A Median Activist Theorem for Two-Stage Spatial Models

Abstract

The spatial model of electoral competition has for decades been a staple of formal political theory. As part of this field, a number of authors have developed two-stage spatial models in which electoral candidates must first win intra-party primary elections, and then compete in a general inter-party election. A universal result in these two-stage models is that party selectorates, and in particular the “median party activist”, exert a centrifugal pull on party platforms. The current paper brings this basic finding into question, suggesting that party voters only exert this centrifugal force under fairly strict conditions; and in particular only if candidates attach fairly high value to the outcome. The paper’s primary result, a “Median-Activist Theorem”, suggests that if candidates place little value on winning the nomination in and of itself, primaries are necessary, but not sufficient, for generating more extreme electoral platforms.

Many thanks to Jim Adams, Jon Eguia, Bernie Grofman, Herbert Kitschelt, Emerson Niou, Gilles Serra, David Soskice, Camber Warren, and participants in the Workshop on Contemporary Applications of the Spatial Model (Juan March Institute, Madrid; April 27–28, 2012) for valuable feedback on previous versions.

Theoretical Appendix

1.1 1 Preliminary Results

Proof of Lemma 1

If $\left \vert x_{L} - x_{m}\right \vert <\left \vert x_{R} - x_{m}\right \vert$, then the median voter chooses L. Since all voters choose according to spatial proximity in the general election, the electorate’s preferences are single-peaked. This in turn implies that either: (a) all voters with ideal points $x_{i} <x_{m}$ also choose P, or (b) all voters with ideal points $x_{i}> x_{m}$ also choose P. In turn, since the contest is determined by plurality rule, P wins.   QED

Proof of Lemma 2

Begin with a strategy vector at which $x_{L2} <x_{L1} <x_{m} <x_{R1} <x_{R2}$, and in which no two candidates are equidistant from the median voter’s ideal point. Given that no two platforms are equidistant from x _m , there are six possible strategic scenarios associated with a strategy vector where $x_{L2} <x_{L1} <x_{m} <x_{R1} <x_{R2}$:

• a) $(x_{m} - x_{L1}),(x_{m} - x_{L2})\; <\; (x_{R1} - x_{m}),(x_{R2} - x_{m})$, i.e. both of party L’s candidates would win a general election contest against either of party R’s candidates.

• b) $(x_{m} - x_{L1}),(x_{m} - x_{L2})\;>\; (x_{R1} - x_{m}),(x_{R2} - x_{m})$, i.e. both of party L’s candidates would lose a general election contest against either of party R’s candidates.

• c) $(x_{m} - x_{L1})\; <\; (x_{R1} - x_{m}),(x_{R2} - x_{m})$ but $(x_{m} - x_{L2})\;>\; (x_{R1} - x_{m}),(x_{R2} - x_{m})$, i.e. candidate L1 (L2) would defeat (lose to) both R1 and R2 in a general election contest.

• d) $(x_{m} - x_{L1}),(x_{m} - x_{L2})\;>\; (x_{R1} - x_{m})$ but $(x_{m} - x_{L1}),(x_{m} - x_{L2})\; <\; (x_{R2} - x_{m})$, i.e. candidate R1 (R2) would defeat (lose to) both L1 and L2 in a general election contest.

• e) $(x_{m} - x_{L1}) <(x_{R1} - x_{m}),(x_{R2} - x_{m})$, and $(x_{m} - x_{L2}) <(x_{R2} - x_{m})$, and $(x_{m} - x_{L2})\;>\; (x_{R1} - x_{m})$, i.e. candidate L1 would defeat both R1 and R2 in a general election contest, but candidate L2 would defeat (lose to) candidate R2 (R1) in a general election contest.

• f) $(x_{m} - x_{L2})> (x_{R1} - x_{m}),(x_{R2} - x_{m})$, and $(x_{m} - x_{L1}) <(x_{R2} - x_{m})$, and $(x_{m} - x_{L1})\;>\; (x_{R1} - x_{m})$, i.e. candidate L2 would lose both R1 and R2 in a general election contest, but candidate L1 would defeat (lose to) candidate R2 (R1) in a general election contest.

These six scenarios exhaustively describe the possible strategic configurations when $x_{L2} <x_{L1} <x_{m} <x_{R1} <x_{R2}$ and no two candidates are equidistant from the median voter’s ideal point. I now demonstrate that none of these six scenarios can constitute a Nash Equilibrium.

* In scenario (a) primary voters in party L know they will win the election regardless of which candidate emerges victorious in the opposing party’s primary, and are thus free to choose the primary candidate whose platform is closest to their ideal point, i.e. the candidate who maximizes their utility from “representation”.

• If $x_{L,m} \in \left [0, \frac{x_{L1}+x_{L2}} {2} \right )$, then $\left \vert x_{L2} - x_{L,m}\right \vert <\left \vert x_{L1} - x_{L,m}\right \vert$ and L’s median activist will choose L2. Since preferences for representation are single-peaked, candidate L2 thus wins the primary, and then by construction the general election. Candidate L1 thus has the incentive to deviate to a position slightly closer to x _L, m than x _L2, so as to take over from L2 the position of winning both the primary and the general election.⁴

• If $x_{L,m} \in \left (\frac{x_{L1}+x_{L2}} {2},x_{m}\right )$, then $\left \vert x_{L2} - x_{L,m}\right \vert> \left \vert x_{L1} - x_{L,m}\right \vert$, and by the same reasoning candidate L2 will have the incentive to deviate.

• If $x_{L,m} = \frac{x_{L1}+x_{L2}} {2}$, then the two primary candidates’ platforms are equidistant from (i.e. symmetric around) x _L, m , and each wins g with probability one-half: primary voters to the left (right) of x _L, m choose L2 (L1) and the median activist randomizes. In turn, both primary candidates will have the incentive to deviate by moving slightly closer to x _L, m so as to receive g with certainty.

• As such, scenario (a) cannot constitute a Nash Equilibrium. Note that party R’s two candidates face a strategically identical situation in scenario (b) to that faced by party L’s two candidates in scenario (a). As such, scenario (b) also cannot be a Nash Equilibrium.   QED

* In scenario (c) primary voters in party L know that, regardless of which candidate emerges victorious in party R, candidate L1 would win the general election and candidate L2 would lose the general election. There are two possible situations:

• If $x_{L,m} \in \left [\frac{x_{L1}+x_{L2}} {2},x_{m}\right )$, then L’s median activist will face no conflict between electoral viability and representation, and will choose L1. By single-peakededness, all primary voters to the right of the party median will also choose L1, who thus wins the primary and then by construction the general election. Candidate L2 thus has the incentive to deviate to a position slightly closer to x _L, m than x _L1, so as to take over from L the position of winning both the primary and the general election.⁵

• If $x_{L,m} \in \left [0, \frac{x_{L1}+x_{L2}} {2} \right )$, then by definition $x_{L1}> x_{L,m}$ (since by construction $x_{L2} <x_{L1}$). In turn, L2 would have an optimal deviation to any point in the range $\left ((1 - x_{R1}),x_{L1}\right )$: by moving to any point in this range, L2 would join L1 in being able to defeat both R1 and R2 in the general election, and would be preferred on representational grounds by the median activist. By single-peakedness, she would also be preferred on representational grounds by all primary voters to the left of the median activist. This deviation would thus allow L2 to win the primary and general election with certainty.

• As such, scenario (c) cannot constitute a Nash Equilibrium. Note that party R’s two candidates face a strategically identical situation in scenario (d) to that faced by party L’s two candidates in scenario (c). As such, scenario (d) also cannot be a Nash Equilibrium.   QED

* In scenario (e) candidate L1 would defeat both R1 and R2 in a general election, but candidate L2 would defeat only R2 in a general election. The proof proceeds identically to the proceeding analysis of scenario (c). There are two possible situations:

• If $x_{L,m} \in \left [\frac{x_{L1}+x_{L2}} {2},x_{m}\right )$, then L’s median activist will face no conflict between electoral viability and representation, and will choose L1. By single-peakedness, all primary voters to the right of the party median will also choose L1, who thus wins the primary and then by construction the general election. Candidate L2 thus has the incentive to deviate to a position slightly closer to x _L, m than x _L1, so as to take over from L the position of winning both the primary and the general election.⁶

• If $x_{L,m} \in \left [0, \frac{x_{L1}+x_{L2}} {2} \right )$, then by definition $x_{L1}> x_{L,m}$ (since by construction x _L2 < x _L1). In turn, L2 would have an optimal deviation to any point in the range $\left ((1 - x_{R1}),x_{L1}\right )$: by moving to any point in this range, L2 would join L1 in being able to defeat both R1 and R2 in the general election, and would be preferred on representational grounds by the median activist. By single-peakedness, she would also be preferred on representational grounds by all primary voters to the left of the median activist. This deviation would thus allow L2 to win the primary and general election with certainty.

• As such, scenario (e) cannot constitute a Nash Equilibrium. Note that party R’s two candidates face a strategically identical situation in scenario (f) to that faced by party L’s two candidates in scenario (e). As such, scenario (f) also cannot be a Nash Equilibrium.   QED

This establishes that any strategy vector at which $x_{L2} <x_{L1} <x_{m} <x_{R1} <x_{R2}$, and in which no two candidates are equidistant from the median voter’s ideal point, can be a Nash Equilibrium. Using identical reasoning, one can easily extend these proofs to situations in which one or more candidates are equidistant from the median voter’s ideal point, and to situations in which one party’s candidates differentiate while the other party’s do not. These extensions are omitted for reasons of redundancy. This establishes that, if the game has a pure strategy Nash equilibrium, it must be the case that x _L1 = x _L2 and x _R1 = x _R2.    QED

1.2 2 Nomination-Seeking Candidates

Lemma 1

As long as n > 0, in equilibrium, candidates in party L(R) will never choose a platform in the range $\left [0,x_{L,m}\right )(\left (x_{R,m},1\right ])$ .

Lemma 2

As long as n > 0, in equilibrium, candidates in party L(R) will never choose platforms $x_{L}> x_{m}(x_{R} <x_{m})$ .

Proof of Lemma 1

Consider the problem from the perspective of candidates L1 and L2. When x _L  < x _L, m there are three possibilities: either $\left \vert x_{L,m} - x_{m}\right \vert <\left \vert x_{R} - x_{m}\right \vert$, $\left \vert x_{L,m} - x_{m}\right \vert> \left \vert x_{R} - x_{m}\right \vert$, or $\left \vert x_{L,m} - x_{m}\right \vert = \left \vert x_{R} - x_{m}\right \vert$. If the first, then both candidates in L receive $\frac{g} {2}$, and could move to some platform in the range $\left (x_{L},x_{L,m}\right ]$ and receive g with certainty, since both the median activist and all primary voters to the right of the median activist would prefer the deviating candidate on representational grounds. If the second, then both candidates in L receive $\frac{n} {2}$, and could move to some platform in the range $\left (x_{L},x_{L,m}\right ]$ and receive n with certainty, since both the median activist and all primary voters to the right of the median activist would prefer the deviating candidate on representational grounds. If the third, then both candidates in L receive $\frac{n+g} {4}$, and could move to some platform in the range $\left (x_{L},x_{L,m}\right ]$ and receive g with certainty, since both the median activist and all primary voters to the right of the median activist would prefer the deviating candidate on representational grounds. The proof of Lemma 2 is identical, and omitted.    QED

Lemma 3

$\hat{x}_{L,m}(x_{R}) \in \left \{x_{L,m},(1 - x_{R}+\epsilon )\right \}$

Proof of Lemma 3

If $x_{L,m}> (1 - x_{R})$, then (trivially…) the median activist in L can choose his or her own ideal point, win the election with their most-preferred policy, and suffer no loss in ideological purity. If $x_{L,m} = (1 - x_{R})$, then the median activist can choose $(1 - x_{R}+\epsilon )$ where ε → 0, and win the election with a policy which is only infinitesimally different from his or her ideal point, and only suffer an infinitesimal cost in ideological purity.⁷ Any platform in the ranges $\left [0,(1 - x_{R}+\epsilon )\right ]$ and $\left ((1 - x_{R}+\epsilon ),x_{m}\right ]$ will be strictly less-preferred by the median activist than $1 - x_{R}+\epsilon$.

If $x_{L,m} <(1 - x_{R})$, then the median activist’s best response depends on the size of α _L, m . Define $\overline{\alpha }_{L,m}$ as the critical value for representation-seeking which leaves the median activist perfectly indifferent between the platforms x _L, m and $(1 - x_{R}+\epsilon$): if $\alpha _{L,m}> \overline{\alpha }_{L,m}$, the median activist prefers the former, and vice versa. A simple expected utility comparison employing equation (3) from the text demonstrates that:

$$\displaystyle{ \overline{\alpha }_{L,m} = \left \{ \frac{(x_{L,m} - x_{R})^{2}} {(1 - x_{L,m} - x_{R})^{2}} - 1\right \}\;. }$$

(5)

If $\alpha _{L,m} <\overline{\alpha }_{L,m}$, then the median activist’s optimal response will be $(1 - x_{R}+\epsilon )$: she prefers winning the election with the minimum sacrifice of ideological purism possible to choosing her own ideal point and forfeiting the election R. Any platform between this response and x _m represents an unnecessary sacrifice of ideological purism, and any platform between x _L, m forfeits the election to R, which is suboptimal if $\alpha _{L,m} <\overline{\alpha }_{L,m}$.

If $\alpha _{L,m}> \overline{\alpha }_{L,m}$, then the median activist’s optimal response will be x _L, m : she prefers forfeiting the election R and choosing her own ideal point, thus making no sacrifice of ideological purism. Any platform between x _L, m and ($1 - x_{R}+\epsilon$) represents an unnecessary sacrifice of ideological purism without winning the election, and any platform between ($1 - x_{R}+\epsilon$) and the median voter’s ideal point x _m wins the election, but since $\alpha _{L,m}> \overline{\alpha }_{L,m}$ this is suboptimal. An identical analysis applies to $\hat{x}_{R,m}(x_{L})$.    QED

Proof of Theorem 1

Without loss of generality consider the problem from the perspective of candidates L1 and L2.

Case 1: :

$x_{L,m}> (1 - x_{R})$

Given some platform x _R adopted by party R, if $x_{L,m}> (1 - x_{R})$, then the median activist’s optimal response will be her ideal point (Lemma 3). At any platform $x_{L}> x_{L,m}$ both candidates L1 and L2 would receive an expected payoff of $\frac{g} {2}$, since they both win the nomination with 50 % probability, and since by construction they would defeat x _R in the general election. In turn each could deviate to the platform x _L, m and receive g with certainty: at this platform they would still defeat x _R in a general election, and would secure the median activist’s support, along with that of all primary voters to the left of x _L, m , and thus win the nomination with certainty.

At any platform $x_{L} <x_{L,m}$ both candidates L1 and L2 would receive an expected payoff of $\frac{g} {2}$ if $x_{L}> (1 - x_{R})$, of $\frac{n} {2}$ if $x_{L} <(1 - x_{R})$, and of $\frac{n+g} {4}$ if $x_{L} = (1 - x_{R})$. In turn each could deviate to the platform x _L, m and receive g with certainty: at this platform they would defeat R in a general election, and would secure the median activist’s support, along with that of all primary voters to the right of x _L, m , and thus win the nomination with certainty.

Case 2: :

$x_{L,m} = (1 - x_{R})$

The median activist’s best response to any platform x _R such that $x_{L,m} = (1 - x_{R})$ will be $\hat{x}_{L,m}(x_{R}) = (1 - x_{R}+\epsilon )$ (Lemma 3).

At any platform $x_{L}> (1 - x_{R}+\epsilon )$ both candidates L1 and L2 would receive an expected payoff of $\frac{g} {2}$, since they both win the nomination with 50 % probability, and since by construction they would defeat x _R in the general election. In turn each could deviate to the platform $(1 - x_{R}+\epsilon )$ and receive g with certainty: at this platform they would still defeat x _R in a general election, and would secure the median activist’s support, along with that of all primary voters to the left of x _L, m , and thus win the nomination with certainty.

At any platform $x_{L} \leq x_{L,m}$ both candidates L1 and L2 would receive an expected payoff of $\frac{n} {2}$ if $x_{L} <x_{L,m}$, and of $\frac{n+g} {4}$ if x _L  = x _L, m . In turn each could deviate to the platform $(1 - x_{R}+\epsilon )$ and receive g with certainty: at this platform they would defeat R in a general election, and would secure the median activist’s support, along with that of all primary voters to the right of x _L, m , and thus win the nomination with certainty.

Case 3: :

$x_{L,m} <(1 - x_{R})$

Recalling that $\epsilon \rightarrow 0$, the point $(\frac{x_{L,m}+1-x_{R}} {2} )$ is the “indifference” point (i.e. midpoint) between x _L, m and $(1 - x_{R}+\epsilon$). Trivially, any primary voter in L with ideal point $x_{i} \geq (\frac{x_{L,m}+1-x_{R}} {2} )$ will prefer the platform $(1 - x_{R}+\epsilon )$ on grounds of both representation and viability, since they are closer in space to said platform. Primary voters with ideal points $x_{i} <(\frac{x_{L,m}+1-x_{R}} {2} )$ face a conflict between representation-seeking and policy-seeking. Define $\overline{\alpha }_{i}$ as the critical value for representation-seeking which leaves such a primary voter perfectly indifferent between the platforms x _L, m and $(1 - x_{R}+\epsilon )$: if $\alpha _{i}> \overline{\alpha }_{i}$, they prefer the former, and vice versa. A simple expected utility comparison employing equation (3) from the text demonstrates that:

$$\displaystyle{ \overline{\alpha }_{i} = \left \{ \frac{(1 - x_{i} - x_{R})^{2} - (x_{i} - x_{R})^{2}} {(x_{i} - x_{L,m})^{2} - (1 - x_{i} - x_{R})^{2}}\right \}\;. }$$

(6)

Substituting x _L, m for x _i in (6) and rearranging yields (5).

Lemma 4

Over the range $\left [0, \frac{x_{L,m}+1-x_{R}} {2} \right )$ the critical value $\overline{\alpha }$ is increasing in $x_{i}$ ( $\frac{\Delta \overline{\alpha }} {\Delta x_{i}}> 0)$ and $\overline{\alpha }_{i} \rightarrow \infty$ as $x_{i} \rightarrow \frac{x_{L,m}+1-x_{R}} {2}$ .

Lemma 4 tells us that this critical value increases asymptotically up to the midpoint $\left \{\frac{x_{L,m}+1-x_{R}} {2} \right \}$. I omit the somewhat tedious proof, which employs the quotient rule to extract the derivative of (6) with respect to x _i and then demonstrates that, over the specified range, this derivative must be greater than 0.

The intuition behind Lemma 4 is quite simple: primary voters who are fairly indifferent between the policies x _L, m and $(1 - x_{R}+\epsilon )$ are only willing to forfeit the election to x _R and choose x _L, m if the parameter α _i is very large. Conversely, the further a primary voter is from $(1 - x_{R}+\epsilon )$, the greater the greater is the cost (in terms of ideological purity) of choosing $(1 - x_{R}+\epsilon )$, and the more likely he or she will be willing to forfeit the election to party R and choose x _L, m . With Lemma 1 and Assumption 1 in hand, we can prove Theorem 1 for case 3, i.e. situations in which x _L, m  < (1 − x _R ).

Case 3a: :

x _L, m  < (1 − x _R ) and $\alpha _{L,m} <\overline{\alpha }_{L,m}$

If x _L, m  < (1 − x _R ) and $\alpha _{L,m} <\overline{\alpha }_{L,m}$ then the median activist’s best response is $\hat{x}_{L,m}(x_{R}) = (1 - x_{R})+\epsilon$.

At any platform $x_{L}> (1 - x_{R})+\epsilon$ both candidates L1 and L2 would receive an expected payoff of $\frac{g} {2}$, since they both win the nomination with 50 % probability, and since by construction L’s candidate would defeat x _R in the general election. In turn each could deviate to the platform $(1 - x_{R})+\epsilon$ and receive g with certainty: at this platform they would still defeat R in a general election, but would secure the median activist’s support, along with that of all primary voters to the left of x _L, m (by single-peakedness of preferences for representation), and thus win the nomination with certainty.

At any platform $x_{L,m} \leq x_{L} \leq (1 - x_{R})$ both candidates L1 and L2 would receive an expected payoff of $\frac{n} {2}$ if $x_{L} <(1 - x_{R})$, and of $\frac{n+g} {4}$ if $x_{L} = (1 - x_{R})$. In turn, they could deviate to the position $(1 - x_{R})+\epsilon$ and secure the median activist’s support, since by construction $\alpha _{L,m} <\overline{\alpha }_{L,m}$. As well, the deviating candidate would also secure the support of all primary voters to the right of x _L, m . To see this, note first that any primary voters in the range $\left [x_{L,m}\frac{x_{L,m}+1-x_{R}} {2} \right )$ support the deviating candidate by Assumption 1 and Lemma 2: these primary voters have values of $\alpha _{i} \leq \alpha _{L,m}$ (Assumption 1), and have critical values of $\overline{\alpha }_{i}> \overline{\alpha }_{L,m}$ (Lemma 2). As such, if the condition for preferring $(1 - x_{R})+\epsilon$ to x _L, m is met for the median activist, it is also met for all primary voters with ideal points in $\left [x_{L,m}\frac{x_{L,m}+1-x_{R}} {2} \right )$. Secondly, any primary voters in the range $\left [\frac{x_{L,m}+1-x_{R}} {2},\overline{x}_{L}\right ]$ support the deviating candidate because they prefer her position on grounds of both representation and electoral viability. Thus a deviation to $(1 - x_{R})+\epsilon$ would secure the nomination, and would then allow the deviating candidate to win the general election and receive g with certainty.

At any platform $x_{L} <x_{L,m}$ both candidates L1 and L2 would receive an expected payoff of $\frac{n} {2}$. In turn each could deviate to the platform $(1 - x_{R})+\epsilon$ and receive g with certainty: at this platform they would defeat R in a general election, but would secure the median activist’s support, along with that of all primary voters to the right of x _L, m (by single-peakedness of preferences for representation), and thus win the nomination with certainty. This concludes the proof that, if $x_{L,m} <(1 - x_{R})$ and $\alpha _{L,m} <\overline{\alpha }_{L,m}$, no policy other than the median activist’s best response $\hat{x}_{L,m}(x_{R}) = (1 - x_{R})+\epsilon$ can be a Nash Equilibrium.

Case 3b: :

$x_{L,m} <(1 - x_{R})$ and $\alpha _{L,m} = \overline{\alpha }_{L,m}$

If $x_{L,m} <(1 - x_{R})$ and $\alpha _{L,m} = \overline{\alpha }_{L,m}$, then the median activist is indifferent between the policies $(1 - x_{R})+\epsilon$ and x _L, m .

The proof that any platform other than one of these two cannot be a Nash Equilibrium is identical to that above for case 3a.

As well, if x _L  = x _L, m , both candidates L1 and L2 would receive an expected payoff of $\frac{n} {2}$, since they both win the nomination with 50 % probability, and since by construction L’s candidate would lose to x _R in the general election. In turn each could deviate to the platform $(1 - x_{R})+\epsilon$ and receive g with certainty: at this platform they would secure the median activist’s support, along with that of all primary voters to the right of x _L, m (see analysis of case 3a), and thus win the nomination with certainty. As a result, the only possible Nash Equilibrium in case 3b is $x_{L}^{{\ast}} = (1 - x_{R})+\epsilon$.

Case 3c: :

$x_{L,m} <(1 - x_{R})$ and $\alpha _{L,m}> \overline{\alpha }_{L,m}$

If $x_{L,m} <(1 - x_{R})$ and $\alpha _{L,m}> \overline{\alpha }_{L,m}$, then the median activist’s best response is $\hat{x}_{L,m}(x_{R}) = x_{L,m}$.

At any platform x _L  > (1 − x _R ) both candidates L1 and L2 would receive an expected payoff of $\frac{g} {2}$, since they both win the nomination with 50 % probability, and since by construction L’s candidate would defeat x _R in the general election. In turn each could deviate to a platform in the range $\left [(1 - x_{R})+\epsilon,x_{L}\right )$ and receive g with certainty: at this platform they would still defeat R in a general election, but would secure the median activist’s support, along with that of all primary voters to the left of x _L, m (by single-peakedness of preferences for representation), and thus win the nomination with certainty.

At any platform $x_{L,m} <x_{L} <(1 - x_{R})$ both candidates L1 and L2 would receive an expected payoff of $\frac{n} {2}$, since they both win the nomination with 50 % probability, and since by construction L’s candidate would lose to x _R in the general election. If n > 0, then either candidate could deviate to x _L, m and win n with certainty (by single-peakedness of preferences for representation); if n < 0, then either candidate could deviate to a position in the range $\left (x_{L},(1 - x_{R})+\epsilon \right )$ and lose n with certainty: by the single-peakedness of preferences for representation she would lose the support of both the median activist and all primary voters to the “left” of the median activist.

At any platform $x_{L} <x_{L,m}$, then both candidates L1 and L2 would receive an expected payoff of $\frac{n} {2}$, since they both win the nomination with 50 % probability, and since by construction L’s candidate would lose to x _R in the general election. If n > 0, then either candidate could deviate to x _L, m and win n with certainty (by single-peakedness of preferences for representation); if n < 0 then either candidate could deviate to a position “left” of x _L and lose n with certainty: by the single-peakedness of preferences for representation she would lose the support of both the median activist and all primary voters to the “right” of the median activist.

If $\alpha _{L,m}> \overline{\alpha }_{L,m}$ and $x_{L} = (1 - x_{R})$, then both candidates L1 and $L2$ would receive an expected payoff of $\frac{n+g} {4}$ (see footnote 4). In turn, they could deviate to the position x _L, m and secure the median activist’s support, since by construction $\alpha _{L,m}> \overline{\alpha }_{L,m}$. As well, the deviating candidate would also secure the support of all primary voters to the “left” of x _L, m : by Assumption 1 these primary voters have values of $\alpha _{i} \geq \alpha _{L,m}$, and by Lemma 2 these primary voters have critical values of $\overline{\alpha }_{i} <\overline{\alpha }_{L,m}$. As such, if the condition for preferring x _L, m to 1 − x _R is met for the median activist, it is also met for all primary voters to her left. Thus a deviation to x _L, m would allow the deviating candidate to secure the nomination, and then by construction lose the general election, thus receiving n with certainty. This deviation will be optimal as long as $n> \frac{n+g} {4}$, which can be simplified to $n> \frac{g} {3}$, which was the condition for “nomination-seeking” established at the outset of Sect. 3 in the text.

Theorem  1: Summary

The preceding analysis establishes that, as long as $n> \frac{g} {3}$, in any Nash Equilibrium candidates from party L must choose their median-activist’s best response to x _R . An identical analysis applies to candidates in party R. This concludes the proof of Theorem 1, that as long as $n> \frac{g} {3}$ any Nash Equilibrium must involve candidates from both parties choosing their median activists’ mutual best responses.     QED