Proof of Theorem

The proof proceeds in four steps.

Step 1: For all ε > 0, there exist $$$a \in (0,{\epsilon})$$$ and $$$\hat{\delta } \in (0,1)$$$ such that for all $$$\delta \in (\hat{\delta },1)$$$ ,

we have $$${{{\bf {{x}}}}^a} \in P(\delta )$$$ .

Fix ε > 0. I begin by choosing parameters a,δ,β ∈ (0,1) with a < ε and a natural number k ≥ 1 to satisfy three inequalities (Inequalities 5–7 below) that are critical for the equilibrium construction. Following these choices, I will specify $$$\hat{\delta }$$$ to satisfy three further inequalities, namely 12–14, and I determine a number m of periods used in punishment phases. First, choose k sufficiently large that for all $$$\tau \ne {{\tau }_0}$$$ , we have

$$\matrix{(1\,{\rm{ - }}\,{{p}_0})\left[ {{{u}_\tau }({{x}_{{{\tau }_1}}})\,{\rm{ - }}\,\frac{{k{{u}_\tau }({{x}_{{{\tau }_1}}})\, + \,{{u}_\tau }(0)}}{{k\, + \,1}}} \right] \cr \; \; \lt {{p}_0}\left[ {\frac{{k{{u}_\tau }({{x}_{{{\tau }_1}}})\, + \,{{u}_\tau }(0)}}{{k\, + \,1}}\,{\rm{ - }}\,{{u}_\tau }(0)} \right]. \\ $$

For each $$$\tau $$$ , define

$${{W}_\tau }\, = \,\frac{1}{{1\,{\rm{ - }}\,{{\delta }^{k\, + \,1}} }}[(1\,{\rm{ - }}\,{{\delta }^k} ){{u}_\tau }({{x}_{{{\tau }_1}}})\, + \,{{\delta }^k} (1\,{\rm{ - }}\,\delta ){{u}_\tau }(0)],$$

which implicitly depends on k and δ. This is the discounted utility (normalized by 1 − δ) of alternating sequences of $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ for k periods and x = 0 for one period. Note that by l'Hôpital's rule, we have for all $$$\tau $$$ ,

$$\mathop {\lim }\limits_{\delta \uparrow 1} {{W}_\tau } = \frac{{k{{u}_\tau }({{x}_{{{\tau }_1}}})\, + \,{{u}_\tau }(0)}}{{k\, + \,1}},$$

and in addition that

$$\mathop {\lim }\limits_{\beta \downarrow 0} \frac{{(1\,{\rm{ - }}\,{{p}_0})\beta (1\,{\rm{ - }}\,\beta )}}{{(1\,{\rm{ - }}\,{{p}_0})\beta \, + \,{{p}_0}{{\beta }^2} }}\, = \,1.$$

Using Inequalities 1–3, there exists $$$\overline{\beta } \in (0,1)$$$ such that two conditions hold: (1) for all $$$\beta \in (0,\overline{\beta } )$$$ , we have

$$2(1\,{\rm{ - }}\,{{p}_0})\beta (1\,{\rm{ - }}\,\beta )\, \gt \,(1\,{\rm{ - }}\,{{p}_0})\beta \, + \,{{p}_0}{{\beta }^2}, $$

and (2) there exists $$$\underline{\delta } \, \in \,(0,1)$$$ such that for all $$$\delta \in (\underline{\delta }, 1)$$$ and all $$$\beta \in (0,\overline{\beta } )$$$ , we have for all $$$\tau \ne {{\tau }_0}$$$ ,

$$(1\,{\rm{ - }}\,{{p}_0})({{u}_\tau }({{x}_{{{\tau }_1}}})\,{\rm{ - }}\,{{W}_\tau })\, \lt \,{{p}_0}(1\,{\rm{ - }}\,\beta )({{W}_\tau }\,{\rm{ - }}\,{{u}_\tau }(0)).$$

Inequality 4 is used in the selection of a, below, and Inequality 5 is used to give voters proper incentives in the equilibrium construction.

Given $$$\delta \in (\underline{\delta }, 1)$$$ and $$$\beta \in (0,\overline{\beta } )$$$ , the next inequality is used to address the incentives of type $$${{\tau }_0}$$$ office holders along the path of play:

$$\matrix{(1\,{\rm{ - }}\,\delta )({{u}_{{{\tau }_0}}}(0)\,{\rm{ - }}\,{{u}_{{{\tau }_0}}}(a)) \hfill \cr \leq \delta [{{p}_0}[(1\,{\rm{ - }}\,{{\beta }^2} )({{u}_{{{\tau }_0}}}(a)\,{\rm{ - }}\,{{u}_{{{\tau }_0}}}(0))\, + \,{{\beta }^2} ({{u}_{{{\tau }_0}}}(a)\,{\rm{ - }}\,{{W}_{{{\tau }_0}}})] \hfill \cr + (1\,{\rm{ - }}\,{{p}_0})[(1\,{\rm{ - }}\,\beta )({{u}_{{{\tau }_0}}}(a)\,{\rm{ - }}\,{{u}_{{{\tau }_0}}}(0))\, + \,\beta ({{u}_{{{\tau }_0}}}(a)\,{\rm{ - }}\,{{u}_{{{\tau }_0}}}({{x}_{{{\tau }_1}}}))]]. \\ $$

The following inequality addresses incentives for type $$$\tau \ne {{\tau }_0}$$$ office holders along the path of play: for all $$$\tau \ne {{\tau }_0}$$$ ,

$${ (1\,{\rm{ - }}\,\delta )\rDelta \leq &amp;#x0026; \delta [{{p}_0}[(1\,{\rm{ - }}\,{{\beta }^2} )({{u}_\tau }(a)\,{\rm{ - }}\,{{u}_\tau }(0))\, + \,{{\beta }^2} ({{u}_\tau }(a)\,{\rm{ - }}\,{{W}_\tau })] \cr + &amp;#x0026; (1\,{\rm{ - }}\,{{p}_0})[(1\,{\rm{ - }}\,\beta )({{u}_\tau }(a)\,{\rm{ - }}\,{{u}_\tau }(0))\, + \,\beta ({{u}_\tau }(a)\,{\rm{ - }}\,{{u}_\tau }({{x}_{{{\tau }_1}}}))], \cr}$$

where $$$\rDelta \, = \,{\max }_{\tau, x,y} [{{u}_\tau }(x)\,{\rm{ - }}\,{{u}_\tau }(y)]$$$ . To see that Inequalities 6 and 7 are compatible for some $$$\delta \in (\underline{\delta }, 1)$$$ and some $$$\beta \in (0,\overline{\beta } )$$$ and for sufficiently small $$$a \in (0,{\epsilon})$$$ , define

$$\matrix{{{A}_\tau }\, = \,{{u}_\tau }(a)\,{\rm{ - }}\,{{u}_\tau }(0) \cr A_{\tau }^{\ast} \, = \,{{u}_\tau }({{x}_{{{\tau }_1}}})\,{\rm{ - }}\,{{u}_\tau }(a), \\ $$

where these two quantities implicitly depend on a. It will be convenient to define $$$\underline{A} \, = \,{\min }_{\tau \ne {{\tau }_0}} {{A}_\tau }$$$ and $$${{A}^\ast} \, = \,{\max }_{\tau \ne {{\tau }_0}} A_{\tau }^{\ast} $$$ . Of course, $$$\underline{A} \rightarrow 0$$$ and $$${{A}^\ast} \rightarrow {\max }_{\tau \ne {{\tau }_0}} [{{u}_\tau }({{x}_{{{\tau }_1}}})\,{\rm{ - }}\,{{u}_\tau }(0)]\, \gt \,0$$$ as a→0. Assuming without loss of generality that $$$a\, \lt \,{{x}_{{{\tau }_1}}}$$$ , the right-hand side of Inequality 6 is positive if

$$\frac{{(1\,{\rm{ - }}\,{{p}_0})\beta }}{{1\, + \,{{p}_0}(\beta \,{\rm{ - }}\,{{\beta }^2} )\,{\rm{ - }}\,\beta }}\, \gt \,\frac{{{{u}_{{{\tau }_{\rm{0}}}}}(0)\,{\rm{ - }}\,{{u}_{{{\tau }_0}}}(a)}}{{{{u}_{{{\tau }_{\rm{0}}}}}(a)\,{\rm{ - }}\,{{u}_{{{\tau }_0}}}({{x}_{{{\tau }_1}}})}},$$

and using $$${{u}_\tau }({{x}_{{{\tau }_1}}})\,\geq \,{{W}_\tau }$$$ , the right-hand side of Inequality 7 is positive if for all $$$\tau \ne {{\tau }_0}$$$ ,

$$\frac{{{{u}_\tau }(a)\,{\rm{ - }}\,{{u}_\tau }(0)}}{{{{u}_\tau }({{x}_{{{\tau }_1}}})\,{\rm{ - }}\,{{u}_\tau }(a)}}\, \gt \,\frac{{(1\,{\rm{ - }}\,{{p}_0})\beta \, + \,{{p}_0}{{\beta }^2} }}{{1\, + \,{{p}_0}(\beta \,{\rm{ - }}\,{{\beta }^2} )\,{\rm{ - }}\,\beta }}.$$

Note that

$${ \mathop {\lim }\limits_{a \downarrow 0} &amp;#x0026; \left( {\frac{{{{u}_{{{\tau }_0}}}(0)\,{\rm{ - }}\,{{u}_{{{\tau }_0}}}(a)}}{{{{u}_{{{\tau }_0}}}(a)\,{\rm{ - }}\,{{u}_{{{\tau }_0}}}({{x}_{{{\tau }_1}}})}}} \right)\left( {\frac{{{{A}^\ast} }}{{\underline{A} }}} \right) \cr &amp;#x0026; \leq \mathop {\lim }\limits_{a \downarrow 0} \left( {\frac{{\frac{{{{u}_{{{\tau }_0}}}(0)\,{\rm{ - }}\,{{u}_{{{\tau }_0}}}(a)}}{a}}}{{{{u}_{{{\tau }_0}}}(a)\,{\rm{ - }}\,{{u}_{{{\tau }_0}}}({{x}_{{{\tau }_1}}})}}} \right)\left( {\frac{{{{{\max }}_{\tau \ne {{\tau }_0}}}{{u}_\tau }({{x}_{{{\tau }_1}}})\,{\rm{ - }}\,{{u}_\tau }(a)}}{{{{{\min }}_{\tau \ne {{\tau }_0}}}\frac{{{{u}_\tau }(a)\,{\rm{ - }}\,{{u}_\tau }(0)}}{a}}}} \right) \cr &amp;#x0026; = \frac{{u_{{{{\tau }_0}}}^{\prime} (0)}}{{{{u}_{{{\tau }_0}}}(0)\,{\rm{ - }}\,{{u}_{{{\tau }_0}}}({{x}_{{{\tau }_1}}})}}\left( {\frac{{{{{\max }}_{\tau \ne {{\tau }_0}}}{{u}_\tau }({{x}_{{{\tau }_1}}})\,{\rm{ - }}\,{{u}_\tau }(0)}}{{{{{\min }}_{\tau \ne {{\tau }_0}}}{{u}_\tau }&#x0027;(0)}}} \right) \cr &amp;#x0026; = 0. \cr}$$

Therefore, we can choose $$$a \in (0,{\epsilon})$$$ and $$$\beta \in (0,\overline{\beta } )$$$ such that

$$\frac{{(1\,{\rm{ - }}\,{{p}_0})\beta }}{{1\, + \,{{p}_0}(\beta \,{\rm{ - }}\,{{\beta }^2} )\,{\rm{ - }}\,\beta }}\, \gt \,\frac{{{{u}_{{{\tau }_0}}}(0)\,{\rm{ - }}\,{{u}_{{{\tau }_0}}}(a)}}{{{{u}_{{{\tau }_0}}}(a)\,{\rm{ - }}\,{{u}_{{{\tau }_0}}}({{x}_{{{\tau }_1}}})}}\, \gt \,\frac{1}{2} \cdot \frac{{(1\,{\rm{ - }}\,{{p}_0})\beta }}{{1\, + \,{{p}_0}(\beta \,{\rm{ - }}\,{{\beta }^2} )\,{\rm{ - }}\,\beta }},$$

automatically fulfilling Inequality 8, and such that for all $$$\tau \ne {{\tau }_0}$$$ ,

$$\frac{{{{u}_{{{\tau }_0}}}(0)\,{\rm{ - }}\,{{u}_{{{\tau }_0}}}(a)}}{{{{u}_{{{\tau }_0}}}(a)\,{\rm{ - }}\,{{u}_{{{\tau }_0}}}({{x}_{{{\tau }_1}}})}}\, \lt \,\frac{1}{4} \cdot \frac{{{{u}_\tau }(a)\,{\rm{ - }}\,{{u}_\tau }(0)}}{{{{u}_\tau }({{x}_{{{\tau }_1}}})\,{\rm{ - }}\,{{u}_\tau }(a)}}.$$

Combining Inequalities 10 and 11 and using Inequality 4, we have

$${ \frac{{{{u}_\tau }(a)\,{\rm{ - }}\,{{u}_\tau }(0)}}{{{{u}_\tau }({{x}_{{{\tau }_1}}})\,{\rm{ - }}\,{{u}_\tau }(a)}}\, \gt \, 4\left( {\frac{{{{u}_{{{\tau }_0}}}(0)\,{\rm{ - }}\,{{u}_{{{\tau }_0}}}(a)}}{{{{u}_{{{\tau }_0}}}(a)\,{\rm{ - }}\,{{u}_{{{\tau }_0}}}({{x}_{{{\tau }_1}}})}}} \right) \cr \gt 2\left( {\frac{{(1\,{\rm{ - }}\,{{p}_0})\beta (1\,{\rm{ - }}\,\beta )}}{{1\, + \,{{p}_0}(\beta \,{\rm{ - }}\,{{\beta }^2} )\,{\rm{ - }}\,\beta }}} \right) \cr \gt \frac{{(1\,{\rm{ - }}\,{{p}_0})\beta \, + \,{{p}_0}{{\beta }^2} }}{{1\, + \,{{p}_0}(\beta \,{\rm{ - }}\,{{\beta }^2} )\,{\rm{ - }}\,\beta }}, \cr}$$

fulfilling Inequality 9, as required. It follows that for these choices of a and β, the right-hand sides of Inequality 6 and 7 are positive, and we can therefore choose $$$\delta \in (\underline{\delta }, 1)$$$ sufficiently close to one to satisfy Inequalities 6 and 7.

Fixing a as above, note that there is some latitude in the choice of β, as Inequalities 8 and 9 involve strict inequalities. Next, I exploit this flexibility to choose $$$\hat{\delta }$$$ close to one and fulfill the statement of Step 1. To do so, let $$${{\beta }_1},{{\beta }_2} \in (0,\overline{\beta } )$$$ be such that β ₁ < β ₂ and for all $$$\beta \in [{{\beta }_1},{{\beta }_2}]$$$ , Inequalities 8 and 9 hold. Then choose $$$\hat{\delta } \in (\underline{\delta }, 1)$$$ sufficiently high that for all $$$\delta \in (\hat{\delta },1)$$$ and all $$$\beta \in [{{\beta }_1},{{\beta }_2}]$$$ , Inequalities 6 and 7 hold in addition to Inequality 5, so that we have for all $$$\tau \ne {{\tau }_0}$$$ ,

$$(1\,{\rm{ - }}\,\delta )\rDelta \, \lt \,\delta {{p}_0}[\beta (1\,{\rm{ - }}\,\beta )({{u}_\tau }({{x}_{{{\tau }_1}}})\,{\rm{ - }}\,{{u}_\tau }(0))\, + \,{{\beta }^2} ({{u}_\tau }({{x}_{{{\tau }_1}}})\,{\rm{ - }}\,{{W}_\tau })]$$

$$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!(1\,{\rm{ - }}\,\delta ){\rm{\rDelta }}\, \lt \,\delta {{p}_0}({{u}_\tau }({{x}_{{{\tau }_1}}})\,{\rm{ - }}\,{{W}_\tau })$$

$$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!(1\,{\rm{ - }}\,\delta )\rDelta \, \lt \,\delta (1\,{\rm{ - }}\,{{p}_0})({{W}_{{{\tau }_0}}}\,{\rm{ - }}\,{{u}_{{{\tau }_0}}}({{x}_{{{\tau }_1}}})),$$

and so that $$$\hat{\delta }{{\beta }_2}\, \gt \,{{\beta }_1}$$$ . Inequalities 12 and 13 are used to address incentives of type $$$\tau \ne {{\tau }_0}$$$ office holders, and Inequality 14 is used to address type $$${{\tau }_0}$$$ office holders, in the punishment phases.

Next, I determine a number m of periods to be used in punishment phases, as a function of $$$\delta \in (\hat{\delta },1)$$$ . Using the inequality $$$\hat{\delta }{{\beta }_2} \gt {{\beta }_1}$$$ , I claim that for each $$$\delta \, \gt \,\hat{\delta }$$$ , there exists m such that $$${{\delta }^m} \in ({{\beta }_1},{{\beta }_2})$$$ . Indeed, we have $$$\delta \, \gt \,{{\beta }_1}$$$ by assumption, and $$${{\delta }^m} \, \lt \,\beta $$$ for sufficiently high m. The claim holds with m = 1 if $$$\delta \, \lt \,{{\beta }_2}$$$ . Otherwise, $$$\delta \,\geq \,{{\beta }_2}$$$ , and we can let m − 1 be the smallest natural number such that $$${{\delta }^{m\,{\rm{ - }}\,1}} \,\geq \,{{\beta }_2}$$$ . Then we have $$${{\delta }^m} \, \lt \,{{\beta }_2}$$$ and $$${{\delta }^m} \, = \,\delta {{\delta }^{m\,{\rm{ - }}\,1}} \,\geq \,\delta {{\beta }_2}\, \gt \,{{\beta }_1}$$$ _, so that $$${{\delta }^m} \in ({{\beta }_1},{{\beta }_2})$$$ , as claimed. The remainder of the construction takes such m as given.

Now, given discount factor $$$\delta \, \gt \,\hat{\delta }$$$ and m as above, I specify strategies and beliefs that support the constant path $$${{{\bf {{x}}}}^a} \, = \,(a,a, \ldots )$$$ . Note that by the preceding argument, we can choose m so that Inequalities 6 and 7 are satisfied by specifying $$$\beta \, = \,{{\delta }^m} \, \in \,({{\beta }_1},{{\beta }_2})$$$ . In the construction, I assume that voter beliefs about an untried challenger are given by the prior p, and that beliefs along an office holder's personal path of play are derived from Bayesian updating, and that off the politician's personal path of play, voters believe that with probability one the incumbent is type $$${{\tau }_0}$$$ . Strategies and beliefs are specified by assigning each history to one of five phases below, three of which are parameterized by a natural number j, and one of those is partitioned into three sub-phases, depending on voter beliefs. Phases 2–4 are used to punish deviations along the path of play, which takes place in Phase 1, and Phase 5 is used to punish deviations in the punishment phases.

Phase 1. All types of office holder choose x = a. If the incumbent chooses x = a, then all voters’ beliefs are given by the prior p, and all voters vote for the incumbent; and if the incumbent chooses $$$x \ne a$$$ , then all voters believe that with probability one the incumbent is type $$${{\tau }_0}$$$ , and all type $$$\tau \,\geq \,{{\tau }_1}$$$ voters vote for the challenger, and type $$${{\tau }_0}$$$ voters vote for the incumbent.

Phase 2(j). All types of office holder choose x = 0. If the incumbent chooses x = 0, then all voters’ beliefs are given by the prior p, and all type $$$\tau \,\geq \,{{\tau }_1}$$$ voters vote for the incumbent and type $$${{\tau }_0}$$$ voters vote for the challenger; and if the incumbent chooses $$$x\, \ne \,0$$$ , then all voters believe that with probability one the incumbent is type $$${{\tau }_0}$$$ , and all type $$$\tau \,\geq \,{{\tau }_1}$$$ voters vote for the challenger, and type $$${{\tau }_0}$$$ voters vote for the incumbent.

Phase 3(j). The type $$${{\tau }_0}$$$ office holder chooses x = 0, and all type $$$\tau \ne {{\tau }_0}$$$ office holders choose $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ . If the incumbent chooses x = 0, then all voters believe that with probability one the incumbent is type $$${{\tau }_0}$$$ , and all type $$$\tau \,\geq \,{{\tau }_1}$$$ voters vote for the incumbent, and type $$${{\tau }_0}$$$ voters vote for the challenger; if $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ , then all voters’ beliefs are given by $$$\hat{p}$$$ , the updated prior conditional on the politician's type being not equal to $$${{\tau }_0}$$$ ; and all type $$$\tau \,\geq \,{{\tau }_1}$$$ voters vote for the incumbent, and type $$${{\tau }_0}$$$ voters vote for the challenger; and otherwise, if $$$x \, \notin \, \{ 0,{{x}_{{{\tau }_1}}}\} $$$ , then all voters believe that with probability one the incumbent is type $$${{\tau }_0}$$$ , and all type $$$\tau \,\geq \,{{\tau }_1}$$$ voters vote for the challenger, and type $$${{\tau }_0}$$$ voters vote for the incumbent.

Phase 4.0(j). Type $$${{\tau }_0}$$$ office holders choose $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ if $$$j \in \{ 1, \ldots, k\} $$$ , they choose x = 0 if j = k+1, and type $$$\tau \ne {{\tau }_0}$$$ office holders choose $$$x = {{x}_{{{\tau }_1}}}$$$ for all $$$j\, = \,1, \ldots, k\, + \,1$$$ . All voters believe that with probability one the incumbent is type $$${{\tau }_0}$$$ . If $$$j \in \{ 1, \ldots, k\} $$$ and the incumbent chooses $$$x = {{x}_{{{\tau }_1}}}$$$ , or if j = k+1 and $$$x \in \{ 0,{{x}_{{{\tau }_1}}}\} $$$ , then all type $$$\tau \,\geq \,{{\tau }_1}$$$ voters vote for the incumbent, and type $$${{\tau }_0}$$$ voters vote for the challenger; and otherwise, type $$$\tau \,\geq \,{{\tau }_1}$$$ voters vote for the challenger, and type $$${{\tau }_0}$$$ voters vote for the incumbent.

Phase 4.1. Type $$$\tau \ne {{\tau }_0}$$$ office holders choose $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ and type $$${{\tau }_0}$$$ office holders choose x = 0. If the incumbent chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ , then all voters’ beliefs are given by $$$\hat{p}$$$ , the updated prior conditional on the politician's type being not equal to $$${{\tau }_0}$$$ , and all type $$$\tau \,\geq \,{{\tau }_1}$$$ voters vote for the incumbent, and type $$${{\tau }_0}$$$ voters vote for the challenger; and otherwise, if $$$x \ne {{x}_{{{\tau }_1}}}$$$ , then all voters believe that with probability one the incumbent is type $$${{\tau }_0}$$$ , and all type $$$\tau \,\geq \,{{\tau }_1}$$$ voters vote for the challenger, and type $$${{\tau }_0}$$$ voters vote for the incumbent.

Phase 4.2. Type $$$\tau \ne {{\tau }_0}$$$ office holders choose $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ and type $$${{\tau }_0}$$$ office holders choose x = 0. If the incumbent chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ , then all voters’ beliefs are given by $$$\hat{p}$$$ , the updated prior conditional on the politician's type being not equal to $$${{\tau }_0}$$$ , and all type $$$\tau \,\geq \,{{\tau }_1}$$$ voters vote for the incumbent and type $$${{\tau }_0}$$$ voters vote for the challenger; and if the incumbent chooses x = 0, then all voters believe that with probability one the politician is type $$${{\tau }_0}$$$ , and all type $$$\tau \,\geq \,{{\tau }_1}$$$ voters vote for the incumbent and type $$${{\tau }_0}$$$ voters vote for the challenger; and if the incumbent chooses $$$x \, \notin \, \{ 0,{{x}_{{{\tau }_1}}}\} $$$ , then voters believe that with probability one the politician is type $$${{\tau }_0}$$$ , all type $$$\tau \,\geq \,{{\tau }_1}$$$ voters vote for the challenger, and type $$${{\tau }_0}$$$ voters vote for the incumbent.

Phase 5. All types of office holder choose their ideal policies. All voters believe that with probability one the incumbent is type $$${{\tau }_0}$$$ , and all type $$$\tau \,\geq \,{{\tau }_1}$$$ voters vote for the challenger, and all type $$${{\tau }_0}$$$ voters vote for the incumbent.

The specification is completed by assigning the initial history Ø to Phase 1, and by defining the rule for transitioning between the phases.

• • In Phase 1, if the office holder chooses x = a, then remain in Phase 1; if the office holder chooses x ≠ a and is re-elected, then move to Phase 5; and if the office holder chooses x ≠ a and is not re-elected, then move to Phase 2(1).

• • In Phase 2(j), if the office holder chooses x = 0 and is re-elected, or if x ≠ 0 and the incumbent is not re-elected, then move to Phase 2 (j+1) if j < m and to Phase 3(1) if j = m; if the office holder chooses x = 0 and is not re-elected, then move to Phase 2(1); and if the office holder chooses x ≠ 0 and is re-elected, then move to Phase 5.

• • In Phase 3(j), if the office holder chooses x = 0 and is re-elected, then move to Phase 3(j+1) if j < m and to Phase 4.0(1) if j = m; if the office holder chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ and is re-elected, then move to Phase 3(j+1) if j < m and to Phase 4.1 if j = m; if the office holder chooses $$$x \, \notin \, \{ 0,{{x}_{{{\tau }_1}}}\} $$$ and is re-elected, then move to Phase 5; if the office holder chooses $$$x \in \{ 0,{{x}_{{{\tau }_1}}}\} $$$ and is not re-elected, then move to Phase 2(1); and if the office holder chooses $$$x \, \notin \, \{ 0,{{x}_{{{\tau }_1}}}\} $$$ and is not re-elected, then move to Phase 4.2.

• • In Phase 4.0(j), first suppose the incumbent is re-elected. If $$$j \in \{ 1, \ldots, k\} $$$ and the incumbent chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ , then move to Phase $$$4.0(j\, + \,1)$$$ ; if $$$j \in \{ 1, \ldots, k\} $$$ and the incumbent chooses $$$x \ne {{x}_{{{\tau }_1}}}$$$ , then move to Phase 5; if j = k+1 and $$$x \in \{ 0,{{x}_{{{\tau }_1}}}\} $$$ , then move to Phase 4.0(1); and if j = k+1 and $$$x \, \notin \, \{ 0,{{x}_{{{\tau }_1}}}\} $$$ , then move to Phase 5. Supposing the incumbent is not re-elected, if $$$j \in \{ 1, \ldots, k\} $$$ and the incumbent chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ , or if j = k+1 and $$$x\, = \,\{ 0,{{x}_{{{\tau }_1}}}\} $$$ , then move to Phase 2(1); and otherwise, move to Phase 4.2.

• • In Phase 4.1, if the office holder chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ and is re-elected, then remain in Phase 4.1; if the office holder chooses $$$x\, \ne \,{{x}_{{{\tau }_1}}}$$$ and is re-elected, then move to Phase 5; if the office holder chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ and is not re-elected, then move to Phase 2(1); and if the office holder chooses $$$x\, \ne \,{{x}_{{{\tau }_1}}}$$$ and is not re-elected, then move to Phase 4.2.

• • In Phase 4.2, if the office holder chooses x = 0 and is re-elected, then move to Phase 4.0(1); if the office holder chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ and is re-elected, then move to Phase 4.1; if the office holder chooses $$$x \, \notin \, \{ 0,{{x}_{{{\tau }_1}}}\} $$$ and is re-elected, then move to Phase 5; if the office holder chooses $$$x \in \{ 0,{{x}_{{{\tau }_1}}}\} $$$ and is not re-elected, then move to Phase 2(1); and if the office holder chooses $$$x \, \notin \, \{ 0,{{x}_{{{\tau }_1}}}\} $$$ and is not re-elected, then remain in Phase 4.2.

• • In Phase 5, if the office holder is re-elected, then remain in Phase 5; and otherwise, if the incumbent is not re-elected, then move to Phase 2(1).

This assignment of histories to phases entails a specification of strategies and beliefs. In order to verify that they form a perfect Bayesian equilibrium, I will focus on the most critical best-response conditions, addressing the more straightforward conditions with less formal arguments. I do not consider the votes of type $$${{\tau }_0}$$$ voters, as they do not affect electoral outcomes.

It is useful to describe the paths of play at the beginning of Phases 1–5. Of course, the path of play beginning in Phase 1 is simply policy choice a and re-election, ad infinitum. Beginning in Phase 2(1), policy x = 0 is chosen for m periods and the incumbent is re-elected. Beginning in Phase 3(1), if the office holder is type $$${{\tau }_0}$$$ , then she chooses x = 0 and is re-elected for m periods; and otherwise, if the office holder is type $$$\tau \ne {{\tau }_0}$$$ , then she chooses $$${{x}_{{{\tau }_1}}}$$$ and is re-elected for m periods. We then move to Phase 4.0 or 4.1, depending on whether the office holder is type $$${{\tau }_0}$$$ or $$$\tau \ne {{\tau }_0}$$$ . In the latter case, the office holder chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ and is re-elected ad infinitum; and in the former case, the office holder alternately chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ for k periods and x = 0 for one period, and is re-elected ad infinitum. In Phase 5, all types of office holder choose their ideal policies. When this phase is reached, voters believe the office holder is type $$${{\tau }_0}$$$ , the politician is removed from office, and we move to Phase 2 with the challenger assuming office.

Next, I calculate the continuation value (normalized by 1 − δ) of each citizen type at the beginning of Phase 2(1) after electing an untried challenger:

$${{V}_\tau }\, = \,{{p}_0}[(1\,{\rm{ - }}\,{{\delta }^{2m}} ){{u}_\tau }(0)\, + \,{{\delta }^{2m}} {{W}_\tau }]\, + \,(1\,{\rm{ - }}\,{{p}_0})[(1\,{\rm{ - }}\,{{\delta }^m} ){{u}_\tau }(0)\, + \,{{\delta }^m} {{u}_\tau }({{x}_{{{\tau }_1}}})].$$

This reflects the fact that with probability p ₀ the office holder is type $$${{\tau }_0}$$$ and chooses x = 0 for 2 m periods and then alternately chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ for k periods and x = 0 for one period; and with probability 1 − p ₀ the office holder is type $$$\tau \ne {{\tau }_0}$$$ and chooses x = 0 for m periods and chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ thereafter. I employ the one-shot deviation principle to confirm that no office holders of any type can gain by deviating, and that all voter types (in particular, types $$$\tau \geq {{\tau }_1}$$$ in Phases 2 and 3) vote as though pivotal.

Phase 1: First, consider the choice of a type $$${{\tau }_0}$$$ office holder. Following the above strategy, the politician chooses x = a and is continually re-elected. Following any deviation, the politician is removed from office, so the best one-shot deviation is to choose x = 0 and to receive $$${{V}_{{{\tau }_0}}}$$$ in expectation after that. The one-shot deviation is unprofitable if

$${{u}_{{{\tau }_0}}}(a)\,\geq \,(1\,{\rm{ - }}\,\delta ){{u}_{{{\tau }_0}}}(0)\, + \,\delta {{V}_{{{\tau }_0}}},$$

or equivalently,

$$\hskip-12.5pc(1\,{\rm{ - }}\,\delta )({{u}_{{{\tau }_0}}}(0)\,{\rm{ - }}\,{{u}_{{{\tau }_0}}}(a))$$

$$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\leq \delta [{{p}_0}[(1\,{\rm{ - }}\,{{\delta }^{2m}} )({{u}_{{{\tau }_0}}}(a)\,{\rm{ - }}\,{{u}_{{{\tau }_0}}}(0))\, + \,{{\delta }^{2m}} ({{u}_{{{\tau }_0}}}(a)\,{\rm{ - }}\,{{W}_{{{\tau }_0}}})]$$

$$ + (1\,{\rm{ - }}\,{{p}_0})[(1\,{\rm{ - }}\,{{\delta }^m} )({{u}_{{{\tau }_0}}}(a)\,{\rm{ - }}\,{{u}_{{{\tau }_0}}}(0))\, + \,{{\delta }^m} ({{u}_{{{\tau }_0}}}(a)\,{\rm{ - }}\,{{u}_{{{\tau }_0}}}({{x}_{{{\tau }_1}}}))]],$$

which after substituting $$$\beta = {{\delta }^m} $$$ is equivalent to Inequality 6. Next, consider a type $$$\tau \ne {{\tau }_0}$$$ office holder. The best one-shot deviation is to choose $$$x = {{x}_\tau }$$$ followed by $$${{V}_\tau }$$$ . The one-shot deviation is unprofitable if

$${{u}_\tau }(a)\,\geq \,(1\,{\rm{ - }}\,\delta ){{u}_\tau }({{x}_\tau })\, + \,\delta {{V}_\tau },$$

which is implied by

$$(1\,{\rm{ - }}\,\delta )\rDelta \,\leq \,\delta [{{p}_0}[(1\,{\rm{ - }}\,{{\delta }^{2m}} )({{u}_\tau }(a)\,{\rm{ - }}\,{{u}_\tau }(0))\, + \,{{\delta }^{2m}} ({{u}_\tau }(a)\,{\rm{ - }}\,{{W}_\tau })]$$

$$ + (1\,{\rm{ - }}\,{{p}_0})[(1\,{\rm{ - }}\,{{\delta }^m} )({{u}_\tau }(a)\,{\rm{ - }}\,{{u}_\tau }(0))\, + \,{{\delta }^m} ({{u}_\tau }(a)\,{\rm{ - }}\,{{u}_\tau }({{x}_{{{\tau }_1}}}))]]$$

which after substituting β = δ ^m is equivalent to Inequality 7.

Now consider a voter of any type. In case the incumbent chooses x = a, if the incumbent is re-elected, then she will continue to choose a and be re-elected. If a challenger is elected, then we remain in Phase 1, and the newly elected office holder continues to choose a. Thus, all voters are indifferent between the incumbent and a challenger, and voting for the incumbent is consistent with being pivotal. In case the incumbent chooses x ≠ a, then all voters believe that with probability one the politician is type $$${{\tau }_0}$$$ . If the incumbent is re-elected, then we move to Phase 5, and voters expect that the office holder will choose $$$x\, = \,{{x}_{{{\tau }_0}}}\, = \,0$$$ , and then an untried challenger is elected, and we move to Phase 2(1). If the challenger is elected, then we move directly to Phase 2(1). Thus, the expected discounted payoff from electing a challenger exceeds that from re-electing the incumbent for a type $$$\tau \,\geq \,{{\tau }_1}$$$ voter if

$${{V}_\tau }\,\geq \,(1\,{\rm{ - }}\,\delta ){{u}_\tau }(0)\, + \,\delta {{V}_\tau },$$

which indeed holds.

Phase 2 (j). First, consider the choice of a type $$${{\tau }_0}$$$ office holder. Following the above strategy, the politician chooses x = 0 and is re-elected for m − j periods, after which we move to Phase 3(1) and she chooses x = 0 for m more periods, after which we move to Phase 4.0(1) and she alternately chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ for k periods and x = 0 for one period. Following any deviation, the politician is removed from office, we continue to Phase 2 (j+1) if j < m or to Phase 3(1) if j = m, and clearly the office holder receives a higher payoff by choosing her ideal policy and avoiding the election of a type $$$\tau \ne {{\tau }_0}$$$ challenger. Next, consider the choice of a type $$$\tau \ne {{\tau }_0}$$$ office holder. Following the above strategy, the politician chooses $$$x\, = \,0$$$ and is re-elected for m − j periods, and then chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ and is continually re-elected. The best one-shot deviation is to choose $$$x\, = \,{{x}_\tau }$$$ , after which the politician is removed from office, and we move to Phase 2 (j+1) or 3(1) with a new office holder. The incentives to deviate are maximal when j = 1, as then the cost of deviating is realized with probability p ₀ after m − 1 periods. The one-shot deviation is unprofitable if

$$(1\,{\rm{ - }}\,\delta ){{u}_\tau }(0)\, + \,\delta [(1\,{\rm{ - }}\,{{\delta }^{m\,{\rm{ - }}\,1}} ){{u}_\tau }(0)\, + \,{{\delta }^{m\,{\rm{ - }}\,1}} {{u}_\tau }({{x}_{{{\tau }_1}}})]$$

$$\quad\ \ \ \ \geq (1\,{\rm{ - }}\,\delta ){{u}_\tau }({{x}_\tau })\, + \,\delta [{{p}_0}[(1\,{\rm{ - }}\,{{\delta }^{2m\,{\rm{ - }}\,1}} ){{u}_\tau }(0)\, + \,{{\delta }^{2m\,{\rm{ - }}\,1}} {{W}_\tau }]$$

$$ \!\!\!\!\!\!\!\!\!\!\!\!+ (1\,{\rm{ - }}\,{{p}_0})[(1\,{\rm{ - }}\,{{\delta }^{m\,{\rm{ - }}\,1}} ){{u}_\tau }(0)\, + \,{{\delta }^{m\,{\rm{ - }}\,1}} {{u}_\tau }({{x}_{{{\tau }_1}}})]],$$

which is implied by

$$(1\,{\rm{ - }}\,\delta )\rDelta \,\leq \,\delta {{p}_0}[{{\delta }^{m\,{\rm{ - }}\,1}} (1\,{\rm{ - }}\,{{\delta }^m} )({{u}_\tau }({{x}_{{{\tau }_1}}})\,{\rm{ - }}\,{{u}_\tau }(0))\, + \,{{\delta }^{2m\,{\rm{ - }}\,1}} ({{u}_\tau }({{x}_{{{\tau }_1}}})\,{\rm{ - }}\,{{W}_\tau })],$$

which after substituting $$$\beta \, = \,{{\delta }^m} $$$ is implied by Inequality 12.

Now consider a type $$$\tau \,\geq \,{{\tau }_1}$$$ voter. In case the incumbent chooses x = 0, then the voters’ beliefs about the incumbent's type are given by the prior p. If the incumbent is re-elected, then she will choose x = 0 for m − 1 more periods, and we move to Phase 3(1). If the challenger is elected, then we move back to Phase 2(1). Thus the expected discounted payoff to the type $$$\tau \,\geq \,{{\tau }_1}$$$ voter from re-electing the incumbent is greater than from electing a challenger. In case the incumbent chooses $$$x\, \ne \,0$$$ , then the voters believe that with probability one the politician is type $$${{\tau }_0}$$$ . If the incumbent is re-elected, then we move to Phase 5, and she chooses x = 0, is removed from office, and we move to Phase 2(1). If the challenger is elected, then we remain in Phase 2 for m − 1 more periods, and we move to Phase 3(1). Thus the expected discounted payoff of electing the challenger is greater for the type $$$\tau \ne {{\tau }_0}$$$ voter than that of re-electing the incumbent.

Phase 3(j). First, consider the choice of a type $$${{\tau }_0}$$$ office holder. Following the above strategy, the politician chooses x = 0 and is re-elected for m − j more periods, after which we move to Phase 4.0(1) and she alternately chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ for k periods and x = 0 for one period. Following any deviation $$$x^{\prime} \, \notin \, \{ 0,{{x}_{{{\tau }_1}}}\} $$$ , the politician is removed from office, and we move to Phase 4.2, where a challenger is elected and chooses policy. If the newly elected office holder is type $$${{\tau }_0}$$$ , then she chooses x = 0, and we move to Phase 4.0(1); and if she is type $$$\tau \ne {{\tau }_0}$$$ , then she chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ , and we move to Phase 4.1. The gains from the one-shot deviation are highest when j = m, and the deviation is unprofitable in this case if

$$(1\,{\rm{ - }}\,\delta ){{u}_{{{\tau }_0}}}(0)\, + \,\delta {{W}_{{{\tau }_0}}}\, \gt \,{{p}_0}[(1\,{\rm{ - }}\,\delta ){{u}_{{{\tau }_0}}}(0)\, + \,\delta {{W}_{{{\tau }_0}}}]\, + \,(1\,{\rm{ - }}\,{{p}_0}){{u}_{{{\tau }_0}}}({{x}_{{{\tau }_1}}}),$$

which is implied by

$$(1\,{\rm{ - }}\,\delta ){{u}_{{{\tau }_0}}}({{x}_{{{\tau }_1}}})\, + \,\delta {{W}_{{{\tau }_0}}}\, \gt \,(1\,{\rm{ - }}\,\delta ){{u}_{{{\tau }_0}}}(0)\, + \,\delta [{{p}_0}{{W}_{{{\tau }_0}}}\, + \,(1\,{\rm{ - }}\,{{p}_0}){{u}_{{{\tau }_0}}}({{x}_{{{\tau }_1}}})],$$

which is implied by Inequality 14. Following the deviation $$$x^{\prime}\, = \,{{x}_{{{\tau }_1}}}$$$ , the office holder is re-elected, and we move to Phase 3(j+1) if j < m and to Phase 4.1 if j = m. In the latter case, the office holder then chooses x = 0, is removed from office, and we move to Phase 4.2, where a type $$$\tau \ne {{\tau }_0}$$$ challenger always leads to Phase 4.1, and a type $$${{\tau }_0}$$$ leads to Phase 4.0(1). The office holder's expected discounted payoff is greater following the initial strategy and choosing her ideal policy x = 0 before entering Phase 4.0(1). In the former case, after moving to Phase 3 (j+1), the office holder continues to choose x = 0 for m − j periods before moving to Phase 4.0(1), and she does not gain from the deviation.

Next, consider a type $$$\tau \ne {{\tau }_0}$$$ office holder. Following the above strategy, the politician chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ and is re-elected for m − j more periods, after which we move to Phase 4.1, and she continues to choose $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ and be re-elected. The best one-shot deviation is to choose $$$x = {{x}_\tau }$$$ after which the politician is removed from office, and we move to Phase 4.2. The one-shot deviation is unprofitable if

$${{u}_\tau }({{x}_{{{\tau }_1}}})\, \gt \,(1\,{\rm{ - }}\,\delta ){{u}_\tau }({{x}_\tau })\, + \,\delta [{{p}_0}[(1\,{\rm{ - }}\,\delta ){{u}_\tau }(0)\, + \,\delta {{W}_\tau }]\, + \,(1\,{\rm{ - }}\,{{p}_0}){{u}_\tau }({{x}_{{{\tau }_1}}})],$$

which is implied by

$${{u}_\tau }({{x}_{{{\tau }_1}}})\, \gt \,(1\,{\rm{ - }}\,\delta ){{u}_\tau }({{x}_\tau })\, + \,\delta [{{p}_0}{{W}_\tau }\, + \,(1\,{\rm{ - }}\,{{p}_0}){{u}_\tau }({{x}_{{{\tau }_1}}})],$$

which is implied by Inequality 13.

Now consider a type $$$\tau \,\geq \,{{\tau }_1}$$$ voter. In case the office holder chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ , then the voters’ posterior puts probability zero on type $$${{\tau }_0}$$$ . If the incumbent is re-elected, then we move either to Phase 3(j+1) or to Phase 4.1, and the politician continues to choose $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ and be re-elected. If the incumbent is removed from office, then we move to Phase 2(1) with a challenger, leading to a lower expected discounted payoff.Footnote ⁹ In case the office holder chooses x = 0, then the voters believe that with probability one the incumbent is type $$${{\tau }_0}$$$ . If the incumbent is re-elected, then we move either to Phase 3(j+1) or to Phase 4.0(1). If the incumbent is removed from office, then we move to Phase 2(1). Thus the expected discounted payoff from re-electing the incumbent exceeds that from electing the challenger for a type $$$\tau \, \ne \,{{\tau }_0}$$$ voter if

$$(1\,{\rm{ - }}\,{{\delta }^{m\,{\rm{ - }}\,j}} ){{u}_\tau }(0)\, + \,{{\delta }^{m\,{\rm{ - }}\,j}} {{W}_\tau }\, \gt \,{{V}_\tau }.$$

This inequality is most restrictive when j = 1 and actually holds with j = 0. Indeed, substituting β = δ ^m, the inequality then becomes

$$\hskip-14pc(1\,{\rm{ - }}\,\beta ){{u}_\tau }(0)\, + \,\beta {{W}_\tau }$$

$$ \ \ \gt {{p}_0}[(1\,{\rm{ - }}\,{{\beta }^2} ){{u}_\tau }(0)\, + \,{{\beta }^2} {{W}_\tau }]\, + \,(1\,{\rm{ - }}\,{{p}_0})[(1\,{\rm{ - }}\,\beta ){{u}_\tau }(0)\, + \,\beta {{u}_\tau }({{x}_{{{\tau }_1}}})],$$

which is equivalent to Inequality 5 after manipulation.Footnote ¹⁰ In case the office holder chooses $$$x \ne {{x}_{{{\tau }_1}}}$$$ , then all voters believe that with probability one the incumbent is type $$${{\tau }_0}$$$ . If the incumbent is re-elected, then we move to Phase 5, she chooses x = 0, is removed from office, and we move to Phase 2(1). If the challenger is elected, then we move to Phase 4.2, and if the newly elected office holder is type $$${{\tau }_0}$$$ , then she chooses x = 0, and we move to Phase 4.0(1); and if she is type $$$\tau \ne {{\tau }_0}$$$ , then she chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ , and we move to Phase 4.1. Since re-electing the incumbent leads to policy x = 0 in the subsequent period, whereas the election of a challenger leads to $$$x \in \{ 0,{{x}_{{{\tau }_1}}}\} $$$ , it suffices to compare the stream of payoffs beginning two periods hence. The expected discounted payoff from electing the challenger exceeds that from re-electing the incumbent for a type $$$\tau \geq {{\tau }_1}$$$ voter if

$$\matrix{{{p}_0}{{W}_\tau } + (1\,{\rm{ - }}\,{{p}_0}){{u}_\tau }({{x}_{{{\tau }_1}}}) \gt (1\,{\rm{ - }}\,{{\delta }^m} ){{u}_\tau }(0)\, + \,{{\delta }^m} [{{p}_0}[(1\,{\rm{ - }}\,{{\delta }^m} ){{u}_\tau }(0)\, + \,{{\delta }^m} {{W}_\tau }] \cr \hskip-17pc+ (1\,{\rm{ - }}\,{{p}_0})[{{u}_\tau }({{x}_{{{\tau }_1}}})]], \\ $$

where we use the fact that in Phase 2(1), all types of office holder choose x = 0 for the first m periods. The above inequality holds if

$${{W}_\tau }\, \gt \,(1\,{\rm{ - }}\,\delta ){{u}_\tau }(0)\, + \,\delta [{{p}_0}[(1\,{\rm{ - }}\,{{\delta }^m} ){{u}_\tau }(0)\, + \,{{\delta }^m} {{W}_\tau }]\, + \,(1\,{\rm{ - }}\,{{p}_0}){{u}_\tau }({{x}_{{{\tau }_1}}})],$$

which is implied by setting j = 1 and deleting the first m − 1 realizations of x = 0 from both sides of Inequality 15.

Phase 4.0(j). First, consider the choice of a type $$${{\tau }_0}$$$ office holder. Following the above strategy, the politician alternately chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ for k periods and x = 0 for one period and is continually re-elected. The gains from deviating are highest when j = 1, and the best one-shot deviation is to choose $$$x^{\prime}\, = \,0$$$ , after which the incumbent is replaced by a challenger and we move to Phase 4.2. The one-shot deviation is unprofitable if

$${{W}_{{{\tau }_0}}}\, \gt \,(1\,{\rm{ - }}\,\delta ){{u}_{{{\tau }_0}}}(0)\, + \,\delta [{{p}_0}{{W}_{{{\tau }_0}}}\, + \,(1\,{\rm{ - }}\,{{p}_0}){{u}_{{{\tau }_0}}}({{x}_{{{\tau }_1}}})],$$

which is implied by

$$(1\,{\rm{ - }}\,\delta ){{u}_{{{\tau }_0}}}({{x}_{{{\tau }_1}}})\, + \,\delta {{W}_{{{\tau }_0}}}\, \gt \,(1\,{\rm{ - }}\,\delta ){{u}_{{{\tau }_0}}}(0)\, + \,\delta [{{p}_0}{{W}_{{{\tau }_0}}}\, + \,(1\,{\rm{ - }}\,{{p}_0}){{u}_{{{\tau }_0}}}({{x}_{{{\tau }_1}}})],$$

which is implied by Inequality 14. Next, consider a type $$$\tau \ne {{\tau }_0}$$$ office holder. Following the above strategy, the politician chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ and is re-elected in every period. The gains from deviating are the same as in Phase 3, and it follows from Inequality 13 and the above arguments that one-shot deviations are unprofitable.

Now consider a type $$$\tau \,\geq \,{{\tau }_1}$$$ voter. All voters believe that with probability one the incumbent is type $$${{\tau }_0}$$$ . In case $$$j \in \{ 1, \ldots, k\} $$$ and the incumbent chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ , or in case j = k+1 and $$$x \in \{ 0,{{x}_{{{\tau }_1}}}\} $$$ , if the incumbent is re-elected, then we remain in Phase 4.0, and the office holder alternately chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ for k periods and x = 0 for one period. If the incumbent is removed from office, then we move to Phase 2(1). The expected discounted payoff from re-electing the incumbent is greater than in Phase 3(1), whereas the expected discounted payoff from electing the challenger is the same as in Phase 3(1), so Inequality 5 and the arguments above imply that voting for the incumbent is consistent with the voter being pivotal. In the complementary case, if the incumbent is re-elected, then we move to Phase 5, the office holder chooses x = 0 and is removed from office, and we move to Phase 2(1). If the challenger is elected, then we move to Phase 4.2, where the new office holder chooses x = 0, and we move to Phase 4.0(1) if she is type $$${{\tau }_0}$$$ , and she chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ ad infinitum otherwise. The expected discounted payoff from electing the challenger exceeds that from re-electing the incumbent, because we have

$$\matrix{{{p}_0}[(1\,{\rm{ - }}\,\delta ){{u}_\tau }(0)\, + \,\delta {{W}_\tau }]\, + \,(1\,{\rm{ - }}\,{{p}_0}){{u}_\tau }({{x}_{{{\tau }_1}}}) \cr \hskip-6pc \!\!\!\gt (1\,{\rm{ - }}\,\delta ){{u}_\tau }(0)\, + \,\delta {{W}_\tau } \cr \hskip-3pc \!\!\!\!\!\gt {{p}_0}[(1\,{\rm{ - }}\,{{\delta }^{2m}} ){{u}_\tau }(0)\, + \,{{\delta }^{2m}} {{W}_\tau }] \cr \!\!\!\!\!\!\!\!\!\!\!+ (1\,{\rm{ - }}\,{{p}_0})[(1\,{\rm{ - }}\,{{\delta }^m} ){{u}_\tau }(0)\, + \,{{\delta }^m} {{u}_\tau }({{x}_{{{\tau }_1}}}), \\ $$

where the second inequality above follows by setting j=m−1 in Inequality 15.

Phase 4.1. First, consider the choice of a type $$${{\tau }_0}$$$ office holder. Following the above strategy, the politician chooses x = 0, is removed from office, and we move to Phase 4.2. Deviating to $$$x^{\prime} \, \notin \, \{ 0,{{x}_{{{\tau }_1}}}\} $$$ again leads to election of a challenger and Phase 4.2, so this cannot be profitable. If the politician deviates to $$$x^{\prime}\, = \,{{x}_{{{\tau }_1}}}$$$ , then she is re-elected and we remain in Phase 4.1, she then chooses x = 0, is removed from office, and we move to Phase 4.2. The office holder's expected discounted payoff is higher following the initial strategy and obtaining her ideal policy immediately before moving to Phase 4.2. Next, consider a type $$$\tau \ne {{\tau }_0}$$$ office holder. Following the above strategy, the politician chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ and is re-elected in every period. The best one-shot deviation is as in Phase 3, and it follows from Inequality 13 and the above arguments that one-shot deviations are unprofitable.

Now consider a type $$$\tau \geq {{\tau }_1}$$$ voter. In case the incumbent chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ , then all voters’ beliefs are given by $$$\hat{p}$$$ . If the incumbent is re-elected, then she continues to choose $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ and be re-elected. If the incumbent is removed from office, then we move to Phase 2(1). The expected discounted payoff from re-electing the incumbent exceeds that from electing a challenger if $$${{u}_\tau }({{x}_{{{\tau }_1}}})\, \gt \,{{V}_\tau }$$$ , which is implied by Inequality 15. In case the incumbent chooses $$$x \ne {{x}_{{{\tau }_1}}}$$$ , then all voters believe that with probability one the incumbent is type $$${{\tau }_0}$$$ . If the incumbent is re-elected, then we move to Phase 5, she chooses x = 0 and is removed from office, and we move to Phase 2(1). If the challenger is elected, then we move to Phase 4.2. By the arguments for Phase 3, the expected discounted payoff from electing the challenger exceeds that from re-electing the incumbent.

Phase 4.2. First, consider the choice of a type $$${{\tau }_0}$$$ office holder. Following the above strategy, the politician chooses x = 0, is re-elected, and we move to Phase 4.0(1), where she alternately chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ for k periods and x = 0 for one period. If the politician deviates to $$$x^{\prime}\, = \,{{x}_{{{\tau }_1}}}$$$ , then she is re-elected, we move to Phase 4.1, she then chooses x = 0 and is removed from office, and we move to Phase 4.2. The one-shot deviation is unprofitable if

$$\matrix {(1\,{\rm{ - }}\,\delta ){{u}_{{{\tau }_0}}}(0)\, + \,\delta {{W}_{{{\tau }_0}}}\, \gt \,(1\,{\rm{ - }}\,\delta ){{u}_{{{\tau }_0}}}({{x}_{{{\tau }_1}}})\, + \,(1\,{\rm{ - }}\,\delta )\delta {{u}_{{{\tau }_0}}}(0) \cr + \,{{\delta }^2} [{{p}_0}[(1\,{\rm{ - }}\,\delta ){{u}_{{{\tau }_0}}}(0)\, + \,\delta {{W}_{{{\tau }_0}}}]\, + \,(1\,{\rm{ - }}\,{{p}_0}){{u}_{{{\tau }_0}}}({{x}_{{{\tau }_1}}})] \\ $$

Using Inequality 14 and the arguments of Phase 3, the above inequality is implied by

$$(1\,{\rm{ - }}\,{{\delta }^2} )[(1\,{\rm{ - }}\,\delta ){{u}_{{{\tau }_0}}}(0)\, + \,\delta {{W}_{{{\tau }_0}}}]\, \gt \,(1\,{\rm{ - }}\,\delta ){{u}_{{{\tau }_0}}}({{x}_{{{\tau }_1}}})\, + \,(1\,{\rm{ - }}\,\delta )\delta {{u}_{{{\tau }_0}}}(0).$$

Since $$${{W}_{{{\tau }_0}}}\, \gt \,{{u}_{{{\tau }_0}}}({{x}_{{{\tau }_1}}})$$$ , this in turn holds if $$${{u}_{{{\tau }_0}}}(0)\, \gt \,{{u}_{{{\tau }_0}}}({{x}_{{{\tau }_1}}})$$$ , which holds by assumption. If the politician deviates to $$$x^{\prime} \, \notin \, \{ 0,{{x}_{{{\tau }_1}}}\} $$$ , then she is removed from office and we remain in Phase 4.2, which is not profitable. Next, consider a type $$$\tau \ne {{\tau }_0}$$$ office holder. Following the above strategy, the politician chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ and is re-elected in every period. The best one-shot deviation is as in Phase 3, and it follows from Inequality 13 and the above arguments that one-shot deviations are unprofitable.

Now consider a type $$$\tau \,\geq \,{{\tau }_1}$$$ voter. In case the office holder chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ , then all voters’ beliefs are given by $$$\hat{p}$$$ . If the incumbent is re-elected, then she continues to choose $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ and be re-elected. If the incumbent is removed from office, then we move to Phase 2(1), and again Inequality 15 implies that the voter's expected discounted payoff from re-electing the incumbent exceeds that from electing a challenger. In case the incumbent chooses x = 0, then all voters believe that with probability one the incumbent is type $$${{\tau }_0}$$$ . If the incumbent is re-elected, then we move to Phase 4.0(1), and if the incumbent is not re-elected, then we move to Phase 2(1). Again, Inequality 15 implies that the voter's expected discounted payoff from re-electing the incumbent exceeds that from electing a challenger. In case the office holder chooses $$$x \, \notin \, \{ 0,{{x}_{{{\tau }_1}}}\} $$$ , then all voters believe the incumbent is type $$${{\tau }_0}$$$ . If the incumbent is re-elected, then we move to Phase 5, she chooses x = 0, is removed from office, and we move to Phase 2(1). If the challenger is elected, then we remain in Phase 4.2. By the arguments for Phase 3, the expected discounted payoff from electing the challenger exceeds that from re-electing the incumbent.

Phase 5. Since the incumbent is always removed from office, it is optimal for office holder's of all types to choose their ideal policies. Now consider a type $$$\tau \geq {{\tau }_1}$$$ voter. All voters believe that with probability one the incumbent is type $$${{\tau }_0}$$$ . If the incumbent is re-elected, then we remain in Phase 5, the politician chooses x = 0, is removed from office, and we move to Phase 2(1). If the challenger is elected, then we move directly to Phase 2(1). Thus the expected discounted payoff from electing the challenger exceeds that from re-electing the incumbent.

To conclude, for all ε > 0, we can choose $$$a \in (0,{\epsilon})$$$ , natural number k, and $$$\hat{\delta } \in (0,1)$$$ such that for all $$$\delta \in (\hat{\delta },1)$$$ , there exists m such that Inequalities 5–7 and 12–14 are satisfied by a, δ, and β = δ ^m. Given these parameters, I have specified strategies and beliefs such that the policy a is chosen along the path of play by all types of office holder, voter beliefs satisfy Bayes’ rule along each office holder's personal path of play, and no politician has a profitable one-shot deviation, and such that all voter types vote as though pivotal. This completes Step 1.

An argument symmetric to that for Step 1 yields the next step of the proof.

Step 2: For all ε > 0, there exist $$$b \in (1\,{\rm{ - }}\,{\epsilon},1)$$$ and $$$\hat{\delta } \in (0,1)$$$ such that for all $$$\delta \in (\hat{\delta },1)$$$ ,

we have $$${{{\bf {{x}}}}^b} \in P(\delta )$$$ .

Given the argument of Step 1, I do not provide the (redundant) details of Step 2. But it will be important to note that a and b can be chosen close enough to zero and one, respectively, to allow the same choice of k and m in both steps. To be more explicit, choose k such that for all $$$\tau \ne {{\tau }_0}$$$ , Inequality 1 holds, and such that for all $$$\tau \ne {{\tau }_n}$$$ , the corresponding inequality for Step 2 holds. Choose $$$\overline{\beta } $$$ such that for all $$$\beta \in (0,\overline{\beta } )$$$ , Inequality 4 and the corresponding inequality for Step 2 hold, and such that there exists $$$\underline{\delta } \in (0,1)$$$ such that for all $$$\delta \in (\underline{\delta }, 1)$$$ and all $$$\beta \in (0,\overline{\beta } )$$$ , we have Inequality 5 for all $$$\tau \ne {{\tau }_0}$$$ and the corresponding inequality for Step 2 for all $$$\tau \ne {{\tau }_n}$$$ . Furthermore, define

$$\matrix{ \!\!\!\!\!\!\!\!\!{{B}_\tau }\, = \,{{u}_\tau }(b)\,{\rm{ - }}\,{{u}_\tau }(1) \cr B_{\tau }^{\ast} \, = \,{{u}_\tau }({{x}_{{{\tau }_{n\,{\rm{ - }}\,1}}}})\,{\rm{ - }}\,{{u}_\tau }(b) \cr \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\underline{B} \, = \,\mathop {\min }\limits_{\tau \ne {{\tau }_n}} {{B}_\tau } \cr \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{{B}^\ast} \, = \,\mathop {\max }\limits_{\tau \ne {{\tau }_n}} B_{\tau }^{\ast}, \\ $$

and note that by the arguments in Step 1, we have

$$\mathop {\lim }\limits_{b \uparrow 1} \left( {\frac{{{{u}_{{{\tau }_n}}}(1)\,{\rm{ - }}\,{{u}_{{{\tau }_n}}}(b)}}{{{{u}_{{{\tau }_n}}}(b)\,{\rm{ - }}\,{{u}_{{{\tau }_n}}}({{x}_{{{\tau }_{n\,{\rm{ - }}\,1}}}})}}} \right)\left( {\frac{{{{B}^\ast} }}{{\underline{B} }}} \right)\, = \,0.$$

Thus we can choose $$$a \in (0,{\epsilon})$$$ , $$$b \in (1\,{\rm{ - }}\,{\epsilon},1)$$$ and $$$\beta \in (0,\overline{\beta } )$$$ such that Inequalities 10 and 11 hold, along with the corresponding inequalities for Step 2. Then the arguments in Step 1 allow us to specify $$$\hat{\delta } \in (\underline{\delta }, 1)$$$ so that for all $$$\delta \in (\hat{\delta },1)$$$ , there exists m such that Inequalities 12–14 hold, along with the corresponding inequalities for Step 2. This permits the equilibrium construction from Step 1 to be applied in Step 2 with the same choice of k and m.

Next I define a procedure for splicing the equilibria from Steps 1 and 2 to produce a path of policies that alternates between a and b, and I use the construction recursively to generate paths that cycle between the alternatives to approximate any empirical frequency.

Step 3: For all ε > 0, there exist $$$a \in (0,{\epsilon})$$$ and $$$b \in (1\,{\rm{ - }}\,{\epsilon},1)$$$ such that for all natural numbers ℓ and all k with $$$k = 0,1, \ldots {{,2}^\ell } $$$ , there exists $$$\tilde{\delta } \in (0,1)$$$ such that for all $$$ \delta \in (\tilde{\delta },1) $$$ , there exists $$$ {\bf {{x}}} \in P(\delta ) $$$ such that in consecutive runs of $$$ {{2}^\ell } $$$ periods, the path x visits a for k periods and b for $$${{2}^\ell } \,{\rm{ - }}\,k$$$ periods; that is, for all $$$ m = 0,1,2, \ldots $$$ ,

$$\matrix{ \!\!\!\!\!\!\!\!\!\!\!\!\!\!\# \{ t&#x007C;m{{2}^\ell } + 1\,\leq \,t\,\leq \,(m\, + \,{{1)2}^\ell } \:\;{\rm and}\;\:{{x}_t}\, = \,a\} \, = \,k \cr \# \{ t&#x007C;m{{2}^\ell } \, + \,1\,\leq \,t\,\leq \,(m\, + \,{{1)2}^\ell } \:\;{\rm and}\;\:{{x}_t}\, = \,b\} \, = \,{{2}^\ell } \,{\rm{ - }}\,k. \\ $$

The proof is by induction. Fix ε > 0. By Step 1, there exist $$$a \in (0,{\epsilon})$$$ and $$$\delta ^{\prime}\, \lt \,1$$$ such that for all $$$\delta \in (\delta ^{\prime},1)$$$ , we have $$${{{\bf {{x}}}}^a} \in P(\delta )$$$ . Similarly, by Step 2, there exist $$$b \in (1\,{\rm{ - }}\,{\epsilon},1)$$$ and $$$\delta ^{\prime\prime}\, \lt \,1$$$ such that for all $$$\delta \in (\delta ^{\prime\prime},1)$$$ , we have $$${{{\bf {{x}}}}^b} \in P(\delta )$$$ . For the basis step, consider ℓ = 1. For k = 0 and k = 2, the desired result follows from Steps 2 and 1, respectively. Consider k = 1, and set $$${\bf {{x}}} = (b,a,b,a, \ldots )$$$ . I focus on the former case without loss of generality. Set $$$\tilde{\delta } = \sqrt {\max \{ \delta ^{\prime},\delta ^{\prime \prime}\} } $$$ , choose $$$\delta \in (\tilde{\delta },1)$$$ , and define strategies and beliefs by ‘splicing’ the specifications in Steps 1 and 2. That is, we specify actions and beliefs according to the equilibrium from Step 1 in odd periods and using the equilibrium from Step 2 in even periods.

To be more precise, we begin in Phase 1 in both odd and even periods, so that the path of play from the initial history is the sequence $$$(a,b,a,b, \ldots )$$$ , and we specify strategies from Step 1 in odd periods and from Step 2 in even periods. Following any deviation in an odd period t, for example, the transition between phases in subsequent odd periods follows the protocol from Step 1 (applied to odd periods), and we specify that the phase transitions in the next even period t+1 as though an analogous deviation had occurred in period t − 1; from that point on, the transition between phases in even periods follows the protocol from Step 2 (applied to even periods). So, if along the equilibrium path the office holder deviates to $$$x^{\prime} \ne a$$$ in an odd period t, then the politician is removed from office, and we move to Phase 2(1) for the next m even periods (with actions determined by Step 2) and for the next m odd periods (with actions determined by Step 1). After that span of 2 m periods has elapsed, we move to Phase 3 for the next m even periods and for the next m odd periods. We then enter Phase 4.0 or Phase 4.1, depending on policy choices in odd and even periods. Subsequent deviations are treated the same way. Suppose in an odd period t′, we are in Phase 3, and the office holder chooses $$$x^{\prime} \, \notin \, \{ 0,{{x}_{{{\tau }_1}}}\} $$$ and is not re-elected. Then we transition to Phase 4.2 in the next even period t′+1 and the next odd period t′+2. If the newly elected office holder is type $$${{\tau }_n}$$$ , for example, then she chooses $$$x\, = \,{{x}_{{{\tau }_n}}}$$$ in period t′+1, we transition to Phase 4.0 in all subsequent even periods, she chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ in period t′+2, and we transition to Phase 4.1 in all subsequent odd periods. On the other hand, if the new office holder is type $$$\tau $$$ with $$${{\tau }_1}\,\leq \,\tau \,\leq \,{{\tau }_{n\,{\rm{ - }}\,1}}$$$ , then she chooses $$${{x}_{{{\tau }_{n\,{\rm{ - }}\,1}}}}$$$ in period t′+1 and $$${{x}_{{{\tau }_1}}}$$$ in period t′+2, and we transition to Phase 4.1 in all subsequent periods.

Policy and voting choices are given by the strategies from Steps 1 and 2, although now we must take care in specifying voter beliefs when we enter Phase 5 and when information is revealed in Phases 3(1) and 4.2. To simplify exposition, we focus on the case in which the transition first occurs in an odd period, so that information is revealed to voters in the next even period. First, note that Phase 5 is entered only off a politician's personal path of play and only when she is re-elected, so that voter beliefs may be specified arbitrarily. Let $$$\overline{x} \in \{ 0,1\} $$$ be the policy that is weakly preferred among {0,1} by a majority of voters, and let $$$\underline{x} \in \{ 0,1\} $$$ be the remaining policy, so that

$$ \mathop{\sum} \{ {{q}_j}&#x007C;{{u}_{{{\tau }_j}}}(\overline{x} )\geq {{u}_{{{\tau }_j}}}(\underline{x} )\} \geq \frac{1}{2}. $$

Thus by concavity, the policy x is the worst possible policy for a majority of voters. Let $$$\underline{\tau } \in \{ {{\tau }_0},{{\tau }_n}\} $$$ be the type such that $$${{x}_{\underline{\tau } }}\, = \,\underline{x} $$$ . In Phase 5, whether in an odd or even period, we specify that voters believe with probability one that the office holder is type $$$\underline {\tau} $$$ . This ensures that a majority of voters prefers to replace the incumbent with a challenger, as dictated in Steps 1 and 2.

Second, after entering Phase 3(1) in an odd period t, if the office holder chooses x = 0, then voters believe that with probability one the office holder is type $$${{\tau }_0}$$$ , and if she is re-elected, then we move to Phase 3(1) in the following even period t+1. As noted in footnote 10, the information revealed by this choice does not affect the incentives of type $$$\tau \ne {{\tau }_n}$$$ voters to re-elect the incumbent prior to period t+1. If the office holder chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ , then voters’ beliefs are given by the posterior $$$\hat{p}$$$ conditional on the office holder's type belonging to $$$\{ {{\tau }_1}, \ldots, {{\tau }_n}\} $$$ , and if she is re-elected, then we move to Phase 3(1) in the following even period. As noted in footnote 9, this revelation of information does not affect the incentives of voters to re-elect the incumbent. If the office holder chooses $$$x \, \notin \, \{ 0,{{x}_{{{\tau }_1}}}\} $$$ , then voters believe she is type $$$\underline {\tau }$$$ , and if she is re-elected, then we move to Phase 5; thus, the office holder will be removed from office, as required. Supposing the incumbent is re-elected in period t, voter beliefs are updated in period t+1 and phases transition as follows. If voter beliefs are that the incumbent is type $$${{\tau }_0}$$$ and she chooses $$$x\, = \,{{x}_{{{\tau }_{n\,{\rm{ - }}\,1}}}}$$$ in period t+1, then beliefs are unchanged, and if she is re-elected, then we move to Phase 3(2) in the next odd period t+2; again, footnote 10 implies that the incumbent will be re-elected. If voter beliefs are that the incumbent is type $$${{\tau }_0}$$$ and she chooses x = 1, then voters believe she is type $$$\underline {\tau }$$$ , and if she is re-elected, then we move to Phase 5, and she is removed from office. If voter beliefs are $$$\hat{p}$$$ and she chooses $$$x\, = \,{{x}_{{{\tau }_{n\,{\rm{ - }}\,1}}}}$$$ , then voter beliefs are given by the posterior conditional on the office holder's type belonging to $$$\{ {{\tau }_1}, \ldots, {{\tau }_{n - 1}}\} $$$ , and if she is re-elected, then we move to Phase 3(2) in the next odd period t+2; and if voter beliefs are $$$\hat{p}$$$ and the incumbent chooses x = 1, then voters believe she is type $$${{\tau }_n}$$$ , and if she is re-elected, then we move to Phase 3(2). Again, footnote 9 implies that the incumbent will be re-elected following either choice. And if the office holder chooses $$$x \, \notin \, \{ {{x}_{{{\tau }_{n - 1}}}},1\} $$$ , then voters believe she is type $$$\underline {\tau }$$$ , and the politician is removed from office. Supposing the incumbent is re-elected in t+1, voter beliefs are again given by Bayes’ rule, as applicable.

Third, after entering Phase 4.2 in an odd period t, if the office holder chooses x = 0, then voters believe that with probability one the office holder is type $$${{\tau }_0}$$$ , and if she is re-elected, then we move to Phase 4.2 in the following even period period t+1. If the office holder chooses $$$x\, = \,{{x}_{{{\tau }_1}}}$$$ , then voter beliefs are given by $$$\hat{p}$$$ , and if she is re-elected, then we also move to Phase 4.2 in the following even period t+1. Again, in either case, type $$$\tau \,\geq \,{{\tau }_1}$$$ voters will avoid transitioning to Phase 2(1) by re-electing the incumbent in period t. Then voter beliefs are updated in period t+1 and phases transition as follows. If voter beliefs are that the incumbent is type $$${{\tau }_0}$$$ and she chooses $$$x\, = \,{{x}_{{{\tau }_{n\,{\rm{ - }}\,1}}}}$$$ in period t+1, then beliefs are unchanged, and when she is re-elected, we move to Phase 4.0 in all subsequent odd periods and to Phase 4.1 in all subsequent even periods, until a further deviation occurs. If voter beliefs are that the incumbent is type $$${{\tau }_0}$$$ and she chooses x = 1, then voters believe she is type $$$\underline\tau $$$ , and if she is re-elected, then we move to Phase 5, and she is removed from office. If voter beliefs are $$$\hat{p}$$$ and she chooses $$$x\, = \,{{x}_{{{\tau }_{n\,{\rm{ - }}\,1}}}}$$$ , then voter beliefs are given by the posterior conditional on the office holder's type belonging to $$$\{ {{\tau }_1}, \ldots, {{\tau }_{n\,{\rm{ - }}\,1}}\} $$$ , and when she is re-elected, we move to Phase 4.1 in all subsequent odd periods and to Phase 4.0 in all subsequent even periods, until a further deviation occurs. If voter beliefs are $$$\hat{p}$$$ and she chooses x = 1, then voters believe that she is type $$${{\tau }_n}$$$ , and when she is re-elected, we move to Phase 4.1 in all subsequent odd periods and to Phase 4.0 in all subsequent even periods. And if the office holder chooses $$$x \, \notin \, \{ {{x}_{{{\tau }_{n\,{\rm{ - }}\,1}}}},1\} $$$ , then voters believe she is type $$$\underline {\tau }$$$ , and she is removed from office. Supposing the incumbent is re-elected in t+1, voter beliefs are again updated via Bayes’ rule, as applicable.

In effect, this splicing of strategies punishes a deviation in an odd period t by following the punishments from Step 1 in all future odd periods, and by following Step 2 in all future even periods, simulating punishments as though an analogous deviation had occurred in the previous even period t − 1. A similar construction is used following a deviation in an even period. This establishes a perfect Bayesian electoral equilibrium with path of play $$$(a,b,a,b, \ldots )$$$ and provides a splicing procedure that can be iterated generally.

For the induction step, assume the claim is true for ℓ − 1, choose any $$$k\, = \,0,1, \ldots {{,2}^\ell } $$$ , and write $$$k\, = \,k^{\prime}\, + \,k^{\prime\prime}$$$ , where $$$k^{\prime},k^{\prime\prime}\leq {{2}^{\ell \,{\rm{ - }}\,1}} $$$ . By the induction hypothesis, there exists $$$\delta ^{\prime} \in (0,1)$$$ such that for all $$$\delta \in (\delta ^{\prime},1)$$$ , there exists $$${\bf {{y}}} \in P(\delta )$$$ such that for all $$$m = 0,1,2, \ldots $$$ ,

$$\matrix{ \!\!\!\!\!\!\!\!\!\!\!\!\!\# \{ t&#x007C;m{{2}^{\ell \,{\rm{ - }}\,1}} \, + \,1\,\leq \,t\,\leq \,(m\, + \,{{1)2}^{\ell \,{\rm{ - }}\,1}} \;{\rm and}\;\:{{y}_t}\, = \,a\} \, = \,k^{\prime} \cr \# \{ t&#x007C;m{{2}^{\ell \,{\rm{ - }}\,1}} \, + \,1\,\leq \,t\,\leq \,(m\, + \,{{1)2}^{\ell \,{\rm{ - }}\,1}} \:\;{\rm and}\;\:{{y}_t}\, = \,b\} \, = \,{{2}^\ell } \,{\rm{ - }}\,k^{\prime}. \\ $$

And there exists $$$\delta ^{\prime\prime} \in (0,1)$$$ such that for all $$$\delta \in (\delta ^{\prime\prime},1)$$$ , there exists $$${\bf {{z}}} \in P(\delta )$$$ such that for all $$$m\, = \,0,1,2, \ldots $$$ ,

$$\matrix {\!\!\!\!\!\!\!\!\!\!\!\!\!\# \{ t&#x007C;m{{2}^{\ell \,{\rm{ - }}\,1}} \, + \,1\leq t\leq (m\, + \,{{1)2}^{\ell \,{\rm{ - }}\,1}} \:\;{\rm and}\;\:{{z}_t}\, = \,a\} \, = \,k^{\prime\prime} \cr \# \{ t&#x007C;m{{2}^{\ell \,{\rm{ - }}\,1}} \, + \,1\leq t\leq (m\, + \,{{1)2}^{\ell \,{\rm{ - }}\,1}} \:\;{\rm and}\;\:{{z}_t}\, = \,b\} \, = \,{{2}^\ell } \,{\rm{ - }}\,k^{\prime\prime}. \\ $$

We then set $$$\tilde{\delta }\, = \,\sqrt {\max \{ \delta ^{\prime},\delta ^{\prime\prime}\} } $$$ , and for every $$$\delta \, \gt \,\tilde{\delta }$$$ , we splice the equilibria supporting paths x and y according to the above procedure. This yields a perfect Bayesian electoral equilibrium that generates the path $$${\bf {{x}}}\, = \,({{x}_1},{{x}_2}, \ldots )\, = \,({{y}_1},{{z}_1},{{y}_2},{{z}_2}, \ldots )$$$ from the initial history. Formally, we define x by

$${{x}_t}\, = \,\left\{ \matrix {{{{y}_{\frac{{t\, + \,1}}{2}}}} \hfill &amp;#x0026; {\:{\rm if}\;t{\rm \;is\;odd},\thinsp} \hfill \cr {{{z}_{\frac{t}{2}}}} \hfill &amp;#x0026; {\:{\rm if}\;t{\rm \;is\;even}.\:} \hfill \\$$

Then for all $$$m\, = \,0,1,2, \ldots $$$ , we have

$$\matrix{ \hskip-5pc \! \# \{ t&#x007C;m{{2}^\ell } \, + \,1\leq t\leq (m\, + \,{{1)2}^\ell } \:\;{\rm and}\;\:{{x}_t}\, = \,a\} \cr = \# \{ t&#x007C;m{{2}^{\ell \,{\rm{ - }}\,1}} \, + \,1\leq t\leq (m\, + \,{{1)2}^{\ell \,{\rm{ - }}\,1}}, \:\;t\;{\rm odd},\;{\rm and}\;{{y}_t}\, = \,a\:\} \cr + \# \{ t&#x007C;m{{2}^{\ell \,{\rm{ - }}\,1}} \, + \,1\leq t\leq (m\, + \,{{1)2}^{\ell \,{\rm{ - }}\,1}}, \:\;t\;{\rm even},\;{\rm and}\;{{z}_t}\, = \,a\:\} \cr \hskip-16pc \ \!\! = k^{\prime}\, + \,k^{\prime\prime} \cr \hskip-18pc = k, \\ $$

where the first equality follows by noting that when t (odd) ranges from $$$m{{2}^\ell } \, + \,1$$$ to $$$(m\, + \,{{1)2}^\ell } \,{\rm{ - }}\,1$$$ , the ratio $$$\frac{{t\, + \,1}}{2}$$$ ranges from $$$m{{2}^{\ell \,{\rm{ - }}\,1}} \, + \,1$$$ to $$$(m\, + \,{{1)2}^{\ell \,{\rm{ - }}\,1}} $$$ , and similarly, when t (even) ranges from $$$m{{2}^\ell } \, + \,2$$$ to $$$(m\, + \,{{1)2}^\ell } $$$ , the ratio $$$\frac{t}{2}$$$ ranges from $$$m{{2}^{\ell \,{\rm{ - }}\,1}} \, + \,1$$$ to $$$(m\, + \,{{1)2}^{\ell \,{\rm{ - }}\,1}} $$$ . This establishes the desired perfect Bayesian equilibrium.

The final step of the proof uses the equilibria from Step 3 to punish deviations from arbitrary paths through a region of the policy space that becomes arbitrarily large as citizens become patient.

Step 4: For all $$$a,b \in X$$$ , with $$$0\, \lt \,a\, \lt \,b\, \lt \,1$$$ , there exists $$$\overline{\delta } \in (0,1)$$$ such that for all $$$\delta \in (\overline{\delta }, 1)$$$ , we have $$${{[a,b]}^\infty } \subseteq P(\delta )$$$ .

Consider $$$a,b \in X$$$ with $$$0\, \lt \,a\, \lt \,b\, \lt \,1$$$ . Fix $$${\epsilon}\, = \,\frac{1}{2}\min \{ a,1\,{\rm{ - }}\,b\} $$$ , and let $$$a^{\prime} \in (0,{\epsilon})$$$ and $$$b^{\prime} \in (1\,{\rm{ - }}\,{\epsilon},1)$$$ be as in Step 3. For each $$$x \in [a,b]$$$ , we can write $$$x\, = \,\alpha (x)a^{\prime}\, + \,(1\,{\rm{ - }}\,\alpha (x))b^{\prime}$$$ , where

$$\alpha (x)\, = \,\frac{{b^{\prime}\,{\rm{ - }}\,x}}{{b^{\prime}\,{\rm{ - }}\,a^{\prime}}},$$

and then we have

$$0\,\, \lt \,\,\frac{{b^{\prime}\,{\rm{ - }}\,b}}{{b^{\prime}\,{\rm{ - }}\,a^{\prime}}}\,\,\leq \,\,\alpha (x)\,\,\leq \,\,\frac{{b^{\prime}\,{\rm{ - }}\,a}}{{b^{\prime}\,{\rm{ - }}\,a^{\prime}}}\,\, \lt \,\,1.$$

Since each $$${{u}_\tau }$$$ is strictly concave, it follows that for all $$$x \in [a,b]$$$ , we have $$${{u}_\tau }(x)\, \gt \,\alpha (x){{u}_\tau }(a^{\prime})\, + \,(1\,{\rm{ - }}\,\alpha (x)){{u}_\tau }(b^{\prime})$$$ . More generally, let Λ denote the set of probability measures λ with support contained in [a, b], and note that for all λ ∈ Λ, we have

$$\matrix{ \int {{u}_\tau }(x)\lambda (dx) \gt \int [\alpha (x){{u}_\tau }(a^{\prime})\, + \,(1\,{\rm{ - }}\,\alpha (x)){{u}_\tau }(b^{\prime})]\lambda (dx) \cr \hskip-5pc= \alpha (E[\lambda ]){{u}_\tau }(a^{\prime})\, + \,(1\,{\rm{ - }}\,\alpha (E[\lambda ])){{u}_\tau }(b^{\prime}). \\ $$

Therefore, since $$${{u}_\tau }$$$ is continuous and Λ is compact in the weak* topology, we have

$$\mathop {\min }\limits_{\lambda \in \Lambda } \left[ {\int {{u}_\tau }(x)\lambda (dx)\,{\rm{ - }}\,\alpha (E[\lambda ]){{u}_\tau }(a^{\prime})\,{\rm{ - }}\,(1\,{\rm{ - }}\,\alpha (E[\lambda ])){{u}_\tau }(b^{\prime})} \right] \equiv \eta \, \gt \,0.$$

For all λ ∈ Λ, we can approximate α(E[λ]) by $$$\frac{k}{{{{2}^\ell } }}$$$ through choice of the natural numbers ℓ and $$$k\, = \,0,1, \ldots {{,2}^\ell } $$$ . Thus, there exist k and ℓ such that

$$\int {{u}_\tau }(x)\lambda (dx)\, \gt \,\left( {\frac{k}{{{{2}^\ell } }}} \right){{u}_\tau }(a^{\prime})\, + \,\left( {1\,{\rm{ - }}\,\frac{k}{{{{2}^\ell } }}} \right){{u}_\tau }(b^{\prime})\, + \,\frac{\eta }{2}.$$

Define Λ^k,ℓ as the set of probability measures λ ∈ Λ such that Inequality 16 holds. It follows that $$${{\{ {{\Lambda }^{k,\ell }} \} }_{k,\ell }}$$$ is an open covering of Λ, and since the latter set is compact, it possesses a finite subcover. Let F be a finite set of (k, ℓ) pairs, with ℓ a natural number and $$$k \in \{ 0,1,2, \ldots {{,2}^\ell } \} $$$ , such that $$$\Lambda \subseteq \bigcup\nolimits_{(k,\ell ) \in F} {{\Lambda }^{k,\ell }} $$$ . By repeated application of Step 3, we can choose $$$\tilde{\delta } \in (0,1)$$$ such that for all $$$\delta \in (\tilde{\delta },1)$$$ and all $$$(k,\ell ) \in F$$$ , there exists $$${{{\bf {{x}}}}^{k,\ell }} \in P(\delta )$$$ such that for all $$$m = 0,1,2, \ldots $$$ ,

$$\matrix {\!\!\!\!\!\!\!\!\!\!\!\!\!\# \{ t&#x007C;m{{2}^\ell } \, + \,1\,\leq \,t\,\leq \,(m\, + \,{{1)2}^\ell } \:\;{\rm and}\;\:{{x}_t}\, = \,a^{\prime}\} \, = \,k \cr \# \{ t&#x007C;m{{2}^\ell } \, + \,1\,\leq \,t\,\leq \,(m\, + \,{{1)2}^\ell } \:\;{\rm and}\;\:{{x}_t}\, = \,b^{\prime}\} \, = \,{{2}^\ell } \,{\rm{ - }}\,k. \\ $$

Given any pair (k, ℓ), discount factor δ < 1, and type $$$\tau $$$ , the discounted sum of utility from $$${{{\bf {{x}}}}^{k,\ell }} $$$ , denoted $$$U_{\tau }^{{k,\ell }} $$$ , satisfies

$$ U_{\tau }^{{k,\ell }} \, = \,\mathop{\sum}\limits_{t\, = \,1}^{{{2}^\ell } } {{\delta }^{t\,{\rm{ - }}\,1}} [{{I}_{a^{\prime}}}(x_{t}^{{k,\ell }} ){{u}_\tau }(a^{\prime})\, + \,{{I}_{b^{\prime}}}(x_{t}^{{k,\ell }} ){{u}_\tau }(b^{\prime})]\, + \,{{\delta }^{{{2}^\ell } }} U_{\tau }^{{k,\ell }}, $$

and therefore, after normalizing by (1 − δ), we have

$$(1\,{\rm{ - }}\,\delta )U_{\tau }^{{k,\ell }} = \frac{{1\,{\rm{ - }}\,\delta }}{{1\,{\rm{ - }}\,{{\delta }^{{{2}^\ell } }} }}\mathop{\sum}\limits_{t\, = \,1}^{{{2}^\ell } } {{\delta }^{t\,{\rm{ - }}\,1}} [{{I}_{a^{\prime}}}(x_{t}^{{k,\ell }} ){{u}_\tau }(a^{\prime})\, + \,{{I}_{b^{\prime}}}(x_{t}^{{k,\ell }} ){{u}_\tau }(b^{\prime})],$$

where $$${{I}_{a^{\prime}}}$$$ and $$${{I}_{b^{\prime}}}$$$ are the indicator functions for a′ and b′. Taking the limit as δ goes to one and applying l'Hôpital's rule, we have

$$\matrix{ \mathop {\lim }\limits_{\delta \uparrow 1} (1\,{\rm{ - }}\,\delta )U_{\tau }^{{k,\ell }} \, = \,\frac{1}{{{{2}^\ell } }}\mathop{\sum}\limits_{t\, = \,1}^{{{2}^\ell } } [{{I}_{a^{\prime}}}(x_{t}^{{k,\ell }} ){{u}_\tau }(a^{\prime})\, + \,{{I}_{b^{\prime}}}(x_{t}^{{k,\ell }} ){{u}_\tau }(b^{\prime})] \cr = \left( {\frac{k}{{{{2}^\ell } }}} \right){{u}_\tau }(a^{\prime})\, + \,\left( {1\,{\rm{ - }}\,\frac{k}{{{{2}^\ell } }}} \right){{u}_\tau }(b^{\prime}). \\ $$

Now let Λ^δ consist of all probability measures λ ∈ Λ such that for some $$$(k,\ell ) \in F$$$ , we have

$$\int {{u}_\tau }(x)\lambda (dx)\, \gt \,(1\,{\rm{ - }}\,\delta )U_{\tau }^{{k,\ell }} \, + \,\frac{\eta }{4}.$$

Since $$$\Lambda \subseteq \bigcup\nolimits_{(k,\ell ) \in F} {{\Lambda }^{k,\ell }} $$$ and $$$(1\,{\rm{ - }}\,\delta )U_{\tau }^{{k,\ell }} \rightarrow \left( {\frac{k}{{{{2}^\ell } }}} \right){{u}_\tau }(a^{\prime})\, + \,\left( {1\,{\rm{ - }}\,\frac{k}{{{{2}^\ell } }}} \right){{u}_\tau }(b^{\prime})$$$ , it follows from Inequality 16 that for each λ ∈ Λ, we have $$$\lambda \in {{\Lambda }^\delta } $$$ for δ sufficiently close to one. That is, $$$\{ {{\Lambda }^\delta } &#x007C;\delta \in (0,1)\} $$$ is an open covering of Λ. Again, since Λ is compact, this collection possesses a finite subcover, indexed by a finite set $$$D \subseteq (0,1)$$$ . Therefore, setting $$$\delta ^{\prime}\, = \,\max D\, \lt \,1$$$ , it follows that for all $$$\delta \in (\delta ^{\prime},1)$$$ and all $$$\lambda \in \Lambda $$$ , there exists $$$(k,\ell ) \in F$$$ such that Inequality 17 holds. Furthermore, choose $$$\overline{\delta } \in (\delta ^{\prime},1)$$$ such that

$$\rDelta \leq \left( {\frac{{\overline{\delta } }}{{1\,{\rm{ - }}\,\overline{\delta } }}} \right)\left( {\frac{\eta }{4}} \right).$$

Finally, consider any $$$\delta \in (\overline{\delta }, 1)$$$ and any $$${\bf {{x}}} \in {{[a,b]}^\infty } $$$ . I specify a perfect Bayesian equilibrium that generates x from the initial history by again defining actions and beliefs in each of a number of phases and by specifying a rule for transitioning between phases. Now, however, the procedure is simpler. Phase 0 will represent the path of play from the initial history, while a countable number of phases, $$$t = 1,2, \ldots $$$ , are used as punishments. Specifically, Phase t is used to punish deviations from Phase 0 in period t, while further deviations from phase t are handled implicitly in Step 3. Note that for each t, we can write the discounted utility of a type $$$\tau $$$ citizen from $$$({{x}_t},{{x}_{t + 1}},{{x}_{t + 2}}, \ldots )$$$ as the integral with respect to a probability measure $$${{\lambda }_t} \in \rLambda $$$ ; that is,

$$\mathop{\sum}\limits_{t^{\prime}\, = \,t}^\infty {{\delta }^{t^{\prime}\,{\rm{ - }}\,t}} {{u}_\tau }({{x}_{t^{\prime}}})\, = \,\frac{1}{{1\,{\rm{ - }}\,\delta }}\int {{u}_\tau }(x){{\lambda }_t}(dx),$$

where λ_t is defined by

$${{\lambda }_t}(Y)\, = \,(1\,{\rm{ - }}\,\delta )\mathop{\sum}\limits_{t^{\prime}:{{x}_{t^{\prime}}} \in Y} {{\delta }^{t^{\prime}\,{\rm{ - }}\,t}} $$

for all measurable $$$Y \subseteq [0,1]$$$ . It follows that there exists $$$(k,\ell ) \in F$$$ such that Inequality 17 holds with λ = λ_t. To define the equilibrium, we specify that in period t of Phase 0, all types of office holder choose x_t, and voters’ beliefs along the path of play are given by the prior p, and all voter types vote to re-elect the incumbent. If the office holder chooses $$$x^{\prime} \ne {{x}_t}$$$ , then voter beliefs are again given by the prior. In Phase $$$t\, = \,1,2, \ldots $$$ , let $$$(k,\ell ) \in F$$$ satisfy Inequality 17 with λ = λ_{t +1}, and use Step 3 to specify strategies and beliefs so that they form a perfect Bayesian electoral equilibrium that generates the path $$${{{\bf {{x}}}}^{k,\ell }} $$$ beginning in period t+1; that is, the path beginning in period t+1 is $$$(x_{1}^{{k,\ell }}, x_{2}^{{k,\ell }}, \ldots )$$$ . The initial history is labeled Phase 0, and in period t of Phase 0, if it is not the case that the office holder chooses x_t and is re-elected, then transition to the punishment Phase t+1 occurs in the next period t+1.

To verify that this specification constitutes a perfect Bayesian electoral equilibrium, note that the discounted payoff to a type $$$\tau $$$ citizen from a one-shot deviation in period t of Phase 0 is bounded above by $$${\max }_{x \in X} {{u}_\tau }(x) + \delta U_{\tau }^{{k,\ell }} $$$ . Furthermore,

$$\matrix{\mathop {\max }\limits_{x \in X} {{u}_\tau }(x)\, + \,\delta U_{\tau }^{{k,\ell }} \hfill \cr \leq \mathop {\min }\limits_{x \in X} {{u}_\tau }(x)\, + \,\left( {\frac{\delta }{{1\,{\rm{ - }}\,\delta }}} \right)\left( {\frac{\eta }{4}} \right)\, + \,\delta U_{\tau }^{{k,\ell }} \hfill \cr = \mathop {\min }\limits_{x \in X} {{u}_\tau }(x)\, + \,\frac{\delta }{{1\,{\rm{ - }}\,\delta }}\left[ {\frac{\eta }{4}\, + \,(1\,{\rm{ - }}\,\delta )U_{\tau }^{{k,\ell }} } \right] \hfill \cr \!\lt \mathop {\min }\limits_{x \in X} {{u}_\tau }(x)\, + \,\frac{\delta }{{1\,{\rm{ - }}\,\delta }}\int {{u}_\tau }(x){{\lambda }_{t\, + \,1}}(dx) \cr \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\leq \mathop{\sum}\limits_{t^{\prime}\, = \,t}^\infty {{\delta }^{t^{\prime}\,{\rm{ - }}\,t}} {{u}_\tau }({{x}_{t^{\prime}}}), \\ $$

where the inequalities follow from 18, 17 and 19, respectively. Therefore, one-shot deviations are unprofitable, and we conclude that there exists a perfect Bayesian equilibrium that generates the path x from the initial history.