Supplement to “Alternating-offer bargaining with the global games information structure”
(Theoretical Economics, Vol. 13, No. 2, May 2018, 869–931) A nton Tsoy
Department of Economics, Einaudi Institute for Economics and Finance
B.6 Interim versions of Theorems 1 and 3
I first introduce interim counterparts of sets E, E , and IR. Define U
ηS(s |τ ρ) = E
η[e
−rτ(sb)(ρ(s b) − c(s))|s] and U
ηB(b |τ ρ) = E
η[e
−rτ(sb)(v(b) − ρ(s b))|b] players’ ex- pected payoffs from the bargaining outcome (τ ρ) at the interim stage, i.e., after types s and b are realized. For any x ∈ [0 1], let
E(x) =
η
lim
→0U
ηS(x |τ
ηρ
η) U
ηB(x |τ
ηρ
η)
: (τ
ηρ
η) →
η→0
(τ ρ) ∈ DL
be the set of all interim expected payoff profiles of types on the diagonal s = x and b = x generated by double limits. Let E (x) be the convex hull of the closure of E(x). Note that when players have access to the public randomization device in the beginning of the game, E (x) is the set of interim expected payoff profiles of types s = x and b = x that can be approximated by double limits of my model.
I put some preliminary restrictions on E (x). Clearly, the feasibility constraint holds:
U
ηS(x |τ
ηρ
η) + U
ηB(x |τ
ηρ
η) ≤ (x)
Moreover, Lemma 1 implies that the seller’s utility is at least U
S(s) ≡ max{y
∗(0) − c(s) 0}
in any frequent-offer PBE limit, and, symmetrically, the buyer’s utility is at least U
B(b) ≡ max {v(b) − y
∗(1) 0 }. Hence, for any η, any outcome of the frequent-offer PBE limit (τ
ηρ
η) satisfies the interim individual rationality constraints
U
ηS(s |τ
ηρ
η) ≥ U
S(s) and U
ηB(b |τ
ηρ
η) ≥ U
B(b)
Denote by
IR(x) =
U
SU
B: U
S+ U
B≤ (x) U
S≥ U
S(x) and U
B≥ U
B(x)
the set of feasible, interim individually rational payoffs of types s = x and b = x, and denote by
PF (x) =
U
SU
B: U
S+ U
B= (x) U
S≥ U
S(x) and U
B≥ U
B(x) its Pareto frontier. Then I have E (x) ⊆ IR(x).
Anton Tsoy:
tsoianton.ru@gmail.com
Copyright © 2018 The Author. This is an open access article under the terms of the Creative Commons Attribution-Non Commercial License, available athttp://econtheory.org.https://doi.org/10.3982/TE2543
The following theorem is the interim counterpart of Theorem 1. The proof follows directly from the argument in the proof of Theorem 1.
T heorem 4. If (U
SU
B) ∈ PF(x) for some x ∈ [0 1], then (U
SU
B) ∈ E (x).
I now turn to the interim version of the folk theorem. Suppose that the buyer’s value is given by v
0(b) + ξ and the seller’s cost is c(s), where v
0, c, and ξ are introduced in Section 3.4. I am interested in the limit of the set E (x) as ξ → 0, which I denote by E
0(x).
Denote by IR
0(x), x ∈ [0 1] the limit of IR(x) as ξ → 0. It is easy to see that IR
0(x) =
U
SU
B: U
S+ U
B≤
0(x) U
S≥ 0 and U
B≥ 0
for x ∈ [0 1]
where
0(x) ≡ v
0(x) − c(x). The following theorem is the interim counterpart of Theo- rem 3.
T heorem 5. For all x ∈ [0 1], E
0(x) = IR
0(x).
P roof. The argument for the Pareto frontier of E
0(x) is analogous to that for E in the proof of Theorem 3. Note that for x = 0 or x = 1, IR
0(x) = {(0 0)}. Then in the double limit that I constructed in Theorem 1, types s = 0 and b = 0 trade at a price close to y
∗(0).
Since v(0) − c(0) →
ξ→0
0, the expected utilities of those types converge to 0 as ξ → 0. Sim- ilarly, the expected utilities of types s = 1 and b = 1 converge to 0 as ξ → 0, which proves that E
0(x) = IR
0(x), x ∈ {0 1}. Now consider x ∈ (0 1). There exists ξ small enough such that y
∗(0) < c(x) < v(x) < y
∗(1). In the double limit constructed in the proof of Theo- rem 3, the utility of types s = x and b = x converges to 0 as ξ → 0, which proves that
E
0(x) = IR
0(x).
B.7 Proof of Lemma 12
Let V
B≡ v(b
∞) −q
S, V
S≡ q
B−c(s
∞), and P ≡ q
S−q
B. Denote φ ≡
1−δδ2 P2, α
B≡ α
B(0) = φV
S, and α
S≡ α
S(0) = φV
B. By (24), α
B> 0 and α
S> 0. Note that φ →
δ→1
0. I choose δ sufficiently large so that α
Sand α
Bare less than 1.
System (25) has steady states (z −z), z ∈ R. I am interested in the positive trajectory that approaches the steady state (0 0). Around this steady state, the linearized system can be written in the matrix form
x
k+1y
k+1=
1 − α
B+ α
Sα
B−α
B(1 − α
S)
−α
S1 − α
Sx
ky
kThe matrix has eigenvalues 1 and λ ≡ (1 − α
B)(1 − α
S) ∈ (0 1). Since one of the eigen- values is equal to 1, the steady state is unstable, and I cannot conclude that in the neigh- borhood of the steady state, the nonlinear system will converge to the steady state or that the trajectory will stay positive. Therefore, I construct a particular trajectory that satisfies the desired properties. The proof proceeds in three steps.
Step 1: Conjectured solution. In the first step, I conjecture the form of solution and
use the method of indeterminate coefficients to derive it. The following preliminary
claim gives the Taylor expansion of α
B(y) and α
S(x).
C laim 1. For any x ∈ (0 1) and y ∈ (0 1),
α
B(y) ≡ α
B+ φ
∞ l=1γ
lBy
lα
S(x) ≡ α
S+ φ
∞l=1
γ
Blx
lwhere γ
Bl≡
dlc(s∞l!)/dxland γ
lS≡
dlv(b∞l!)/dxl.
Note that by the regularity conditions on v and c, γ
lB< D and γ
lS< D for all l.
I conjecture that there exists (μ
xiμ
yi)
∞i=1such that the solution (25) takes the form
17x
ky
k=
∞i=1
λ
ikλ
i/2μ
xiμ
yifor k = 1 2 (74)
and in addition, satisfies for all i = 1 2 ,
μ
xi≤u
δM
iand μ
yi≤u
δM
i(75) for some positive M and u
δsuch that
M < 1 < 1
λ(1 + u
δ) (76)
Given this conjecture, I next derive expressions for coefficients μ
xiand μ
yi, and in the next step, I will verify that for δ sufficiently close to 1, upper bounds (75) indeed hold.
Series (74) defining (x
ky
k) are absolutely convergent, as they are dominated by the absolutely convergent series u
δ∞i=1
λ
ikM
i. Plugging the conjectured solution (74) into system (25), I get
⎧ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎨
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎩
∞ i=1λ
i(k+12)μ
xi− μ
xiλ
i− α
Bμ
xi+ μ
yiλ
i2= φ
∞l=1
γ
Bl ∞i=1
μ
yiλ
i(k+1)l
∞
i=1
λ
i(k+12)μ
xi+ μ
yiλ
2i∞
i=1
λ
ikμ
yi− μ
yiλ
i− α
Sμ
xiλ
2i+ μ
yi= φ
∞l=1
γ
Sl ∞i=1
μ
xiλ
ikl
∞
i=1
λ
ikμ
xiλ
2i+ μ
yi(77)
17This is a natural guess given the eigenvalues of the linearized system.
Consider the first equation in system (77). Since λ ∈ (0 1) and |μ
yi| ≤ u
δM
i, by Mertens’
theorem,
∞ l=1γ
Bl ∞i=1
μ
yiλ
i(k+1)l
=
∞ l=1γ
Bl ∞i=l
i1+···+il=i
μ
yi1
· · μ
yilλ
i(k+1)(78)
The series in (78) is absolutely convergent by
∞l=1
∞ i=lλ
i(k+1)γ
Bli1+···+il=i
μ
yi1
· · μ
yil≤ D
∞ l=1 ∞ i=lλ
i(k+1)i1+···+il=i
μ
yi1
· · μ
yil≤ D
∞l=1
∞ i=lλ
i(k+1)i1+···+il=i
u
lδM
i= D
∞l=1
u
lδ ∞i=l
λ
i(k+1)M
ii − 1 l − 1
= D
∞l=1
u
lδλ
k+1M 1 − λ
k+1M
l≤ D
∞l=1
u
lδλM 1 − λM
lwhere the first inequality arises via the triangle inequality and |γ
lB| < D, the second inequality follows from (75), the first equality arises from the fact that the number of compositions of i into exactly l parts is
i−1l−1
, the second equality is by summing over i, and the third inequality is by λ
k+1< λ < 1. The resulting series is convergent whenever u
δ λM1−λM
< 1, which holds by (76). Therefore, by Fubini’s theorem, exchanging the order of summation in (78) results in
∞ l=1γ
lB ∞i=1
μ
yiλ
i(k+1)l
=
∞l=1
γ
Bl ∞i=l
i1+···+il=i
μ
yi1· · μ
yilλ
i(k+1)=
∞i=1
λ
i(k+1) i l=1i1+···+il=i
γ
lBμ
yi1
· · μ
yilBy the absolute convergence of both series on the right-hand side of (77), the product
on the right-hand side is equal to the Cauchy product, and so I can rewrite system (77)
as ⎧
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎨
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎩
∞ i=1λ
i(k+12)μ
xi− μ
xiλ
i− α
Bμ
xi+ μ
yiλ
i2− φ
i−1 j=1μ
xi−jλ
j2+ μ
yi−jλ
i2j
l=1
γ
Blj1+···+jl=j
μ
yj1
· · μ
yjl= 0
∞ i=1λ
ikμ
yi− μ
yiλ
i− α
Sμ
xiλ
2i+ μ
yi− φ
i−1
j=1
μ
xi−jλ
i−j2+ μ
yi−jj
l=1
γ
lSj1+···+jl=j
μ
xj1· · μ
xjl= 0
Setting all coefficients at λ
i(k+1/2)and λ
ikequal to zero results in the system
⎧ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎨
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎪ ⎩
μ
xi− μ
xiλ
i− α
Bμ
xi+ μ
yiλ
i2= φ
i−1
j=1
μ
xi−jλ
j2+ μ
yi−jλ
2ij
l=1
γ
lBj1+···+jl=j
μ
yj1
· · μ
yjlμ
yi− μ
yiλ
i− α
Sμ
xiλ
2i+ μ
yi= φ
i−1
j=1
μ
xi−jλ
i−j2+ μ
yi−jj
l=1
γ
lSj1+···+jl=j
μ
xj1
· · μ
xjl(79)
Using notation
A
i≡
1 − λ
i− α
B−α
Bλ
i/2−α
Sλ
i/21 − λ
i− α
Sμ
i≡
μ
xiμ
yiand
ϕ
i=
ϕ
xiϕ
yi≡
⎛
⎜ ⎜
⎜ ⎜
⎜ ⎝ φ
i−1
j=1
μ
xi−jλ
j/2+ μ
yi−jλ
i/2j
l=1
γ
Blj1+···+jl=j
μ
yj1
· · μ
yjlφ
i−1
j=1
μ
xi−jλ
(i−j)/2+ μ
yi−jj
l=1
γ
lSj1+···+jl=j
μ
xj1· · μ
xjl⎞
⎟ ⎟
⎟ ⎟
⎟ ⎠
(80)
I can write the system in matrix form as
A
iμ
i= ϕ
iSince det(A
i) = (1 − λ
i)(λ − λ
i) > 0, for i ≥ 2, matrix A
iis invertible, and I can solve for all μ
i(with the exception of i = 1):
μ
i= A
−1iϕ
i(81)
Equation (81) expresses μ
ithrough μ
1μ
i−1. For i = 1, the equalities in ( 79) are lin- early dependent (as det(A
i) = 0) and the relation between μ
x1and μ
y1is given by
μ
x1= λ
−12α
Bα
S(1 − α
S)μ
y1(82)
Equations (81) and (82) give the desired expressions for μ
xiand μ
yithrough the parame- ters of the model.
Step 2: Verify bounds. In this step, I verify that for μ
igiven by (82) and (81), bounds (75) and (76) indeed hold, and so my derivation in Step 1 is justified. Fix any M < 1. Let
u
δ= λ
−1/4− 1
Then
λ(1+u1δ)
= λ
−3/4> 1 and so (76) holds. If
ααBS
(1 − α
S) ≤ λ
12, then I set μ
y1= u
δM
μ
x1= λ
−12α
Bα
S(1 − α
S)μ
y1(83)
and otherwise, set
μ
x1= u
δM
μ
y1= λ
12α
Sα
B(1 − α
S) μ
x1(84)
The next claim verifies (75).
C laim 2. There exists ˆδ ∈ (0 1) such that for any δ ∈ ( ˆδ 1) such that for μ
x1and μ
y1defined above in (83) and (84) and μ
xiand μ
yidefined in (81), bounds (75) hold.
P roof. The proof is by induction on i. By ( 83) and (84), |μ
x1| ≤ u
δM and |μ
yi| ≤ u
δM , which proves the base of induction. Now, I prove the inductive step. Suppose that the statement is true for all j < i. I show that |μ
xi| < u
δM
iand |μ
yi| < u
δM
i. I can find the closed-form solution to system (81),
μ
xi= 1 − λ
i− α
Sϕ
xi+ α
Bλ
i/2ϕ
yi1 − λ
iλ − λ
i≤ 4 max
1 − λ
iα
Sα
B· max ϕ
xiϕ
yi1 − λ
iλ − λ
iand the same upper bound holds for |μ
yi|. Thus, it is sufficient to show that 4 max
1 − λ
iα
Sα
B· max ϕ
xiϕ
yi1 − λ
iλ − λ
i< u
δM
iNotice that
1−λαSi<
1α−λSfor i ≥ 2, and by l’Hospital rule, lim
δ→1 αS1−λ
= lim
δ→1 αS αS+αB−αSαB=
VS
VS+VB
< 1. Hence, for sufficiently large δ and all i ≥ 2, I have
1−λαSi< 1, and by an anal-
ogous argument,
1−λαBi< 1. Therefore,
max{1−λ1−λiαiSαB}≤ 1 for sufficiently large δ and it
remains to show that
4 max ϕ
xiϕ
yiλ − λ
i< u
δM
ifor sufficiently large δ. I will show that
(λ−λ|ϕi)uxi|δMi
<
14, and by symmetric argument,
|ϕyi|
(λ−λi)uδMi
<
14. Recall from (80) that
ϕ
xi= φ
i−1 j=1μ
xi−jλ
j2+ μ
yi−jλ
2ij
l=1
γ
lBj1+···+jl=j
μ
yj1
· · μ
yjl≤ φ
i−1 j=1λ
2j jl=1
γ
Blj1+···+jl=j
μ
xi−jμ
yj1
· · μ
yjl+ φλ
2ii−1
j=1
j l=1γ
Blj1+···+jl=j
μ
yi−jμ
yj1· · μ
yjl≤ φ
i−1
j=1
λ
2j jl=1
γ
lBj1+···+jl=j
u
lδ+1M
i+ φλ
i2i−1
j=1
j l=1γ
lBj1+···+jl=j
u
lδ+1M
i= φ
i−1 j=1λ
2j jl=1
γ
lBu
lδ+1M
ij − 1 l − 1
+ φλ
2i i−1 j=1 j l=1γ
Blu
lδ+1M
ij − 1 l − 1
≤ 2φu
δM
iD
i−1
j=1
λ
j2 jl=1
u
lδj − 1 l − 1
≤ 2φu
δM
iD
i−1
j=1
λ
j2u
δ(1 + u
δ)
j−1= 2φu
δM
iD u
δλ
121 − λ
i−12(1 + u
δ)
i−11 − λ
12(1 + u
δ)
= 2φu
δM
iDλ
121 − λ
i−12where the first inequality is due to the triangle inequality, the second inequality arises via the inductive hypothesis, the first equality makes use of the fact that the number of compositions of j into exactly l parts is
j−1l−1
, the fourth inequality is by λ
j> λ
ifor j < i and |γ
lB| < D, the fifth inequality is by summing over l, the second equality is the summation over j, and the last equality is plugging in u
δ= λ
−1/4− 1. Thus, I need to show that
2φD λ
141 − λ
i−14λ − λ
i< 1
4
This inequality holds for sufficiently large δ, as φ → 0 as δ → 1 and
δ→1
lim λ
141 − λ
i−14λ − λ
i= lim
δ→1
1 − λ
i−141 − λ
i−1= lim
δ→1
− i − 1 4 λ
i−54−iλ
i−2= i − 1 4i
This completes the proof of the inductive step and the claim. Step 3: Check solution. In this step, I verify that the candidate trajectories (x
ky
k) given by (74) that I have constructed indeed satisfy all conditions of Lemma 12. The convergence to (0 0) follows immediately by taking the limit k → ∞ of ( 74) and noting that λ < 1. Note that I still have one free parameter left: M that pins down μ
x1and μ
y1in (83) or (84). I choose M so that
x
1+ y
1= λ
3/2μ
xi+ λμ
yi= 2η
and so the initial condition in (25) is satisfied. In follows from (74) that x
kand y
kare decreasing in k and so
x
k+ y
k+1≤ x
k+ y
k≤ x
1+ y
1= 2η
verifying inequalities (28) and (29).
Finally, I show that x
kand y
kare positive. Observe that x
k=
∞i=1
λ
i(k+1/2)μ
xi= λ
k+1/2μ
x1+
∞i=2
λ
i(k+1/2)μ
xi≥ λ
k+1/2μ
x1−
∞i=2