Strategy-proofness of the randomized Condorcet voting system

Abstract

In this paper, we study the strategy-proofness properties of the randomized Condorcet voting system (RCVS). Discovered at several occasions independently, the RCVS is arguably the natural extension of the Condorcet method to cases where a deterministic Condorcet winner does not exists. Indeed, it selects the always-existing and essentially unique Condorcet winner of lotteries over alternatives. Our main result is that, in a certain class of voting systems based on pairwise comparisons of alternatives, the RCVS is the only one to be Condorcet-proof. By Condorcet-proof, we mean that, when a Condorcet winner exists, it must be selected and no voter has incentives to misreport his preferences. We also prove two theorems about group-strategy-proofness. On one hand, we prove that there is no group-strategy-proof voting system that always selects existing Condorcet winners. On the other hand, we prove that, when preferences have a one-dimensional structure, the RCVS is group-strategy-proof.

Appendices

Appendix A: Proof of Theorem 3

First, we notice that $\varDelta ({\mathcal {O}})$ can be regarded as the simplex of ${\mathbb {R}}^{{\mathcal {O}}}$, and that the map ${\mathcal {R}} : \varDelta ({\mathcal {O}}) \rightarrow {\mathbb {R}}^{X \times X}$ can be uniquely extended to a linear map ${\mathcal {R}} : {\mathbb {R}}^{\mathcal O} \rightarrow {\mathbb {R}}^{X \times X}$. This linear map has the two following properties.

Lemma 2

The image of ${\mathcal {R}}$ coincides with the space of skew-symmetric matrices.¹¹

Proof

It is straightforward to see that the image of ${\mathcal {R}}$ is included in the space of skew-symmetric matrices. Reciprocally, to show the equality, we need only show that any element of the canonical basis of skew-symmetric matrices is in the image of $\mathsf {referendum}$. Such an element $R \in {\mathbb {R}}^{X \times X}$ is of the form $R_{xy} = - R_{yx} = 1$ for some two different alternatives $x,y \in X$, and $R_{vw} = 0$ in any other case. Denote $v_1, \ldots , v_n$ the other alternatives. We define ${{\tilde{\theta }}} \in {\mathbb {R}}^{{\mathcal {O}}}$ by ${{\tilde{\theta }}} = (\theta _1 + \theta _2)/2$, where

$$\begin{aligned} \theta _1 : x \succ y \succ v_1 \succ \cdots \succ v_n \quad \text {and} \quad \theta _2 : v_n \succ \cdots \succ v_1 \succ x \succ y. \end{aligned}$$

We then have ${\mathcal {R}}_{vw} ({{\tilde{\theta }}}) = 0$ if $(v,w) \notin \{ (x,y), (y,x) \}$ and ${\mathcal {R}}_{xy} ({{\tilde{\theta }}}) = - {\mathcal {R}}_{yx} ({{\tilde{\theta }}}) = 1$, which proves that $R = {\mathcal {R}} ({{\tilde{\theta }}})$. $\square$

Lemma 3

${\mathcal {R}}(\varDelta ({\mathcal {O}}))$ contains all skew-symmetric matrices of $\ell _1$-norm at most 2.

Proof

In the proof of Lemma 2, we showed that each element of the canonical basis is the image ${\mathcal {R}} (\tilde{\theta })$ of some preferences ${{\tilde{\theta }}} \in \varDelta ({\mathcal {O}})$ of the people. Yet, such elements of the canonical basis are of $\ell _1$-norm 2. Since ${\mathcal {R}}$ is linear and $\varDelta (\mathcal O)$ convex, ${\mathcal {R}} (\varDelta ({\mathcal {O}}))$ therefore contains all convex combinations of the elements of the canonical basis of skew-symmetric matrices. Plus, since, for any two distinct matrices of the canonical basis, the norm of the sum is the sum of the norms, all skew-symmetric matrices of $\ell _1$-norm 2 are images by ${\mathcal {R}}$ of some preferences of $\varDelta ({\mathcal {O}})$. Plus, the uniform distribution ${{\tilde{u}}} \in \varDelta ({\mathcal {O}})$ on all ballots is trivially in the kernel of ${\mathcal {R}}$. A convex combination involving ${{\tilde{u}}}$ then enables to obtain any skew-symmetric matrix of $\ell _1$-norm at most 2. $\square$

We still need another lemma before proving the theorem.

Lemma 4

Let the preferences ${{\tilde{\theta }}} \in \varDelta ({\mathcal {O}})$, the ballots ${{\tilde{a}}} \in \varDelta ({\mathcal {O}})$ and a subset ${\mathcal {C}} \subset {\mathcal {O}}$ of manipulators. Then, there exists a strategy $s \in S$ such that $s({{\tilde{\theta }}}) = {{\tilde{a}}}$ and $\mathsf {Manipulator}(s) = {\mathcal {C}}$ if and only if $\mathbb P_{{{\tilde{\theta }}}} [{\mathcal {D}}] \le {\mathbb {P}}_{{{\tilde{a}}}} [\mathcal D]$ for all subsets ${\mathcal {D}} \subset {\mathcal {O}} - {\mathcal {C}}$. In particular, if X is finite, this condition amounts to $\mathbb P_{{{\tilde{\theta }}}} ({{\tilde{\theta }}} = a) \le {\mathbb {P}}_{{{\tilde{a}}}} ({{\tilde{a}}} = a)$ for all $a \notin {\mathcal {C}}$.

Proof

First notice that ${{\tilde{\theta }}}$ could have produced ${{\tilde{a}}}$ with a set ${\mathcal {C}}$ of manipulators if and only if there exists a strategy s such that $\mathsf {Manipulator}(s) = {\mathcal {C}}$ and $s({{\tilde{\theta }}}) = {{\tilde{a}}}$. Consider that we indeed have $s({{\tilde{\theta }}}) = {{\tilde{a}}}$, and let us prove the direct implication of the lemma. Let ${\mathcal {D}} \subset {\mathcal {O}} - {\mathcal {C}}$. Then,

$$\begin{aligned} {\mathbb {P}}_{{{\tilde{a}}}} [ {\mathcal {D}} ]= & {} {\mathbb {E}}_{{{\tilde{\theta }}}} \left[ {\mathbb {P}}_{s({{\tilde{\theta }}})} [ {\mathcal {D}}] \right] = \mathbb E_{{{\tilde{\theta }}}} \left[ {\mathbb {P}}_{s({{\tilde{\theta }}})} [ {\mathcal {D}} ] \Big | {{\tilde{\theta }}} \in {\mathcal {D}} \right] {\mathbb {P}}_{{{\tilde{\theta }}}} [ {\mathcal {D}} ]\\&+\, {\mathbb {E}}_{{{\tilde{\theta }}}} \left[ \mathbb P_{s({{\tilde{\theta }}})} [ {\mathcal {D}}] \Big | {{\tilde{\theta }}} \notin {\mathcal {D}} \right] {\mathbb {P}}_{{{\tilde{\theta }}}} [ {\mathcal {O}} - \mathcal D]. \end{aligned}$$

But since ${\mathcal {D}} \cap {\mathcal {C}} = \emptyset$, for all $\theta \in {\mathcal {D}}$, we have $\theta \notin {\mathcal {C}}$. Thus, $s(\theta ) = s^{truth}(\theta ) = \delta _\theta$. Thus, $\mathbb P_{s(\theta )} [ {\mathcal {D}} ] = 1$. Thus, the expression above simplifies to

$$\begin{aligned} {\mathbb {P}}_{{{\tilde{a}}}} [ {\mathcal {D}} ] = {\mathbb {P}}_{{{\tilde{\theta }}}} [ {\mathcal {D}} ] + {\mathbb {E}}_{{{\tilde{\theta }}}} \left[ {\mathbb {P}}_{s(\tilde{\theta })} [ {\mathcal {D}} ] \Big | {{\tilde{\theta }}} \notin {\mathcal {D}} \right] {\mathbb {P}}_{{{\tilde{\theta }}}} [ {\mathcal {O}} - {\mathcal {D}} ]. \end{aligned}$$

Therefore, ${\mathbb {P}}_{{{\tilde{a}}}} [ {\mathcal {D}} ] \ge \mathbb P_{{{\tilde{\theta }}}} [ {\mathcal {D}}]$, hence proving the direct implication.

Reciprocally, if ${\mathbb {P}}_{{{\tilde{\theta }}}} [ {\mathcal {C}} ] = 0$, then the inequality ${\mathbb {P}}_{{{\tilde{\theta }}}} [ {\mathcal {D}} ] \ge {\mathbb {P}}_{{{\tilde{a}}}} [ {\mathcal {D}} ]$ implies ${{\tilde{\theta }}} = \tilde{a}$. Thus, $s^{truth}({{\tilde{\theta }}}) = {{\tilde{a}}}$, which proves that ${{\tilde{\theta }}}$ could have produced ${{\tilde{a}}}$ with the set ${\mathcal {C}}$ of manipulators. Otherwise, ${\mathbb {P}}_{{{\tilde{\theta }}}} [ {\mathcal {C}} ] \ne 0$. We define $s : {\mathcal {C}} \rightarrow \varDelta ({\mathcal {O}})$ by ${\mathbb {P}}_{s(\theta )} [ {\mathcal {C}} - \{ \theta \} ] = 0$,

• 1. $\forall {\mathcal {E}} \subset {\mathcal {C}}, {\mathbb {P}}_{s(\theta )} [ {\mathcal {E}} ] = {\mathbb {P}}_{{{\tilde{a}}}} [ {\mathcal {E}} ] / {\mathbb {P}}_{{{\tilde{\theta }}}} [ {\mathcal {C}} ]$.

• 2. $\forall {\mathcal {D}} \subset {\mathcal {O}} - {\mathcal {C}}, {\mathbb {P}}_{s(\theta )} [ {\mathcal {D}} ] = \left( {\mathbb {P}}_{{{\tilde{a}}}} [ {\mathcal {D}} ] - {\mathbb {P}}_{{{\tilde{\theta }}}} [ {\mathcal {D}} ] \right) / {\mathbb {P}}_{{{\tilde{\theta }}}} [ {\mathcal {C}} ]$.

Assuming ${\mathbb {P}}_{{{\tilde{\theta }}}} [ {\mathcal {D}} ] \le \mathbb P_{{{\tilde{a}}}} [ {\mathcal {D}} ]$ for all supsets ${\mathcal {D}} \supset {\mathcal {O}} - {\mathcal {C}}$, the probabilities we have defined here are all non-negative. It is straightforward to see that the additivity of the probability is satisfied. Plus,

$$\begin{aligned} {\mathbb {P}}_{{{\tilde{\theta }}}} [ {\mathcal {O}} ] = {\mathbb {P}}_{{{\tilde{\theta }}}} [ {\mathcal {C}} ] + {\mathbb {P}}_{{{\tilde{\theta }}}} [ {\mathcal {O}} - \mathcal C) = \frac{{\mathbb {P}}_{{{\tilde{a}}}} [ {\mathcal {C}} ]}{{\mathbb {P}}_{\tilde{\theta }} [ {\mathcal {C}} ]} + \frac{{\mathbb {P}}_{{{\tilde{a}}}} [ {\mathcal {O}} -{\mathcal {C}}) - {\mathbb {P}}_{{{\tilde{\theta }}}} [ {\mathcal {O}} - {\mathcal {C}} ]}{{\mathbb {P}}_{{{\tilde{\theta }}}} [ {\mathcal {C}} ]} = 1. \end{aligned}$$

Therefore, $s(\theta )$ is a well-defined probability, and s a well-defined strategy of support ${\mathcal {C}}$. Plus, for ${\mathcal {D}} \subset {\mathcal {O}} - {\mathcal {C}}$,

$$\begin{aligned} {\mathbb {P}}_{s({{\tilde{\theta }}})} [ {\mathcal {D}} ]&= {\mathbb {P}}_{{{\tilde{\theta }}}} [ {\mathcal {D}} ] + {\mathbb {E}}_{{{\tilde{\theta }}}} \left[ {\mathbb {P}}_{s({{\tilde{\theta }}})} [ {\mathcal {D}} ] \Big | {{\tilde{\theta }}} \in {\mathcal {C}} \right] {\mathbb {P}}_{{{\tilde{\theta }}}} [ {\mathcal {C}} ]\\&\quad + {\mathbb {E}}_{{{\tilde{\theta }}}} \left[ {\mathbb {P}}_{s({{\tilde{\theta }}})} [ {\mathcal {D}} ] \Big | {{\tilde{\theta }}} \notin {\mathcal {C}} \cup {\mathcal {D}} \right] {\mathbb {P}}_{{{\tilde{\theta }}}} [ {\mathcal {O}} - {\mathcal {C}} \cup {\mathcal {D}} ] \\&= {\mathbb {P}}_{{{\tilde{\theta }}}} [ {\mathcal {D}} ] + \frac{\mathbb P_{{{\tilde{a}}}} [ {\mathcal {D}} ] - {\mathbb {P}}_{{{\tilde{\theta }}}} [ \mathcal D ]}{{\mathbb {P}}_{{{\tilde{\theta }}}} [ {\mathcal {C}} ]} {\mathbb {P}}_{\tilde{\theta }} [ {\mathcal {C}} ] = {\mathbb {P}}_{{{\tilde{a}}}} [ {\mathcal {D}} ], \end{aligned}$$

and, similarly for any ${\mathcal {E}} \subset {\mathcal {C}}$, we have

$$\begin{aligned} {\mathbb {P}}_{s({{\tilde{\theta }}})} [ {\mathcal {E}} ] = {\mathbb {E}}_{\tilde{\theta }} \left[ {\mathbb {P}}_{s({{\tilde{\theta }}})} [ {\mathcal {E}} ] \Big | {{\tilde{\theta }}} \in {\mathcal {C}} \right] {\mathbb {P}}_{{{\tilde{\theta }}}} [ {\mathcal {C}} ] = \frac{{\mathbb {P}}_{{{\tilde{a}}}} [ {\mathcal {E}} ]}{\mathbb P_{{{\tilde{\theta }}}} [ {\mathcal {C}} ]} {\mathbb {P}}_{{{\tilde{\theta }}}} [ {\mathcal {C}} ] = {\mathbb {P}}_{{{\tilde{a}}}} [ {\mathcal {E}}]. \end{aligned}$$

These two equalities prove that $s({{\tilde{\theta }}}) = {{\tilde{a}}}$, which is what we had to prove. $\square$

We can now prove Theorem 3.

Proof

(Theorem 3) By contradiction, assume that ${\mathscr {V}}$ disagrees with the RCVS ${\mathscr {C}}$ for some referendum matrix R of $\ell _1$-norm less than 2. Let us consider ballots $a_{vw}^+, a_{vw}^- \in {\mathcal {O}}$ defined by

$$\begin{aligned} a_{vw}^+ : v \succ w \succ x_1 \succ \cdots \succ x_n \quad \text {and} \quad a^-_{vw} : x_n \succ \cdots \succ x_1 \succ v \succ w. \end{aligned}$$

Like in the proof of Lemma 2, the choices of $x_1, \ldots , x_n$ are irrelevant. What matters to us is the fact that ${\mathcal {R}}(a_{vw}^+ + a^-_{vw})$ is a skew-symmetric matrix of the canonical basis. In particular, the referendum matrix R that the pairwise voting system is based on can be obtained from the ballots ${{\tilde{a}}}$ defined by

$$\begin{aligned} {{\tilde{a}}} = \frac{1}{2} \sum \limits _{v,w : R_{vw}> 0} R_{vw} a_{vw}^+ + \frac{1}{2} \sum \limits _{v,w : R_{vw} > 0} R_{vw} a_{vw}^- + \delta {{\tilde{u}}}, \end{aligned}$$

where $\delta = 1- \frac{1}{2} ||R||_1$. By assumption on R, we have $\delta > 0$.

Since, by assumption, $R_{xy} \ne 0$ for all $x \ne y$, the ballots ${{\tilde{a}}}$ yields a unique randomized Condorcet winner. Therefore, the fact that ${\mathscr {V}}$ disagrees with ${\mathscr {C}}$ on R implies that ${\mathscr {V}}({{\tilde{a}}})$ is not the randomized Condorcet winner of ${{\tilde{a}}}$. In other words, there must be some alternative $x \in X$ that the majority $M_{{{\tilde{a}}}}$ strictly prefers to ${\mathcal {V}}({{\tilde{a}}})$. Denoting $Z = \{ z_1, \ldots , z_n \}$ and $Y = \{ y_1, \ldots , y_m \}$ the sets of alternatives that the majority $M_{{{\tilde{a}}}}$ respectively prefers x to ($M_{\tilde{a}} : x \gg z$) and prefers to x ($M_{{{\tilde{a}}}} : y \gg x$), this means that ${\mathbb {P}}_{{\mathscr {V}} ({{\tilde{a}}})} \left[ Z \right] > {\mathbb {P}}_{{\mathscr {V}} ({{\tilde{a}}})} \left[ Y \right]$. Evidently, by definitions of sets Y and Z, we have $R_{yx} > 0$ and $R_{xz} > 0$ for all $y \in Y$ and $z \in Z$.

We then define the preference $\theta \in {\mathcal {O}}$ of manipulators by

$$\begin{aligned} \theta : z_1 \succ \cdots \succ z_n \succ x \succ y_1 \succ \cdots \succ y_m. \end{aligned}$$

Notice that we have $\theta : {\mathscr {V}} ({{\tilde{a}}}) \succ x$. We can now define the preferences ${{\tilde{\theta }}} \in \varDelta ({\mathcal {O}})$ of the people by

$$\begin{aligned} {{\tilde{\theta }}}= & {} \frac{1}{2} \sum \limits _{\begin{array}{c} M_{{{\tilde{a}}}}: v \gg w \\ (v, w) \notin Y \times \{ x \} \end{array}} R_{vw} (a^+_{vw} + a^-_{vw}) + \frac{1}{2} \sum \limits _{y \in Y} R_{yx}a^+_{yx} \\&+\left( \epsilon + \frac{1}{2} \sum \limits _{y \in Y} R_{yx} \right) \theta + (\delta -\epsilon ) {{\tilde{u}}}, \end{aligned}$$

with $0< \epsilon < \delta$. Importantly, any ballot $a \in {\mathcal {O}}$ except $\theta$ is more frequent in ${{\tilde{a}}}$ than in ${{\tilde{\theta }}}$. Thus, by Lemma 4, manipulators $\theta$ can have produced the ballots ${{\tilde{a}}}$. Moreover, we have:

$$\begin{aligned} \begin{array}{rllclccl} \forall z \in Z, \quad {\mathcal {R}}_{xz} ({{\tilde{\theta }}}) &{}= &{} R_{xz} + \displaystyle \frac{1}{2} \sum \limits _{y \in Y} R_{yx} - \epsilon \displaystyle - \frac{1}{2} \sum \limits _{y \in Y} R_{yx} = R_{xz} &{}- \epsilon , \\ \forall y \in Y, \quad {\mathcal {R}}_{xy} ({{\tilde{\theta }}}) &{}= &{} \displaystyle - \frac{1}{2} \sum \limits _{y \in Y} R_{yx} + \epsilon + \displaystyle \frac{1}{2} \sum \limits _{y \in Y} R_{yx} = +\epsilon . \end{array} \end{aligned}$$

Recall that $R_{xz} > 0$ for all $z \in Z$, hence we can choose $\epsilon$ smaller than all $R_{xz}$. By doing so, we guarantee that ${\mathcal {R}}_{xz} ({{\tilde{\theta }}})$ and ${\mathcal {R}}_{xy} (\tilde{\theta })$ are positive for all $z \in Z$ and $y \in Y$. Thus, x is a Condorcet winner for ${{\tilde{\theta }}}$. Yet, by creating ballots ${{\tilde{a}}}$, manipulators $\theta$ have obtained strictly better, as we have seen that $\theta : {\mathscr {V}}({{\tilde{a}}}) \succ x$. This shows that ${\mathscr {V}}$ is not Condorcet-proof. This proves that a Condorcet-proof pairwise voting system must agree with the RCVS whenever $||{\mathcal {R}}({{\tilde{\theta }}})||_1 < 2$.

The uniqueness of Condorcet-proof tournament-based voting system is then immediately derived by considering any skew-symmetric matrix $R^0$, and by dividing each entry by the $\ell _1$-norm of the matrix, hence yielding a skew-symmetric matrix R of norm 1 to which the previous part applies. $\square$

Appendix B: Proof of Theorem 5

We first prove the following lemma about the structure of the set of manipulators when preferences are one-dimensional.

Lemma 5

When preferences are single-peaked or single-crossing with a unique median voter, manipulators must either be all strictly on the left or all strictly on the right of the median voter.

Proof

Let ${\mathscr {V}}$ a voting system. Denote x the Condorcet winner of the single-peaked preferences ${{\tilde{\theta }}} \in \varDelta (\mathcal O)$. Denote Z and Y the sets of alternatives that are respectively on the left and on the right of x. Now, consider a strategy s. We denote $s({{\tilde{\theta }}}) = {{\tilde{a}}}$. The strict incentive to conspire means that

$$\begin{aligned} \forall \theta \in \mathsf {Manipulator}(s), \quad \theta : \mathscr {V} ({{\tilde{a}}}) \succ x. \end{aligned}$$

If preferences are single-peaked or single-crossing and $\theta \in \mathsf {Manipulator}(s)$ is on the right of the median voter, we know that $\theta : x \succ z$ for all $z \in Z$. Indeed, if preferences are single-peaked, this is due to $\theta$’s ideal point being on the right of x. And if preferences are single-crossing, this is because x cannot be switched with a left alternative as we look preferences on the right of the median voter. Since the median voter ranked x better than any $z \in Z$, all voters on its right must do so too.

Thus, $\{ z \in X \; | \; \theta : x \succ z \} \supset Z$, and, as a result, $\{ y \in X \;\Vert \; \theta : y \succ x \} \subset Y$. Therefore,

$$\begin{aligned} 0 < {\mathbb {P}}_{v \sim {\mathscr {V}} ({{\tilde{a}}})} \left[ \theta : v \succ x \right] - {\mathbb {P}}_{v \sim {\mathscr {V}} ({{\tilde{a}}})} \left[ \theta : v \prec x \right] \le {\mathbb {P}}_{{\mathscr {V}} ({{\tilde{a}}})} [Y] - {\mathbb {P}}_{{\mathscr {V}} ({{\tilde{a}}})} [Z]. \end{aligned}$$

Therefore, we have ${\mathbb {P}}_{{\mathscr {V}} ({{\tilde{a}}})} [Y] > \mathbb P_{{\mathscr {V}} ({{\tilde{a}}})} [Z]$. But if $\theta \in \mathsf {Manipulator}(s)$ is on the left of the median voter, we must have the opposite inequality. Both cases cannot occur simultaneously, which proves that all manipulators must be on the same side of the left-right spectrum with regards to the median voter. This proves the lemma. $\square$

We can now prove Theorem 5. The proof slightly differs depending on the assumption of one-dimensionality of preferences that is considered. For clarity, we write the proofs of the two cases in separate blocks.

Proof

(Theorem 5 for single-peakedness preferences) Let ${\mathscr {C}}$ denote the randomized Condorcet voting system. We use the same notations $x, {{\tilde{\theta }}}, Y, Z, s$ and ${{\tilde{a}}}$ as in the proof of Lemma 5. Without loss of generality, we can assume that manipulators are all strictly on the right of the median voter.

Let $z \in Z$. As we have seen in the previous proof, we have $\theta : x \succ z$ for all $\theta \in \mathsf {Manipulator}(s)$. Since manipulators agree with $M_{{{\tilde{\theta }}}} : x \gg z$ for $z \in Z$, according to Lemma 1, they cannot invert these pairwise comparisons. Therefore, $M_{{{\tilde{a}}}} : x \gg z$. Yet, for manipulators to gain by conspiring, ${\mathscr {C}} ({{\tilde{a}}})$ must differ from x, which means that there must be some $y \in Y$ such that $M_{{{\tilde{a}}}} : y \gg x$. Let $y^*$ the most leftist alternative that the majority of ballots prefers to x, i.e.

$$\begin{aligned} y^* = \min \{ y \in Y \; | \; M_{{{\tilde{a}}}} : y \gg x \}, \end{aligned}$$

where the minimum corresponds to the order relation “<” on alternatives. We denote $Y_- \triangleq \{ y_- \in Y \;|\; x< y_- < y^* \}$ and $Y_+ \triangleq Y - Y_- = \{ y_+ \in Y \;|\; y^* \le y_+ \}$.

Since x is Condorcet winner of ${{\tilde{\theta }}}$, we know that $M_{{{\tilde{\theta }}}} : x \gg y^*$. Thus, manipulators must have inverted the majority preference of x over $y^*$. Since, according to Lemma 1, manipulators can only invert arcs they agree with, this means that there must be a manipulator $\theta \in \mathsf {Manipulator}(s)$ who agrees with $M_{{{\tilde{\theta }}}} : x \gg y^*$. This manipulator thus thinks $\theta : x \succ y^*$. We will show that assuming that he had incentive to conspire leads to a contradiction.

On one hand, by definition of $y^*$, we have $M_{{{\tilde{a}}}} : x \gg y$ for all $y \in Y_-$, i.e.

$$\begin{aligned} Y_- \subset \big \{ w \in X \; | \; M_{{{\tilde{a}}}} : x \gg w \big \} \quad \text {and} \quad \{ w \in X \; | \; M_{{{\tilde{a}}}} : w \gg x \} \subset Z \cup Y_+. \end{aligned}$$

Combining this with the property $M_{{{\tilde{a}}}} : {\mathscr {C}} (\tilde{a}) \; {\underline{\gg }} \; x$ satisfied by the randomized Condorcet voting system yields

$$\begin{aligned} {\mathbb {P}}_{{\mathscr {C}} ({{\tilde{a}}})} [ Y_-]&\le {\mathbb {P}}_{w \sim {\mathscr {C}} ({{\tilde{a}}})} \left[ M_{{{\tilde{a}}}} : x \gg w \right] \nonumber \\&\le {\mathbb {P}}_{w \sim {\mathscr {C}} ({{\tilde{a}}})} \left[ M_{{{\tilde{a}}}} : w \gg x \right] \nonumber \\&\le {\mathbb {P}}_{{\mathscr {C}} ({{\tilde{a}}})} \left[ Z \cup Y_+ \right] . \end{aligned}$$

(6)

On the other hand, strict incentives to conspire for $\theta$ imply that

$$\begin{aligned} {\mathbb {P}}_{w \sim {\mathscr {C}} ({{\tilde{a}}})} \left[ \theta : w \succ x \right] > {\mathbb {P}}_{w \sim {\mathscr {C}} ({{\tilde{a}}})} \left[ \theta : x \succ w \right] . \end{aligned}$$

Since $\theta : x \succ y^*$, we know that the ideal point of $\theta$ is necessarily on the left of $y^*$. As a result, for $y \in Y_+$, we have $\theta : x \succ y^* \succ y$. Therefore,

$$\begin{aligned} \big \{ w \in X \; | \; \theta : x \succ w \big \} \supset Z \cup Y_+ \quad \text {and} \quad Y_- \supset \big \{ w \in X \; | \; \theta : w \succ x \big \}, \end{aligned}$$

which leads to ${\mathbb {P}}_{{\mathscr {C}} ({{\tilde{a}}})} [ Y_- ] > \mathbb P_{{\mathscr {C}} ({{\tilde{a}}})} \left[ Z \cup Y_+ \right]$. This contradicts equation (6), and proves the theorem for single-peaked preferences. $\square$

Proof

(Theorem 5 for single-crossing preferences) We reuse the same notations $x, {{\tilde{\theta }}}, Y, Z, s$ and $\tilde{a}$ as in the proof of Lemma 5. Let $\theta _m$ the median voter. Lemma 5 allows us to assume without loss of generality that the manipulators are all on the right of $\theta _m$, i.e. $\theta > \theta _m$ for all $\theta \in \mathsf {Manipulator}(s)$. Then, we have, once again, $M_{{{\tilde{a}}}} : x \gg z$ for all $z \in Z$.

Let now $\theta = \min \; \mathsf {Manipulator}(s)$ the most leftist manipulator. Denote $Y^+$ and $Y^-$ defined by

$$\begin{aligned} Y^+ = \{ y \in Y \; | \; \theta : y \succ x \} \quad \text {and} \quad Y^- = \{ y \in Y \; | \; \theta : x \succ y \}. \end{aligned}$$

Contrary to the proof for single-peakedness, $Y^+$ now corresponds to the alternatives some manipulators prefer to x, as the sign $``+''$ now refers to $\theta$’s preference rather than the left-right line of alternatives.

Let $y^+ \in Y^+$. Since for any $\theta ' \in \mathsf {Manipulator}(s)$, we have $\theta < \theta '$, single-crossing implies that $\theta ' : y^+ \succ x$. Therefore, manipulators all disagree with the majority preference $M_{\tilde{\theta }} : x \gg y^+$, and hence, by Lemma 5, cannot invert it. Therefore, $M_{{{\tilde{a}}}} : x \gg y^+$. Since this holds for all $y^+ \in Y^+$, we have

$$\begin{aligned} Y^+ \subset \{ w \in X \; | \; M_{{{\tilde{a}}}} : x \gg w \} \quad \text {and} \quad \{ w \in X \; | \; M_{{{\tilde{a}}}} : w \gg x \} \subset Z \cup Y^-. \end{aligned}$$

Yet, the fundamental property of the RCVS applied to x then implies

$$\begin{aligned} {\mathbb {P}}_{{\mathscr {C}} ({{\tilde{a}}})} [ Y^+ ]&\le {\mathbb {P}}_{w \sim {\mathscr {C}} ({{\tilde{a}}})} \left[ M_{{{\tilde{a}}}} : x \gg w \right] \\&\le {\mathbb {P}}_{w \sim {\mathscr {C}} ({{\tilde{a}}})} \left[ M_{{{\tilde{a}}}} : w \gg x \right] \\&\le {\mathbb {P}}_{{\mathscr {C}} ({{\tilde{a}}})} \left[ Z \cup Y^- \right] . \end{aligned}$$

But this contradicts the strict incentives for $\theta$ to conspire, i.e. ${\mathbb {P}}_{{\mathscr {C}} ({{\tilde{a}}})} [ Y^+ ] > \mathbb P_{{\mathscr {C}} ({{\tilde{a}}})} \left[ Z \cup Y^- \right]$. Thus, we reach the same conclusion for single-crossing preferences as we did for single-peaked ones. $\square$