Majority rip-off in referendum voting

Abstract

Many democracies complement a parliamentarian system with elements of direct democracy, where the electorate decides on single issues by majority voting. A well-known paradox states that in a sequence of referenda one can get from an arbitrary original income distribution to one in which one player gets almost all the cake. In this paper we design a three-player game modelling the sequential modification mechanism. The strategic analysis reveals that the paradox survives even with rational strategic voters and though the right to propose is allocated to each player once: the last player receives almost the entire cake. The result can be extended to the three-party n-voter case and is for some cases similar when we consider a random rather than fixed sequence of proposers.

Appendices

Appendix A. Proof of Lemma 1

We first look at the players’ decisions at the voting sub-stages.

1.1 A.1 Third stage voting behaviour

Since the game ends after the third stage, as-if-sincere voting immediately implies that v _i ³ = yes if p _i ³ > s _i ², and v _i ³ = no otherwise. If p _i ³ = s _i ², then the changing costs will induce v _i ³ = no. (A1)

1.2 A.2 Player 3’s proposal

At this stage, we need to consider two cases. If the greater of the two shares of players 1 and 2 does not exceed 1, then player 3 has no proposal that improves his payoff and finds a majority. Otherwise, the status quo will always be changed.

Case A.1 max{s ₁ ², s ₂ ²} ≤ 1

If max{s ₁ ², s ₂ ²} ≤ 1 then there is no proposal P ³ such that p ₃ ³ > s ₃ ² and v _i ³ = yes for i∈{1, 2}. v ₁ ² = yes requires p ₁ ² ≥ s ₁ ² + 1, v ₂ ² = yes requires p ₂ ² ≥ s ₂ ² + 1. If any of those is true, then p ₃ ³ ≤ s ₃ ². Thus, P ³ would be worse for player 3 by at least the changing costs. Therefore S ³ = S ². (A2).

Case A.2 max{s ₁ ², s ₂ ²} > 1

If max{s ₁ ², s ₂ ²} > 1 then P ³ ≠ S ² and S ³ ≠ S ², since there exists a proposal P ³ such that v ₃ ³ = yes and (v ₁ ³ = yes or v ₂ ³ = yes), and p ₃ ³ > s ₃ ². If min{s ₁ ², s ₂ ²} = s ₁ ², then p ₁ ³ = s ₁ ² + 1 and p ₂ ³ = 0 implies v ₁ ³ = yes and v ₃ ³ = yes and therefore s ₃ ³ > s ₃ ². From max{s ₁ ², s ₂ ²} = s ₂ ² ≥ 2, it follows that s ₃ ³ = p ₃ ³ = C − s ₁ ² − 1 > C − s ₁ ² − s ₂ ² = s ₃ ². The analogous argument holds for min{s ₁ ², s ₂ ²} = s ₂ ². Thus, there exists a proposal that improves player 3 and gets a majority, therefore S ² cannot sustain in an SPE. (A3)

Further, in an SPE either p ₁ ³ = 0 or p ₂ ³ = 0 and S ³ = P ³ ≠ S ². (A3) implies that p ₃ ³ > s ₃ ², thus from (A1) follows v ₃ ³ = yes. Suppose there is an SPE with p ₁ ³ > 0 and p ₂ ³ > 0. (A1) and (A3) imply either p ₂ ³ > s ₂ ² or p ₁ ³ > s ₁ ². Suppose p ₁ ³ > s ₁ ². Thus v ₁ ³ = yes and v ₃ ³ = yes and S ³ = P ³. Now consider P ³′ with p ₂ ³′ = 0 and p ₁ ³′ = p ₁ ³. Still v ₁ ³ = yes and v ₃ ³ = yes, but p ₃ ³′ > p ₃ ³. Thus, P ³ cannot be an equilibrium proposal. The analogous argument holds for p ₂ ³ > 0. (A4)

In an SPE p ₁ ³ ≤ s ₁ ² + 1 and p ₂ ³ ≤ s ₂ ² + 1 holds. Suppose there is an SPE with p ₁ ³ > s ₁ ² + 1. (A4) implies that then p ₂ ³ = 0, which from (A1) implies v ₂ ³ = no. (A4) and the supposition imply that then v ₁ ³ = yes and v ₃ ³ = yes. Now consider P ³′ with p ₁ ³′ = s ₁ ² + 1 and p ₂ ³′ = 0. Still v ₁ ³ = yes and v ₃ ³ = yes, but p ₃ ³′ > p ₃ ³. Thus, P ³ cannot be an equilibrium proposal. The analogous argument holds for p ₂ ³ > s ₂ ² + 1. (A5)

(A1), (A4), and (A5) imply that if max{s ₁ ², s ₂ ²} > 1, in an SPE, either P ³ = (s ₁ ² + 1, 0, C − s ₁ ² − 1) or P ³ = (0, s ₂ ² + 1, C − s ₂ ² − 1).

This implies that if s ₁ ² < s ₂ ², then P ³ = (s ₁ ² + 1, 0, C − s ₁ ² −1), and if s ₁ ² > s ₂ ², then P ³=(0, s ₂ ² + 1, 0, C − s ₂ ² −1). (A6)

If s ₁ ² = s ₁ ³, then both P ³ = (s ₁ ² + 1, 0, C − s ₁ ² − 1) and P ³ = (0, s ₂ ² + 1, 0, C − s ₂ ² − 1) are SPE options. So are all behaviour strategies involving any randomisation between the two. □

Appendix B. Proof of Lemma 2.

For convenience, denote by q(H) the probability for player 3 choosing P ³ = (s ₁ ² + 1, 0, C − s ₁ ² − 1) for a given history of play H = {P ¹, V ¹, S ¹, P ², V ², S ²}. q(H) is defined for all histories of play S ² in which s ₁ ² = s ₂ ². Notice that player 3 can choose a different randomisation for every possible history of play.

1.1 B.1 Second stage voting behaviour

At the second stage, all players will vote strategically, i.e. they vote for a proposal if the proposal offers them a higher final payoff than the status quo. We shall now derive the conditions for voting for or against a proposal.

If p ₁ ² > p ₂ ² then v ₁ ² = no, since s ₁ ² > s ₂ ² implies s ₁ ³ = 0. Thus, changing costs ensure that player 1 will always prefer S ² = S ¹. The analogous argument yields that if p ₂ ² > p ₁ ² then v ₂ ² = no.

If s ₁ ¹ > s ₂ ¹ and p ₁ ² < p ₂ ², then v ₁ ² = yes, since S ² = S ¹ would imply s ₁ ³ = 0, S ² = P ² would imply s ₁ ³ = p ₁ ² + 1. The analogous argument yields that if s ₁ ¹ > s ₁ ² and p ₁ ² < p ₂ ², then v ₁ ² = yes.

If s ₁ ¹ < s ₂ ¹ and p ₁ ² < p ₂ ² then v ₁ ² = yes if p ₁ ² > s ₁ ¹, v ₁ ² = no otherwise. Changing costs ensure that if p ₁ ² = s ₁ ¹, player 1 will prefer S ² = S ¹. The analogous argument yields that if s ₂ ¹ < s ₁ ¹ and p ₂ ² < p ₁ ² then v ₂ ² = yes if p ₂ ² > s ₂ ¹, v ₂ ² = no otherwise.

If s ₁ ¹ = s ₂ ¹ and p ₁ ² < p ₂ ² then lemma 1 implies that v ₁ ² = yes if q(H)(s ₁ ¹ + 1) > p ₁ ¹ + 1 − ε, and v ₁ ² = no if q(H)(s ₁ ¹ + 1) < p ₁ ¹ + 1 − ε. Notice that the former is always true if p ₁ ² > s ₁ ¹. If q(H)(s ₁ ¹ + 1) < p ₁ ¹ + 1 − ε, then any randomisation between v ₁ ² = yes and v ₁ ²=no is a best response for player 1. The analogous holds for player 2’s voting behaviour.

Now suppose s ₁ ¹ > s ₂ ¹ and p ₁ ² = p ₂ ². v ₁ ² = yes if q(p ₁ ²)(p ₁ ² + 1) > ε, since if S ² = S ¹ then (A6) implies that s ₁ ³ = 0. v ₁ ² = no if q(H)(p ₁ ² + 1) < ε. If q(H)(p ₁ ² + 1) = ε then lemma 1 implies any randomisation between v ₁ ² = yes and v ₁ ² = no is a best response. v ₂ ² = yes if (1 − q(H))(p ₂ ² + 1) − ε > s ₂ ¹ + 1. This is always false if s ₂ ¹ > p ₂ ². v ₂ ² = no if (1 − q(H))(p ₂ ² +1) −ε < s ₂ ¹ + 1. If (1 − q(H))(p ₂ ² + 1) − ε = s ₂ ¹ + 1, then any randomisation between v ₂ ² = yes and v ₂ ² = no is a best response. q(H) denotes the history of play given S ² = P ².

Analogously, suppose s ₁ ¹ < s ₂ ¹ and p ₂ ² = p ₁ ². v ₂ ² = yes if (1 − q(H))(p ₂ ² + 1) > ε, since if S ² = S ¹ then (A6) implies that s ₂ ³ = 0. v ₂ ² = no if (1 − q(H))(p ₂ ² + 1) < ε. If q(H)(p ₂ ² + 1) = ε then lemma 1 implies any randomisation between v ₂ ² = yes and v ₂ ² = no is a best response. v ₂ ² = yes if (1 − q(H))(p ₂ ² + 1) − ε > s ₂ ¹ + 1. This is always false if s ₂ ¹ > p ₂ ². v ₁ ² = no if q(H)(p ₁ ² + 1) − ε < s ₁ ¹ + 1. If q(H)(p ₁ ² + 1) − ε = s ₁ ¹ + 1, then any randomisation between v ₁ ² = yes and v ₁ ² = no is a best response. q(H) denotes the history of play given S ² = P ². (B1)

If s ₁ ¹ = s ₂ ¹ and p ₁ ² > p ₂ ² then v ₁ ² = no. S ² = P ² implies that s ₁ ³ = 0. Regardless of q, changing costs will already ensure that player 1 will prefer S ² = S ¹. The analogous argument yields that if s ₁ ¹ = s ₂ ¹ and p ₂ ² > p ₁ ² then v ₂ ² = no.

If s ₁ ¹ = s ₂ ¹ and p ₁ ² = p ₂ ² then v ₁ ² = yes if q(H′)(s ₁ ¹ +1 ) < q(H)(p ₁ ² + 1) − ε. v ₁ ² = no if q (H′)(s ₁ ¹ + 1) > q(H)(p ₁ ² + 1) − ε. If v ₁ ² = yes if q(H′)(s ₁ ¹ + 1) = q(H)(p ₁ ² + 1) − ε, then any randomisation between v ₂ ² = yes and v ₂ ² = no is a best response for player 1. The analogous holds for player 2’s voting behaviour. q(H) denotes the history of play given S ² = P ²; q(H′) denotes the history of play given S ² = S ¹. (B2)

(A6) together with changing costs implies that v ₃ ² = yes if min{p ₁ ², p ₂ ²} < min{s ₁ ¹, s ₂ ¹}, v ₃ ² = no otherwise. (B3)

We can summarise the results of second stage voting as follows. In any SPE, player 3 votes yes whenever the minimum of the others’ shares is smaller in P ² than in S ¹. Players 1 votes yes if his share in P ² is smaller than that of player 2, except if it is already smaller in S ¹ and his share in P ² is smaller than in S ¹. If the shares of players 1 and 2 are equal, then the decision depends on q(H). Player 2 votes in the analogous way, comparing his shares to those of player 1.

1.2 B.2 Player 2’s proposal

To find necessary conditions for an SPE in the second stage subgame, we have to consider several cases separately. The history of play has lead to a second stage status quo S ¹. We will obtain that the structure of P ² depends crucially on whether s ₁ ¹ < s ₂ ¹, s ₁ ¹ = s ₂ ¹, or s ₁ ¹ > s ₂ ¹. Secondly, we get a special case when s ₁ ¹ = s ₂ ¹ = 1 or s ₁ ¹ = s ₂ ¹ = 0.

Case B.1. s ₁ ¹ = s ₂ ¹ = 0

If S ² = S ¹, then s ₂ ³ = C.

First suppose q(H) = 1.

If S ² = S ¹, then (A2) implies that s ₃ ³ = C. Therefore, changing costs already imply v ₃ ² = no. Therefore, P ² can be an SPE proposal with S ² = P ² ≠ S ¹ only if v ₁ ² = yes and v ₂ ² = yes. (B4)

(A6) implies that if p ₁ ² < p ₂ ², then s ₂ ³ = 0. Thus, changing costs induce v ₂ ² = no. Accordingly, if p ₁ ² > p ₂ ², then s ₁ ³ = 0. Thus, changing costs induce v ₁ ² = no. Therefore, p ₁ ¹ = p ₂ ¹. (B5)

p ₁ ² = p ₂ ² > 1 implies max{s ₁ ², s ₂ ²} > 1. Since q(H)=1, this implies s ₂ ³ = 0. Therefore, changing costs imply that v ₂ ² = no. Therefore p ₁ ² = p ₂ ² ≤ 1.

p ₁ ² = p ₂ ² = 0 means P ² = S ¹. (B6)

Thus, the remaining candidate for an SPE proposal with S ² = P ² and P ² ≠ S ¹ is p ₁ ² = p ₂ ² = 1. If S ² = P ², (A2) implies that S ³ = S ². Thus, s ₁ ³ = 1 > 0 and s ₂ ³ = 1 > 0, therefore v ₁ ² = yes and v ₂ ² = yes, and it also follows that P ² is an SPE proposal.

It remains to show that all path perfect equilibria of the subgame are characterised by P ² = (1, 1, C − 2). A necessary condition for further SPE is that q(H) < 1. (B4), (B5), and (B6) hold without restriction. Thus, further SPE may exist only with proposals in which p ₁ ² = p ₂ ² > 1. If p ₁ ² = p ₂ ² > 1, then if S ² = P ² lemma 1 implies that s ₃ ³ = C − p ₁ ² − 1 ≤ C − 2. Therefore, if there are SPE with P ²′ ≠ P ², in all these SPE necessarily s ₃ ³′ < C − 2. This is the payoff player 3 gets in all SPE with P ² = (1, 1, C − 2). Necessary condition for these equilibria to be selected is that q(H) < 1 for at least one H. This implies that player 3 has an influential option against P ²′.¹⁴ Thus, these equilibria cannot be path perfect.¹⁵

Case B.2 s ₁ ¹ = s ₂ ¹ = 1

First, suppose q(H) = 1 ∀H.

Suppose there is an SPE proposal such that S ² = P ² ≠ S ¹. Suppose in P ² p ₁ ² < p ₂ ² holds. Then (A6) implies that s ₂ ³ = 0. If S ² = S ¹, (A2) implies that S ³ = S ¹, thus s ₂ ³ = 1. Since P ²′ = S ¹ is available, P ² cannot be an SPE proposal. (B7)

Suppose in P ² p ₁ ² = p ₂ ² holds. Then (A6) together with q(H) = 1 implies that s ₂ ³ = 0. If S ² = S ¹, (A2) implies that S ³ = S ¹, thus s ₂ ³ = 1. Since P ²′ = S ¹ is available, P ² cannot be an SPE proposal.

Now suppose p ₁ ² < p ₂ ². If S ² = P ², then (A6) implies that s ₁ ³ = 0, thus v ₁ ² = no. Therefore, P ² can only be an SPE proposal if v ₂ ² = yes and v ₂ ³ = yes. (B3) states that if max {p ₁ ², p ₂ ²} > 1, then v ₂ ³ = no if min {p ₁ ², p ₂ ²} ≥ min{s ₁ ¹, s ₂ ¹}. Both is true if p ₂ ² ≥ 1. Therefore p ₂ ² = 0. If p ₂ ² = 0, then either p ₁ ² = 1 or p ₁ ² > 1. In the former case (A2) implies that if S ² = P ² then S ³ = P ² and thus s ₂ ³ = 0. Thus p ₁ ² = 1 and p ₂ ² = 0 cannot be an SPE proposal. If p ₁ ² > 1 and p ₂ ² = 0, then (A6) implies that if S ² = P ², then s ₂ ³ = 1. Thus, the changing costs will induce that v ₂ ² = no. (B8)

Therefore, if s ₁ ¹ = s ₂ ¹ = 1 and q(H) = 1 ∀H, then S ² = S ¹.

It remains to show that all path perfect equilibria of the subgame are characterised by S ² = S ¹. A necessary condition for further SPE is that q(H) < 1. (B7) and (B8) hold without restriction. Thus, further SPE may exist only with proposals in which p ₁ ² = p ₂ ² > 1. If p ₁ ² = p ₂ ² > 1, then if S ² = P ² lemma 1 implies that s ₃ ³ = C−p ₁ ² − 1 ≤ C − 2. Therefore, if there are SPE with P ² ≠ S ¹, in all these SPE necessarily s ₃ ³′ < C − 2. This is the payoff player 3 gets in all SPE with S ² = S ¹. Necessary condition for these equilibria to be selected is that q(H) < 1 for at least one H. This implies that player 3 has an influential option against P ²′. Thus, these equilibria cannot be path perfect.

Case B.3 s ₁ ¹ > s ₂ ¹

First suppose there is an SPE proposal P ² such that S ² = P ² ≠ S ¹. Let us suppose that in P ², p ₁ ² < p ₂ ². If S ² = S ¹, then (A6) implies that s ₂ ³ = s ₂ ¹ + 1. If S ² = P ², then (A6) implies that s ₂ ³ = 0. Thus, P ² cannot be an equilibrium proposal since P ² = S ² is an available option. (B9)

Now suppose that in P ², p ₁ ² > p ₂ ². v ₁ ¹ = no, since if S ² = S ¹, s ₁ ³ = 0, and if S ² = P ², s ₁ ³ = 0, thus changing costs induce v ₁ ¹ = no. (A6) implies that to be an SPE proposal p ₂ ² ≥ s ₂ ¹ must hold, changing costs induce that the inequality is strict, thus p ₂ ² > s ₂ ¹. But then min{p ₂ ², p ₁ ²} > min {s ₁ ¹, s ₂ ¹}, thus from (B3) it follows that v ₃ ² = no. Therefore, there is no SPE proposal P ² with p ₁ ² > p ₂ ² when s ₁ ¹ > s ₂ ¹. (B10)

Finally, we need to check whether there can be an SPE proposal P ² with p ₁ ² = p ₂ ² and S ² = P ² ≠ S ¹. We shall consider two cases separately.

First, suppose q(H) = 1∀H.

Since s ₁ ¹ > s ₂ ¹, if S ² = S ¹, s ₂ ³ = s ₂ ¹ + 1. If S ² = P ², lemma 1 and q(H) = 0 imply that s ₂ ³ = 0. Therefore, there is no SPE proposal with p ₁ ² = p ₂ ² when q(H) = 1 and s ₁ ¹ > s ₂ ¹. (B11)

(B9), (B10), and (B11) together imply that if q(H) = 1 ∀H, then in an SPE S ² = S ¹ if s ₁ ¹ > s ₂ ¹. This can result from player 2 setting P ² = S ¹, or from a proposal that does not find a majority.

Now, allow for q(H) < 1 for arbitrary H.

If S ² = S ¹, then (A6) implies that s ₂ ³ = s ₁ ¹ + 1. If S ² = P ², then s ₂ ³ = p ₂ ² with probability 1 − q(H). This implies that P ² cannot be an SPE proposal if p ₂ ² ≤ s ₂ ¹, changing costs induce that we require the strict inequality p ₂ ²>s ₂ ¹. (B12)

Since min{s ₁ ¹, s ₂ ¹} = s ₂ ¹ and min{p ₁ ², p ₂ ²} = p ₂ ², (B3) implies that v ₃ ³ = no. Therefore, S ² = P ² requires v ₁ ² = yes and v ₂ ² = yes.

This already allows to show that all path perfect equilibria of the subgame are characterised by S ² = S ¹. A necessary condition for further SPE is that q(H) < 1. (B9) and (B10) hold without restriction. Thus, further SPE may exist only with proposals in which p ₁ ² = p ₂ ² > 1. (B12) implies that if there are SPE with P ² ≠ S ¹, in all these SPE necessarily s ₃ ³ < C − s ₁ ¹ − 1. This is the payoff player 3 gets in all SPE with S ² = S ¹. Necessary condition for these equilibria to be selected is that q(H) < 1 for at least one H. This implies that player 3 has an influential option against P ²′. Thus, these equilibria cannot be path perfect.

Case B.4 s ₁ ¹ = s ₂ ¹ > 1

First suppose q(H) = 0 ∀H.

Suppose there is an SPE proposal P ² such that S ² = P ² ≠ S ¹. Let us suppose that in P ², p ₁ ² < p ₂ ². If S ² = S ¹, then (A6) together with q(H) = 0 implies that s ₂ ³ = 0. If S ² = P ², then (A6) implies that s ₂ ³ = 0. Changing costs already induce that P ² cannot be an equilibrium proposal since P ² = S ² is an available option (though not necessarily the best response).

Therefore p ₁ ² ≥ p ₂ ². Now suppose that in P ², p ₁ ² = p ₂ ². If S ² = S ¹, then (A6) implies that s ₂ ³ = 0. If S ² = P ², then (A6) together with q(H) = 0 implies that s ₂ ³ = 0. Changing costs already induce that P ² cannot be an equilibrium proposal since P ² = S ² is an available option (though not necessarily the best response). (B13)

Therefore necessary condition for P ² to be an SPE proposal with S ² = P ² ≠ S ¹ is p ₁ ² > p ₂ ². (B14)

Now suppose p ₂ ² < s ₁ ¹ − 1. P ² can only be an SPE proposal if p ₂ ² ≥ s ₁ ¹ − 1. If p ₂ ² < s ₁ ¹ − 1, then consider P ²′ with p ₂ ²′ = s ₁ ¹ − 1 and p ₁ ²′ = p ₁ ². Still v ₂ ² = yes and v ₃ ² = yes, but s ₂ ³′ > s ₂ ³. Thus, P ² can only be an SPE proposal if p ₂ ² ≥ s ₁ ¹ − 1. (B15)

If p ₂ ² ≥ s ₁ ¹, min{p ₁ ², p ₂ ²} ≥ min{s ₁ ¹, s ₂ ¹}. (B3) implies that v ₃ ² = no. If p ₁ ² > p ₂ ², (A6) implies that if S ² = P ², then s ₁ ³ = 0, while if S ² = S ¹, then (A6) and q(H) = 1 imply s ₁ ³ = s ₁ ¹ + 1. Thus, v ₁ ² = no. (B16)

(B14), (B15), and (B16) imply that if there is an SPE proposal P ² when s ₁ ¹ = s ₂ ¹ and q(H) = 0, it will involve p ₂ ² = s ₁ ¹ − 1 and p ₁ ² > p ₂ ². We now need to show that p ₂ ² = s ₁ ¹ − 1 and p ₁ ² > p ₂ ² indeed establish an SPE proposal, hence this proposal improves player 2’s payoff compared to a situation with S ² = S ¹. As stated in (B13), if S ² = S ¹, then (A6) together with q(H) = 1 implies that s ₂ ³ = 0. If S ² = P ², then (A6) implies s ₂ ³ = s ₁ ¹. Since in the case analysed here s ₁ ¹ > 1 holds by assumption, P ² is indeed an SPE proposal. (B17)

Now, allow for q(H) < 1 for arbitrary H.

Suppose there is an SPE proposal P ² such that S ² = P ² ≠ S ¹. Let us suppose that in P ², p ₁ ² < p ₂ ². Therefore if S ² = P ² then (A6) implies that s ₂ ³ = 0. Regardless of q(H), changing costs will induce that player 2 will prefer S ² = S ¹. Since he has the option P ²′ = S ¹, P ² cannot be an equilibrium proposal.

If there are only SPE with p ₁ ² > p ₂ ², then (B15), (B16), and (B17) already characterise them. Thus, SPE apart from (B17) require p ₁ ² = p ₂ ². If s ₁ ² = s ₂ ² then (B2) implies that, first, v ₁ ² = no if q(H(s ₁ ¹))(s ₁ ¹ + 1) > q(H(p ₁ ²))(p ₁ ² + 1) − ε, and second, v ₂ ² = no if (1 − q(H(s ₂ ¹)))(s ₂ ¹ + 1) > (1 − q(H(p ₂ ²)))(p ₂ ² + 1) − ε. (B17) requires that p ₂ ² ≥ s ₁ ¹ − 1, since the proposal characterised in (B17) is available. (B18)

Suppose p ₂ ² > s ₁ ¹− 1. This implies min{p ₁ ², p ₂ ²} > min{s ₁ ¹, s ₁ ²} and (B3) implies that v ₃ ² = no. If there is an SPE with p ₁ ² = p ₂ ² > s ₁ ¹ − 1 and S ² = P ² ≠ S ¹, then this equilibrium cannot be path perfect, because necessary condition for this to be an SPE is that q(H) > 0 for some H, such that player 3 has an influential option against it. Therefore, an SPE with p ₁ ² = p ₂ ² can only be path perfect if p ₂ ² = s ₁ ¹ − 1. The conditions in (B18) imply that this is the case if q(H(s ₁ ¹ − 1)) = 0.¹⁶

Case B.5 0 < s ₁ ¹ < s ₂ ¹

Suppose there is an SPE proposal P ² such that S ² = P ² ≠ S ¹. Let us suppose that in P ², p ₁ ² < p ₂ ². If S ² = S ¹, then (A6) implies that s ₂ ³ = 0. If S ² = P ², then (A6) implies that s ₂ ³ = 0. Changing costs already induce that P ² cannot be an equilibrium proposal since P ² = S ² is an available option (though not necessarily the best response).

Therefore p ₁ ² ≥ p ₂ ². Now consider the cases for different q(H) separately. First suppose q(H) = 0 ∀H.

Suppose that in P ², p ₁ ² = p ₂ ². If S ² = S ¹, then (A6) implies that s ₂ ³ = 0. If S ² = P ², then (A6) together with q(H) = 0 implies that s ₂ ³ = 0. Changing costs already induce that P ² cannot be an equilibrium proposal since P ² = S ² is an available option (though not necessarily the best response). (B19)

Therefore necessary condition for P ² to be an SPE proposal with S ² = P ² ≠ S ¹ is p ₁ ² > p ₂ ². (B20)

Now suppose p ₂ ² < s ₁ ¹ − 1. can only be an SPE proposal if p ₂ ² ≥ s ₁ ¹ − 1. If p ₂ ² < s ₁ ¹ − 1, then consider P ²′ with p ₂ ²′ = s ₁ ¹ − 1 and p ₁ ²′ = p ₁ ². Still v ₂ ² = yes and v ₃ ² = yes, but s ₂ ³′ > s ₂ ³. Thus, P ² can only be an SPE proposal if p ₂ ² ≥ s ₁ ¹ − 1. (B21)

If p ₂ ² ≥ s ₁ ¹, min{p ₁ ², p ₂ ²} ≥ min{s ₁ ¹, s ₂ ¹}. (B3) implies that v ₃ ² = no. Since p ₁ ² > p ₂ ², (A6) implies that if S ² = P ², then s ₁ ³ = 0, while if S ² = S ¹, then (A6) and q(H) = 1 imply s ₁ ³ = s ₁ ¹ + 1. Thus, v ₁ ² = no. (B22)

(B20), (B21), and (B22) imply that if there is an SPE proposal P ² when s ₁ ¹ = s ₂ ¹ and q(H) = 0, it will involve p ₂ ² = s ₁ ¹ − 1 and p ₁ ² > p ₂ ². We now need to show that p ₂ ² = s ₁ ¹ − 1 and p ₁ ² > p ₂ ² indeed establish an SPE proposal, hence this proposal improves player 2’s payoff compared to a situation with S ² = S ¹. As stated in (B19), if S ² = S ¹, then (A6) together with q(H) = 1 implies that s ₂ ³ = 0. If S ² = P ², then (A6) implies s ₂ ³ = s ₁ ¹. Since in the case analysed here s ₁ ¹ > 1 holds by assumption, P ² is indeed an SPE proposal. (B23)

Now, allow for q(H) < 1 for arbitrary H.

Suppose there is an SPE proposal P ² with p ₁ ² > p ₂ ². Then (B21), (B22), and (B23) hold analogously. Thus, we have to analyse the case that p ₁ ² = p ₂ ².

Denote by P ²′ the proposal derived in (B23). If S ² = P ², then s ₂ ³ = (1 − q(H(P ²)))(p ₂ ² + 1) − ε. If S ² = P ²′, then s ₂ ³ = s ₁ ¹. Necessary condition for P ² being an SPE proposal is that (1 − q(H(P ²)))(p ₂ ² + 1) − ε > s ₁ ¹ − ε. This can only be true if p ₂ ² ≥ s ₁ ¹ − 1. Suppose p ₂ ² > s ₁ ¹ − 1. This implies min{p ₁ ², p ₂ ²} > min{s ₁ ¹, s ₁ ²} and (B3) implies that v ₃ ² = no. If there is an SPE with p ₁ ² = p ₂ ² > s ₁ ¹ − 1 and S ² = P ² ≠ S ¹, then this equilibrium cannot be path perfect, because necessary condition for this to be an SPE is that q(H) > 0 for some H, such that player 3 has an influential option against it. Therefore, an SPE with p ₁ ² = p ₂ ² can only be path perfect if p ₂ ² = s ₁ ¹ − 1. The conditions in (B23) imply that this is the case if q(H(s ₁ ¹ − 1)) = 0.

Case B.6 0 = s ₁ ¹ < s ₂ ¹

Suppose there is an SPE proposal P ² such that S ² = P ² ≠ S ¹. Let us suppose that in P ², p ₁ ² < p ₂ ². If S ² = S ¹, then (A6) implies that s ₂ ³ = 0. If S ² = P ², then (A6) implies that s ₂ ³ = 0. Changing costs already induce that P ² cannot be an equilibrium proposal since P ² = S ² is an available option (though not necessarily the best response).

Now consider the cases for different q(H) separately. First suppose q(H) = 0 ∀H.

Suppose that in P ², p ₁ ² = p ₂ ². If S ² = S ¹, then (A6) implies that s ₂ ³ = 0. If S ² = P ², then (A6) implies that s ₂ ³ = 0. Changing costs already induce that P ² cannot be an equilibrium proposal since P ² = S ² is an available option (though not necessarily the best response). (B24)

Therefore necessary condition for P ² to be an SPE proposal with S ² = P ² ≠ S ¹ is p ₁ ² ≥ p ₂ ².

If p ₁ ² > p ₂ ², (A6) implies that if S ² = P ², then s ₁ ³ = 0, while if S ² = S ¹, then (A6) implies s ₁ ³ = s ₁ ¹ + 1. Thus, v ₁ ² = no. (B26)

If p ₁ ² = p ₂ ², then (A6) together with q(H) = 1 implies that if S ² = P ², then s ₂ ³ = 0. Changing costs induce that v ₂ ² = no.

If p ₂ ² ≥ s ₁ ¹, min{p ₁ ², p ₂ ²} ≥ min{s ₁ ¹, s ₂ ¹}. (B3) implies that v ₃ ² = no.

Since s ₁ ¹ = 0, p ₂ ² cannot be smaller than s ₁ ¹. Therefore S ² = S ¹. (B27)

It remains to show that all path perfect equilibria of the subgame are characterised by S ² = S ¹. A necessary condition for further SPE is that q(H) < 1. (B24), (B26), and (B27) hold without restriction. Thus, further SPE may exist only with proposals in which p ₁ ² = p ₂ ² > 1. If p ₁ ² = p ₂ ² > 1, then if S ² = P ² lemma 1 implies that s ₃ ³ = C − p ₁ ² − 1 ≤ C − 1. Therefore, if there are SPE with P ² ≠ S ¹, in all these SPE necessarily s ₃ ³′ < C − 2. This is the payoff player 3 gets in all SPE with S ² = S ¹. Necessary condition for these equilibria to be selected is that q(H) < 1 for at least one H. This implies that player 3 has an influential option against P ²′. Thus, these equilibria cannot be path perfect. □