Proof of Lemma 1. Suppose it is optimal for every buyer to choosed = 1. The optimal offer pair{qs,b, τs,b}solves the Nash program
qs,bmax,τs,b [u(qs,b)−τs,b(Vb−Vb−1)]θ[−qs,b+τs,b(Vs+1−Vs)]1−θ s.tτs,b≤1 Suppose thatVs+1> Vs, otherwise no trade would take place. Substituting forτs,b, consider the Lagrangian
λs,bmax,qs,b,τs,b
[u(qs,b)−τs,b(Vb−Vb−1)]θ[−qs,b+τs,b(Vs+1−Vs)]1−θ +λs,b(1−τs,b)
whereλs,bis the multiplier onτs,b≤1, independent ofdbecause in the equilibrium conjectured every buyer offersd= 1.The equilibriumqs,b, τs,bandλs,bmust satisfy three sufficient and necessary first-order conditions
u(qs,b) θ
1−θ =u(qs,b)−τs,b(Vb−Vb−1)
−qs,b+τs,b(Vs+1−Vs) Vb−Vb−1
Vs+1−Vs θ
1−θ =u(qs,b)−τs,b(Vb−Vb−1)
−qs,b+τs,b(Vs+1−Vs) −λs,bA λs,b(1−τs,b) = 0
whereA=
−qs,b+τs,b(Vs+1−Vs) u(qs,b)−τs,b(Vb−Vb−1)
θ
≥0.Note howqs,bandτs,bgenerally depend on both the seller’s and the buyer’s wealth positions, via their reservation values Vs+1−VsandVb−Vb−1, and their relative bargaining powers,θ/(1−θ).Two cases might arise, depending on whether the constraintτs,b≤1is binding or not.
1. Ifτs,b = 1thenλs,b>0. Henceu(qs,b) =1−θθ ·u(q−qs,bs,b)−(V+Vs+1b−V−Vb−1s). 2. Ifτs,b ≤1thenλs,b = 0. Henceu(qs,b) = VVbs+1−V−Vb−1s ⇒qs,b =
)Vs+1−Vs
Vb−Vb−1
*1
γ. That is the marginal utility from consumption from spendingd= 1,with prob- ability τs,b, must be equal to the ratios of the value of that unit of money to seller and buyer. Notice that since VVs+1−Vs
b−Vb−1 = 1,in general, the quantity trade will be generally inefficient, unless buyer and seller ‘swap’ wealth positions (i.e.
s=b−1).
From the first order conditions we obtain τs,b= (1−θ) u(qs,b)
Vb−Vb−1 +θ qs,b Vs+1−Vs
= qs,b Vs+1−Vs
(1−θ) u(qs,b) qs,bu(qs,b)+θ
= qs,b Vs+1−Vs
1−θγ 1−γ so that we see that, givenV qs,b
s+1−Vs, τs,bdecreases inθ. Proof of Lemma 2. Ifd= 1andτs,b∈(0,1), then forn= 0, N
ρVn= γ 1−γ
N−1 s=0
ms Vs+1−Vs Vn−Vn−1
1−γ
γ
Note that
ρVn(Vn−Vn−1)1−γγ = γ 1−γ
N−1
s=0
ms(Vs+1−Vs)1−γγ
is independent ofn.It follows that forn≥2 : Vn(Vn−Vn−1)1−γγ
Vn−1(Vn−1−Vn−2)1−γγ = 1⇒ Vn
Vn−1 = Vn−1−Vn−2 Vn−Vn−1
1−γγ
⇒ VN V1
= V1−V0 VN−VN−1
1−γγ
because of a telescoping product.
If we leta1= 1,a0= 0, andVn =anV1then VVn
n−1 = aan−1n andVn−Vn−1= (an−an−1)V1.Therefore we can find{an}Nn=1recursively:
a2
a1 = )V1−V0
V2−V1
*1−γγ
⇒ a
1−γγ
2 (a2−1) = 1(sincea1= 1) a
1−γγ
n (an−an−1) = 1∀2< n≤N
Thusanis a function solely ofγ,hinging ona2=a(γ).It is easy to see thata2>1 anda2<2becausea
1−γγ
2 (a2−1)increases ina2and ata2= 2does not satisfy the equality above. Also,{an−an−1}is a positive but decreasing sequence (because an−an−1 = 1/a
1−γγ
n ,an−an−1must be decreasing inn).ThereforeVn is an increasing function ofn,and{Vn−Vn−1}is a decreasing sequence.
Use the result thatVn−Vn−1= (an−an−1)V1.Then:
ρV1(V1−V0)1−γγ = γ 1−γ
N−1 s=0
ms(Vs+1−Vs)1−γγ
ρV1(V1)1−γγ = γ 1−γ
N−1
s=0
ms(as+1−as)1−γγ V
1−γ
1 γ
(useVs−Vs−1= (as−as−1)V1) ρV1= γ
1−γ
N−1
s=0
ms(as+1−as)1−γγ
ρV1= γ 1−γ
N−1
s=0
ms
as+1 (use(as+1−as)1−γγ =a−1s+1) where we notice thatV1<∞since{as+1−as}is a converging sequence.
Using the definition ofan,τs,b= a
1γ b
as+1V1.Hence,{τs,b}is a sequence increasing
inband decreasing ins.
Proof of Lemma 3. To start we notice that (10) - (12) imply mn
N b=1
τn,bmb=mn+1
N−1
s=0
τs,n+1ms ∀n=N (17)
which means that the expected money flow to sellers withnunits of money, must be equal to the expected money outflow of buyers withn+ 1units of money. To see why this holds, start with (10), and then use it in (11) withn= 1. Observe that only the summations to the extreme left and extreme right of (11) are left (the inner summations cancel out). Then repeat it recursively, for eachn < N.
Now use (17) replacing the lotteries by their expressions given in (9) to get
mn
an+1
N
b=1ab1−γ1 mb=mn+1an+11−γ1 N−1
s=0 ms
as+1 ∀n=N
⇒ mmn+1
n a
2−γ1−γ
n+1=
N
b=1a
1−γ1 b mb
N−1 s=0 ms
as+1
⇒ mmn+1n =mm01a
2−γ1−γ
n+1 ∀n=N
(18)
since (10) impliesmm1
0 =
N
b=1a
1−γ1 b mb
N−1
s=0 ms as+1
after one substitutes for (9).
We can use the last line of (18) for any two adjacentnandn+ 1to obtain m2n
mn+1mn−1 = an+1 an
2−γ1−γ
∀n= 0, N (19)
This tells us that 'an+1
an
(N−1
n=1 is a decreasing sequence so that ' mn
mn+1
mn
mn−1
(N−1
n=1
is a decreasing sequence also. It follows that ' mn
mn+1
(N−1
n=1 cannot be an increasing sequence, i.e.mn > mn+1∀n= 0, N cannot be an equilibrium. Now use the last line of (18). We see thatm0> m1is not possible (it would implymn> mn+1∀n= 0, N). Thusm0 < m1must hold. Sincemn > mn+1∀n = 0, N is not possible, then the only equilibrium ismn < mn+1for some1 ≤n < n∗andmn > mn+1 forn≥n∗.That is,{mn}is hump-shaped.
Since mmn+1
0 = mmn+1n ×mmn−1n ×...×mm10 andmmn
n+1 = mm01a
2−γ1−γ
n+1 then mn+1
m0 = m1 m0
n+1
Πj=1n+1a−
2−γ1−γ
j for all n=N.
LetAn+1=Πj=1n+1a−
2−γ1−γ
j forn=N, and notice thatA0= 1.Then, the stationary distribution solves the system ofN+ 1non-linear equations inN+ 1unknowns.:
mn+1=m0 )m1
m0
*n+1
An+1 ∀n= 0, N m0+N−1
n=0 m0 )m1
m0
*n+1
An+1= 1 N−1
n=0 (n+ 1)m0 )m1
m0
*n+1
An+1=M These expressions can be rewritten to yield (13) and (14).
We next show uniqueness of the stationary distribution for anyN and money supplyM ∈(0, N). The first thing to note is thatm0+N−1
n=0 m0 )m1
m0
*n+1
An+1= 1 implies ∂m∂m1
0 < 0. Thus, for any N andm0 there is a uniqueM that satisfies mn+1=m0
)m1
m0
*n+1
An+1andm0+N−1
n=0 m0 )m1
m0
*n+1
An+1= 1. Next, note thatN−1
n=0 (n+ 1)m0 )m1
m0
*n+1
×An+1=M implies thatm0is monotonically decreasing inM (recall that ∂m∂m1
0 < 0). Accordingly, for anynandM ∈ (0, N)
there is a unique{mn}satisfying (13) and (14).
Proof of Lemma 4.Letθ= 1.Supposed= 1andτs,b∈(0,1)∀b, sis an equilibrium.
Consider the strategy of a representative buyerbin a match with a sellers.To prove individual optimality of the strategy proposed we take three steps. Finally, we prove that every single coincidence match will result in a trade. That is the buyer always puts a positive probability on the transfer ofd= 1.
1. First, we prove that if a buyer offers a lottery on the transfer of d ∈ Ds,b units of money, thend = 1is individually optimal. The proof is by means of contradiction. Pick any feasible offerdand suppose that the buyer wants to offer a lottery. Since the buyer extracts the seller’s entire surplus, then the optimal transfer probability must satisfyτs,b(d)∈[0,1]and
τs,b(d) = qs,b(d) Vs+d−Vs.
Given this probability, the buyer choosesqs,b(d)to maximize his surplus u[qs,b(d)]−τs,b(d)(Vb−Vb−d) =u[qs,b(d)]−qs,b(d)Vb−Vb−d
Vs+d−Vs. Sinceu(q) = q1−γ1−γ,then optimal consumption is
qs,b(d) = Vs+d−Vs Vb−Vb−d
1γ
. (20)
The implication is that, givend∈Ds,b,whenτs,b(d)andqs,b(d)are optimally chosen then the buyer’s surplus isu[qs,b(d)]−qs,b(d)qs,b(d)γ,or
γu[qs,b(d)]. (21)
Clearly thedthat maximizes (21) must generate the largest quantity, i.e. it must maximizeqs,b(d) =
)Vs+d−Vs
Vb−Vb−d
*1
γ.It is easily proved that Vb−Vb−d
Vs+d−Vs = (Vb−Vb−1) + (Vb−1−Vb−2) +...+ (Vb−d+1−Vb−d) (Vs+d−Vs+d−1) + (Vs+d−1−Vs+d−2) +...+ (Vs+1−Vs)
≥ Vb−Vb−1
Vs+1−Vs ∀d≥1
sinceVb−Vb−1 < Vb−1−Vb−2 < .. < Vb−d+1−Vb−d,whileVs+1−Vs >
Vs+2−Vs+1> ... > Vs+d−Vs+d−1,because{Vn+1−Vn}is a decreasing se- quence, in equilibrium. That is, raisingdabove one, increases the numerator and decreases the denominator of the ratioVVb−Vb−d
s+d−Vs.SinceVVb−Vb−d
s+d−Vs ≥ VVbs+1−V−Vb−1s∀d≥ 1,then it follows that settingd ≥2is worse than offeringd = 1.Offering a lottery ond≥2is suboptimal because, in the equilibrium conjectured, it simply reduces the quantity consumed by the buyer, hence his surplus.
2. Now we provide a condition guaranteeing thatτs,b<1is individually optimal.
That is, offeringd = 1with certainty is suboptimal. In the conjectured equi- libriumτs,b(1) = τs,b, defined by (3) forθ = 1.Because{τs,b}is increasing inband decreasing ins,it follows that a sufficient condition forτs,b(1)<1is τ0,N <1.Using the results in the prior Lemmas, this amounts to the inequality
a1/γN
V1 <1that, substituting forV1can be rearranged as ρ <ρ¯= γ
1−γ 1 a
1 γ
N N−1
s=0
ms as+1.
It is seen that the sequences{an}and{ms}only depend onγ,M,andN. 3. Finally, we prove that every buyer offers to spendsomethinginevery single-
coincidence match, i.e. τs,b > 0∀s, b is individually optimal. Since τs,b =
qs,b
Vs+1−Vs in equilibrium, the buyer’s expected surplus is positive in every possi- ble match, i.e.u(qs,b)−τs,b(Vb−Vb−1)≡γu(qs,b)>0∀s, b.In equilibrium qs,b=
)Vs+1−Vs
Vb−Vb−1
*1
γ.Thereforeτs,b >0.
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