Proof of Proposition 3.2. The elements in Ω(P P) will be selected in the long run if and only if C(Ω(P P)) has the minimum cost among all the absorbing sets. One can immediately obtain the condition using Table 3.2.
Proof of Proposition 3.3. The elements in Ω(RR) will be the LRE if and only ifC(Ω(RR))≤C(Ω(P R)). Rearranging the inequality, one can obtain
d≤2q∗+∆A(d) N , where ∆A(d)∈(−1,2) is a function ofd.
To select the elements in Ω(P P) as stochastically stable requires⌊dN⌋= 2N−1 orC(Ω(P R)) =C(Ω(P P)). The first equation can be rearranged as
d= 2−∆B(d)
N , (3.14)
where ∆B(d)∈(0,1]. The second requirement can be presented as d= 2 +∆C(d)
N , (3.15)
where ∆C(d) has a negative lower bound and a positive upper bound.
Consider the function f(d) = 2q∗+∆AN(d), (3.14) and (3.15). Let b1 be the maximum of the absolute values among the lower bounds and upper bounds of ∆A(d), ∆B(d) and ∆C(d). Then, the following conditions hold simultaneously.
2q∗− b1
N ≤ 2q∗+∆AN(d) ≤2q∗+ b1
N (3.16)
2− b1
N ≤ 2− ∆BN(d) ≤2 (3.17)
2− b1
N ≤ 2 +∆CN(d) ≤2 + b1
N (3.18)
Hence, for any η > 0, there exist an integer ¯N > b1/η, such that for all N >N¯, bN1 < η, and hence,
2q∗−η < 2q∗+∆AN(d) <2q∗+η (3.19) 2−η < 2− ∆BN(d) <2 (3.20) 2−η < 2 +∆CN(d) <2 +η (3.21) Therefore, for any η > 0, there exist an integer ¯N, such that for all N >N¯,
(1) the elements in Ω(RR) will be selected if d≤2q∗−η;
(2) the elements in Ω(P R) and Ω(RP) will be selected if 2q∗+η ≤ d≤ 2−η.
Proof of Proposition 3.4. If c1≤c2, TP1 has the minimum cost for the transition from Ω(P R) to Ω(RR). Hence, the elements in Ω(P P) will be
selected if and only ifC(Ω(P P))≤C(Ω(P P)) andC(Ω(P P))≤C(Ω(RR)), which implies ⌈(2N − ⌊dN⌋)(1−q∗)⌉=⌈(2N − ⌊dN⌋)q∗⌉ ≤c1. If c1 > c2, TP2 has the minimum cost for the transition. The same argument indicates
⌈(2N − ⌊dN⌋)(1−q∗)⌉=⌈(2N − ⌊dN⌋)q∗⌉ ≤c2. Combining the two cases
gives the result in the statement.
Proof of Proposition 3.5. Let CT P1(Ω(RR)) be the minimum cost of Ω(RR) through TP1, and CT P2(Ω(RR)) that of Ω(RR) through TP2. For
must be in the range offd˜(d). A straightforward computation shows that if d¯≥(<)2q∗, then ˜d≤(>)2q∗.
Selecting the elements in Ω(P P) in the long run entails⌊dN⌋= 2N−1 orC(Ω(P P)) =C(Ω(P R)), which requires (3.14) or (3.15).
Letb2 be the maximum of the absolute value of lower bounds and upper bounds of ∆X(d) for all X ∈ {B, C, D, E}. The following conditions must
N >N¯ the following conditions hold.
d¯−η < fd¯(d) <d¯+η (3.28) d˜−η < fd˜(d) <d˜+η (3.29) 2−η < 2− ∆BN(d) <2 (3.30) 2−η < 2 +∆CN(d) <2 +η (3.31) Consequently, given any η > 0, there exists an ¯N, such that for all N >N¯,
(1). CT P1(Ω(RR))< CT P2(Ω(RR)) ifd≤d¯−η,CT P1(Ω(RR))> CT P2(Ω(RR)) ifd≥d¯+η;
(2). CT P1(Ω(RR))< C(Ω(P R)) ifd≤2q∗−η,CT P1(Ω(RR))> C(Ω(P R)) ifd≥2q∗+η.
(3). CT P2(Ω(RR))< C(Ω(P R)) if d≤d˜−η,CT P2(Ω(RR))> C(Ω(P R)) ifd≥d˜+η.
(4). C(Ω(P P))> C(Ω(P R)), if d≤2−η.
Case 1. d¯≥2q∗ if and only if ˆq ≤1/q∗−2 + 2q∗. Then, ˜d ≤2q∗ ≤ d.¯ Hence, if d ≤ 2q∗ −η, CT P1(Ω(RR)) < C(Ω(P R)), hence, the elements in Ω(RR) are selected. If d ≥ 2q∗ +η, C(Ω(P R)) < CT P1(Ω(RR)) and C(Ω(P R)) < CT P2(Ω(RR)). Hence, the elements in Ω(P R) and Ω(RP) will be selected.
Case 2. d <¯ 2q∗ if and only if ˆq >1/q∗−2 + 2q∗. Then, ¯d <2q∗ <d. if˜ d≤d˜−η,CT P2(Ω(RR))< C(Ω(P R)). Hence, the elements in Ω(RR) will be selected. Ifd≥d˜+η,CT P1(Ω(RR))> C(Ω(P R)) andCT P2(Ω(RR))>
C(Ω(P R)). Hence, the elements in Ω(P R) and Ω(RP) will be selected.
Proof of Lemma 3.1. Appendix II exhibits the absorbing sets in the case with asymmetric capacity and mobility constraints. The case with sym-metric constraints is simply a particular situation and is contained in the asymmetric case. All the results involving the symmetric constraints still
hold, which gives the statement in this lemma.
Proof of Theorem 3.1. Case III.1: ⌊dkN⌋ < 2N −1, for all k = 1,2.
In this case, there are two absorbing sets and two non-singleton absorbing sets, and, for each of them, there are four candidates for the minimum-cost transition tree. For each class of Ω(φ)-trees (φ∈Φ ={RR, P R, RP, P P}), one can compare the costs of the four candidates presented in Table 3.5, which gives the following result.
Then, for eachξ∈ {1,2,3,4}, we compare the minimum costs among the Ω(φ)ξ-trees for allφ∈Φ. The element(s) in the absorbing sets which have the lowest cost among the minimum costs of Ω(φ)ξ-trees will be selected as stochastically stable. A straightforward computation shows that Ω(RP) has the minimum cost if the condition in (2a) holds, Ω(RR) has the minimum cost if the conditions in (2b) and (2c) hold, and Ω(P R) has the minimum cost if the condition in (2d) hold. For the area between the sets in (2b) and (2c), that is{(d1, d2)|d1 ≥d2−η, d1≤d2+η, d1 ≤2−1−q∗q∗d2−η, andd1 ≥
2q∗
1−q∗ −1−q∗q∗d2+η}, one can see that Ω(RR) has the minimum cost.
Case III.2: ⌊dkN⌋ = 2N −1 and ⌊dℓN⌋ < 2N −1, for k, ℓ = 1,2, k 6= ℓ.
Consider first the case wherek= 1. According to Appendix I and II, there are two absorbing sets: Ω(RP) and Ω(P R, P P). Further, C(Ω(RP)) =
⌈(2N− ⌊d2N⌋)(1−q∗)⌉,C(Ω(P R, P P)) =⌈(2N− ⌊d2N⌋)q∗⌉. C(Ω(RP))<
C(Ω(P R, P P)) if
d2 <2 +∆C(d2)
N . (3.32)
Since ∆C(d2) are bounded, for any η > 0, there exists an integer ¯N such that for allN >N¯,|∆CN(d2)|< η, and hence,
2−η <2 +∆C(d2)
N .
Therefore, if ⌊d1N⌋ = 2N −1 and ⌊d2N⌋ < 2N −1, for any η > 0, there exists an integer ¯N such that for allN >N¯, the element in Ω(RP) is selected as stochastically stable ifd2 ≤2−η.
The same argument holds for k = 2. Hence, for ⌊d2N⌋ = 2N −1 and
⌊d1N⌋<2N −1, for any η >0, there exists an integer ¯N such that for all N >N¯, Ω(P R) is selected as stochastically stable ifd1≤2−η.
If ⌊dkN⌋ = 2N −1 for both k = 1,2, there is a unique absorbing set Ω(P R, P P, RP). Rearranging⌊dkN⌋= 2N−1, we have
dk = 2−∆B(dk)
N .
Because ∆B(dk) is bounded, for any η >0, there exists an integer ¯N, such that for all N > N¯, |∆BN(dk)|< η, and hence 2−η < 2− ∆BN(dk), for both k= 1,2. Therefore, the elements in Ω(P R, P P, RP) can be selected only if dk >2−η for both k= 1,2. Equivalently, the elements in Ω(P R, P P, RP)
cannot be selected if d1 ≤2−η ord2 ≤2−η.
Case III.3: dk = 2 and ⌊dℓN⌋ < 2N −1, for k, ℓ = 1,2 and k 6=ℓ. Con-sider the case where k = 1 first. There are three absorbing sets, Ω(P O), Ω(RO) and Ω(RP). According to Table 3.6, the minimum costs of the each absorbing set are
C(Ω(P O)) = 1 +⌈(2N − ⌊d2N⌋)q∗⌉ C(Ω(RP)) = 1 +⌈(2N − ⌊d2N⌋)(1−q∗)⌉ C(Ω(RO)) = ⌈2N(1−q∗)⌉+⌈⌊d2N⌋(1−q∗)⌉
A straightforward comparison shows that C(Ω(RO)) > C(Ω(RP)) and C(Ω(RO))> C(Ω(P O)) ifN > 2(1−1q∗). Hence, ifN is large enough, Ω(RO) can never be selected.
C(Ω(RP)) < C(Ω(P O)) if d2 ≤ 2 + ∆CN(d2). Again, since ∆C(d2) is bounded, for anyη >0, there exists an integer ¯N, such that for allN >N¯,
|∆CN(d2)|< η, and hence 2−η <2 +∆CN(d2). Therefore, ifN is large enough, the element in Ω(RP) is the unique LRE in this case. The same argument holds for the case wherek= 2, hence, Ω(P R) is the unique LRE there.
Case III.4: dk = 2 and ⌊dℓN⌋ = 2N −1. Consider first the case k = 1.
According to Lemma 3.1, there are only two absorbing sets, Ω(P O) and Ω(RO). The analysis in Appendix II shows that the transition Ω(P O) → Ω(RO) only requires one mutant. The transition in the reverse direction needs ⌈2N(1−q∗)⌉ mutants. Hence, ifN > 2(1−1q∗), the element in Ω(P O) is the unique LRE. The condition ⌊d2N⌋ = 2N −1 can be rearranged as d2 = 2− ∆BN(d2). Since ∆B(d2) is bounded, for any η > 0, there exists an integer ¯N, such that for all N > N¯, 2−η < 2− ∆BN(d2), and hence, the element in Ω(P O) can be selected only if d2 >2−η. The same argument holds for the case where k = 2. Therefore, for any η > 0, the element in Ω(OP) can be selected only if d1 >2−η.
Case III.5: d1 = d2 = 2. There are four absorbing sets, Ω(RO), Ω(P O), Ω(OR) and Ω(OP). As explained in section 3.3.2, the minimum cost of Ω(P O)- or Ω(OP)-tree is 3.
We have shown in Appendix II (Case II.5 ) that, a minimum-cost Ω(RO) (Ω(OR))-tree must involve a direct transition from Ω(P O)(Ω(OP)) to Ω(RO) (Ω(OR)), which requires⌈2N(1−q∗)⌉ mutants.
Hence, ifN > 2(1−1q∗), the minimum transition cost of Ω(RO)- or Ω(OR)-tree must be larger than 3. Therefore, Ω(P O) and Ω(OP) are LRE for N large enough.
Consider all the five cases above together. For any η > 0, there exists an integer ¯N′, which is the maximum of all the ¯Ns in all the cases, such that when N > N¯′, all the results above hold simultaneously. Renaming ¯N′ as
N¯, we have the statement in the theorem.
Proof of Proposition 3.6. We show in the proof of Theorem 3.1 that (1) the LRE are the elements in Ω(P P) if d1 = d2 = 2, and (2) the LRE form a subset of Ω(RP)∪Ω(P R)∪Ω(OP)∪Ω(P O)∪Ω(P P), if (d1, d2)∈ Vc(η)\ {(2,2)}.
Then we are looking for the LRE in the remaining area of V(η). Con-sidering a large enough N, we use a similar approach to that in the proof of Theorem 3.1. After comparing the minimum costs of different transition trees presented in Table 3.5, one can see that for anyη >0, there exists an integer ¯N, such that for all N >N¯, the following results hold.
AreaVa(η). In this area, Ω(RR)1- or Ω(RR)2-tree has the minimum transi-tion cost among all Ω(RR)-trees. Similarly, Ω(RP)1- or Ω(RP)2-tree has the minimum transition cost among all Ω(RP)-trees. LetZ1={Ω(RR)1,Ω(RR)2, Ω(RP)1,Ω(RP)2}. Then, one can obtain that, for (d1, d2)∈Va(η),C(z)<
C(Ω(P R)ξ) andC(z)< C(Ω(P P)ξ) for allz∈Z1and for allξ ∈ {1,2,3,4}. Hence, the LRE inVa(η) form a subset of Ω(RR)∪Ω(RP).
Area Vb(η). In this area, Ω(RR)3- or Ω(RR)4-tree has the minimum tran-sition cost among all Ω(RR)-trees. Meanwhile, Ω(P R)3- or Ω(P R)4-tree has the minimum cost among all Ω(P R)-trees. Denote by Z2 ={Ω(RR)3, Ω(RR)4,Ω(P R)3, Ω(P R)4}. After a series of comparisons, one can obtain that, for (d1, d2) ∈ Vb(η), C(z) < C(Ω(RP)ξ) and C(z) < C(Ω(P P)ξ) for allz∈Z2 and for allξ ∈ {1,2,3,4}. Hence, the LRE in Va(η) form a subset of Ω(RR)∪ {Ω(P R)}.
Area Vc(η). In this area, Ω(RP)1- or Ω(RP)4-tree has the minimum tran-sition cost among all Ω(RP)-trees. Meanwhile, Ω(P R)1- or Ω(P R)4-tree has the minimum cost among all Ω(P R)-trees. Denote byZ3 ={Ω(RP)1,Ω(RP)4, Ω(P R)1,Ω(P R)4}. After a series of comparisons, one can obtain that, for (d1, d2)∈Vc(η), C(z) < C(Ω(RR)ξ) and C(z)< C(Ω(P P)ξ) for all z∈Z3
and for all ξ ∈ {1,2,3,4}. Hence, the LRE in Va(η) form a subset of
Ω(RP)∪Ω(P R).
Proof of Theorem 3.2. When h < r, the costs for the transition from the co-existence of conventions to global coordination on the risk-dominant equilibrium may have less cost through TP2 than those in the case where h ≥ r through TP1. For the transition from Ω(P R) to either Ω(RR) or Ω(RO), TP2 cost less than TP1 if
⌈⌊d1N⌋(1−q)ˆ⌉+⌈(2N − ⌊d2N⌋)(1−q∗)⌉ ≤ ⌈⌊d1N⌋(1−q∗)⌉ (3.33) Similarly, for the transition from Ω(RP) to either Ω(RR) or Ω(OR), the cost through TP2 is lower than that of TP1 if
⌈⌊d2N⌋(1−q)ˆ⌉+⌈(2N − ⌊d1N⌋)(1−q∗)⌉ ≤ ⌈⌊d2N⌋(1−q∗)⌉ (3.34) Denote Λ(dk) = 2− q1ˆ−−qq∗∗dk, for k = 1,2. Rearranging the conditions above, we obtain respectively
d2 ≥ Λ(d1) +∆2(d1, d2)
N , (3.35)
d1 ≥ Λ(d2) +∆1(d1, d2)
N , (3.36)
where both ∆1(d1, d2) and ∆2(d1, d2) are bounded. Hence, for any η > 0, there exists an integer ¯N, such that, for all N > N¯, |∆1(dN1,d2)| < η and
|∆2(dN1,d2)|< η.
If both (3.35) and (3.36) hold, TP2 leads to the minimum cost for both Ω(RP)→Ω(RR) and Ω(P R)→Ω(RR), hence we have to replace the costs of both transitions in the case h ≥r by those through TP2. If only (3.35) or (3.36) holds, TP2 has the minimum cost only for Ω(RP) → Ω(RR) or Ω(P R)→ Ω(RR) respectively. Hence, we have to use the minimum cost of Ω(RP) or Ω(P R) generated by TP2 in each of the corresponding cases. If neither of them holds, the costs are the same as in the case h ≥r. Hence, comparing the minimum costs of all the transition trees in each of the areas mentioned above, we can find the LRE. There are two cases involving the transitions mentioned above,⌊dkN⌋<2N−1 for bothk= 1,2, and dk= 2 and ⌈dℓ⌉<2N −1 fork, ℓ= 1,2, k 6=ℓ.
Case 1: qˆ≤1/q∗−2 + 2q∗.
Case1.1: ⌊dkN⌋<2N−1for both k= 1,2. In this case, Λ(dk)≥Ψ(dk) for both k= 1,2,dk∈[1,2]. If neither (3.36) nor (3.35) holds, the LRE is the same as stated in Theorem 3.1.
If (3.36) holds, the element in Ω(RP) will be selected if and only if d1 ≥d2+ ∆a(dN1,d2) and d1 ≥Υ(d2) +∆b(dN1,d2).
Similarly, if (3.35) holds, Ω(P R) will be selected if and only if d1 ≤ d2+∆a(dN1,d2) and d2 ≥Υ(d1) +∆c(dN1,d2).
∆y(d1, d2) are bounded for all y ∈Y ={a, b, c}. Hence, for any η >0, there exists an integer ¯N, such that, for allN >N¯,|∆y(d1, d2)|< η, for all y∈Y. Note that, if ˆq ≤1/q∗−2 + 2q∗, Υ(dk)≤Ψ(dk)≤Λ(k) fordk ∈[1,2]
and for bothk= 1,2. Building on the notations introduced in the symmetric case, let CT P1(Ω(·)) be the minimum cost of Ω(·)-tree where T P1 leads to the minimum cost for the transition frombothΩ(RP) and Ω(P R) to Ω(RR), and CT P2(Ω(·)) be the minimum cost of Ω(·)-tree where T P2 leads to the minimum cost either from Ω(RP) to Ω(RR), or from Ω(P R) to Ω(RR), or both. After a series of comparisons of minimum transition costs of different absorbing sets in respective areas, one can obtain that, for anyη >0, for N large enough, the following results hold.
(1.1) CT P1(Ω(RP)) orCT P2(Ω(RP)) is the minimum among the costs for all the transition trees of absorbing sets ifd1 ≥Ψ(d2) +ηandd1≥d2+η;
(1.2) CT P1(Ω(P R)) orCT P2(Ω(P R)) is the minimum among the costs for all the transition trees of absorbing sets ifd2 ≥Ψ(d1) +ηandd1≤d2+η.
(1.3) CT P1(Ω(RR)) is the minimum if d1≤Ψ(d2)−η andd2 ≤Ψ(d1)−η.
Case 1.2: dk = 2 and ⌈dℓN⌉ <2N −1 for k, ℓ = 1,2, k 6=ℓ. Consider the case k= 1. Using Table 3.6, if TP2 lead to the minimum cost, the cost of Ω(RO)-tree will change to⌈2N(1−q∗)⌉+⌈⌊d2N⌋(1−q)ˆ⌉+⌈(2N−⌊d1N⌋)(1− q∗)⌉. As long asN is large enough, this cost is still larger than the minimum cost of Ω(RP)-tree. Hence, the element in Ω(RP) is still selected. The same argument holds fork= 2. If N is large enough, Ω(P R) will still be selected fork= 2.
In all the other situations, the results are the same as in the caseh≥r.
Consider all the cases altogether, we obtain the same results as in Theorem 3.1 for ˆq <1/q∗−2 + 2q∗.
Case 2: q >ˆ 1/q∗−2 + 2q∗.
Case 2.1: ⌊dkN⌋<2N −1 for both k= 1,2. The analysis in Case 1.1 still holds. If (3.36) holds, the element in Ω(RP) will be selected if and only if d1 ≥Υ(d2) + ∆d(dN1,d2) and d1 ≥d2+∆e(dN1,d2). If (3.35) holds, Ω(P R) will be selected if and only if d2 ≥Υ(d1) +∆f(dN1,d2) and d1 ≤d2+∆g(dN1,d2).
However, the difference in this case is Υ(dk) >Ψ(dk) >Λ(dk) for dk ∈ [1,2] and for bothk= 1,2, which lead to the different predictions.
The approach is the same as in Case 1. One has to compare the minimum transition costs of different absorbing sets in different areas. Then, one can obtain that, for any η > 0, there exists an integer ¯N, such that for all N ≥N¯, the following results hold.
(2.1) CT P2(Ω(RP)) is the minimum ifd1≥Υ(d2) +η and d1 ≥d2+η;
(2.2) CT P2(Ω(P R)) is the minimum ifd2≥Υ(d1) +η and d1 ≤d2+η;
(2.3) CT P1(Ω(RR)) orCT P2(Ω(RR)) is the minimum ifd1≤Υ(d2)−ηand d2 ≤Υ(d1)−η.
Case 2.2: dk= 2 and ⌈dℓ⌉<2N−1 for k, ℓ= 1,2, k6=ℓ. All the results in Case 1.2 hold. That is, forN large enough, if k= 1, the element in Ω(RP) is the unique LRE; ifk= 2, Ω(P R) is the unique LRE.
In all the other cases, the result is the same as in the case h ≥ r.
Combine the results in all the cases together, we have the statement in the
theorem.
Proof of Proposition 3.7. The proof is similar to that of Proposition 3.6.
The LRE inUc(η) =Vc(η) and on (2,2) are presented in the proof of Theo-rem 3.1. The LRE in the Theo-remaining area of U(η) are derived by comparing the minimum costs of different transition trees. The only difference is that one has to use the costs generated by TP2 when they are proven to be the minimum. After a series of comparisons of the minimum costs of different transition trees, one can obtain the result in Proposition 3.7.
Proof of Theorem 3.3. Consider first the case that h≥r or ˆq ≤1/q∗− 2−2q∗. The LRE in this case are presented in Theorem 3.1. (d1, d2)∈D1(η) leads to the selection of the element in Ω(RP) in the long run. Planner 2 has no incentive to changed1 ord2 by changing (c2, p2), because the individuals in location 2 are coordinating on the Pareto-efficient equilibrium. Planner 1 has no incentive to only change d1, because changing d1 can only move
the LRE from Ω(RP) to Ω(RR) or Ω(RP) and the elements in Ω(RR).
In either case, the individuals in location 1 would coordinate on the risk-dominant equilibrium. However, planner 1 has an incentive to increase d2, because if d2 is large enough, the LRE would become Ω(P R). Note that planner 1 cannot directly change d2. The only possible way to changed2 is to change p1, because d2 = min{c2,2−p1}. Any intention to decrease d1 is always feasible, because planner 1 can increase p1 and make 2−p1 =d2. However, the effort to increased2 isnotalways effective. 2−p1 will increase by decreasing p1, but, as long as 2−p1 > c2, d2 = c2, and decreasing p1 cannot increase d2 any more. Hence, let c2 = d2 and 2−p1 ≥ d2. Then planner 1 has no incentive to change d2. Therefore, any strategy profile ((c1, p1),(c2, p2)) projected onD1(η) such thatd1= min{c1,2−p2},d2=c2 and 2−p1 ≥d2 is a Nash equilibrium.
(d1, d2) ∈ D2(η) leads to the selection of Ω(P R) in the long run. The argument is the same as above. Planner 1 has no incentive to change his strategy. By settingd1=c1<2−p2, planner 2 has no incentive to deviate either. Hence, for (d1, d2) ∈ D2(η), any strategy profile ((c1, p1),(c2, p2)) projected onD2(η) such that d2 = min{c2,2−p1},d1 =c1 and 2−p2 ≥d1 is a Nash equilibrium.
If (d1, d2)∈A1(η), the LRE are the elements in Ω(RR). Each plannerk would have an incentive to increase dℓ (ℓ 6=k) by decreasing pk. However, this effort would be ineffective ifdℓ=cℓ<2−pk. Hence, any strategy profile ((c1, p1),(c2, p2)) projected on A1(η) such that dk =ck and 2−pℓ ≥dk for bothk= 1,2,ℓ6=k, is a Nash equilibrium.
Lastly, we consider the area D3(η). If (d1, d2) ∈ (D3(η)∩A2(η)), the LRE is the element in Ω(RP). In this case, planner 1 will have an incentive to decreasec1 to the extent thatd1 < d2−η. It changes the LRE to Ω(P R) and increases the social welfare of location 1.
If (d1, d2) ∈ (D3(η)∩A3(η)), the LRE is the element in Ω(RP). In this case, planner 2 will have an incentive to decreasec2 to the extent that d2 < d1−η. It leads to the selection of Ω(RP) in the long run, and increases the social welfare of location 2.
If (d1, d2)∈(D3(η)∩Vd(η)), the LRE form a subset of Ω(RP)∪Ω(P R).
We have shown that any strategy profile leading to the element in Ω(RP) or Ω(P R) is not a NE. Hence, any strategy profile leading to the selection of the elements in both Ω(RP) and Ω(P R) in the long run is not stable
either, because each social planner k will have an incentive to decrease ck fork= 1,2.
Ifd1 =d2 = 2, the elements in both Ω(P O) and Ω(OP) will be selected in the long run. Each of them will occur with probability 1/2. Hence, each social planner k will have an incentive to decrease ck so that dk < dℓ−η (k 6= ℓ). Then, the players in location k will coordinate on the efficient equilibrium with probability one, and the social welfare of location k will increase.
If (d1, d2)∈Vc(η)\ {(2,2)}, the LRE form a subset of Ω(OP)∪Ω(RP)∪ Ω(P R)∪Ω(P O)∪Ω(P P). Note that there are no LRE which only consist of the elements in Ω(P P). If the elements in Ω(P P) are selected, the element in either Ω(RP) or Ω(P R) (or both) will be selected as well. Then, in any possible subset of the set above, with a positive probability, at least one location will either have players coordinating on the less efficient equilibrium or have no players at all. Then, the social planner in this location k can always improve the social welfare by decreasing ck so that dk < dℓ −η (k6=ℓ). Therefore, any strategy profile projected on this area is not stable.
An analogous argument holds for the case with h < r and ˆq >1/q∗ − 2 + 2q∗. We only have to replaceD1(η),D2(η), D3(η), A2(η), andA3(η) by G1(η), G2(η), G3(η), B2(η), and B3(η) respectively in the analysis above.
Hence we obtain the result in the theorem.
Proof of Theorem 3.4. Consider a strategy profile ((c1, p1),(c2, p2)) such that c1 = c2 = 1 and p1 = p2 = 1. Planner k = 1,2 has no incentive to deviate from his strategy. Changing ck has no effect, because all the individuals in locationℓ6=kare immobile, hence cannot move to locationk.
Changingp1 has no effect either, because the maximum capacity of location ℓ is N, hence, the mobile players in location k cannot move to location ℓ. Hence, this strategy profile is a Nash equilibrium, which corresponds to d1 =d2= 1.
Forh≥ror ˆq ≤1/q∗−2 + 2q∗, consider any (d1, d2)∈[1,2]2\({(1,1)}∪
Ve(η)). We first claim that any strategy profile projected onA1(η)\ {(1,1)} is not a NE. In this area, the LRE are the elements in Ω(RR). The social planner of location k will always have incentive to decrease dℓ (ℓ 6= k) by increasing pk. The reason is that the population in location k fluctuates between 2N − ⌊dℓN⌋ and ⌊dkN⌋. Decreasing dℓ will increase the lower
bound of the population in location k, hence improving the social welfare.
For (d1, d2)∈A2(η), the LRE is the element in Ω(RP), and the popula-tion in locapopula-tion 1 is 2N− ⌊d2N⌋. The social planner in location 1 will have an incentive to decrease d1 by setting a lower c1, so that the LRE form a subset of Ω(RR)∪Ω(P R). Denote byd′k the parameter of effective capacity of locationk= 1,2 after a deviation. If such a deviation leads the elements in Ω(RR) to be selected, the population in location 1 will fluctuate between 2N− ⌊d2N⌋and ⌊d′1N⌋, hence the social welfare will increase. If the devia-tion leads Ω(P R) to be selected, the players in locadevia-tion 1 will coordinate on P, and the population will increase to ⌊d′1N⌋, which improving the social welfare of location 1. Based on the results above, any deviation that results in the selection of the elements in Ω(RR) and Ω(P R) also increase the social welfare of location 1. Hence, any strategy profile projected on A2(η) is not a NE.
Symmetrically, any strategy profile projected onA3(η) is not a NE. The argument is analogous to the case above. Here, the social planner of location 2 will always have an incentive to decreased2. Such a deviation can at least increase the population in location 2, hence improving the social welfare.
If (d1, d2)∈Va(η), Ψ(d2)−η < d1<Ψ(d2)+η. The LRE form a subset of Ω(RR)∪Ω(RP). We have argued that if the LRE are the elements in Ω(RR) or Ω(RP), the social planner of location 1 will always have an incentive to deviate. If the LRE are the elements in Ω(RR) and Ω(RP), with a positive probability the population in location 1 will fluctuate between 2N − ⌊d2N⌋ and⌊d1N⌋, and with the remaining probability the population in location 1 is 2N− ⌊d2N⌋. In this case, the social planner of location 1 will always have an incentive to decreasec1at most to the extent thatd′1= Ψ(d2)−η. Then, the LRE are the elements in Ω(RR) and the population in location 1 will at worst fluctuate between 2N− ⌊d2N⌋ and⌊(d1−η)N⌋with probability one.
Forη small enough, this will increase the social welfare of location 1.
Symmetrically, for (d1, d2) ∈ Vb(η), the same argument holds for the social planners of location 2. Hence, he will have incentive to decreased2.
If (d1, d2)∈Vd(η), the LRE form a subset of Ω(RP)∪Ω(P R). We have shown above that the strategy profiles which lead to the selection of the element in either Ω(RP) or Ω(P R) are not NE. If the LRE are Ω(RP) and Ω(P R), the expected population in each locationkshould fall in the interval ]2N − ⌊dℓN⌋,⌊dkN⌋[ where dℓ−η < dk < dℓ +η, and the individuals in
locationkwill either coordinate onRor onP. The social planner of location kwill have an incentive to decreaseck, at most to the extent thatd′k=dℓ−η.
Then, the players in location k will coordinate on P with probability one, and the population will be at at least⌊(dk−η)N⌋. Forη small enough, this deviation will increase the social welfare of location k.
Ifd1=d2 = 2, the elements in both Ω(P O) and Ω(OP) will be selected in the long run. Each of them will occur with probability 1/2. Hence, the average expected payoff of location kis lower than the payoff of the Pareto-efficient equilibrium, and the expected population will fall in the interval ]0,2N[. The social planner of location k = 1,2 will have an incentive to decrease ck at most to the extent that d′k = 2−η. Then, the players in location kwill coordinate on P with probability one, and the population of locationk will be at least⌊(2−η)N⌋. Forη small enough, this will increase the social welfare of locationk.
If (d1, d2)∈(Vc(η)\ {(2,2)}), dk ∈(2−η,2) for both k= 1,2, and the LRE form a subset of Ω(OP)∪Ω(RP)∪Ω(P R)∪Ω(P O)∪Ω(P P). The same argument applies to show that the strategy profiles leading to the selection of the element in an singleton absorbing set (Ω(RP),Ω(P R),Ω(OP) or Ω(P O)) are not NE. We have pointed out in the proof of Theorem 3.3 that if the elements in Ω(P P) are selected, the element(s) in either Ω(RP) or Ω(P R) (or both) must be selected as well in the long run. Hence, in any possible subset of the set above, with a positive probability, at least one location k will either have players coordinating onRor have no players at all. In such a case, the average expected payoff for the players in locationk will be less than that of the Pareto-efficient equilibrium, and the expected population will be in the interval ]0,2N[. Hence, the social planner of location k will have an incentive to decreaseckat most to the extent thatd′k = 2−η. Then,
If (d1, d2)∈(Vc(η)\ {(2,2)}), dk ∈(2−η,2) for both k= 1,2, and the LRE form a subset of Ω(OP)∪Ω(RP)∪Ω(P R)∪Ω(P O)∪Ω(P P). The same argument applies to show that the strategy profiles leading to the selection of the element in an singleton absorbing set (Ω(RP),Ω(P R),Ω(OP) or Ω(P O)) are not NE. We have pointed out in the proof of Theorem 3.3 that if the elements in Ω(P P) are selected, the element(s) in either Ω(RP) or Ω(P R) (or both) must be selected as well in the long run. Hence, in any possible subset of the set above, with a positive probability, at least one location k will either have players coordinating onRor have no players at all. In such a case, the average expected payoff for the players in locationk will be less than that of the Pareto-efficient equilibrium, and the expected population will be in the interval ]0,2N[. Hence, the social planner of location k will have an incentive to decreaseckat most to the extent thatd′k = 2−η. Then,