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The fragmentation of views in a democracy

Abstract

Are voters in democracies more competent if there are more media outlets? To answer this question, I provide a game-theoretic model of media capture and political persuasion in democratic countries. In the model, there are two politicians, the Incumbent and the Challenger. They co-opt the media by offering them access to information. In exchange, the media support politicians who are available for interviews or include journalists in press pools. Voters choose like-minded media. I show that if the Incumbent is sufficiently popular, then media bias in her favor weakly increases in the number of media outlets. Otherwise, media bias in the Incumbent’s favor weakly decreases in the number of media outlets. The welfare of voters weakly increases and decreases in the respective cases. The intuition is that, in equilibrium, the Incumbent can co-opt only one media outlet and ensure that enough voters read it. In this case, media outlets compete for access to the Incumbent and agree for a higher bias as their number increases.

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Correspondence to Arseniy Samsonov.

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Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

I would like to thank my graduate advisor Michael Chwe, my committee members Leslie Johns, Barry O’Neill, and Daniel Treisman; Stepan Alexenko, Kemal Kivanc Akoz, Pasha Andreyanov, Ashley Blum, Matthew Kerbel, Konstantin Sonin, Maxim Senkov, Alexei Zakharov, the conference and seminar participants, an anonymous referee, the editor Amihai Glazer and an anonymous referee for helpful comments. All errors are my own.

Appendix

Appendix

Proof of the Proposition

Step 1

Fix the levels of bias \(p_{I}\) and \(p_{C}\). A voter with position x who follows a media outlet that contacts the Incumbent gets:

$$\begin{aligned} {{\mathbb {P}}}[\Theta =1](1-x)+{{\mathbb {P}}}[\Theta =0]{{\mathbb {P}}}[S_{I}=0\mid \Theta =0]x+{{\mathbb {P}}}[\Theta =0]{{\mathbb {P}}}[S_{I}=1\mid \Theta =0](-x) \end{aligned}$$

Recall that

$$\begin{aligned} {{\mathbb {P}}}[\Theta =1]=1-{{\mathbb {P}}}[\Theta =0]=\frac{1}{2} \end{aligned}$$

and

$$\begin{aligned} {{\mathbb {P}}}[S_{I}=1\mid \Theta =0]=1-{{\mathbb {P}}}[S_{I}=0\mid \Theta =0]=p_{I} \end{aligned}$$

Substituting these values yields

$$\begin{aligned}&{{\mathbb {P}}}[\Theta =1](1-x)+{{\mathbb {P}}}[\Theta =0]{{\mathbb {P}}}[S_{I}=0\mid \Theta =0]x+{{\mathbb {P}}}[\Theta =0]{{\mathbb {P}}}[S_{I}=1\mid \Theta =0](-x)= \\&\quad =\frac{1-x}{2}+\frac{(1-p_{I})x}{2}+\frac{p_{I}(-x)}{2}=\frac{1}{2}-p_{I}x \end{aligned}$$

Define

$$\begin{aligned} f(p_{I},x)=\frac{1}{2}-p_{I}x \end{aligned}$$

A voter who follows an outlet that contacts the Challenger gets

$$\begin{aligned}&{{\mathbb {P}}}[\Theta =0]x+{{\mathbb {P}}}[\Theta =1]{{\mathbb {P}}}[\Theta _{C}=1\mid \Theta =1]{{\mathbb {P}}}[S_{C}=1\mid \Theta _{C}=1](1-x)\\&\quad +(1-{{\mathbb {P}}}[\Theta =0]-{{\mathbb {P}}}[\Theta =1]{{\mathbb {P}}}[\Theta _{C}=1\mid \Theta =1]{{\mathbb {P}}}[S_{C}=1\mid \Theta _{C}=1])(x-1) \end{aligned}$$

Recall that

$$\begin{aligned} {{\mathbb {P}}}[\Theta _{C}=1\mid \Theta =1]=r \end{aligned}$$

and

$$\begin{aligned} {{\mathbb {P}}}[S_{C}=1\mid \Theta _{C}=1]=1-p_{C} \end{aligned}$$

Substituting these values yields

$$\begin{aligned}&\frac{x}{2}+\frac{r(1-p_{C})(1-x)}{2}+(1-\frac{1+r(1-p_{C})}{2})(x-1)=\\&\quad =(1-p_{C})r(1-x)+x-\frac{1}{2} \end{aligned}$$

Define the value above as

$$\begin{aligned} g(p_{C},x)=(1-p_{C})r(1-x)+x-\frac{1}{2} \end{aligned}$$

Solving

$$\begin{aligned} f(p_{I},x)=g(p_{C},x) \end{aligned}$$

yields

$$\begin{aligned} x=\frac{1-(1-p_{C})r}{1+p_{I}-(1-p_{C})r} \end{aligned}$$

Define the value above as

$$\begin{aligned} {\hat{x}}(p_{I},p_{C})=\frac{1-(1-p_{C})r}{1+p_{I}-(1-p_{C})r} \end{aligned}$$

The interpretation of \({\hat{x}}(p_{I},p_{C})\) is the voter who is indifferent between following a media outlet that contacts the Incumbent and a media outlet that contacts the Challenger.

Observe that

$$\begin{aligned} 1+p_{I}-(1-p_{C})r\ne 0 \end{aligned}$$

and

$$\begin{aligned} {\hat{x}}(p_{I},p_{C})\in (0,1) \end{aligned}$$

Observe also that

$$\begin{aligned} \frac{d}{dp_{I}}{\hat{x}}(p_{I},p_{C})=\frac{r-p_{C}r-1}{(p_{I}+(p_{C}-1)r+1)^{2}}<0 \end{aligned}$$

and

$$\begin{aligned} \frac{d}{dp_{C}}{\hat{x}}(p_{I},p_{C})=\frac{p_{I}r}{(p_{I}+(p_{C}-1)r+1)^{2}}>0 \end{aligned}$$

Finally, notice that

$$\begin{aligned} \frac{d}{dx}(g(p_{C},x)-f(p_{I},x))=p+1-(1-q)r>0 \end{aligned}$$

Hence, voters \(x<{\hat{x}}(p_{I},p_{C})\) prefer following a media outlet that contacts the Incumbent and voters \(x>{\hat{x}}(p_{I},p_{C})\) prefer following a media outlet that contacts the Challenger.

A voter who always votes for the Incumbent gets

$$\begin{aligned} f(1,x)=\frac{1}{2}-x<f(p_{I},x) \end{aligned}$$

for any \(p_{I}<1\) and \(x\in (0,1).\) A voter who always votes for the Challenger gets

$$\begin{aligned} g(1,x)=x-\frac{1}{2}<g(p_{C},x) \end{aligned}$$

for any \(p_{C}<1\) and \(x\in (0,1)\). Notice that voters \(x<{\hat{x}}(1,p_{C})\) prefer to always vote for the Incumbent over following an outlet that contacts the Challenger and voters \(x>{\hat{x}}(1,p_{C})\) prefer following an outlet that contacts the Challenger over always voting for the Incumbent. Similarly, voters \(x<{\hat{x}}(p_{I},1)\) prefer following an outlet that contacts the Incumbent over always voting for the Challenger while voters \(x>{\hat{x}}(p_{I},1)\) prefer to always vote for the Challenger.

Observe that

$$\begin{aligned} 0<{\hat{x}}(1,p_{C})<{\hat{x}}(p_{I},p_{C})<{\hat{x}}(p_{I},1)<1 \end{aligned}$$

Finally, it is clear that all voters \(x<0\) always vote for the Incumbent and all voters \(x>1\) always vote for the Challenger. Because of this, voters \(x<0\) and \(x>1\) never read any media outlets.

Step 2

Suppose that in some equilibrium

$$\begin{aligned} \frac{{\hat{x}}(p_{I},1)}{m}>1-{\hat{x}}(p_{I},p_{C}) \end{aligned}$$

If all media outlets contact the Incumbent, voters \(x\in (0,{\hat{x}}(p_{I},1))\) will follow the media outlets while voters \(x>{\hat{x}}(p_{I},1)\) will always vote for the Challenger. Hence, each media outlet will obtain

$$\begin{aligned} \frac{{\hat{x}}(p_{I},1)}{m} \end{aligned}$$

None will deviate and contact the Challenger. Indeed, if a media outlet deviates, then voters \(x\in (0,{\hat{x}}(p_{I},p_{C}))\) will follow outlets that contact the Incumbent while voters \(x\in ({\hat{x}}(p_{I},p_{C}),1)\) will follow the outlet that contacts the Challenger. Hence, the outlet that deviated will obtain

$$\begin{aligned} 1-{\hat{x}}(p_{I},p_{C})\le \frac{{\hat{x}}(p_{I},1)}{m} \end{aligned}$$

and therefore will not deviate.

If in equilibrium all media outlets contact the Challenger, each gets

$$\begin{aligned} \frac{1-{\hat{x}}(1,p_{C})}{m} \end{aligned}$$

while if one media outlet deviates and contacts the Incumbent, it gets \({\hat{x}}(p_{I},p_{C})\). Notice that

$$\begin{aligned} \frac{{\hat{x}}(p_{I},1)}{m}>1-{\hat{x}}(p_{I},p_{C}) \end{aligned}$$

implies

$$\begin{aligned} \frac{1-{\hat{x}}(1,p_{C})}{m}<{\hat{x}}(p_{I},p_{C}) \end{aligned}$$

Hence, a media outlet will indeed deviate and contact the Incumbent.

Finally, suppose that in some equilibrium, \(k\in [1,m-1]\) media outlets contact the Challenger and \(m-k\) contact the Incumbent. Clearly, for \(k=1\), the inequality

$$\begin{aligned} \frac{{\hat{x}}(p_{I},1)}{m}>1-{\hat{x}}(p_{I},p_{C}) \end{aligned}$$

rules out such an equilibrium because the outlet that contacts the Challenger will deviate and contact the Incumbent. If \(k\in [2,m-1]\), then it must be that no outlet wants to deviate from contacting the Challenger to contacting the Incumbent, hence

$$\begin{aligned} \frac{{\hat{x}}(p_{I},p_{C})}{m-k+1}\le \frac{1-{\hat{x}}(p_{I},p_{C})}{k} \end{aligned}$$

which gives a contradiction given that

$$\begin{aligned} \frac{{\hat{x}}(p_{I},1)}{m}>1-{\hat{x}}(p_{I},p_{C}) \end{aligned}$$

Hence, if

$$\begin{aligned} \frac{{\hat{x}}(p_{I},1)}{m}>1-{\hat{x}}(p_{I},p_{C}) \end{aligned}$$

then in equilibrium, all media outlets contact the Incumbent.

Step 3

Suppose that

$$\begin{aligned} {\hat{x}}(p_{I},p_{C})<\frac{1-{\hat{x}}(1,p_{C})}{m} \end{aligned}$$

Hence, in equilibrium, it is possible that all the media outlets contact the Challenger. Notice that

$$\begin{aligned} {\hat{x}}(p_{I},p_{C})<\frac{1-{\hat{x}}(1,p_{C})}{m} \end{aligned}$$

implies

$$\begin{aligned} \frac{{\hat{x}}(p_{I},1)}{m}<1-{\hat{x}}(p_{I},p_{C}) \end{aligned}$$

Hence, in equilibrium, it cannot be the case that m media outlets contact the Incumbent. Notice also that

$$\begin{aligned} {\hat{x}}(p_{I},p_{C})<\frac{1-{\hat{x}}(1,p_{C})}{m} \end{aligned}$$

rules out equilibria when exactly one media outlet contacts the Incumbent and others contact the Challenger. Finally, if \(k\in [2,m-1]\) media outlets contact the Incumbent and \(m-k\) contact the Challenger, then

$$\begin{aligned} \frac{{\hat{x}}(p_{I},p_{C})}{k}\ge \frac{1-{\hat{x}}(p_{I},p_{C})}{m-k+1} \end{aligned}$$

which gives a contradiction given that

$$\begin{aligned} {\hat{x}}(p_{I},p_{C})<\frac{1-{\hat{x}}(1,p_{C})}{m} \end{aligned}$$

Step 4

Suppose that

$$\begin{aligned} {\hat{x}}(p_{I},p_{C})>\frac{1-{\hat{x}}(1,p_{C})}{m} \end{aligned}$$

and

$$\begin{aligned} 1-{\hat{x}}(p_{I},p_{C})>\frac{{\hat{x}}(p_{I},1)}{m} \end{aligned}$$

Then there are no equilibria in which all the media outlets contact the Incumbent or all contact the Challenger.

Define

$$\begin{aligned} h(k)=\frac{{\hat{x}}(p_{I},p_{C})}{k}-\frac{1-{\hat{x}}(p_{I},p_{C})}{m-k+1} \end{aligned}$$

where \(k\in [1,m]\). Suppose that in equilibrium, k media outlets contact the Incumbent. If \(k\in [1,m-1]\), then

$$\begin{aligned} h(k)>0 \end{aligned}$$

implies that no media outlet will deviate from contacting the Incumbent to contacting the Challenger. Notice further that

$$\begin{aligned} h(k+1)=\frac{{\hat{x}}(p_{I},p_{C})}{k+1}-\frac{1-{\hat{x}}(p_{I},p_{C})}{m-k}<0 \end{aligned}$$

implies that no media outlet will deviate from contacting the Challenger to contacting the Incumbent. Hence, if such a k exists, then in equilibrium k media outlets contact the Incumbent and \(m-k\) contact the Challenger.

Notice that

$$\begin{aligned} h(1)={\hat{x}}(p_{I},p_{C})-\frac{1-{\hat{x}}(p_{I},p_{C})}{m}\ge {\hat{x}}(p_{I},p_{C})-\frac{1-{\hat{x}}(1,p_{C})}{m}>0 \end{aligned}$$

and

$$\begin{aligned} h(m)=\frac{{\hat{x}}(p_{I},p_{C})}{m}-(1-{\hat{x}}(p_{I},p_{C}))\le \frac{{\hat{x}}(p_{I},1)}{m}-(1-{\hat{x}}(p_{I},p_{C}))<0 \end{aligned}$$

Notice further that h(k) is continuous and decreasing in k. Hence, there exists a unique \(k^{*}\in (1,m)\) such that

$$\begin{aligned} h(k^{*})=0 \end{aligned}$$

Moreover,

$$\begin{aligned} h(\max \{k\in {{\mathbb {N}}}:k\le k^{*}\})>0 \end{aligned}$$

and

$$\begin{aligned} h(\min \{k\in {{\mathbb {N}}}:k\ge k^{*}\})<0 \end{aligned}$$

Setting

$$\begin{aligned} k=\max \{k\in {{\mathbb {N}}}:k\le k^{*}\} \end{aligned}$$

yields

$$\begin{aligned} k+1=\min \{k\in {{\mathbb {N}}}:k\ge k^{*}\} \end{aligned}$$

and

$$\begin{aligned} h(k)>0 \end{aligned}$$

and

$$\begin{aligned} h(k+1)<0 \end{aligned}$$

Hence, in equilibrium, exactly \(k\in [1,m-1]\) media outlets contact the Incumbent.

Step 5

Notice that for any \(p_{C}\)

$$\begin{aligned} \lim _{p_{I}\rightarrow 0}\left( \frac{{\hat{x}}(p_{I},1)}{m}-(1-{\hat{x}}(p_{I},p_{C}))\right) =\frac{1}{m}>0 \end{aligned}$$

Hence, for any \(p_{C}\), the Incumbent can set \(p_{I}\) sufficiently close to 0 and ensure that all the media outlets contact her. Further, notice that

$$\begin{aligned} \lim _{p_{I}\rightarrow 0}{\hat{x}}(p_{I},p_{C})=1 \end{aligned}$$

Hence, by setting \(p_{I}\) sufficiently close to 0, the Incumbent can ensure that the median voter \(x^{*}\) and all voters \(x<x^{*}\) follow a media outlet that contacts her. As a result, the Incumbent wins iff

$$\begin{aligned} S_{I}=1 \end{aligned}$$

which occurs with probability

$$\begin{aligned} \frac{1+p_{I}}{2}>\frac{1}{2} \end{aligned}$$

Hence, the Incumbent can guarantee winning with probability more than \(\frac{1}{2}\).

Step 6

Suppose that the median voter \(x^{*}\) does not follow any media outlet and always votes for the Incumbent. Suppose that \(p_{I}=1\). If

$$\begin{aligned} {\hat{x}}(1,0)<x^{*} \end{aligned}$$

then the Challenger can deviate and set \(p_{C}=0\), in which case all the media outlets will contact her and the median voter will read one of the media outlets. As a result, the Challenger will win with probability \(\frac{1}{2}\) instead of 0, so she will deviate. If

$$\begin{aligned} {\hat{x}}(1,0)=\frac{1-r}{2-r}\ge x^{*} \end{aligned}$$

then this is not true and the situation can be a part of an equilibrium. Moreover, because 1 is the highest payoff that the Incumbent can obtain in the game and she cannot obtain it otherwise, the Incumbent will set \(p_{I}=1\). In this case, the Challenger is indifferent between any \(p_{C}\). If \(p_{C}=1\), then by Assumption all the media outlets contact the Incumbent. If \(p_{C}<1\), all the media outlets contact the Challenger.

For the same reason, if \(x^{*}>1\), then the Challenger can set \(p_{C}=1\). In this case, the median voter will always vote for the Challenger, who always wins. By assumption, all the media outlets contact the Incumbent in this case.

If \(p_{I}<1\), the situation is impossible because, by Step 1

$$\begin{aligned} f(x,p_{I})>\frac{1}{2}-x \end{aligned}$$

Suppose that, in equilibrium, the median voter reads an outlet that contacts the Challenger. In this case, the Challenger wins iff \(S_{C}=0\), which occurs with probability

$$\begin{aligned} \frac{1+p_{C}}{2}\ge \frac{1}{2} \end{aligned}$$

By Step 5, this is not possible in equilibrium because the Incumbent can guarantee winning with probability higher than \(\frac{1}{2}\).

Therefore, if

$$\begin{aligned} {\hat{x}}(1,0)<x^{*} \end{aligned}$$

then, in equilibrium, the median voter follows an outlet that contacts the Incumbent.

Step 7

Suppose that

$$\begin{aligned} {\hat{x}}(1,0)<x^{*} \end{aligned}$$

and that in equilibrium all the media outlets contact the Incumbent and the median voter follows a media outlet. Then

$$\begin{aligned} {\hat{x}}(p_{I},1)\ge x^{*} \end{aligned}$$

and

$$\begin{aligned} \frac{{\hat{x}}(p_{I},1)}{m}\ge 1-{\hat{x}}(p_{I},p_{C}) \end{aligned}$$

Notice that if

$$\begin{aligned} {\hat{x}}(p_{I},1)>x^{*} \end{aligned}$$

and

$$\begin{aligned} \frac{{\hat{x}}(p_{I},1)}{m}>1-{\hat{x}}(p_{I},p_{C}) \end{aligned}$$

then, by continuity of \({\hat{x}}(p_{I},p_{C}),\) the Incumbent can increase \(p_{I}\) by \(\epsilon\) such that

$$\begin{aligned} {\hat{x}}(p_{I}+\epsilon ,1)>x^{*} \end{aligned}$$

and

$$\begin{aligned} \frac{{\hat{x}}(p_{I}+\epsilon ,1)}{m}>1-{\hat{x}}(p_{I}+\epsilon ,p_{C}) \end{aligned}$$

in which case her payoff increases from

$$\begin{aligned} \frac{1+p_{I}}{2} \end{aligned}$$

to

$$\begin{aligned} \frac{1+p_{I}+\epsilon }{2} \end{aligned}$$

Hence, either

$$\begin{aligned} {\hat{x}}(p_{I},1)=x^{*} \end{aligned}$$

or

$$\begin{aligned} \frac{{\hat{x}}(p_{I},1)}{m}=1-{\hat{x}}(p_{I},p_{C}) \end{aligned}$$

Similarly, if \(k\in [1,m-1]\) media outlets contact the Incumbent, then it must be that

$$\begin{aligned} {\hat{x}}(p_{I},p_{C})\ge x^{*} \end{aligned}$$

and

$$\begin{aligned} {\hat{x}}(p_{I},p_{C})\ge \frac{1-{\hat{x}}(1,p_{C})}{m} \end{aligned}$$

with at least one expression holding as an equality.

Step 8

Suppose that

$$\begin{aligned} {\hat{x}}(1,0)<x^{*} \end{aligned}$$

and

$$\begin{aligned} {\hat{x}}(p_{I},p_{C})=x^{*} \end{aligned}$$

and

$$\begin{aligned} {\hat{x}}(p_{I},p_{C})\ge \frac{1-{\hat{x}}(1,p_{C})}{m} \end{aligned}$$

Notice that the median voter follows an outlet that contacts the Incumbent, so the Challenger wins with probability smaller than \(\frac{1}{2}\). Suppose that \(p_{I}>0\) and

$$\begin{aligned} {\hat{x}}(p_{I},0)<\frac{1-{\hat{x}}(1,0)}{m} \end{aligned}$$

Notice that

$$\begin{aligned} {\hat{x}}(p_{I},0)<{\hat{x}}(p_{I},p_{C})=x^{*} \end{aligned}$$

Hence, if the Challenger deviates and sets

$$\begin{aligned} p_{I}=0 \end{aligned}$$

then at least one media outlet will contact her and the median voter will follow such a media outlet. Hence, the Challenger will win with probability \(\frac{1}{2}\). Therefore, she will deviate. Hence, if, in equilibrium

$$\begin{aligned} {\hat{x}}(p_{I},p_{C})=x^{*} \end{aligned}$$

and

$$\begin{aligned} {\hat{x}}(p_{I},p_{C})\ge \frac{1-{\hat{x}}(p_{I},1)}{m} \end{aligned}$$

then

$$\begin{aligned} {\hat{x}}(p_{I},0)=x^{*} \end{aligned}$$

and

$$\begin{aligned} {\hat{x}}(p_{I},0)\ge \frac{1-{\hat{x}}(p_{I},1)}{m} \end{aligned}$$

Notice that

$$\begin{aligned} {\hat{x}}(p_{I},0)=x^{*}={\hat{x}}(p_{I},p_{C})\Rightarrow p_{C}=0 \end{aligned}$$

Suppose now that

$$\begin{aligned} p_{C}>0 \end{aligned}$$

and

$$\begin{aligned} {\hat{x}}(p_{I},p_{C})>x^{*} \end{aligned}$$

and

$$\begin{aligned} {\hat{x}}(p_{I},p_{C})=\frac{1-{\hat{x}}(p_{I},1)}{m} \end{aligned}$$

Notice that by construction

$$\begin{aligned} {\hat{x}}(1,0)<x^{*} \end{aligned}$$

and

$$\begin{aligned} {\hat{x}}(p_{I},p_{C})=\frac{1-{\hat{x}}(p_{I},p_{C})}{m}\Rightarrow {\hat{x}}(1,0)<\frac{1-{\hat{x}}(1,0)}{m} \end{aligned}$$

Thus, if the Challenger deviates and sets

$$\begin{aligned} p_{C}=0 \end{aligned}$$

then all the media outlets will contact her and the median voter will follow a media outlet, so the Challenger will win with probability \(\frac{1}{2}\). Hence, the Challenger will deviate. As in the previous case, it must be that

$$\begin{aligned} p_{C}=0 \end{aligned}$$

Step 9

Suppose that

$$\begin{aligned} {\hat{x}}(1,0)<x^{*} \end{aligned}$$

and

$$\begin{aligned} p_{C}>0 \end{aligned}$$

and

$$\begin{aligned} {\hat{x}}(p_{I},1)=x^{*} \end{aligned}$$

and

$$\begin{aligned} \frac{{\hat{x}}(p_{I},1)}{m}\ge 1-{\hat{x}}(p_{I},p_{C}) \end{aligned}$$

and

$$\begin{aligned} \frac{{\hat{x}}(p_{I},1)}{m}<1-{\hat{x}}(p_{I},0) \end{aligned}$$

If the Challenger deviates and sets

$$\begin{aligned} p_{C}=0 \end{aligned}$$

then at least one media outlet will contact her. Moreover

$$\begin{aligned} {\hat{x}}(p_{I},0)<{\hat{x}}(p_{I},1)=x^{*} \end{aligned}$$

and therefore the median voter will read an outlet that contacts the Challenger. Hence, the Challenger will win with probability \(\frac{1}{2}\) instead of 0, so she will deviate. Therefore, it must be that

$$\begin{aligned} p_{C}=0 \end{aligned}$$

Suppose that

$$\begin{aligned} p_{C}>0 \end{aligned}$$

and

$$\begin{aligned} {\hat{x}}(p_{I},1)>x^{*} \end{aligned}$$

and

$$\begin{aligned} \frac{{\hat{x}}(p_{I},1)}{m}=1-{\hat{x}}(p_{I},p_{C}) \end{aligned}$$

Notice that

$$\begin{aligned} \frac{{\hat{x}}(p_{I},1)}{m}=1-{\hat{x}}(p_{I},p_{C}) \end{aligned}$$

implies

$$\begin{aligned} {\hat{x}}(p_{I},p_{C})\ge \frac{1-{\hat{x}}(p_{I},1)}{m} \end{aligned}$$

Hence,

$$\begin{aligned} {\hat{x}}(p_{I},1)>x^{*} \end{aligned}$$

and

$$\begin{aligned} \frac{{\hat{x}}(p_{I},1)}{m}=1-{\hat{x}}(p_{I},p_{C}) \end{aligned}$$

implies

$$\begin{aligned} {\hat{x}}(p_{I},1)>x^{*} \end{aligned}$$

and

$$\begin{aligned} {\hat{x}}(p_{I},p_{C})\ge \frac{1-{\hat{x}}(1,p_{C})}{m} \end{aligned}$$

which we have already considered in the previous Step.

Step 10

We have shown in the previous steps that if

$$\begin{aligned} {\hat{x}}(1,0)<x^{*} \end{aligned}$$

then, in equilibrium,

$$\begin{aligned} p_{C}=0 \end{aligned}$$

and either

$$\begin{aligned} \left( {\hat{x}}(p_{I},0)=x^{*}\text {and}\, {{\hat{x}}(p_{I},0)\ge \frac{1-{\hat{x}}(1,0)}{m}}\right) \end{aligned}$$

or

$$\begin{aligned} \left( {\hat{x}}(p_{I},0)>x^{*}\text {and}\, {{\hat{x}}(p_{I},0)=\frac{1-{\hat{x}}(1,0)}{m}}\right) \end{aligned}$$

or

$$\begin{aligned} \left( {\hat{x}}(p_{I},1)=x^{*}\text {and}\, {\frac{{\hat{x}}(p_{I},1)}{m}\ge 1-{\hat{x}}(p_{I},0)}\right) \end{aligned}$$

or

$$\begin{aligned} \left( {\hat{x}}(p_{I},1)>x^{*}\text {and}\, {\frac{{\hat{x}}(p_{I},1)}{m}=1-{\hat{x}}(p_{I},0)}\right) \end{aligned}$$

Notice that

$$\begin{aligned} {\hat{x}}(p_{I},0)\ge x^{*}\Leftrightarrow p_{I}\le \frac{(1-r)(1-x^{*})}{x^{*}} \end{aligned}$$

and

$$\begin{aligned} {\hat{x}}(p_{I},0)\ge \frac{1-{\hat{x}}(1,0)}{m} \Leftrightarrow \\ {\left\{ \begin{array}{ll} p_{I}\le (1-r)(m(2-r)-1) & \text {if}\, {\frac{m-1}{m}<r}\\ p_{I}\in [0,1] & \text {if }{{\frac{m-1}{m}\ge r}} \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned}&{\hat{x}}(p_{I},1)\ge x^{*}\Leftrightarrow \\&{\left\{ \begin{array}{ll} p_{I}\in [0,1] & \text {if}\, {x^{*}\in [0,\frac{1}{2}]}\\ p_{I}\le \frac{1-x^{*}}{x^{*}} & \text {if}\, {x^{*}\in [\frac{1}{2},1]} \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned}&\frac{{\hat{x}}(p_{I},1)}{m}\ge 1-{\hat{x}}(p_{I},0)\Leftrightarrow \\&p_{I}\le \frac{m\left( \sqrt{\frac{m^{2}-4mr+2m+1}{m^{2}}}-1\right) +1}{2m} \end{aligned}$$

Define

$$\begin{aligned} p_{1}=\frac{(1-r)(1-x^{*})}{x^{*}} \end{aligned}$$

and

$$\begin{aligned} p_{2}=(1-r)(m(2-r)-1) \end{aligned}$$

And observe that if

$$\begin{aligned} p_{I}=\min \{p_{1},p_{2}\} \end{aligned}$$

then at least one media outlet contacts the Incumbent and the median voter reads such a media outlet. Moreover, if

$$\begin{aligned} p_{I}=p_{2} \end{aligned}$$

then exactly one media outlet contacts the Incumbent. Define

$$\begin{aligned} p_{3}=\frac{1-x^{*}}{x^{*}} \end{aligned}$$

and

$$\begin{aligned} p_{4}=\frac{m\left( \sqrt{\frac{m^{2}-4mr+2m+1}{m^{2}}}-1\right) +1}{2m} \end{aligned}$$

Observe that if

$$\begin{aligned} p_{I}=\min \{p_{3},p_{4}\} \end{aligned}$$

then all the media outlets contact the Incumbent and the median voter reads such a media outlet. Hence, the Incumbent maximizes her payoff by setting

$$\begin{aligned} p_{I}=\max \{\min \{p_{1},p_{2},1\},\min \{p_{3},p_{4},1\}\} \end{aligned}$$

Solving the corresponding inequalities shows that for \(j\in \{1,2,3,4\}:\)

$$\begin{aligned} p_{j}=\max \{\min \{p_{1},p_{2},1\},\min \{p_{3},p_{4},1\}\}\Rightarrow p_{j}\in (0,1) \end{aligned}$$

and so we can write

$$\begin{aligned} p_{I}=\max \{\min \{p_{1},p_{2}\},\min \{p_{3},p_{4}\}\} \end{aligned}$$

Solving the equation

$$\begin{aligned} p_{1}=\max \{\min \{p_{1},p_{2}\},\min \{p_{3},p_{4}\}\} \end{aligned}$$

yields

$$\begin{aligned} \frac{1}{m(2-r)}\le x^{*}\le \frac{m\left( r\left( \sqrt{\frac{m^{2}-4mr+2m+1}{m^{2}r^{2}}}+2\right) -1\right) -1}{2mr} \end{aligned}$$

Solving

$$\begin{aligned} p_{2}=\max \{\min \{p_{1},p_{2}\},\min \{p_{3},p_{4}\}\} \end{aligned}$$

yields

$$\begin{aligned} 0<x^{*}\le \frac{1}{m(2-r)} \end{aligned}$$

Notice that

$$\begin{aligned} {\hat{x}}(p_{I},0)=\frac{1}{m(2-r)} \end{aligned}$$

Solving

$$\begin{aligned} p_{3}=\max \{\min \{p_{1},p_{2}\},\min \{p_{3},p_{4}\}\} \end{aligned}$$

yields

$$\begin{aligned} \frac{-r\sqrt{\frac{m^{2}-4mr+2m+1}{r^{2}}}+m+1}{2r}\le x^{*}<1 \end{aligned}$$

and solving

$$\begin{aligned} p_{4}=\max \{\min \{p_{1},p_{2}\},\min \{p_{3},p_{4}\}\}\ne p_{3} \end{aligned}$$

yields

$$\begin{aligned} \frac{m\left( r\left( \sqrt{\frac{m^{2}-4mr+2m+1}{m^{2}r^{2}}}+2\right) -1\right) -1}{2mr}\le x^{*}\le \frac{m+1-r\sqrt{\frac{m^{2}-4mr+2m+1}{r^{2}}}}{2r} \end{aligned}$$

Notice that

$$\begin{aligned} {\hat{x}}(p_{4},1)=\frac{2m}{m\sqrt{\frac{m^{2}-4mr+2m+1}{m^{2}}}+m+1} \end{aligned}$$

To summarize, if

$$\begin{aligned} 0<x^{*}<\frac{1}{m(2-r)} \end{aligned}$$

then

$$\begin{aligned} p_{I}=p_{2}=(1-r)(m(2-r)-1) \end{aligned}$$

exactly one media outlet contacts the Incumbent, voters

$$\begin{aligned} x\le \frac{1}{m(2-r)} \end{aligned}$$

follow a media outlet that contacts the Incumbent and voters

$$\begin{aligned} x>\frac{1}{m(2-r)} \end{aligned}$$

follow a media outlet that contacts the Challenger.

If

$$\begin{aligned} \frac{1}{m(2-r)}\le x^{*}\le \frac{m\left( r\left( \sqrt{\frac{m^{2}-4mr+2m+1}{m^{2}r^{2}}}+2\right) -1\right) -1}{2mr} \end{aligned}$$

then

$$\begin{aligned} p_{I}=p_{1}=\frac{(1-r)(1-x^{*})}{x^{*}} \end{aligned}$$

\(k\in [1,m-1]\) media outlets contact the Incumbent, voters \(x\le x^{*}\) follow media outlets that contact the Incumbent, and voters \(x>x^{*}\) follow media outlets that contact the Challenger. If

$$\begin{aligned} \frac{m\left( r\left( \sqrt{\frac{m^{2}-4mr+2m+1}{m^{2}r^{2}}}+2\right) -1\right) -1}{2mr}\le x^{*}\le \frac{m+1-r\sqrt{\frac{m^{2}-4mr+2m+1}{r^{2}}}}{2r} \end{aligned}$$

then

$$\begin{aligned} p_{I}=p_{4}=\frac{m\left( \sqrt{\frac{m^{2}-4mr+2m+1}{m^{2}}}-1\right) +1}{2m} \end{aligned}$$

all media outlets contact the Incumbent, voters

$$\begin{aligned} x\le \frac{2m}{m\sqrt{\frac{m^{2}-4mr+2m+1}{m^{2}}}+m+1} \end{aligned}$$

follow outlets that contact the Incumbent and voters

$$\begin{aligned} x>\frac{2m}{m\sqrt{\frac{m^{2}-4mr+2m+1}{m^{2}}}+m+1} \end{aligned}$$

always vote for the Challenger.

Finally, if

$$\begin{aligned} \frac{m+1-r\sqrt{\frac{m^{2}-4mr+2m+1}{r^{2}}}}{2r}<x^{*}<1 \end{aligned}$$

then

$$\begin{aligned} p_{I}=p_{3}=\frac{1-x^{*}}{x^{*}} \end{aligned}$$

all the media outlets contact the Incumbent, voters \(x\le x^{*}\) follow outlets that contact the Incumbent and voters \(x>x^{*}\) always vote for the Incumbent.

Step 11

Let \(m=1\). As before, the Incumbent can obtain a payoff strictly higher than \(\frac{1}{2}\). If

$$\begin{aligned} {\hat{x}}(1,0)<x^{*} \end{aligned}$$

the Incumbent sets \(p_{I}\) such that the only media outlet contacts her and the median voter reads such a media outlet. Hence, similarly to the previous steps

$$\begin{aligned} {\hat{x}}(p_{I},1)\ge 1-{\hat{x}}(1,p_{C}) \end{aligned}$$

and

$$\begin{aligned} {\hat{x}}(p_{I},1)\ge x^{*} \end{aligned}$$

with at least one expression being an equality. Notice that it must be that

$$\begin{aligned} {\hat{x}}(p_{I},1)\ge 1-{\hat{x}}(1,0) \end{aligned}$$

because otherwise the Challenger can deviate by setting \(p_{C}=0\), in which case the media outlet will contact the Challenger and the median voter will follow it. Hence, the Incumbent sets the highest \(p_{I}\) such that

$$\begin{aligned} \left( {\hat{x}}(p_{I},1)=1-{\hat{x}}(1,0)\text { and}\, {{\hat{x}}(p_{I},1)\ge x^{*}}\right) \end{aligned}$$

or

$$\begin{aligned} \left( {\hat{x}}(p_{I},1)>1-{\hat{x}}(1,0)\text { and}\, {{\hat{x}}(p_{I},1)\ge x^{*}} \right) \end{aligned}$$

Solving

$$\begin{aligned} {\hat{x}}(p_{I},1)=1-{\hat{x}}(1,0) \end{aligned}$$

yields

$$\begin{aligned} p_{I}=\frac{1-x^{*}}{x^{*}} \end{aligned}$$

and solving

$$\begin{aligned} {\hat{x}}(p_{I},1)=1-{\hat{x}}(1,0) \end{aligned}$$

yields

$$\begin{aligned} p_{I}=1-r \end{aligned}$$

Similarly to the previous steps, the Incumbent sets

$$\begin{aligned} p_{I}=\min \{\frac{1-x^{*}}{x^{*}},1-r\} \end{aligned}$$

If

$$\begin{aligned} x^{*}>\frac{1}{2-r} \end{aligned}$$

then

$$\begin{aligned} 1-r>\frac{1-x^{*}}{x^{*}}\Rightarrow p_{I}=\frac{1-x^{*}}{x^{*}} \end{aligned}$$

In this case, voters \(x\le x^{*}\)follow the media outlet and voters \(x>x^{*}\) always vote for the Challenger. Notice that solving the inequalities

$$\begin{aligned} x^{*}>\frac{1}{2-r} \end{aligned}$$

and

$$\begin{aligned} {\hat{x}}(1,0)<x^{*} \end{aligned}$$

together yields

$$\begin{aligned} x^{*}>\frac{1}{2-r} \end{aligned}$$

If

$$\begin{aligned} \frac{1-r}{2-r}<x^{*}<\frac{1}{2-r} \end{aligned}$$

then

$$\begin{aligned} 1-r<\frac{1-x^{*}}{x^{*}}\Rightarrow p_{I}=1-r \end{aligned}$$

In this case, voters

$$\begin{aligned} x\le {\hat{x}}(1-r,1)=\frac{1}{2-r} \end{aligned}$$

follow the media outlet and voters

$$\begin{aligned} x>\frac{1}{2-r} \end{aligned}$$

always vote for the Challenger. In both cases, the media outlet contacts the Incumbent.

Similarly to the previous cases, if

$$\begin{aligned} x^{*}<{\hat{x}}(1,0) \end{aligned}$$

the Incumbent sets

$$\begin{aligned} p_{I}=1 \end{aligned}$$

and voters

$$\begin{aligned} x\le {\hat{x}}(1,0) \end{aligned}$$

always vote for the Incumbent while voters

$$\begin{aligned} x>{\hat{x}}(1,0) \end{aligned}$$

always vote for the Challenger. This completes the proof. \(\square\)

Proof of Corollary 1

1. Clearly, if

$$\begin{aligned} x^{*}\le \frac{1-r}{2-r} \end{aligned}$$

then

$$\begin{aligned} p_{I}=1 \end{aligned}$$

and therefore \(p_{I}\) is constant in m.

In the rest of the proof, I first prove the statement for \(m\in \{1,2\}\) and then prove it for \(m\ge 2\). Suppose that \(m=1\) and

$$\begin{aligned} \frac{1-r}{2-r}<x^{*}<\frac{1}{2(2-r)} \end{aligned}$$

Then

$$\begin{aligned} p_{I}=1-r \end{aligned}$$

Notice that

$$\begin{aligned} \frac{1-r}{2-r}<x^{*}<\frac{1}{2(2-r)} \end{aligned}$$

implies

$$\begin{aligned} p_{I}=p_{2}=(1-r)(m(2-r)-1) \end{aligned}$$

Observe that

$$\begin{aligned} (1-r)(m(2-r)-1)\mid _{m=2}=(1-r)(3-2r)>1-r \end{aligned}$$

Suppose now that \(m=1\) and

$$\begin{aligned} \frac{1}{2(2-r)}<x^{*}<\frac{1}{2} \end{aligned}$$

Notice that

$$\begin{aligned} \frac{1}{2(2-r)}<x^{*}<\frac{1}{2}<\frac{1}{2-r} \end{aligned}$$

Therefore, again

$$\begin{aligned} p_{I}=1-r \end{aligned}$$

Notice that \(m=2\) and

$$\begin{aligned} \frac{1}{2(2-r)}<x^{*}<\frac{1}{2} \end{aligned}$$

imply

$$\begin{aligned} p_{I}=p_{1}=\frac{(1-r)(1-x^{*})}{x^{*}}>1-r \end{aligned}$$

Hence, if

$$\begin{aligned} 0<x^{*}<\frac{1}{2} \end{aligned}$$

and m increases from 1 to 2, then \(p_{I}\) increases.

Assume now that \(m=1\) and

$$\begin{aligned} \frac{1}{2}<x^{*}<\frac{1}{2-r} \end{aligned}$$

Then

$$\begin{aligned} p_{I}=1-r \end{aligned}$$

If \(m=2\) then

$$\begin{aligned} p_{I}=p_{1}=\frac{(1-r)(1-x^{*})}{x^{*}}<1-r \end{aligned}$$

or

$$\begin{aligned} p_{I}=p_{4}=\frac{m\left( \sqrt{\frac{m^{2}-4mr+2m+1}{m^{2}}}-1\right) +1}{2m}\mid _{m=2}=\frac{1}{4}\left( \sqrt{9-8r}-1\right) <1-r \end{aligned}$$

Finally, let \(m=1\) and

$$\begin{aligned} x^{*}>\frac{1}{2-r} \end{aligned}$$

Then

$$\begin{aligned} p_{I}=\frac{1-x^{*}}{x^{*}} \end{aligned}$$

If \(m=2\) and

$$\begin{aligned} x^{*}>\frac{1}{2-r} \end{aligned}$$

then

$$\begin{aligned} p_{I}=p_{3}=\frac{1-x^{*}}{x^{*}} \end{aligned}$$

or

$$\begin{aligned} p_{I}=p_{1}=\frac{(1-r)(1-x^{*})}{x^{*}}<\frac{1-x^{*}}{x^{*}} \end{aligned}$$

or

$$\begin{aligned} p_{I}=p_{4}=\frac{m\left( \sqrt{\frac{m^{2}-4mr+2m+1}{m^{2}}}-1\right) +1}{2m}\mid _{m=2}=\frac{1}{4}\left( \sqrt{9-8r}-1\right) \end{aligned}$$

which is smaller than

$$\begin{aligned} \frac{1-x^{*}}{x^{*}} \end{aligned}$$

given that

$$p_{I}=p_{4}$$

Hence, if

$$\begin{aligned} \frac{1}{2}<x^{*}<1 \end{aligned}$$

and m increases from 1 to 2, then \(p_{I}\) does not increase.

Suppose now that \(m\ge 2\). Write

$$\begin{aligned} p_{2}(\mu )=(1-r)(1-\mu (2-r)) \end{aligned}$$

and

$$\begin{aligned} p_{4}(\mu )=\frac{\mu \left( \sqrt{\frac{\mu ^{2}-4\mu r+2\mu +1}{\mu ^{2}}}-1\right) +1}{2\mu } \end{aligned}$$

where \(\mu \in {{\mathbb {R}}}\) and \(\mu \ge 2\). Write

$$\begin{aligned} p_{I}(\mu )=\max \{\min \{p_{1},p_{2}(\mu )\},\{p_{3},p_{4}(\mu )\}\} \end{aligned}$$

Notice that \(\mu \ge 2\) and \(0<r<1\) imply

$$\begin{aligned} \frac{\mu ^{2}-4\mu r+2\mu +1}{\mu ^{2}}>0 \end{aligned}$$

Hence, \(p_{4}(\mu )\) is continuous in \(\mu\). It is clear that \(p_{1},p_{2},\) and \(p_{3}(\mu )\) are also continuous in \(\mu\). Hence, \(p_{I}(\mu )\) is continuous in \(\mu\). Notice that

$$\begin{aligned} 0<x^{*}<\frac{1}{2} \end{aligned}$$

implies that

$$\begin{aligned} x^{*}<\frac{\mu \left( r\left( \sqrt{\frac{\mu ^{2}-4\mu r+2\mu +1}{\mu ^{2}r^{2}}}+2\right) -1\right) -1}{2\mu r} \end{aligned}$$

and therefore either

$$\begin{aligned} p_{I}(\mu )=p_{1} \end{aligned}$$

or

$$\begin{aligned} p_{I}(\mu )=p_{2}(\mu ) \end{aligned}$$

Now,

$$\begin{aligned} \frac{d}{d\mu }p_{1}=0 \end{aligned}$$

and

$$\begin{aligned} \frac{d}{d\mu }p_{2}(\mu )=\frac{d}{d\mu }(1-r)(1-\mu (2-r))=(2-r)(1-r)>0 \end{aligned}$$

Hence, if

$$\begin{aligned} 0<x^{*}<\frac{1}{2} \end{aligned}$$

then \(p_{I}(\mu )\) weakly increases in \(\mu .\) Hence, for any \(m\in {{\mathbb {N}}},m\ge 2\) it must be that

$$\begin{aligned} p_{I}(m+1)\ge p_{I}(m) \end{aligned}$$

Suppose instead that

$$\begin{aligned} \frac{1}{2}<x^{*}<1 \end{aligned}$$

Then for any \(\mu \ge 2\)

$$\begin{aligned} x^{*}>\frac{1}{\mu (2-r)} \end{aligned}$$

and therefore

$$\begin{aligned} p_{I}(\mu )=p_{1} \end{aligned}$$

or

$$\begin{aligned} p_{I}(\mu )=p_{3} \end{aligned}$$

or

$$\begin{aligned} p_{I}(\mu )=p_{4}(\mu ) \end{aligned}$$

Notice that

$$\begin{aligned} \frac{d}{d\mu }p_{1}=\frac{d}{d\mu }p_{3}=0 \end{aligned}$$

and

$$\begin{aligned}&\frac{d}{d\mu }p_{4}(\mu )=\frac{d}{d\mu }\frac{\mu \left( \sqrt{\frac{\mu ^{2}+2\mu -4\mu r+1}{\mu ^{2}}}-1\right) +1}{2\mu }\\&\quad =\frac{-\mu \left( \sqrt{\frac{\mu ^{2}+2\mu -4\mu r+1}{\mu ^{2}}}-2r+1\right) -1}{2\mu ^{3}\sqrt{\frac{\mu ^{2}+2\mu -4\mu r+1}{\mu ^{2}}}}<0 \end{aligned}$$

Therefore, if

$$\begin{aligned} \frac{1}{2}<x^{*}<1 \end{aligned}$$

then \(p_{I}(\mu )\) is non-increasing in \(\mu\). Hence, for any \(m\in {{\mathbb {N}}},m\ge 2\)

$$\begin{aligned} p_{I}(m+1)\le p_{I}(m) \end{aligned}$$

This proves part 1.

Notice that for any \(m\ge 1\) assuming that

$$\begin{aligned} \frac{1-r}{2-r}<x^{*}<\frac{1}{(m+1)(2-r)} \end{aligned}$$

implies

$$\begin{aligned} \frac{1-r}{2-r}<x^{*}<\frac{1}{m(2-r)} \end{aligned}$$

Thus, if m increases by 1, \(p_{I}\) strictly increases. Further, notice that solving the inequality

$$\begin{aligned} \frac{1-r}{2-r}<\frac{1}{(m+1)(2-r)} \end{aligned}$$

yields

$$\begin{aligned} \frac{m}{m+1}<r<1 \end{aligned}$$

Hence, the conditions

$$\begin{aligned} \frac{1-r}{2-r}<x^{*}<\frac{1}{(m+1)(2-r)} \end{aligned}$$

and

$$\begin{aligned} \frac{m}{m+1}<r<1 \end{aligned}$$

imply that \(p_{I}\) strictly increases if m increases by 1.

To show that the Social Welfare of Voters increases and decreases in the respective cases, note that, if

$$\begin{aligned} \frac{1-r}{2-r}<x^{*}<1 \end{aligned}$$

and

$$\begin{aligned} {{\mathbb {P}}}[d=C\mid \Theta =0]=1-p_{I} \end{aligned}$$

The Social Welfare of Voters is given by

$$\begin{aligned}&P[\Theta =1]\times \left( P[d=I\mid \Theta =1]\times \int _{{{{\underline{x}}}}}^{{\bar{x}}}(\theta -x)dx+P[d=C\mid \Theta =1]\times \int _{{{{\underline{x}}}}}^{{\bar{x}}}(x-\theta )dx\right) \\&\quad +{{\mathbb {P}}}[\Theta =0]\times \left( P[d=I\mid \Theta =0]\times \int _{{{{\underline{x}}}}}^{{\bar{x}}}(\theta -x)dx+P[d=C\mid \Theta =0]\times \int _{{{{\underline{x}}}}}^{{\bar{x}}}(x-\theta )dx\right) \end{aligned}$$

and only depends on m through \({{\mathbb {P}}}[d=C\mid \Theta =0]\). Hence, increasing (decreasing) \(p_{I}\) decreases (increases) the Social Welfare of Voters. \(\square\)

Proof of Corollaries 2 and 3

If

$$\begin{aligned} x^{*}\le \frac{1-r}{2-r} \end{aligned}$$

then the statement is trivial because \(p_{I}=1\).

Suppose that \(m=1\) and

$$\begin{aligned} \frac{1-r}{2-r}<x^{*}<\frac{1}{2-r} \end{aligned}$$

Then

$$\begin{aligned} p_{I}=1-r \end{aligned}$$

which clearly decreases in r and is constant in \(x^{*}\). If \(m=1\) and

$$\begin{aligned} x^{*}>\frac{1}{2-r} \end{aligned}$$

then

$$\begin{aligned} p_{I}=\frac{1-x^{*}}{x^{*}} \end{aligned}$$

which is constant in r and decreases in \(x^{*}\). For the case when \(m\ge 2\), notice that

$$\begin{aligned} \frac{d}{dr}p_{1}=\frac{d}{dr}\frac{(1-r)(1-x^{*})}{x^{*}}=-\frac{(1-x^{*})}{x^{*}}<0 \end{aligned}$$

and

$$\begin{aligned} \frac{d}{dx^{*}}p_{1}=\frac{d}{dx^{*}}\frac{(1-r)(1-x^{*})}{x^{*}}=-\frac{1-r}{x^{*2}}<0 \end{aligned}$$

and

$$\begin{aligned} \frac{d}{dr}p_{2}=\frac{d}{dr}(1-r)(1-m(2-r))=1-m(3-2r)<0 \end{aligned}$$

and

$$\begin{aligned} \frac{d}{dx^{*}}p_{2}=\frac{d}{dx^{*}}(1-r)(1-m(2-r))=0 \end{aligned}$$

and

$$\begin{aligned} \frac{d}{dr}p_{3}=\frac{d}{dr}\frac{1-x^{*}}{x^{*}}=0 \end{aligned}$$

and

$$\begin{aligned} \frac{d}{dx^{*}}p_{3}=\frac{d}{dx^{*}}\frac{1-x^{*}}{x^{*}}=-\frac{1}{x^{*2}}<0 \end{aligned}$$

and

$$\begin{aligned} \frac{d}{dr}p_{4}=\frac{d}{dr}\frac{m\left( \sqrt{\frac{m^{2}-4mr+2m+1}{m^{2}}}-1\right) +1}{2m}=-\frac{1}{m\sqrt{\frac{m^{2}-4mr+2m+1}{m^{2}}}}<0 \end{aligned}$$

and

$$\begin{aligned} \frac{d}{dx^{*}}p_{4}=\frac{d}{dx^{*}}\frac{m\left( \sqrt{\frac{m^{2}-4mr+2m+1}{m^{2}}}-1\right) +1}{2m}=0 \end{aligned}$$

Notice that

$$\begin{aligned} p_{I}=\max \{\min \{p_{1},p_{2}\},\{p_{3},p_{4}\}\} \end{aligned}$$

is continuous in r and \(x^{*}\). Hence, \(p_{I}\) weakly decreases in rand \(x^{*}\).

To prove the welfare implication of increasing r, notice that

$$\begin{aligned} {{\mathbb {P}}}[d=C\mid \Theta =0]=1-p_{I} \end{aligned}$$

and the Social Welfare of Voters only depends on r through \({{\mathbb {P}}}[d=C\mid \Theta =0]\). Hence, the Social Welfare of Voters increases in r. \(\square\)

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Samsonov, A. The fragmentation of views in a democracy. Econ Gov 22, 215–256 (2021). https://doi.org/10.1007/s10101-021-00258-7

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Keywords

  • Polarization
  • Persuasion
  • Signaling
  • Media market
  • Voting
  • Regulation

JEL Classification

  • C72
  • D72
  • D80