Incumbent-challenger and open-seat elections in a spatial model of political competition

Abstract

In Groseclose (Am J Political Sci 45:862–886, 2001), candidates with asymmetric valence scores and varying degrees of policy motivation simultaneously choose divergent policies. I take a version of the Groseclose model with policy-motivated candidates and extend it to allow for sequential policy announcements. This may be a suitable approach for incumbent-challenger elections because the incumbent’s policy is typically known before the challenger’s. I show that policy divergence is greater when candidates announce policies sequentially as opposed to simultaneously. I also show that policy-motivated incumbents benefit from a first-mover advantage.

Appendix

Lemma 1

In equilibrium $\frac{\partial F(m_\beta )}{\partial v}>0$, $\frac{\partial F(m_\beta )}{\partial P_L}>0$ and $\frac{\partial F(m_\beta )}{\partial P_R}>0$.

Proof

If $P_L<{P_R}$ , then $\frac{\partial F(m_\beta )}{\partial v}=\frac{-1}{(4P_L-4P_R)}>0$ . While I impose no constraints on the candidates’ policy choices, apart from $P_c\in [-1,1]$ , it is always the case that $P_L<{P_R}$ . The policy-motivated left-wing candidate never chooses a policy further right than the ideologically opposed right-wing candidate. Differentiating $F(m_\beta )$ with respect to $P_L$ yields $\frac{\partial F(m_\beta )}{\partial P_L}=\frac{v}{4(P_L-P_R)^2}+\frac{1}{4}>0$ . If $v<{(P_L-P_R)^2}$ , then $\frac{\partial F(m_\beta )}{\partial P_R}=\frac{1}{4}-\frac{v}{4(P_L-P_R)^2}>0$ . Given $m_\beta =\frac{\displaystyle v-P_L^2+P_R^2}{\displaystyle 2(P_R-P_L)}$ , it is straightforward to show that the inequality $v<{(P_L-P_R)^2}$ is the same as $P_R>{m_\beta }$ . It cannot be that $P_R<m_\beta$ because then $\frac{\partial F(m_\beta )}{\partial P_R}<0$ , meaning that R could choose a more preferable right-wing position while at the same time increasing her probability of victory. As such, R always locates to the right of the indifferent voter and, therefore, $\frac{\partial F(m_\beta )}{\partial P_R}=\frac{1}{4}-\frac{v}{4(P_L-P_R)^2}>0$. $\square$

Lemma 2

If the valence advantage exceeds a critical value $\underline{v}$ , there is a unique equilibrium at which L wins with probability 1 in the incumbent-challenger and open-seat elections. L policy choice always lies within’s $P_L\in [-1,1-\sqrt{v}]$.

Proof

L has a 100 % probability of victory when $m_\beta =1$. From Lemma 1 we know that R never locates to the left of the indifferent voter, so if $m_\beta =1$ then $P_R=1$. Using $P_R=m\beta =1$ in Eq. (4) gives the policy point at which L can locate and guarantee victory, which is a function of valence, $P_L^{win}=1-\sqrt{v}$. Clearly L will never move further right than that point because her victory already is guaranteed; therefore, $P_L\in [-1,1-\sqrt{v}]$. Using $P_R=1$ in L’s utility function and taking the first-order condition $\frac{\partial EU_L}{\partial P_L}$ yields $\frac{v}{4}-\frac{5P_L}{2}-\frac{3P_L^2}{4}-\frac{3}{4}=0$. Using $P_L=1-\sqrt{v}$ and solving for v gives $v=1.37$. Therefore, when $v=1.37$, L locates at $P_L=1-\sqrt{1.37}=-0.17$ forcing R to locate at the edge of the policy space, i.e., at $P_R=1$. When $v\ge {1.37}$, L is guaranteed victory in equilibrium in both open-seat and incumbent-challenger elections, so $\underline{v}=1.37$. $\square$

Lemma 3

When $v>0$ , then $\frac{\partial ^2 F(m_\beta )}{\partial P_L^2}>0$. The valence-advantaged candidate’s probability of victory increases at an increasing rate the closer she locates to her opponent: $\frac{\partial ^2 F(m_\beta )}{\partial P_L^2}$ is increasing in v, meaning that the greater is L’s valence advantage, the greater is the associated increase in her probability of victory for a given movement along the policy line.

Proof

Given that $P_L<0$ and $P_R>0$, then $\frac{\partial ^2 F(m_\beta )}{\partial P_L^2}=\frac{-v}{2(P_L-P_R)^3}>0$ when $v>0$. As v increases $\frac{-v}{2(P_L-P_R)^3}>0$ increases. $\square$

Proposition 1

If $v\le {\underline{v}}$ then,

• (i) A move along the policy line by a candidate causes her opponent to react by moving in the same direction. That is, $\frac{\partial R_L(P_R,v)}{\partial P_R}>0$ and $\frac{\partial R_R(P_L,v)}{\partial P_L}>0$.

• (ii) An increase in L’s valence advantage causes L to moderate her position and R to move to the right. That is, $\frac{\partial R_L(P_R,v)}{\partial v}>0$ and $\frac{\partial R_R(P_L,v)}{\partial v}>0$.

Proof

If $v<\underline{v}$, then $\frac{\partial R_L(P_R,v)}{\partial P_R}=\frac{4P_R+4}{3\sqrt{4P_R^2+8P_R+3v+4}}-\frac{1}{3}>0$. If $v<\underline{v}$ and $P_L\in [-1,0]$, then $\frac{\partial R_R(P_L,v)}{\partial P_L}=-\frac{4P_L-4}{3\sqrt{4P_L^2-8P_L-3v+4}}-\frac{1}{3}>0$ . I do not impose any constraints such that the left-wing candidate is forced to implement a left-wing policy (where $P_L\in {[-1,0]}$); however, this always occurs. As $P_L<\, {P_R}$, then if L ever chose a right-wing policy, she could always improve by moving leftwards to the center and in doing so adopt a more preferable policy position while also improving her chances of winning. (ii) Differentiating the candidates’ reaction functions with respect to v yields $\frac{\partial R_L(P_R,v)}{\partial v}=\frac{1}{2\sqrt{4P_R^2+8P_R+3v+4}}>0$ and $\frac{\partial R_R(P_L,v)}{\partial v}=\frac{1}{2\sqrt{4P_L^2-8P_L-3v+4}}>0$. $\square$

Proposition 2

$P_L^I$ and $P_R^C$ are the equilibrium positions for the incumbent-challenger election and $P_L*$ and $P_R*$ are the equilibrium positions for the open-seat election. If $v=0$ , then $P_L^I=x_L=-1,$ $P_R^C=\frac{1}{3}$ and $P_L*=-\frac{1}{2}$, $P_R*=\frac{1}{2}$.

Proof

If $v=0$, the reaction functions simplify to $R_L(P_R,v)=\frac{-2 + P_R}{3}$ and $R_R(P_L,v)=\frac{2 + P_L}{3}$. Solving the equations simultaneously gives the policy locations for the open-seat election as $P_L*=-0.5$ and $P_R*=0.5$. For the incumbent-challenger election, I substitute $v=0$ into the incumbent’s first-order condition, Eq. (14), and solve to give $P_L^I=-1$. Using this result in R’s reaction function gives $R_R(P_L,v)=\frac{2 + (-1)}{3}=\frac{1}{3}$, so $P_R^C=\frac{1}{3}$. $\square$

Proposition 3

When v increases from zero to a small value, the valence-advantaged candidate moderates towards the center and the valence-disadvantaged candidate adopts a more extreme policy in both open-seat and incumbent-challenger elections. That is, $\frac{\partial P_L{*}}{\partial v}\big |_{v=0}>0$, $\frac{\partial P_R{*}}{\partial v}\big |_{v=0}>0$ and $\frac{\partial P_L^I}{\partial v}\big |_{v=0}>0$, $\frac{\partial P_R^C}{\partial v}\big |_{v=0}>0$.

Proof

From Eqs. (9) and (10), $P_L*=R_L(P_R*,v)=R_L(R_R(P_L*,v),v)$. Taking the partial derivative with respect to v gives $\frac{\partial P_L*}{\partial v}=\frac{\partial R_L}{\partial R_R} \frac{\partial R_R}{\partial P_L*} \frac{\partial P_L*}{\partial v}+\frac{\partial R_L}{\partial R_R} \frac{\partial R_R}{\partial v}+\frac{\partial R_L}{\partial v}$. Solving for $\frac{\partial P_L*}{\partial v}$ gives the expression, $\frac{\partial P_L*}{\partial v}=\frac{\frac{\partial R_L}{\partial P_R}\frac{\partial R_R}{\partial v}+\frac{\partial R_L}{\partial v}}{1-\frac{\partial R_L}{\partial P_R}\frac{\partial R_R}{\partial P_L}}$. I determine the sign of $\frac{\partial P_L*}{\partial v}\big |_{v=0}$ by analyzing the RHS terms. This yields $\frac{\partial R_L}{\partial P_R}\big |_{v=0}=\frac{2P_R+2}{3\sqrt{P_R^2+2P_R+1}}-\frac{1}{3}>0$, $\frac{\partial R_R}{\partial v}\big |_{v=0}=\frac{1}{4\sqrt{P_L^2-2P_L+1}}>0$, $\frac{\partial R_L}{\partial v}\big |_{v=0}=\frac{1}{4\sqrt{P_R^2+2P_R+1}}>0$ and $\frac{\partial R_R}{\partial P_L}\big |_{v=0}=-\frac{2P_L-2}{3\sqrt{P_L^2-2P_L+1}}-\frac{1}{3}>0$. Finally, $\frac{\partial R_L}{\partial P_R}\frac{\partial R_R}{\partial P_L}<1$, so $1-\frac{\partial R_L}{\partial P_R}\frac{\partial R_R}{\partial P_L}>0$ and, hence, $\frac{\partial P_L{*}}{\partial v}\big |_{v=0}>0$. In the same way it can be shown that $\frac{\partial P_R{*}}{\partial v}\big |_{v=0}>0$. For the incumbent-challenger election, recall from Proposition 2 that when $v=0$, $P_L^I=-1$. Using this in the incumbent’s first-order condition, Eq. (14), gives $\frac{\partial EU_L^I}{\partial P_L^I}\big |_{P_L^I=-1}=\frac{1}{6}(v(\frac{4}{\sqrt{16-3v}}+1))$. Clearly, when $v=0$ this equation equals zero as utility is maximized. However, notice that when v goes from zero to a positive value we have $\frac{\partial EU_L^I}{\partial P_L^I}>0$, meaning that candidate L moves to the right to increase her utility. Therefore, $\frac{\partial P_L^I}{\partial v}\big |_{v=0}>0$. We have already seen that $\frac{\partial R_R}{\partial P_L}>0$, so as the incumbent L moves right so too does the challenger R and, as such, $\frac{\partial P_R^C}{\partial v}\big |_{v=0}>0$. $\square$

Proposition 4

If $0\le {v}<\underline{v}$ then,

• (i) $P_L^I<P_L*$ and $P_R^C<P_R*$

• (ii) $|P_L^I-P_R^C|>|P_L*-P_R*|$

Proof

(i) The incumbent’s expected utility function is written as $EU_L^I(P_L^I,R_R^C$ $(P_L^I,v),v)$. The first-order condition for the expected utility optimization problem is $\frac{\partial EU_L^I}{\partial P_L^I}=\frac{\partial EU_L^I}{\partial P_L^I}+\frac{\partial EU_L^I}{\partial R_R^C}\frac{\partial R_R^C}{\partial P_L^I}$. Evaluating the FOC at the open-seat equilibrium position $P_L*$, means that the first term on the RHS of the FOC satisfies L’s open-seat utility optimization problem, $\frac{\partial EU_L^I}{\partial P_L^I}\big |_{P_L*}$. Therefore, the sign of the second term, $\frac{\partial EU_L^I}{\partial R_R^C}\frac{\partial R_R^C}{\partial P_L^I}$ tells us whether, as an incumbent, L can improve her utility by locating at a point that differs from the open-seat election. If $-1\le {P_L^I}<1-\sqrt{v}$, then $\frac{\partial EU_L^I}{\partial R_R^C}=-\frac{{P_L^I}^2}{4}+\frac{P_L^I P_R^C}{2}+\frac{3{P_R^C}^2}{4}+\frac{v}{4}-1<0$ . In the proof of Proposition 1, I show that $\frac{\partial R_R}{\partial P_L}>0$. Therefore, $\frac{\partial EU_L^I}{\partial R_R^C}\frac{\partial R_R^C}{\partial P_L^I}<0$, meaning that when L is the incumbent she does better by locating to the left of her open-seat position, so $P_L^I<P_L*$. From Proposition 1, $\frac{\partial R_R(P_L,v)}{\partial P_L}>0$, meaning that $P_L^I<P_L*$ implies $P_R^C<P_R*$. (ii) It is straightforward to show that $0<\frac{\partial R_R}{\partial P_L}<1$, so a move to the left by L causes R to move to the left by a smaller amount. As such, given $P_L^I<P_{L}{*}$ it must be that $|P_L^I-P_R^C|>|P_L{*}-P_R{*}|$ $\square$

Proposition 5

If $-\underline{v}<v<0$ then,

• (i) $P_L^I=x_L=-1$

• (ii) $|P_L^I-P_R^C|$ is higher compared to when $v=v_L-v_R>0$.

• (iii) $P_L^I<P_L*$

Proof

If $v=0$, then $P_L^I=-1$. The incumbent’s expected utility function is $EU_L^I(P_L^I,R_R^C(P_L^I,v),v)$. From this, $\frac{\partial EU_L^I}{\partial P_L^I}\big |_{P_L^I=-1}=\frac{v}{6}+\frac{32}{9\sqrt{16-3v}}-\frac{2\sqrt{16-3v}}{9}$. In this equation, $v=v_L-v_R$ so that when the incumbent is at a disadvantage, $v<0$. If $v<0$ then $\frac{\partial EU_L^I}{\partial P_L^I}\big |_{P_L^I=-1}<0$. As the incumbent becomes disadvantaged she would like to move further left than $P_L^I=-1$, but $P_L\in [-1,1-\sqrt{v}]$. Therefore, L stays at the extreme left of the policy line. For (ii), note that $\frac{\partial R_R}{\partial P_L^I}=\frac{4P_R^C+4}{3 \sqrt{4{P_R^C}^2+8P_R^C+3v+4}}-\frac{1}{3}$, giving $0<\frac{\partial R_R^C}{\partial P_L^I}<1$. A move to the left by L causes R to move left by less than L. Given that the disadvantaged incumbent locates further left than when she is advantaged, it is the case that $|P_L^{inc}-P_R^{cha}|$ is higher for $v=v_L-v_R<0$ than when $v=v_L-v_R>0$. For (iii), note that Proposition 3 shows $\frac{\partial P_L*}{\partial v}=\frac{\frac{\partial R_L}{\partial P_R}\frac{\partial R_R}{\partial v}+\frac{\partial R_L}{\partial v}}{1-\frac{\partial R_L}{\partial P_R}\frac{\partial R_R}{\partial P_L}}$. I determine the sign of $\frac{\partial P_L*}{\partial v}$ by analyzing the RHS terms of this equation. This yields $\frac{\partial R_L}{\partial P_R}=\frac{2P_R+2}{3\sqrt{P_R^2+2P_R+.75v+1}}-\frac{1}{3}>0$, $\frac{\partial R_R}{\partial v}=\frac{1}{4\sqrt{P_L^2-2P_L-.75v+1}}>0$, $\frac{\partial R_L}{\partial v}=\frac{1}{4\sqrt{P_R^2+2P_R+.75v+1}}>0$ when $-\underline{v}<v<0$. Note that $\frac{\partial \widehat{P_L}}{\partial P_R}\frac{\partial \widehat{P_R}}{\partial P_L}<1$, so that $1-\frac{\partial \widehat{P_L}}{\partial P_R}\frac{\partial \widehat{P_R}}{\partial P_L}>0$. Therefore, $\frac{\partial P_L{*}}{\partial v}>0$ when $-\underline{v}<v<0$. As such, $P_L^I<P_L*$. $\square$

Proposition 6

If $-\underline{v}\,<v\, <\underline{v}$ , then $E(U_L^I)>E(U_L*)>E(U_L^C)$ . The incumbent benefits from moving first irrespective of whether the incumbent or challenger benefits from a valence advantage.

Proof

This follows directly from Propositions 2, 4 and 5. The incumbent can always do at least as well as if she were a candidate in an open seat. However, the incumbent as first-mover chooses a policy position further left than the open-seat equilibrium in order to increase her expected utility. Given that the reaction functions are upward sloping, L’s iso-utility curve is concave to her policy axis, and lower iso-utility curves are associated with higher utility, it follows that if $P_L^I<P_L{*}<P_L^C$ then $E(U_L^I)>E(U_L*)>E(U_L^C)$. $\square$