Appendix A: First-best Joint Effort
In this appendix, we show that the maximization problem (3.1) has a unique solution
(
e
i
,
t
*
)
i
∈
N
′
; in addition
e
i
,
t
*
=
θ
0
,
t
θ
i
,
t
(
1
+
δ
γ
β
N
′
,
t
+
1
)
for some
β
N
′
,
t
+
1
∈
R
+
. The proof is by induction. By definition, U
N
′,T+1(θ
T+1) = 0; hence,
E
θ
−
0
,
T
+
1
U
N
′
,
T
+
1
(
θ
T
+
1
)
=
θ
0
,
T
+
1
β
N
′
,
T
+
1
for β
N
′,T+1 = 0. Fix some period t. Assume that
E
θ
−
0
,
t
+
1
U
N
′
,
t
+
1
(
θ
t
+
1
)
=
θ
0
,
t
+
1
β
N
′
,
t
+
1
for some
β
N
′
,
t
+
1
∈
R
+
. We have
E
U
N
′
,
t
+
1
(
θ
t
+
1
)
|
θ
t
,
(
e
i
,
t
)
i
∈
N
′
=
E
θ
0
,
t
+
1
E
θ
−
0
,
t
+
1
U
N
′
,
t
+
1
(
θ
t
+
1
)
|
θ
t
,
(
e
i
,
t
)
i
∈
N
′
=
E
θ
0
,
t
+
1
θ
0
,
t
+
1
β
N
′
,
t
+
1
|
θ
t
,
(
e
i
,
t
)
i
∈
N
′
=
E
[
θ
0
,
t
+
1
|
θ
t
,
(
e
i
,
t
)
i
∈
N
′
]
β
N
′
,
t
+
1
=
θ
0
,
t
+
γ
∑
i
∈
N
′
e
i
,
t
β
N
′
,
t
+
1
.
The first order condition for maximization problem (3.1) is: for each i ∈ N′,
1
+
δ
∂
E
U
N
′
,
t
+
1
(
θ
t
+
1
)
|
θ
t
,
(
e
i
,
t
)
i
∈
N
′
∂
e
i
,
t
=
e
i
,
t
θ
0
,
t
θ
i
,
t
,
which holds if and only if e
i,t
= θ
0,t
θ
i,t
(1 + δγβ
N
′,t+1). Thus, this maximization problem has a unique solution. We have
E
θ
−
0
,
t
U
N
′
,
t
(
θ
t
)
=
E
θ
−
0
,
t
∑
i
∈
N
′
θ
0
,
t
θ
i
,
t
(
1
+
δ
γ
β
N
′
,
t
+
1
)
−
1
2
θ
0
,
t
θ
i
,
t
(
1
+
δ
γ
β
N
′
,
t
+
1
)
2
+
E
θ
−
0
,
t
δ
θ
0
,
t
+
γ
∑
i
∈
N
′
θ
0
,
t
θ
i
,
t
(
1
+
δ
γ
β
N
′
,
t
+
1
)
β
N
′
,
t
+
1
= θ
0,t
β
N
′,t
, where
β
N
′
,
t
=
E
θ
−
0
,
t
1
2
(
1
+
δ
γ
β
N
′
,
t
+
1
)
2
∑
i
∈
N
′
θ
i
,
t
+
δ
β
N
′
,
t
+
1
≥
0
.
Lemma A.1
β
N
′,t
and β
N,t
− β
{i},t
are strictly increasing in T − t.
Proof
Let
θ
̄
≡
E
θ
i
=
∑
k
=
1
K
p
(
θ
̃
k
)
θ
̃
k
.
Step 1. To show that β
N
′,t
is strictly increasing in T − t, it suffices to show that β
N
′,t
> β
N
′,t+1 for each t ∈ {1, …, T}. The proof is by induction. It is obvious that β
N
′,T
> β
N
′,T+1 = 0. Suppose β
N
′,t
> β
N
′,t+1 for some t ∈ {2, …, T}. Then
β
N
′
,
t
−
1
=
|
N
′
|
2
(
1
+
δ
γ
β
N
′
,
t
)
2
θ
̄
+
δ
β
N
′
,
t
>
|
N
′
|
2
(
1
+
δ
γ
β
N
′
,
t
+
1
)
2
θ
̄
+
δ
β
N
′
,
t
+
1
=
β
N
′
,
t
.
Step 2. To show that β
N,t
− β
{i},t
is strictly increasing in T − t, it suffices to show that β
N,t
− β
{i},t
> β
N,t+1 − β
{i},t+1 for each t ∈ {1, …, T}. The proof is by induction. It is obvious that β
N,T
− β
{i},T
> β
N,T+1 − β
{i},T+1 = 0. Suppose β
N,t
− β
{i},t
> β
N,t+1 − β
{i},t+1 for some t ∈ {2, …, T}. Then x
t
≡ nβ
N,t
− β
{i},t
> nβ
N,t+1 − β
{i},t+1 ≡ x
t+1 and
y
t
≡
n
β
N
,
t
2
−
β
{
i
}
,
t
2
>
n
β
N
,
t
+
1
2
−
β
{
i
}
,
t
+
1
2
≡
y
t
+
1
, which implies
β
N
,
t
−
1
−
β
{
i
}
,
t
−
1
=
1
2
(
n
−
1
+
2
δ
γ
x
t
+
δ
2
γ
2
y
t
)
θ
̄
+
δ
(
β
N
,
t
−
β
{
i
}
,
t
)
>
1
2
(
n
−
1
+
2
δ
γ
x
t
+
1
+
δ
2
γ
2
y
t
+
1
)
θ
̄
+
δ
(
β
N
,
t
+
1
−
β
{
i
}
,
t
+
1
)
=
β
N
,
t
−
β
{
i
}
,
t
.
□
B Design
B.1 Bonuses
For constructing bonus
b
h
0
,
t
, for each
e
i
,
t
∈
R
+
, each x ∈ {0, 1}, and each θ
i,t
∈ Θ
i,t
, define
V
(
e
i
,
t
,
x
,
θ
i
,
t
)
=
−
c
i
,
t
(
e
i
,
t
,
θ
0
,
t
,
θ
i
,
t
)
+
q
h
0
,
t
s
i
,
t
(
h
0
,
t
)
π
(
e
i
,
t
,
e
−
i
,
t
*
,
r
)
+
δ
ξ
i
,
t
+
1
(
h
t
,
e
t
*
,
r
,
D
,
x
)
(
θ
0
,
t
+
γ
π
(
e
i
,
t
,
e
−
i
,
t
*
,
r
)
)
+
(
1
−
q
h
0
,
t
)
×
π
(
e
i
,
t
,
e
−
i
,
t
i
,
r
)
+
δ
ξ
i
,
t
+
1
(
h
t
,
e
t
i
,
r
,
D
,
x
)
(
θ
0
,
t
+
γ
π
(
e
i
,
t
,
e
−
i
,
t
i
,
r
)
)
.
Inequality (6.1) holds if
V
(
e
i
,
t
*
,
r
,
1
,
θ
i
,
t
)
+
(
1
−
q
h
0
,
t
)
b
h
0
,
t
≥
max
e
i
,
t
∈
[
0
,
e
i
,
t
*
,
r
]
V
(
e
i
,
t
,
0
,
θ
i
,
t
)
. We solve the maximization problem on the right hand side below. The marginal gain of e
i,t
is
M
G
≡
q
h
0
,
t
s
i
,
t
(
h
0
,
t
)
+
δ
γ
ξ
i
,
t
+
1
(
h
t
,
e
t
*
,
r
,
D
,
0
)
+
(
1
−
q
h
0
,
t
)
1
+
δ
γ
ξ
i
,
t
+
1
(
h
t
,
e
t
i
,
r
,
D
,
0
)
,
whereas the marginal cost of e
i,t
is MC(e
i,t
) ≡ e
i,t
/(θ
0,t
θ
i,t
). If
M
G
>
M
C
(
e
i
,
t
*
,
r
)
, then the solution is
e
i
,
t
*
,
r
; otherwise, the solution is θ
0,t
θ
i,t
MG. Finally, let
b
h
0
,
t
=
1
1
−
q
h
0
,
t
max
0
,
max
(
i
,
θ
i
,
t
)
max
e
i
,
t
∈
[
0
,
e
i
,
t
*
,
r
]
V
(
e
i
,
t
,
0
,
θ
i
,
t
)
−
V
(
e
i
,
t
*
,
r
,
1
,
θ
i
,
t
)
.
Fix a history
h
0
,
t
≡
(
h
t
,
θ
0
,
t
,
θ
̂
i
,
t
,
(
θ
j
,
t
)
j
≠
i
)
. If
π
τ
<
π
(
e
τ
r
)
for some τ < t, then it is obviously optimal for each agent to exert the recommended effort. In the following, we assume either t = 1 or
π
τ
≥
π
(
e
τ
r
)
for each τ < t. If agent i is recommended not to work, then she expects to receive no profit share from period t onwards; thus, her optimal effort is
e
i
,
t
*
=
0
. Suppose agent i receives recommended effort
e
i
,
t
*
,
r
. Due to the bonus
b
h
0
,
t
, she prefers exerting some
e
i
,
t
≥
e
i
,
t
*
,
r
to exerting any
e
i
,
t
<
e
i
,
t
*
,
r
. Thus, her optimal effort
e
i
,
t
*
is the solution to
max
e
i
,
t
≥
e
i
,
t
*
,
r
V
(
e
i
,
t
,
1
,
θ
i
,
t
)
, where V is defined in Appendix B.1. The marginal gain of e
i,t
is
M
G
′
≡
q
h
0
,
t
s
i
,
t
(
h
0
,
t
)
+
δ
γ
ξ
i
,
t
+
1
(
h
t
,
e
t
*
,
r
,
D
,
1
)
+
(
1
−
q
h
0
,
t
)
1
+
δ
γ
ξ
i
,
t
+
1
(
h
t
,
e
t
i
,
r
,
D
,
1
)
,
whereas the marginal cost of e
i,t
is MC(e
i,t
) ≡ e
i,t
/(θ
0,t
θ
i,t
). If
M
G
′
<
M
C
(
e
i
,
t
*
,
r
)
, then the solution is
e
i
,
t
*
,
r
; otherwise, the solution is θ
0,t
θ
i,t
MG′. Since the maximal aggregate payoff of a
|
N
h
0
,
t
|
–member partnership is
β
N
h
0
,
t
,
t
+
1
θ
0
,
t
+
1
, we have
ξ
i
,
t
+
1
(
h
t
,
e
t
*
,
r
,
D
,
1
)
≤
β
N
h
0
,
t
,
t
+
1
. Since the maximal aggregate payoff of a
|
N
h
0
,
t
|
–member partnership is increasing in
|
N
h
0
,
t
|
, we have
ξ
i
,
t
+
1
(
h
t
,
e
t
i
,
r
,
D
,
1
)
=
β
{
i
}
,
t
+
1
≤
β
N
h
0
,
t
,
t
+
1
. Hence, if
θ
̂
i
,
t
=
θ
i
,
t
, then
M
G
′
≤
M
C
(
e
i
,
t
*
,
r
)
=
1
+
δ
γ
β
N
h
0
,
t
,
t
+
1
. It follows that agent i’s optimal effort is
e
i
,
t
*
=
e
i
,
t
*
,
r
. □
B.3 Increasing Differences
Fix some history
h
0
,
t
≡
(
h
t
,
θ
0
,
t
,
θ
̂
i
,
t
,
(
θ
j
,
t
)
j
≠
i
)
, where
h
t
≡
(
θ
0
,
τ
,
θ
̂
−
0
,
τ
,
e
τ
r
,
π
τ
,
m
τ
)
τ
<
t
. Let
N
h
0
,
t
be the set of agents that were recommended to work in period t − 1. If either
|
N
h
0
,
t
|
=
1
or
π
τ
<
π
(
e
τ
r
)
for some τ < t, then agent i’s expected lifetime payoff
W
̃
i
,
t
(
h
t
,
θ
t
;
θ
̂
i
,
t
;
D
′
)
is independent of her report
θ
̂
i
,
t
; thus,
W
̃
i
,
t
(
h
t
,
⋅
,
θ
−
i
,
t
;
⋅
;
D
′
)
satisfies increasing differences. In the following, we assume
|
N
h
0
,
t
|
>
1
, and either t = 1 or
π
τ
≥
π
(
e
τ
r
)
for each τ < t, and show that
W
̃
i
,
t
(
h
t
,
⋅
,
θ
−
i
,
t
;
⋅
;
D
′
)
satisfies increasing differences.
Step 1. By construction,
s
i
,
t
(
h
0
,
t
)
=
θ
̂
i
,
t
+
δ
γ
β
N
h
0
,
t
,
t
+
1
θ
̂
i
,
t
−
∑
j
∈
N
h
0
,
t
θ
̂
j
,
t
/
|
N
h
0
,
t
|
∑
j
∈
N
h
0
,
t
θ
̂
j
,
t
.
It is easy to show that s
i,t
(
h
0,t
) is strictly increasing in
θ
̂
i
,
t
.
Step 2. As showed in Appendix B.2, if agent i is recommended not to work, then her optimal effort is
e
i
,
t
*
=
0
; if agent i receives recommended effort
e
i
,
t
*
,
r
, then her optimal effort is either
e
i
,
t
*
=
e
i
,
t
*
,
r
(corner solution) or
e
i
,
t
*
=
θ
0
,
t
θ
i
,
t
M
G
′
(interior solution). Thus, agent i’s maximal payoff from reporting
θ
̂
i
,
t
is
W
̃
i
,
t
(
h
t
,
θ
t
;
θ
̂
i
,
t
;
D
′
)
=
1
−
ϵ
*
+
ϵ
*
|
N
h
0
,
t
|
V
(
e
i
,
t
*
,
1
,
θ
i
,
t
)
where V is defined in Appendix B.1. If
e
i
,
t
*
is a corner solution, then
∂
V
(
e
i
,
t
*
,
1
,
θ
i
,
t
)
/
∂
θ
i
,
t
=
−
∂
c
i
,
t
(
e
i
,
t
*
,
θ
0
,
t
,
θ
i
,
t
)
/
∂
θ
i
,
t
(as all other terms of
V
(
e
i
,
t
*
,
1
,
θ
i
,
t
)
are independent of θ
i,t
). If
e
i
,
t
*
=
θ
0
,
t
θ
i
,
t
M
G
′
, then
∂
V
(
e
i
,
t
*
,
1
,
θ
i
,
t
)
∂
θ
i
,
t
=
−
∂
c
i
,
t
(
e
i
,
t
*
,
θ
0
,
t
,
θ
i
,
t
)
∂
θ
i
,
t
+
∂
e
i
,
t
*
∂
θ
i
,
t
M
G
′
−
e
i
,
t
*
θ
0
,
t
θ
i
,
t
=
−
∂
c
i
,
t
(
e
i
,
t
*
,
θ
0
,
t
,
θ
i
,
t
)
∂
θ
i
,
t
.
Consequently,
∂
W
̃
i
,
t
(
h
t
,
θ
t
;
θ
̂
i
,
t
;
D
′
)
∂
θ
i
,
t
=
1
−
ϵ
*
+
ϵ
*
|
N
h
0
,
t
|
(
e
i
,
t
*
)
2
2
θ
0
,
t
θ
i
,
t
2
.
Step 3. With a slight abuse of notation, we let
e
i
,
t
*
(
θ
̂
i
,
t
)
be agent i’s optimal effort conditional on reporting
θ
̂
i
,
t
and receiving recommended effort
e
i
,
t
*
,
r
≡
θ
0
,
t
θ
̂
i
,
t
(
1
+
δ
γ
β
N
h
0
,
t
,
t
+
1
)
. We show that
e
i
,
t
*
(
θ
̂
i
,
t
)
is increasing in
θ
̂
i
,
t
. Let
θ
̂
i
,
t
′
>
θ
̂
i
,
t
. If both
e
i
,
t
*
(
θ
̂
i
,
t
)
and
e
i
,
t
*
(
θ
̂
i
,
t
′
)
are corner solutions, then
e
i
,
t
*
(
θ
̂
i
,
t
′
)
=
θ
0
,
t
θ
̂
i
,
t
′
(
1
+
δ
γ
β
N
h
0
,
t
,
t
+
1
)
>
θ
0
,
t
θ
̂
i
,
t
(
1
+
δ
γ
β
N
h
0
,
t
,
t
+
1
)
=
e
i
,
t
*
(
θ
̂
i
,
t
)
. If both
e
i
,
t
*
(
θ
̂
i
,
t
)
and
e
i
,
t
*
(
θ
̂
i
,
t
′
)
are interior solutions, then
e
i
,
t
*
(
θ
̂
i
,
t
′
)
=
θ
0
,
t
θ
i
,
t
M
G
′
(
θ
̂
i
,
t
′
)
>
θ
0
,
t
θ
i
,
t
M
G
′
(
θ
̂
i
,
t
)
=
e
i
,
t
*
(
θ
̂
i
,
t
)
[the inequality is due to the fact that s
i,t
(
h
0,t
) is strictly increasing in
θ
̂
i
,
t
]. If
e
i
,
t
*
(
θ
̂
i
,
t
)
is corner and
e
i
,
t
*
(
θ
̂
i
,
t
′
)
is interior, then
e
i
,
t
*
(
θ
̂
i
,
t
′
)
≥
θ
0
,
t
θ
̂
i
,
t
′
(
1
+
δ
γ
β
N
h
0
,
t
,
t
+
1
)
>
θ
0
,
t
θ
̂
i
,
t
(
1
+
δ
γ
β
N
h
0
,
t
,
t
+
1
)
=
e
i
,
t
*
(
θ
̂
i
,
t
)
[the first inequality is due to the fact that agent i’s optimal effort weakly exceeds her recommended effort]. If
e
i
,
t
*
(
θ
̂
i
,
t
)
is interior and
e
i
,
t
*
(
θ
̂
i
,
t
′
)
is corner, then
e
i
,
t
*
(
θ
̂
i
,
t
′
)
>
θ
0
,
t
θ
i
,
t
M
G
′
(
θ
̂
i
,
t
′
)
>
θ
0
,
t
θ
i
,
t
M
G
′
(
θ
̂
i
,
t
)
=
e
i
,
t
*
(
θ
̂
i
,
t
)
[the first inequality is due to the fact that
M
G
′
(
θ
̂
i
,
t
′
)
<
M
C
(
e
i
,
t
*
(
θ
̂
i
,
t
′
)
)
].
Step 4. Without loss of generality, assume
θ
i
,
t
′
−
θ
i
,
t
=
ϵ
is sufficiently small. We have
W
̃
i
,
t
(
h
t
,
θ
i
,
t
′
,
θ
−
i
,
t
;
θ
̂
i
,
t
′
;
D
′
)
−
W
̃
i
,
t
(
h
t
,
θ
i
,
t
′
,
θ
−
i
,
t
;
θ
̂
i
,
t
;
D
′
)
−
W
̃
i
,
t
(
h
t
,
θ
i
,
t
,
θ
−
i
,
t
;
θ
̂
i
,
t
′
;
D
′
)
−
W
̃
i
,
t
(
h
t
,
θ
i
,
t
,
θ
−
i
,
t
;
θ
̂
i
,
t
;
D
′
)
=
W
̃
i
,
t
(
h
t
,
θ
i
,
t
′
,
θ
−
i
,
t
;
θ
̂
i
,
t
′
;
D
′
)
−
W
̃
i
,
t
(
h
t
,
θ
i
,
t
,
θ
−
i
,
t
;
θ
̂
i
,
t
′
;
D
′
)
−
W
̃
i
,
t
(
h
t
,
θ
i
,
t
′
,
θ
−
i
,
t
;
θ
̂
i
,
t
;
D
′
)
−
W
̃
i
,
t
(
h
t
,
θ
i
,
t
,
θ
−
i
,
t
;
θ
̂
i
,
t
;
D
′
)
=
∂
W
̃
i
,
t
(
h
t
,
θ
i
,
t
,
θ
−
i
,
t
;
θ
̂
i
,
t
′
;
D
′
)
∂
θ
i
,
t
−
∂
W
̃
i
,
t
(
h
t
,
θ
i
,
t
,
θ
−
i
,
t
;
θ
̂
i
,
t
;
D
′
)
∂
θ
i
,
t
ϵ
=
1
−
ϵ
*
+
ϵ
*
|
N
h
0
,
t
|
1
2
θ
0
,
t
θ
i
,
t
2
[
e
i
,
t
*
(
θ
̂
i
,
t
′
)
]
2
−
[
e
i
,
t
*
(
θ
̂
i
,
t
)
]
2
ϵ
≥
0
.
The last inequality is due to Step 3.
B.4 Transfers
Let
M
i
,
t
(
h
0
,
t
,
D
′
)
=
0
for each
i
∉
N
h
0
,
t
. We compute
M
i
,
t
(
h
0
,
t
,
D
′
)
for each
i
∈
N
h
0
,
t
below. If
N
h
0
,
t
=
{
i
}
, let
M
i
,
t
(
h
0
,
t
,
D
′
)
=
0
. In the following, we assume
|
N
h
0
,
t
|
>
1
.
Step 1. For each
i
∈
N
h
0
,
t
and each k ∈ {1, …, K − 1}, let
E
Δ
W
̃
i
,
t
(
h
t
,
θ
0
,
t
,
θ
̃
k
;
D
′
)
≡
E
(
θ
j
,
t
)
j
≠
i
W
̃
i
,
t
h
t
,
θ
0
,
t
,
θ
̃
k
,
(
θ
j
,
t
)
j
≠
i
;
θ
̃
k
+
1
;
D
′
−
W
̃
i
,
t
h
t
,
θ
0
,
t
,
θ
̃
k
,
(
θ
j
,
t
)
j
≠
i
;
θ
̃
k
;
D
′
be the expected welfare difference for an agent who has type
θ
̃
k
but raises her report from
θ
̃
k
to
θ
̃
k
+
1
(taking expectation over other agents’ current types).
Step 2. We let
M
t
[
h
t
,
θ
0
,
t
,
θ
̂
−
0
,
t
,
D
′
]
≠
(
0
,
…
,
0
)
only if the reported type profile
θ
̂
−
0
,
t
satisfies the following condition: one agent in
N
h
0
,
t
reports
θ
̃
k
and all other agents in
N
h
0
,
t
report
θ
̃
k
+
1
. Fix some agent
i
∈
N
h
0
,
t
. Let
χ
i
,
t
+
(
θ
̃
1
)
be a type profile in ∏
j∈N
Θ
j,t
such that agent i has type
θ
̃
1
and all other agents in
N
h
0
,
t
have type
θ
̃
2
. For each k ∈ {2, …, K − 1}, let
χ
i
,
t
−
(
θ
̃
k
)
be a type profile in ∏j∈N
Θ
j,t
such that some agent
j
∈
N
h
0
,
t
\
{
i
}
has type
θ
̃
k
−
1
and all other agents in
N
h
0
,
t
have type
θ
̃
k
; let
χ
i
,
t
+
(
θ
̃
k
)
be a type profile in ∏
j∈N
Θ
j,t
such that agent i has type
θ
̃
k
and all other agents in
N
h
0
,
t
have type
θ
̃
k
+
1
. Let
χ
i
,
t
−
(
θ
̃
K
)
be a type profile in ∏
j∈N
Θ
j,t
such that some agent
j
∈
N
h
0
,
t
\
{
i
}
has type
θ
̃
K
−
1
and all other agents in
N
h
0
,
t
have type
θ
̃
K
.
Let
M
i
,
t
+
(
θ
̃
1
)
=
p
|
N
h
0
,
t
|
−
1
(
θ
̃
2
)
M
i
,
t
[
h
t
,
θ
0
,
t
,
χ
i
,
t
+
(
θ
̃
1
)
,
D
′
]
.
For each k ∈ {2, …, K − 1}, let
M
i
,
t
−
(
θ
̃
k
)
=
p
(
θ
̃
k
−
1
)
p
|
N
h
0
,
t
|
−
2
(
θ
̃
k
)
M
i
,
t
[
h
t
,
θ
0
,
t
,
χ
i
,
t
−
(
θ
̃
k
)
,
D
′
]
and
M
i
,
t
+
(
θ
̃
k
)
=
p
|
N
h
0
,
t
|
−
1
(
θ
̃
k
+
1
)
M
i
,
t
[
h
t
,
θ
0
,
t
,
χ
i
,
t
+
(
θ
̃
k
)
,
D
′
]
.
Let
M
i
,
t
−
(
θ
̃
K
)
=
p
(
θ
̃
K
−
1
)
p
|
N
h
0
,
t
|
−
2
(
θ
̃
K
)
M
i
,
t
[
h
t
,
θ
0
,
t
,
χ
i
,
t
−
(
θ
̃
K
)
,
D
′
]
.
Fix some k ∈ {2, …, K − 1}. If agent i reports
θ
̃
k
, then
M
t
[
h
t
,
θ
0
,
t
,
θ
̂
−
0
,
t
,
D
′
]
≠
(
0
,
…
,
0
)
only in the following cases:
the reported type profile is
χ
i
,
t
−
(
θ
̃
k
)
: some agent
j
∈
N
h
0
,
t
\
{
i
}
reports
θ
̃
k
−
1
and all other agents in
N
h
0
,
t
report
θ
̃
k
[this case occurs with probability
p
(
θ
̃
k
−
1
)
p
|
N
h
0
,
t
|
−
2
(
θ
̃
k
)
], or
reported type profile is
χ
i
,
t
+
(
θ
̃
k
)
: each agent
j
∈
N
h
0
,
t
\
{
i
}
reports
θ
̃
k
+
1
[this case occurs with probability
p
|
N
h
0
,
t
|
−
1
(
θ
̃
k
+
1
)
].
Thus, agent i’s expected transfer from reporting
θ
̃
k
is
M
i
,
t
−
(
θ
̃
k
)
+
M
i
,
t
+
(
θ
̃
k
)
. Similar interpretations apply to
M
i
,
t
+
(
θ
̃
1
)
and
M
i
,
t
−
(
θ
̃
K
)
.
Remark B.1
For each k ∈ {1, …, K − 1},
p
(
θ
̃
k
)
M
i
,
t
+
(
θ
̃
k
)
+
(
|
N
h
0
,
t
|
−
1
)
p
(
θ
̃
k
+
1
)
M
i
,
t
−
(
θ
̃
k
+
1
)
=
p
(
θ
̃
k
)
p
|
N
h
0
,
t
|
−
1
(
θ
̃
k
+
1
)
∑
j
∈
N
M
j
,
t
[
h
t
,
θ
0
,
t
,
χ
i
,
t
+
(
θ
̃
k
)
,
D
′
]
=
0
.
Step 3. We construct
M
i
,
t
−
and
M
i
,
t
+
below.
For each k ∈ {1, …, K − 1}, let
ρ
k
≡
p
(
θ
̃
k
+
1
)
p
(
θ
̃
k
)
.
Let
ρ
̃
1
≡
(
|
N
h
0
,
t
|
−
1
)
ρ
1
and
ρ
̃
k
≡
(
|
N
h
0
,
t
|
−
1
)
ρ
̃
k
−
1
ρ
k
1
+
ρ
̃
k
−
1
for each k ∈ {2, …, K − 1}.
Set
M
i
,
t
−
(
θ
̃
K
)
=
−
∑
k
=
1
K
−
1
∏
l
=
k
K
−
1
1
1
+
ρ
̃
l
E
Δ
W
̃
i
,
t
(
h
t
,
θ
0
,
t
,
θ
̃
k
;
D
′
)
.
For each k ∈ {2, …, K − 1}, set
M
i
,
t
+
(
θ
̃
k
)
=
−
(
|
N
h
0
,
t
|
−
1
)
ρ
k
M
i
,
t
−
(
θ
̃
k
+
1
)
(see Remark B.1) and
M
i
,
t
−
(
θ
̃
k
)
=
−
∑
k
′
=
1
k
−
1
∏
l
=
k
′
k
−
1
1
1
+
ρ
̃
l
E
Δ
W
̃
i
,
t
(
h
t
,
θ
0
,
t
,
θ
̃
k
′
;
D
′
)
−
1
1
+
ρ
̃
k
−
1
M
i
,
t
+
(
θ
̃
k
)
.
Set
M
i
,
t
+
(
θ
̃
1
)
=
−
(
|
N
h
0
,
t
|
−
1
)
ρ
1
M
i
,
t
−
(
θ
̃
2
)
(see Remark B.1).
Given
M
i
,
t
−
and
M
i
,
t
+
, computing
M
t
is straightforward.
B.5 Linearity and Independence
In this appendix, we show that
(
σ
τ
*
,
μ
τ
*
)
τ
≥
t
induces linearity and independence. Let
D
′
′
be an arbitrary direct mechanism that contains
(
σ
τ
*
,
μ
τ
*
)
τ
≥
t
. Fix some history
h
0
,
t
≡
(
h
t
,
θ
0
,
t
,
θ
̂
−
0
,
t
)
, where
h
t
≡
(
θ
0
,
τ
,
θ
̂
−
0
,
τ
,
e
τ
r
,
π
τ
,
m
τ
)
τ
<
t
. Let
N
h
0
,
t
be the set of agents that were recommended to work in period t − 1. If
N
h
0
,
t
=
{
i
}
, then agent i is the sole member from period t onwards; thus
E
θ
−
0
,
t
W
̃
i
,
t
(
h
t
,
θ
t
,
D
′
′
)
=
θ
0
,
t
β
{
i
}
,
t
and
E
θ
−
0
,
t
W
̃
j
,
t
(
h
t
,
θ
t
,
D
′
′
)
=
0
for each j ≠ i. If
π
τ
<
π
(
e
τ
r
)
for some τ < t, then each agent
i
∈
N
h
0
,
t
becomes the sole member from period t onwards with probability
1
/
|
N
h
0
,
t
|
; thus,
E
θ
−
0
,
t
W
̃
i
,
t
(
h
t
,
θ
t
,
D
′
′
)
=
θ
0
,
t
β
{
i
}
,
t
/
|
N
h
0
,
t
|
for each
i
∈
N
h
0
,
t
and
E
θ
−
0
,
t
W
̃
i
,
t
(
h
t
,
θ
t
,
D
′
′
)
=
0
for each
i
∉
N
h
0
,
t
. In the following, we assume
|
N
h
0
,
t
|
>
1
, and either t = 1 or
π
τ
≥
π
(
e
τ
r
)
for each τ < t. For each
i
∉
N
h
0
,
t
, we have
E
θ
−
0
,
t
W
̃
i
,
t
(
h
t
,
θ
t
,
D
′
′
)
=
0
. Fix some
i
∈
N
h
0
,
t
. We will show that
E
θ
−
0
,
t
W
̃
i
,
t
(
h
t
,
θ
t
,
D
′
′
)
=
ξ
̃
i
,
t
(
D
′
′
)
θ
0
,
t
for some
ξ
̃
i
,
t
(
D
′
′
)
∈
R
+
(see Lemma B.3).
Agent i’s maximal expected lifetime payoff from reporting some
θ
i
,
t
′
∈
Θ
i
,
t
(taking expectation over other agents’ current types) is
Z
̃
i
,
t
(
h
t
,
θ
0
,
t
,
θ
i
,
t
;
θ
i
,
t
′
;
D
′
′
)
≡
E
(
θ
j
,
t
)
j
≠
i
W
̃
i
,
t
h
t
,
θ
0
,
t
,
θ
i
,
t
,
(
θ
j
,
t
)
j
≠
i
;
θ
i
,
t
′
;
D
′
+
M
i
,
t
−
(
θ
i
,
t
′
)
+
M
i
,
t
+
(
θ
i
,
t
′
)
,
where
M
i
,
t
−
(
θ
̃
1
)
=
M
i
,
t
+
(
θ
̃
K
)
≡
0
.
Lemma B.1
Z
̃
i
,
t
(
h
t
,
θ
0
,
t
,
θ
i
,
t
;
θ
i
,
t
;
D
′
′
)
≥
Z
̃
i
,
t
(
h
t
,
θ
0
,
t
,
θ
i
,
t
;
θ
i
,
t
′
;
D
′
′
)
Proof
The transfer rule
M
t
ensures local Bayesian incentive compatibility: for each k ∈ {1, …, K − 1}, an agent who has type
θ
̃
k
is indifferent between reporting
θ
̃
k
and
θ
̃
k
+
1
; formally,
Z
̃
i
,
t
(
h
t
,
θ
0
,
t
,
θ
̃
k
;
θ
̃
k
+
1
;
D
′
′
)
=
Z
̃
i
,
t
(
h
t
,
θ
0
,
t
,
θ
̃
k
;
θ
̃
k
;
D
′
′
)
. For each θ
−i,t
∈ Θ−i,t
, the payoff function
W
̃
i
,
t
(
h
t
,
⋅
,
θ
−
i
,
t
;
⋅
;
D
′
)
satisfies increasing differences; thus, standard arguments apply to show that local Bayesian incentive compatibility implies global Bayesian incentive compatibility. □
Lemma B.2
Z
̃
i
,
t
(
h
t
,
θ
0
,
t
,
θ
i
,
t
;
θ
i
,
t
;
D
′
′
)
≥
0
Proof
Step 1. We show that there exists k* ∈ {1, …, K} such that
M
i
,
t
−
(
θ
̃
k
*
)
+
M
i
,
t
+
(
θ
̃
k
*
)
≥
0
.
The proof is by contradiction. Suppose that
M
i
,
t
−
(
θ
̃
k
)
+
M
i
,
t
+
(
θ
̃
k
)
<
0
for each k ∈ {1, …, K}. Then
M
i
,
t
+
(
θ
̃
1
)
<
0
. Fix k ∈ {1, …, K − 2}. If
M
i
,
t
+
(
θ
̃
k
)
<
0
, then
M
i
,
t
−
(
θ
̃
k
+
1
)
>
0
, which implies
M
i
,
t
+
(
θ
̃
k
+
1
)
<
0
. Hence,
M
i
,
t
+
(
θ
̃
K
−
1
)
<
0
. Since
M
i
,
t
+
(
θ
̃
K
−
1
)
+
(
n
−
1
)
ρ
K
−
1
M
i
,
t
−
(
θ
̃
K
)
=
0
, we have
M
i
,
t
−
(
θ
̃
K
)
>
0
, which is a contradiction.
Step 2. We show that
Z
̃
i
,
t
(
h
t
,
θ
0
,
t
,
θ
i
,
t
;
θ
̃
k
*
;
D
′
′
)
≥
0
.
Since
M
i
,
t
−
(
θ
̃
k
*
)
+
M
i
,
t
+
(
θ
̃
k
*
)
≥
0
, it suffices to show that
W
̃
i
,
t
h
t
,
θ
0
,
t
,
θ
i
,
t
,
(
θ
j
,
t
)
j
≠
i
;
θ
̃
k
*
;
D
′
≥
0
for each
(
θ
j
,
t
)
j
≠
i
. Let
h
0
,
t
≡
(
h
t
,
θ
0
,
t
,
θ
̃
k
*
,
(
θ
j
,
t
)
j
≠
i
)
. We have
W
̃
i
,
t
h
t
,
θ
0
,
t
,
θ
i
,
t
,
(
θ
j
,
t
)
j
≠
i
;
θ
̃
k
*
;
D
′
≥
1
−
ϵ
*
+
ϵ
*
|
N
h
0
,
t
|
V
(
e
i
,
t
*
,
r
,
1
,
θ
i
,
t
)
where the right hand side is agent i’s expected lifetime payoff from exerting
e
i
,
t
*
,
r
conditional on receiving recommended effort
e
i
,
t
*
,
r
=
θ
0
,
t
θ
̃
k
*
(
1
+
δ
γ
β
N
h
0
,
t
,
t
+
1
)
and exerting no effort conditional on being recommended not to work. Due to symmetry and ϵ–efficiency,
|
N
h
0
,
t
|
ξ
i
,
t
+
1
(
h
t
,
e
t
*
,
r
,
D
,
1
)
is arbitrarily close to
β
N
h
0
,
t
,
t
+
1
. In addition,
q
h
0
,
t
is arbitrarily close to 1. Hence, we can find sufficiently small ϵ such that
(B.1)
V
(
e
i
,
t
*
,
r
,
1
,
θ
i
,
t
)
>
s
i
,
t
(
h
0
,
t
)
π
(
e
t
*
,
r
)
+
δ
β
N
h
0
,
t
,
t
+
1
|
N
h
0
,
t
|
θ
0
,
t
+
γ
π
(
e
t
*
,
r
)
−
c
i
,
t
(
e
i
,
t
*
,
r
,
θ
0
,
t
,
θ
i
,
t
)
−
ϵ
.
Substituting
s
i
,
t
(
h
0
,
t
)
=
e
i
,
t
*
,
r
+
δ
γ
β
N
h
0
,
t
,
t
+
1
(
e
i
,
t
*
,
r
−
π
(
e
t
*
,
r
)
/
|
N
h
0
,
t
|
)
∑
j
∈
N
h
0
,
t
e
j
,
t
*
,
r
+
δ
γ
β
N
h
0
,
t
,
t
+
1
(
e
j
,
t
*
,
r
−
π
(
e
t
*
,
r
)
/
|
N
h
0
,
t
|
)
into (B.1) gives
V
(
e
i
,
t
*
,
r
,
1
,
θ
i
,
t
)
>
1
2
θ
0
,
t
θ
̃
k
*
(
1
+
δ
γ
β
N
h
0
,
t
,
t
+
1
)
2
+
δ
β
N
h
0
,
t
,
t
+
1
|
N
h
0
,
t
|
θ
0
,
t
−
ϵ
≥
0
.
This implies
W
̃
i
,
t
h
t
,
θ
0
,
t
,
θ
i
,
t
,
(
θ
j
,
t
)
j
≠
i
;
θ
̃
k
*
;
D
′
≥
0
.
Step 3. By Lemma B.1,
Z
̃
i
,
t
(
h
t
,
θ
0
,
t
,
θ
i
,
t
;
θ
i
,
t
;
D
′
′
)
≥
Z
̃
i
,
t
(
h
t
,
θ
0
,
t
,
θ
i
,
t
;
θ
̃
k
*
;
D
′
′
)
≥
0
. □
Lemma B.3
E
θ
−
0
,
t
W
i
,
t
(
h
t
,
θ
t
,
D
′
′
)
=
θ
0
,
t
ξ
̃
i
,
t
(
D
′
′
)
, where
ξ
̃
i
,
t
(
D
′
′
)
∈
R
+
.
Proof
The proof is by induction. Assume that for each
h
t
+
1
≡
(
θ
0
,
τ
,
θ
̂
−
0
,
τ
,
e
τ
r
,
π
τ
,
m
τ
)
τ
<
t
+
1
such that
π
τ
≥
π
(
e
τ
r
)
for each τ < t + 1 and
e
j
,
t
r
>
0
for each
j
∈
N
h
0
,
t
, we have
E
θ
−
0
,
t
+
1
W
̃
i
,
t
(
h
t
+
1
,
θ
t
+
1
,
D
′
)
=
ξ
̃
i
,
t
+
1
(
D
′
)
θ
0
,
t
+
1
for some
ξ
̃
i
,
t
+
1
(
D
′
)
∈
R
+
.
Step 1. Let
h
0
,
t
≡
(
h
t
,
θ
̂
i
,
t
,
θ
−
i
,
t
)
. Let
e
i
,
t
*
be agent i’s optimal effort conditional on receiving recommended effort
e
i
,
t
*
,
r
.
If
e
i
,
t
*
=
e
i
,
t
*
,
r
, then
W
̃
i
,
t
(
h
t
,
θ
t
;
θ
̂
i
,
t
;
D
′
)
=
−
1
2
θ
0
,
t
θ
̂
i
,
t
(
1
+
δ
γ
β
N
h
0
,
t
,
t
+
1
)
2
+
q
h
0
,
t
s
i
,
t
(
h
0
,
t
)
+
δ
γ
ξ
̃
i
,
t
+
1
(
D
′
)
θ
0
,
t
(
1
+
δ
γ
β
N
h
0
,
t
,
t
+
1
)
θ
̂
i
,
t
+
∑
j
∈
N
h
0
,
t
\
{
i
}
θ
j
,
t
+
θ
0
,
t
δ
ξ
̃
i
,
t
+
1
(
D
′
)
+
(
1
−
q
h
0
,
t
)
1
+
δ
γ
β
{
i
}
,
t
+
1
θ
0
,
t
θ
̂
i
,
t
(
1
+
δ
γ
β
N
h
0
,
t
,
t
+
1
)
+
θ
0
,
t
δ
β
{
i
}
,
t
+
1
.
If
e
i
,
t
*
=
θ
0
,
t
θ
i
,
t
M
G
′
, where
M
G
′
≡
q
h
0
,
t
s
i
,
t
(
h
0
,
t
)
+
δ
γ
ξ
̃
i
,
t
+
1
(
D
′
)
+
(
1
−
q
h
0
,
t
)
1
+
δ
γ
β
{
i
}
,
t
+
1
,
then
W
̃
i
,
t
(
h
t
,
θ
t
;
θ
̂
i
,
t
;
D
′
)
=
−
1
2
θ
0
,
t
θ
i
,
t
(
M
G
′
)
2
+
q
h
0
,
t
s
i
,
t
(
h
0
,
t
)
+
δ
γ
ξ
̃
i
,
t
+
1
(
D
′
)
θ
0
,
t
θ
i
,
t
M
G
′
+
(
1
+
δ
γ
β
N
h
0
,
t
,
t
+
1
)
∑
j
θ
j
,
t
+
θ
0
,
t
δ
ξ
̃
i
,
t
+
1
(
D
′
)
+
(
1
−
q
h
0
,
t
)
1
+
δ
γ
β
{
i
}
,
t
+
1
θ
0
,
t
θ
i
,
t
M
G
′
+
θ
0
,
t
δ
β
{
i
}
,
t
+
1
.
By construction, s
i,t
(
h
0,t
) depends only on
(
θ
̂
i
,
t
,
θ
−
i
,
t
)
. Hence, we can write
W
̃
i
,
t
(
h
t
,
θ
t
;
θ
̂
i
,
t
;
D
′
)
=
θ
0
,
t
ζ
i
,
t
(
θ
−
0
,
t
;
θ
̂
i
,
t
;
D
′
)
, where
ζ
i
,
t
(
θ
−
0
,
t
;
θ
̂
i
,
t
;
D
′
)
<
∞
.
Step 2. For each k ∈ {1, …, K − 1}, define
ζ
̃
i
,
t
(
θ
̃
k
,
D
′
)
≡
E
(
θ
j
,
t
)
j
≠
i
ζ
i
,
t
θ
̃
k
,
(
θ
j
,
t
)
j
≠
i
;
θ
̃
k
+
1
;
D
′
−
ζ
i
,
t
θ
̃
k
,
(
θ
j
,
t
)
j
≠
i
;
θ
̃
k
;
D
′
.
Then,
E
Δ
W
̃
i
,
t
(
h
t
,
θ
0
,
t
,
θ
̃
k
,
D
′
)
=
θ
0
,
t
ζ
̃
i
,
t
(
θ
̃
k
,
D
′
)
. Thus, for each k ∈ {1, …, K}, we can write
M
i
,
t
+
(
θ
̃
k
)
≡
θ
0
,
t
μ
i
,
t
+
(
θ
̃
k
,
D
′
)
and
M
i
,
t
−
(
θ
̃
k
)
≡
θ
0
,
t
μ
i
,
t
−
(
θ
̃
k
,
D
′
)
for some
μ
i
,
t
−
(
θ
̃
k
,
D
′
)
<
∞
and
μ
i
,
t
+
(
θ
̃
k
,
D
′
)
<
∞
.
Step 3. We have
E
(
θ
j
,
t
)
j
≠
i
W
i
,
t
(
h
t
,
θ
t
,
D
′
′
)
=
E
(
θ
j
,
t
)
j
≠
i
W
̃
i
,
t
(
h
t
,
θ
t
;
θ
i
,
t
;
D
′
)
+
M
i
,
t
−
(
θ
i
,
t
)
+
M
i
,
t
+
(
θ
i
,
t
)
=
E
(
θ
j
,
t
)
j
≠
i
θ
0
,
t
ζ
i
,
t
θ
i
,
t
,
(
θ
j
,
t
)
j
≠
i
;
θ
i
,
t
;
D
′
+
θ
0
,
t
μ
i
,
t
−
(
θ
i
,
t
,
D
′
)
+
θ
0
,
t
μ
i
,
t
+
(
θ
i
,
t
,
D
′
)
=
θ
0
,
t
ξ
̂
i
,
t
(
θ
i
,
t
,
D
′
′
)
for some
ξ
̂
i
,
t
(
θ
i
,
t
,
D
′
′
)
<
∞
.
By Lemma B.2, we have
Z
̃
i
,
t
(
h
t
,
θ
0
,
t
,
θ
i
,
t
;
θ
i
,
t
;
D
′
′
)
=
E
(
θ
j
,
t
)
j
≠
i
W
i
,
t
(
h
t
,
θ
t
,
D
′
′
)
≥
0
, which implies
ξ
̂
i
,
t
(
θ
i
,
t
,
D
′
′
)
≥
0
. It follows that
E
θ
−
0
,
t
W
i
,
t
(
h
t
,
θ
t
,
D
′
′
)
=
θ
0
,
t
ξ
̃
i
,
t
(
D
′
′
)
for some
ξ
̃
i
,
t
(
D
′
′
)
∈
R
+
. □
By construction, at each history
h
0
,
t
≡
(
h
t
,
θ
t
)
∈
H
0
,
t
*
, the mechanism designer assigns probability 1 − ϵ* to the first-best joint effort
σ
̃
t
F
(
θ
t
,
N
)
. With ϵ* < ϵ, the direct mechanism
D
*
satisfies ϵ-efficiency. Condition (a) of periodic Bayesian incentive compatibility holds due to Lemma 6.1 and the fact that the introduction of
M
t
does not affect optimal efforts. Condition (b) of periodic Bayesian incentive compatibility holds due to Lemma B.1 and the fact that the mechanism designer’s decision is independent of reports at any history
h
0,t
such that either
|
N
h
0
,
t
|
=
1
or
π
τ
<
π
(
e
τ
r
)
for some τ < t. Finally, interim individual rationality holds due to Lemma B.2 and the fact that each agent either gets excluded permanently or becomes a permanent sole member at any history
h
0,t
such that either
|
N
h
0
,
t
|
=
1
or
π
τ
<
π
(
e
τ
r
)
for some τ < t (it is easy to show that a sole member partnership always earns a non-negative expected lifetime payoff). □
Let
G
V
*
be the partnership game associated with voting mechanism
V
*
. For each agent i and each period t, a stage-1 history is some
h
i
,
t
1
≡
⟨
(
θ
i
,
τ
,
e
i
,
τ
)
τ
<
t
,
h
t
,
θ
0
,
t
,
θ
i
,
t
⟩
at which she votes, a stage-2 history is some
h
i
,
t
2
≡
(
h
i
,
t
1
,
v
t
,
e
i
,
t
r
)
at which she supplies effort. We construct a perfect Bayesian equilibrium of the partnership game
G
V
*
below.
Beliefs. Fix agent i ∈ N and agent j ≠ i. At any stage-1 history
h
i
,
t
1
, if agent i observes that agent j has selected item k
τ
in each period τ < t, then agent i assigns probability one to the event that agent j’s past types are
(
θ
̃
k
τ
)
τ
<
t
; formally,
β
i
j
*
(
θ
̃
k
τ
)
τ
<
t
|
h
i
,
t
1
=
1
. Likewise, at any stage-2 history
h
i
,
t
2
, if agent i observes that agent j has selected item k
τ
in each period τ ≤ t, then agent i assigns probability one to the event that agent j’s past types are
(
θ
̃
k
τ
)
τ
≤
t
; formally,
β
i
j
*
(
θ
̃
k
τ
)
τ
≤
t
|
h
i
,
t
2
=
1
.
Strategies. Fix agent i ∈ N. A strategy
λ
i
*
assigns a voting strategy to a stage-1 history and an effort to a stage-2 history:
λ
i
*
(
h
i
,
t
1
)
∈
V
i
,
t
*
(
h
t
,
θ
0
,
t
)
and
λ
i
*
(
h
i
,
t
2
)
∈
E
i
,
t
. For each
h
i
,
t
1
≡
⟨
(
θ
i
,
τ
,
e
i
,
τ
)
τ
<
t
,
h
t
,
θ
0
,
t
,
θ
̃
k
⟩
, let
λ
i
*
(
h
i
,
t
1
)
be a voting strategy that selects item k. For each
h
i
,
t
2
≡
(
h
i
,
t
1
,
v
t
,
e
i
,
t
r
)
such that each agent j ∈ N has selected item k
j
in period t, let
λ
i
*
(
h
i
,
t
2
)
∈
arg
max
e
i
,
t
W
̂
i
,
t
(
h
t
,
θ
t
,
θ
̃
k
i
,
e
i
,
t
r
,
e
i
,
t
,
D
*
)
, where
θ
j
,
t
=
θ
̃
k
j
for each j ≠ i.
Fix period t with history h
t
, experience θ
0,t
, and true types
(
θ
̃
k
1
,
…
,
θ
̃
k
n
)
. By construction of
λ
i
*
(
h
i
,
t
1
)
, each agent i selects item k
i
on her menu. Since
α
t
*
(
h
t
,
θ
0
,
t
,
θ
̃
k
1
,
…
,
θ
̃
k
n
)
is contained in item k
i
for each agent i, it gets n votes. If an arrangement
α
t
*
(
h
t
,
θ
0
,
t
,
θ
−
0
,
t
′
)
has
θ
i
,
t
′
≠
θ
̃
k
i
for some agent i, then it is not contained in item k
i
for agent i and gets at most n − 1 votes. As a result,
α
t
*
(
h
t
,
θ
0
,
t
,
θ
̃
k
1
,
…
,
θ
̃
k
n
)
is the winning arrangement. By construction of
λ
i
*
(
h
i
,
t
2
)
, it is clear that after each agent i selects item k
i
on her menu, agents supply recommended efforts. In the following, we show that
(
β
i
j
*
,
λ
i
*
)
i
∈
N
is a perfect Bayesian equilibrium.
Bayes’ Rule. Without loss of generality, fix a stage–1 history
h
i
,
t
1
at which agent i observes that agent j has selected item k
τ
in each period τ < t. By construction of
β
i
j
*
, at
h
i
,
t
1
, agent i assigns probability one to the event that agent j’s past types are
(
θ
̃
k
τ
)
τ
<
t
. By construction of
λ
j
*
, agent j has selected item k
τ
in each period τ < t if and only if agent j’s past types are
(
θ
̃
k
τ
)
τ
<
t
. Thus,
β
i
j
*
satisfies Bayes’ rule given
λ
j
*
.
Sequential Rationality. Fix some period t with history h
t
and experience θ
0,t
, and some agent i with type
θ
i
,
t
=
θ
̃
k
i
*
. Assume the following: (a) from period t + 1 onwards, each agent j ∈ N with type
θ
̃
k
j
selects item k
j
and supplies recommended effort, (b) in period t, each agent j ≠ i with type
θ
̃
k
j
selects item k
j
and exerts recommended effort.
Suppose agent i selects item k
i
in period t. Then
α
t
*
(
h
t
,
θ
0
,
t
,
θ
̃
k
1
,
…
,
θ
̃
k
n
)
is the winning arrangement in period t. It is easy to see that if agent i receives recommended effort
e
i
,
t
r
and exerts effort e
i,t
, then her expected payoff is
W
̂
i
,
t
(
h
t
,
θ
t
,
θ
̃
k
i
,
e
i
,
t
r
,
e
i
,
t
,
D
*
)
, where
θ
j
,
t
=
θ
̃
k
j
for each j ≠ i. Thus, exerting effort
λ
i
*
(
h
i
,
t
2
)
is sequentially rational.
It is easy to see that if agent i selects item k
i
in period t, then her expected payoff is
E
(
θ
j
,
t
)
j
≠
i
W
̃
i
,
t
(
h
t
,
θ
t
;
θ
̃
k
i
;
D
*
)
. Due to periodic Bayesian incentive compatibility of
D
*
, selecting item
k
i
*
is sequentially rational. □
