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Preference intensity representation: strategic overstating in large elections

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Abstract

If voters vote strategically, is it useful to offer them the possibility of expressing nuanced opinions, or would they always overstate the intensity of their preferences? For additive voting rules, say that a ballot is extremal if it is neither abstention-like nor can be expressed as a mixture of the available ballots. We give a sufficient condition for strategic equivalence: if two rules share the same set of extremal ballots (up to an homothetic transformation), they are strategically equivalent in large elections. This condition is also necessary for the strategic equivalence of positional rules. These results do not hold for small electorates.

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Notes

  1. EV is obviously related to Utilitarianism; see Karni (1998), Dhillon and Mertens (1999), Segal (2000), d’Aspremont and Gevers (2002) and Gaertner and Xu (2012) who called it Range Voting, for axiomatic analyses. \(AV\) is often advocated since it emerged in the literature in the mid 70s; see Laslier and Sanver (2010) for a detailed account. In an election held under \(PV\), a voter is allowed to give at most one point to at most one candidate and the candidate with the most votes wins the election. The most common rule for direct presidential elections is Plurality with a Runoff (Blais et al. 1997), but we here restrict attention to one-round voting systems.

  2. Other models of large electorates have been proposed: Palfrey (1989), Laslier (2000), Myerson (2000), and McKelvey and Patty (2006).

  3. See Sawyer and MacRae (1962), Brams (1975), Nitzan (1985), Cox (1990) and Gerber et al. (1998).

  4. This is roughly equivalent to assume that the probability of candidates \(c_i\) and \(c_j\) being tied for first place is the same as the probability of candidate \(c_i\) being in first place one point ahead of candidate \(c_j\) (and both candidates above the rest of the candidates), which is in turn the same one as the probability of candidate \(c_j\) being in first place one vote ahead of candidate \(c_i\). Myerson and Weber (1993) justify this assumption by arguing that it seems reasonable when the electorate is large enough. This is not verified in Poisson games, a formal model of large elections in which the pivot probabilities are derived endogenously from the structure of the game.

  5. See Theorem 1, page 105 in Myerson and Weber 1993.

  6. See Milgrom 2009, 2010 and Perez-Richet (2011).

  7. See Myerson (2002), Laslier (2009), Núñez (2010), Bouton et al. (2012), Goertz and Maniquet (2011), and Núñez (2010).

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Acknowledgments

This work has benefited from useful discussions with Francis Bloch, Seok-ju Cho, Francesco De Sinopoli, Santiago Oliveros and Eduardo Perez-Richet. This work has also benefited from comments of conference and seminar participants in the American Political Science Association Meeting 2011—Seattle, OECD—New Directions in Welfare Conference 2011, the Workshop Advances on Approval Voting 2010, Université Paris II, THEMA—Université de Cergy-Pontoise and Ecole Polytechnique. We would also like to thank the editor John Duggan and a referee for very useful suggestions that substantially upgraded the quality of this work.

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Correspondence to Matías Núñez.

Appendix: Proof of Proposition 4

Appendix: Proof of Proposition 4

The first step of the proof consists in showing that \(g\) is a mixed strategy equilibrium. To do so, we compute the probability, under \(g\), of each pivot outcome a player can face and, from these probabilities, the expected utility derived from each undominated strategy.

Voters 1,2,3

Even though the best responses are explained for the voter 1, the reasoning is analogous for voters 2 and 3.

$$\begin{aligned} p((4,4,6)\,|\,g_{-1})=1/3\\ p((4,5,6)\,|\,g_{-1})=1/3\\ p((4,6,6)\,|\,g_{-1})=1/3. \end{aligned}$$

From the pivot probabilities previously described, we have

$$\begin{aligned} \pi _\mathbf{1 }(2,0,0)=25/9\\ \pi _\mathbf{1 }(2,1,0)=19/9\\ \pi _\mathbf{1 }(2,2,0)=13/9. \end{aligned}$$

which entails that \((2,0,0)\) is the unique best response for voter 1.

Voters 4,5

Voter 4’s best responses are analyzed, the reasoning being analogous for the voter 5.

$$\begin{aligned} p((6,2,5)\,|\,g_{-4})=1/3\\ p((6,3,5)\,|\,g_{-4})=1/3\\ p((6,4,5)\,|\,g_{-4})=1/3. \end{aligned}$$

From the pivot probabilities previously described, we have

$$\begin{aligned} \pi _\mathbf{4 }(0,2,0)=1\\ \pi _\mathbf{4 }(0,2,1)=10/9\\ \pi _\mathbf{4 }(0,2,2)=1. \end{aligned}$$

implying that \((0,2,1)\) is the unique best response for voter 4.

Voter 6

The most probable pivot outcomes faced by voter 6 are as follows

$$\begin{aligned} p((6,4,4)\,|\,g_{-6})=1/3\\ p((6,5,4)\,|\,g_{-6})=1/3\\ p((6,6,4)\,|\,g_{-6})=1/3 \end{aligned}$$

From the pivot probabilities previously described, we have

$$\begin{aligned} \pi _\mathbf{6 }(0,0,2)=25/9\\ \pi _\mathbf{6 }(0,1,2)=19/9\\ \pi _\mathbf{6 }(0,2,2)=13/9. \end{aligned}$$

implying that \((0,0,2)\) is the unique best response for voter 6.

Voter 7

The most probable pivot outcome faced by voter 7 is the event (6,4,4). Due to her utility profile, voter 7 strictly prefers to use an undominated strategy and is indifferent among all of them: that is \( (0,0,2),(0,1,2),(0,2,2)\). Hence, the mixed strategy \(g_{7}\) is a best response.

The second step of the proof consists in showing that \(g\) is a perfect equilibrium. To do so, consider the following completely mixed strategy combination \(g^{\varepsilon }\), where \(\eta _{i}\) denotes the mixed strategy of voter \(i\) which assigns equal probability to all his pure strategies.

$$\begin{aligned} i&= \mathbf 1 ,\mathbf 2 ,\mathbf 3 \, \, g_{i}^{\varepsilon }=(1-27\varepsilon ^{2})(2,0,0)+27\varepsilon ^{2}\eta _{i} \\ i&= \mathbf 4 ,\mathbf 5 \, \, g_{i}^{\varepsilon }=(1-27\varepsilon ^{2})(0,2,1)+27\varepsilon ^{2}\eta _{i} \\ i&= \mathbf 6 \, \, g_{i}^{\varepsilon }=(1-\varepsilon -27\varepsilon ^{2})(0,0,2)+ \varepsilon (0,1,2)+27\varepsilon ^{2}\eta _{i}, \\ i&= \mathbf 7 \, \, g_{i}^{\varepsilon }=g_{7}+27\varepsilon ^{2}\eta _{i}. \end{aligned}$$

It is easy to see that, for \(\varepsilon \) sufficiently close to zero, this is an \(\varepsilon \)-perfect equilibrium. Suppose all voters other than \(i\) choose the strategies prescribed by \(g\). Since for \(\varepsilon \) going to zero, the probability of voter 6 to tremble towards \((0, 1, 2)\) is infinitely greater than the probability of any other mistake, it is enough to check that the limiting strategy is preferred to the other undominated strategy when either this mistake or no mistake at all occurs.

For voters 1 to 5, the relevant contingency is the one described by the limiting strategy \(g\). Indeed, as has been shown, their unique best response is the one depicted by \(g\) as when the trembles tends towards, they have a unique best response. For voter 6, the same argument applies.

Finally, one can deduce that for voter \(i=\mathbf 7 \) casting the mixed strategy ballot \(g_{7}\) is a best response, against \(g^{\varepsilon }\). Indeed, for voter 7, the relevant contingency are summarized by the vectors \((6,4,4)\) and \((6,5,4)\). Let us denote their probabilities by \( p((6,4,4)\,|\,f^{\varepsilon }_{-i})\) and \(p((6,6,4)\,|\,f^{\varepsilon }_{-i}) \). Since

$$\begin{aligned} \pi _\mathbf{7 }(0,0,2)&= \, 3p((6,4,4)\, |\,f^{\varepsilon }_{-i})+\,3p((6,5,4)\,|f^{\varepsilon }_{-i}) \\&= \,\pi _\mathbf{7 }(0,1,2),\pi _\mathbf{7 }(0,2,2). \end{aligned}$$

the mixed strategy \(g_{7}\) is a best reply to \(g^{\varepsilon }_{-i}\).

Hence, \(\{g^{\varepsilon }\}\) is a sequence of \(\varepsilon \)-perfect equilibria. Since \(g\) is the limit of \(g^{\varepsilon }\), it is a perfect equilibrium in which voters’ best responses are not extremal.

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Núñez, M., Laslier, J.F. Preference intensity representation: strategic overstating in large elections. Soc Choice Welf 42, 313–340 (2014). https://doi.org/10.1007/s00355-013-0728-0

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