Onlne Appendx to Asset Prcng n Large Informaton Networks Han N. Ozsoylev and Johan Walden The proofs of Propostons 1-12 have been omtted from the man text n the nterest of brevty. Ths Onlne Appendx contans these proofs. Equaton numberng contnues n sequence wth that establshed n the man text. See the man text for references to Theorem 1 and equatons 1)-46). Proof of Proposton 1: We construct a growng sequence of caveman networks that converge to a gven degree dstrbuton. A caveman network s one whch parttons the set of agents n the sense that f agent s connected wth j and j s connected wth k, then s connected wth k see Watts [2]). We proceed as follows: Frst we observe that for d1) 1, the result s trval, so we assume that d1) 1. For a gven d S, defne k mn { 1: supp[d]}. Form>k, we defne ˆd m S m by ˆd m ) d)/ m j1 dj). Clearly, lm m m ˆd m ) d) 0. For an arbtrary n k 3, choose m n 1/3.For1<l m, l k, choose zl n ˆd m l) n/l, and zk n n l k zm l l)/k. Now, defne G n, wth degree dstrbuton d n, as a network n whch there are zl n clusters of tghtly connected sets of agents, wth l members, 1 <l m and n m l2 lzn l sngletons. Wth ths constructon, zl nl/n ˆd m ) l/n for l>2and l k. Moreover, z1 n /n ˆd m 1) k +1)/n, and zk nk/n ˆd m k) k +1)/n + m 2 /n, so m l1 zn l l ˆd m l) 2k +1)/n +2m 2 /n On 1/3 ). Thus, n 1/3 d n ) ˆd n1/3 ) 0, when n and snce n 1/3 ˆd n1/3 ) d) 0, when n, ths sequence of caveman networks ndeed provdes a constructve example for whch the degree dstrbuton converges to d. Moreover, t s straghtforward to check that f d) O α ), α>1, then 10) s satsfed n the prevously constructed sequence of caveman networks, and that f α>2, then 11) s satsfed. If d) α, α 2, on the other hand, then clearly d), so 11) wll fal. 1
Proof of Proposton 2: Wefrstshowtheformforβ. Wehave: n k1 lm Wn ) lm k c n s 2 n αk α k ζα) 1 k α 1) ζα) 1 ζα 1). k1 For 10), we notce that for a network wth n m α nodes, the maxmum degree, W n ) wll not be larger than m. However, snce each of the neghbors to that node has no more than m neghbors, W n j Wn ) j m 2 n 2/α on) whenα>2. Proof of Proposton 3: It follows from Theorem 1 that 47) mples that π ) 2 σ 2 γ ) 2 Δ 2 β 2 β +Δ 2 ) 2 σ 6 Δ 2 + β β +Δ 2 ) σ 2 ) 2, 47) Δ 2 β +Δ 2 ) 2 σ 4 Δ 2 + β β +Δ 2 ) σ 2 ) 2, 48) var p) β +Δ2 ) 2 σ 4 Δ 2 + β 2 σ 2 ) Δ 2 + β β +Δ 2 ) σ 2 ) 2. 49) π ) 2 σ 2 β 2βΔ2 β +Δ 2 )2β +Δ 2 ) σ 6 Δ 2 + β β +Δ 2 ) σ 2 ) 3 > 0, and ths proves part a). 48) mples that γ ) 2 Δ 2 β 2Δ4 β +Δ 2 ) σ 4 2Δ 2 β +Δ 2 ) 3 σ 6 Δ 2 + β β +Δ 2 ) σ 2 ) 3. The expresson above s strctly negatve f and only f β> Δ σ Δ2. Ths proves part b). Fnally, 49) mples that var p) β 2Δ4 β +Δ 2 ) σ 4 2Δ 2 β 3 +2βΔ 4 +Δ 6 ) σ 6 Δ 2 + β β +Δ 2 ) σ 2 ) 3. The expresson above s strctly postve f and only f Δ 2 < 1 βσ2 + 1 1 2βσ 2 +5β 2 σ 4. Ths 2σ 2 2 σ 4 2
proves part c). Proof of Proposton 4: It s straghtforward from Theorem 1 and the projecton theorem that ) 2 σ ) β 2 Δ 2 +σ 2 β σ 2 var X p σ 2 βσ 2 Δ 2 +Δ 2 +β 2 σ ) 2 2 ) 2 σ β 2 Δ 2 +σ 2 β σ βσ 2 Δ 2 +Δ 2 +β 2 σ 2 σ2 + 2 Δ 2 +σ 2 β βσ 2 Δ 2 +Δ 2 +β 2 σ Δ 2 2 Hence the result follows. Δ 2 σ 2 Δ 2 + β 2 σ 2. Proof of Proposton 5: From 23), we know that agent s demand wll take the form ψ x, p) α 0 + α 1 α2 x + 1 ) p. Smlar arguments as n the proof of Theorem 1 shows that α 0 X ) σ π0 2 γn Z A, 1 T π)s 2 πt s ) where A γns 2 πt Sπ+γ 2 n 2 Δ 2 ) π T s, converges to β ) 2 Δ 2 for large n. Therefore α 0 α 1 XΔ 2 + Zβσ 2 σ 2 Δ 2 + σ 2 β, α 2 π T Sπ + γ 2 n 2 Δ 2 1 T π)π T s ) s 2 πt Sπ + γ 2 n 2 Δ 2 ) π T s ) 2 1 W s 2 s, 2 1 T π)s 2 π T s ) s 2 πt Sπ + γ 2 n 2 Δ 2 ) π T s ) 2 β Δ 2 γ. 3
Smlarly, we have 1 σ 2 + s 2 ) π T Sπ + n 2 Δ 2 γ 2 +1 T π) 2 σ 2) 1 T π)σ 2 +π T s ) ) 2 σ 2 s 2 πt Sπ + γ 2 n 2 Δ 2 ) π T s ) 2 ) σ 2 + s 2 ) q T Sq/n 2 +Δ 2 +1 T q) 2 σ 2 /n 2) 1 T q)σ 2 +q T s ) ) 2 /n 2 σ 2 s 2 qt Sq/n 2 +Δ 2 ) q T s ) 2 /n 2 ) σ2 + s 2 )Δ 2 + β 2 σ 2 ) βσ 2 ) 2 1 σ 2 s 2 Δ2 s 2 + 1 σ 2 + β2 Δ 2. 50) Thus, ψ x, p) XΔ 2 + Zβσ 2 σ 2 Δ 2 + σ 2 β + W β s x 2 p)+ Δ 2 γ 1 ) σ β2 p. 2 Δ 2 Snce β Δ 2 γ 1 σ β2 ββσ2 Δ 2 +Δ 2 + β 2 σ 2 ) 2 Δ 2 Δ 2 σ 2 Δ 2 + σ 2 β) Δ 2 σ 2 Δ 2 + β), the expresson for the demand functon reduces to Δ 4 + βδ 2 Δ 2 σ 2 Δ 2 + σ 2 β) β2 σ 2 Δ 2 + β) Δ 2 σ 2 Δ 2 + σ 2 β) ψ x, p) XΔ 2 + Zβσ 2 σ 2 Δ 2 + σ 2 β Δ 2 σ 2 Δ 2 + β) p + W s 2 x p). 51) Expected profts are of the form E[ψ x, p) X p)], and therefore 17) mmedately follows. Proof of Proposton 6: We defne the average expected proft n economy n, Π n [ ] n E X p n) ψ n x n, p n ). n From Theorem 1, we know that the market clearng condton n ψ x, p)/n Z n. We 4
therefore have [ ] Π n E X p n) Zn [ ) ] E X π0 n π n X + η n )+γ n Zn Z n 1 1 π n π n ) [ ] [ ] [ ] E X Zn π0 n E Zn + γ n E Zn Zn ) X Z π n 0 Z + γ n Δ 2 + Z 2 ) 1 π ) X Z π 0 Z + γ Δ 2 + Z 2 ). Now, snce X Z 0 t follows that We also have Π Πγ Δ 2 Δ2 β +Δ 2 ) σ 2 Δ 2 + β β +Δ 2 ) σ 2. 52) Δ 2 γ ) 2 Δ 2 π 1 π )σ 2) + W 1 π ) 2 σ 2 +γ ) 2 Δ 2) σ 2 Δ 2 + β) s 2 Δ4 W + s 2 Δ 2 ) σ 2 + W Δ 2 β +Δ 2 ) 2 σ 4 s 2 Δ 2 + β β +Δ 2 ) σ 2 ) 2. 53) It then follows from 53) that Π W Δ4 σ 2 +Δ 2 β +Δ 2 ) 2 σ 4 s 2 Δ 2 + β β +Δ 2 ) σ 2 ) 2 > 0, Π β Hence the proposton follows. 2Δ4 s 2 Δ 4 + β W +2s 2 Δ 2 )) σ 4 +2W Δ 2 β +Δ 2 ) 3 σ 6 s 2 Δ 2 + β β +Δ 2 ) σ 2 ) 3 < 0. Proof of Proposton 7: It follows from 52) that Π β Δ4 σ 2 Δ 2 β +Δ 2 ) 2 σ 4 Δ 2 + β β +Δ 2 ) σ 2 ) 2. Observe that the above s strctly negatve f and only f σ < 1 Δ and β < Δ σ Δ2. 5
Proof of Proposton 8: Followng Theorem 1 and 51), we can rewrte agent s demand functon as follows: ψ x, p) c + Δ2 βs 2 +W ) X + s2 Δ 2 +β+δ 2 )σ 2 W Z k W ) + ɛ k, 54) s 2 Δ 2 +ββ+δ 2 )σ 2 ) s 2 Δ 2 +ββ+δ 2 )σ 2 ) s 2 where c s a constant scalar. Thus, cov ψ x, p),ψ j x j, p)) ) Δ 2 βs 2 +W ) s 2 Δ 2 +ββ+δ 2 )σ 2 ) ) + s 2 Δ 2 +β+δ 2 )σ 2 W s 2 Δ 2 +ββ+δ 2 )σ 2 ) Δ 2 βs 2 +W j) s 2 Δ 2 +ββ+δ 2 )σ 2 ) ) σ 2 s 2 Δ 2 +β+δ 2 )σ 2 W j s 2 Δ 2 +ββ+δ 2 )σ 2 ) ) Δ 2 + W j. 55) On the other hand, observe from 54) that the varance of agent s demand, var ψ x, p)), does not depend on W j. Therefore, followng 55) we have corrψ x, p),ψ j x j, p)) W j Hence we have the desred result. 1 var ψ x, p)) var ψ j x j, p)) > 0. Proof of Proposton 9: a) From 8) and 51), ) t follows that ψ N0,a 1 + a 2 W + a 3 W 2 ), where a 1 Δ6 +β 2 Δ 4 σ 2, a a 2 2 1 1+ 2Δ6 σ 2, a 4 s 2 a 2 3 Δ2 σ 2 β 2 σ 2 +Δ 4 σ 2 +Δ 2 +2Δ 2 βσ 2 ), and a 4 s 4 a 2 4 4 β 2 σ 2 +Δ 2 +Δ 2 βσ 2 2A. Snce, E[ z ] for a general normally dstrbuted random varable, π z N0,A), t follows that ψ unsgned 2a1 + a 2 W + a 3 W 2 π ). 56) It mmedately follows that ths functon s ncreasng and concave, wth the gven asymptotcs. It s also clear that E[ψ 2 ]varψ )+E[ψ ] 2 varψ )a 1 + a 2 W + a 3 W 2 π 2 ψunsgned ) 2, so t s ndeed the case that ψ unsgned 2 π E[ψ2 ]. b,c) We have [ ] E ψ 2 W )d) E[ψ 2 W )]d) a 1 + a 2 W + a 3 W 2 )d) a 1 + a 2 s 2 β + a 3 s 4 β 2 + σ 2 β) β 2 +Δ 2 + a 3 σ 2 β. 6
Therefore, ψ market π 2 β2 +Δ 2 + a 3 σβ 2), and ψmarket s ncreasng n σ β. Moreover, for small σ β, ψ market s ncreasng n β. Also, t s easy to show that a 3 < 0, so for large σ β β, ψ market s decreasng n β. Proof of Proposton 10: The followng lemma ensures that the lmt of average certanty equvalents s equal to the average certanty equvalent n the large economy. Lemma 1. If Assumpton 1 and the condtons of Theorem 1 are satsfed, and the functon f : N R s concave and ncreasng, then lm n dn )f) d)f) wth probablty one. Proof: Snce f s concave, t s clear that f g, whereg) def f1)+f2) f1)) def c 0 +c 1. From 11), and snce f s ncreasng, t s therefore clear that n dn )f) [c 0,c 0 +c 1 β+ɛ], for arbtrary small ɛ>0, for large n. Now, for arbtrary m and ɛ>0, by Assumpton 1, for large enough n 0, for all n n 0, d n ɛ ) d). Also, for large enough m and mc 0 +c 1 ) n 0, for all n n 0, n m+1 dn )f) ɛ, from 11). Fnally, from Assumpton 1, for large enough m, m+1 d)f) ɛ. Thus, for an arbtrary ɛ>0, a large enough m can be chosen and n 0 maxm, n 0,n 0) such that for all n n 0, d n m m )f) d)f) d n )f) d)f) + d n )f) d)f) + d)f) m+1 m+1 n+1 ɛ + ɛ + ɛ, and snce ɛ>0 s arbtrary, convergence follows. The expected utlty n the large economy of an agent wth W connectons s UW ) E [ ] e ψ x, p) X p) 1 8π3 σ 2 Δ 2 W/s 2 e ψ X+η,p)X+η p) X 2 2σ 2 Z2 sβ 2 σ 2 +Δ 2 +Δ 2 βσ 2 ) Δ2 +β +Δ 2 ) 2 σ 2 )β 2 s 2 σ 2 +Δ 2 σ 2 +Δ 2 σ 2 W ), 2Δ 2 η2 2W/s 2 dxdzdη where the last equalty follows by usng 12-15,51). Snce UW ) e CEW ), condton a) mmedately follows. 7
Moreover, snce the functon CEW ) s ncreasng and concave n W, from Lemma 1, t s clear that the average certanty equvalent s as defned n b). Proof of Proposton 11: a) Ths follows mmedately from Jensen s nequalty, snce CEW ) s a strctly convex functon of W 1. b) We frst note that the two-pont dstrbuton, for whch a fracton β β of the agents has β + 1 connectons and the rest has β connectons, has connectedness β β ) β +1)+1 β + β ) β β, so the two-pont dstrbuton s ndeed a canddate for an optmal dstrbuton. Clearly, ths s the only two-pont dstrbuton wth support on {n, n +1} that has connectedness β, and for β/ N, there s no one-pont dstrbuton wth connectedness β. We defne n β, q n 1 β + β, q n+1 β β. We ntroduce some new notaton. We wsh to study a larger space of dstrbutons than the ones wth support on the natural numbers. Therefore, we ntroduce the space of dscrete dstrbutons wth fnte frst moment, D { 0 r δ x }, where r 0, and 0 x for all, 0 < r < and r x <. 1 The subset, D 1 D, n addton satsfes r 1. The c.d.f. of a dstrbuton n D s a monotone functon, F d : R + R +, defned as F d x) 0 r θx x ), where R + {x R : x 0}. Here, θ s the Heavsde step functon. Clearly, F d s bounded: sup x 0 F d x) r <. We use the Lévy metrc to separate dstrbutons n D, Dd 1,d 2 ) nf{ɛ >0:F d1 x ɛ) ɛ F d2 x) F d1 x + ɛ) for all x R + }. We thus dentfy d 1 d 2 ff Dd 1,d 2 )0. For d D, we defne the operaton of addton and multplcaton: d 1 r1 δ x 1, d 2 r2 δ x 2 leads to d 1 + d 2 r1 δ x 1 + r2 δ x 2 and αd 1 αr1 δ x 1,forα>0. The two-pont dstrbuton can then be expressed as ˆd q n δ n + q n+1 δ n+1. The support of a dstrbuton d r x δ x n D s now supp[d] {x : r > 0}. A subset of D s the set of dstrbutons wth support on the ntegers, D N {d D : supp[d] N}. For ths space, we can wthout loss of generalty assume that the x s are ordered, x. The expectaton of a dstrbuton s E[d] r x and the total mass s Sd) r. Both the total mass and expectatons operators are lnear. Another subset of D, gven β>0, s D β {d D : E[d] β}. Gven a strctly concave, functon f : R + R, we defne the operator V f : D D, such that V f d) r δ fx ). The functon fx) CEx), s, of course, strctly concave R +. 1 Dstrbuton here s n the sense of a functonal on the space of nfntely dfferentable functons wth compact support, C0 see Hörmander [ ]), and δ x s the Drac dstrbuton, defned by δ x f) fx) for f C0. 8
Clearly, V f s a lnear operator, V f d 1 + d 2 )V f d 1 )+V f d 2 ). The second part of the theorem, whch we wsh to prove, now states that for all d D 1 D N D β,wthβ / N, fd ˆd, t s the case that E[V f d)] >E[V f ˆd)]. It turns out that the nequalty holds for any strctly concave functon on f : R + R. To prove ths, we use Jensen s nequalty, whch n our notaton reads: Lemma 2. Jensen): For any d D, wth support on more than one pont, and for a strctly concave functon, f : R + R, the followng nequalty holds: E[V f d)] < Sd)E[V f δ E[d]/Sd) )] E[V f Sd)δ E[d]/Sd) )]. Now, let us take a canddate functon for an optmal soluton, d ˆd, such that d D 1 D N D β. Clearly, snce ˆd s the only two-pont dstrbuton n D 1 D N D β,andthere s no one-pont dstrbuton n D 1 D N D β, the support of d s at least on three ponts. Also, snce q n + q n+1 1,andd D 1, t must ether be the case that r n <q n,or r n+1 < q n+1, or both. We wll now decompose d nto three parts, dependng on whch stuaton holds: Frst, let s assume that r n+1 q n+1. If, n addton, r n+1 >q n+1,then t must be that r n < q n, and r > 0 for at least one < n. Otherwse, t could not be that E[d] β. In ths case, we defne d 1 <n r δ, d 2 r n δ n + q n+1 δ n+1 and d 3 r n+1 q n+1 )δ n+1 + >n+1 r δ. If, on the other hand, r n+1 q n+1, then there must be an <nsuch that r > 0andalsoaj>n+1 such that r j > 0, snce otherwse t would not be possble to have E[d] β. In ths case, we defne, d 1 <n r δ, d 2 r n δ n + q n+1 δ n+1 and d 3 >n+1 r δ. Exactly the same technque can be appled n the case of r n q n and r n+1 <q n+1. Fnally, n the case of r n <q n and r n+1 <q n+1, there must, agan, be an <nsuch that r > 0andaj>n+ 1, such that r j > 0, otherwse E[d] β would not be possble. In ths case, we decompose d 1 <n r δ, d 2 r n δ n + q n+1 δ n+1 and d 3 r n+1 q n+1 )δ n+1 + >n+1 r δ. These decompostons mply that E[V f d)] E[V f d 1 )] + E[V f d 2 )] + E[V f d 3 )] Sd 1 )E[V f δ E[d1 ]/Sd 1 ))] + E[V f d 2 )] + Sd 3 )E[V f δ E[d3 ]/Sd 3 ))] E [ )] V f Sd1 )δ E[d1 ]/Sd 1 ) + d 2 + Sd 3 )δ E[d3 ]/Sd 3 ) E[V f d m )], where d m d L + d 2 + d R, d L Sd 1 )δ E[d1 ]/Sd 1 ) and d R Sd 3 )δ E[d3 ]/Sd 3 ). Clearly, d m 9
D 1 D β. Now, f r n+1 q n+1, snce d D 1, t must be that Sd 1 )+Sd 3 )q n r n, and snce E[d L + d 2 + d R ]β E[q n δ n + q n+1 δ n+1 ] t must be that E[d L + d R ]q n r n )E[δ n ] E[Sd 1 )+Sd 2 ))δ n ]E[d a ], where d a Sd 1 )+Sd 2 ))δ n. Moreover, snce d a + d 2 has support on {n, n +1} and E[d a + d 2 ]β, t s clear that d a + d 2 ˆd. From Jensen s nequalty, t s furthermore clear that E[V f d L + d R )] <E[V f d a )], and therefore E[V f d m )] E[V f d L + d R + d 2 )] <E[V f d a + d 2 )] E[V f ˆd)]. Thus,allnall, E[V f d)] E[V f d m )] <E[V f ˆd)]. A smlar argument can be appled f r n q n. Fnally, n the case n whch r n <q n and r n+1 <q n+1, we defne α E[d 1 ]/Sd 1 )and β E[d 3 ]/Sd 3 ). Obvously, α<n<n+1 <β. Now, we can defne g 1 β n q β α n r n )δ α + n αq β α n r n )δ β and g 2 β n 1 q β α n+1 r n+1 )δ α + n+1 αq β α n r n )δ β. Clearly, g 1 D and g 2 D and, moreover, g 1 + g 2 + d 2 d 1 + d 2 + d 3 d. Also, Jensen s nequalty mples that E[V f g 1 )] <E[V f q n r n )δ n )] and E[V f g 2 )] <E[V f q n+1 r n+1 )δ n+1 )], so E[V f d)] E[V f g 1 + g 2 + d 2 )] <E[V f q n r n )δ n +q n+1 r n+1 )δ n+1 + d 2 )] E[V f ˆd)]. We are done. c) From a,b,18,19) t follows that CE β) softheformce β) 1 logvβ)), where 2 vβ) Δ2 +β+δ 2 ) 2 σ 2 )β 2 s 2 σ 2 +Δ 2 s 2 +βδ 2 σ 2 ). It mmedately follows that v β) softheform s 2 β 2 σ 2 +Δ 2 +Δ 2 βσ 2 ) 2 v 2 β)c 4 β 4 + c 3 β 3 + c 2 β 2 + c 1 β + c 0 ), where v 2 β) > 0 for all β>0, c 4 > 0, c 3 > 0and c 2 > 0, and where c 1 Δ 2 +4s 2 3andc 0 2Δ 2 s 2 σ 2 Δ 2 σ 2 1. Moreover, snce c 4 > 0, t follows that v β) < 0forlargeβ. From Descartes rule of sgns, t follows that the maxmum number of roots to v β) 0 s two, and there can only be two roots f c 1 < 0andc 0 > 0. The condton c 0 > 0s equvalent to 2s 2 1 > 1, whch n partcular mples that s 2 > 1. Smlarly, c Δ 2 σ 2 2 1 < 0ff 3 4s 2 > Δ 2 σ 2, whch n partcular mples that s 2 < 3. Multplyng these two condtons, we 4 get that a necessary condton for the two roots to be possble s that 3 4s 2 )2s 2 1) > 1, for s 1/2, 3/4), but t s easy to check that 3 4s 2 )2s 2 1) s n fact less than one n ths regon. Therefore, t can not be the case that c 1 < 0andc 0 > 0 at the same tme, and there can be at most one root to the equaton v β) 0. Snce v β) < 0forlargeβ, tmust therefore be the case that vβ) s ether decreasng for all β, or ntally ncreasng and then decreasng, wth a unque maxmum. It s easy to check numercally that both cases are n fact possble. We are done. Proof of Proposton 12: Snce f s weakly convex and CE s concave, CE f s a concave functon of W, and an dentcal argument as n the proofs of Proposton 11 a),b) can be 10
made to show that a degenerate.e., unform) network s optmal. Now, snce CE,as defned n Proposton 11, s decreasng for large β, t follows that CE f s decreasng for large β, so the optmal β must be nteror. References [1] [2] L. Hörmander, The Analyss of Lnear Partal Dfferental Operators, I-IV, Sprnger Verlag, 1983. D.J. Watts, Small Worlds, Prnceton Unversty Press, 1999. 11