Stochastic same-sidedness in the random voting model

Abstract

We study the implications of stochastic same-sidedness (SSS) axiom in the random voting model. At a given preference profile if one agent changes her preference ordering to an adjacent one by swapping two consecutively ranked alternatives, then SSS imposes two restrictions on the lottery selected by a voting rule before and after the swap. First, the sum of probabilities of the alternatives which are ranked strictly higher than the swapped pair should remain the same. Second, the sum of probabilities assigned to the swapped pair should also remain the same. We show that every random social choice function (RSCF) that satisfies efficiency and SSS is a random dictatorship provided that there are two voters or three alternatives. For the case of more than two voters and atleast four alternatives, every RSCF that satisfies efficiency, tops-onlyness and SSS is a random dictatorship.

Appendix

1.1 The Proof of Proposition 1

Proof

First we show that $\text {Conv}[{\mathbb {F}}^{BR}]\subset \text {Conv}[{\mathbb {F}}^{SS}]$. By Muto and Sato (2017), ${\mathbb {F}}^{BR}\subseteq {\mathbb {F}}^{SS}$. Now we provide an example of DSCF that belongs to ${\mathbb {F}}^{SS}$ but not in ${\mathbb {F}}^{BR}$. Let $|A|\ge 6$ and $N=\{1,2\}$. Fix $a,b\in A$, $a\ne b$. We define $f^{a,b}: {\mathbb {P}}^2\longrightarrow A$ as follows: For any $P\in {\mathbb {P}}^2$,

$$\begin{aligned} f^{a,b}(P) = \left\{ \begin{array}{l l} r_3(P_1) &{} \quad \text {if } r_{m-1}(P_1)=a \text { and } r_m(P_1)=b,\\ r_1(P_1) &{} \quad \text {otherwise}. \\ \end{array} \right. \end{aligned}$$

It can be verified that $f^{a,b}$ satisfies same-sidedness but not BR. Therefore, we have that ${\mathbb {F}}^{BR}\subset {\mathbb {F}}^{SS}$. Hence, $\text {Conv}[{\mathbb {F}}^{BR}]\subset \text {Conv}[{\mathbb {F}}^{SS}]$.

Next we show that $\text {Conv}[{\mathbb {F}}^{SS}] \subseteq \Phi ^{SSS}$. Pick any $\varphi \in \text {Conv}[{\mathbb {F}}^{SS}]$. Let $\varphi =\sum _{j=1}^t \lambda _j f_j$ where $f_j\in {\mathbb {F}}^{SS}$ for all $j\in \{1,\ldots ,t\}$ and $\lambda _j\ge 0$, $\sum _{j=1}^{t}\lambda _j=1$.¹³ Consider an $i \in N$, $P = (P_i, P_{-i})\in {\mathbb {P}}^n$ and $P^\prime _i \in {\mathbb {P}}$ such that $P_i$ and $P^\prime _i$ are adjacent. Let $A(P_i,P^\prime _i) = \left\{ x,y\right\}$. Without loss of generality, assume that $r(P_i,x)=r(P^\prime _i,y)=k$ and $r(P_i,y)=r(P^\prime _i,x)=k+1$. First, we will show that

$$\begin{aligned} \sum \limits _{a\in U(P_i,\{x,y\})}\varphi _a(P)=\sum \limits _{a\in U(P_i,\{x,y\})}\varphi _a(P'_i,P_{-i}). \end{aligned}$$

Note that

$$\begin{aligned} \sum \limits _{a\in U(P_i,\{x,y\})}\varphi _a(P)=\sum _{\{j:j\in \{1,\ldots ,t\} \textit{ and }f_j(P)\in U(P_i,\{x,y\})\}}\lambda _j \end{aligned}$$

and

$$\begin{aligned} \sum \limits _{a\in U(P_i,\{x,y\})}\varphi _a(P'_i,P_{-i})=\sum _{\{j:j\in \{1,\ldots ,t\} \textit{ and }f_j(P'_i,P_{-i})\in U(P_i,\{x,y\})\}}\lambda _j. \end{aligned}$$

Note that $f_j$ satisfies same-sidedness. Therefore, $f_j(P)\in U(P_i,\{x,y\})$ if and only if $f_j(P'_i,P_{-i})\in U(P_i,\{x,y\})$ for any $j\in \{1,\ldots ,t\}$. Hence,

$$\begin{aligned} \sum \limits _{a\in U(P_i,\{x,y\})}\varphi _a(P)=\sum \limits _{a\in U(P_i,\{x,y\})}\varphi _a(P'_i,P_{-i}). \end{aligned}$$

Again, same-sidedness implies that $f_j(P)\in \{x,y\}$ if and only if $f_j(P'_i,P_{-i})\in \{x,y\}$ for any $j\in \{1,\ldots ,t\}$. Therefore, we can establish that

$$\begin{aligned} \sum \limits _{a\in \{x,y\}}\varphi _a(P)=\sum \limits _{a\in \{x,y\}}\varphi _a(P'_i,P_{-i}). \end{aligned}$$

Hence, $\varphi$ satisfies SSS. Therefore, we have that $\text {Conv}[{\mathbb {F}}^{SS}] \subseteq \Phi ^{SSS}$.

Finally, we show that the rule in Example 2 belongs to $\Phi ^{SSS}$ but not in $\text {Conv}[{\mathbb {F}}^{SS}]$. Note that the rule satisfies SSS and efficiency but not random dictatorship. Since it satisfies SSS, it belongs to $\Phi ^{SSS}$. A simple variant of the Birkhoff-von Neumann Theorem on doubly stochastic matrices establishes that any RSCF can be represented as a convex combination some DSCFs.¹⁴So, the rule in Example 2 can be represented as a convex combination some DSCFs and let ${\mathbb {F}}^*$ denote the collection of these DSCFs. Since the rule in Example 2 satisfies efficiency, for all $f\in {\mathbb {F}}^*$, f satisfies efficiency. We argue that there exists $f\in {\mathbb {F}}^*$ such that f does not satisfy same-sidedness. Suppose not, i.e., for all $f\in {\mathbb {F}}^*$, f satisfies same-sidedness. Then by Muto and Sato (2017), for all $f\in {\mathbb {F}}^*$, f satisfies dictatorship. This implies that the rule in Example 2 must satisfy random dictatorship - a contradiction. Therefore, there exists $f\in {\mathbb {F}}^*$ such that f does not satisfy same-sidedness and the rule in Example 2 does not belong to $\text {Conv}[{\mathbb {F}}^{SS}]$. Hence, $\text {Conv}[{\mathbb {F}}^{SS}] \subset \Phi ^{SSS}$. $\square$

1.2 The Proof of Proposition 2

Proof

Consider an $i \in N$, $P = (P_i, P_{-i})\in {\mathbb {P}}^n$ and $P^\prime _i \in {\mathbb {P}}$ such that $P_i$ and $P^\prime _i$ are adjacent. Let $A(P_i,P^\prime _i) = \left\{ x,y\right\}$. So applying sd-strategy-proofness for the deviation from $P_i$ to $P^\prime _i$ and from $P^\prime _i$ to $P_i$, we have

$$\begin{aligned} \sum \limits _{a\in U(P_i, A(P_i,P'_i))}\varphi _a(P_i,P_{-i})=\sum \limits _{a\in U(P'_i, A(P_i,P'_i))}\varphi _a(P'_i,P_{-i}). \end{aligned}$$

Again, applying sd-strategy-proofness for the deviation from $P_i$ to $P^\prime _i$ and from $P^\prime _i$ to $P_i$ we have

$$\begin{aligned} \sum \limits _{a\in U(P_i, A(P_i,P'_i))\cup A(P_i,P'_i)}\varphi _a(P_i,P_{-i})=\sum \limits _{a\in U(P'_i, A(P_i,P'_i))\cup A(P_i,P'_i)}\varphi _a(P'_i,P_{-i}). \end{aligned}$$

Now, it follows that

$$\begin{aligned} \sum _{a\in U(P_i, A(P_i,P'_i))\cup A(P_i,P'_i)}\varphi _a(P_i,P_{-i})= & {} \left\{ \sum _{a\in U(P_i, A(P_i,P'_i))}\varphi _a(P_i,P_{-i})\right\} \\{} & {} + \varphi _x(P_i,P_{-i}) + \varphi _y(P_i,P_{-i}), \text { and} \\ \sum _{a\in U(P'_i, A(P_i,P'_i))\cup A(P_i,P'_i)}\varphi _a(P^\prime _i,P_{-i})= & {} \left\{ \sum _{a\in U(P'_i, A(P_i,P'_i))}\varphi _a(P^\prime _i,P_{-i})\right\} \\{} & {} + \varphi _x(P^\prime _i,P_{-i}) + \varphi _y(P^\prime _i,P_{-i}). \end{aligned}$$

Combining we have

$$\begin{aligned} \varphi _x(P_i,P_{-i}) + \varphi _y(P_i,P_{-i})= & {} \varphi _x(P^\prime _i,P_{-i}) + \varphi _y(P^\prime _i,P_{-i}). \end{aligned}$$

Hence, $\varphi$ satisfies SSS. $\square$

1.3 The Proof of Theorem 1

Proof

It is well known that random dictatorship satisfies efficiency and sd-strategy-proofness. By Lemma 1, random dictatorship satisfies SSS. Therefore we show the only if part. Let $\varphi : {\mathbb {P}}^2\longrightarrow \Delta (A)$ satisfies efficiency and SSS. We will show that $\varphi$ is a random dictatorship.

We complete the proof by showing following two lemmas.

Lemma 2

For any $P,P'\in {\mathbb {P}}^2$ and $a,b\in A$ such that $r_1(P_1)=r_1(P'_1)=a$ and $r_1(P_2)= r_1(P'_2)=b$, we have

• 1. $\varphi _a(P)+\varphi _b(P)=\varphi _a(P')+\varphi _b(P')=1$, and

• 2. $\varphi _a(P)=\varphi _a(P')$ and $\varphi _b(P)=\varphi _b(P')$.

Proof

If $a=b$, then we are done by efficiency.

Consider any two alternatives $a,b\in A$ and a preference profile ${\bar{P}}\in {\mathbb {P}}^2$ such that $r_1(\bar{P_1})=r_2(\bar{P_2})=a\ne b= r_1(\bar{P_2})=r_2(\bar{P_1})$. By efficiency, $\varphi _a({\bar{P}})+\varphi _b({\bar{P}})=1$. We assume that $\varphi _a({\bar{P}})=\beta _1$ and $\varphi _b({\bar{P}})=\beta _2$ where $0\le \beta _1,\beta _2\le 1$ and $\beta _1+\beta _2=1$. We will show that for any $P\in {\mathbb {P}}^2$ such that $r_1(P_1)=a\ne b= r_1(P_2)$, we have $\varphi _a(P)=\beta _1$ and $\varphi _b(P)=\beta _2$. We complete the proof by showing following claims.

Claim 1

For any $P\in {\mathbb {P}}^2$ such that $r_1(P_1)=a\ne b= r_1(P_2)= r_2(P_1)$ or $r_1(P_1)=r_2(P_2)=a\ne b= r_1(P_2)$, we have $\varphi _a(P)=\beta _1$ and $\varphi _b(P)=\beta _2$.

Proof

Case 1: $P\in {\mathbb {P}}^2$ be such that $r_1(P_1)=a\ne b= r_1(P_2)= r_2(P_1)$.

Let $P'_1$ be an ordering where $\bar{P_1}(A{\setminus } \{b\})=P'_1(A{\setminus } \{b\})$ and $r_m(P'_1)=b$. Let $\bar{P_1}=P^1,P^2,\ldots ,P^k=P'_1$ be a path from $\bar{P_1}$ to $P'_1$ where $P^j(A\setminus \{b\})=P^{j+1}(A\setminus \{b\})$ for all $j\in \{1,2,\ldots ,k-1\}$. By efficiency and SSS, we have $\varphi ({\bar{P}})=\varphi (P^2,\bar{P_2})=\varphi (P^3,\bar{P_2})=\ldots = \varphi (P'_1,\bar{P_2})$.

Let $P''_1$ be an ordering where $P_1(A{\setminus } \{b\})=P''_1(A{\setminus } \{b\})$ and $r_m(P''_1)=b$. Let $P'_1=P^1,P^2,\ldots ,P^l=P''_1$ be a path from $P'_1$ to $P''_1$ where $r_1(P^j)=a$ and $r_m(P^j)=b$ for all $j\in \{1,2,\ldots ,l\}$. Again, by efficiency and SSS, we have $\varphi (P'_1,\bar{P_2})=\varphi (P^2,\bar{P_2})=\varphi (P^3,\bar{P_2})=\ldots = \varphi (P''_1,\bar{P_2})$.

Let $P''_1=P^1,P^2,\ldots ,P^r=P_1$ be a path from $P''_1$ to $P_1$ where $P^j(A\setminus \{b\})=P^{j+1}(A\setminus \{b\})$ for all $j\in \{1,2,\ldots ,r-1\}$. Applying efficiency and SSS, we get $\varphi (P''_1,\bar{P_2})=\varphi (P^2,\bar{P_2})=\varphi (P^3,\bar{P_2})=\ldots = \varphi (P_1,\bar{P_2})$. Therefore, we have that $\varphi ({\bar{P}})=\varphi (P_1,\bar{P_2})$. We will conclude this case by showing that $\varphi (P_1,\bar{P_2})=\varphi (P_1,P_2)$.

Let $P'_2$ be an ordering where $\bar{P_2}(A{\setminus } \{a\})=P'_2(A{\setminus } \{a\})$ and $r_m(P'_2)=a$. Let $\bar{P_2}=P^1,P^2,\ldots ,P^{k'}=P'_2$ be a path from $\bar{P_2}$ to $P'_2$ where $P^j(A\setminus \{a\})=P^{j+1}(A\setminus \{a\})$ for all $j\in \{1,2,\ldots ,k'-1\}$. By efficiency and SSS, we have $\varphi (P_1,\bar{P_2})=\varphi (P_1,P^2)=\varphi (P_1,P^3)=\ldots = \varphi (P_1,P'_2)$.

Let $P''_2$ be an ordering where $P_2(A{\setminus } \{a\})=P''_2(A{\setminus } \{a\})$ and $r_m(P''_2)=a$. Let $P'_2=P^1,P^2,\ldots ,P^{l'}=P''_2$ be a path from $P'_2$ to $P''_2$ where $r_1(P^j)=b$ and $r_m(P^j)=a$ for all $j\in \{1,2,\ldots ,l'\}$. Again, by efficiency and SSS, we have $\varphi (P_1,P'_2)=\varphi (P_1,P^2)=\varphi (P_1,P^3)=\ldots = \varphi (P_1,P''_2)$.

Let $P''_2=P^1,P^2,\ldots ,P^{r'}=P_2$ be a path from $P''_2$ to $P_2$ where $P^j(A\setminus \{a\})=P^{j+1}(A\setminus \{a\})$ for all $j\in \{1,2,\ldots ,r'-1\}$. Applying efficiency and SSS, we get $\varphi (P_1,P''_2)=\varphi (P_1,P^2)=\varphi (P_1,P^3)=\ldots = \varphi (P_1,P_2)$. Therefore, we have that $\varphi (P_1,\bar{P_2})=\varphi (P_1,P_2)$. The Fig. 1 provides a concise visual representation of all the necessary steps involved in the proof of Case 1 of Claim 1.

Case 2: $P\in {\mathbb {P}}^2$ be such that $r_1(P_1)= r_2(P_2)=a\ne b= r_1(P_2)$. Using same arguments as in Case 1, we can get $= \varphi ({\bar{P}}_1,{\bar{P}}_2)= \varphi (P_1,P_2)$. $\square$

Fig. 1

A graphical illustration of Case 1 of Claim 1

Claim 2

For any $P\in {\mathbb {P}}^2$ and $x\in A$ such that $r_1(P_1)=a\ne b= r_1(P_2)$ and $r_2(P_1)=r_2(P_2)=x$, we have $\varphi _a(P)=\beta _1$ and $\varphi _b(P)=\beta _2$.

Proof

Consider any $P\in {\mathbb {P}}^2$ and $x\in A$ such that $r_1(P_1)=a\ne b= r_1(P_2)$ and $r_2(P_1)=r_2(P_2)=x$. We show that $\varphi _a(P)= \beta _1$ (a symmetric argument holds for b).

Let $P'_2$ be an ordering where $P_2(A{\setminus } \{a\})=P'_2(A{\setminus } \{a\})$ and $r_3(P'_2)=a$. Let $P_2=P^1,P^2,\ldots ,P^{k}=P'_2$ be a path from $P_2$ to $P'_2$ where $P^j(A{\setminus } \{a\})=P^{j+1}(A{\setminus } \{a\})$ for all $j\in \{1,2,\ldots ,k-1\}$. By efficiency and SSS, $\varphi _a(P_1,P^j)=\varphi _a(P_1,P^{j+1})$ for all $j\in \{1,2,\ldots ,k-1\}$. Therefore we have $\varphi _a(P_1,P_2)=\varphi _a(P_1,P'_2)$.

Let $P'_1$ be an ordering where $P_1(A{\setminus } \{x\})=P'_1(A{\setminus } \{x\})$ and $r(P'_1,x)=r(P'_1,b)+1$. Let $P_1=P^1,P^2,\ldots ,P^l=P'_1$ be a path from $P_1$ to $P'_1$ where $P^j(A{\setminus } \{x\})=P^{j+1}(A{\setminus } \{x\})$ for all $j\in \{1,2,\ldots ,l-1\}$. By efficiency and SSS, $\varphi _a(P^j,P'_2)=\varphi _a(P^{j+1},P'_2)$ for all $j\in \{1,2,\ldots ,l-1\}$. Therefore we have $\varphi _a(P_1,P_2)=\varphi _a(P_1,P'_2)=\varphi _a(P'_1,P'_2)$.

Let $P''_2$ be an adjacent ordering to $P'_2$ such that $A(P'_2,P''_2)=\{x,a\}$. By efficiency and SSS, $\varphi _a(P'_1,P'_2)=\varphi _a(P'_1,P''_2)$. By Claim 1, $\varphi _a(P'_1,P''_2)= \beta _1$. This immediately implies with our previous equations that $\varphi _a(P_1,P_2)= \beta _1$. $\square$

Claim 3

For any $P\in {\mathbb {P}}^2$ such that $r_1(P_1)=a\ne b= r_1(P_2)$, we have $\varphi _a(P)=\beta _1$ and $\varphi _b(P)=\beta _2$.

Proof

Consider any profile $P\in {\mathbb {P}}^2$ such that $r_1(P_1)=a\ne b= r_1(P_2)$. First we show that $\varphi _a(P)+\varphi _b(P)=1$. We assume for contradiction that $\varphi _a(P)+\varphi _b(P)<1$. So, there exists an alternative $y\in A$ such that $y\ne a,b$ and $\varphi _y(P)>0$. Note that by efficiency, $a\;P_1\;y\;P_1\;b$ and $b\;P_2\;y\;P_2\;a$.

Let $P'_1$ be an ordering where $P_1(A{\setminus } \{y\})=P'_1(A{\setminus } \{y\})$ and $r_2(P'_1)=y$. Let $P_1=P^1,P^2,\ldots ,P^l=P'_1$ be a path from $P_1$ to $P'_1$ where $P^j(A{\setminus } \{y\})=P^{j+1}(A{\setminus } \{y\})$ for all $j\in \{1,2,\ldots ,l-1\}$. Fix a $j\in \{1,2,\ldots ,l-1\}$ and let $A(P^j,P^{j+1})=\{y,z\}$. Note that $z\;P^j \; y$ and $y\;P^{j+1} \; z$. Also note that by efficiency, if $\varphi _y(P^j,P_2)>0$ then $y\;P_2\;z$ and $\varphi _z(P^{j+1},P_2)=0$. Therefore, by efficiency and SSS, $\varphi _y(P^j,P_2)\le \varphi _y(P^{j+1},P_2)$ for all $j\in \{1,2,\ldots ,l-1\}$. Therefore, we have $0<\varphi _y(P_1,P_2)\le \varphi _y(P'_1,P_2)$.

Let $P'_2$ be an ordering where $P_2(A{\setminus } \{y\})=P'_2(A{\setminus } \{y\})$ and $r_2(P'_2)=y$. Let $P_2=P^1,P^2,\ldots ,P^k=P'_2$ be a path from $P_2$ to $P'_2$ where $P^j(A{\setminus } \{y\})=P^{j+1}(A{\setminus } \{y\})$ for all $j\in \{1,2,\ldots ,k-1\}$. By efficiency and SSS, $\varphi _y(P'_1,P^j)\le \varphi _y(P'_1,P^{j+1})$ for all $j\in \{1,2,\ldots ,k-1\}$. Therefore, we have $0<\varphi _y(P_1,P_2)\le \varphi _y(P'_1,P_2)\le \varphi _y(P'_1,P'_2)$. This contradicts Claim 2. Therefore, we have that $\varphi _a(P)+\varphi _b(P)=1$.

Finally, we show that $\varphi (P)= \varphi ({\bar{P}})$ i.e., $\varphi _a(P)=\beta _1$ and $\varphi _b(P)=\beta _2$. Let $P'_2$ be an ordering where $P_2(A\setminus \{a\})=P'_2(A\setminus \{a\})$ and $r_2(P'_2)=a$. Let $P_2=P^1,P^2,\ldots ,P^l=P'_2$ be a path from $P_2$ to $P'_2$ where $P^j(A{\setminus } \{a\})=P^{j+1}(A{\setminus } \{a\})$ for all $j\in \{1,2,\ldots ,l-1\}$. By the result from the previous paragraph, $\varphi _a(P_1,P^j)+\varphi _b(P_1,P^j)=1$ for all $j\in \{1,2,\ldots ,l\}$. Therefore, by SSS and the fact that $\varphi _a(P_1,P^j)+\varphi _b(P_1,P^j)=1$ for all $j\in \{1,2,\ldots ,l\}$, we have $\varphi _a(P)=\varphi _a(P_1,P^1)=\varphi _a(P_1,P^2)=\ldots = \varphi _a(P_1,P^l)=\varphi _a(P_1,P'_2)$. By Claim 1, we have that $\varphi _a(P_1,P'_2)=\beta _1$. Therefore, $\varphi _a(P)=\beta _1$. Finally, this immediately implies with our previous equation $\varphi _a(P)+\varphi _b(P)=1$ that $\varphi _b(P)=\beta _2$. $\square$

Claims 1, 2 and 3 establish Lemma 2. $\square$

Lemma 3

For any $P,{\bar{P}}\in {\mathbb {P}}^2$ such that $r_1(P_1)\ne r_1(P_2)$ and $r_1(\bar{P_1})\ne r_1(\bar{P_2})$, we have $\varphi _{r_1(P_1)}(P_1,P_2)= \varphi _{r_1(\bar{P_1})}(\bar{P_1},\bar{P_2})$ and $\varphi _{r_1(P_2)}(P_1,P_2)= \varphi _{r_1(\bar{P_2})}(\bar{P_1},\bar{P_2})$.

Proof

We consider $P,{\bar{P}}\in {\mathbb {P}}^2$ such that $r_1(P_1)=a\ne b=r_1(P_2)$ and $r_1(\bar{P_1})=c\ne d=r_1(\bar{P_2})$. We consider the following two cases.

Case 1: $c\ne b$. Now we consider following four sub-cases.

Sub-case 1.1: $a=c$ and $b=d$. By Lemma 2, we are done with this sub-case.

Sub-case 1.2: $a=c$ and $b\ne d$. W.l.o.g., we consider $P,{\bar{P}}\in {\mathbb {P}}^2$ such that $P_1={\bar{P}}_1$, $P_2$ and ${\bar{P}}_2$ are adjacent and $A(P_2, {\bar{P}}_2)=\{b,d\}$. Applying Lemma 2 and SSS, we have $\varphi _a(P_1,P_2)= \varphi _c(\bar{P_1},\bar{P_2})$ and $\varphi _b(P_1,P_2)= \varphi _d(\bar{P_1},\bar{P_2})$.

Sub-case 1.3: $a\ne c$ and $b=d$. W.l.o.g., we consider $P,{\bar{P}}\in {\mathbb {P}}^2$ such that $P_2={\bar{P}}_2$, $P_1$ and ${\bar{P}}_1$ are adjacent and $A(P_1, {\bar{P}}_1)=\{a,c\}$. Again applying Lemma 2 and SSS, we have $\varphi _a(P_1,P_2)= \varphi _c(\bar{P_1},\bar{P_2})$ and $\varphi _b(P_1,P_2)= \varphi _d(\bar{P_1},\bar{P_2})$.

Sub-case 1.4: $a\ne c$ and $b\ne d$. By sub-case 1.3, for $P',P''\in {\mathbb {P}}^2$ such that $r_1(P'_1)=a\ne b=r_1(P'_2)$ and $r_1(P''_1)=c\ne b=r_1(P''_2)$, we have $\varphi _a(P')= \varphi _c(P'')$ and $\varphi _b(P')= \varphi _b(P'')$. W.l.o.g., we assume $r_2(P''_2)=d$ and consider $P,{\bar{P}}\in {\mathbb {P}}^2$ such that $P=P'$, ${\bar{P}}_1=P''_1$, ${\bar{P}}_2$ and $P''_2$ are adjacent and $A(P''_2, {\bar{P}}_2)=\{b,d\}$. Applying Lemma 2 and SSS, we have $\varphi _a(P)=\varphi _c({\bar{P}})$ and $\varphi _b(P)=\varphi _d({\bar{P}})$.

Case 2: $c= b$. Now we consider following two sub-cases.

Sub-case 2.1: $d=a$. Since $m \ge 3$, there exists $x\in A$ distinct from a and b. Let $\hat{P_1}$ be such that $r_1(\hat{P_1})=x$. From Sub-case 1.4, $\varphi _a(P_1,P_2)=\varphi _x(\hat{P_1},\bar{P_2})$ and $\varphi _b(P_1,P_2)=\varphi _a(\hat{P_1},\bar{P_2})$. From Sub-case 1.3, $\varphi _x(\hat{P_1},\bar{P_2})= \varphi _b(\bar{P_1},\bar{P_2})$ and $\varphi _a(\hat{P_1},\bar{P_2})=\varphi _a(\bar{P_1},\bar{P_2})$. Therefore, we have $\varphi _a(P_1,P_2)= \varphi _b(\bar{P_1},\bar{P_2})$ and $\varphi _b(P_1,P_2)= \varphi _a(\bar{P_1},\bar{P_2})$.

Sub-case 2.2: $d\ne a$. Let $\tilde{P_2}$ be such that $r_1(\tilde{P_2})=a$, $r_2(\tilde{P_2})=d$ and $\tilde{P_2}$ and $\bar{P_2}$ are adjacent. From Sub-case 2.1, $\varphi _a(P_1,P_2)= \varphi _b(\bar{P_1},\tilde{P_2})$ and $\varphi _b(P_1,P_2)= \varphi _a(\bar{P_1},\tilde{P_2})$. Applying Lemma 2 and SSS, we have $\varphi _b(\bar{P_1},\tilde{P_2})=\varphi _b(\bar{P_1},\bar{P_2})$ and $\varphi _a(\bar{P_1},\tilde{P_2})=\varphi _d(\bar{P_1},\bar{P_2})$. Therefore, we have $\varphi _a(P_1,P_2)= \varphi _b(\bar{P_1},\bar{P_2})$ and $\varphi _b(P_1,P_2)= \varphi _d(\bar{P_1},\bar{P_2})$. $\square$

Lemmas 2 and 3 establish that $\varphi$ is a random dictatorship. $\square$

1.4 The Proof of Theorem 2

We will complete the proof of Theorem 2 with the help of Proposition 3 and Proposition 4.

Proposition 3

Let $N = \{1,2,\ldots ,n\}$, $n \ge 3$ and $|A|\ge 3$. If $\varphi : {\mathbb {P}}^n \rightarrow \Delta (A)$ satisfies efficiency, tops-onlyness and SSS, then it is a random dictatorship.

Proof

Let $N = \{1,2,\ldots ,n\}$, $n \ge 3$ and $|A|\ge 3$. Let $\varphi : {\mathbb {P}}^n \longrightarrow \Delta (A)$ satisfies efficiency, tops-onlyness and SSS. It will be shown that $\varphi$ is a random dictatorship.

We will use induction arguments. Assume that for all integers $k < n$, the following statement is true:

Induction Hypothesis (IH): If $\phi : {\mathbb {P}}^k \rightarrow \Delta (A)$ satisfies efficiency, tops-onlyness and SSS, then it is a random dictatorship.

Let ${\hat{N}} = \{{\hat{1}},3,\ldots ,n\}$ be a set of voters where $3,\ldots ,n \in N$. A RSCF $g: {\mathbb {P}}^{n-1} \rightarrow \Delta (A)$ for the set of voters ${\hat{N}}$ is defined as follows: For all $(P_{{\hat{1}}}, P_3,\ldots ,P_n) \in {\mathbb {P}}^{n-1},$

$$\begin{aligned} g(P_{{\hat{1}}},P_3,\ldots ,P_n ) = \varphi (P_1, P_1, P_3,\ldots ,P_n). \end{aligned}$$

Voter ${\hat{1}}$ in the RSCF g is obtained by “cloning" voters 1 and 2 in N. Thus if voters 1 and 2 in N have a common ordering $P_1$, then voter ${\hat{1}}$ in ${\hat{N}}$ has ordering $P_{{\hat{1}}}$.

Lemma 4

The RSCF g is a random dictatorship.

Proof

We use the induction hypothesis to prove that g is a random dictatorship. It is straightforward to verify that g satisfies efficiency and tops-onlyness. We will show that g satisfies SSS. Consider any $P\in {\mathbb {P}}^{n-1}$, $i\in {\hat{N}}$ and $P'_i\in {\mathbb {P}}$ such that $P_i$ and $P'_i$ are adjacent. Let $A(P_i,P'_i)=\{x,y\}$. Let voter i be the voter ${\hat{1}}$. Since $\varphi$ satisfies SSS, we have

$$\begin{aligned} \sum _{a\in U(P_{{\hat{1}}},\{x,y\})}g_a(P_{{\hat{1}}},P_3,\ldots ,P_n)&= \sum _{a\in U(P_1,\{x,y\})}\varphi _a(P_1,P_1,P_3,\ldots ,P_n)\\&=\sum _{a\in U(P_1,\{x,y\})}\varphi _a(P'_1,P_1,P_3,\ldots ,P_n)\\&=\sum _{a\in U(P_1,\{x,y\})}\varphi _a(P'_1,P'_1,P_3,\ldots ,P_n)\\&= \sum _{a\in U(P_{{\hat{1}}},\{x,y\})}g_a(P'_{{\hat{1}}},P_3,\ldots ,P_n) \end{aligned}$$

and

$$\begin{aligned} \sum _{a\in \{x,y\}}g_a(P_{{\hat{1}}},P_3,\ldots ,P_n)&= \sum _{a\in \{x,y\}}\varphi _a(P_1,P_1,P_3,\ldots ,P_n)\\&=\sum _{a\in \{x,y\}}\varphi _a(P'_1,P_1,P_3,\ldots ,P_n)\\&=\sum _{a\in \{x,y\}}\varphi _a(P'_1,P'_1,P_3,\ldots ,P_n)\\&= \sum _{a\in \{x,y\}}g_a(P'_{{\hat{1}}},P_3,\ldots ,P_n). \end{aligned}$$

For $i \in \{3,\ldots ,n\}$ it is straightforward that above qualities holds because $\varphi$ satisfies SSS. Hence, g satisfies SSS. Therefore, the IH will imply that g is a random dictatorship. $\square$

Let $\beta , \beta _3,\ldots , \beta _n$ be the weights associated with the random dictatorship g; i.e., $\beta _i$ is the weight associated with voter i, $i=3,\ldots ,n$ and $\beta$ is the weight associated with voter ${\hat{1}}$.

Lemma 5

Let $P\in {\mathbb {P}}^n$ be an arbitrary profile. Let $a=r_1(P_1)$ and $b=r_1(P_2)$ and let $\beta ^x=\sum _{\{i\in \{3,\ldots ,n\}: r_1(P_i)=x\}}\beta _i$ for all $x\in A$. Then,

• (i) $\varphi _a(P)= \beta + \beta ^a$ if $a=b$,

• (ii) $\varphi _c(P)= \beta ^c$ for all $c\ne a=b$, and

• (iii) $\varphi _a(P) + \varphi _b(P)= \beta + \beta ^a + \beta ^b$ if $a\ne b$.

Proof

Proof of Part (i): By tops-onlyness and the fact that $r_1(P_1)=r_1(P_2)=a$, we have

$$\begin{aligned} \varphi _a(P_1,P_2,P_3,\ldots ,P_n)&= \varphi _a(P_1,P_1,P_3,\ldots ,P_n)\\&=g_a(P_{{\hat{1}}},P_3,\ldots ,P_n)\\&=\beta + \beta ^a. \end{aligned}$$

This establishes (i).

Proof of Part (ii): By tops-onlyness and the fact that $r_1(P_1)=r_1(P_2)=a$, we have

$$\begin{aligned} \varphi _c(P_1,P_2,P_3,\ldots ,P_n)&= \varphi _c(P_1,P_1,P_3,\ldots ,P_n)\\&=g_c(P_{{\hat{1}}},P_3,\ldots ,P_n)\\&=\beta ^c. \end{aligned}$$

This establishes the proof of part (ii).

Proof of Part (iii): Note that a and b are distinct. Let $P'_2$ and $P''_2$ be two adjacent orderings where $r_1(P'_2)=a=r_2(P''_2)$ and $r_2(P'_2)=b=r_1(P''_2)$. In the view of part (i) and (ii), it can be deduced that $\varphi _a(P_1,P'_2,P_3,\ldots ,P_n)+\varphi _b(P_1,P'_2,P_3,\ldots ,P_n)= \beta + \beta ^a + \beta ^b$. By tops-onlyness and SSS, we can conclude that

$$\begin{aligned}&\varphi _a(P_1,P_2,P_3,\ldots ,P_n)+\varphi _b(P_1,P_2,P_3,\ldots ,P_n)\\ {}&= \varphi _a(P_1,P''_2,P_3,\ldots ,P_n)+\varphi _b(P_1,P''_2,P_3,\ldots ,P_n)\\&=\varphi _a(P_1,P'_2,P_3,\ldots ,P_n)+\varphi _b(P_1,P'_2,P_3,\ldots ,P_n)\\&=\beta + \beta ^a + \beta ^b.\\ \end{aligned}$$

This completes the proof of part (iii). $\square$

Lemma 6

Fix $a,b\in A$, $a\ne b$. For any $P\in {\mathbb {P}}^n$ where $r_1(P_1)=a$ and $r_1(P_2)=b$ and $x\in A$, let $\beta ^x(P)=\sum _{\{i\in \{3,\ldots ,n\}: r_1(P_i)=x\}}\beta _i$. Then, there exist $0\le \beta _1^a,\beta _2^b\le \beta$, $\beta _1^a +\beta _2^b=\beta$ such that for any $P\in {\mathbb {P}}^n$ where $r_1(P_1)=a$ and $r_1(P_2)=b$, $\varphi _a(P)= \beta _1^a + \beta ^a(P)$, $\varphi _b(P)= \beta _2^b + \beta ^b(P)$ and $\varphi _c(P)= \beta ^c(P)$ for all $c\ne a,b$.

Proof

Fix an alternative $d\ne a,b$. Let $P^*\in {\mathbb {P}}^n$ be such that $r_1(P^*_1)=a$, $r_1(P^*_2)=b$ and for all $j\in \{3,\ldots ,n\}$, $r_1(P^*_j)=d$. In the view of Part (iii) of Lemma 5, we have that $\varphi _a(P^*) + \varphi _b(P^*)=\beta$. W.l.o.g., we assume that $\varphi _a(P^*)=\beta _1^a$ and $\varphi _b(P^*)=\beta _2^b$ where $0\le \beta _1^a,\beta _2^b\le \beta$ and $\beta _1^a +\beta _2^b=\beta$. We will show that for any $P\in {\mathbb {P}}^n$ where $r_1(P_1)=a$ and $r_1(P_2)=b$, $\varphi _a(P)= \beta _1^a + \beta ^a(P)$, $\varphi _b(P)= \beta _2^b + \beta ^b(P)$ and $\varphi _c(P)= \beta ^c(P)$ for all $c\ne a,b$. We use the following simple fact in the subsequent reasoning without providing a proof. $\square$

Fact 1

Let $\varphi : {\mathbb {P}}^n\rightarrow \Delta (A)$ be an efficient and tops-only RSCF. Then, for any $P\in {\mathbb {P}}^n$ and $a\in A$, $\varphi _a(P)>0$ implies $a=r_1(P_i)$ for some $i\in N$.

Now we show the following Claim.

Claim 4

Let $P\in {\mathbb {P}}^n$ be a preference profile where $r_1(P_1)=a$, $r_1(P_2)=b$ and $r_1(P_j)\in \{a,b\}$ for all $j\in \{3,\ldots ,n\}$. Then $\varphi _a(P)= \beta _1^a +\beta ^a(P)$, $\varphi _b(P)= \beta _2^b +\beta ^b(P)$ and $\varphi _c(P)= 0$ for all $c\ne a,b$.

Proof

Consider a preference profile $P\in {\mathbb {P}}^n$ such that $r_1(P_1)=a$, $r_1(P_2)=b$ and $r_1(P_j)\in \{a,b\}$ for all $j\in \{3,\ldots ,n\}$. By Fact 1, $\varphi _c(P)= 0$ for all $c\ne a,b$. We complete the proof by showing that $\varphi _a(P)= \beta _1^a +\beta ^a(P)$ and $\varphi _b(P)= \beta _2^b +\beta ^b(P)$. W.l.o.g., we assume that $r_1(P_j)=a$ for all $j\in \{3,\ldots ,k\}$ and $r_1(P_j)=b$ for all $j\in \{k+1,\ldots ,n\}$, $k\le n$. We assume for contradiction that $\varphi _a(P) \ne \beta _1^a +\beta ^a(P)$.

Let ${\bar{P}}\in {\mathbb {P}}^n$, such that $r_1({\bar{P}}_i)=r_1(P_i)$ for all for all $i\in N$ and $r_2({\bar{P}}_i)=d$ for all $i\in \{3,\ldots ,n\}$. By tops-onlyness, $\varphi ({\bar{P}})=\varphi (P)$. Also note that $\beta ^a(P)=\beta ^a({\bar{P}})=\sum _{i=3}^k\beta _i$. For all $i\in \{3,\ldots ,k\}$, let ${\bar{P}}_i$ and ${\hat{P}}_i$ be two adjacent ordering where $r_1({\bar{P}}_i)=r_2({\hat{P}}_i)=a$ and $r_2({\bar{P}}_i)=r_1({\hat{P}}_i)=d$. Also, For all $i\in \{k+1,\ldots ,n\}$, let ${\bar{P}}_i$ and ${\hat{P}}_i$ be two adjacent ordering where $r_1({\bar{P}}_i)=r_2({\hat{P}}_i)=b$ and $r_2({\bar{P}}_i)=r_1({\hat{P}}_i)=d$. By SSS and the fact that $\varphi _a({\bar{P}})\ne \beta _1^a +\beta ^a({\bar{P}})$ and $\varphi _d({\bar{P}})=0$, we have

$$\begin{aligned}&\varphi _a({\bar{P}})+\varphi _d({\bar{P}})\\&= \varphi _a({\bar{P}}_1,{\bar{P}}_2,{\hat{P}}_3,{\bar{P}}_4,\ldots ,{\bar{P}}_k,{\bar{P}}_{k+1},\ldots ,{\bar{P}}_n)+\varphi _d({\bar{P}}_1,{\bar{P}}_2,{\hat{P}}_3,{\bar{P}}_4\ldots ,{\bar{P}}_k,{\bar{P}}_{k+1},\ldots ,{\bar{P}}_n)\\&\ \vdots \\&=\varphi _a({\bar{P}}_1,{\bar{P}}_2,{\hat{P}}_3,\ldots ,{\hat{P}}_k,{\bar{P}}_{k+1},\ldots ,{\bar{P}}_n)+\varphi _d({\bar{P}}_1,{\bar{P}}_2,{\hat{P}}_3,\ldots ,{\hat{P}}_k,{\bar{P}}_{k+1},\ldots ,{\bar{P}}_n)\\&\ne \beta _1^a +\beta ^a({\bar{P}})\\&=\beta _1^a + \sum _{i=3}^k\beta _i. \end{aligned}$$

By Fact 1 and part (iii) of Lemma 5,

$$\begin{aligned} \varphi _d({\bar{P}}_1,{\bar{P}}_2,{\hat{P}}_3,\ldots ,{\hat{P}}_k,{\bar{P}}_{k+1},\ldots ,{\bar{P}}_n)=\sum _{i=3}^k\beta _i. \end{aligned}$$

Therefore, we can conclude that

$$\begin{aligned} \varphi _a({\bar{P}}_1,{\bar{P}}_2,{\hat{P}}_3,\ldots ,{\hat{P}}_k,{\bar{P}}_{k+1},\ldots ,{\bar{P}}_n)\ne \beta _1^a. \end{aligned}$$

By Fact 1 and SSS,

$$\begin{aligned} \varphi _a({\bar{P}}_1,{\bar{P}}_2,{\hat{P}}_3,\ldots ,{\hat{P}}_k,{\bar{P}}_{k+1},\ldots ,{\bar{P}}_n)&= \varphi _a({\bar{P}}_1,{\bar{P}}_2,{\hat{P}}_3,\ldots ,{\hat{P}}_k,{\hat{P}}_{k+1},\ldots ,{\bar{P}}_n)\\&\ \vdots \\&=\varphi _a({\bar{P}}_1,{\bar{P}}_2,{\hat{P}}_3,\ldots ,{\hat{P}}_k,{\hat{P}}_{k+1},\ldots ,{\hat{P}}_n)\\&\ne \beta _1^a. \end{aligned}$$

However, by tops-onlyness, $\varphi _a(P^*)=\varphi _a({\bar{P}}_1,{\bar{P}}_2,{\hat{P}}_3,\ldots ,{\hat{P}}_k,{\hat{P}}_{k+1},\ldots ,{\hat{P}}_n)=\beta _1^a$ - a contradiction. Hence, $\varphi _a(P) = \beta _1^a +\beta ^a(P)$. Since $\varphi _a(P) +\varphi _b(P)=1$, we have $\varphi _b(P) = \beta _1^b +\beta ^b(P)$. $\square$

We suppose that voter 1 top-ranks a, voter 2 top-ranks b and proceed inductively on the size of the set of top-ranked alternatives. The induction basis holds by Claim 4. Now, suppose that the lemma holds for some $k \in \{2,\ldots ,m-1\}$ and consider a profile P in which precisely $k+1$ alternatives are top-ranked. Let $c \in A \setminus \{a, b\}$ be an alternative which is top-ranked by at least one voter in P. First we show that $\varphi _a(P)+\varphi _c(P)= \beta _1^a +\beta ^a(P)+\beta ^c(P)$. We consider the profile $P'$ which agrees with P except that all voters who top-rank c place a second. Finally, consider the profile $P''$ derived from $P'$ by letting all voters who top-rank c swap a and c. SSS and Fact 1 imply that $\varphi _a(P'')=\varphi _a(P)+\varphi _c(P)$. Moreover, the induction hypothesis requires that $\varphi _a(P'')=\beta _1^a +\beta ^a(P'')=\beta _1^a +\beta ^a(P)+\beta ^c(P)$. Therefore, we have that $\varphi _a(P)+\varphi _c(P)=\beta _1^a +\beta ^a(P)+\beta ^c(P)$. Clearly, a symmetric arguments holds for b and we have that $\varphi _b(P)+\varphi _c(P)= \beta _1^b +\beta ^b(P)+\beta ^c(P)$. Finally, we know from Lemma 5 (iii) that $\varphi _a(P)+\varphi _b(P)=\beta _1^a+\beta _1^b +\beta ^a(P)+\beta ^b(P)$. Hence, we can now consider $(\varphi _a(P)+\varphi _c(P))+(\varphi _b(P)+\varphi _c(P))-(\varphi _a(P)+\varphi _b(P))$ and compute that $\varphi _c(P)=\beta ^c(P)$. Finally, from equations $\varphi _a(P)+\varphi _c(P)= \beta _1^a +\beta ^a(P)+\beta ^c(P)$, $\varphi _b(P)+\varphi _c(P)= \beta _1^b +\beta ^b(P)+\beta ^c(P)$ and $\varphi _c(P)=\beta ^c(P)$, we have that $\varphi _a(P)= \beta _1^a +\beta ^a(P)$ and $\varphi _b(P)= \beta _1^b +\beta ^b(P)$. This completes the proof of Lemma 6.

The proof of the proposition is now completed by considering following two cases.

Case 1: $\beta > 0$. Fix $(P_3,\ldots ,P_n) \in {\mathbb {P}}^{n-2}$ and let $\beta ^x=\sum _{\{i\in \{3,\ldots ,n\}: r_1(P_i)=x\}}\beta _i$ for all $x\in A$. We define the function $h: {\mathbb {P}}^2 \rightarrow {\mathbb {R}}^m$ below: for all $(P_1,P_2)\in {\mathbb {P}}^2$ and $a \in A$,

$$\begin{aligned} h_a(P_1,P_2)=\frac{1}{\beta }[\varphi _a(P_1,P_2,P_3\ldots ,P_n) - \beta ^a]. \end{aligned}$$

Lemma 7

The function h is a RSCF and satisfies efficiency and SSS.

Proof

Pick an arbitrary profile $(P_1,P_2)\in {\mathbb {P}}^2$. Let $a \in A$. If $r_1(P_1)=r_1(P_2) =a$, then $\varphi _a(P_1,P_2,P_3,\ldots ,P_n)=\beta + \beta ^a$ according to Lemma 5 part (i). Hence $h_a(P_1,P_2)=1$. Suppose $r_1(P_1)=a\ne b = r_1(P_2)$. From Lemma 6, $\varphi _a(P_1,P_2, P_3,\ldots ,P_n)\ge \beta ^a$ and $\varphi _b(P_1,P_2, P_3,\ldots ,P_n)\ge \beta ^b$. Hence $h_a(P_1,P_2)\ge 0$ and $h_b(P_1,P_2)\ge 0$. If $a \notin \{ r_1(P_1)\cup r_1(P_2)\}$, then $\varphi _a(P_1,P_2,P_3,\ldots ,P_n)= \beta ^a$ and $h_a(P_1,P_2)=0$. Therefore, we have that $h_a(P_1,P_2)\ge 0$ for all $a \in A$. Note also that $\sum _{a\in A} \varphi _a(P_1,P_2,P_3,\ldots ,P_n)=\beta + \sum _{a\in A}\beta ^a=1$, so that $\sum _{a\in A} h_a(P_1,P_2)=1$. Therefore, h is a RSCF.

First we show that h satisfies efficiency. Pick an arbitrary profile $(P_1,P_2)\in {\mathbb {P}}^2$. Let $a,b \in A$ such that $aP_ib$ for all $i\in \{1,2\}$. From Lemma 6, $\varphi _b(P_1,P_2, P_3,\ldots ,P_n)= \beta ^b$. Hence $h_b(P_1,P_2)=0$. This concludes that h is efficient.

Next we show that h satisfies SSS. Pick an arbitrary profile $(P_1,P_2)\in {\mathbb {P}}^2$. Let $P_i$ and $P'_i$ be two adjacent orderings where $i\in \{1,2\}$. W.l.o.g., we assume that $i=1$. Let $A(P_1,P'_1)=\{x,y\}$. We will show that

• (i). $\sum \limits _{a\in U(P_i,\{x,y\})}h_a(P_1,P_2)=\sum \limits _{a\in U(P'_i,\{x,y\})}h_a(P'_i,P_2)$ and

• (ii). $\sum \limits _{a\in \{x,y \}}h_a(P_1,P_2)=\sum \limits _{a\in \{x,y\}}h_a(P'_1,P_2)$.

Note that for any $a\in A$, we have

$$\begin{aligned} \varphi _a(P_1,P_2,P_3,\ldots ,P_n)= \beta \cdot h_a(P_1,P_2) +\beta ^a \end{aligned}$$

and

$$\begin{aligned} \varphi _a(P'_1,P_2,P_3,\ldots ,P_n)= \beta \cdot h_a(P'_1,P_2) +\beta ^a. \end{aligned}$$

Since $\varphi$ satisfies SSS and $\beta > 0$, it can be easily verified that equalities in (i) and (ii) hold. Hence, h satisfies SSS. $\square$

Since h satisfies efficiency and SSS, it follows from Theorem 1 that h is a random dictatorship. Assume that the weights associated with h are $\gamma _1$ and $\gamma _2$ for voters 1 and 2 respectively. It follows from the definition of h that for all $(P_1,P_2)\in {\mathbb {P}}^2$ and $a \in A$,

$$\begin{aligned} \varphi _a(P_1,P_2,P_3,\ldots ,P_n)= \sum _{\{i\in \{1,2\}: r_1(P_i)=a\}}\beta \gamma _i + \sum _{\{i\in \{3,\ldots ,n\}: r_1(P_i)=a\}}\beta _i. \end{aligned}$$

Therefore $\varphi$ is a random dictatorship with weights $\beta \gamma _1,\beta \gamma _2,\beta _3,\ldots ,\beta _n$ if the weights for the random dictatorship do not depend on the initial choice of the profile $(P_3,\ldots ,P_n)$ for voters $3,\ldots ,n$. Lemma 6 establishes that this is indeed the case.

This completes the proof of random dictatorship in Case 1.

Case 2: $\beta = 0$. For any $P\in {\mathbb {P}}^n$ and $x\in A$, let $\beta ^x(P)=\sum _{\{i\in \{3,\ldots ,n\}: r_1(P_i)=x\}}\beta _i$. Applying Lemma 5 and 6, it follows that for any $P\in {\mathbb {P}}^n$, $\varphi _a(P) = \beta ^a(P)$ for all $a \in A$. But this implies that $\varphi$ is a random dictatorship with weights $\beta _1 = \beta _2 =0$ and $\beta _i, i =3,\ldots ,n$. This concludes the proof. $\square$

Proposition 4

Let $N = \{1,2,\ldots ,n\}$, $n \ge 3$ and $|A|=3$. If RSCF $\varphi : {\mathbb {P}}^n\longrightarrow \Delta (A)$ satisfies efficiency and SSS, then it satisfies tops-onlyness.

Proof

Let $A=\{a,b,c\}$. For any preference profile $P\in {\mathbb {P}}^n$, let $\tau (P)$ be the set of top ranked alternatives at P i.e., $\tau (P)=\{x\in A | x=r_1(P_i) \textit{ for some } i\in N\}$. Let $\varphi : {\mathbb {P}}^n \longrightarrow \Delta (A)$ satisfy efficiency and SSS. Pick any $P^1,P^2\in {\mathbb {P}}^n$ such that $r_1(P^1_i)=r_1(P^2_i)$ for all $i\in N$. We will show that $\varphi _x(P^1)=\varphi _x(P^2)$ for all $x\in A$. Note that $\tau (P^1)=\tau (P^2)$. We complete the proof by considering following three cases.

Case 1: $|\tau (P^1)|=1$. Let $\tau (P^1)=\tau (P^2)=\{x\}$. By efficiency, we have that $\varphi _x(P^1)=\varphi _x(P^2)=1$ and $\varphi _y(P^1)=\varphi _y(P^2)=0$ for any $y\in A{\setminus } \{x\}$.

Case 2: $|\tau (P^1)|=2$. First we will show that for any $P\in {\mathbb {P}}^n$ such that $|\tau (P)|=2$, we have $\sum _{x\in \tau (P)}\varphi _x(P)=1$. Suppose not. Let $P\in {\mathbb {P}}^n$ be such that $|\tau (P)|=2$ and $\sum _{x\in \tau (P)}\varphi _x(P)<1$. W.l.o.g., we assume that at P, $r_1(P_1)=\ldots =r_1(P_k)=a$ and $r_1(P_{k+1})=\ldots =r_1(P_n)=b$, where $k\in \{1,\dots ,n-1\}$. Therefor, $\varphi _a(P)+\varphi _b(P)<1$ and $\varphi _c(P)>0$. Now we construct the preference profile ${\bar{P}}$ from P as follows. At ${\bar{P}}$, $a\; {\bar{P}}_i \; b \; {\bar{P}}_i\; c$ for all $i\in \{1,\ldots ,k\}$ and ${\bar{P}}_i=P_i$ for all $i\in \{k+1,\ldots ,n\}$. By SSS, $\varphi _a({\bar{P}})=\varphi _a(P)$. By efficiency and SSS, $\varphi _b({\bar{P}})=\varphi _b(P)+\varphi _c(P)$. Next we construct the preference profile ${\tilde{P}}$ from ${\bar{P}}$ as follows. At ${\tilde{P}}$, $\tilde{P_i}={\bar{P}}_i$ for all $i\in \{1,\ldots ,k\}$ and $b\; {\tilde{P}}_i \; a \; {\tilde{P}}_i\; c$ for all $i\in \{k+1,\ldots ,n\}$. By SSS, $\varphi _b({\tilde{P}})=\varphi _b({\bar{P}})=\varphi _b(P)+\varphi _c(P)$. By efficiency and SSS, $\varphi _a({\tilde{P}})=\varphi _a({\bar{P}})=\varphi _a(P)$. Similarly, we construct the preference profile $P'$ from P as follows. At $P'$, $P'_i=P_i$ for all $i\in \{1,\ldots ,k\}$ and $b\; P'_i \; a \; P'_i\; c$ for all $i\in \{k+1,\ldots ,n\}$. By SSS, $\varphi _b(P')=\varphi _b(P)$. By efficiency and SSS, $\varphi _a(P')=\varphi _a(P)+ \varphi _c(P)$. Finally, we construct the preference profile $P''$ from $P'$ as follows. At $P''$, $a\; P''_i \; b \; P''_i\; c$ for all $i\in \{1,\ldots ,k\}$ and $P''_i=P'_i$ for all $i\in \{k+1,\ldots ,n\}$. By SSS, $\varphi _a(P'')=\varphi _a(P')=\varphi _a(P)+ \varphi _c(P)$. By efficiency and SSS, $\varphi _b(P'')=\varphi _b(P')=\varphi _b(P)$. Note that ${\tilde{P}}\equiv P''$. However, $\varphi _a(P'')=\varphi _a(P)+ \varphi _c(P)\ne \varphi _a(P)=\varphi _a({\tilde{P}})$. This contradicts the fact that $\varphi$ is a function. Therefore, we conclude that for any $P\in {\mathbb {P}}^n$ such that $|\tau (P)|=2$, we have $\sum _{x\in \tau (P)}\varphi _x(P)=1$.

Now we are ready to show that $\varphi _x(P^1)=\varphi _x(P^2)$ for all $x\in A$. W.l.o.g., we assume that at $P^1$, $r_1(P^1_1)=\ldots =r_1(P^1_k)=a$ and $r_1(P^1_{k+1})=\ldots =r_1(P^1_n)=b$, where $k\in \{1,\dots ,n-1\}$. Note that at $P^2$, $r_1(P^2_1)=\ldots =r_1(P^2_k)=a$ and $r_1(P^2_{k+1})=\ldots =r_1(P^2_n)=b$. By SSS,

$$\begin{aligned} \varphi _a(P^1)&= \varphi _a(P^2_1,P^1_2,\ldots ,P^1_n)\\&\ \vdots \\&=\varphi _a(P^2_1,P^2_2,\ldots ,P^2_k,P^1_{k+1},\ldots ,P^1_n). \end{aligned}$$

By the result from the previous paragraph which tells $\varphi _c(P^2_1,P^2_2,\ldots ,P^2_k,P^1_{k+1},\ldots ,P^1_n)= \varphi _c(P^1)=0$, we conclude that $\varphi _b(P^2_1,P^2_2,\ldots ,P^2_k,P^1_{k+1},\ldots ,P^1_n)= \varphi _b(P^1)$. Similarly, by SSS,

$$\begin{aligned} \varphi _b(P^1)&= \varphi _b(P^2_1,P^2_2,\ldots ,P^2_k,P^1_{k+1},\ldots ,P^1_n)\\&=\varphi _b(P^2_1,P^2_2,\ldots ,P^2_k,P^2_{k+1},\ldots ,P^1_n)\\&\ \vdots \\&=\varphi _b(P^2_1,P^2_2,\ldots ,P^2_n)\\&=\varphi _b(P^2). \end{aligned}$$

By the result from the previous paragraph which tells $\varphi _c(P^1)= \varphi _c(P^2)=0$, we conclude that $\varphi _a(P^1)= \varphi _a(P^2)$. Therefore, $\varphi _x(P^1)=\varphi _x(P^2)$ for all $x\in A$.

Case 3: $|\tau (P^1)|=3$. We assume for contradiction that $\varphi _x(P^1)\ne \varphi _x(P^2)$ for some $x\in A$. W.l.o.g., we assume that $\varphi _a(P^1)\ne \varphi _a(P^2)$. Further, w.l.o.g., we assume that at $P^1$, $r_1(P^1_1)=\ldots =r_1(P^1_k)=a$, $r_1(P^1_{k+1})=\ldots =r_1(P^1_l)=b$ and $r_1(P^1_{l+1})=\ldots =r_1(P^1_n)=c$, where $1\le k<l<n$. Note that at $P^2$, $r_1(P^2_1)=\ldots =r_1(P^2_k)=a$, $r_1(P^2_{k+1})=\ldots =r_1(P^2_l)=b$ and $r_1(P^2_{l+1})=\ldots =r_1(P^2_n)=c$.

First we construct ${\bar{P}}$ and $P^*$ from $P^1$ and $P^2$ respectively as follows. At ${\bar{P}}$, $a\;{\bar{P}}_i\; b \; {\bar{P}}_i \; c$ for all $i\in \{1,\ldots , k\}$ and ${\bar{P}}_i= P^1_i$ for all $i\notin \{1,\ldots , k\}$. At $P^*$, $a\;P^*_i\; b \; P^*_i \; c$ for all $i\in \{1,\ldots , k\}$ and $P^*_i= P^2_i$ for all $i\notin \{1,\ldots , k\}$. By SSS,

$$\begin{aligned} \varphi _a(P^1)&= \varphi _a({\bar{P}}_1,P^1_2,\ldots ,P^1_n)\\&\ \vdots \\&=\varphi _a({\bar{P}}_1,{\bar{P}}_2,\ldots ,{\bar{P}}_k,P^1_{k+1},\ldots ,P^1_n)\\&=\varphi _a({\bar{P}}), \end{aligned}$$

and

$$\begin{aligned} \varphi _a(P^2)&= \varphi _a(P^*_1,P^1_2,\ldots ,P^1_n)\\&\ \vdots \\&=\varphi _a(P^*_1,P^*_2,\ldots ,P^*_k,P^1_{k+1},\ldots ,P^1_n)\\&=\varphi _a(P^*). \end{aligned}$$

Therefore, we conclude that $\varphi _a({\bar{P}})=\varphi _a(P^1)\ne \varphi _a(P^2)=\varphi _a(P^*)$. Note that $\varphi _a({\bar{P}})\ne \varphi _a(P^*)$ implies $\varphi _x({\bar{P}})\ne \varphi _x(P^*)$ for some $x\in \{b,c\}$. We will complete the proof of Case 3 by considering the following two sub-cases that lead to a contradiction.

Sub-case 3.1: $\varphi _b({\bar{P}})\ne \varphi _b(P^*)$. We construct $\bar{{\bar{P}}}$ and $P^{**}$ from ${\bar{P}}$ and $P^*$ respectively as follows. At $\bar{{\bar{P}}}$, $b\;\bar{{\bar{P}}}_i\; c \; \bar{{\bar{P}}}_i \; a$ for all $i\in \{k+1,\ldots , l\}$ and $\bar{{\bar{P}}}_i={\bar{P}}$ for all $i\notin \{k+1,\ldots , l\}$. At $P^{**}$, $b\;P^{**}_i\; c \; P^{**}_i \; a$ for all $i\in \{k+1,\ldots , l\}$ and $P^{**}_i= P^*_i$ for all $i\notin \{k+1,\ldots , l\}$. Applying SSS, we conclude that $\varphi _b(\bar{{\bar{P}}})=\varphi _b({\bar{P}})\ne \varphi _b(P^*)=\varphi _b(P^{**})$. Note that $\varphi _b(\bar{{\bar{P}}})\ne \varphi _b(P^{**})$ implies $\varphi _x(\bar{{\bar{P}}})\ne \varphi _x(P^{**})$ for some $x\in \{a,c\}$.

First we consider that $\varphi _a(\bar{{\bar{P}}})\ne \varphi _a(P^{**})$. We construct ${\tilde{P}}$ and $P'$ from $\bar{{\bar{P}}}$ and $P^{**}$ respectively as follows. At ${\tilde{P}}$, $c\;{\tilde{P}}_i\; b \; {\tilde{P}}_i \; a$ for all $i\in \{k+1,\ldots , l\}$ and ${\tilde{P}}=\bar{{\bar{P}}}_i$ for all $i\notin \{k+1,\ldots , l\}$. At $P'$, $c\;P'_i\; b \; P'_i \; a$ for all $i\in \{k+1,\ldots , l\}$ and $P'_i=P^{**}_i$ for all $i\notin \{k+1,\ldots , l\}$. Applying SSS, we conclude that $=\varphi _a({\tilde{P}})=\varphi _a(\bar{{\bar{P}}})\ne \varphi _a(P^{**})=\varphi _a(P')$. However, this contradicts Case 2, since $r_1({\tilde{P}}_i)=r_1(P'_i)$ for all $i\in N$ and $|\tau ({\tilde{P}})|=|\tau (P')|=2$.

Now we consider that $\varphi _c(\bar{{\bar{P}}})\ne \varphi _c(P^{**})$. We construct $\tilde{{\tilde{P}}}$ and $P''$ from $\bar{{\bar{P}}}$ and $P^{**}$ respectively as follows. At $\tilde{{\tilde{P}}}$, $b\;\tilde{{\tilde{P}}}_i\; a \; \tilde{{\tilde{P}}}_i \; c$ for all $i\in \{1,\ldots , k\}$ and $\tilde{{\tilde{P}}}=\bar{{\bar{P}}}_i$ for all $i\notin \{1,\ldots , k\}$. At $P''$, $b\;P''_i\; a \; P''_i \; c$ for all $i\in \{1,\ldots , k\}$ and $P''_i=P^{**}_i$ for all $i\notin \{1,\ldots , k\}$. Applying SSS, we conclude that $=\varphi _c(\tilde{{\tilde{P}}})=\varphi _c(\bar{{\bar{P}}})\ne \varphi _c(P^{**})=\varphi _c(P'')$. However, this contradicts Case 2, since $r_1(\tilde{{\tilde{P}}}_i)=r_1(P''_i)$ for all $i\in N$ and $|\tau (\tilde{{\tilde{P}}})|=|\tau (P'')|=2$.

Sub-case 3.2: $\varphi _c({\bar{P}})\ne \varphi _c(P^*)$. We construct $\hat{{\bar{P}}}$ and $\hat{P^*}$ from ${\bar{P}}$ and $P^*$ respectively as follows. At $\hat{{\bar{P}}}$, $b\;\hat{{\bar{P}}}_i\; a \; \hat{{\bar{P}}}_i \; c$ for all $i\in \{1,\ldots , k\}$ and $\hat{{\bar{P}}}_i={\bar{P}}_i$ for all $i\notin \{1,\ldots , k\}$. At $\hat{P^*}$, $b\;\hat{P^*}_i\; a \; \hat{P^*}_i \; c$ for all $i\in \{1,\ldots , k\}$ and $\hat{P^*}_i= P^*_i$ for all $i\notin \{1,\ldots , k\}$. Applying SSS, we conclude that $\varphi _c(\hat{{\bar{P}}})=\varphi _c({\bar{P}})\ne \varphi _c(P^*)=\varphi _c(\hat{P^*})$. However, this contradicts Case 2, since $r_1(\hat{{\bar{P}}}_i)=r_1(\hat{P^*}_i)$ for all $i\in N$ and $|\tau (\hat{{\bar{P}}})|=|\tau (\hat{P^*})|=2$. $\square$

The proof of Theorem 2

It is well known that random dictatorship satisfies efficiency, tops-onlyness and sd-strategy-proofness. By Lemma 1, it satisfies SSS. Therefore, Theorem 2a follows from Proposition 3 and Proposition 4 and Theorem 2b follows from Proposition 3. $\square$