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Endogenous party platforms: ‘stochastic’ membership

Abstract

We propose a model of endogenous party platforms with stochastic membership. The parties’ proposals depend on their membership, while the membership depends both on the proposals of the parties and on the unobserved idiosyncratic preferences of citizens over parties. An equilibrium of the model obtains when the members of each party prefer the proposal of the party to which they belong to, rather than the proposal of the other party. We prove the existence of such an equilibrium and study its qualitative properties. For the cases in which parties use either the average or the median to aggregate the preferences of their members, we show that if the unobserved idiosyncratic characteristics of the parties are similar, then parties make different proposals in the stable equilibria. Conversely, we argue that if parties differ substantially in their unobserved idiosyncratic characteristics, then the unique equilibrium is convergent.

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Notes

  1. 1.

    As already mentioned, our theoretical model makes the following two assumptions: that the policy proposed by each political party can be seen as an aggregation of the policy preferences of its members, and that party membership is determined by both policy and non-policy characteristics of parties. In an appendix to the unpublished working paper version of the paper (see Gomberg et al. 2013), we provide some empirical evidence, suggesting that these might be realistic assumptions. In particular, we first analyze the political platforms of the main parties and the average ideal policy of their supporters. We find that, in countries with only two major parties, such as US and UK, the political platform of the party and the average ideal policy of its supporters are strikingly similar. This is not the case in other countries with more than two major parties or in a clearly unstable political period. In the same appendix, we also analyze data about the self-reported ideological position of citizens and which party they vote for. As expected, the ideological position is not sufficient to perfectly determine the voting behavior of citizens (some left-wing citizens vote for the right-wing party, and some right-wing citizens vote for the left-wing party). Moreover, we show that such data seem consistent with the existence of a non-policy variable such that agents’ preferences over it are independent of their preferences on the left–right ideological space.

  2. 2.

    The authors in Dziubiński and Roy (2011) provide a somehow related result. In their case, if parties strongly differ in their fixed policies in a given dimension, there is full convergence in the other dimension. In our case, in general there is no full convergence and the explanation and logic behind our result is very different from theirs.

  3. 3.

    The case \(y_{1}=y_{2}\) being the one we considered in Gomberg et al. (2004).

  4. 4.

    In most real cases, only a small fraction of the population is a member of a party. Our results remain true if we assume that the set of citizens who become members of parties is a random sample of the whole population.

  5. 5.

    Note that the convergent equilibrium in this example relies on full indifference of every agent between the two parties, requiring a slight modification of the definitions along the lines of Gomberg et al. (2004). In the uncorrelated case, this equilibrium is properly defined for any \(y_{1}<\frac{1 }{2}\). In the correlated case, we rely on complete population indifference in the centrist equilibrium even if \(y_{1}<\frac{1}{2}\). However, as long as the correlation is imperfect, so that the population distribution is not concentrated on the hyperplane, this would no longer be the case, while, as we show below, the structure of equilibrium set remains the same.

  6. 6.

    Note that we parametrize \(f_b\) by the endpoints of \(B=[-b,b]\).

  7. 7.

    The expressions of \(\phi _{1}, \phi _{2}\) are much more complicated for values of \(t,x_1,x_2\in [0,1]\) for which the graph of z(tx) intersects the lines \(y=\pm b\).

  8. 8.

    Of course, there is a corresponding symmetric equilibrium with \(x_1 > x_2\).

  9. 9.

    Ansolabehere et al. (2012) assume that parties represent the median of their elected officials.

  10. 10.

    Though, strictly speaking, the function \(\phi \) is not differentiable (not even necessarily continuous) in this case, the instability of the convergent equilibrium and the consequent existence of a divergent equilibrium can be easily shown using standard approximation techniques.

  11. 11.

    See Fauli-Oller et al. (2003) for a different view on party polarization.

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Corresponding author

Correspondence to Andrei M. Gomberg.

Additional information

We thank G. Caruana, A. Loeper, and the participants at the ICER Conference on Constitutional Political Economy and at the seminars at U. Carlos III and UC Davis. We feel especially indebted to the anonymous referee of this journal for his/her careful and insightful reading of the manuscript. Financial support from the Asociación Mexicana de Cultura and from Project No. ECO2013-42710-P from the Spanish Ministry of Science and Innovation is gratefully acknowledged.

Appendix: Proofs

Appendix: Proofs

Proof of Lemma 5

We will show that the mappings \(\sigma _1, \sigma _2:X\times X\rightarrow \varSigma \) are Fréchet differentiable (which, of course, implies continuity). We do the proof for \(i=1\). Let \(x=(x_1,x_2)\in X\times X\). Note that

$$\begin{aligned} \frac{\partial g_1}{\partial x_1}=f(\alpha ,z(\alpha ;x))\frac{\alpha -x_1}{y_2-y_1} \end{aligned}$$

and

$$\begin{aligned} \frac{\partial g_1}{\partial x_2}=f(\alpha ,z(\alpha ;x))\frac{x_2-\alpha }{y_2-y_1} \end{aligned}$$

From the above expressions, the Fréchet derivative can be easily computed. We use the notation \(e=(h,k)\) and \(\Vert e\Vert = \sqrt{h^2 + k^2}\). Fix \(x\in X\times X\). A simple computation shows that

$$\begin{aligned} z(\alpha ;x+e) - z(\alpha ;x)= & {} \frac{h (\alpha -x_1) + k (x_2 - \alpha ) -(k^2 - h^2)/2}{y_2- y_1} \\= & {} \frac{(h-k)\alpha -x_1 h + x_2 k}{y_2- y_1} + \frac{h^2 - k^2}{2(y_2-y_1)} \end{aligned}$$

Note that there is \(M>0\) such that

$$\begin{aligned} \frac{\left| z(\alpha ;x+e)-z(\alpha ;x)\right| }{\Vert e\Vert }\le M \end{aligned}$$

for any \(\alpha \in A\) and e such that \(\Vert e\Vert \le 1\). Let \( \varepsilon >0\). Note that

$$\begin{aligned} g_{1}(\alpha ;x+e)-g_{1}(\alpha ;(x))=\int _{z(\alpha ;x)}^{z(\alpha ;x+e)}f(\alpha ,\beta )\,{\text {d}}\beta \end{aligned}$$

We are ready now to compute \(D\sigma (e;x):{\mathbb {R}}^{2}\rightarrow C(A)\), the derivative of \(\sigma \) at the point x. We continue to use the notation \(e=(h,k)\). We will show that \(D\sigma (e;x)(\alpha )\) takes the following value

$$\begin{aligned} D\sigma (e;x)(\alpha )= & {} \frac{f(\alpha ,z(\alpha ;x)}{y_{2}-y_{1}}\left( h(\alpha -x_{1})+k(x_{2}-\alpha )\right) \\= & {} \int _{z(\alpha ;x)}^{z(\alpha ;x+e)}f(\alpha ,z(\alpha ;x))\,{\text {d}}\beta -\frac{k^{2}-h^{2}}{2|y_{2}-y_{1}|} \end{aligned}$$

Note that \(D\sigma (e;x)(\alpha )\), as defined in the above equation, is linear in e and continuous in all the variables. Note that,

$$\begin{aligned}&g_{1}(\alpha ;x+e)-g_{1}(\alpha ;x)-D\sigma (e;x)(\alpha )\\&\quad = \int _{z(\alpha ;x)}^{z(\alpha ;x+e)}\left( f(\alpha ,\beta )-f(\alpha ,z(\alpha ;x))\right) \,{\text {d}}\beta + \frac{k^{2}-h^{2}}{2|y_{2}-y_{1}|} \end{aligned}$$

Since z is continuous and \(f(\alpha ,\beta )\) is uniformly continuous in \( A\times [z(\alpha ;x),z(\alpha ;x+e)]\), given \(\varepsilon >0\), there is a \(0<\delta <1\) such that if \(\Vert e\Vert \le \delta \), then

$$\begin{aligned} |f(\alpha ,\beta )-f(\alpha ,z(\alpha ;x))|\le \frac{\varepsilon }{M} \end{aligned}$$

Thus, as long as \(\Vert e\Vert \le \delta \) we have that

$$\begin{aligned} \left| g_{1}(\alpha ;x+e)-g_{1}(\alpha ;(x))-D\sigma (e;x)(\alpha )\right| \le \frac{\varepsilon }{M}\left| z(\alpha ;x+e)-z(\alpha ;x)\right| +\frac{k^{2}+h^{2}}{2|y_{2}-y_{1}|} \end{aligned}$$

So, for any \(\alpha \in A\), the following holds

$$\begin{aligned} \frac{\left| g_{1}(\alpha ;x+e)-g_{1}(\alpha ;(x))-D\sigma (e;x)(\alpha )\right| }{\Vert e\Vert }\le \varepsilon +\frac{k^{2}+h^{2}}{2\Vert e\Vert }=\varepsilon +\frac{\Vert e\Vert }{2|y_{2}-y_{1}|} \end{aligned}$$

Hence, we have that

$$\begin{aligned} \sup _{\alpha \in A}\frac{\left| g_{1}(\alpha ;x+e)-g_{1}(\alpha ;(x))-D\sigma (e;x)(\alpha )\right| }{\Vert e\Vert }\le \varepsilon + \frac{\Vert e\Vert }{2|y_{2}-y_{1}|} \end{aligned}$$

Therefore,

$$\begin{aligned} \frac{\Vert g_{1}(\alpha ;x+e)-g_{1}(\cdot ;x)-D\sigma (e;x)\Vert }{\Vert e\Vert }\le \varepsilon +\frac{\Vert e\Vert }{2|y_{2}-y_{1}|} \end{aligned}$$

Since \(\varepsilon >0\) is arbitrary, we see that

$$\begin{aligned} \lim _{\Vert e\Vert \rightarrow 0}\frac{\Vert \sigma _{1}(x+e)-\sigma _{1}(x)-D\sigma (e;x)\Vert }{\Vert e\Vert }=0 \end{aligned}$$

and the Lemma is proved.

Proof of Lemma 12

Lemma 12 is consequence of Lemma 5 and of Lemma 24 below.

Lemma 24

The map \(P^{\mu }: \varSigma \times \varSigma \rightarrow A\times A\) is continuous and differentiable.

Proof of Lemma 24

For each \(i=1,2\), the maps \(\nu _{i}\mapsto \nu _{i}(A)\) and \(\nu _{i}\mapsto \int _{A}\alpha {\text {d}}\,\nu _{i}(\alpha )\) are linear. It is easy to see that they are also continuous. For example, given \(\varepsilon >0\), let

$$\begin{aligned} \delta =\frac{\varepsilon }{\int _{A}\alpha \,{\text {d}}\alpha } \end{aligned}$$

If \(\sup \{|f(\alpha )-g(\alpha )|:\alpha \in A\}\le \delta \) then

$$\begin{aligned} \left| \int _{A}\alpha f(\alpha )\,{\text {d}}\alpha -\int _{A}\alpha g(\alpha )\,{\text {d}}\alpha \right| \le \int _{A}\alpha |f(\alpha )-g(\alpha )|\,{\text {d}}\alpha \le \delta \int _{A}\alpha \,{\text {d}}\alpha =\varepsilon \end{aligned}$$

So \(\nu _{i}\mapsto \int _{A}\alpha {\text {d}}\nu _{i}(\alpha )\) is continuous. The proof that \(\nu _{i}\mapsto \nu _{i}(A)\) is continuous is similar. Hence, it follows that the maps \(\nu _{i}\mapsto \nu _{i}(A)\) and \(\nu _{i}\mapsto \int _{A}\alpha {\text {d}}\nu _{i}(\alpha )\) are differentiable. Since for every \(i=1,2\) we have that \(\nu _{i}(A)\ne 0\), the mapping \(\nu _{i}\mapsto \frac{ \int _{A}\alpha {\text {d}}\,\nu _{i}(\alpha )}{\nu _{i}(A)}\) is also differentiable. Therefore, so is \(P^{\mu }\).

For the proof of Proposition 13, we will make use of the following result.

Lemma 25

$$\begin{aligned} \underset{x\rightarrow (\mu ,\mu )}{\lim } \frac{\partial \phi _{j}}{\partial x_ i}(x) =\frac{1}{ \lambda _{j}}\int _{A}(\alpha -\mu ) \frac{\partial g_j (\alpha ;(\mu ,\mu )}{\partial x_ i}\, {\text {d}}\alpha \end{aligned}$$

Proof of Lemma 25

Note that

$$\begin{aligned} \frac{\partial \phi _{j}}{\partial x_ i}(x)&=\frac{1}{(\int _{A}g_{j}(\alpha ;x)\, {\text {d}}\alpha )^2} \left( \int _{A}\alpha \frac{\partial g_j }{\partial x_ i}(\alpha ;(\mu ,\mu )\, {\text {d}}\alpha \int _{A}g_{j}(\alpha ;x)\, {\text {d}}\alpha \right. \\&\quad \,\, \left. - \int _{A}\alpha g_{j}(\alpha ;x)\, {\text {d}}\alpha \int _{A}\frac{\partial g_j }{\partial x_ i}(\alpha ;(\mu ,\mu )\, {\text {d}}\alpha \right) \\&=\frac{1}{(\int _{A}g_{j}\,(\alpha ;x){\text {d}}\alpha )}\int _{A}\left( \alpha - \phi _{j}(x)\right) \frac{\partial g_j }{\partial x_ i}(\alpha ;(\mu ,\mu )\, {\text {d}}\alpha \end{aligned}$$

Thus,

$$\begin{aligned} \underset{x\rightarrow (\mu ,\mu )}{\lim }\frac{\partial \phi _{j}}{\partial x_ i}(x)= & {} \frac{1}{ \lambda _{j}}\int _{A}(\alpha -\mu ) \frac{\partial g_j }{\partial x_ i}(\alpha ;(\mu ,\mu )\,{\text {d}}\alpha \end{aligned}$$

and the Lemma follows.

Proof of Proposition 13

Recall that, given the proposals \(x=(x_1,x_2)\) of the parties, we use the notation

$$\begin{aligned} g_{j}(\alpha ;x)=\int _{\left\{ (\alpha ,\beta ):||(x_{j},y_{j})-(\alpha ,\beta )||\ge ||(x_i,y_i)-(\alpha , \beta )||,i\ne j\right\} }f(\alpha ,\beta ){\text {d}}\beta \end{aligned}$$

when we want to make explicit the dependence of the density functions that describe the induced population partitions on the policies proposed by the parties. We have seen in the proof of Lemma 5 that

$$\begin{aligned} \frac{\partial g_{1}(\alpha ;x)}{\partial x_1}=\left. f(\alpha ,z(\alpha ;x))\partial _{1}z(t;x)\right| _{t=\alpha }=f(\alpha ,z(\alpha ;x))\frac{\alpha -x_{1} }{y_{2}-y_{1}} \end{aligned}$$

Furthermore,

$$\begin{aligned} \frac{\partial g_{1}(\alpha ;x)}{\partial x_2}=\left. f(\alpha ,z(\alpha ;x))\partial _{2}z(t;x)\right| _{t=\alpha }=f(\alpha ,z(\alpha ;x))\frac{x_{2}-\alpha }{y_{2}-y_{1}} \end{aligned}$$
(11)

which implies that

$$\begin{aligned} \frac{\partial g_1}{\partial x_1}(\alpha ;(\mu ,\mu ))=- \frac{\partial g_1}{\partial x_2}(\alpha ;(\mu ,\mu ))=f\left( \alpha ,\frac{y_{1}+y_{2}}{2}\right) \frac{\alpha -\mu }{ y_{2}-y_{1}} \end{aligned}$$

Since for party 2 the relevant population density is

$$\begin{aligned} g_{2}(\alpha ;x)=\int _{z(\alpha ;x)}^{\infty }f(\alpha ,\beta )\,{\text {d}}\beta \end{aligned}$$

we get that

$$\begin{aligned} \frac{\partial g_2}{\partial x_2}(\alpha ;(\mu ,\mu ))= & {} - \frac{\partial g_2}{\partial x_1}(\alpha ;(\mu ,\mu ))=\frac{\partial g_1}{\partial x_1}(\alpha ;(\mu ,\mu )) \\= & {} f\left( \alpha ,\frac{y_{1}+y_{2}}{2}\right) \frac{\alpha -\mu }{ y_{2}-y_{1}} \nonumber \end{aligned}$$
(12)

To ease the notation, we will write \(g_i=g_i(\alpha ;(\mu ,\mu ))\). Applying Lemma 25 and using the above formulae for \(\frac{\partial g_j}{\partial x_i}\), we have to establish conditions under which

$$\begin{aligned} \lim _{x\rightarrow (\mu ,\mu )} |B(x)|= & {} \left| \begin{array}{ll} \frac{1}{\lambda _1}\int _{A}(\alpha -\mu ) \frac{\partial g_1}{\partial x_1}\, {\text {d}}\alpha -1 &{} \frac{1}{ \lambda _1}\int _{A}(\alpha -\mu ) \frac{\partial g_1}{\partial x_2}\, {\text {d}}\alpha -1 \\ &{} \\ \frac{1}{\lambda _2}\int _{A}(\alpha -\mu ) \frac{\partial g_2}{\partial x_1}\, {\text {d}}\alpha -1 &{} \frac{1}{ \lambda _2}\int _{A}(\alpha -\mu ) \frac{\partial g_2}{\partial x_2}\, {\text {d}}\alpha -1 \end{array} \right| \\&\hbox { }&\\= & {} \left| \begin{array}{ll} \frac{1}{\lambda _1}\int _{A}(\alpha -\mu ) \frac{\partial g_1}{\partial x_1}\, {\text {d}}\alpha -1 &{} \frac{1}{ \lambda _1}\int _{A}(\alpha -\mu ) \frac{\partial g_1}{\partial x_1}\, {\text {d}}\alpha -1 \\ &{} \\ \frac{1}{\lambda _2}\int _{A}(\alpha -\mu ) \frac{\partial g_1}{\partial x_1}\, {\text {d}}\alpha -1 &{} \frac{1}{ \lambda _2}\int _{A}(\alpha -\mu ) \frac{\partial g_1}{\partial x_1}\, {\text {d}}\alpha -1 \end{array} \right| \\= & {} 1-\frac{1}{\lambda _1 \lambda _2}\int _{A}(\alpha -\mu ) \frac{\partial g_1}{\partial x_1}\, {\text {d}}\alpha \\= & {} 1-\frac{1}{\lambda _1 \lambda _2\left( y_2-y_1\right) }\int _{A}(\alpha - \mu )^2 f\left( \alpha ,\frac{y_1+y_2}{2}\right) \, {\text {d}}\alpha <0 \end{aligned}$$

which clearly holds if

$$\begin{aligned} |y_2-y_1|< \frac{1}{\lambda _1 \lambda _2} \int _{A}(\alpha -\mu )^2 f\left( \alpha ,\frac{y_1+y_2}{2}\right) \,{\text {d}}\alpha \end{aligned}$$

The Proposition follows from the above inequality.

Proof of Proposition 16

Fix a pair of \(x_{1}\ne x_{2}\) and for any \(s\in \left[ 0,1\right] \) define \(y_{1}^s=y_{1}+t\left( y_{2}-y_{1}\right) \). Note that as \(s\rightarrow 0\) the induced partition hyperplane will converge to perfect sorting in the policy dimension, as in Remark 1. This, of course, implies that the induced population partitions for \(g_1^s(\alpha ;x),g_2^s(\alpha ;x)\) converge uniformly to those given in Remark 1. Hence, for any \(x\notin \Delta \) the induced measures \(\sigma ^s_1 (x), \sigma ^s_2(x)\) will converge uniformly in \(L^{1}\) as \(s\rightarrow 0\) to the measures obtained in Remark 1. (Note that this convergence is not uniform for \(x\in X\times X\), as, in fact, the functions will not converge on the diagonal.) It follows that the stable fixed points of \(\phi ^s\) converge to some stable fixed points of the limiting model.

Now, in the proof of Proposition 13 we have seen that

$$\begin{aligned} |B^s(\mu ,\mu )|= & {} 1-\frac{1}{\lambda _1 \lambda _2\left( y_2^s-y_1^2\right) }\int _{A}(\alpha - \mu )^2 f\left( \alpha ,\frac{y_1^s+y_2^s}{2}\right) \,{\text {d}}\alpha \\= & {} 1-\frac{1}{s}\left( \frac{1}{2 \lambda _1 \lambda _2\left( y_2-y_1\right) }\int _{A}(\alpha - \mu )^2 f\left( \alpha ,y_1+y_2\right) \,{\text {d}}\alpha \right) \end{aligned}$$

The term inside the brackets does not depend on s. Hence, there is a \(\varepsilon >0\) such that \(|B^s(\mu ,\mu )| <0\) for any \(x\in X\times X\) and \(0<s<\varepsilon \). It follows that the stable fixed points of \(\phi ^s\) cannot converge to the diagonal.

Proof of Proposition 17

Since \(y_{1}^u=-\frac{u}{2}\), \(y_{2}^u=\frac{u}{2}\), we have that \(\frac{ y_{1}^u+y_{2}^u}{2}=0\). To make explicit the dependence on u, we will write \(z(t;x;u)=z(t;x)=\frac{(x_{1}-x_{2})(2\text {t}-x_{1}-x_{2})}{2u} \), \(\phi _{j}(x;u)=\phi _{j}(x)\) and let \(g_{i}(\alpha ;x;u)=g_{i}(\alpha ;x)\), \( i=1,2\), defined by (3) with f replaced by \(f_b\). Note that

$$\begin{aligned} |z(t;x;u)|\le \frac{1}{u} \end{aligned}$$

for every \(x=(x_{1},x_{2})\in A\times A\), \(t\in A\). Hence, \(\lim _{u\rightarrow +\infty }z(t;x;u)=0\) and

$$\begin{aligned} \lim _{u\rightarrow \infty }\int _{A}g_{i}(\alpha ;x;u)\,{\text {d}}\alpha = \lambda _i > 0, \quad i=1,2 \end{aligned}$$

uniformly on \(x\in A\times A\). That is, there is a real number \(b_{1}>0\) such that if \(u\ge b_{1}\), then

$$\begin{aligned} \int _{A}g_{i}(\alpha ;x;u)\,{\text {d}}\alpha \ge \frac{\lambda _{i}}{2} \end{aligned}$$

for every \(x\in X\times X\). Let M be the supremum of \(f_0\) on the set \( A\times [-\infty ,\infty ] \). From the proof of Lemma  24, we see that

$$\begin{aligned} \left| \frac{\partial g_{j}(\alpha ;x;u)}{\partial x_i} \right| =f(\alpha ,z(\alpha ;x))\frac{|\alpha -x_{i}|}{|y_{2}-y_{1}|}=f(\alpha ,z(\alpha ;x))\frac{|\alpha -x_{i}|}{u}\le M\frac{|\alpha -x_{i}|}{u} \end{aligned}$$

And, since \(\alpha ,\phi _{j}(x;u)\) and \(x_{i}\) belong to \(A=[0,1]\) we have that \(|\alpha -\phi _{j}(x;u)|\le 1\) and \(|\alpha -x_{i}|\le 1\). Hence,

$$\begin{aligned} \left| \frac{\partial \phi _{j}(x;u)}{\partial x_{i}} \right| \le \frac{1}{(\int _{A}g_{j}\,(\alpha ;x){\text {d}}\alpha )}\int _{A}\left| \alpha -\phi _{j}(x)\right| \left| \frac{\partial g_{j}(\alpha ;x;u)}{\partial x_i} \right| \,{\text {d}}\alpha \le \frac{M}{u\lambda _{j}} \end{aligned}$$

for every \(x\in X\times X\) and \(u>b_1\). It follows that there is \(\bar{b} \ge b_1\) such that \( \left| \frac{\partial \phi _{j}(x;u)}{\partial x_{i}} \right| < 1/4\) for every \(u\ge \bar{b}\) and for every \(x\in X\times X\).

Let \(b \ge u\ge \bar{b}\). The proof of Lemma  24 shows that \( \det (B(x))=1+\partial _{1}\phi _{1}(x;u)\partial _{2}\phi _{2}(x;u)-\partial _{1}\phi _{2}(x;u)\partial _{2}\phi _{2}(x;u)-\partial _{1}\phi _{2}(x;u)-\partial _{2}\phi _{2}(x;u)>0 \). Since the sum of the indices of the fixed points must add to the Euler Characteristics of the rectangle \(A\times B\), which is \(+1\), there can be at most a fixed point.

Proof of Lemma 19

Let \(j=1,2\) and consider the function \(F_{j}:X\times X\times X\rightarrow {\mathbb {R}}\) defined by

$$\begin{aligned} F(x_{1},x_{2},y)=\int _{-\infty }^{y}g_{j}(\alpha ;x)\,{\text {d}}\alpha \end{aligned}$$

Since \(f(\alpha ,\beta )\) is equivalent to Lebesgue measure, for each \( x=(x_{1},x_{2})\in X\times X\), the equation

$$\begin{aligned} F(x_{1},x_{2},y)=\frac{\lambda _{j}}{2} \end{aligned}$$

has a unique solution \(y=\phi _{j}(x)\) in the interior of X. And since

$$\begin{aligned} \frac{\partial F}{\partial y}=g_{j}(y;x)>0 \end{aligned}$$

we may apply the implicit function theorem to conclude that there is an open neighborhood U of x in \(X\times X\), an open neighborhood V of y in X and a continuously differentiable function \(\phi _{j}:U\rightarrow V\) such that

$$\begin{aligned} F(x_{1},x_{2},\phi _{j}(x))=\frac{\lambda _{j}}{2} \end{aligned}$$

Hence, the function \(\phi =(\phi _{1},\phi _{2}):V\times V\rightarrow X\) is continuously differentiable.

Proof of Proposition 20

Differentiating \(\phi \) implicitly with respect to \(x_i\) in Equation (9), we obtain

$$\begin{aligned} g_j(\phi _j(x);x) \frac{ \partial \phi _j}{\partial x_i}(x) +\int _{-\infty }^{\phi _j(x_j)} \frac{\partial g_j(\alpha ;x)}{\partial x_i}\, {\text {d}}\alpha = \frac{1}{2}\int _{-\infty }^{\infty } \frac{\partial g_j(\alpha ;x)}{\partial x_i}\, {\text {d}}\alpha \end{aligned}$$

Taking the limit of the expression as \(x\rightarrow (m,m)\), we obtain

$$\begin{aligned} f_1(m) \frac{ \partial \phi _j}{\partial x_i}(m) + \int _{-\infty }^{m} \frac{\partial g_j}{\partial x_i} (\alpha ;m)\,{\text {d}}\alpha = \frac{1}{2}\int _{-\infty }^{\infty } \frac{\partial g_j}{\partial x_i} (\alpha ;m)\,{\text {d}}\alpha \end{aligned}$$
(13)

with

$$\begin{aligned} f_1(m)=\int _{-\infty }^{\frac{y_1+ y_2}{2}} f(m,\beta )\, {\text {d}}\beta \end{aligned}$$

Now, taking into account Eqs. (11) and (12) the formulas for \(\frac{\partial g_j(\alpha ;x)}{\partial x_i}\), we see that

$$\begin{aligned} \frac{\partial \phi _1}{\partial x_1} (m)= \frac{\partial \phi _2}{\partial x_2} (m)= - \frac{\partial \phi _2}{\partial x_1} (m)= - \frac{\partial \phi _1}{\partial x_2} (m) \end{aligned}$$

Thus, the determinant in (7) becomes,

$$\begin{aligned} B(x)=\left| \begin{array}{ll} \frac{\partial \phi _1}{\partial x_1} (m)-1 &{}\quad \frac{\partial \phi _1}{\partial x_2} (m) \\ \frac{\partial \phi _2}{\partial x_1} (m) &{}\quad \frac{\partial \phi _2}{\partial x_2} (m)-1 \end{array} \right| = \left| \begin{array}{ll} \frac{\partial \phi _1}{\partial x_1} (m)-1 &{}\quad - \frac{\partial \phi _1}{\partial x_1} (m) \\ - \frac{\partial \phi _1}{\partial x_1} (m) &{}\quad \frac{\partial \phi _1}{\partial x_1} (m)-1 \end{array} \right| =1- 2 \frac{\partial \phi _1}{\partial x_1} (m) \end{aligned}$$

Finally, from equations (13), (11) and (12) we get that

$$\begin{aligned} \frac{\partial \phi _1}{\partial x_1}(x)= & {} \frac{1}{(y_2-y_1)f_1(m)} \left( \frac{1}{2} \int _{-\infty }^{\infty }(\alpha - m)f\left( \alpha ,\frac{y_{_1}+y_{_2}}{2}\right) \, {\text {d}}\alpha \right. \\&- \left. \int _{-\infty }^{m}(\alpha - m)f\left( \alpha ,\frac{y_{_1}+y_{_2}}{2}\right) \, {\text {d}}\alpha \right) . \end{aligned}$$

And the proposition follows.

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Gomberg, A.M., Marhuenda, F. & Ortuño-Ortín, I. Endogenous party platforms: ‘stochastic’ membership. Econ Theory 62, 839–866 (2016). https://doi.org/10.1007/s00199-015-0936-1

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Keywords

  • Political parties
  • Stochastic membership
  • Divergent equilibria

JEL Classification

  • D72