper. The set of states we use here as Le-Van and Truong-Xuan, the set [, 1], the consumpton set s L p ([, 1]), 1 p, each agent has an endowment e (s), utlty functon under the form U (x ) := We defne A and U as n the secton above. u (x (s))h (s)ds 12
Assumpton 3 For all, j, a < b j. Assumpton 4 < m < nf [,1] h (s) sup [,1] h (s) M < + Assumpton 5 For all, u s concave and u ( ) = +. Theorem 4 Under the Assumpton 4 and the Assumpton 5, there exsts equlbrum. Lemma 3 Assume that x L p, = 1, m s.t m =1 x (s) = m =1 e (s) for all s, U (x ) U (e ) for all, then there exst C > such that for all : x (s) h (s)ds < C Proof: We wll usng the same method as the secton 1. Note d = h (s)ds. Choose a > b such that c 1 = max u (a) < c 2 = mn u (b). We have: u (b) [mn j u (a)h (s)ds u (b)h (s)ds u (x + (s))h (s)ds u (a) u ( x (s))h (s)ds u (b) (a x + (s))h (s)ds (b + x (s))h (s)ds x (s)h (s)ds [u (a) au (a) bu (b)]d U (e ) + max u j (a) j u j (b) max u j (a)] j x (s)h (s)ds < m =1 So we have for all, x (s)h (s)ds C + max u j (a) j x (s)h (s)ds C mn j u j (b) max j u j (a) + max j u j (a) mn j u j (b) max j u j (a) < C 1 + C 2 x (s)h (s)ds x (s)h (s)ds < mc 1 + C 2 e (s)h (s)ds =: X x (s)h (s)ds < mc 1 + C 2 e (s)h (s)ds =: X x + (s)h (s)ds x (s)h (s)ds 13
m =1 x + (s)h (s)ds = x + (s)h (s)ds = Then we have x (s) h (s)ds = e (s)h (s)ds + e (s)h (s)ds + m =1 m x + (s)h (s)ds + =1 x (s)h (s)ds < Y x (s)h (s)ds < Y x (s)h (s)ds < C Lemma 4 U s bounded. Proof: U s bounded below, from the defnton of U. We wll prove that U s bounded above. Suppose (x 1,..., x m ) A, u s concave, ncreasng, so we have: u (x (s))h (s)ds < d u ( x (s) h (s) d ds) < d u ( C d ) Theorem 5 U s closed. Proof: Suppose that there exsts a sequence (x 1 n, x 2 n,..., x m n ) A, lm n U (x n) = v, we have to prove that (v 1, v 2,..., v m ) U. Frstly, we show that (x n) s weakly compact n σ(l 1, L ). Suppose the converse, then there exsts a sequence X n [, 1] wth the Lebegue measure µ(x n ) and lm nf n X n x n(s) h (s) > for some. Wth each s, there exsts j such that x j n(s) x n(s) e(s) m 1. Wthout loosng the generalty, we can fxe, and suppose that on the sequence E n, x n(s) > and x j n(s) x n(s) e(s) m 1. Then we can fxe, j and a subset E n such that: x j n(s) x n(s) e(s) for all n and all s E n lm x n n(s)h (s)ds = c > E n Wth each M very large, we defne the set S n E n = {s : x n (s) e(s) m 1 > M}. Note that lm n S n x n(s)h (s)ds = c. Defne the new sequence (yn(s)) k as below: yn(s) = x n(s) x n(s) e(s) + M on S n yn(s) j = x j n(s) + xj n(s) + e(s) M on S n yn(s) k = x k n(s) wth other k or s 14
Note that k yk = e. As n the secton above, we wll estmatng U k (y k )) wth k =, j. U (yn) U (x n) u (yn(s))(y n(s) x n(s))h (s)ds S n u (M)( x n(s) e(s) )h (s)ds S n U j (y j n) U j (x j n) lm nf n U (y n) v u (M)c u j (yn(s))(y j n(s) j x j n(s))h j (s)ds S n u j ( M)( x n(s) e(s) )h j (s)ds S n lm nf n U j (y j n) v j + vj ( M)c j and lm n U k (yn) k = v k for others k. So we have constructed the sequence (yn) k wth U k (yn) k s bounded below, and f we let M, our sequence s unbounded above because u j ( ) =, that leads us to a contradcton. Then the sequence (x 1 n, x 2 n,..., x m n ) s σ(l 1, L ) compact. Wth each M, denote the set T n = {s : x n(s) < M for all }. We can choose M suffcently large such that Lebesgue measure µ(t n ) > 1 2. Choose M very large such that for all, u ( M)h 2 < h 1 u ( M). Defne En = {s : x n(s) e(s) m 1 > M}. Frstly, we consder the case that there exsts, lm nf n µ(en) >. Suppose not, then we can fnd such that lm n µ(en) = c >. Wthout loosng the generalty, we can assume x n(s) > on En. Usng the same argument as above, we assume that there exst j and S n En satsfy: Construct the sequence (y k n) as: x j n(s) x n(s) e(s) for all s S n lm µ(s n) = c > n y n(s) = x n(s) + α on T n yn(s) = x n(s) x n(s) e(s) + M on S n yn(s) j = x j n(s) α on T n yn(s) j = x j n(s) + x n(s) e(s) M on S n yn(s) k = x k n(s) for others k or s 15
Now we estmate U (yn) and U j (yn): j U (yn) U (x n) α u (x n(s) + α)h (s)ds T n u (x n(s) x n(s) e(s) + M)( x n(s) e(s) + M) S n αu (M) h (s)ds u (M) ( x n(s) e s M) S n S n then we have lm nf n U (yn) > lm n U (x n) = v. U j (yn) j U j (x j n) α u j (x j n(s) α)h (s)ds + T n u j (x j n(s) + x n(s) e(s) M)( x n(s) e(s) M) S n α u j ( M)h 2 ds + h 1 u j ( M)ds T n S n We have u j ( M) > h 2 /h 1 u j ( M), then we have lm nf U j (y j n) > v j. We have constructed the sequence (y k n) such that k yk n = e, lm nf U k (y k n) v k wth the strct nequalty when k =, j. Choose ɛ > such that lm nf n U (y n) ɛu ( M)h 2 > v. Fx k, defne a new sequence (z l n) as: z n(s) = y n(s) ɛ z k n(s) = y k n(s) + ɛ z l n(s) = y l n(s) n others cases Wth the sequence (zn), l we have: U (zn) U (yn) ɛ u (yn(s) ɛ)h (s)ds T n ɛu ( M)h 2 lm nf n U (z (n)) lm nf n U (yn) ɛu ( M)h 2 > v. U k (zn) k U k (yn) k ɛ u k (yn(s) k + ɛ)h k (s)ds T n 1 2 ɛuk (M)h 1 > lm nf n U k (zn) k > v k. By nducton, we can construct the sequence (zn) k such that for all k, lm nf n U k (zn) k > v k there exsts n such that for all k, U k (zn) k > v k (v 1, v 2,..., v m ) U. Now we consder the case for all, lm n µ(en) =. In ths case, we wll construct a sequence satsfy the propertes: lm nf n U (yn) = v and sup n sup s yn(s) < +. If those propertes are true for a sequence (x n), we 16
have nothng to do, n the converse case, there exst such that for all M, there exst an nfnte n wth s En, µ(en) >, s.t x n(s) > M. Defne, M, S n, as above, remarks that j xj n(s) = e s x n(s) <. Then we have j xj+ n (s) < j xj n (s). Then there exsts a sequence zn(s) x j n (s) such that j zj n(s) = j xj+ n (s). We defne the sequence (yn): yn(s) = e(s) f s Sn yn(s) = x n(s) f s / Sn yn(s) j = x j n(s) + zs(n) j f s Sn yn(s) j = x j n(s) f s / Sn We can check that m k=1 yk n = e. We have µ(sn) +, so from the Lemma 2 U (yn) U (x n) u (yn) u (x n) h (s)ds S n and U j (yn) j U j (x j n) = [u j (yn(s)) j u j (x j n(s))]h (s)ds Sn u j (x j n(s) + zn(s))z j n(s)h j j (s)ds Sn u j () zn(s)h j j (s)ds S n So lm n U (y (n)) = v and for n great enough, for all s, we have y n(s) M() e(s). By nducton, n applyng the same method, we can construct our sequence wth the propertes desred. We have the sequence (y (n)) A satsfy: lm U (y n n) = v M > such that yn < M Then we have the sequence (y n) s σ(l, L 1 ) compact. We can suppose that y n y L. And U (y ) v for all (v 1, v 2,..., v m ) U. Theorem 6 U s compact. Proof: From Lemma 4 and Theorem 5. 17
4 The case of fnte countable states There are m agents ndexed by 1,..., m, each agent has a consumpton set X R k, a vector of endowment e and a contnuous concave utlty functon u : R k R. We frst recall some standard concepts of general equlbrum theory. The set of ndvdually ratonal attanable allocatons A s defned by A = {(x ) (R k ) m m x = =1 m =1 e and u (x ) u (e ) for all.} Defnton 1 A par ((x )m =1, p ) A R k s a contngent Arrow - Debreu equlbrum f 1. for each agent and x R k, u (x ) > u (x ) mples p x > p x, 2. for each agent, p x = p e. For x R k, let P (x) = {y R s u (y) u (x)} and let R be ts recesson cone. R s called the set of useful vectors for and s defned as R = {w R S u (x + λw) u (x), for all λ } The lnealty space of s defned by L = {w R l u (x + λw) u (x), for all λ R} = R R Elements n L wll be called useless vectors. The no unbounded arbtrage condton denoted from now on by NUBA s ntroduced by Page (1987). Defnton 2 The economy satsfes the NUBA condton f m =1 w = and w R for all mples w = for all. There exsts a weaker condton, called the weak no market arbtrage condton (WNMA), ntroduced by Hart[1974]. Defnton 3 The economy satsfes the WNMA condton f m =1 w = and w R for all mples w L for all. 18
We wll prove the propostons that gve us the smlarty under the NUBA condton and the WNMA condton. Choose θ suffcently large such that ê θ for all. Defne T θ := {t L t θ}. We defne the new economy Ẽ θ = ( X θ, ũ, e ) such that X θ := L T θ, ũ : R k R defned as the restrcton of u on X θ. Evdently, we have e X θ for all. Proposton 2 If (( x )m =1, p ) s an equlbrum of Ẽ then (( x )m =1, p ) s equlbrum of E. Proof: We frst prove that p m =1 L. For each, there exst ɛ such that u ( x +ɛ ) > u ( x ). y T, u ( x +ɛ +y ) > u ( x ) p.( x +ɛ +y ) > p. x. Let ɛ, we have p.y. Wth the smlar argument, we found that p.( y ) p.y = y T θ p L. Observe that (( x )m =1, p ) s equlbrum of Ẽ x = e. Now let u (x ) > u ( x ) u (x ) > u ( x ) p.x > p. x p.(x + ˆx ) > p. x. So (( x )m =1, p ) s equlbrum of E. Proposton 3 If ((x )m =1, p ) s an equlbrum of E, then there exsts θ > such that ((x )m =1, p ) s equlbrum of Ẽθ. Proof: Choose θ max{ x, ê }. Proposton 4 The economy E satsfes Weak No Market Arbtrage condton f and only f Ẽ satsfes No Unbounded Arbtrage condton. Proof: Frstly, suppose that E satsfes WNMA condton. In the economy Ẽ, L θ = {}, so Rθ = R L. Suppose that w R θ such that w = w L for all, w L L w =. Suppose that Ẽ satsfes NUBA condton. If w R such that w =, then we have w = w = for all from the NUBA propertes w L. Now we defne the noton of no-arbtrage prce as n Allouch, Le Van, Page (22) and the NAPS noton: { } {p L p.w >, w (R L Defnton 4 S = )\{} f R \L } L f R = L Defnton 5 The economy E satsfes the NAPS condton f S. 19
Proposton 5 (Page and Wooders, 1996) Assume L = {},, then NUBA NAPS. Proof: In [5] Proposton 6 (Allouch, LeVan and Page (22)) WNMA S. Proof: In [1] References [1] Allouch, N., C.Le Van and F.H.Page (22): The geometry of arbtrage and the exstence of compettve equlbrum. Journal of Mathematcal Economcs 38, 373-391. [2] Dana, R.A, C.Le Van and F.Magnen (1999): On the dfferent notons of arbtrage and exstence of equlbrum. Journal of Economc Theory 86, 169-193. [3] Hart, O. (1974): On the Exstence of an Equlbrum n a Securtes Model. Journal of Economc Theory 9, 293-311. [4] C. Le Van, D.H. Truong Xuan (21):Asset market equlbrum n L p spaces wth separable utltes. Journal of Mathematcal Economcs 36, 241-254. [5] Page, F.H. Jr, Wooders M.H, (1996): A necessary and suffcent condton for compactness of ndvdually ratonal and feasble outcomes and exstence of an equlbrum. Economcs Letters 52, 153-162. [6] Page, F.H. Jr, Wooders M.H and P.K. Montero (2): Inconsequental arbtrage. Journal of Mathematcal Economcs 34, 439-469. [7] Page, F.H (1987): On equlbrum n Hart s securtes exchange model. Journal of Economc Theory 41, 392-44. [8] Rockafellar, R.T (197): Convex Analyss, Prnceton Unversty Press, Prnceton, New-Jersey. [9] Werner, J.(1987): Arbtrage and the Exstence of Compettve Equlbrum. Econometrca 55, 143-1418. 2