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Voting in small committees

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Abstract

We analyze the voting behavior of a small committee that has to approve or reject a project proposal whose return is uncertain. Members have diverse preferences: some of them want to maximize the expected value, while others have a bias towards project approval and ignore their information on the project value. We focus on the most efficient use of scarce information when members cannot communicate prior to voting, and we provide insights on the optimal composition of the committee. Our main result is that the presence of biased members can improve the voting outcome, by simplifying the strategies of unbiased members. Thus, committees with diverse members perform as well as homogeneous committees, and even better in some cases. In particular, when value-maximizing members outnumber biased members by one vote, the optimal equilibrium becomes unique.

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Notes

  1. Additional examples are special juries as Supreme or Constitutional Courts, and technical committees, where politicians, bureaucrats and experts meet to provide advice.

  2. See, for instance, Duggan and Martinelli (2001), Myerson (1998) and McLennan (1998). See also Piketty (1999) for a brief review of recent contributions about the information-aggregation role of political institutions.

  3. See Austen-Smith and Banks (1996) and an experiment on the use of strategic voting by Eckel and Holt (1989).

  4. See for instance Blinder (2006), Jung (2011), Riboni and Ruge-Murcia (2007) for the case on Monetary Policy Committees.

  5. This strand of literature has been reviewed by Austen-Smith and Feddersen (2009). See also Adams and Ferreira (2007), Harris and Raviv (2008), and Raheja (2005) for communication in boards of directors.

  6. Berk and Bierut (2004) suggest that often in small committees it is technologically or politically unfeasible to implement optimal voting rules, thus binding voting to be based on simple majority.

  7. Committee members have heterogeneous preferences also in the model of Cai (2009). The focus of Cai, however, is the optimal size of the committee rather than its composition.

  8. Alternatively, we can assume that every member observes a signal that is perfectly informative with probability \(\alpha \) and is totally noisy with probability \(1-\alpha \). Referring to this interpretation, it can be shown that our results would not qualitatively change if the signal were only partially noisy (Balduzzi 2005).

  9. Of course we acknowledge some exceptions, such as Morton and Tyran (2008). Note also that in many committees abstention is explicitly or implicitly ruled out (juries, the European Courts of Human Rights and the Italian Constitutional Court are some examples).

  10. This rules out “uninteresting” equilibria where each member is never pivotal. For instance when every member votes “yes” independently of his information. However it can be easily verified that all these equilibria yield \(E(v)=0.\)

  11. Obviously, nothing substantial in our results would change if a biased member always supported rejection.

  12. On this point, see footnote 10.

  13. That \((1-\alpha )^{n+1}\) represents the probability of making the wrong decision is shown in the Proof of Proposition 1, point (v).

  14. For example, the need to represent different stakeolders or to balance different powers or just plain political criteria.

  15. We do not develop this aspect in the present paper because it would distort the attention away from our main focus. Costly information acquisition can be studied along the lines set in Persico (2004) who analyzes the optimal voting rule (in terms of both incentives to acquire information and information aggregation) in a similar setting. See also Harris and Raviv (2008) on the effects of costly information acquisition on the optimal size of boards of directors.

  16. Clearly our comment on the optimal size of the committee following Corollary 1 still applies.

  17. For simplicity, we only consider pure strategies for the \(M\) members.

  18. Alternatively we can assume that messages are exchanged among all the members and enter everybody’s information set. Notice however that biased members cannot commit to send truthful messages because of their strong bias. Thus, they would never be believed. This is equivalent to assuming that they do not send any message, i.e. \(\sigma _{B}=0.\) On the other hand members of type \(B,\) given their preferences, would not change their strategies even if they received a message revealing that the state of nature is \(L\). For these reasons we focus on the message strategies of the \(M \) members.

  19. The introduction of communication results in the expansion of the set of equilibria. When no information is revealed, \(M\) members now know that nobody is actually informed and thus have no strategic reason to contrast biased members and make other unbiased members pivotal. They can indifferently cast any vote and thus multiple equilibria arise: there now also exist equilibria where some unbiased members vote “no” after observing \(\sigma _{i}=H\). These additional equilibria may entail an unconvincing behavior on the part of the \(M\) members, nonetheless they all yield \(E(v)^{*}=\frac{1}{2}[1-(1-\alpha )^{n+1}]\). Notice however that, contrary to what happens in the type (ii) equilibria of Proposition 1, this multiplicity does not entail any coordination problem: whatever the choice of the \(M\) members receiving message \(\sigma _{i}=H\), an equilibrium with expected value \(E(v)^{*}\) is reached.

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Acknowledgments

We thank two anonymous referees, Ferdinando Colombo, Luigi Vannucci, and Giulio Pianigiani for very helpful suggestions. We also thank participants at the ASSET Conference 2008, the CEPET Workshop 2009, the EALE Conference 2009, the CESifo Area Conference in applied Micro 2010 and at seminars held at Stony Brook University, Catholic University of Milan, University of Milan-Bicocca for comments on previous versions of the paper. All mistakes are the authors’ own responsibility. Financial help from IEF (Catholic University) and PRIN 2008 is gratefully acknowledged.

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Correspondence to Paolo Balduzzi.

Appendix

Appendix

1.1 Proof of Proposition 1

Recall that value-maximizing members choose their strategies conditioning on being pivotal. Then, each informed member votes according to his information, as this maximizes the probability of making the correct decision. Thus, in what follows we only focus on the voting strategies of uninformed members.

Considering a committee composed of \(2n+1\) members of type \(M\), we prove that there only exist (i) multiple equilibria where \(2n\) individuals compensate in pure strategies, (ii) a unique equilibrium where all members compensate in mixed strategies playing \(m_{j}=1/2\), and multiple equilibria where \(2n-2k,\) \(k=1,2, \ldots ,n-1,\) individuals compensate in pure strategies and \(2k+1\) individuals play \(m_{k}=1/2\), (iii) multiple equilibria where \((n-k_{1}) \) members choose \(m_{j}=0\) (\(m_{z}=1\)) and \((n-k_{2})\) members choose \(m_{z}=1\) (\(m_{j}=0\)) with \(k_{1}<k_{2}\le n\) while all the others, \(k\ne j,z\), choose the same mixed strategy \(m_{k}<\frac{1}{2}\) (\(m_{k}>\frac{1}{2}).\) Finally, we prove that equilibria of type (i) are optimal in that they maximize \(E(v).\)

  1. (i)

    There exist multiple equilibria where \(n\) members choose \(m_{j}=1,\) \(n\) members choose \(m_{z}=0,\) and one member, denoted by \(M_{i}\) \(i\ne j,z\) , chooses \(m_{i}\in \left[ 0,1\right] \). There cannot exist an equilibrium where more than \(n+1\) members choose either \(m_{j}=1\) or \(m_{z}=0.\) We prove the existence of these equilibria in four steps. First we prove that player \(i\) is voting optimally, given the strategies of the other \(2n\) players; then we prove that the other \(2n\) members are voting optimally as well (steps 2 and 3). Finally, we prove that these are the only equilibria where at least \(n+1\) members choose either \(m_{j}=1\) or \(m_{z}=0.\)

    1. 1.

      If \(n\) members choose \( \,m_{j}\!=\!1,\) \(n\) members choose \( \,m_{z}\!=\!0,\) the best response of \(M_{i},\) \(i\ne j,z\), is to choose \(m_{i}\in \left[ 0,1\right].\) When \(n\) members are voting “yes” and \(n\) members are voting “no” \(M_{i}\) is pivotal in both states of the world with the same probability. Indeed, when \(v=1,\) \(M_{i}\) is pivotal when everybody else is uninformed or when the only informed members are those \(n\) members who would vote “yes” even if uninformed (thus not changing their votes whether informed or not). As the former case (everybody is uninformed) can be considered as a sub case of the latter, the probability that \(M_{i}\) is pivotal is

      $$\begin{aligned} (1-\alpha )^{n}\left[ \sum \limits _{j=0}^{n}\frac{n!}{j!(n-j)!}\alpha ^{n-j}(1-\alpha )^{j}\right] =(1-\alpha )^{n}. \end{aligned}$$

      where \(\frac{n!}{j!(n-j)!}\) represents the number of combination with \(n-j\) informed members among the \(n\) members who vote “yes” if uninformed, and the term in brackets is equal to \(1 \) from the binomial theorem. When \(v=-1,\) \(M_{i}\) is pivotal when everybody else is uninformed or when the only informed members are those who would vote “no” even if uninformed. Then the probability that \(M_{i}\) is pivotal is again \((1-\alpha )^{n}.\) Hence, \(M_{i}\) is indifferent between the possible values of \(m_{i}\in \left[ 0,1 \right].\)

    2. 2.

      If \(n\) members choose \(m_{j}=1,\) \( n-1\) members choose \(m_{z}=0,\) and member \(M_{i},\) \(i\ne j,z,\) chooses \(m_{i}=0\) the best response of the remaining member (denoted by \(k,\) \(k\ne i,j,z\) ) is to choose \(m_{k}\in \left[ 0,1\right] ;\) if \(M_{i}\) chooses \( m_{i}>0\) the best response of \(M_{k}\) is to choose \(m_{k}=0.\) If \(M_{i}\) chooses \(m_{i}=0,\) we are back to point 1. So the optimal response of the remaining \(M_{k}\) is \(m_{k}\in \left[ 0,1\right].\) If instead \(M_{i}\) chooses \(m_{i}=1,\) then \(M_{k}\) is pivotal only when \( v=-1.\) Consequently, \(M_{k}\) chooses \(m_{k}=0\). But \(M_{k}\) chooses \(m_{k}=0\) even if \(1>m_{i}>0\) because \(M_{k}\) is pivotal with a higher probability in \(v=-1\) than in \(v=1.\)

    3. 3.

      If \(n-1\) members choose \(m_{j}=1\) , \(n\) members choose \(m_{z}=0,\) and member \(M_{i},\) \(i\ne j,z,\) chooses \(m_{i}=1\) the best response of the remaining member (denoted by \(k,\) \(k\ne i,j,z\) ) is to choose \(m_{k}\in \left[ 0,1\right] ;\) if \(M_{i},\) chooses \(m_{i}<1\) the best response of \(M_{k}\) is to choose \(m_{k}=1.\) The argument is symmetric to the one used at point 2. Finally, note that any member can be in the position of \(M_{i},\) or in that of an \(M_{j}\) voting “yes”, or also in that of an \(M_{z}\) voting “no” or randomizing when uninformed. Thus, there is a multiplicity of equilibria such as the one we are considering.

    4. 4.

      There cannot exist an equilibrium where more than \(n+1\) members choose either \(m_{j}=1\) or \(m_{z}=0.\) Consider what happens if more than \(n+1\) members vote “yes”, i.e. suppose \(n+1+k\) members (\( k\in \left\{ 1,2,3, \ldots ,n-1\right\} \)) choose \(m_{j}=1\). Then every remaining member knows that he is pivotal with a higher probability when \(v=-1\). Hence, the remaining \(n-k\) members choose \(m_{z}=0\). However, this cannot be an equilibrium. Also members voting \(m_{j}=1\) know that they are pivotal with a higher probability when \(v=-1\). Hence, as long as more than \(n+1\) members still vote “yes” when uninformed (\(k\ne 0\)), they have an incentive to change their strategy and vote “no” when uninformed. A symmetric argument can be used to analyze what happens if more than \(n+1\) members vote “no” when uninformed, i.e. if \(n+1+k\) members (\(k\in \left\{ 1,2,3, \ldots ,n-1\right\} \)) choose \(m_{z}=0\) and consequently to rule out the existence of such equilibria.

  2. (ii)

    There exist multiple equilibria where \(n-k\) members choose \(m_{j}=1,\) \(n-k\) members choose \( m_{z}=0, \) and \(2k+1\) members choose \(m_{k}=1/2\) for \( k=1,2, \ldots ,n-1.\) If \(2k+1\) choose \(m_{k}=1/2\) for \( k=1,2, \ldots ,n-1\) for an equilibrium to exist, members choosing pure strategies must compensate each other. If \(2(n-k)\) members compensate in pure strategies, there cannot exist an equilibrium with at least one member choosing \(m_{k}\ne 1/2\). There exists a unique equilibrium where all the \(2n+1\) members play mixed strategies ( \(k=n\)). Such an equilibrium is symmetric with \(m_{k}=\frac{1}{2}.\) We prove the existence of these equilibria in three steps.

    1. 1.

      If \(n-k\) members choose \(\,m_{j}=1,\) \(n-k\) members choose \(m_{z}=0,\) and \(2k\) members choose \(m_{k}=1/2\), \(k=1,2, \ldots ,n\), the best response of the remaining member (denoted by \(i\) , \(i\ne j,k,z\) ) is to choose \(m_{i}=\frac{1}{2}.\) Both when \(v=1\) and when \(v=-1,\) \(M_{i}\) is pivotal if (a) everybody is uninformed and \(n\) members vote “yes” while the other \(n\) members vote “no”, or (b) no more than \(n\) members are informed and vote accordingly, while uninformed members vote in such a way that results in \(n\) members voting “yes” and \(n\) members voting “no”. Given that the other members either compensate in pure strategies or choose \(m_{k}=\frac{1}{2}\), \(M_{i}\) is pivotal with the same probability in both states of the world. Since both states are equally possible, \(M_{i}\) is then indifferent among any \(m_{i}\in \left[ 0,1\right].\) This holds true for every member playing a mixed strategy. If nobody plays a pure strategy \((k=n)\) it immediately follows that \(m_{k}=\frac{1}{2}\) for \( j=1, \ldots ,2n+1,\) sustains an equilibrium of the game.

    2. 2.

      If there are more members choosing \(m_{j}=1\) ( \(m_{z}=0\) ) than members choosing \(m_{z}=0\) ( \(m_{j}=1\) ), while the other members but one choose \(m_{k}=\frac{1}{2}\), the remaining member has a pure strategy as his best response. If \(2k+1\) members choose \(m_{k}=\frac{1}{2},\) the other members must compensate each other in pure strategies. If there are more members choosing \(m_{j}=1\) (\(m_{z}=0\)) than members choosing \(m_{z}=0\) (\(m_{j}=1\)) while all the others but member \(M_{i},\) \( i\ne j,z,k\), choose \(m_{k}=\frac{1}{2}\), the best response of \(M_{i}\) is to choose \(m_{i}=0\) (\(m_{i}=1\)) because \(M_{i}\) is pivotal with a higher probability when \(v=-1\) (\(v=1\)) than when \(v=1\) (\(v=-1).\) From point 1. and 2., it follows that a situation where \(n-k\) members choose \(m_{j}=1,\) \(n-k\) members choose \(m_{z}=0,\) and \(2k+1\) members choose \( m_{k}=1/2,\) \(k=1,2, \ldots ,n-1,\) constitutes an equilibrium of the game. Moreover it follows from 2., that if \(2k+1\) choose \(m_{k}=1/2,\) \(k=1,2, \ldots , n-1,\) for an equilibrium to exist, members choosing pure strategies must compensate each other. Given that \(k\) can take values \(1,2, \ldots , n-1\) and that any member can be in the position of a \(j\), a \(z\) or a \(k\) member, there exists a multiplicity of such equilibria.

    3. 3.

      There cannot exist either an equilibrium in mixed strategies with \(m_{k}\ne \frac{1}{2}\) for one or more members or an equilibrium with at least one member choosing \(m_{k}\ne 1/2\) when \(n-k\) members choose \(m_{j}=1,\) \(n-k\) members choose \(m_{z}=0\) for \(k=1,2, \ldots ,n-1\) and the other \( 2k\) members choose mixed strategies. If \(M_{i}\) were to choose \(m_{i}>\frac{1}{2}\) (\(m_{i}<\frac{1}{2}\)) while \( 2n-1\) members choose \(m_{k}=\frac{1}{2},\) the best response of the remaining member denoted by \(j,\) \(j\ne i,k\), would be \(m_{j}=0\) (\(m_{j}=1\))\(,\) because \(M_{j}\) would be pivotal with a higher probability when \(v=-1\) (\(v=1\) ) than when \(v=1\) (\(v=-1\))\(.\) With \(2n-1\) members choosing \(m_{k}=\frac{1}{2} \) and \(M_{j}\) choosing \(m_{j}=0\) (\(m_{j}=1\))\(,\) however the best response of \(M_{i}\) becomes \(m_{i}=1\) (\(m_{i}=0\))\(,\) because \(M_{i}\) would be pivotal with a higher probability in \(v=1\) (\(v=-1\)) than in \(v=-1\) (\(v=1\))\(.\) A similar argument holds true if member \(M_{i},\) choosing \(m_{i}>\frac{1}{2} , \) were compensated by another member, denoted by \(h\), choosing \(m_{h}< \frac{1}{2}\) and such that \(m_{i}+m_{h}=1\). In this case no other member has an incentive to deviate from \(m_{j}=\frac{1}{2}\), but it immediately appears that \(m_{h}<\frac{1}{2}\) is not a best response. Given that \(M_{h}\) is pivotal with a higher probability in \(v=-1\) than in \(v=1\), his best response is \(m_{h}=0.\) More generally, by applying the same line of reasoning, it can be verified that there cannot exist an equilibrium with \(m_{i}\ne \frac{1}{2}\) for at least one \(i\), because as soon as one or more agents choose \(m_{i}\ne \frac{ 1}{2},\) there is at least one agent (possibly one of those choosing \( m_{i}\ne \frac{1}{2}\)) who has a pure strategy as his best response. Hence the only equilibrium in mixed strategies is the one with \(m_{i}=\frac{1}{2}\) for \(i=1,2, \ldots ,2n+1.\) Exactly the same argument can be applied in order to rule out equilibria where \(2k+1\) members choose mixed strategies \(m_{k}\ne \frac{1}{2},\) when there are \(2(n-k)\) members, \(k=1,2, \ldots , n-1\), compensating each other in pure strategies.

  3. (iii)

    There may exist multiple equilibria where \(\mathbf (n-k _{1}\mathbf ) \) members choose \(m_{j}=0\) ( \(m_{z}=1\) ) and \(\mathbf (n-k _{2}\mathbf ) \) members choose \( m_{z}=1\) ( \(m_{j}=0\) ) with \(\mathbf k _{1}<\mathbf k _{2}\le n\) while all the others, \(k\ne j,z\), choose the mixed strategy \(m_{k}>\frac{1}{2}\) (\(m_{k}<\frac{1}{2}).\) We do not need to characterize all possible equilibria of this type. To our purpose, it is sufficient to show that there may exist equilibria where some members use a mixed strategy \(m\in (0,1)\) and the other members choose pure strategies. More precisely, we build an example for the case of \(n=3\) and \( k_{1}=1<k_{2}=2.\) This is the smallest board where some members may compensate in pure strategies, while another member chooses a pure strategy and the rest choose mixed strategies (if \(n=2\) there could be an equilibrium with one member choosing a pure strategy while the rest are choosing the same mixed strategy). First of all by the argument used at point (ii)1., for this to be an equilibrium, members choosing mixed strategies must choose the same value of \(m\). Hence, consider the case where \(m_{1}=m_{2}=0,\) \(m_{3}=1,\) \( m_{4}=m_{5}=m_{6}=m_{7}=\overline{m}.\) The value of \(\overline{m}\) can be derived by maximizing the function \(V(m)\) defined as follows for \(m\in [0,1]\)

    $$\begin{aligned} V(m)&= E(v)=\frac{1}{2}[Y(\cdot |v=1)-Y(\cdot |v=-1)]\quad \mathrm{when}\\ m_{1}&= m_{2}=0 \\ m_{3}&= 1 \\ m_{4}&= m_{5}=m_{6}=m_{7}=\overline{m}. \end{aligned}$$

    Before showing that the solution to the maximization of \(V(m)\) identifies the equilibrium mixed strategy\(,\) we show that such maximization has an interior solution. Given that (a) all \(M\) players have identical payoffs equal to \(E(v)\) and (b) the game is symmetric with respect to the \(M\) players, neither \(m=0\) nor \(m=1\) maximize \(V(m).\) To check this, suppose that \(V(m)\) is maximized by \(m=1\) and consider \(M_{4}.\) Given \(m_{5}=m_{6}=m_{7}=m=1,\) \(M_{4}\) is more likely to be pivotal in \(v=-1\) than in \(v=1,\) then his best reply is not \(m_{4}=1\) but it is \(m_{4}=0,\) because with such a choice he can obtain the highest possible value of \(E(v).\) Since a player’s expected payoff is linear in his own mixed strategy, we can then replace \(m_{4}=1\) with \(m_{4}^{\prime }=0+\varepsilon \) for any \(\varepsilon \in (0,1)\) and raise \(E(v).\) Given (a) and (b), this is true also for \(M_{5},\) \(M_{6}\) and \(M_{7}.\) Since \(E(v)\) is a polynomial in (\(m_{1},m_{2}, \ldots , m_{7}\)), first-order effects dominate for sufficiently small \(\varepsilon >0,\) if \(m=1\) is replaced by \(m^{\prime }=0+\varepsilon \) in \(V(m)\) so that \(V(0+\varepsilon )>V(1),\) contradicting that \(V(m)\) is maximized by \(m=1.\) An analogous argument rules out that \(V(m)\) is maximized by \(m=0.\) Since \(V(m)\) is continuos in \( m\), there then exists a value \(\overline{m}\in (0,1)\) that maximizes \(V(m),\) and, considering that if it were \(\overline{m}\le 1/2\) the above argument (on possible increases of \(E(v)\) by changing \(m\)) could be applied again, we can conclude that \(\overline{m}>1/2.\) In order to show that \(m_{1}=m_{2}=0,\) \(m_{3}=1,\) \(m_{4}=m_{5}=m_{6}=m_{7}= \overline{m}\) is actually an equilibrium, consider first of all one of the members choosing \(\overline{m},\) e.g. \(M_{4}\). For this to be an equilibrium, \(\overline{m}\) must represent the best reply of \(M_{4}.\) Suppose instead that \(M_{4}\) can raise his payoff, \(E(v),\) by choosing \( m_{4}^{\prime }\ne \overline{m}\) and let \(m_{4}^{\prime }=\overline{m} +\Delta .\) Since a member’s payoff is linear in his mixed strategy, we can replace \(\overline{m}\) with \(\overline{m}+\varepsilon \Delta \) for any \( \varepsilon \in (0,1)\) and so raise \(E(v).\) Then the above argument can be repeated to show that for sufficiently low \(\varepsilon \), it is possible to replace \(\overline{m}\) with \(\overline{m}+\varepsilon \Delta \) for \(M_{4},\) \(M_{5},\) \(M_{6},\) \(M_{7}\) and obtain \(V(\overline{m}+\varepsilon \Delta )>V(\overline{m}),\) thus contradicting that \(\overline{m}\) maximizes \(V(m)\). To check that also \(M_{1},\) \(M_{2},\) and \(M_{3}\) are choosing their best replies, consider one of the members choosing \(m_{i}=0,\) e.g. \(M_{1}\). As \( \overline{m}>1/2,\) the probability that \(M_{1}\) is pivotal is higher in \( v=-1\) than in \(v=1\), consequently his best reply is to choose \(m_{1}=0.\) The same clearly holds for \(M_{2}\). Consider instead \(M_{3}.\) Given \( m_{1}=m_{2}=0\) and the fact that \(\overline{m}\) is calculated by maximizing \(E(v)\) in a case where only one member is choosing \(m=1\), \(M_{3}\) is pivotal with a higher probability in \(v=1\) than in \(v=-1.\) Consequently his best response is to actually choose \(m_{3}=1\). Since any of the members can be in the position of \(M_{1},\) \(M_{2},\) \(M_{3},\) \(M_{4},\) \(M_{5},\) \(M_{6},\) \(M_{7},\) there exist a multiplicity of such equilibria. By solving this problem for \(\alpha ={\frac{1}{2}}\) we find:

    $$\begin{aligned} m_{4}&= m_{5}=m_{6}=m_{7}=0.656537; \\ E(v)&= 0.448531 \end{aligned}$$

    Clearly an analogous equilibrium can be constructed where \(m_{1}=m_{2}=1,\) \( m_{3}=0,\) \(m_{4}=m_{5}=m_{6}=m_{7}=m_{k}<1/2.\)

  4. (iv)

    There cannot exist other equilibria than those considered at points (i)–(iii). From points (i)–(iii) it immediately follows that all possible combinations of strategies have been considered.

  5. (v)

    The equilibria with compensation in pure strategies obtained at point (i) maximize \(E(v)\) In order to compare the expected value obtained in different equilibria, consider that \(E(v)\) can be written as

    $$\begin{aligned} E(v)&= \frac{1}{2}[Y(\cdot |v=1)-Y(\cdot |v=-1)] \\&= \frac{1}{2}\left[ 1-Y(\cdot |v=-1)-(1-Y(\cdot |v=1)\right]. \end{aligned}$$

where \(Y(\cdot |v=-1)+(1-Y(\cdot |v=1)\) is the probability of making the wrong decision.

Consider the equilibrium where every member plays the mixed strategy \(m_{k}=1/2.\) We know from point (ii)1. that when there are \(2n\) members choosing \(m_{k}=1/2,\) the \(2n+1th\) member, denoted by \(M_{i},\) is in fact indifferent among any \(m_{i}\in \left[ 0,1 \right].\) This means that he cannot raise \(E(v)\) by playing one instead of another strategy because as \(Y(\cdot |v=-1)\) is lowered (raised) by choosing a specific strategy, \((1-Y(\cdot |v=1)\) is raised (lowered) by exactly the same amount. Then the expected value is the same for any \(m_{i}\in \left[ 0,1 \right].\) This implies that if \(M_{i}\) were to choose \(m_{i}=1,\) \(E(v)\) would stay at \(E(v)^{MS},\) even if this would not be an equilibrium situation. Consider another member, denoted by \(M_{h}.\) From point (ii)2. we know that in such non equilibrium situation, his best response is to choose \(m_{h}=0\), because he is pivotal with a higher probability in \(v=-1\) than in \(v=1\) and by choosing \(m_{h}=0\), he can raise \( E(v)\) above \(E(v)^{MS}\). We know from point (ii) that the situation where \( M_{i}\) chooses \(m_{i}=1,\) \(M_{h} \) chooses \(m_{h}=0\) and all the other members choose \(m_{k}=1/2\) constitutes an equilibrium. Then, such an equilibrium yields \(E(v)>E(v)^{MS}.\) The above argument, however, can be applied again by singling out one of the members choosing \(m_{k}=1/2\) who is in fact indifferent among any \(m\in \left[ 0,1\right] \), and letting him choose \(m=1\) (\(m=0\)). The new situation will not constitute an equilibrium but there will again be another member who can raise \(E(v)\) by choosing \( m=0 \) (\(m=1\)), and this new situation will constitute an equilibrium. Clearly the argument can be recursively repeated to the point where an equilibrium in which \(2n\) members compensate in pure strategies while the remaining member chooses \(m_{k}\in \left[ 0,1\right]\) is reached. Consequently, an equilibrium with compensation in pure strategies yields a higher \(E(v)\) than the equilibrium in mixed strategies, as well as than the equilibria with \(2(n-k)\) agents compensating in pure strategies and \(k+1\) agents choosing \(m_{k}=1/2.\)

Consider now the case where \((n-k_{1})\) members choose \(m_{j}=0\) and \( (n-k_{2})\) members choose \(m_{z}=1\) with \(k_{1}<k_{2}\le n\), while all the others, \(k\ne j,z\), choose the mixed strategy \(m_{k}>\frac{1}{2}.\) In particular, consider the equilibrium characterized at point iii. The same argument as above can be applied with slight modifications.

Consider \(M_{4}\) who is choosing the mixed strategy \(m_{k}.\) \(M_{4}\) is in fact indifferent among any \(m_{4}\in \left[ 0,1\right] \) meaning that \(E(v)\) is not modified if he changes his strategy (provided that the others do not change theirs). So let him switch to \(m_{4}=1\). \(E(v)\) is still \(0.448531\) but this is no longer an equilibrium because \(m_{k}=0.656537\) was calculated so as to make any of \(M_{4},\) \(M_{5},\) \(M_{6},\) \(M_{7}\) indifferent among \( m_{k}\in \left[ 0,1\right] \) when the four of them where playing \(m_{k},\) but now there are only three of them choosing \(m_{k}\) implying that each is pivotal with a higher probability in \(v=-1\) than in \(v=1.\) Consider then \( M_{5}\). His best response is to choose \(m_{5}=0\), meaning that, by doing so, he can raise \(E(v)\) above \(0.448531\). This is not an equilibrium, because if \(m_{1}=m_{2}=m_{5}=0\) and \(m_{3}=m_{4}=1\) and \(m_{7}=m_{k},\) the best response of \(M_{6}\) is to choose \(m_{6}=1\) as he is pivotal with a higher probability in \(v=1\) than in \(v=-1.\) But this means that, by doing so, \( M_{6} \) can further raise \(E(v).\) Moreover as now \(m_{1}=m_{2}=m_{5}=0\) and \( m_{3}=m_{4}=\) \(m_{6}=1\), an equilibrium with compensation in pure strategies has been established.

This argument can clearly be generalized to any equilibrium where \((n-k_{1})\) members choose \(m_{j}=0\) (\(m_{j}=1\)) and \((n-k_{2})\) members choose \( m_{z}=1 \) (\(m_{z}=0\)) with \(k_{1}<k_{2}\le n\), while all the others, \( k\ne j,z\), choose the mixed strategy \(m_{k}>\frac{1}{2}.\) It is sufficient to single out one of the members choosing the mixed strategy and let him switch to a pure strategy. Then, by having the the others successively switch to their best responses, the equilibrium with compensation in pure strategies is reached as an improvement upon the starting point.

1.2 Proof of Corollary 1

The Corollary follows immediately considering that

$$\begin{aligned} \frac{\partial [E(v)^{*}]}{\partial \alpha }=\frac{n+1}{2} (1-\alpha )^{n}>0, \frac{\partial [E(v)^{*}]^{2}}{ \partial \alpha ^{2}}=-\frac{n(n+1)}{2}(1-\alpha )^{n-1}; \end{aligned}$$

and that the marginal expected value when \(n\) increases is

$$\begin{aligned} \Delta _{2n+1}E(v)^{PS}\equiv \frac{1}{2}\left[1-(1-\alpha )^{(n+1)+1}\right]-\frac{1}{2 }\left[1-(1-\alpha )^{n+1}\right]=\frac{\alpha }{2}(1-\alpha )^{n+1} \end{aligned}$$

which is clearly decreasing in \(n.\)

1.3 Proof of Proposition 2

Recall that value-maximizing members choose their strategies conditioning on being pivotal. Then, each informed \(M\) member votes according to his information, as this maximizes the probability of making the correct decision. Thus, in what follows we only focus on the voting strategies of uninformed members. The proof is organized as follows. We prove that:

  • (i) if the committee is composed of \(n+1\) value-maximizing members and \(n\) biased members, there exists a unique equilibrium where each \(M\) member votes “no” when uninformed;

  • (ii\({_1}\)) if the committee is composed of \(n+1+k\) value-maximizing members and \(n-k\) biased members, there always exist equilibria where \(n-k\) value-maximizing members vote “no” when uninformed and the remaining \(2k\) value-maximizing members compensate for each other in pure strategies.

  • (ii\(_{2}\)) if the committee is composed of \(n+1+k\) value-maximizing members and \(n-k\) biased members \((n>k>0)\), there may exist equilibria where some or all the value-maximizing members play the same mixed strategy;

  • (iii) the equilibria sub (i) and sub (ii\(_{1})\) are optimal while the equilibria sub \((\mathrm{ii}_2)\) are suboptimal.

 

(i) In a committee with \(n+1\) value-maximizing members and \(n\) biased members there exists a unique equilibrium where all the \(M\) members vote “no” whenever uninformed (that is, \(m_{i}=0;i=1,2, \ldots ,n+1\)).

Consider member \(M_{n+1}\). When \(v=1,\) \(M_{n+1}\) is pivotal only if all the other \(M\) members are uninformed and vote “ no”, which happens with probability:

$$\begin{aligned} (1-\alpha )^{n}\prod \limits _{i=1}^{n}(1-m_{i}) \end{aligned}$$

When \(v=-1,\) \(M_{n+1}\) is pivotal if:

  • all the other \(M\) members are uninformed and vote “no”, which happens with probability

    $$\begin{aligned} (1-\alpha )^{n}\prod \limits _{i=1}^{n}(1-m_{i}), \end{aligned}$$
  • all the other \(M\) members are informed, which happens with probability

    $$\begin{aligned} \alpha ^{n}, \end{aligned}$$
  • at least one (but not all) of the other \(M\) members is informed and the others vote “no” when uninformed, which happens with probability

$$\begin{aligned} \sum \limits _{h=1}^{n}\frac{n!}{h!(n-h)!}\alpha ^{n-h}(1-\alpha )^{h}\prod \limits _{i=1}^{h}(1-m_{i}). \end{aligned}$$

where \(\frac{n!}{h!(n-h)!}\) represents the number of combination with \(h\) uninformed value-maximizing members and \(n-h\) informed value-maximizing members. It is straightforward that \(M_{n+1}\) is pivotal with a higher probability when \(v=-1\). Hence \(M_{n+1}\) chooses \(m_{n+1}=0.\) As the same reasoning holds for any other value-maximizing member \(i\ne n+1\), it follows that every \(M\) member will vote “no” when uninformed.

Finally, note that we have not restricted \(m_{i},\) \(i\ne n+1,\) to any particular value, so the result also proves that this equilibrium is unique.

(ii\(_{1})\) In the case of \(n-k\) biased members ( \(n>k>0)\) and \(n+1+k\) value-maximizing members, there exist multiple equilibria with \(n-k+1\) value-maximizing members voting against the project and \(2k\) value-maximizing members compensating for each other in pure strategies.

We prove the existence of these equilibria in three steps. In the first step, we prove that when \(n-k\) value-maximizing members vote against the project and \(2k\) value-maximizing members compensate for each other, the remaining \(M\) member has still an incentive to vote against the project; in the second step, we prove that when \(n-k\) value-maximizing members vote against the project to contrast the \(n-k\) biased members, and a majority of the other value-maximizing members also vote against the project, the remaining \(M\) member has an incentive to compensate, voting “yes”. Finally, we show that there are no other equilibria in pure strategies.

  1. 1.

    If \(n\) value-maximizing members choose \( m_{z}=0,\) and \(k\) value-maximizing members choose \( m_{j}=1,\) the best response of \(M_{i},\) \(i\ne j,z,\) is to choose \(m_{i}=0.\) When \(v=1,\) \(M_{i}\) is pivotal if all the value-maximizing members are uninformed or if at least one of those \(k\) value-maximizing members who choose \(m_{j}=1\) when uninformed, is in fact informed. Thus, \( M_{i}\) is pivotal with probability

    $$\begin{aligned} (1-\alpha )^{n}\left[ \sum \limits _{j=0}^{k}\frac{k!}{j!(k-j)!}\alpha ^{k-j}(1-\alpha )^{j}\right] =(1-\alpha )^{n} \end{aligned}$$

    where \(\frac{k!}{j!(k-j)!}\) represents the number of combination with \(j\) uninformed value-maximizing members, \(k-j\) informed \(M\) members and the term in brackets is equal to \(1\) from the binomial theorem. When \(v=-1,\) \(M_{i}\) is pivotal if all the \(M\) members are uninformed or if at least one of those \(n\) value-maximizing members who choose \(m_{z}=0\) when uninformed, is in fact informed. Then \(M_{i}\) is pivotal with probability

    $$\begin{aligned} (1-\alpha )^{k}\left[ \sum \limits _{z=0}^{n}\frac{n!}{z!(n-z)!}\alpha ^{n-z}(1-\alpha )^{z}\right] =(1-\alpha )^{k} \end{aligned}$$

    Given that \((1-\alpha )^{k}>(1-\alpha )^{n},\) the probability that \(M_{i}\) is pivotal is higher when \(v=-1\) than when \(v=1.\) Hence \(M_{i}\) chooses \( m_{i}=0.\)

  2. 2.

    If \(n+1\) value-maximizing members choose \( m_{z}=0\) and \(k-1\) value-maximizing members choose \( m_{j}=1,\) the best response of \(M_{i},\) \(i\ne j,z\) is to choose \(m_{i}=1\) When \(v=1,\) \(M_{i}\) is pivotal if only one of the \(n+1\) value-maximizing members choosing \(m_{z}=0\) is informed and votes “yes”. This happens with probability

    $$\begin{aligned} (n+1)(1-\alpha )^{n}\alpha . \end{aligned}$$

    On the contrary, \(M_{i}\) is never pivotal when \(v=-1.\) Hence, he chooses \( m_{i}=1.\) Finally, note that any \(M\) member can be in the position of \(M_{i}\) or in that of an \(M_{j}\) voting “yes”, or also in that of an \(M_{z}\) voting “no”. Thus, there is a multiplicity of equilibria such as the one we are considering.

  3. 3.

    There cannot exist other equilibria in pure strategies than those characterized at points 1. and 2. We must now consider what happens if either a) more than \(n\) value-maximizing members vote “no” and the others vote “yes”, or b) more than \(k\) value-maximizing members vote “yes” and the rest vote “no”.

 

  1. (a)

    If \(n-h\) value-maximizing members choose \(m_{z}=0,\) and \(k+h\) value-maximizing members choose \(m_{j}=1,\) \( n\ge h>0\), the best response of \(M_{i},\) \(i\ne j,z,\) is to choose \(m_{i}=0\) because \(M_{i}\) is never pivotal when \(v=1\) while he may be pivotal when \(v=-1.\) This happens in the case where \(h\) of those \(n+h\) value-maximizing members who choose \(m_{j}=1\) if uninformed, are in fact informed. As this is true for any \(h>0\), we are back to the case examined at point 1 above.

  2. (b)

    If \(n+h\) value-maximizing members choose \(m_{z}=0,\) and \(k-h\) value-maximizing members choose \(m_{j}=1,\) \(k\ge h>1,\) the best response of \(M_{i},\) \( i\ne j,z,\) is to choose \(m_{i}=1\) because \(M_{i}\) is never pivotal when \(v=-1\) while he may be pivotal when \(v=1.\) This happens in the case where \(h\) of those \(n+h\) value-maximizing members who choose \(m_{z}=0\) if uninformed, are in fact informed. As this is true for any \(h>1\), we are back to the case examined at point 2 above.

(ii\(_{2}\)) In the case of \(n-k\) biased members ( \(n>k>0)\) and \(n+1+k\) value-maximizing members, there may exist equilibria where some or all of the \(M\) members choose the same mixed strategy.

In an equilibrium where all the \(M\) members choose the same mixed strategy, the value of such strategy, \(\overline{m}\), can be derived by maximizing the function \(V(m)\) defined as follows for \(m\in [0,1]\)

$$\begin{aligned}&V(m) =E(v)=\frac{1}{2}[Y(\cdot |v=1)-Y(\cdot |v=-1)]\quad \text{ when} \\&\quad m_{i} =m_{j}=\overline{m},\quad \forall j\ne i i,j=1,2, \ldots ,n+k+1 \\&\quad B_{z} \text{ always} \text{ votes} \text{``yes''} z =1, \ldots ,n-k \end{aligned}$$

Given that (a) all \(M\) players have identical payoffs equal to \(E(v),\) (b) the game is symmetric with respect to the \(M\) players, and (c) the number of the \(M\) members exceeds that of the \(B\) members by more than one, the same argument as in Proposition 1, point iii can be used to prove that the maximization of \(V(m)\) has an interior solution \(\overline{m}<1/2,\) and that \(\overline{m}\) represents the best reply of the \(M\) members. Considering that any \(B\) member is following his dominant strategy this then represents an equilibrium.

We have solved this problem for the case of \(n=2\), with one member of type \( B \) (hence, \(B_{1}\) always votes “yes”) and the remaining four members of type \(M.\) The solution of the problem, evaluated at \(\alpha =\frac{1}{2},\) is:

$$\begin{aligned} m_{1}=m_{2}=m_{3}=m_{4}=\frac{5-\sqrt{13}}{6}, \end{aligned}$$

yielding \(E(v)=.384973.\)

Analogously to the third and fourth case of proposition \(1\), there may also exist equilibria in which some of the \(M\) members choose \(m_{j}=0,\) and the others choose the same mixed strategy, or even equilibria where part of the \(M\) members choose \(m_{z}=1.\) We do not characterize such equilibria but we show at point (iii) that, whenever they exist, they are suboptimal.

(iii) The equilibria in pure strategies are optimal, other equilibria are suboptimal

Recall that

$$\begin{aligned} E(v)&= \frac{1}{2}[Y(\cdot |v=1)-Y(\cdot |v=-1)] \\&= \frac{1}{2}\left[ 1-Y(\cdot |v=-1)-(1-Y(\cdot |v=1)\right]. \end{aligned}$$

where \(Y(\cdot |v=-1)+(1-Y(\cdot |v=1)\) is the probability of making the wrong decision.

In the unique equilibrium of the case with \(n\) biased members and \(n+1\) value-maximizing members (point i), as well as in the pure strategy equilibrium of the case with \(n-k\) biased members (\(n>k>0)\) and \(n+1+k\) value-maximizing members (point ii\(_{1})\), \(Y(\cdot |v=-1)=0\) and \(Y(\cdot |v=1)\) is equal to the probability that at least one of the \(n+1\) members who choose \(m_{i}=0\) is informed. Then in both cases \(E(v)\) is equal to

$$\begin{aligned} \frac{1}{2}\sum \limits _{i=0}^{n}\frac{\left( n+1\right) !}{i!(n+1-i)!}\alpha ^{n+1-i}(1-\alpha )^{i}=\frac{1}{2}\left[ 1-(1-\alpha )^{n+1}\right] =E(v)^{*} \end{aligned}$$

implying that these equilibria are optimal.

In order to prove that the equilibria where unbiased members choose mixed strategies are suboptimal, an analogous argument to the one used in point (v) of Proposition 1 can be applied. Consider the equilibrium introduced as an example at point (ii\(_{2}\)) above. Each the \(M\) members, individually taken, is in fact indifferent among any \(m_{i}\in \left[ 0,1\right] \). Let \(M_{1}\) switch to \(m_{1}=0\). \(E(v)\) is not modified but this is no longer an equilibrium because \(m_{j}=\frac{5-\sqrt{13}}{6}\) was calculated so as to make unbiased members playing mixed strategies indifferent among \(m_{j}\in \left[ 0,1\right] \) when there were four of them, while now there are only three \(M\) members playing \(m_{j}=\frac{5-\sqrt{13}}{6}\). This implies that one of the members who are still choosing \(m_{j}<\frac{1}{2},\) say \(M_{2}\), has as his best response \(m_{2}=1\) because he is now pivotal with a higher probability in \(v=1\) than in \(v=-1.\) This in turn implies that, by switching to \(m_{2}=1\), he can raise \(E(v)\) above \(.384973\). The resulting situation will not be an equilibrium (as \(m_{j}\) is unaltered for those who still play the mixed strategy, the probability that one of them is pivotal is higher in \(v=-1\) than in \(v=1\)) meaning that there is another member, say \(M_{3}\) who can further raise \(E(v)\) by switching to \(m_{3}=0\). Since now there is a \( B_{1}\) always voting “yes”, \(M_{1}\) and \( M_{3}\) choosing \(m_{1}=m_{3}=0\) and \(M_{2}\) choosing \(m_{2}=1,\) we know from point (ii\(_{1})\) 1 that the best response of \(M_{4}\) is to choose \( m_{4}=0.\) But, by doing so, \(M_{4}\) further raises \(E(v)\) while reaching an equilibrium of type (ii\(_{2}).\)

Clearly the same procedure applies to any equilibrium where the \(M\) members adopt the same mixed strategy, independently of the size of the board, and of the proportion to the \(B\) members. It also applies to possible equilibria in which some of the \(M\) members choose \(m_{j}=0,\) and the others choose the same mixed strategy, or to equilibria where part of the \(M\) members choose \(m_{z}=1.\) First of all observe that for such an equilibrium to be established it must be at least \(n=2\) and that the agents adopting a mixed strategy must choose the same \(m_{k}\). Moreover there must be at least three agents choosing the mixed strategy. It is immediate to verify that there cannot be only two of them, because if there is at least one \(M\) member choosing a pure strategy and only one \(M\) member choosing a mixed strategy, the best response of the remaining member is a pure strategy. Whenever there are more than one agent using the same mixed strategy, however, we can consider the situation where one of these agents switches from his mixed equilibrium strategy to a pure one. This would not represent an equilibrium and \(E(v)\) could be improved by having another agent moving to his best response. By applying this method recursively the equilibria in pure strategies are reached.

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Balduzzi, P., Graziano, C. & Luporini, A. Voting in small committees. J Econ 111, 69–95 (2014). https://doi.org/10.1007/s00712-012-0321-5

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