Inequality, participation, and polarization

Abstract

The upward co-movement of income inequality and partisan polarization in the U.S. is typically attributed to intensified class conflict or a political wealth bias. This paper formalizes a theory of polarization where changes in the income distribution do not affect citizens’ policy preferences, but instead change their patterns of political participation: aggregate voting decreases relative to aggregate giving, reducing the electoral penalty for partisan policies. By endogenizing party composition the model captures both the ideological and compositional dimensions of polarization, and addresses less-discussed polarization features, such as intra-party homogeneity and the increase in safe seats. According to the model, observed polarization patterns imply that parties have diverged more than candidates, and that the gap between party and candidate divergence has increased with income inequality.

Appendix

Proof of Proposition 1

Consider two income distributions $Y_{\lambda _{1}}(y)$ and $Y_{\lambda _{2}}(y)$ with inequality levels $\lambda _{1},\lambda _{2}$ and equal means: $\int _{0}^{\infty }ydY_{\lambda _{1}}(y)=\int _{0}^{\infty }ydY_{\lambda _{2}}(y)=1$. It is known that under these conditions, $\lambda _{1}<\lambda _{2}$ if and only if $Y_{\lambda _{1}}(y)$ second-order stochastically dominates $Y_{\lambda _{2}}(y).$ Then, by SOSD, because $\nu \left( y\right)$ is strictly concave in income y,  it follows that $v(\lambda _{1})=\int _{0}^{\infty }\nu \left( y\right) dY_{\lambda _{1}}\left( y\right) >\int _{0}^{\infty }\nu \left( y\right) dY_{\lambda _{2}}\left( y\right) =v(\lambda _{2}),$ implying that $\frac{dv(\lambda )}{ d\lambda }<0.$ Because $\gamma \left( y\right)$ is strictly convex in income y,  we have $g(\lambda _{1})=\int _{0}^{\infty }\gamma \left( y\right) dY_{\lambda _{1}}\left( y\right) <\int _{0}^{\infty }\gamma \left( y\right) dY_{\lambda _{2}}\left( y\right) =g(\lambda _{2}),$ implying that $\frac{dg(\lambda )}{d\lambda }>0.$ $\square$

Proof of Proposition 2

Solve for an L candidate’s winning probability:

$$\begin{aligned} w_{L_{s}}\left( {\overline{x}}_{s},\lambda \right)= & {} {{\mathbb {P}}}\left\{ \frac{ \left[ Z_{s}\left( {\overline{x}}_{s}\right) \right] ^{\frac{v\left( \lambda \right) }{v\left( \lambda \right) +g\left( \lambda \right) }}\left[ D_{s}\left( {\overline{x}}_{s}|{\tilde{\psi }}_{s}\right) \right] ^{\frac{g\left( \lambda \right) }{v\left( \lambda \right) +g\left( \lambda \right) }}}{\left[ 1-Z_{s}\left( {\overline{x}}_{s}\right) \right] ^{\frac{v\left( \lambda \right) }{v\left( \lambda \right) +g\left( \lambda \right) }}\left[ 1-D_{s}\left( {\overline{x}}_{s}|{\tilde{\psi }}_{s}\right) \right] ^{\frac{ g\left( \lambda \right) }{v\left( \lambda \right) +g\left( \lambda \right) }} }\ge 1\right\} \nonumber \\= & {} {{\mathbb {P}}}\left\{ \frac{\left[ \frac{1}{1+e^{-\frac{{\overline{x}}_{s}-s}{ \sigma _{v}}}}\right] ^{\frac{v\left( \lambda \right) }{v\left( \lambda \right) +g\left( \lambda \right) }}\left[ \frac{1}{1+e^{-\frac{{\overline{x}} _{s}-s-{\tilde{\psi }}_{s}}{\sigma _{g}}}}\right] ^{\frac{g\left( \lambda \right) }{v\left( \lambda \right) +g\left( \lambda \right) }}}{\left[ 1- \frac{1}{1+e^{-\frac{{\overline{x}}_{s}-s}{\sigma _{v}}}}\right] ^{\frac{ v\left( \lambda \right) }{v\left( \lambda \right) +g\left( \lambda \right) }} \left[ 1-\frac{1}{1+e^{-\frac{{\overline{x}}_{s}-s-{\tilde{\psi }}_{s}}{\sigma _{g}}}}\right] ^{\frac{g\left( \lambda \right) }{v\left( \lambda \right) +g\left( \lambda \right) }}}\ge 1\right\} \nonumber \\= & {} {{\mathbb {P}}}\left\{ e^{\frac{{\overline{x}}_{s}-s}{\sigma _{v}}\frac{v\left( \lambda \right) }{v\left( \lambda \right) +g\left( \lambda \right) }}e^{ \frac{{\overline{x}}_{s}-s-{\tilde{\psi }}_{s}}{\sigma _{g}}\frac{g\left( \lambda \right) }{v\left( \lambda \right) +g\left( \lambda \right) }}\ge 1\right\} \end{aligned}$$

(18)

$$\begin{aligned}= & {} {{\mathbb {P}}}\left\{ \left[ \frac{v\left( \lambda \right) }{\sigma _{v}}+ \frac{g\left( \lambda \right) }{\sigma _{g}}\right] \left( {\overline{x}} _{s}-s\right) \ge {\tilde{\psi }}_{s}\frac{g\left( \lambda \right) }{\sigma _{g}}\right\} \nonumber \\= & {} {{\mathbb {P}}}\left\{ {\tilde{\psi }}_{s}\le \left[ 1+\frac{v\left( \lambda \right) /\sigma _{v}}{g\left( \lambda \right) /\sigma _{g}}\right] \left( {\overline{x}}_{s}-s\right) \right\} \nonumber \\= & {} \frac{1}{2}+\frac{1}{\psi }\left[ 1+\frac{v\left( \lambda \right) /\sigma _{v}}{g\left( \lambda \right) /\sigma _{g}}\right] \left( \frac{ x_{L_{s}}+x_{R_{s}}}{2}-s\right) \end{aligned}$$

(19)

and $w_{R_{s}}\left( {\overline{x}}_{s},\lambda \right) =1-w_{L_{s}}\left( {\overline{x}}_{s},\lambda \right) .$

The necessary conditions for a candidate equilibrium are, using $w_{L_{s}}+w_{R_{s}}=1$:

$$\begin{aligned} \frac{\partial }{\partial x_{L_{s}}}u_{L_{s}}= & {} -w_{L_{s}}+\left( x_{R_{s}}-x_{L_{s}}\right) \frac{\partial }{\partial x_{L_{s}}}w_{L_{s}}=0 \end{aligned}$$

(20)

$$\begin{aligned} \frac{\partial }{\partial x_{R_{s}}}u_{R_{s}}= & {} \left( 1-w_{L_{s}}\right) -\left( x_{R_{s}}-x_{L_{s}}\right) \frac{\partial }{\partial x_{R_{s}}} w_{L_{s}}=0 \end{aligned}$$

(21)

and adding these two first-order conditions, and observing that $\frac{ \partial }{\partial x_{L_{s}}}w_{L_{s}}=\frac{\partial }{\partial x_{R_{s}}} w_{L_{s}}=\frac{1}{2\psi }\left[ 1+\frac{v\left( \lambda \right) /\sigma _{v} }{g\left( \lambda \right) /\sigma _{g}}\right]$ from Eq. (19), yields $w_{L_{s}}=w_{R_{s}}=\frac{1}{2}.$ Using this in the first-order conditions, one can solve for the candidate equilibrium strategies$\left( x_{L_{s}},x_{R_{s}}\right) =\left( s-\frac{\Delta x}{2},s+ \frac{\Delta x}{2}\right)$ where candidate divergence is:

$$\begin{aligned} \Delta x\equiv x_{R_{s}}-x_{L_{s}}=\frac{w_{L_{s}}}{\frac{\partial }{ \partial x_{L_{s}}}w_{L_{s}}}=\frac{\psi }{1+\frac{v\left( \lambda \right) /\sigma _{v}}{g\left( \lambda \right) /\sigma _{g}}}. \end{aligned}$$

(22)

To establish that there are no global profitable deviations the following properties suffice: (i) $u_{L_{s}}(x_{L_{s}},x_{R_{s}})$ is strictly concave in $x_{L_{s}}$ on $(-\infty ,x_{R_{s}}],$ which implies that the $x_{L_{s}}$ that solves Eqs. (20)–(21) is a global maximizer of $u_{L_{s}}(x_{L_{s}},x_{R_{s}})$ on $(-\infty ,x_{R_{s}}]$, and (ii) a candidate $L_{s}$ deviation from the equilibrium $x_{L_{s}}$ to some$x_{L_{s}}^{\prime }>x_{R_{s}}$ strictly lowers $L_{s}$’s payoff. The argument for candidate $R_{s}$ is analogous. For part (i), note that $\frac{ \partial ^{2}}{\partial x_{L_{s}}^{2}}u_{L_{s}}(x_{L_{s}},x_{R_{s}})=-2\frac{ \partial }{\partial x_{L_{s}}}w_{L_{s}}+\left( x_{R_{s}}-x_{L_{s}}\right) \frac{\partial ^{2}}{\partial x_{L_{s}}^{2}}w_{L_{s}}$ and since by Eq. (19) we have that $w_{L_{s}}$ is linear and strictly increasing in $x_{L_{s}}$, it follows that $\frac{\partial ^{2}}{\partial x_{L_{s}}^{2}} u_{L_{s}}(x_{L_{s}},x_{R_{s}})=-2\frac{\partial }{\partial x_{L_{s}}} w_{L_{s}}<0.$ For part (ii), note that the policy lottery induced by the deviation is first-order stochastically dominated for $L_{s}$ since $u_{L_{s}}(x_{L_{s}}^{\prime },x_{R_{s}})<u_{L_{s}}(x_{R_{s}},x_{R_{s}})=-x_{R_{s}}\le -\frac{ x_{L_{s}}+x_{R_{s}}}{2}=u_{L_{s}}(x_{L_{s}},x_{R_{s}}).$ $\square$

Proof of Proposition 3

A candidate’s marginal electoral penalty for moving away from the district mean ideology s, namely $\frac{1}{2\psi }\left[ 1+\frac{v\left( \lambda \right) /\sigma _{v}}{g\left( \lambda \right) /\sigma _{g}}\right]$ is lower the lower the ratio $v\left( \lambda \right) /g\left( \lambda \right) ,$ namely the more important is giving relative to voting. Thus, a candidate policy-motivated deviation from the district mean ideology is more profitable the lower the ratio $v\left( \lambda \right) /g\left( \lambda \right)$. Taking the difference between the two first-order conditions (20)–(21) and solving for $x_{R_{s}}-x_{L_{s}}$ yields $\Delta x= \frac{1}{\frac{\partial }{\partial x_{L_{s}}}w_{L_{s}}+\frac{\partial }{ \partial x_{R_{s}}}w_{L_{s}}}=\frac{\psi }{1+\frac{v\left( \lambda \right) /\sigma _{v}}{g\left( \lambda \right) /\sigma _{g}}}.$ Then, using $v^{\prime }(\lambda )<0$ and $g^{\prime }(\lambda )>0$ from Proposition 1 gives that $\frac{\partial }{\partial \lambda }\Delta x>0.$ The comparative statics with respect to $\psi ,\sigma _{v},\sigma _{g}$ follow immediately from the expression for $\Delta x$. $\square$

Proof of Proposition 4

Parties anticipate that candidates will play the strategies characterized in Proposition 2. Given candidate strategies, parties calculate their seat shares and expected donations, which together will determine their policy weights. To show equilibrium existence and uniqueness, the proof consists of the following steps. (i) Necessary conditions for a party equilibrium in the case $X_{{\mathcal {L}}}\le -\frac{\Delta x}{2}\le \frac{\Delta x}{2}\le X_{{\mathcal {R}}}$. (ii) Ruling out global deviations. (iii) Ruling out a party equilibrium with$-\frac{\Delta x}{2}<X_{{\mathcal {L}}}\le X_{\mathcal {R }}<\frac{\Delta x}{2}.$

• (i) Consider an equilibrium with the property $X_{{\mathcal {L}}}\le -\frac{ \Delta x}{2}\le \frac{\Delta x}{2}\le X_{{\mathcal {R}}}$. To solve for the parties’ policy weights, it is necessary to derive the parties’ seat shares. See Eq. (12) and Fig. 3.

  $$\begin{aligned} h_{{\mathcal {L}}}= & {} \frac{1}{\mu }\left\{ \frac{1}{2}\left[ \left( X_{ {\mathcal {L}}}\!+\!\frac{\mu }{2}\right) -\left( -\frac{\Delta x}{2}-\frac{\mu }{2} \right) \right] +\frac{1}{2}\left[ \left( X_{{\mathcal {R}}}\!-\!\frac{\mu }{2} \right) \!-\!\left( \frac{\Delta x}{2}-\frac{\mu }{2}\right) \right] \right\} \nonumber \\= & {} \frac{1}{\mu }\left( \frac{\mu }{2}+\frac{X_{{\mathcal {L}}}+X_{{\mathcal {R}}} }{2}\right) \end{aligned}$$

  (23)
  $$\begin{aligned} h_{{\mathcal {R}}}= & {} \frac{1}{\mu }\left\{ \frac{1}{2}\left[ \left( \frac{ \Delta x}{2}+\frac{\mu }{2}\right) -\left( X_{{\mathcal {R}}}\!-\!\frac{\mu }{2} \right) \right] \!+\!\frac{1}{2}\left[ \left( -\frac{\Delta x}{2}+\frac{\mu }{2} \right) -\left( X_{{\mathcal {L}}}+\frac{\mu }{2}\right) \right] \right\} \nonumber \\= & {} \frac{1}{\mu }\left( \frac{\mu }{2}-\frac{X_{{\mathcal {L}}}+X_{{\mathcal {R}}} }{2}\right) \end{aligned}$$

  (24)

  and thus $h_{{\mathcal {L}}}-h_{{\mathcal {R}}}=\frac{1}{\mu }\left( X_{{\mathcal {L}} }+X_{{\mathcal {R}}}\right) =\frac{2{\bar{X}}}{\mu }.$ Policy weights can then be calculated as follows.

  $$\begin{aligned} W_{{\mathcal {L}}}\left( {\bar{X}},\lambda \right)= & {} {{\mathbb {P}}}\left\{ \frac{ \left[ Z\left( \frac{2{\bar{X}}}{\mu }\right) \right] ^{\frac{1}{1+g\left( \lambda \right) }}\left[ D\left( {\bar{X}}|{\tilde{\Psi }}\right) \right] ^{\frac{ g\left( \lambda \right) }{1+g\left( \lambda \right) }}}{\left[ 1-Z\left( \frac{2{\bar{X}}}{\mu }\right) \right] ^{\frac{1}{1+g\left( \lambda \right) }} \left[ 1-D\left( {\bar{X}}|{\tilde{\Psi }}\right) \right] ^{\frac{g\left( \lambda \right) }{1+g\left( \lambda \right) }}}\ge 1\right\} \end{aligned}$$

  (25)
  $$\begin{aligned}= & {} {{\mathbb {P}}}\left\{ \frac{\left[ \frac{1}{1+e^{-\frac{{\bar{X}}}{\mu }}} \right] ^{\frac{1}{1+g\left( \lambda \right) }}\left[ \frac{1}{1+e^{-\frac{ {\bar{X}}-{\tilde{\Psi }}}{\Sigma _{g}}}}\right] ^{\frac{g\left( \lambda \right) }{1+g\left( \lambda \right) }}}{\left[ 1-\frac{1}{1+e^{-\frac{{\bar{X}}}{\mu }} }\right] ^{\frac{1}{1+g\left( \lambda \right) }}\left[ 1-\frac{1}{1+e^{- \frac{{\bar{X}}-{\tilde{\Psi }}}{\Sigma _{g}}}}\right] ^{\frac{g\left( \lambda \right) }{1+g\left( \lambda \right) }}}\ge 1\right\} \nonumber \\= & {} {{\mathbb {P}}}\left\{ e^{\frac{{\bar{X}}}{\mu }\frac{1}{1+g\left( \lambda \right) }}e^{\frac{{\bar{X}}-{\tilde{\Psi }}}{\Sigma _{g}}\frac{g\left( \lambda \right) }{1+g\left( \lambda \right) }}\ge 1\right\} ={{\mathbb {P}}}\left\{ \frac{{\bar{X}}}{\mu }+\frac{{\bar{X}}-{\tilde{\Psi }}}{\Sigma _{g}}g\left( \lambda \right) \ge 0\right\} \nonumber \\= & {} {{\mathbb {P}}}\left\{ \frac{{\bar{X}}}{\mu }+\frac{{\bar{X}}}{\Sigma _{g}}g\left( \lambda \right) \ge g\left( \lambda \right) \frac{{\tilde{\Psi }}}{\Sigma _{g}} \right\} ={{\mathbb {P}}}\left\{ {\tilde{\Psi }}\le \left[ 1+\frac{1/\mu }{g\left( \lambda \right) /\Sigma _{g}}\right] {\bar{X}}\right\} \nonumber \\= & {} \frac{1}{2}+\frac{1}{\Psi }\left[ 1+\frac{1/\mu }{g\left( \lambda \right) /\Sigma _{g}}\right] \frac{X_{{\mathcal {L}}}+X_{{\mathcal {R}}}}{2} \end{aligned}$$

  (26)

  and $W_{{\mathcal {R}}}\left( {\bar{X}},\lambda \right) =1-W_{{\mathcal {L}}}\left( {\bar{X}},\lambda \right) .$ The necessary conditions for a party equilibrium are, using $W_{{\mathcal {L}}}+W_{{\mathcal {R}}}=1$:

  $$\begin{aligned} \frac{\partial }{\partial X_{{\mathcal {L}}}}U_{{\mathcal {L}}}= & {} -W_{{\mathcal {L}} }+\left( X_{{\mathcal {R}}}-X_{{\mathcal {L}}}\right) \frac{\partial }{\partial X_{ {\mathcal {L}}}}W_{{\mathcal {L}}}=0 \end{aligned}$$

  (27)
  $$\begin{aligned} \frac{\partial }{\partial X_{{\mathcal {R}}}}U_{{\mathcal {R}}}= & {} \left( 1-W_{ {\mathcal {L}}}\right) -\left( X_{{\mathcal {R}}}-X_{{\mathcal {L}}}\right) \frac{ \partial }{\partial X_{{\mathcal {R}}}}W_{{\mathcal {L}}}=0 \end{aligned}$$

  (28)

  and adding together these two first-order conditions, and observing that$\frac{\partial }{\partial X_{{\mathcal {L}}}}W_{{\mathcal {L}}}=\frac{\partial }{ \partial X_{{\mathcal {R}}}}W_{{\mathcal {L}}}=\frac{1}{2\Psi }\left[ 1+\frac{ 1/\mu }{g\left( \lambda \right) /\Sigma _{g}}\right]$ from Eq. (26), yields $W_{{\mathcal {L}}}=W_{{\mathcal {R}}}=\frac{1}{2}.$ Using this in the first-order conditions, one can solve for the party equilibrium strategies $\left( X_{{\mathcal {L}}},X_{{\mathcal {R}}}\right) =\left( -\frac{ \Delta X}{2},\frac{\Delta X}{2}\right)$ where:

  $$\begin{aligned} \Delta X\equiv X_{{\mathcal {R}}}-X_{{\mathcal {L}}}=\frac{W_{{\mathcal {L}}}}{\frac{ \partial }{\partial X_{{\mathcal {L}}}}W_{{\mathcal {L}}}}=\frac{\Psi }{1+\frac{ 1/\mu }{g\left( \lambda \right) /\Sigma _{g}}} \end{aligned}$$

  (29)

• (ii) To establish that the party strategies above constitute an equilibrium, not only local but also global deviations have to be unprofitable. The argument is made for party ${\mathcal {L}}$ only, since by symmetry it also applies to party ${\mathcal {R}}$. First note that parties’ payoffs in the conjectured equilibrium are zero by symmetry. Party ${\mathcal {L}}$ deviations in the interval $\left( -\infty ,-\frac{\Delta x}{2}\right]$ are not profitable because party ${\mathcal {L}}$’s payoff function in this interval is strictly concave in $X_{{\mathcal {L}}}.$ To see this, note that$\frac{ \partial ^{2}}{\partial X_{{\mathcal {L}}}^{2}}U_{{\mathcal {L}}}=-2\frac{\partial }{\partial X_{{\mathcal {L}}}}W_{{\mathcal {L}}}+\left( X_{{\mathcal {R}}}-X_{ {\mathcal {L}}}\right) \frac{\partial ^{2}}{\partial X_{{\mathcal {L}}}^{2}}W_{ {\mathcal {L}}}=-2\frac{\partial }{\partial X_{{\mathcal {L}}}}W_{{\mathcal {L}}}<0$ because $W_{{\mathcal {L}}}$ is strictly increasing and linear in $X_{\mathcal {L }}.$ See Eq. (26). Party ${\mathcal {L}}$ deviations in the interval $\left[ 0,\infty \right)$ are not profitable because they produce at best a non-negative policy, which gives party ${\mathcal {L}}$ at most a zero payoff. Party ${\mathcal {L}}$ deviations in the interval $\left( -\frac{ \Delta x}{2},0\right)$ yield a negative payoff. To see this, note that as party ${\mathcal {L}}$ moves its position $X_{{\mathcal {L}}}$ beyond $-\frac{ \Delta x}{2},$ its policy weight $W_{{\mathcal {L}}}$ drops faster than a commensurate rightward deviation between $-\frac{\Delta X}{2}$ and $-\frac{ \Delta x}{2},$ because in the former deviation its seat share shrinks as the other party’s stays constant, whereas in the latter deviation its seat share increases as the other party’s declines. See Fig. 3. Thus, party $\mathcal { L}$’s payoff drops faster than if it were to gain seats at the same rate.

• (iii) To establish uniqueness, an equilibrium with the property $-\frac{ \Delta x}{2}<X_{{\mathcal {L}}}\le X_{{\mathcal {R}}}<\frac{\Delta x}{2}$ has to be ruled out. Suppose parties pick positions with these properties. To solve for the parties’ policy weights, it is necessary to first derive the parties’ seat shares:

  $$\begin{aligned} h_{{\mathcal {L}}}= & {} \frac{1}{\mu }\left\{ \frac{1}{2}\left[ \left( -\frac{ \Delta x}{2}+\frac{\mu }{2}\right) -\left( X_{{\mathcal {L}}}-\frac{\mu }{2} \right) \right] \right\} =\frac{1}{2\mu }\left( \mu -\frac{\Delta x}{2}-X_{ {\mathcal {L}}}\right) \nonumber \\ \end{aligned}$$

  (30)
  $$\begin{aligned} h_{{\mathcal {R}}}= & {} \frac{1}{\mu }\left\{ \frac{1}{2}\left[ \left( X_{ {\mathcal {R}}}+\frac{\mu }{2}\right) -\left( \frac{\Delta x}{2}-\frac{\mu }{2} \right) \right] \right\} =\frac{1}{2\mu }\left( \mu -\frac{\Delta x}{2}+X_{ {\mathcal {R}}}\right) \nonumber \\ \end{aligned}$$

  (31)

  and thus $h_{{\mathcal {L}}}-h_{{\mathcal {R}}}=-\frac{1}{2\mu }\left( X_{\mathcal { L}}+X_{{\mathcal {R}}}\right) =-\frac{{\bar{X}}}{\mu }.$ Following a computation as in equation (26), policy weights are as follows:

  $$\begin{aligned} W_{{\mathcal {L}}}\left( {\bar{X}},\lambda \right) =\frac{1}{2}+\frac{1}{\Psi } \left[ 1-\frac{1/2\mu }{g\left( \lambda \right) /\Sigma _{g}}\right] \frac{ X_{{\mathcal {L}}}+X_{{\mathcal {R}}}}{2} \end{aligned}$$

  (32)

  and $W_{{\mathcal {R}}}\left( {\bar{X}},\lambda \right) =1-W_{{\mathcal {L}}}\left( {\bar{X}},\lambda \right) .$ Solving the necessary equilibrium conditions in Eqs. (27)–(28), gives the party equilibrium strategies $\left( X_{{\mathcal {L}}},X_{{\mathcal {R}}}\right) =\left( -\frac{\Delta X}{2}, \frac{\Delta X}{2}\right)$ where:

  $$\begin{aligned} \Delta X\equiv X_{{\mathcal {R}}}-X_{{\mathcal {L}}}=\frac{W_{{\mathcal {L}}}}{\frac{ \partial }{\partial X_{{\mathcal {L}}}}W_{{\mathcal {L}}}}=\frac{\Psi }{1-\frac{ 1/2\mu }{g\left( \lambda \right) /\Sigma _{g}}} \end{aligned}$$

  (33)

  However, since $\frac{\Psi }{1-\frac{1/2\mu }{g\left( \lambda \right) /\Sigma _{g}}}>\frac{\Psi }{1+\frac{1/\mu }{g\left( \lambda \right) /\Sigma _{g}}},$ this larger party divergence is not consistent with the restriction $- \frac{\Delta x}{2}<X_{{\mathcal {L}}}\le X_{{\mathcal {R}}}<\frac{\Delta x}{2},$ which implies a smaller party divergence, thus ruling out an equilibrium in this range. $\square$

Proof of Proposition 5

A party’s marginal electoral penalty for moving away from the district mean, namely $\frac{1}{2\Psi }\left[ 1+\frac{1/\mu }{g\left( \lambda \right) /\Sigma _{g}}\right]$ is lower the higher is aggregate giving $g\left( \lambda \right)$. Thus, a party’s policy-motivated deviation from the mean of district means is more profitable the higher is $g\left( \lambda \right)$ . Taking the difference between the two first-order conditions (27)-(28) and solving for $X_{{\mathcal {R}}}-X_{{\mathcal {L}}}$ yields $\Delta X=\frac{1}{\frac{\partial }{\partial X_{{\mathcal {L}}}}W_{{\mathcal {L}}}+\frac{ \partial }{\partial X_{{\mathcal {R}}}}W_{{\mathcal {L}}}}=\frac{\Psi }{1+\frac{ 1/\mu }{g\left( \lambda \right) /\Sigma _{g}}}.$ Then, using $g^{\prime }(\lambda )>0$ from Proposition 1 gives that $\frac{\partial }{\partial \lambda }\Delta X>0.$ The comparative statics with respect to $\Psi ,\mu ,\Sigma _{g}$ follow immediately from the expression for $\Delta X.$ $\square$

Proof of Proposition 6

Equilibrium partisan polarization can be expressed as a function of candidate divergence and party divergence.

$$\begin{aligned} \Delta {{\mathbb {E}}}\left( x\right)= & {} \frac{1}{\mu }\left( \int _{k_{s}\in {\mathcal {R}}}x_{k_{s}}ds-\int _{k_{s}\in {\mathcal {L}}}x_{k_{s}}ds\right) =\frac{ 2}{\mu }\int _{k_{s}\in {\mathcal {R}}}x_{k_{s}}ds \nonumber \\= & {} \frac{2}{\mu }\left[ \left( \frac{\mu }{2}+\frac{\Delta x}{2}\right) -\left( X_{{\mathcal {R}}}-\frac{\mu }{2}\right) \right] \left[ \left( X_{ {\mathcal {R}}}-\frac{\mu }{2}\right) +\left( \frac{\mu }{2}+\frac{\Delta x}{2} \right) \right] /2 \nonumber \\&+\frac{2}{\mu }\left[ \left( \frac{\mu }{2}-\frac{\Delta x}{2}\right) -\left( X_{{\mathcal {L}}}+\frac{\mu }{2}\right) \right] \left[ \left( X_{ {\mathcal {L}}}+\frac{\mu }{2}\right) +\left( \frac{\mu }{2}-\frac{\Delta x}{2} \right) \right] /2 \nonumber \\= & {} \frac{1}{\mu }\left\{ \left[ \left( \frac{\mu }{2}+\frac{\Delta x}{2} \right) ^{2}-\left( X_{{\mathcal {R}}}-\frac{\mu }{2}\right) ^{2}\right] +\left[ \left( \frac{\mu }{2}-\frac{\Delta x}{2}\right) ^{2}-\left( X_{{\mathcal {L}}}+ \frac{\mu }{2}\right) ^{2}\right] \right\} \nonumber \\= & {} \frac{1}{\mu }\left[ \frac{\mu \Delta x}{2}+\frac{\left( \Delta x\right) ^{2}}{4}-X_{{\mathcal {R}}}^{2}+X_{{\mathcal {R}}}\mu -\frac{\mu \Delta x}{2}+ \frac{\left( \Delta x\right) ^{2}}{4}-X_{{\mathcal {L}}}^{2}-X_{{\mathcal {L}}}\mu \right] \nonumber \\= & {} \frac{1}{\mu }\left[ \frac{\left( \Delta x\right) ^{2}}{2}-X_{{\mathcal {R}} }^{2}+X_{{\mathcal {R}}}\mu -X_{{\mathcal {L}}}^{2}-X_{{\mathcal {L}}}\mu \right] \nonumber \\= & {} \frac{1}{\mu }\left[ \frac{\left( \Delta x\right) ^{2}}{2}-\frac{\left( \Delta X\right) ^{2}}{2}+\mu \Delta X\right] \nonumber \\= & {} \frac{1}{\mu }\left[ \frac{\left( \Delta x\right) ^{2}}{2}+\frac{\Delta X\left( 2\mu -\Delta X\right) }{2}\right] \nonumber \\= & {} \Delta x+\left( \Delta X-\Delta x\right) \left( 1-\frac{\Delta X+\Delta x}{2\mu }\right) \end{aligned}$$

(34)

Ideological polarization is $\Delta x$ and compositional polarization is $\left( \Delta X-\Delta x\right) \left( 1-\frac{\Delta X+\Delta x}{2\mu } \right) ,$ which is positive if and only if $\Delta X>\Delta x.$ If $\Delta X=\Delta x,$ then $\Delta {{\mathbb {E}}}\left( x\right) =\Delta x$ is pure ideological polarization. See Fig. 4, middle panel. If $\Delta x=0,$ then $\Delta {{\mathbb {E}}}\left( x\right) =\frac{\Delta X\left( 2\mu -\Delta X\right) }{2\mu }$ is pure compositional polarization. See Fig. 4, bottom panel. $\square$

Proof of Proposition 7

The derivations of the different features of equilibrium polarization are as follows. See also Fig. 3. Inter-party heterogeneity is the policy difference between the rightmost ${\mathcal {R}}$ party member and the leftmost ${\mathcal {L}}$ party member:

$$\begin{aligned} \max _{k_{s}\in {\mathcal {R}}}x_{k_{s}}-\min _{k_{s}\in {\mathcal {L}} }x_{k_{s}}=\left( \frac{\Delta x}{2}+\frac{\mu }{2}\right) -\left( -\frac{ \Delta x}{2}-\frac{\mu }{2}\right) =\mu +\Delta x \end{aligned}$$

(35)

Intra-party heterogeneity is the policy difference between the rightmost and leftmost member within each party, and by equilibrium symmetry, is the same in both parties:

$$\begin{aligned} \max _{k_{s}\in {\mathcal {R}}}x_{k_{s}}-\min _{k_{s}\in {\mathcal {R}}}x_{k_{s}}= & {} \left( \frac{\Delta x}{2}+\frac{\mu }{2}\right) -\left( X_{{\mathcal {R}}}- \frac{\mu }{2}\right) =\mu -\frac{1}{2}\left( \Delta X-\Delta x\right) \nonumber \\ \end{aligned}$$

(36)

$$\begin{aligned} \max _{k_{s}\in {\mathcal {L}}}x_{k_{s}}-\min _{k_{s}\in {\mathcal {L}}}x_{k_{s}}= & {} \left( X_{{\mathcal {L}}}+\frac{\mu }{2}\right) -\left( -\frac{\Delta x}{2}- \frac{\mu }{2}\right) =\mu -\frac{1}{2}\left( \Delta X-\Delta x\right) \nonumber \\ \end{aligned}$$

(37)

as $X_{{\mathcal {R}}}=-X_{{\mathcal {L}}}=\frac{\Delta X}{2}$ in equilibrium. Party overlap is the policy difference between the rightmost ${\mathcal {L}}$ party member and the leftmost ${\mathcal {R}}$ party member:

$$\begin{aligned} \max _{k_{s}\in {\mathcal {L}}}x_{k_{s}}-\min _{k_{s}\in {\mathcal {R}} }x_{k_{s}}=\left( X_{{\mathcal {L}}}+\frac{\mu }{2}\right) -\left( X_{\mathcal {R }}-\frac{\mu }{2}\right) =\mu -\Delta X \end{aligned}$$

(38)

The fraction of safe seats is the fraction of districts where both candidates are affiliated with the same party:

$$\begin{aligned} \frac{1}{\mu }\left( \int _{L_{s},R_{s}\in {\mathcal {L}}}ds+\int _{L_{s},R_{s} \in {\mathcal {R}}}ds\right)= & {} \frac{1}{\mu }\left\{ \left[ \left( X_{\mathcal { R}}-\frac{\mu }{2}\right) -\left( \frac{\Delta x}{2}-\frac{\mu }{2}\right) \right] \right. \nonumber \\&\quad \left. +\left[ \left( -\frac{\Delta x}{2}+\frac{\mu }{2}\right) -\left( X_{ {\mathcal {L}}}+\frac{\mu }{2}\right) \right] \right\} \nonumber \\= & {} \frac{1}{\mu }\left[ \left( X_{{\mathcal {R}}}-X_{{\mathcal {L}}}\right) -\Delta x\right] =\frac{1}{\mu }\left( \Delta X-\Delta x\right) \nonumber \\ \end{aligned}$$

(39)

According to Propositions 3 and 5, both $\Delta x$ and $\Delta X$ are strictly increasing in inequality $\lambda .$ Condition (17) implies that $\frac{\partial }{\partial \lambda }\Delta X>\frac{\partial }{\partial \lambda }\Delta x,$ which means that $\Delta X-\Delta x$ is strictly increasing in inequality. $\square$