Asymptotic utilitarianism in scoring rules

Abstract

Given a large enough population of voters whose utility functions satisfy certain statistical regularities, we show that voting rules such as the Borda rule, approval voting, and evaluative voting have a very high probability of selecting the social alternative which maximizes the utilitarian social welfare function. We also characterize the speed with which this probability approaches one as the population grows.

Appendices

Appendix 1: Background

The proofs in this paper depend on some results from Pivato (2016c). In this appendix, we briefly review these results.

Pivato (2016c) considers the problem of a utilitarian social planner who can only make noisy observations of the utility functions of the individuals in society and the correct system of interpersonal comparisons. For every i in ${\mathcal { I}}$, let $u_i:{\mathcal { A}}{{\longrightarrow }}{\mathbb {R}}$ be the true cardinal utility function for voter i, and let $c_i>0$ be a calibration constant, which we will use to make cardinal interpersonal utility comparisons. We suppose that the social planner wants to maximize the utilitarian social welfare function $U_{\mathcal { I}}:{\mathcal { A}}{{\longrightarrow }}{\mathbb {R}}$ defined by Formula (1); however, she faces the following informational problems.

(U1) :

$\{c_i\}_{i\in {\mathcal { I}}}$ are unknown. The planner regards $\{c_i\}_{i\in {\mathcal { I}}}$ as independent (but not necessarily identically distributed) real-valued random variables. There are constants ${\overline{c}}>0$ and $\sigma _c^2\ge 0$ (independent of I) such that ${\mathbb {E}}[c_i]={\overline{c}}$ and $\mathrm{var}[c_i]\le \sigma _c^2$, for all $i\in {\mathcal { I}}$.

(U2) :

The utility functions $\{u_i\}_{i\in {\mathcal { I}}}$ are not precisely observable. Instead, for each i in ${\mathcal { I}}$, the planner can only observe a function $v_i:=u_i+\epsilon _i$, where $\epsilon _i:{\mathcal { A}}{{\longrightarrow }}{\mathbb {R}}$ is a random “error” term. For each alternative a in ${\mathcal { A}}$, the random errors $\{\epsilon _i(a)\}_{i\in {\mathcal { I}}}$ are independent¹⁶ (but not necessarily identically distributed), and they all have an expected value of 0 and a variance less than or equal some constant $\sigma _\epsilon ^2>0$ (which is independent of I). Finally, the random variables $\{c_i\}_{i\in {\mathcal { I}}}$ are independent of the random functions $\{\epsilon _i\}_{i\in {\mathcal { I}}}$.

We assume that the utility profile $\{u_i\}_{i\in {\mathcal { I}}}$ satisfies one or both of the following conditions.

(U3) :

There is a constant $\Delta >0$ (independent of I) such that $\max _{{\mathcal { A}}} (U_{\mathcal { I}})-U_{\mathcal { I}}(a)>\Delta$ for every $a\not \in {{\mathrm{\mathrm{argmax}}}}_{{\mathcal { A}}} (U_{\mathcal { I}})$.

(U4) :

There is a constant $M>0$ (independent of I) such that for every a in ${\mathcal { A}}$.

Note that, while we assume that $\{v_i\}_{i\in {\mathcal { I}}}$ and $\{c_i\}_{i\in {\mathcal { I}}}$ are random variables, we make no assumptions about the mechanism generating the underlying profile of utility functions $\{u_i\}_{i\in {\mathcal { I}}}$. These utility functions might be fixed in advance, or they might themselves be generated by some other random process, as long as they satisfy (U4) and (U3). Define the function $V_{\mathcal { I}}:{\mathcal { A}}{{\longrightarrow }}{\mathbb {R}}$ as in Eq. (2), and define the SWF $U_{\mathcal { I}}:{\mathcal { A}}\,{{\longrightarrow }}\,{\mathbb {R}}$ as in Eq. (1). Here is Theorem 1 of Pivato (2016c).

Theorem 6.1

For every i in ${\mathcal { I}}$, let $u_i:{\mathcal { A}}\,{{\longrightarrow }}\, {\mathbb {R}}$ be a utility function. Suppose the profile $\{u_i\}_{i\in {\mathcal { I}}}$ satisfies (U3) and (U4). If $\{c_i\}_{i\in {\mathcal { I}}}$, $\{\epsilon _i\}_{i\in {\mathcal { I}}}$ and $\{v_i\}_{i\in {\mathcal { I}}}$ are randomly generated according to rules (U1) and (U2), then

$$\begin{aligned} \lim \nolimits _{I{\rightarrow }{\infty }} \mathrm{Prob} \left[ {{\mathrm{\mathrm{argmax}}}}_{{\mathcal { A}}} (V_{\mathcal { I}})\subseteq {{\mathrm{\mathrm{argmax}}}}_{\mathcal { A}}(U_{\mathcal { I}})\right] = 1. \end{aligned}$$

For any $\delta >0$ and $p\in (0,1)$, we define

$$\begin{aligned} {\overline{I}}(\delta ,p):= 4\,|{\mathcal { A}}|\,\frac{M^2\, \sigma _c^2+ \sigma _\epsilon ^2}{p\,\delta ^2}. \end{aligned}$$

(11)

Define $U^*_{\mathcal { I}}:=\max {\left\{ U_{\mathcal { I}}(a) \; ; \; a\in {\mathcal { A}} \right\} }$. Here is Theorem 2 of Pivato (2016c).

Theorem 6.2

Suppose $\{u_i\}_{i\in {\mathcal { I}}}$, $\{c_i\}_{i\in {\mathcal { I}}}$, $\{\epsilon _i\}_{i\in {\mathcal { I}}}$ and $\{v_i\}_{i\in {\mathcal { I}}}$ satisfy (U1), (U2) and (U4). For any $\delta >0$ and $p\in (0,1)$, if $I\ge {\overline{I}}(\delta ,p)$, then $\mathrm{Prob}\left[ U_{\mathcal { I}}(a)< U^*_{\mathcal { I}}-\delta \right]\ < \ p$, for all a in ${{\mathrm{\mathrm{argmax}}}}_{{\mathcal { A}}} (V_{\mathcal { I}})$.

Appendix 2: Proofs

Proof of Theorem 3.1

We will derive this from Theorem 6.1. Recall that $u_i:=w_i/c_i$, so that $w_i=c_i\,u_i$; with this substitution, formulae (3) and (1) are equivalent. Observe that hypothesis (C) implies (U1) (with ${\overline{c}}:=1$), hypothesis (E) implies (U2), and hypothesis $(\Delta )$ implies (U3). Meanwhile, hypothesis (U4) is true automatically, with $M=1$, because the functions $\{u_i\}_{i\in {\mathcal { I}}}$ range over [0, 1]. The asymptotic probability claim now follows from Theorem 6.1. $\square$

Proof of Proposition 3.2

If we set $M:=1$ in formula (11), we obtain formula (5). The asymptotic probability inequality now follows from Theorem 6.2. $\square$

Proof of Theorem 4.1

Let $g:={\mathbb {E}}[u^i_a\,|\, u^i_a\ge \theta _i]$ and let $b:={\mathbb {E}}[u^i_a\,|\, u^i_a< \theta _i]$; thus $g>b$. By $(\Theta 1)$ and $(\Theta 2)$, these values do not depend on i or a. For all $i\in {\mathcal { I}}$, define $v_i:{\mathcal { A}}\,{{\longrightarrow }}\,{\mathbb {R}}$ by

$$\begin{aligned} \begin{array}{ll} v_i(a) := \left\{ \begin{array}{ll} g &{} \quad \text{ if } \; u^i_a\ge \theta _i; \\ b &{} \quad \text{ if } \; u^i_a< \theta _i. \end{array} \right. &{}= \left\{ \begin{array}{ll} g &{} \quad \text{ if } a\in {\mathcal { G}}_i; \\ b &{} \quad \text{ otherwise. } \end{array}\right. \end{array} \end{aligned}$$

[The second equality is by Eq. (6).] Then define $\epsilon _i(a):=v_i(a)-u_i(a)$ for all $a\in {\mathcal { A}}$. Thus, $v_i = u_i+\epsilon _i$. By construction, ${\mathbb {E}}[u_i(a)\,|\,v_i(a)]=v_i(a)$, and thus ${\mathbb {E}}[\epsilon _i(a)]=0$, for all $i\in {\mathcal { I}}$ and $a\in {\mathcal { A}}$. Let ${\overline{c}}:=g-b$; then ${\overline{c}}>0$. For all $i\in {\mathcal { I}}$, let ${\widetilde{u}}_i:=u_i/{\overline{c}}$ and ${\widetilde{c}}_i:={\overline{c}}\cdot c_i$; thus, $c_i\, u_i = {\widetilde{c}}_i\,{\widetilde{u}}_i$. Thus, $U_{\mathcal { I}}= \frac{1}{I}\sum _{i\in {\mathcal { I}}} {\widetilde{c}}_i{\widetilde{u}}_i$. For all $i\in {\mathcal { I}}$, let ${\widetilde{v}}_i:=v_i/{\overline{c}}$ and ${\widetilde{\epsilon }}_i := \epsilon _i/{\overline{c}}$; thus, ${\widetilde{v}}_i = {\widetilde{u}}_i+{\widetilde{\epsilon }}_i$. Let ${\widetilde{V}}_{\mathcal { I}}=\sum _{i\in {\mathcal { I}}} {\widetilde{v}}_i$. Then ${\widetilde{V}}_{\mathcal { I}}=V_{\varvec{{\mathcal { G}}}}+$(a constant). Thus, ${{\mathrm{\mathrm{argmax}}}}_{\mathcal { A}}({\widetilde{V}}_{\mathcal { I}})={{\mathrm{\mathrm{argmax}}}}_{\mathcal { A}}(V_{\varvec{{\mathcal { G}}}})$. But ${{\mathrm{\mathrm{argmax}}}}_{\mathcal { A}}(V_{\varvec{{\mathcal { G}}}})=\mathrm{Appr}({\varvec{{\mathcal { G}}}})$; thus, it suffices to compute the asymptotic probability that ${{\mathrm{\mathrm{argmax}}}}_{\mathcal { A}}({\widetilde{V}}_{\mathcal { I}})\subseteq {{\mathrm{\mathrm{argmax}}}}_{\mathcal { A}}(U_{\mathcal { I}})$, using Theorem 6.1. To do this, we must verify hypotheses (U1)–(U4). First, let ${\overline{u}}:={\mathbb {E}}[u^i_a]$ and let $\sigma ^2_u:=\mathrm{var}[u^i_a]$ for any $i\in {\mathcal { I}}$ and $a\in {\mathcal { A}}$. By hypothesis $(\Theta 1)$, these values are finite and independent of i and a. Let $M:={\overline{u}}^2+\sigma _u^2$. $\square$

Claim A

$\displaystyle \lim _{I{\rightarrow }{\infty }} \mathrm{Prob}\quad (M \quad \text {and the profile} \{u_i\}_{i\in {\mathcal { I}}} \ \text {satisfy condition } \; \mathrm{(}U4)) = 1$.

Proof

Fix $a\in {\mathcal { A}}$. For all $i\in {\mathcal { I}}$, we have ${\mathbb {E}}[u_i^2(a)] ={\overline{u}}^2 + \sigma ^2_u = M^2$. Thus, $\frac{1}{I}\sum _{i\in {\mathcal { I}}} u_i(a)^2$ is an average of I independent random variables (by $(\Theta 1)$), each with expected value $M^2$. Thus, the Law of Large Numbers implies that

$$\begin{aligned} \lim _{I{\rightarrow }{\infty }} \mathrm{Prob}\left[\frac{1}{I}\sum _{i\in {\mathcal { I}}} u_i(a)^2 \ < \ M^2\right] = 1. \end{aligned}$$

Thus, since ${\mathcal { A}}$ is finite, the claim follows.$\diamond$

Hypotheses $(\Theta 1)$ and $(\Theta 2)$ imply that $\{{\widetilde{u}}_i\}_{i\in {\mathcal { I}}}$, $\{{\widetilde{v}}_i\}_{i\in {\mathcal { I}}}$ and $\{{\widetilde{\epsilon }}_i\}_{i\in {\mathcal { I}}}$ satisfy (U2). Hypothesis (C) implies that $\{{\widetilde{c}}_i\}_{i\in {\mathcal { I}}}$ satisfies (U1), and hypothesis $(\Delta )$ implies that $\{{\widetilde{u}}_i\}_{i\in {\mathcal { I}}}$ satisfies (U3) (with $\widetilde{\Delta }:=\Delta /{\overline{c}}$). Now apply Claim A and Theorem 6.1 to $\{{\widetilde{u}}_i\}_{i\in {\mathcal { I}}}$, $\{{\widetilde{v}}_i\}_{i\in {\mathcal { I}}}$, $\{{\widetilde{\epsilon }}_i\}_{i\in {\mathcal { I}}}$ and $\{{\widetilde{c}}_i\}_{i\in {\mathcal { I}}}$ to derive the claimed asymptotic probability. $\square$

Proof of Theorem 4.2

The strategy is very similar to the proof of Theorem 4.1. Let g be the mean value of $\gamma$, and let b be the mean value of $\beta$; thus $g>b$. For all $i\in {\mathcal { I}}$, define $v_i:{\mathcal { A}}{{\longrightarrow }}{\mathbb {R}}$ by

$$\begin{aligned} v_i(a) :=\left\{ \begin{array}{ll} g &{} \quad \text{ if } \; a\in {\mathcal { G}}_i;\\ b &{} \quad \text{ if } \; a\in {\mathcal { B}}_i. \end{array}\right. \end{aligned}$$

Then define $\epsilon _i(a):=v_i(a)-u_i(a)$ for all $a\in {\mathcal { A}}$. Thus, $v_i = u_i+\epsilon _i$. By construction, ${\mathbb {E}}[u_i(a)]=v_i(a)$, and thus ${\mathbb {E}}[\epsilon _i(a)]=0$, for all $i\in {\mathcal { I}}$ and $a\in {\mathcal { A}}$. Let ${\overline{c}}:=g-b$; then ${\overline{c}}>0$. For all $i\in {\mathcal { I}}$, let ${\widetilde{u}}_i:=u_i/{\overline{c}}$ and ${\widetilde{c}}_i:={\overline{c}}\cdot c_i$; thus, $c_i\, u_i = {\widetilde{c}}_i\,{\widetilde{u}}_i$. Thus, $U_{\mathcal { I}}= \frac{1}{I}\sum _{i\in {\mathcal { I}}} {\widetilde{c}}_i{\widetilde{u}}_i$. For all $i\in {\mathcal { I}}$, let ${\widetilde{v}}_i:=v_i/{\overline{c}}$ and ${\widetilde{\epsilon }}_i := \epsilon _i/{\overline{c}}$; thus, ${\widetilde{v}}_i = {\widetilde{u}}_i+{\widetilde{\epsilon }}_i$. Let ${\widetilde{V}}_{\mathcal { I}}=\sum _{i\in {\mathcal { I}}} {\widetilde{v}}_i$. Then ${\widetilde{V}}_{\mathcal { I}}=V_{\varvec{{\mathcal { G}}}}+$(a constant). Thus, ${{\mathrm{\mathrm{argmax}}}}_{\mathcal { A}}({\widetilde{V}}_{\mathcal { I}})={{\mathrm{\mathrm{argmax}}}}_{\mathcal { A}}(V_{\varvec{{\mathcal { G}}}})$. But ${{\mathrm{\mathrm{argmax}}}}_{\mathcal { A}}(V_{\varvec{{\mathcal { G}}}})=\mathrm{Appr}({\varvec{{\mathcal { G}}}})$; thus, it suffices to compute the asymptotic probability that ${{\mathrm{\mathrm{argmax}}}}_{\mathcal { A}}({\widetilde{V}}_{\mathcal { I}})\subseteq {{\mathrm{\mathrm{argmax}}}}_{\mathcal { A}}(U_{\mathcal { I}})$, using Theorem 6.1. $\square$

Let $M_g:=g^2 + \mathrm{var}[\gamma ]$ and $M_b:=b^2 + \mathrm{var}[\beta ]$. Let $M:=\sqrt{\max \{M_g,M_b\}}$ and let ${\widetilde{M}}:=M/{\overline{c}}$.

Claim B

$\displaystyle \lim \limits _{I{\rightarrow }{\infty }} \mathrm{Prob}({\widetilde{M}}\; \text { and the profile } \; \{{\widetilde{u}}_i\}_{i\in {\mathcal { I}}} \; \text { satisfy condition }\; \mathrm{(U4)}) = 1$.

Proof

Fix $a\in {\mathcal { A}}$. For all $i\in {\mathcal { I}}$, if $a\in {\mathcal { G}}_i$, then ${\mathbb {E}}[u_i^2(a)] =M_g$. If $a\in {\mathcal { B}}_i$, then ${\mathbb {E}}[u_i^2(a)] =M_b$. Either way, ${\mathbb {E}}[u_i^2(a)] \le M^2$. Thus, $\frac{1}{I}\sum _{i\in {\mathcal { I}}} u_i(a)^2$ is an average of I independent random variables (by (S)), each with expected value $M^2$. Thus, the Law of Large Numbers implies that

$$\begin{aligned} \lim _{I{\rightarrow }{\infty }} \mathrm{Prob}\left[\frac{1}{I}\sum _{i\in {\mathcal { I}}} u_i(a)^2 \ < \ M^2\right] = 1. \end{aligned}$$

Since ${\widetilde{u}}_i:=u_i/{\overline{c}}$ for all $i\in {\mathcal { I}}$, it follows that

$$\begin{aligned} \lim _{I{\rightarrow }{\infty }} \mathrm{Prob}\left[\frac{1}{I}\sum _{i\in {\mathcal { I}}} {\widetilde{u}}_i(a)^2 \ < \ {\widetilde{M}}^2\right] = 1. \end{aligned}$$

Thus, since ${\mathcal { A}}$ is finite, the claim follows.$\diamond$

Hypothesis (C) implies that $\{{\widetilde{c}}_i\}_{i\in {\mathcal { I}}}$ satisfies (U1). Define $\sigma ^2_\epsilon :=\max \{\mathrm{var}(\gamma ),\mathrm{var}(\beta \}/{\overline{c}}^2$. Since ${\widetilde{\epsilon }}_i={\widetilde{u}}_i-{\widetilde{v}}_i$, it follows that $\mathrm{var}({\widetilde{\epsilon }}_i)\le \sigma ^2_\epsilon$ for all i. Thus, hypothesis (S) implies that $\{{\widetilde{u}}_i\}_{i\in {\mathcal { I}}}$, $\{{\widetilde{v}}_i\}_{i\in {\mathcal { I}}}$ and $\{{\widetilde{\epsilon }}_i\}_{i\in {\mathcal { I}}}$ satisfy (U2). Finally, hypothesis $(\Delta )$ implies that $\{{\widetilde{u}}_i\}_{i\in {\mathcal { I}}}$ satisfy (U3) (with $\widetilde{\Delta }:=\Delta /{\overline{c}}$). Now apply Claim B and Theorem 6.1 to $\{{\widetilde{u}}_i\}_{i\in {\mathcal { I}}}$, $\{{\widetilde{v}}_i\}_{i\in {\mathcal { I}}}$, $\{{\widetilde{\epsilon }}_i\}_{i\in {\mathcal { I}}}$ and $\{{\widetilde{c}}_i\}_{i\in {\mathcal { I}}}$ to derive the claimed asymptotic probability. $\square$

Proof of Proposition 4.3

Define $\{{\widetilde{u}}_i\}_{i\in {\mathcal { I}}}$, $\{{\widetilde{v}}_i\}_{i\in {\mathcal { I}}}$, $\{{\widetilde{\epsilon }}_i\}_{i\in {\mathcal { I}}}$ and ${\widetilde{V}}_{\mathcal { I}}$ as in the proof of Theorem 4.2. From Theorem 6.2, along with Claim B, we immediately obtain the limit Eq. (8). However, to obtain more precise estimates of the convergence speed, we must first estimate the speed of the convergence in Claim B, using the next result. $\square$

Claim C

Suppose the fourth moments of $\gamma$ and $\beta$ are finite. Then there is some $C_1>0$ (determined by $\gamma$ and $\beta$) such that, for any $p\in (0,1)$, if $I>{C_1}/{p}$, then

$$\begin{aligned} \mathrm{Prob}\left( {\widetilde{M}}\;\text { and }\; \{{\widetilde{u}}_i\}_{i\in {\mathcal { I}}} \;\text {violate condition \; (U4) }\right) < \frac{p}{2}. \end{aligned}$$

The proof is very similar to the proof of Claim E in the proof of Proposition 5.3 (below).

Recall that $\{{\widetilde{u}}_i\}_{i\in {\mathcal { I}}}$, $\{{\widetilde{v}}_i\}_{i\in {\mathcal { I}}}$ and $\{{\widetilde{\epsilon }}_i\}_{i\in {\mathcal { I}}}$ satisfy (U2), with $\sigma ^2_\epsilon :=\max \{\mathrm{var}(\gamma ),\mathrm{var}(\beta \}/{\overline{c}}^2$. For any $\delta >0$ and $p\in (0,1)$, define ${\overline{I}}(\delta ,p)$ as in equation (11). Finally, define $C_2:=8\,|{\mathcal { A}}|\,({\widetilde{M}}^2\, \sigma _c^2+ \sigma _\epsilon ^2)$. Thus, for any $p,\delta \in (0,1)$, if $I>C_2/p\,\delta ^2$, then $I>{\overline{I}}(\delta ,p/2)$, so that, for any $a\in \mathrm{Appr}({\varvec{{\mathcal { G}}}})={{\mathrm{\mathrm{argmax}}}}_{\mathcal { A}}({\widetilde{V}}_{\mathcal { I}})$, Theorem 6.2 says

$$\begin{aligned} \mathrm{Prob}\left[ U_{\mathcal { I}}(a)< U^*_{\mathcal { I}}-\delta \left|\right. {\widetilde{M}}\; \text{ and } \; \{{\widetilde{u}}_i\}_{i\in {\mathcal { I}}} \; \text { satisfy (U4)}\right] < \frac{p }{2}. \end{aligned}$$

(12)

If $I>C_1/p$ also, then Claim C applies. This, together with inequality (12), implies that $\mathrm{Prob}\left[ U_{\mathcal { I}}(a)< U^*_{\mathcal { I}}-\delta \right] < \frac{p}{2}+\frac{p}{2} = p$, as desired.

Theorem 5.1 follows from Theorem 5.2, so we will prove that first.

Proof of Theorem 5.2

Since $\mathrm{Score}_{\mathbf { s}}({\mathcal { P}}_{\mathcal { I}})={{\mathrm{\mathrm{argmax}}}}_{\mathcal { A}}(V^{\mathbf { s}}_{{\mathcal { P}}_{\mathcal { I}}})$, it suffices to compute the asymptotic probability that ${{\mathrm{\mathrm{argmax}}}}_{\mathcal { A}}(V^{\mathbf { s}}_{{\mathcal { P}}_{\mathcal { I}}})\subseteq {{\mathrm{\mathrm{argmax}}}}_{\mathcal { A}}(U_{\mathcal { I}})$, as $I{\rightarrow }{\infty }$. As usual, we will use Theorem 6.1. Hypothesis (C) implies (U1). For all $i\in {\mathcal { I}}$ and all $a\in {\mathcal { A}}$, if we know that i ranks a in kth place (in particular, if we know the preference order $\succ _i$), then the expected value of $u_i(a)$, conditional on this information, is $s_k$. But, by definition, $v_i(a)=s_k$. Thus, ${\mathbb {E}}[ u_i(a)|\succ _i]=v_i(a)$. Thus, if we define $\epsilon _i(a):=u_i(a)-v_i(a)$, then ${\mathbb {E}}[ \epsilon _i(a)|\succ _i]=0$. By hypothesis, the variance of the random variable $u_i(a)$ is finite; thus, the variance of $\epsilon _i(a)$ is finite. Finally, by hypothesis (X), the random functions $\{u_i\}_{i\in {\mathcal { I}}}$ are independent of one another and independent of $\{c_i\}_{i\in {\mathcal { I}}}$. Thus, the random functions $\{\epsilon _i\}_{i\in {\mathcal { I}}}$ are independent of one another and independent of $\{c_i\}_{i\in {\mathcal { I}}}$. This establishes (U2). It remains to verify (U4).

Let ${}^{\scriptscriptstyle \uparrow }\!{\mathbf { u}}=({}^{\scriptscriptstyle \uparrow }\!u_1,\ldots ,{}^{\scriptscriptstyle \uparrow }\!u_N)\in {\mathbb {U}}$ be a $\lambda$-random vector. The coordinates ${}^{\scriptscriptstyle \uparrow }\!u_1,\ldots ,{}^{\scriptscriptstyle \uparrow }\!u_N$ are themselves random variables (neither independent, nor identically distributed). Let $\sigma ^2_1,\ldots ,\sigma ^2_N$ denote their variances. Since $\lambda$ has finite variance, it is easy to check that $\sigma ^2_1,\ldots ,\sigma ^2_N$ are all finite. Define $\sigma ^2_\epsilon :=\max \{\sigma ^2_{1},\sigma ^2_{2},\ldots ,\sigma ^2_{N}\}$. Also, let $S:=\max \{|s_1|$, $|s_2|$, $\ldots ,$ $|s_N|\}$, and choose any $M>\sqrt{S^2 + \sigma ^2_\epsilon }$. $\square$

Claim D

$\displaystyle \lim _{I{\rightarrow }{\infty }} \mathrm{Prob}(M \; \text { and the profile }\; \{u_i\}_{i\in {\mathcal { I}}} \; \text { satisfy condition (U4)}) = 1$.

Proof

Fix $a\in {\mathcal { A}}$. For all $i\in {\mathcal { I}}$, if a is ranked kth from the bottom by $\succ _i$, then $u_i(a)$ is a random variable with mean $s_k$ and variance $\sigma ^2_k$. Thus,

$$\begin{aligned} {\mathbb {E}}[u_i^2(a)] = (s_k)^2 + \sigma ^2_k \le S^2 + \sigma ^2_\epsilon < M^2. \end{aligned}$$

(13)

Thus, for any $a\in {\mathcal { A}}$, the sum $\frac{1}{I}\sum \nolimits _{i\in {\mathcal { I}}} u_i(a)^2$ is an average of I independent random variables, each with expected value smaller than $M^2$, by inequality (13). Thus, regardless of how the preferences $\{\succ _i\}_{i\in {\mathcal { I}}}$ are obtained, the Law of Large Numbers implies that

$$\begin{aligned} \lim _{I{\rightarrow }{\infty }} \mathrm{Prob}\left[\frac{1}{I}\sum _{i\in {\mathcal { I}}} u_i(a)^2 \ < \ M^2\right] = 1. \end{aligned}$$

Thus, since ${\mathcal { A}}$ is finite, the claim follows.$\diamond$

Finally, hypothesis $(\Delta )$ implies that $\{u_i\}_{i\in {\mathcal { I}}}$ satisfy (U3). Now apply Claim 1 and Theorem 6.1 to $\{u_i\}_{i\in {\mathcal { I}}}$, $\{v_i\}_{i\in {\mathcal { I}}}$, $\{\epsilon _i\}_{i\in {\mathcal { I}}}$ and $\{c_i\}_{i\in {\mathcal { I}}}$ to derive the limit (9). $\square$

Proof of Theorem 5.1

Define ${\mathbb {U}}$ as in Sect. 5.2, and let $\lambda$ be the conditionalization of $\mu$ on ${\mathbb {U}}$. In the Endogenous Preference model of Sect. 5.1, the ordinal preference profile ${\mathcal { P}}_{\mathcal { I}}=\{\succ _i\}_{i\in {\mathcal { I}}}$ is a random variable (determined by the underlying cardinal utility profile $\{{\mathbf { u}}^i\}_{i\in {\mathcal { I}}}$. (${\mathcal { P}}_{\mathcal { I}}$ is almost surely a profile of strict preferences, because by hypothesis on $\mu$, no two alternatives yield the same utility for any voter, almost surely.) However, if we fix a particular realization of ${\mathcal { P}}_{\mathcal { I}}$, then conditional on this realization, the probability distribution of the cardinal profile $\{{\mathbf { u}}^i\}_{i\in {\mathcal { I}}}$ is described by $\lambda$ via the Exogenous preference model of Sect. 5.2. Thus, for any particular realization of ${\mathcal { P}}_{\mathcal { I}}$, Theorem 5.2 implies that the limit (9) holds.¹⁷ Thus, integrating over all possible realizations of ${\mathcal { P}}_{\mathcal { I}}$, and applying Lebesgue’s dominated convergence theorem, we conclude that the limit (9) holds unconditionally.¹⁸ $\square$

Proof of Proposition 5.3

We will apply Theorem 6.2. Define $\{\epsilon _i\}_{i\in {\mathcal { I}}}$ and M as in the proof of Theorem 5.2. Claim 1 established that $\lim _{I{\rightarrow }{\infty }} \mathrm{Prob}[M$ and the profile $\{u_i\}_{i\in {\mathcal { I}}}$ satisfy condition (U4)]$\,{=}\,$1. However, to obtain the more precise estimate of convergence speed, we need the next observation.

Claim E

Suppose the fourth moment of $\lambda$ is finite. Then there is some $C_1>0$ (determined by $\lambda$) such that, for any $p\in (0,1)$, if $I>{C_1}/{p}$, then

$$\begin{aligned} \displaystyle \mathrm{Prob}\left( M \;\text { and } \; \{u_i\}_{i\in {\mathcal { I}}} \; \text { violate condition (U4)}\right) < \frac{p}{2}. \end{aligned}$$

Proof

If the fourth moment of $\lambda$ is finite, then there is some $C'>0$ such that for any $a\in {\mathcal { A}}$, the fourth moments of each of the random variables $\{u_i(a)\}_{i\in {\mathcal { I}}}$ is less than $C'$. In other words, the second moments of each of the random variables $\{u_i(a)^2\}_{i\in {\mathcal { I}}}$ is less than $C'$. This implies that there is some $C''>0$ such that the variance of each of $\{u_i(a)^2\}_{i\in {\mathcal { I}}}$ is less than $C''$. Also, these random variables are independent. Thus,

$$\begin{aligned} \mathrm{var}\left[\frac{1}{I}\sum _{i\in {\mathcal { I}}} u_i(a)^2\right]<\frac{C''}{I}. \end{aligned}$$

(14)

Next, inequality (13) says each of $\{u_i(a)^2\}_{i\in {\mathcal { I}}}$ has expected value less than $M^2$. Thus,

$$\begin{aligned} {\mathbb {E}}\left[\frac{1}{I}\sum _{i\in {\mathcal { I}}} u_i(a)^2\right] < M^2. \end{aligned}$$

(15)

Thus, Chebyshev’s inequality and inequalities (14) and (15) imply that there is some $C_1>0$ (determined by $C''$) such that, for any $p>0$, if $I>C_1/p$, then

$$\begin{aligned} \mathrm{Prob}\left[\frac{1}{I}\sum _{i\in {\mathcal { I}}} u_i(a)^2 > M^2\right] < \frac{p}{2|{\mathcal { A}}|}. \end{aligned}$$

(16)

Now, if the profile $\{u_i\}_{i\in {\mathcal { I}}}$ and M to violate condition (U4), then $\frac{1}{I}\sum _{i\in {\mathcal { I}}} u_i(a)^2 > M^2$ for some $a\in {\mathcal { A}}$. Thus, adding together $|{\mathcal { A}}|$ copies of inequality (16) proves the claim. $\diamond$

For any $\delta >0$ and $p>0$, let ${\overline{I}}(\delta ,p)$ be as in Eq. (11). Finally, define $C_2:=8\,|{\mathcal { A}}|\,(M^2\, \sigma _c^2+ \sigma _\epsilon ^2)$. Thus, for any $p,\delta \in (0,1)$, if $I>C_2/p\,\delta ^2$, then $I>{\overline{I}}(\delta ,p/2)$, so that, for all $a\in {{\mathrm{\mathrm{argmax}}}}_{\mathcal { A}}({\widetilde{V}}_{\mathcal { I}})$, Theorem 6.2 says that

$$\begin{aligned} \mathrm{Prob}\left[ U_{\mathcal { I}}(a)< U^*_{\mathcal { I}}-\delta \left|\right. M \; \text{ and } \; \{u_i\}_{i\in {\mathcal { I}}} \; \text { satisfy (U4)}\;\right]< \frac{p }{2}. \end{aligned}$$

(17)

If $I>C_1/p$ also, then Claim E applies. This, together with inequality (17), implies that $\mathrm{Prob}\left[ U_{\mathcal { I}}(a)< U^*_{\mathcal { I}}-\delta \right] < \frac{p}{2}+\frac{p}{2} = p$ for any $a\in {{\mathrm{\mathrm{argmax}}}}_{\mathcal { A}}({\widetilde{V}}_{\mathcal { I}})=\mathrm{Score}_{\mathbf { s}}({\mathcal { P}}_{\mathcal { I}})$, as desired.$\square$