Opportunist politicians and the evolution of electoral competition

Abstract

This paper studies a uni-dimensional model of electoral competition between two parties with two types of politicians. ‘Opportunist’ ones care only about the spoils of the office, and ‘militant’ ones have ideological preferences on the policy space. Opportunist politicians review their affiliations and may switch parties, seeking better election prospects. In this framework, we compare a winner-take-all system, where all the spoils of office go to the winner, to a proportional system, where the spoils of office are split among the two parties in proportion to their vote shares. We study the existence of short term political equilibria and the dynamics and stability of policies and of party membership decisions. In the long run, it is possible that proportional systems see opportunist politicians spread over all parties, but this situation is unstable in winner-take-all systems.

Appendix A

1.1 A.1 The two-stage game for the electoral competition

In this section, we analyze the model as a static two-stage game. We consider now that at the first stage, opportunist politicians can all change their membership and at the second stage, party platforms are determined. We analyze the case where all the opportunists are in party L or in party R. Note that the following utilities apply at the second stage:

$$u_{L}(t_{L}(s),t_{R}(s)) = \frac{1}{l+ns}F(t^{eq}(s)) $$

$$u_{R}(t_{L}(s),t_{R}(s))=\frac{1}{r+n(1-s)}(1-F(t^{eq}(s))) $$

where $t^{eq}(s)=\frac {2nsr+lr}{2\left (lr+n(l(1-s)+sr)\right )}$.

The opportunists will choose either party L or party R. In order to see which party the opportunists prefer, we have to analyze the difference of the utility of being in party L when all opportunists affiliate with party L and the utility of being in party R when they all choose party R. Let Δ = u _L (t _L (1), t _R (1))−u _R (t _L (0), t _R (0)). Then we have the following result:

$$\begin{array}{@{}rcl@{}} {\Delta} &=& \frac{1}{l+n}F\left( t^{eq}(1)\right)-\frac{1}{r+n} \left( 1-F(t^{eq}(0))\right))\\ &=& \frac{1}{(1-r)}F\left( \frac{2n+l}{2\left( 1-r\right) } \right)+\frac{1}{(r+n)}F\left( \frac{r}{2(r+n)} \right)-\frac{1}{(r+n)}. \end{array} $$

Drawing the previous function for different distributions of voters, one obtains Fig. 8, which describes the region where the utility of being in party L when all opportunists choose party L is larger than the utility of being in party R when all choose party R. The region is defined according to the number of opportunist politicians and militants in party R. In the region below the solid line in Fig. 8, opportunists will prefer party R when citizens are uniformly distributed along the unit line.

Fig. 8

Gain from switching from party R to party L, for uniform and skewed distributions of voters

For a distribution skewed to the left or to the right, the frontier moves to the left and to the right accordingly. Switching from party R to party L is less likely to be beneficial when voters are distributed more to the right. In Fig. 8, the area below the curves shrinks as the distribution of voters changes from a mostly leftist to a mostly rightist population. Let n = 0.4, r = 0.4, l = 0.2. One can see that all the opportunists will choose to be affiliated with party L only when citizens are distributed uniformly or more to the left.

If the distribution of voters is uniform and there are equal numbers of militants in each party, then Δ = u _L (t _L (1), t _R (1))−u _R (t _L (0), t _R (0))=0. In that case the opportunist politicians will be indifferent between affiliating with one party or the other. Even though the case where they are distributed equally may be one of the equilibrium for the stage game, it is not an evolutionary stable outcome.

1.2 A.2 Proof of Proposition 1

(Proposition 1) In the proportional system, the state where all opportunists are in party L (that is s = 1) is asymptotically stable if G(1)>0. The state where all opportunists are in party R (that is s = 0) is asymptotically stable if G(0)<0.

To prove Proposition 1, we need to first prove the following lemma.

Lemma 1

Given a monotonic selection dynamics ξ, a population state sis asymptotically stable if

$$(2s-1)\left( u_{L}\left( {t}_{L}^{eq}(s),{t}_{R}^{eq}(s)\right)-u_{R}\left( {t}_{L}^{eq}(s),{t}_{R}^{eq}(s)\right)\right)>0. $$

Proof

Let s ^∗=0. If $(2s^{\ast }-1) \left (u_{L}\left ({t}_{L}^{eq} (s^{\ast }), {t}_{R}^{eq} (s^{\ast })\right ) - u_{R} \left ({t}_{L}^{eq} (s^{\ast }), {t}_{R}^{eq} (s^{\ast })\right )\right )>0$ , then $u_{L} \left ({t}_{L}^{eq} (s^{\ast }), {t}_{R}^{eq} (s^{\ast })\right ) < u_{R} \left ({t}_{L}^{eq} (s^{\ast }), {t}_{R}^{eq}(s^{\ast })\right )$. By continuity of the payoffs with respect to the population share, there exists a neighborhood $\mathcal {N}$ of s ^∗ such that, for all $s\in \mathcal {N} \setminus \{s^{\ast }\}$, $u_{L} \left ({t}_{L}^{eq}(s), {t}_{R}^{eq}(s)\right ) < u_{R}\left ({t}_{L}^{eq}(s), {t}_{R}^{eq}(s)\right )$. If the dynamics ξ(s) are monotonic, then $\dot {s}<0$. Let L(s)=1−s. L(s) attains its maximum value of 1 when s = s ^∗, and is positive and increasing in $\mathcal {N} \setminus \{ s^{\ast } \}$. This is a strict Liapunov function for s and s is asymptotically stable by Liapunov’s Stability Theorem. □

Let s ^∗=1, the condition writes: $u_{L} \left ({t}_{L}^{eq} (s^{\ast }), {t}_{R}^{eq} (s^{\ast })\right ) > u_{R} \left ({t}_{L}^{eq} (s^{\ast }), {t}_{R}^{eq} (s^{\ast })\right )$. By continuity, there exists a neighborhood $\mathcal {N}$ of s ^∗ such that, for all $s\in \mathcal {N} \setminus \{s^{\ast } \}$, $u_{L} \left ({t}_{L}^{eq} (s), {t}_{R}^{eq} (s)\right ) > u_{R} \left ({t}_{L}^{eq} (s), {t}_{R}^{eq}(s)\right )$. If the dynamics ξ(s) are monotonic, then $\dot {s} > 0$. Let L(s) = s, L(s) attains its maximum value of 1 when s = s ^∗, and is positive and increasing in $\mathcal {N} \setminus \{ s^{\ast }\}$. This is a strict Liapunov function for s and s is asymptotically stable.

Proof Proof of the Proposition

The lemma simply says that s = 1 is stable if u _L (.)>u _R (.) and s = 0 is stable if u _L (.)<u _R (.). Thus, s = 1 is stable if G(1)>0 and s = 0 is stable if G(0)<0. □

Notice that the conditions are not tight. A more explicit version of these conditions is as follows:

For s = 1 to be asymptotically stable we need the following condition:

$$u_{L} \left( {t}_{L}^{eq}(1), {t}_{R}^{eq}(1)\right) - u_{R} \left( {t}_{L}^{eq}(1), {t}_{R}^{eq}(1)\right) >0. $$

The difference can be written:

$$\frac{1}{l+n}F(t^{eq}(1))-\frac{1}{r}(1-F(t^{eq}(1))) = \frac{F(t^{eq}(1))-l-n}{(l+n)r}, $$

hence the condition:

$$F(t^{eq}(1))>l+n. $$

For s = 0 to be asymptotically stable, we obtain in the same manner the following condition:

$$\frac{F(t^{eq}(0))-l}{(r+n)l}>0\Rightarrow F(t^{eq}(0))>l. $$

1.3 A.3 Proof of Proposition 2

(Proposition 2) In the proportional system, if G ^′(s)<0 in the unit interval, G(0)>0, and G(1)<0, then there is a unique, asymptotically stable mixed state.

Proof

Rest points of the evolutionary dynamics require all surviving strategies to have equal payoffs. Then, an interior rest point s ^∗∈(0,1) requires that $u_{L} \left ({t}_{L}^{eq} (s^{\ast }), {t}_{R}^{eq} (s^{\ast })\right ) = u_{R} \left ({t}_{L}^{eq} (s^{\ast }), {t}_{R}^{eq}(s^{\ast })\right )$. Computation shows that:

$$\begin{array}{@{}rcl@{}} &&u_{L}\left( t_{L}^{eq}(s^{\ast}),t_{R}^{eq}(s^{\ast})\right)-u_{R}\left( t_{L}^{eq}(s^{\ast}),t_{R}^{eq}(s^{\ast})\right)\\ &&\qquad = \frac{1}{l+ns^{\ast}}F\left( t^{eq}(s^{\ast})\right)-\frac{1}{r+n(1-s^{\ast})}\left( 1-F(t^{eq}(s^{\ast}))\right)\\ &&\qquad = \frac{F\left( t^{eq}(s^{\ast})\right)-l-ns^{\ast}}{(r+n(1-s^{\ast}))(l+ns^{\ast})}. \end{array} $$

The condition is thus:

$$F(t^{eq}(s^{\ast}))=l+ns^{\ast}. $$

The interior rest points are roots of this equation.

If payoffs are continuous at any state z and G ^′(⋅)<0 in the unit interval then, for z = s ^∗−𝜖, $u_{L} \left ({t}_{L}^{eq}(z), {t}_{R}^{eq}(z)\right ) > u_{R} \left ({t}_{L}^{eq} (z), {t}_{R}^{eq}(z)\right )$, and for z = s ^∗ + 𝜖, $u_{L} \left ({t}_{L}^{eq}(z), {t}_{R}^{eq}(z)\right ) < u_{R} \left ({t}_{L}^{eq}(z), {t}_{R}^{eq}(z)\right )$. Let $\mathcal {N} (\epsilon ) = \left [ s^{\ast } - \epsilon , s^{\ast } + \epsilon \right ]$. For any 𝜖>0, we have $s \in \mathcal {N}(\epsilon )$. Hence $\mathcal {N}(\epsilon )$ is an invariant set: any trajectory originating in $\mathcal {N}(\epsilon )$ remains there forever. Moreover, for all $s \in \mathcal {N}(\epsilon )$, all trajectories originating in $\mathcal {N}(\epsilon )$ converge to some rest point z. But since G ^′(⋅)<0, G(0)>0 and G(1)<0, by the intermediate value theorem, there is a unique rest point of this type, at z = s ^∗, and all trajectories originating in $\mathcal {N}(\epsilon )$ converge to this point, which is asymptotically stable. □

1.4 A.4 Proof of Proposition 3

(Proposition 3) In the winner-take-all system, the graph of the probability of victory, as we have defined it, is not continuous. There is at most one solution to the Eq. 11. There are three cases to analyze...

Proof

Note that s = 1 is stable if u _L (⋅)>u _R (⋅) and that s = 0 is stable if u _L (⋅)<u _R (⋅). □

Case 1

$t_{min}=\frac {r}{2(r+n)} \leq t^{med}\leq \frac {2n+l}{2(l+n)}=t_{max}$. If all opportunists are in party L, then t ^eq(0) = t _min and $\pi \left (t_{L}^{eq}(0),t_{R}^{eq}(0)\right )=0$, and we have:

$$\begin{array}{lll} u_{L}\left( t_{L}^{eq}(0),t_{R}^{eq}(0)\right) &=& \frac{1}{l}\pi\left( t_{L}^{eq}(0),t_{R}^{eq}(0)\right)=0\\ &<& u_{R}\left( t_{L}^{eq}(0),t_{R}^{eq}(0)\right) =\frac{1}{r+n}\left( 1-\pi \left( t_{L}^{eq}(0),t_{R}^{eq}(0)\right)\right) = \frac{1}{r+n}. \end{array} $$

If all opportunists are in party R, then t ^eq(1) = t _max and $\pi \left (t_{L}^{eq}(1),t_{R}^{eq}(1)\right )=1$ , and we have $u_{L}\left (t_{L}^{eq}(1),t_{R}^{eq}(1)\right )=\frac {1}{l+n}\pi \left (t_{L}^{eq}(1),t_{R}^{eq}(1)\right )=\frac {1}{l+n}>u_{R}\left (t_{L}^{eq}(1),t_{R}^{eq}(1)\right )= \frac {1}{r}\left (1-\pi \left (t_{L}^{eq}(1),t_{R}^{eq}(1)\right )\right )=0$. These states will both be stable. Let s ^M be the solution to the Eq. 11. Then s ^M must satisfy the following conditions: t ^eq(s ^M) = t ^med and $l+ns^{M}=\frac {1}{2}$. From the previous results, s ^M will not be stable.

Case 2

t ^med<t _min . Then t ^med is always smaller than t ^eq(s), $\pi \left (t_{L}^{eq}(s),t_{R}^{eq}(s)\right )=1$ and the utility of being a member of party L is always larger than affiliating with party R. The case where all the opportunists are in party L (that is s = 0) is the only stable state.

Case 3

t ^med>t _max . Then t ^med is always larger than t ^eq(s), $\pi (t_{L}^{eq}(s),t_{R}^{eq}(s))=0$ and being a member of party R has always more utility than being a member of party L. The case where all the opportunists are in the party R (that is s = 1) is the only stable state.