Conflict under the shadow of elections

Abstract

In this article we study whether an elected leader has incentives to withdraw from an ongoing military conflict before elections take place. Remaining active in the conflict increases the chances of a victory, which would boost the political incumbent’s reelection probability. On the other hand, if the rival does not surrender, then the incumbent politician’s reelection probability decreases because of the costly conflict citizens have to endure. We show that when the crisis is costly to voters, the more distant the elections are, the more likely the ruling government withdraws from an ongoing war early. For low-cost conflicts, politicians never give up the fighting. We also show that confrontations in the shadow of elections are socially inefficient (too long or too short) because of the inherent misalignment of preferences between citizens and politicians.

Appendices

Appendix A: Pre-electoral period

From (12), the first-order derivative is

$$\begin{aligned} \frac{\partial {\mathcal {V}}_I}{\partial T_I^G}=-\frac{a}{2} \frac{\partial {\mathbb {E}}[T^b_{W}] }{\partial T_I^G}(\mu -a{\mathbb {E}}[T_W^b]) +\kappa f_R(T_I^G), \end{aligned}$$

where the marginal probability of the opponent declaring defeat, $f_R(.)$ is given by $1/{{\bar{T}}}$, and thus

$$\begin{aligned} {\mathbb {E}}[T_W^b]=\int _{0}^{T_I^G} s f_R(s) ds=\frac{1}{\bar{T}}\int _{0}^{T_I^G} s ds= \frac{(T_I^G)^2}{2{{\bar{T}}}}; \end{aligned}$$

; hence,

$$\begin{aligned} \frac{\partial {\mathbb {E}}[T_W^b] }{\partial T_I^G}=\frac{T_I^G}{\bar{T}}. \end{aligned}$$

The first-order condition is then

$$\begin{aligned} \frac{\partial {\mathcal {V}}_I}{\partial T_I^G}=-\frac{a T_I^G}{2\bar{T}}\left( \mu - a\frac{(T_I^G)^2}{2{{\bar{T}}}}\right) +\frac{\kappa }{\bar{T}}=:\Phi (T_I^G). \end{aligned}$$

(15)

The second-order derivative is

$$\begin{aligned} \frac{\partial ^2{\mathcal {V}}_I}{\partial (T_I^G)^2}=\Phi '(T_I^G)=-\frac{a}{2{{\bar{T}}}}\left( \mu - a\frac{(T_I^G)^2}{2{{\bar{T}}}}\right) +\frac{a^2 (T_I^G)^2}{2{{\bar{T}}}^2}. \end{aligned}$$

Observe that $\Phi (T_I^G)$ admits a minimal value on ${\textbf{R}}^{+}$ at $T_I^G=\sqrt{\frac{2\mu {\bar{T}}}{3a}}$, with $\Phi (0)=h\kappa /{{\bar{T}}}>0$. In addition, $\Phi (\sqrt{\frac{2\mu {\bar{T}}}{3a}}) < 0 \Leftrightarrow {{\bar{T}}}> \frac{27 \kappa ^2}{2ah\mu ^3}$; a condition that we assume throughout the paper since a fairly large time horizon is considered (a high ${{\bar{T}}}$).

As $\Phi (\sqrt{\frac{2\mu {\bar{T}}}{3a}})< 0$, according to the intermediate value theorem, ${\mathcal {V}}_I$ admits a local interior maximum at ${\check{T}}\in ]0,\sqrt{\frac{2\mu {\bar{T}}}{3a}}[$, which will also be the global maximum if ${\mathcal {V}}_I\left( \check{T}\right) >{\mathcal {V}}_I(T_E)$. Imposing that $T_E\le \sqrt{\frac{2\mu {{\bar{T}}}}{3a}}$ enables us to conclude that the maximum of ${\mathcal {V}}_I$ on $[0,T_E]$ will always be such that ${\hat{T}}_I^G:=argmax_{T_I^G \in [0,T_E]} {\mathcal {V}}_I(T_I^G) =\min \left\{ {\check{T}},T_E\right\}$.

By applying the implicit function theorem to $\Phi (T_I^G)=0$, we can then deduce that $\partial T_I^G/\partial \kappa =-\frac{{\partial \Phi (T_I^G)}/{\partial \kappa }}{{\partial \Phi (T_I^G)}/{\partial T_I^G}}=-\frac{1/{{\bar{T}}}}{{\partial ^2{\mathcal {V}}_I}/{\partial (T_I^G)^2}}$. As $T^I_G={\check{T}}$ is a local maximum of ${\mathcal {V}}_I$, the second-order condition implies that $\partial {\check{T}}/\partial \kappa >0.$ In addition, looking at Eq. (15), ${\check{T}}=0<T_E$ if $\kappa =0$, and $\check{T}\rightarrow +\infty >T_E$ as $\kappa \rightarrow \infty$. Hence, there is a critical level ${\underline{\kappa }}>0$, such that: ${\hat{T}}_I^G={\check{T}}<T_E$ if $\kappa <{\underline{\kappa }}$, and ${\check{T}}\ge T_E={\hat{T}}_I^G$ if $\kappa \ge {\underline{\kappa }}$.

Additionally, the critical level ${\underline{\kappa }}$ is such that ${\check{T}}-T_E=0$; hence,

$$\begin{aligned} \frac{\partial {\underline{\kappa }}}{\partial T_E}=\left[ \left. \frac{\partial {\check{T}}}{\partial \kappa }\right| _{\kappa ={\underline{\kappa }}}\right] ^{-1}>0. \end{aligned}$$

Appendix:B: Post-election period

After the election, the first- and second-order derivatives are

$$\begin{aligned} \frac{\partial {\mathcal {V}}_I}{\partial T_I^G}= [1-F_R(T_E)]v'(T^G_I);\text { and }\frac{\partial ^2 {\mathcal {V}}_I}{\partial (T_I^G)^2}= [1-F_R(T_E)]v''(T^G_I). \end{aligned}$$

where

$$\begin{aligned} v(T_j^G)=&\int _0^{T_E} -ve^{-rt}\;\text {d}t\nonumber \\&+[1-F_R(T_j^G)]\left\{ \int _{T_E}^{T_j^G}-ve^{-rt}\;\text {d}t \right\} \nonumber \\&+\int _{x=T_E}^{x=T_j^G} \left\{ \int _{T_E}^x -ve^{-rt}\;\text {d}t+\alpha e^{-rx} \right\} f_R(x)\;\text {d}x. \end{aligned}$$

(16)

Hence,

$$\begin{aligned} v'(T_I^G)=\alpha e^{-rT_I^G}f_R(T_I^G)-ve^{-rT_I^G}[1-F_R(T_I^G)]. \end{aligned}$$

Given the assumed uniform distribution, we obtain

$$\begin{aligned} v'(T_I^G)=\frac{1}{{{\bar{T}}}} e^{-rT_I^G}\left[ \alpha -v(\bar{T}-T_I^G)\right] , \end{aligned}$$

(17)

and

$$\begin{aligned} v''(T_I^G)=\frac{e^{-rT_I^G}}{{{\bar{T}}}}\left[ -\alpha r+rv(\bar{T}-T_I^G)+v\right] . \end{aligned}$$

(18)

There is a unique critical point at

$$\begin{aligned} {\hat{T}}={{\bar{T}}}-\frac{\alpha }{v}. \end{aligned}$$

Substituting in (18), we observe that $v''({\hat{T}})>0$; hence ,${\hat{T}}$ is a global minimum. Therefore, the maximum of the payoff ${\mathcal {V}}_I$ on $[T_E,{{\bar{T}}}]$ is reached at a boundary, either at $t=T_E$ or $t={{\bar{T}}}$.

To then find the maximum, we need to determine the sign of ${\mathcal {V}}_I({{\bar{T}}})-{\mathcal {V}}_I(T_E)$. Using Eqs. (5) and (11), we compute

$$\begin{aligned} \frac{{\mathcal {V}}_I({{\bar{T}}})-{\mathcal {V}}_I(T_E)}{[1-F_R(T_E)]}= & {} v({{\bar{T}}})-v(T_E) \nonumber \\= & {} \int _{x=T_E}^{x=T_j^G} \left\{ \int _{T_E}^x -ve^{-rt}\;\text {d}t+\alpha e^{-rx} \right\} f_R(x)\text {d}x \nonumber \\= & {} \frac{v}{{{\bar{T}}}r}\int _{T_E}^{T_j^G} \left( e^{-rx}-e^{-rT_E}\right) \text {d}x+\frac{\alpha }{{{\bar{T}}}}\int _{T_E}^{T_j^G} e^{-rx} \text {d}x \nonumber \\= & {} \frac{1}{r{{\bar{T}}}}\psi (v), \end{aligned}$$

(19)

where

$$\begin{aligned} \psi (v):= \left( \alpha -\frac{v}{r}\right) \left( e^{-rT_E}-e^{-r\bar{T}}\right) -v({{\bar{T}}} - T_E)e^{-rT_E}. \end{aligned}$$

(20)

We have $\psi '<0$, $\psi (+\infty )=-\infty$, and $\psi (0)=\alpha (e^{-rT_E}-e^{-r{{\bar{T}}}})>0$.

There is a unique critical value ${\underline{v}} >0$, such that $\psi (v)>0\Leftrightarrow {\mathcal {V}}_I({{\bar{T}}})>{\mathcal {V}}_I(T_E)$ for $v<{\underline{v}}$, and $\psi (v)\le 0\Leftrightarrow {\mathcal {V}}_I({{\bar{T}}})\le {\mathcal {V}}_I(T_E)$ for $v\ge \underline{v}$. Consequently, the maximum of the payoff ${\mathcal {V}}_I$ on the post-election period is ${{\bar{T}}}$ if $v<{\underline{v}}$, and $T_E$, if $v\ge {\underline{v}}$.

Appendix C: Global extremum

Proof of Proposition 4

Observe that ${\mathcal {V}}_I(\cdot )$ is a continuous mapping on $[0,{{\bar{T}}}]$. Therefore, Propositions 2 and 3 allow us to distinguish four cases, depending on the values of parameters $\kappa$ and $T_E$.

First: $\kappa \ge \underline{\kappa }$ and $v\ge {\underline{v}}$. In this case, we have $\max _{t \in [0,T_E]} {\mathcal {V}}_I(t)={\mathcal {V}}_I(T_E)$, and $\max _{t \in [T_E,{{\bar{T}}}]} {\mathcal {V}}_I (t)={\mathcal {V}}_I(T_E)$. Consequently, it follows that $\max _{t \in [0,{{\bar{T}}}]} {\mathcal {V}}_I(t)={\mathcal {V}}_I(T_E)$; hence, $T^*=T_E$.

Second: $\kappa < \underline{\kappa }$ and $v \ge {\underline{v}}$. In this case, we have $\max _{t\in [0,T_E]} {\mathcal {V}}_I(t)={\mathcal {V}}_I({\check{T}})$ and $\max _{t\in [T_E,\bar{T}]} {\mathcal {V}}_I(t)={\mathcal {V}}_I(T_E)$. Consequently, it follows that $\max _{t\in [0,{{\bar{T}}}]} {\mathcal {V}}_I(t)={\mathcal {V}}_I(\check{T})$; hence, $T^*={\check{T}}$.

Third: $\kappa \ge \underline{\kappa }$ and $v <{\underline{v}}$. In this case, we have $\max _{t\in [0,T_E]} {\mathcal {V}}_I(t)={\mathcal {V}}_I(T_E)$ and $\max _{t\in [T_E,{{\bar{T}}}]} {\mathcal {V}}_I (t)={\mathcal {V}}_I({{\bar{T}}})$. It follows that $\max _{t\in [0,{{\bar{T}}}]} {\mathcal {V}}_I(t)={\mathcal {V}}_I({{\bar{T}}})$; hence, $T^*={{\bar{T}}}$.

Fourth: $\kappa <{\underline{\kappa }}$, and $v<{\underline{v}}$. In this case, we have $\max _{t\in [0,T_E]} {\mathcal {V}}_I(t)={\mathcal {V}}_I({\check{T}})$ and $\max _{t\in [T_E,\bar{T}]} {\mathcal {V}}_I(t)={\mathcal {V}}_I({{\bar{T}}})$. Since $\left. {\mathcal {V}}_I({\check{T}})\right| _{\kappa< \underline{\kappa }}<\left. {\mathcal {V}}_I(T_E)\right| _{\kappa \ge \underline{\kappa }}$, it follows that $\max _{t\in [0,{{\bar{T}}}]} {\mathcal {V}}_I(t)={\mathcal {V}}_I({{\bar{T}}})$; hence, $T^*={{\bar{T}}}$. $\square$

Proof of Corollary 1

We first establish the relationship between ${\underline{v}}$ and $T_E$ by implicitly differentiating (20):

$$\begin{aligned} \frac{\partial {\underline{v}}}{\partial T_E}=-\left. \frac{\partial \psi (v)/\partial T_E}{\psi '(v)}\right| _{v={\underline{v}}}. \end{aligned}$$

Since $\psi '(\cdot )<0$, the sign of $\partial {\underline{v}}/ \bar{T}_E$ is given by the sign of the numerator, i.e., by the sign of $\partial \psi ({\underline{v}})/\partial T_E$.

From Eq. (20), we obtain

$$\begin{aligned} \frac{\partial \psi ({\underline{v}})}{\partial \bar{T}}=\Lambda _{{\underline{v}}}(T_E)e^{-rT_E}, \end{aligned}$$

(21)

where

$$\begin{aligned} \Lambda _{{\underline{v}}}(T_E):=2{\underline{v}}-\alpha r + \underline{v}r({{\bar{T}}}-T_E). \end{aligned}$$

Observe that $\Lambda _{{\underline{v}}}(T_E)\ge 0 \Leftrightarrow T_E\le {\check{T}}_E:= {{\bar{T}}} - (\alpha r - 2{\underline{v}})/\underline{v}r$. Since $\alpha r >2{\underline{v}}$, if $\Lambda _{\underline{v}}(0)>0$, then ${\check{T}}_E\in ]0,{{\bar{T}}}[$. Consequently, we deduce:

$$\begin{aligned} {\left\{ \begin{array}{ll} {\partial {\underline{v}}}/{\partial T_E}\ge 0 \hspace{2mm}\text {if}\hspace{2mm} \Lambda _{{\underline{v}}}(0)\ge 0 \hspace{2mm}\text {and}\hspace{2mm} T_E \le \check{T}_E \\ {\partial {\underline{v}}}/{\partial T_E}< 0\hspace{2mm}\text {otherwise}\hspace{2mm}.\end{array}\right. } \end{aligned}$$

In words, We thus deduce that the relationship between $\underline{v}$ and $T_E$ is described by an inverted U-shaped curve if $\Lambda _{{\underline{v}}}(0)>0$, and ${\underline{v}}$ monotonically decreases in $T_E$ otherwise.

Let us now assume that values of $T_E$ are very high (formally, $T_E>{\check{T}}_E$). In this case, we have $\partial {\underline{\kappa }} /\partial T_E >0$ (as shown in Appendix A), and $\partial {\underline{v}} /\partial T_E < 0$. Namely, if $T_E$ is high enough, ${\underline{\kappa }}$ is also high but ${\underline{v}}$ is low. Hence, as $T_E$ increases it is increasingly likely that $\kappa <{\underline{\kappa }}$ and $v\ge {\underline{v}}$, i.e., that the equilibrium is described by case 2 in Fig. 1. Under this scenario, the incumbent drops out before elections since $T^*\in (0,T_E)$.

Assume next that the value of $T_E$ declines. Since $T_E$ is initially high, ${\underline{v}}$ will consequently increase, while ${\underline{\kappa }}$ decreases. Hence, as $T_E$ decreases it is increasingly likely that $\kappa \ge {\underline{\kappa }}$ and $v <{\underline{v}}$, i.e., that the equilibrium is described by case 3 in Fig. 1. Under this scenario, the incumbent drops out after elections since $T^*={{\bar{T}}}$.

Further decreases of $T_E$ imply that $T_E$ ultimately falls below the threshold ${\check{T}}_E$, and the relationship between $\underline{v}$ and $T_E$ then reverses (i.e., $\partial {\underline{v}}/\partial T_E>0$). Consequently both bounds ${\underline{\kappa }}$ and ${\underline{v}}$ become smaller and it becomes increasingly likely that $\kappa \ge {\underline{\kappa }}$ and $v\ge {\underline{v}}$, i.e., that the equilibrium is described by case 1 in Fig. 1. Under this scenario, the incumbent drops out at the election date $T^*=T_E$.

Appendix D: Social welfare

By differentiating (14) with respect to $T_S$, we have:

$$\begin{aligned} u^{'}(T_S)=\alpha e^{-rT_S}f_R(T_S)-ve^{-rT_S}[1-F_R(T_S)]. \end{aligned}$$

Given the assumed uniform distribution, we obtain

$$\begin{aligned} u^{'}(T_S)=\frac{1}{{{\bar{T}}}} e^{-rT_S}\left[ \alpha -v(\bar{T}-T_S)\right] , \end{aligned}$$

(22)

, and

$$\begin{aligned} u^{''}(T_S)=\frac{v}{{{\bar{T}}}} e^{-rT_S}-r\frac{1}{{{\bar{T}}}} e^{-rT_S}\left[ \alpha -v({{\bar{T}}}-T_S)\right] . \end{aligned}$$

(23)

Observe that $u^{'}(0)<0$ and that $u^{'}({\bar{T}})>0$. Moreover, when $u^{'}(T_S)=0$, $u^{''}(T_S)>0$ which implies that function u(.) will only change sign once and that it is accordingly U-shaped on $[0,{{\bar{T}}}]$ with a minimum at ${{\bar{T}}}_S=\bar{T}-\alpha /v$. Consequently, $\max u(t)\in \{u(0),u({{\bar{T}}})\}$.

We know, first, that $u(0)=0$ since if the planner were to drop out in $T=0$, in which case the WoA’s duration would be zero and the focal country would lose with certainty and obtain no gains.

Second, $u({{\bar{T}}})=\int _{x=0}^{x={{\bar{T}}}} \left\{ \int _{0}^x -ve^{-rt}\;\text {d}t+\alpha e^{-rx} \right\} f_R(x)\;\text {d}x$. Indeed, if the planner drops out at ${{\bar{T}}}$, the country wins for sure and the representative citizen obtains $\alpha$. The cost of this certain victory is given by the period cost of the WoA, weighted by the number of periods the planner expects the rival to be actively fighting the WoA. This cost is given by $\int _{0}^x -ve^{-rt}\;\text {d}t$.

Computing this payoff, we obtain:

$$\begin{aligned} u({{\bar{T}}}) =\int _{x=0}^{x={{\bar{T}}}} \left\{ \int _{0}^x -ve^{-rt}\;\text {d}t+\alpha e^{-rx} \right\} f_R(x)\;\text {d}x=\frac{1}{{{\bar{T}}} r}\Psi (v), \end{aligned}$$

with,

$$\begin{aligned} \Psi (v)=\left( \alpha - \frac{v}{r}\right) \left( 1-e^{-r\bar{T}}\right) -v{{\bar{T}}}e^{-r {{\bar{T}}} }. \end{aligned}$$

Hence, the social optimum hinges on the sign of $\Psi (v)$ since $\Psi (v)> 0\Rightarrow T_S={{\bar{T}}}$, and $\Psi (v)\le 0\Rightarrow T_S=0$. Following the same reasoning as in Appendix B, we can apply the intermediate value theorem on function $\Psi$ and deduce that there is a unique level ${\hat{v}}>0$, such that: $v < {\hat{v}}\Leftrightarrow \Psi (v) > 0 \Leftrightarrow T_S^*= {{\bar{T}}}$, and $v \ge {\hat{v}}\Leftrightarrow \Psi (v)\le 0 \Leftrightarrow T_S^*=0$.

In Appendix C we show that $v< {\underline{v}} \Leftrightarrow \psi (v)> 0\Leftrightarrow T^*={{\bar{T}}}$, and that otherwise $T^*=T_E$ or $T^*={\check{T}}\in (0,T_E)$.

We next show that ${\underline{v}} <{\hat{v}}$. To establish this result, from Eq. (20), we obtain:

$$\begin{aligned} \Psi (v)-\psi (v)=&\left( \alpha - \frac{v}{r}\right) \left( 1-e^{-rT_E}\right) +v\bar{T}\left( e^{-rT_E}-e^{-r{{\bar{T}}}}\right) -vT_Ee^{-rT_E}=:\Lambda (T_E). \end{aligned}$$

It is clear that $\Lambda '(T_E)\ge 0$ under the condition $\bar{T}\le T_E+(r\alpha -2v)/rv$ that we assume to be true. In this case, $\Psi (v)-\psi (v)\ge \Lambda (0)= v{{\bar{T}}}\left( 1-e^{-r\bar{T}}\right) >0$, so that ${\underline{v}} <{\hat{v}}$. Hence, there are three cases:

• (i) If $v<{\underline{v}} <{\hat{v}}$. In this case, the incumbent and the social planner both decide to drop out at ${{\bar{T}}}$, i.e., $T^*=T_S^*={{\bar{T}}}$. The political equilibrium is socially optimal.

• (ii) If ${\underline{v}} \le v < {\hat{v}}$. In this case, according to Proposition 4, the incumbent chooses either $T^*=T_E$ or $T^*=\check{T} \in (0,T_E)$, while the social planner decides $T_S^*={{\bar{T}}}$. The political equilibrium is socially suboptimal.

• (iii) If $v \ge {\hat{v}}$. In this case, according to Proposition 4, the incumbent chooses either $T^*=T_E$ or $T^*={\check{T}} \in (0,T_E)$, while the social planner decides $T_S^*= 0$. The political equilibrium is socially suboptimal.