Targeted Advertising in Elections

Abstract

KEYWORDS: Persuasion, verifiable information, targeted advertising, elections

JEL CLASSIFICATION: D72, D82, D83

1. Introduction

Targeted advertising, broadly defined as private messaging aimed at specific groups of voters, played a key role in many successful electoral campaigns. In 1960, John F. Kennedy’s campaign distributed two million copies of “the blue bomb”—a pamphlet advertising his support of civil rights—to African American churches across the U.S. Decades later, George W. Bush’s 2004 reelection campaign used direct mail to communicate his opposition to gay marriage and support for “traditional family values” to evangelical Christian households. More recently, the 2016 Brexit referendum and the Trump presidential campaign both employed the services of Cambridge Analytica, a data mining firm, to design and distribute thousands of targeted ads to diverse audiences. Although these examples suggest broad awareness of a correlation between targeted advertising and electoral success, the precise mechanisms by which tailoring messages to different voters influences election outcomes remain poorly understood.

This paper proposes a simple theoretical model that fills this gap. The model is grounded in three stylized facts about electoral campaigns. First, voters have incomplete information and update their beliefs in response to campaign messages (Kendall, Nannicini, and Trebbi, 2015; Spenkuch and Toniatti, 2018; Le Pennec and Pons, 2023). Second, politicians use the strategy of ambiguity to avoid making precise statements about their positions on issues (Page, 1978; Druckman, Kifer, and Parkin, 2009; Fowler et al., 2021). Third, politicians tailor messages to specific groups of voters (Hillygus and Shields, 2014). I identify a novel mechanism by which targeted advertising changes electoral outcomes. In particular, I show that privately revealing different pieces of information to different voters allows politicians to persuade voters with diametrically opposing preferences and win otherwise unwinnable elections, in which such voters are pivotal.

• I. If the median voter's bliss point is left or right of the status quo, then the election is winnable with public advertising.
• II. If the median voter's bliss point is at the status quo, but status quo voters do not form a majority, then the election is unwinable with public advertising but winnable with targeted advertising.
• III. If the median voter's bliss point is at the status quo and status quo voters do form a majority, then the election is unwinable with public or targeted advertising.

The second part of the paper focuses on the optimal targeted advertising strategy that maximizes the challenger's (ex-ante) odds of winning elections that are unwinable with public advertising. I make a further simplification that all left voters have the same bliss point $L < 0$ and all right voters have the same bliss point $R > 0$. Below I use a motivating example in which $L = -0.2$, $R = 0.4$, the voters' preferences are quadratic, the minimal decisive coalition includes left and right voters, and $x \sim U[-1, 1]$, to illustrate the following two results: Proposition 2 identifies the optimal targeted advertising strategy, while Proposition 3 describes the comparative statics as the right voters become more extreme/the electorate becomes more polarized.

To see how this challenger-preferred equilibrium outcome changes as right voters become more extreme, suppose the right voters' bliss point increases from $R = 0.4$ to $R' = 0.5$. Following the same logic as above, we find that the convincing messages are $[-0.4, 0.2]$ for left voters and $[-0.5, 1]$ for right voters. The challenger wins when his policy outcome is between $-0.4$ and $0.2$, exactly as before. However, the challenger's equilibrium odds of winning may be even higher. Specifically, observe that when the challenger's policy outcome is between $-0.5$ and $-0.4$, the strategy described above convinces right but not left voters. However, left voters actually prefer policy outcomes in $[-0.5, -0.4]$ to those in $[0.1, 0.2]$ as the former are closer to their bliss point. Hence, we can recalculate the message that convinces left voters (making them indifferent between approval and rejection), forcing it to start at $-0.5$. That message is $[-0.5, 0.179]$. Figure 2 illustrates the electoral outcome after right voters become more extreme.

In the new equilibrium, the set of winning policy outcomes is $[-0.5, 0.179]$ and the challenger's odds of winning are 33.96%. Proposition 3 confirms that when right voters become more extreme, the set of winning policy outcomes shifts in the opposite direction, to the left; also, the challenger's odds of winning increase. Intuitively, when right voters become more extreme, their dissatisfaction with the status quo grows, which makes them persuadable by wider ranges of policy outcomes.

My findings suggest a novel explanation for why politicians use the strategy of ambiguity: advertising different ranges of policy outcomes to different voters allows politicians to persuade voters with diametrically opposing preferences without lying (by commission) to any of them. Previous explanations for why politicians use the strategy of ambiguity include voters' risk-seeking behavior (Shepsle, 1972), candidates' preference for ambiguity (Aragonès and Neeman, 2000), subsequent elections (Meirowitz, 2005, Alesina and Holden, 2008), resolution of uncertainty after an election (Kartik, Van Weelden, and Walton, 2017). Two previous papers find that ambiguity enables politicians to persuade voters with opposing preferences: in Callander and Wilson (2008), voters have context-dependent preferences, and in Tolvanen (2024), the voters' preferences are correlated with the state of the world. I reach a similar conclusion in a setting where voters have standard single-peaked and single-crossing preferences.

My analysis suggests that targeted advertising is bad for democracy because it elects politicians who are guaranteed to lose when voters possess the same information. For example, under simple majority, targeted advertising allows the challenger to beat the status quo located at the median voter's bliss point, which, according to various versions of the median voter theorem, is unbeatable. The most effective policy to make targeted advertising obsolete is to publicize all ads transmitted during electoral campaigns. While voters may still make mistakes due to incomplete (but public) information, having a common belief would be sufficient for them to collectively make the right choice.

2. Model

I assume that $\alpha_v(x)$ is continuous, measurable, bounded, and satisfies two properties standard in spatial voting models:

ASSUMPTION 1. Voters' preferences are single-peaked (strictly quasiconcave): for each voter $v \in V$, her net payoff from approval $\alpha_v(x)$ is strictly increasing on $[-1, v]$ and strictly decreasing on $[v, 1]$.

One relevant consequence of Assumption 1 is that voter $v$ 's upper contour sets are convex. In particular, the approval set $A_v$ is an interval $[l, 0]$ for left voters with $v < 0$, a point $\{0\}$ for status quo voters with $v = 0$, and an interval $[0, r]$ for right voters with $v > 0$.

ASSUMPTION 2. Voters' preferences are single-crossing: for each belief $\mu \in \Delta X \setminus \delta_0$,

and

DEFINITION 1. A left voter $w$ is more extreme than a left voter $v$ if $w < v < 0$. A right voter $w$ is more extreme than a right voter $v$ if $0 < v < w$.

I illustrate my results for quadratic voter preferences, $\alpha_v(x) = -(v-x)^2 + (v-0)^2 = -x^2 + 2vx$, which is one example of standard preferences that satisfy all the assumptions.

Figure 3 illustrates the preferences of a quadratic voter $v < 0$.

I say that a set of policy outcomes $W \subseteq X$ is implementable if there exists an equilibrium in which the challenger's interim utility is $U_I(x) = \mathbb{1}(x \in W)$. In that equilibrium, the challenger wins if and only if $x \in W$ (i.e., $W$ is the set of “winning” policy outcomes) and his odds of winning are $\mu_0(W)$.

Most of my constructive results involve equilibria in which the challenger uses a simple pure strategy of revealing whether his policy outcome is or is not in some set. Let $M := (M_v)_{v \in V}$ be a collection of subsets of $X$. A direct strategy $\sigma_M$ is defined as $\sigma_M(x) := (M_v \text{ if } x \in M_v, \text{ otherwise } M_v^c)_{v \in V}$. In words, when the challenger's policy outcome is $x$, he tells voter $v$ whether $x \in M_v$ (by sending message $M_v$) or not (by sending message $M_v^c$). A direct public strategy $\sigma_M$ is one with $M_v = M$ for all $v \in V$. When the challenger uses a direct strategy, voter $v$ hears one of two on-path messages, $m \in \{M_v, M_v^c\}$, and thus learns whether $x \in m$ and nothing else; her posterior belief is then $\mu_0(\cdot \mid m)$. I say that voter $v$ is willing to approve a set of policy outcomes $M_v$ (if the challenger uses a direct strategy $\sigma_M$ and $v$ receives message $M_v$) if it satisfies her obedience constraint:

3. Analysis

To begin analysis, let us first classify elections in the following way. I say that coalition $D \subseteq V$ of voters is left (right), denoted $D < 0$ ($D > 0$), if it consists of left (right) voters only; I say that $D$ is a mixed coalition if it has both left and right voters.

DEFINITION 2. An election $(V, \mathcal{D})$ is a

• left- (right-) leaning election if there exists a left (right) decisive coalition;

• status quo-leaning election if every decisive coalition includes a status quo voter;

• polarized election if there exists a mixed decisive coalition and there are no left or right decisive coalitions.

Given a coalition of voters $D \subseteq V$, let $\lambda(D) := \max_{v \in D, v < 0} v$ and $\rho(D) := \min_{v \in D, v > 0} v$ be its left pivot and right pivot, if they exist, respectively. Let us define a representative voter as the voter whose approval set equals the union of the approval sets of all decisive coalitions.

DEFINITION 3. The representative voter of election $(V, \mathcal{D})$ is the voter $v^* \in X$ such that $A_{v^*} = \bigcup_{D \in \mathcal{D}} \bigcap_{v \in D} A_v$.

Voter $v^*$ essentially represents the preferences of the entire electorate because, under complete information, if she prefers to approve, then so does at least one decisive coalition, and if she prefers to reject, then so do all decisive coalitions. The lemma below determines the location of the representative voter in each election type, allowing us to state subsequent results for the public advertising game in terms of a single voter rather than sets of decisive coalitions.

LEMMA 1. Consider an election $(V, \mathcal{D})$. Then,

Proof. Recall that from Assumption 1, the approval set of voter $v \in V$ is $[[A_v], 0]$ if $v < 0$, $A_v = \{0\}$ if $v = 0$, and $A_v = [0, [A_v]]$ if $v > 0$. Therefore, for any decisive coalition $D \in \mathcal{D}$ that is mixed or includes a status quo, we have $\bigcap_{v \in D} A_v = \{0\}$. Consequently, for status quo-leaning and polarized elections, we have $v^* = 0$.

Next, observe that from (SC2), if voter $v$ is less extreme than voter $w$, then $A_v \subseteq A_w$. Then, for any left coalition $D < 0$, we have $\bigcap_{v \in D} A_v = \bigcap_{v \in D} [[A_v], 0] = [[A_{\lambda(D)}], 0] = A_{\lambda(D)}$, because $\lambda(D)$ is the least extreme voter in $D$. Therefore, for any left-leaning election $(V, \mathcal{D})$, we have

where $v^* = \min_{D \in \mathcal{D}, D < 0} \lambda(D) < 0$. The proof is analogous for a right-leaning election. ■

Public Advertising

In the public advertising game, the voters' common prior belief is updated to a common posterior. Therefore, the electorate faces a collective choice problem: whether to choose a safe option (the status quo) or a lottery over the challenger's policy outcomes (represented by their common posterior belief $\mu \in \Delta X$). The first result describes which elections are "unwinnable" for the challenger with public advertising.

THEOREM 1. In the public advertising game,

1. The challenger's odds of winning are zero in every equilibrium if and only if the election is status quo-leaning or polarized, i.e., there is no left or right decisive coalition.
2. A set of policy outcomes is implementable in a left/right-leaning election if it is obedient for the representative voter and includes her approval set.

As the equilibrium constructions from the proofs of Theorems 1 and 2 are used later in Section 4, I provide the proofs of these theorems in the main text.

Proof. I prove sufficiency of part 1 directly. If there is no left or right decisive coalition, then the election is either status quo-leaning or polarized. In both cases, the challenger convinces a decisive coalition only if the public belief is $\delta_0$, or whenever $x = 0$, which has zero prior measure. To see this, observe that for any belief other than $\delta_0$, status quo voters strictly prefer to reject; also, left and right voters never prefer to approve at the same time by SC1. Therefore, the challenger's odds of winning are zero in every equilibrium of the public advertising game unless there is a left or right decisive coalition.

To prove the rest of the theorem, we consider (without loss) a left-leaning election with representative voter $L < 0$ and construct an equilibrium that implements $M \subseteq X$, where $A_L \subseteq M$ and $\int_M \alpha_L(x) d\mu_0(x) \geq 0$. Note that such $M$ exists—for example, we can let $M = A_L = [[A_L], 0]$.

Suppose that the challenger uses the direct public strategy $\sigma_M$. When message $M$ is heard on the path, the public posterior becomes $\mu_0(\cdot \mid M)$; the representative voter prefers to approve because $M$ satisfies her obedience constraint. Since $L$ is the least extreme voter in some left decisive coalition $D \in \mathcal{D}$, by (SC2) we have $\int_M \alpha_v(x) d\mu_0(x) \geq 0$ for all $v \in D$, which means that the entire decisive coalition prefers to approve after message $M$.

We now show that when the public posterior $\mu$ is supported on $M^c$ (for instance, after message $M^c$), the challenger loses the election because every decisive coalition includes a voter who strictly prefers to reject. First, voter $L$ strictly prefers to reject. Indeed, since $A_L \subseteq M$, we have $\alpha_L(x) < 0$ for all $x \in M^c$, and therefore $\mathbb{E}_\mu[\alpha_L(x)] < 0$. Next, all left voters less extreme than $L$ also strictly prefer to reject: by the contrapositive of (SC2), $\mathbb{E}_\mu[\alpha_L(x)] < 0$ implies $\mathbb{E}_\mu[\alpha_v(x)] < 0$ for all $v \in (L, 0)$. In the left-leaning election under consideration, every decisive coalition is left, mixed, or includes a status quo voter. Every left decisive coalition includes a voter $v \in [L, 0)$ by Lemma 1. Every other decisive coalition contains a voter who strictly prefers to reject by the argument in the first paragraph of this proof. Consequently, the challenger loses whenever the public belief is supported on $M^c$ (in particular, when message $M^c$ is heard on the path). Therefore, if the challenger uses strategy $\sigma_M$, he wins if and only if $x \in M$, his interim utility is $U_I(x) = \mathbb{1}(x \in M)$, and his odds of winning are $\mu_0(M) > 0$.

Finally, we specify the voters' off-path beliefs and show that the challenger does not have profitable deviations. When $x \in M$, the challenger receives the highest possible payoff and, therefore, has no profitable deviations. We thus do not restrict voters' beliefs for off-path messages $m \subset M$. Conversely, when the challenger's policy outcome is not in $M$, his on-path payoff is zero; a profitable deviation would require sending a message that results in a win. To deter such deviations, we require voters to hold skeptical beliefs. Specifically, for any off-path message $m \not\subseteq M$, we restrict the voters' posterior to be an element of $\Delta(m \cap M^c)$. This ensures that any message that the challenger can send when $x \notin M$ induces a public posterior supported on $M^c$, after which the challenger loses the election, as shown in the previous paragraph. This completes the equilibrium characterization and the proof. $\blacksquare$

Theorem 1 essentially states that an election is winnable with public advertising if and only if the representative voter is left or right. Furthermore, in winnable elections, the challenger effectively caters to the representative voter: a set of “winning” policy outcomes is implementable if it contains that voter’s approval set and satisfies her obedience constraint. The former condition ensures that the representative voter (and thus some voter in every decisive coalition) strictly prefers to reject after every message available to the challenger when $x \notin M$. The latter condition guarantees that the representative voter (and thus every voter in some decisive coalition) approves after message $M$. In Section 4, we will characterize the largest implementable set of winning policy outcomes for a left-leaning election in which all voters’ bliss points are the same.

Under simple majority, we arrive at a familiar characterization of elections that are unwinnable with public advertising.

COROLLARY 1. Under simple majority, the challenger’s odds of winning are zero in every equilibrium of the public advertising game if and only if the status quo is the median voter’s bliss point.

Targeted Advertising

Some elections are unwinnable for the challenger with public advertising because the status quo beats any lottery over the challenger’s policy outcomes. Targeted advertising allows the challenger to induce different beliefs among different voters and win some of these elections. The next result describes which elections are “unwinnable” for the challenger with targeted advertising.

THEOREM 2. In the targeted advertising game,

1. The challenger's odds of winning are zero in every equilibrium if and only if the election is status quo-leaning, i.e., every decisive coalition includes a status quo voter.
2. An interval $[a, b]$ such that $a < 0 < b$ is implementable in a polarized election with a mixed decisive coalition $D \in \mathcal{D}$ if the set $[a, b] \cup A_v$ is obedient for each pivot $v \in \{\lambda(D), \rho(D)\}$.

Proof. For sufficiency of part 1, recall that status quo voters always reject. I prove necessity of part 1 by contraposition. If not every decisive coalition includes a status quo voter, then two cases are possible: the election is left- (right-) leaning, or it is polarized. In the first case, the public advertising equilibria described in the proof of Theorem 1 are equilibria of the targeted advertising game. In the remainder of this proof, we focus on a polarized election with a mixed decisive coalition $D \in \mathcal{D}$; for brevity, we denote its pivots by $L := \lambda(D)$ and $R := \rho(D)$. We will show that an interval described in part 2 of Theorem 2 exists and construct an equilibrium that implements it.

Skeptical off-path beliefs. For any off-path message $m \notin \{M_v, M_v^c\}$, let voter $v$ 's posterior be an element of $\Delta(m \cap A_v^c)$ whenever $m \cap A_v^c$ is non-empty. That way, when a voter $v \in V$ hears an off-path message that includes any policy outcome outside of her approval set, she rejects.

Electoral outcome. The challenger convinces every voter in the decisive coalition $D$ if $x \in \bigcap_{v \in D} M_v = [a, b]$. In the polarized election under consideration, every decisive coalition is mixed or includes a status quo voter. Since left (right) voter never approve after a message including policy outcomes right of $b$ (left of $a$), and status quo voters always reject, the challenger loses if $x \notin [a, b]$. His interim utility is thus $U_I(x) = \mathbb{1}(x \in [a, b])$, and his odds of winning are $\mu_0([a, b]) > 0$.

No profitable deviations. If $x \in [a, b]$, the challenger receives the highest possible payoff and, therefore, has no profitable deviations. If $x < a < 0$, then any verifiable message (i.e., a message that includes $x$) convinces every right voter to reject. Since every decisive coalition includes a right voter or a status quo voter, the challenger has no profitable deviations. A similar argument applies when $x > b > 0$. This completes the equilibrium characterization and the proof. $\blacksquare$

Theorem 2 describes implementable sets of winning policy outcomes in the targeted advertising game for polarized elections. In the proof, I show how to implement these outcomes in an equilibrium where the challenger uses a direct strategy that effectively caters to the left and right pivots, $L$ and $R$, of some mixed decisive coalition. More precisely, an interval $[a, b]$ is implementable if the set $A_v \cup [a, b]$ is obedient for pivots $L$ and $R$ —that is, the lower bound $a < 0$ cannot be too small, and the upper bound $b > 0$ cannot be too large. In Section 4, we will characterize the largest implementable interval of winning policy outcomes in a polarized election in which all left (right) voters share the same bliss point.

Theorems 1 and 2 describe how the challenger advertises his policy outcome depending on the composition of the electorate. If every decisive coalition includes a status quo voter, he loses with public and targeted advertising. If the election is left- (right-) leaning, he wins by advertising publicly and tailoring his strategy to the representative voter. Finally, if the election is polarized, then the challenger can win with targeted advertising but not public advertising. Under simple majority, targeted advertising allows the challenger to defeat the status quo policy that the median voter theorems deem unbeatable.

4. Baseline Election

While Theorems 1 and 2 characterize which elections are winnable with public disclosure and targeted advertising, they do not make a unique prediction about how the challenger wins these elections. Specifically, the proof of each theorem involves providing an example of an equilibrium in which the challenger's odds of winning are positive. The reason for this is that the model admits multiple equilibria. There are two sources of multiplicity. First, there may be multiple decisive coalitions. Second, the verifiable disclosure game has a range of equilibrium outcomes even if there is only one receiver (Titova and Zhang, 2025). To move forward in the analysis, I consider a class of baseline elections in which the minimal decisive coalition is unique. I focus on the challenger-preferred equilibrium in order to provide the upper bound on the challenger's odds of winning across all equilibria.

DEFINITION 4. A baseline election has electorate $\{L, 0, R\}$, where $-1 \leq L < 0 < R \leq 1$.

In the baseline election, all left voters have the same bliss point, $L < 0$, and all right voters have the same bliss point, $R > 0$. This assumption limits the number of possible decisive coalitions and allows us to focus on the messages to be sent to left (right) voters, all of whom have the same bliss point.

Given our focus on the challenger-preferred equilibrium, it is useful to first find the highest probability of convincing one voter. Consider the following auxiliary problem with parameters $l$ and $r$ such that $-1 \leq l \leq \lfloor A_v \rfloor < \lceil A_v \rceil \leq r \leq 1$:

Roughly speaking, Problem (AUX) identifies the largest (in terms of prior measure) set of policy outcomes $I$ such that, if voter $v$ learns that $x \in I$ and nothing else (i.e., if the challenger uses a direct strategy $\sigma_I$), she prefers to approve. The objective function is the probability of convincing the voter, while the obedience constraint $\int_I \alpha_v(x) d\mu_0(x) \geq 0$ ensures that the voter prefers to approve given her information. Parametrizing the problem with $l$ and $r$ allows us to focus on certain subsets of $X$. For example, if we are interested in the largest set of right policy outcomes that a left voter $L < 0$ is willing to approve, then we let $l = \lfloor A_L \rfloor$ and $r = 1$.

Problem (AUX) comes from the information design literature and provides a theoretical upper bound on the ex-ante probability of convincing a Bayesian voter (see, e.g., Alonso and Câmara, 2016 and Titova and Zhang, 2025). The solution is an interval characterized by a cutoff value for the voter's net payoff from approval: voter $v$ approves every policy outcome $x \in [l, r]$ for which $\alpha_v(x) \geq c_v^*(l, r)$, that is, the net payoff is not too negative. The cutoff value $c_v^*(l, r) < 0$ is obtained from the binding obedience constraint. The set $\{x \in [l, r] \mid \alpha_v(x) \geq c_v^*(l, r)\}$ is the upper contour set of the strictly quasiconcave function $\alpha_v(x)$ (by Assumption 1) and is therefore an interval. Lemma 2 characterizes the solution of the auxiliary problem; the formal proof can be found in the appendix.

LEMMA 2. For $v \neq 0$ and parameters $-1 \leq l \leq \lfloor A_v \rfloor < \lceil A_v \rceil \leq r \leq 1$, Problem (AUX) admits a solution $I_v(l, r)$ that is a closed interval. Specifically,

• if $\int_l^r \alpha_v(x) d\mu_0(x) \geq 0$, then $I_v(l, r) = [l, r]$,

• if $\int_l^r \alpha_v(x) d\mu_0(x) < 0$, then $I_v(l, r) = \{x \in [l, r] \mid \alpha_v(x) \geq c_v^*(l, r)\}$, where $c_v^*(l, r) < 0$ is the unique value satisfying $\int_{I_v(l, r)} \alpha_v(x) d\mu_0(x) = 0$.

Furthermore, all other solutions coincide with $I_v(l, r)$ $\mu_0$ -almost everywhere.

Winning Elections with Public Advertising

Here I find the challenger-preferred equilibrium for a left/right-leaning baseline elections, which are winnable with public advertising. Without loss of generality, I will focus on a left-leaning election, meaning that $\{L\} \in \mathcal{D}$. In this election, targeted advertising is as good as public advertising, because the challenger wins if and only if voter $L$ approves. To maximize his odds of winning, the challenger finds the largest subset of $[-1, 1]$ that voter $L$ is willing to approve: that is, he solves Problem (AUX) for voter $L$ with parameters $l = -1$ and $r = 1$. He then publicly reveals whether his policy outcome is in that interval.

PROPOSITION 1. Consider a left-leaning baseline election, $\{L\} \in \mathcal{D}$. Then the challenger's highest odds of winning across all equilibria of the public advertising and targeted advertising games are $\mu_0(I_L(-1, 1))$. He achieves these odds by publicly revealing to all voters whether his policy outcome is in $I_L(-1, 1)$.

Proof. In a left-leaning baseline election, voter $L$ is trivially the representative voter. By Lemma 2, $I_L(-1, 1)$ includes her approval set and satisfies her obedience constraint. Using the equilibrium construction from the proof of Theorem 1, we conclude that the set $I_L(-1, 1)$ is implementable; in the equilibrium that implements it, the challenger uses a direct public strategy $\sigma_M$ with $M = I_L(-1, 1)$, his interim utility is $U_I(x) = \mathbb{1}(x \in I_L(-1, 1))$, and his odds of winning are $\mu_0(I_L(-1, 1)) > 0$. His odds of winning cannot be higher in any other equilibrium because $\mu_0(I_L(-1, 1))$ is the upper bound on the odds of convincing a Bayesian voter $L$. ■

Swinging Elections with Targeted Advertising

In the remainder of this section, I focus on a baseline election that is unwinable with public advertising but winnable with targeted advertising. From Theorems 1 and 2, that election is polarized, so the unique decisive coalition that does not include a status quo voter is $\{L, R\}$. The challenger wins if and only if left and right voters approve. Let us consider the following problem:

I will shortly show that this problem admits a solution $(M_L^*, M_R^*)$ such that $W^* := M_L^* \cap M_R^*$ is an interval $[a, b]$ with $a < 0 < b$. Using the equilibrium construction from Theorem 2, I will conclude that the set $W^*$ is implementable; in the equilibrium that implements it, the challenger uses a direct strategy $\sigma_{(M_L^*, M_R^*)}$, his interim utility is $U_I(x) = \mathbb{1}(x \in W^*)$ and his odds of winning are $\mu_0(W^*)$.

Let us now show that Problem (AUX-TA) admits a solution $(M_L^*, M_R^*)$ such that $M_L^* \cap M_R^*$ is a closed interval. For that, it is helpful to define the largest asymmetric interval of policy outcomes that each voter is willing to approve:

DEFINITION 5. The largest asymmetric interval of approved policy outcomes $I_v$ for voter $v \in \{L, R\}$ is defined as follows:

One might guess that sending each voter her largest asymmetric interval—specifically, setting $(M_L^*, M_R^*) = (I_L, I_R)$ —maximizes the challenger’s odds of winning. This strategy is indeed optimal if $I_L \cap I_R = [a_R, b_L]$, which occurs when $[A_L] \leq a_R$ and $b_L \leq [A_R]$. It is straightforward to see that the challenger’s odds of winning cannot exceed $\mu_0([a_R, b_L])$: every policy outcome outside $[a_R, b_L]$ lies further from at least one voter’s bliss point, making it more “costly” in terms of that voter’s obedience constraint. Figure 6 illustrates this challenger-preferred equilibrium outcome for quadratic voters $L = -0.2$ and $R = 0.25$, and a uniform prior.

PROPOSITION 2. Consider a polarized baseline election, $\{L, R\} \in \mathcal{D}$ but $\{L\}, \{R\} \notin \mathcal{D}$. Then, the challenger's maximal odds of winning across all equilibria of the targeted advertising game are $\mu_0(M_L^* \cap M_R^*) > 0$:

1. If $\lfloor A_L \rfloor \leq a_R$ and $b_L \leq \lceil A_R \rceil$, then

2. If $a_R < \lfloor A_L \rfloor$ and $b_L \leq \lceil A_R \rceil$, then

3. If $\lfloor A_L \rfloor \leq a_R$ and $\lceil A_R \rceil < b_L$, then

The challenger achieves these odds of winning by using the direct strategy $\sigma_{(M_L^*, M_R^*)}$.

The formal proof of this result is in the appendix and involves three steps. In step 1, I confirm that the pair $(M_L^*, M_R^*)$ solves Problem (AUX-TA). In step 2, I describe the equilibrium using the construction from the proof of Theorem 2. In step 3, I formulate the information design problem and show that the pair $(M_L^*, M_R^*)$ also characterizes an optimal experiment and the challenger's odds of winning are also $\mu_0(M_L^* \cap M_R^*)$.

Comparative Statics

Proposition 2 suggests that the boundaries of the challenger-preferred equilibrium set of winning policy outcomes depend on the voters' bliss points. I explore this relationship here. First, observe that more extreme voters are willing to approve wider ranges of policy outcomes.

LEMMA 3. If $w$ is a more extreme voter than $v$, then $I_w \supseteq I_v$. Furthermore,

• if $0 < v < w$, then $\lfloor I_w \rfloor = a_w \leq a_v = \lfloor I_v \rfloor$, and the inequality is strict if $-1 < a_v$ and $\lceil A_v \rceil < \lceil A_w \rceil$;
• if $w < v < 0$, then $\lceil I_v \rceil = b_v \leq b_w = \lceil I_w \rceil$, and the inequality is strict if $b_v < 1$ and $\lceil A_w \rceil < \lceil A_v \rceil$.

Lemma 3 follows from the assumption (SC2) that more extreme voters are more persuadable. Figure 8 illustrates the intuition for Lemma 3 for right voters $v < w$. First, observe that a more extreme voter $w$ would be persuaded by the most biased message intended for voter $v$, that is, the interval $[a_v, \lceil A_v \rceil]$ is obedient for $w$. Furthermore, a more extreme voter $w$ has a larger approval set (i.e., $\lceil A_v \rceil \leq \lceil A_w \rceil$), so the interval $[a_v, \lceil A_w \rceil]$ is obedient for $w$ also. Therefore, we can decrease the left boundary of that interval $a_v$ to $a_w \leq a_v$ so that the set $[a_w, \lceil A_w \rceil]$ remains obedient for $w$; the inequality is strict if $-1 < a_v$ and $\lceil A_v \rceil < \lceil A_w \rceil$.

Next, let us explore how the challenger-preferred equilibrium outcome of a baseline election, described in Proposition 2, changes as right voters become more extreme. Note that making right voters more extreme makes the (already polarized) electorate more polarized (Esteban and Ray, 1994).

PROPOSITION 3. Consider the targeted advertising game with a polarized baseline election, $\{L, R\} \in \mathcal{D}$ but $\{L\}, \{R\} \notin \mathcal{D}$. Let $(M_L^*, M_R^*)$ be the challenger-preferred equilibrium intervals of approved policy outcomes described in Proposition 2. Suppose that $b_L \leq [A_R]$. Then, as $R$ increases,

• the challenger's odds of winning $\mu_0(M_L^* \cap M_R^*)$ increase;
• the set of winning policy outcomes $M_L^* \cap M_R^*$ shifts to the left, that is, $[M_L^* \cap M_R^*]$ and $[M_L^* \cap M_R^*]$ decrease.

5. Discussion and Conclusion

This paper studied how a challenger advertises his privately known policy outcome to an electorate of voters with single-peaked and single-crossing preferences, using verifiable messages. Below I discuss how the analysis of the main model can be extended to other assumptions common in the political economy literature. I consider the following extensions: the presence of a strategic incumbent, a citizen-candidate challenger, probabilistic voting, instrumental voting, and information spillovers.

Strategic Incumbent

In the model, the incumbent does not advertise, and his policy outcome is known and normalized to zero. This assumption can be relaxed in a number of ways.

First, suppose that the status quo is a lottery, $\nu_0 \in \Delta X$ (independent of $\mu_0$), and the incumbent is still nonstrategic. Then, voter $v$ 's expected payoff from rejection is $\int u_v(y) d\nu_0(y)$. While there are no left and right voters anymore (they were defined relative to 0), voters $v$ and $w$ can still be defined as having diametrically opposing preferences if $\int u_v(x) d\mu(x) \geq \int u_v(y) d\nu_0(y)$ implies $\int u_w(x) d\mu(x) < \int u_w(y) d\nu_0(y)$ for all $\mu \in \Delta X$, meaning that at most one of these voters prefers to approve given the choice between $\nu_0$ and any $\mu$. Then, an election is unwinable for the challenger without targeted advertising if every decisive coalition requires convincing such voters. With targeted advertising, the challenger induces different posteriors among different voters and is still able to convince voters with diametrically opposing preferences with a positive probability.

Next, suppose that the incumbent is strategic and can change $\nu_0$ to a common belief $\nu$ about the status quo policy outcome, perhaps by publicly advertising it. Assuming that the challenger has time to react, he still benefits from targeted advertising for the same reason as in the above paragraph. In fact, even if the incumbent chooses the status quo policy outcome, the challenger can win as long as not every decisive coalition includes a status quo voter.

Finally, if the candidates are symmetric (e.g., they both use targeted advertising to advertise their own and/or their opponent's policy outcome), then full unraveling of information takes place (Janssen and Teteryatnikova, 2017; Schipper and Woo, 2019). Therefore, the key insight of this paper is that having access to better targeted-advertising technology and/or better voter data allows politicians to win otherwise unwinnable elections. Without this advantage, voters choose the same candidate as they would under complete information.

Citizen-candidate Challenger

It is easy to see that the equilibria described in Theorems 1 and 2 remain equilibria of the citizen-candidate games with public advertising and targeted advertising, respectively. In these equilibria, the challenger does not have profitable deviations when he wins because he reaches the highest possible payoff. On the other hand, he does not have profitable deviations when he loses due to voters' skeptical off-path beliefs, which ensure that he receives at most his complete information payoff if he deviates. Thus, as long as the challenger loses under complete information, whether his payoff depends on $x$ is irrelevant for the purpose of equilibrium construction.

Although the equilibria described in Propositions 1 and 2 maximize the challenger's odds of winning in a baseline election, they may not maximize his citizen-candidate ex-ante utility. It may be ex-ante optimal to let the set of winning policy outcomes include the most extreme challengers, as they would obtain the lowest payoff from losing. Characterizing the citizen-candidate's ex-ante optimal set of winning policy outcomes remains an open question.

Probabilistic Voting

In the model, the challenger knows the set $\mathcal{D}$ of decisive coalitions. In this extension, I explore what happens if the challenger experiences uncertainty about which coalition is decisive. For concreteness, suppose that there are two possible voter bliss points, $L < 0$ and $R > 0$, and the unique minimal decisive coalition is $\{L\}$, $\{R\}$, or $\{L, R\}$ with probability $\gamma_L$, $\gamma_R$, or $\gamma_{LR}$, respectively. The timing of the game is the same, except the uncertainty about which coalition is decisive is resolved after the challenger chooses his messages.

It is useful to establish the challenger's payoff from fully revealing his policy outcome, as that provides the lower bound on his equilibrium payoff. Under full revelation (i.e., if the challenger's strategy is to send message $\{x\}$ for all $x \in X$), his expected payoff is $\gamma_L$ if $x \in [[A_L], 0)$, 1 if $x = 0$, and $\gamma_R$ if $x \in (0, [A_R]]$. Furthermore, much like in the main model, deviations to full revelation are the only ones we need to rule out; other deviations can be made unprofitable by imposing voters' skeptical off-path beliefs. Below we analyze pure-strategy equilibria of public and targeted advertising games.

Public Advertising For any pure-strategy equilibrium, consider the mapping from the challenger's realized policy outcome $x \in X$ to the actions taken by voters $L$ and $R$. Since these voters never approve under a common belief, there are three possible outcomes for each $x$: $L$ approves, $R$ approves, or no one approves. We can thus characterize a pure-strategy equilibrium outcome as a partition of the space of policy outcomes into those approved by the left coalition ($W_L$), those approved by the right coalition ($W_R$), and those approved by nobody ($W_\emptyset = X \cap W_L^c \cap W_R^c$). Furthermore, we can implement any such partition $(W_L, W_R, W_\emptyset)$ by letting the challenger reveal which partition element his policy outcome belongs to: his partitional strategy is to send message $W_L$ if $x \in W_L$, message $W_R$ if $x \in W_R$, and message $W_\emptyset$ otherwise. For that to be an equilibrium strategy, $W_v$ must be obedient for voter $v \in \{L, R\}$. Moreover, the challenger must be obtaining at least his full revelation payoff.

One example of an equilibrium partitional strategy is “divide-and-conquer”—getting left (right) voters to approve the largest interval of left (right) policy outcomes (by letting $W_L = I_L(-1, 0)$ and $W_R = I_R(0, 1)$). Another example of a partitional strategy is to maximize the odds of convincing the left voters (by letting $W_L = I_L(-1, 1)$ and $W_R$ be the largest subset of $X \cap W_L^c$ obedient for $R$). In the end, the partitional strategy that maximizes the challenger’s odds of winning among pure-strategy equilibria will depend on parameters $\gamma_L$ and $\gamma_R$.

Targeted Advertising It is easy to see that any direct strategy $\sigma_{(M_L, M_R)}$, where $M_v$ is obedient for voter $v \in \{L, R\}$ and includes her approval set (e.g., the strategy described in Proposition 2), is an equilibrium strategy. The challenger’s ex-ante utility is then $\gamma_L \cdot \mu_0(M_L) + \gamma_R \cdot \mu_0(M_R) + \gamma_{LR} \cdot \mu_0(M_L \cap M_R)$. It is also easy to see that, depending on the values of $\gamma_L$, $\gamma_R$ and $\gamma_{LR}$, letting $(M_L, M_R) = (M_L^*, M_R^*)$ —that is, maximizing the odds of convincing the mixed decisive coalition—may not maximize the challenger’s ex-ante utility. For example, if the left coalition is almost certainly decisive ($\gamma_L \rightarrow 1$), then tailoring the message to that coalition by letting $M_L = I_L(-1, 1)$ increases the odds of winning.

Overall, when there is uncertainty about decisive coalitions, the challenger faces tradeoffs that are not present in the main model—in particular, he may want to cater to coalitions that are more likely to be decisive. The challenger-preferred equilibrium outcome characterization, whether the focus on pure-strategy equilibria is without loss, and whether the challenger reaches the same payoff as in information design, remain open questions.

Instrumental Voting

In the model, the voters have expressive preferences: if voter $v$ votes for policy outcome $y \in X$, then her payoff is $u_v(y)$. While this assumption is reasonable in large elections wherein the probability that an individual vote is pivotal is vanishingly small, in other elections voters may have instrumental concerns and derive utility also from the winning policy outcome. In this extension, I suppose that voters have both expressive (with probability $\beta \in [0, 1]$) and instrumental (with probability $1 - \beta$) concerns. Suppose that voter $v$ 's payoff when she votes for policy outcome $y$ and the winning policy outcome is $y_w$ is $\tilde{u}_v(y, y_w) = \beta u_v(y) + (1 - \beta)u_v(y_w)$. Below I argue that polarized elections remain winnable if and only if $\beta > 0$, that is, whenever the voters do not have purely instrumental concerns.

Observe that the voters' obedience constraints now have two terms. The first term reflects the pivotality event, in which the winning policy outcome is $x$ because both voters approve. The second term is present only if the voters have expressive concerns (i.e., if $\beta > 0$), as in that case, the winning policy outcome remains $0$.

Suppose first that voters are purely instrumental, that is, $\beta = 0$. Then the obedience constraint of voter $v$ in the challenger-preferred equilibrium becomes $\int_{\tilde{M}_L \cap \tilde{M}_R} \alpha_v(x) d\mu_0(x) \geq 0$. Crucially, that means that each voter's net payoff from approval is non-negative when she is pivotal, i.e., when both voters are recommended to approve. However, both voters are pivotal in the same event, whenever $x \in \tilde{M}_L \cap \tilde{M}_R$, and have a common posterior, $\mu_0(\cdot \mid \tilde{M}_L \cap \tilde{M}_R)$, in that event. By (SC1), the unique common belief under which left and right voters prefer to approve is $\delta_0$, so $\tilde{M}_L \cap \tilde{M}_R = \{0\}$. Therefore, the challenger's odds of convincing purely instrumental and jointly pivotal voters $L < 0$ and $R > 0$ are zero even in his most-preferred equilibrium of the targeted advertising game.

Information Spillovers

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Appendix

LEMMA A.1. Let $W \subseteq X$ and suppose that voter $w \in V$ is more extreme than $v \in V$. Then:

Proof. By (SC2), $\alpha_v(x) \geq 0 \implies \alpha_w(x) \geq 0$ for all $x \in X \setminus \{0\}$, so $A_v \subseteq A_w$. Next, write $W \cup A_w$ as a partition into disjoint sets $W \cup A_v$ and $A_w \cap W^c \cap A_v^c$. Therefore,

where the last term is non-negative because $\alpha_w(x) \geq 0$ for all $x \in A_w$. Consequently, $\int_{W \cup A_v} \alpha_v d\mu_0 \geq 0$ implies $\int_{W \cup A_v} \alpha_w d\mu_0 \geq 0$ (by SC2), which implies $\int_{W \cup A_w} \alpha_w d\mu_0 \geq 0$. ■

Proof of Lemma 2

If $\int_l^r \alpha_v d\mu_0 \geq 0$, then $I_v(l, r) = [l, r]$ solves Problem (AUX). We thus assume for the remainder of the proof that $\int_l^r \alpha_v d\mu_0 < 0$.

We first show that the set $I_v(l, r)$ is well-defined. Since $\alpha_v(x)$ is strictly quasiconcave, the set $S(d) := \{x \in [l, r] \mid \alpha_v(x) \geq d\}$ is convex (i.e., is an empty set, a point, or a closed interval) for any $d \in \mathbb{R}$. Furthermore, $S(d)$ expands as $d$ decreases, i.e., if $d_2 < d_1$, then $S(d_1) \subseteq S(d_2)$; the inclusion is strict unless $S(d_1) = [l, r]$. Now, let $F(d) := \int_{S(d)} \alpha_v d\mu_0$. By the dominated convergence theorem, $F(d)$ is continuous ($\alpha_v(x)$ is continuous and bounded, while $\mu_0$ is a finite measure). Observe that $F(0) > 0$ since $S(0) = \{x \in [l, r] \mid \alpha_v(x) \geq 0\} = A_v$; here, $A_v \subseteq [l, r]$ because $l \leq \lfloor A_v \rfloor < \lceil A_v \rceil \leq r$.

Next, let $\underline{d} := \min_{x \in [l, r]} \alpha_v(x) = \min\{\alpha_v(l), \alpha_v(r)\}$; note that $S(d) = [l, r]$ if and only if $d \leq \underline{d}$ and thus $F(\underline{d}) = \int_l^r \alpha_v(x) d\mu_0(x) < 0$. Observe that $F(d)$ is strictly increasing in $d$ for $d \in [\underline{d}, 0]$. Indeed, if $\underline{d} \leq d_2 < d_1 \leq 0$, then

since $\alpha_v(x) < 0$ for all $x \in S(d_2) \setminus S(d_1)$ and $S(d_1) \subset S(d_2)$ if $\underline{d} \leq d_2 < d_1$. By the intermediate value theorem, there exists a unique $d^* \in (\underline{d}, 0)$ such that $F(d^*) = 0$.

We now show that all solutions to Problem (AUX) agree with $I^* := S(d^*) = I_v(l, r)$ $\mu_0$ -almost everywhere. Let $\tilde{I} \subseteq [l, r]$ be an arbitrary solution. The obedience constraint binds for $I^*$ and holds for $\tilde{I}$, therefore:

Now, since $I^* = S(d^*)$, we have $\alpha_v(x_1) < d^* \leq \alpha_v(x_2)$ for all $x_1 \in (I^*)^c$ and $x_2 \in I^*$. We thus obtain:

since $d^* < 0$; these inequalities are strict unless $\mu_0(\tilde{I} \cap (I^*)^c) = \mu_0(I^* \cap \tilde{I}^c) = 0$. Therefore, $\tilde{I}$ is a solution if and only if it agrees with $I^*$ $\mu_0$ -almost everywhere.

LEMMA A.2. Let $I_L(l, 1) = [a^*, b^*]$ be the interval solution to Problem (AUX) for voter $L < 0$ with $l \in [-1, \lfloor A_L \rfloor]$ and $r = 1$. Suppose that $\int_l^1 \alpha_L d\mu_0 < 0$. Let $S(d) := \{x \in [l, 1] \mid \alpha_L(x) \geq d\}$ and $d^* \in (\min\{\alpha_L(l), \alpha_L(1)\}, 0)$ be the unique solution to $\int_{S(d^*)} \alpha_L(x) d\mu_0 = 0$. Then:

1. $a^* \in (l, \lfloor A_L \rfloor)$ and $b^* \in (0, 1)$ if $d^* > \max\{\alpha_L(l), \alpha_L(1)\}$;
2. $a^* \in (l, \lfloor A_L \rfloor)$ and $b^* = 1$ if $\alpha_L(l) < d^* \leq \alpha_L(1)$;
3. $a^* = l$ and $b^* \in (0, 1)$ if $\alpha_L(l) \geq d^* > \alpha_L(1)$.

Proof. Note that we showed the existence of $d^* \in (\min\{\alpha_L(l), \alpha_L(1)\}, 0)$ in the proof of Lemma 2. Also note that $[a^*, b^*] = S(d^*)$. Furthermore, recall that $\alpha_L([A_L]) = 0$; since $d^* < 0$, it follows that $d^* < \alpha_L([A_L])$. Consequently, if $\alpha_L(l) < d^*$ (as in cases 1 and 2), then $\alpha_L(l) < \alpha_L([A_L])$, $l < [A_L]$ (since $\alpha_L$ is strictly increasing on $[-1, [A_L]]$) and thus $a^* \in (l, [A_L])$ from cases 1 and 2 is well-defined. We have:

1. $d^* > \max\{\alpha_L(l), \alpha_L(1)\}$ so that $a^* \in (l, [A_L])$, $b^* \in (0, 1)$ and $\alpha_L(a^*) = \alpha_L(b^*) = d^*$:

• (i) $\alpha_L(l') \leq d^* = \alpha_L(a^*) \iff l' \leq a^*$, so $S(d^*, l) \subseteq [l', 1]$, $S(d', l') = S(d^*, l)$, and thus $[a', b'] = [a^*, b^*]$.
• (ii) $d^* = \alpha_L(a^*) < \alpha_L(l') \iff a^* < l'$. If $\int_{l'}^1 \alpha_L d\mu_0 \geq 0$, then $[a', b'] = [l', 1]$ and the claim holds since $a^* < l'$ and $b^* < 1$. Else, if $\int_{l'}^1 \alpha_L d\mu_0 < 0$, then $d' \in (\min\{\alpha_L(l'), \alpha_L(1)\}, 0)$ exists, and:

so that $d' < d^*$ since the function $\int_{S(d, l')} \alpha_L d\mu_0$ is strictly increasing in $d$ (see the proof of Lemma 2). Therefore, we have

and by Lemma A.2 (case 3), $[a', b'] = [l', b']$, where $\alpha_L(b') = d' < d^* = \alpha_L(b^*) \iff b' > b^*$, and the claim holds.

2. $\alpha_L(l) < d^* \leq \alpha_L(1)$ so that $a^* \in (l, \lfloor A_L \rfloor)$, $b^* = 1$ and $\int_{a^*}^1 \alpha_L d\mu_0 = 0$. Then, if $a^* \leq l'$, we have $\int_{l'}^1 \alpha_L d\mu_0 > 0$ so that $[a', b'] = [l', 1]$. If, on the other hand, $l < l' < a^*$, then $\alpha_L(l') < \alpha_L(a^*) = d^*$, so $S(d^*, l) \subseteq [l', 1]$, $S(d', l') = S(d^*, l)$, and thus $[a', b'] = [a^*, b^*]$. Overall, in case 2 we have $[a', b'] = [\max\{a^*, l'\}, 1]$ and the claim holds.
3. If $\alpha_L(l) \geq d^* > \alpha_L(1)$, then $a^* = l$, $b^* \in (0, 1)$, $\alpha_L(b^*) = d^*$ and $\int_l^{b^*} \alpha_L d\mu_0 = 0$. If $\int_{l'}^1 \alpha_L d\mu_0 \geq 0$, then $[a', b'] = [l', 1]$ and the claim holds. If $\int_{l'}^1 \alpha_L d\mu_0 < 0$, then $\alpha_L(l') > \alpha_L(l) \geq d^* > d' > \alpha_L(1)$ by the same argument as in case 1.(ii). Consequently, by Lemma A.2 (case 3), $[a', b'] = [l', b']$, where $\alpha_L(b') = d' < d^* = \alpha_L(b^*) \iff b' > b^*$. Overall, in case 3 we have $[a', b'] = [l', b']$, where $l' > l$, $b' > b$, and the claim holds. ■

Proof of Proposition 2

Step 1: Show that Problem (AUX-TA) admits a solution

1. $M_L^* = [\lfloor A_L \rfloor, b_L]$ and $M_R^* = [a_R, \lceil A_R \rceil]$ if $\lfloor A_L \rfloor \leq a_R$ and $b_L \leq \lceil A_R \rceil$;
2. $M_L^* = I_L(a_R, 1)$ and $M_R^* = [a_R, \lceil A_R \rceil]$ if $a_R < \lfloor A_L \rfloor$ and $b_L \leq \lceil A_R \rceil$;
3. $M_L^* = [\lfloor A_L \rfloor, b_L]$ and $M_R^* = I_R(-1, b_L)$ if $\lfloor A_L \rfloor \leq a_R$ and $\lceil A_R \rceil < b_L$.

We will use the voters' obedience constraints to show that $\mu_0(W^*) \geq \mu_0(\widehat{W})$, which implies that $(M_L^*, M_R^*)$ is a solution to Problem (AUX-TA). By contradiction, suppose that $\mu_0(\widetilde{W}) > \mu_0(W^*) \iff \mu_0(Y_L) + \mu_0(Y_R) > \mu_0(Z_L) + \mu_0(Z_R)$. We will address cases 1 and 2 (each with multiple subcases), in which $M_R^* = [a_R, \lceil A_R \rceil]$; case 3 is proved analogously to case 2.

1.1 $-1 < \lfloor A_L \rfloor \leq a_R$ and $b_L \leq \lceil A_R \rceil < 1$. In this case, $M_L^* = I_L(\lfloor A_L \rfloor, 1) = [\lfloor A_L \rfloor, b_L]$; also, $Z_L = [a_R, 0] \cap \widetilde{W}^c$, $Y_L = [-1, a_R] \cap \widetilde{W}$, $Z_R = [0, b_L] \cap \widetilde{W}^c$, and $Y_R = (b_L, 1] \cap \widetilde{W}$. First, note that $M_L^*$ binds $L$ 's obedience constraint since $b_L < 1$, we have $\int_{\lfloor A_L \rfloor}^1 \alpha_L d\mu_0 < 0$ and $\int_{\lfloor A_L \rfloor}^{b_L} \alpha_L d\mu_0 = 0$. Similarly, $M_R^*$ binds $R$ 's obedience constraint since $-1 < a_R$.

Similarly, we partition $\widetilde{M}_L$ into four sets $[-1, \lfloor A_L \rfloor) \cap \widetilde{M}_L$, $A_L$, $[0, b_L] \cap \widetilde{M}_L$ and $(b_L, 1] \cap \widetilde{M}_L$, and write down $L$ 's obedience constraint for $\widetilde{M}_L$ as:

where the last inequality holds because $\widetilde{W} \subseteq \widetilde{M}_L$ (so that $[0, b_L] \cap \widetilde{W} \subseteq [0, b_L] \cap \widetilde{M}_L$ and $Y_R \subseteq (b_L, 1] \cap \widetilde{M}_L$) and $\alpha_L$ is negative outside $A_L$. Combining (1) and (2), we get:

Since $Z_R \subseteq [0, b_L]$, $Y_R \subseteq (b_L, 1]$, and $\alpha_L$ is strictly decreasing on $[0, 1]$, we get:

which implies that $\mu_0(Z_R) \geq \mu_0(Y_R)$ since $\alpha_L(b_L) < 0$. Using the same argument (i.e., by comparing the terms in $R$ 's obedience constraints for $M_R^*$ and $\widetilde{M}_R$), we obtain $\mu_0(Z_L) \geq \mu_0(Y_L)$. Therefore, $\mu_0(Z_L) + \mu_0(Z_R) \geq \mu_0(Y_L) + \mu_0(Y_R)$, a contradiction.

1.2 $-1 = \lfloor A_L \rfloor = a_R$ and $b_L \leq \lceil A_R \rceil < 1$. In this case, $L$ 's constraint for $M_L^*$ is binding (since $b_L < 1$), while $R$ 's constraint for $M_R^*$ may or may not bind.

Using the fact that $M_L^*$ binds $L$ 's constraint, we obtain $\mu_0(Z_R) \geq \mu_0(Y_R)$ (see step 1.1). While $R$ 's constraint for $M_R^*$ may not bind, from $-1 = \lfloor A_L \rfloor = a_R$ we have $M_R^* = [-1, \lceil A_R \rceil]$ and $[-1, 0] \cap W^* = [-1, 0]$, so $Y_L = \emptyset$ and $\mu_0(Z_L) \geq 0 = \mu_0(Y_L)$. Therefore, $\mu_0(Z_L) + \mu_0(Z_R) \geq \mu_0(Y_L) + \mu_0(Y_R)$, a contradiction.

1.3 $-1 < \lfloor A_L \rfloor \leq a_R$ and $b_L = \lceil A_R \rceil = 1$. This case is analogous to 1.2.

1.4 $-1 = \lfloor A_L \rfloor = a_R$ and $b_L = \lceil A_R \rceil = 1$. This case is impossible because then $\int_{-1}^1 \alpha_v d\mu_0 \geq 0$ for each $v \in \{L, R\}$, i.e., both voters weakly prefer to approve under the prior $\mu_0$; that contradicts (SC1).

2.2 $\lfloor I_L(-1, 1) \rfloor < a_R < \lfloor A_L \rfloor$ and $b_L \leq \lceil A_R \rceil$. In this case, $M_L^* = I_L(a_R, 1) =: [a_R, b]$, $M_R^* = I_R(-1, \lceil A_R \rceil) = [a_R, \lceil A_R \rceil]$ and $W^* = [a_R, b]$. Also, $Z_L = [a_R, 0] \cap \widetilde{W}^c$, $Y_L = [-1, a_R] \cap \widetilde{W}$, $Z_R = [0, b] \cap \widetilde{W}^c$, and $Y_R = (b, 1] \cap \widetilde{W}$. It is worth mentioning that $b_L$ and $b$ are different objects: $b_L$ is the upper bound of $I_L(\lfloor A_L \rfloor, 1) = [\lfloor A_L \rfloor, b_L]$, while $b$ is the upper bound of $I_L(a_R, 1) = [a_R, b]$. By Lemma A.3, we have $b \leq b_L$. Now, since $-1 \leq \lfloor I_L(-1, 1) \rfloor < a_R$, we have $\int_{-1}^{\lfloor A_R \rceil} \alpha_R d\mu_0 < 0$, so $\int_{a_R}^{\lfloor A_R \rceil} \alpha_R d\mu_0 = 0$ and $R$ 's constraint for $M_R^*$ binds. From that, $\mu_0(Z_L) \geq \mu_0(Y_L)$ (see step 1.1).

For voter $L$, two cases are possible: $b = 1$ and $b < 1$. If $b = 1$, then $[a_R, 1]$ satisfies $L$ and $R$ 's obedience constraints, i.e., both voters prefer to approve under belief $\mu_0(\cdot \mid [a_R, 1])$, which contradicts (SC1). Therefore, $b < 1$.

Now, we have $\int_{a_R}^1 \alpha_L d\mu_0 < 0$ and $\int_{a_R}^b \alpha_L d\mu_0 = 0$, i.e., $M_L^*$ binds $L$ 's obedience constraint. Partition $M_L^*$ into $[a_R, \lfloor A_L \rfloor) \cap \widetilde{W} =: i_1$, $[a_R, \lfloor A_L \rfloor) \cap \widetilde{W}^c = Z_L \cap A_L^c$, $A_L$, $[0, b] \cap \widetilde{W} =: i_2$, $[0, b] \cap \widetilde{W}^c = Z_R$, and write down $L$ 's (binding) obedience constraint for $M_L^*$ as:

Similarly, partition $\widetilde{M}_L$ into $[-1, a_R) \cap \widetilde{M}_L$, $[a_R, [A_L]) \cap \widetilde{M}_L$, $A_L$, $[0, b] \cap \widetilde{M}_L$, $(b, 1] \cap \widetilde{M}_L$. Using the fact that $\widetilde{W} \subseteq \widetilde{M}_L$ and $\alpha_L$ is negative outside $A_L$, from the obedience constraint for $\widetilde{M}_L$ we obtain

Combining (3) and (4), we get

and, since $\alpha_L(y) \leq \alpha_L(a_R) \leq \alpha_L(z) \leq 0$ for all $y \in Y_L$, $z \in Z_L \cap A_L^c$ and $\alpha_L(y) \leq \alpha_L(b) \leq \alpha_L(z) \leq 0$ for all $y \in Y_R$, $z \in Z_R$, we obtain

since $\alpha_L(a_R) < 0$.

Next, observe that if $b < 1$, then $\alpha_L(a_R) \geq \alpha_L(b)$. Indeed, since $[a_R, b] = I_L(a_R, 1)$ and $\int_{a_R}^1 \alpha_L d\mu_0 < 0$, from Lemma 2 we get that $[a_R, b] = \{x \in [a_R, 1] \mid \alpha_L(x) \geq d^*\}$ for some $d^* < 0$. In particular, we have $\alpha_L(x) \geq d^*$ for all $x \in [0, b]$ and $\alpha_L(x) < d^*$ for all $x \in (b, 1]$, which is a non-empty set if $b < 1$. By the continuity and strict monotonicity of $\alpha_L$ on $[0, 1]$, we have $\alpha_L(b) = d^*$ so that $\alpha_L(a_R) \geq d^* = \alpha_L(b)$.

Now, dividing the last inequality by $\alpha_L(b) < 0$, we get:

Next, add $\left(1 - \frac{\alpha_L(a_R)}{\alpha_L(b)}\right) (\mu_0(Y_L) + \mu_0(Z_L))$ to both sides to obtain:

Rearranging terms, we get:

since $1 - \frac{\alpha_L(a_R)}{\alpha_L(b)} \geq 0$ if $\alpha_L(a_R) \geq \alpha_L(b)$ and $\mu_0(Z_L) - \mu_0(Y_L) \geq 0$ from the previous calculations in the beginning of step 2.2. Therefore, $\mu_0(Z_L) + \mu_0(Z_R) \geq \mu_0(Y_L) + \mu_0(Y_R)$, a contradiction.

3 $[A_L] \leq a_R$ and $[A_R] < b_L$. This case is analogous to case 2.

Step 2: Equilibrium Characterization.

Observe that the mixed decisive coalition is $\{L, R\}$; the set $M_L^* \cap M_R^* = [a, b]$ is an interval such that $a < 0$ and $b > 0$; the set $A_v \cup [a, b] = M_v^*$ satisfies the obedience constraint of voter $v \in \{L, R\}$. Thus, $[a, b]$ is implementable and the equilibrium that implements is described in the proof of Theorem 2; the equilibrium strategy of the challenger is $\sigma_{(M_L^*, M_R^*)}$

Step 3: Show that there exists an optimal experiment that is characterized by $(M_L^*, M_R^*)$, for which the challenger's odds of winning are $\mu_0(M_L^* \cap M_R^*)$.

Similarly, for $v$ to reject after signal $s_v = 0$, her expected net payoff from approval must be negative, $\int \alpha_v(x) \psi_v(0 | x) d\mu_0(x) < 0$. An optimal experiment maximizes the challenger's odds of winning and solves

Moreover, we can drop the less-than-zero constraints as letting $\psi_v(1 | x) = 1$ for each $v \in \{L, R\}$ and $x \in A_v$ weakly increases the objective and loosens $v$ 's constraints. Now, since each $\alpha_v$ is bounded, $\mu_0$ is a finite and atomless positive measure, and $X$ is a closed interval, a deterministic optimal experiment $\psi^* : X \rightarrow \{0, 1\}^2$ exists (by an argument similar to one in the proof of Proposition 2 in Titova and Zhang, 2025). Next, let $C_v^* := \{x \in X | \psi_v^*(1 | x) = 1\}$ for each $v \in \{L, R\}$ be the set of policy outcomes that $v$ is recommended to approve. Then, an optimal deterministic experiment is characterized by a pair $(C_L^*, C_R^*)$ that solves Problem (AUX-TA). Hence, $\mu_0(M_L^* \cap M_R^*)$ is the challenger's odds of winning in the information design problem.

Proof of Lemma 3

We prove this statement for left voters $w < v < 0$. The proof for right voters $0 < v < w$ is analogous. From the definition of $I_v$ ($I_w$) as a solution to Problem (AUX) with $l = \lfloor A_v \rfloor$ ($l = \lfloor A_w \rfloor$) and $r = 1$, three cases are possible:

Case 1 $\int_{\lfloor A_v \rfloor}^1 \alpha_v d\mu_0 \geq 0$. Then, $I_v = [\lfloor A_v \rfloor, 1]$. By (SC2), $\int_{\lfloor A_v \rfloor}^1 \alpha_v d\mu_0 \geq 0 \implies \int_{\lfloor A_v \rfloor}^1 \alpha_w d\mu_0 \geq 0 \implies \int_{\lfloor A_w \rfloor}^1 \alpha_w d\mu_0 \geq 0$ so $I_w = [\lfloor A_w \rfloor, 1]$ and $I_w \supseteq I_v$ since $\lfloor A_w \rfloor \leq \lfloor A_v \rfloor$.

Case 2 $\int_{\lfloor A_v \rfloor}^1 \alpha_v d\mu_0 < 0$ and $\int_{\lfloor A_w \rfloor}^1 \alpha_w d\mu_0 \geq 0$. Then, $I_v = [\lfloor A_v \rfloor, b_v]$, where $\int_{\lfloor A_v \rfloor}^{b_v} \alpha_v d\mu_0 = 0$ and $b_v < 1$. Therefore, $I_v = [\lfloor A_v \rfloor, b_v] \subset [\lfloor A_w \rfloor, 1] = I_w$.

Case 3 $\int_{\lfloor A_v \rfloor}^1 \alpha_v d\mu_0 < 0$ and $\int_{\lfloor A_w \rfloor}^1 \alpha_w d\mu_0 < 0$. Then, $I_v = [\lfloor A_v \rfloor, b_v]$, where $\int_{\lfloor A_v \rfloor}^{b_v} \alpha_v d\mu_0 = 0$ and $b_v < 1$; the same is true for $I_w$ —in particular, $\int_{\lfloor A_w \rfloor}^{b_w} \alpha_w d\mu_0 = 0$. By (SC2), $\int_{\lfloor A_v \rfloor}^{b_v} \alpha_v d\mu_0 = 0 \implies \int_{\lfloor A_v \rfloor}^{b_v} \alpha_w d\mu_0 \geq 0 \implies \int_{\lfloor A_w \rfloor}^{b_v} \alpha_w d\mu_0 \geq 0$ (the last inequality is strict unless $\lfloor A_v \rfloor = \lfloor A_w \rfloor$). Now, since the function $\int_{\lfloor A_w \rfloor}^z \alpha_w d\mu_0$ is continuous and strictly decreasing in $z$ for $z \in (0, 1)$, we have

so that $I_v = [\lfloor A_v \rfloor, b_v] \subseteq [\lfloor A_w \rfloor, b_w] = I_w$. Furthermore, unless $\lfloor A_v \rfloor = \lfloor A_w \rfloor$, we have $\lfloor A_w \rfloor < \lfloor A_v \rfloor$ and $b_v < b_w$.

Proof of Proposition 3

Let $\{L, 0, E\}$ be the baseline electorate with the more extreme right voter $E > R$. Let $(\widetilde{M}_L, \widetilde{M}_E)$ be the solution to (AUX-TA) for this electorate (described in Proposition 2). Note that $b_L \leq \lfloor A_R \rfloor$ and $R < E$ imply that $b_L \leq \lfloor A_E \rfloor$, so $(M_L^*, M_R^*)$ and $(\widetilde{M}_L, \widetilde{M}_E)$ are both described by Case 1 or 2 of Proposition 2. Therefore, $M_R^* = [a_R, \lfloor A_R \rfloor]$, $\widetilde{M}_E = [a_E, \lfloor A_E \rfloor]$ and, by Lemma 3, $a_E \leq a_R$. To simplify exposition, let $W^* := M_L^* \cap M_R^*$ and $\widetilde{W} := \widetilde{M}_L \cap \widetilde{M}_E$.

If $\lfloor A_L \rfloor \leq a_E \leq a_R$, then both elections fall into case 1 of Proposition 2. We have $W^* = [a_R, b_L]$, $\widetilde{W} = [a_E, b_L]$, and the claims of the proposition are true because $a_E \leq a_R$.

Next, suppose that $a_E < \lfloor A_L \rfloor$, in which case $\widetilde{M}_L = \widetilde{W} = I_L(a_E, 1) =: [\widetilde{a}_L, \widetilde{b}_L]$. Observe that $\mu_0(I_L(l, 1))$ is decreasing in $l$ (which is a parameter in Problem AUX) on $[-1, \lfloor A_L \rfloor]$: increasing $l$ shrinks the feasible region $[l, 1]$, so the objective value $\mu_0(I_L(l, 1))$ can only go down. Therefore, $\mu_0(I_L(a_E, 1)) \geq \mu_0(I_L(l, 1))$ for all $l \in [a_E, \lfloor A_L \rfloor]$. In particular, if $a_E < \lfloor A_L \rfloor \leq a_R$, then $\mu_0(I_L(a_E, 1)) \geq \mu_0(I_L(\lfloor A_L \rfloor, 1)) \geq \mu_0([a_R, b_L])$, and if $a_E \leq a_R < \lfloor A_L \rfloor$, then $\mu_0(I_L(a_E, 1)) \geq \mu_0(I_L(a_R, 1))$. Either way, $\mu_0(\widetilde{W}) \geq \mu_0(W^*)$.

Finally, from Lemma A.3, both boundaries of $I_L(a_E, 1)$ are left of the corresponding boundaries of $I_L(a_R, 1)$ (often strictly so—see the proof of Lemma A.3). If $a_E \leq a_R < \lfloor A_L \rfloor$, then $\widetilde{W} = I_L(a_E, 1)$ and $W^* = I_L(a_R, 1)$ so $\widetilde{W}$ is left of $W^*$. If, on the other hand, $a_E < \lfloor A_L \rfloor \leq a_R$, then $\widetilde{W} = I_L(a_E, 1)$ is left of $I_L(\lfloor A_L \rfloor, 1) = [\lfloor A_L \rfloor, b_L]$, which is a superset of $[a_R, b_L] = W^*$, so $\widetilde{W}$ is left of $W^*$.