Given parties’ choices
\(x_L\) and
\(x_R\), let
\({\hat{\theta }}_{x_L,x_R}\) be the associated cutpoint outcome,
\({\theta }^*_{x_L,x_R}(n)\) the equilibrium cutpoint of the voting game when the expected number of voters is
n, and
\(X(\theta ^*_{x_L,x_R}(n))\) the corresponding expected outcome, i.e.
\(X(\theta ^*_{x_L,x_R}(n))= F(\theta ^*_{x_L,x_R}(n)) x_L + [1 - F(\theta ^*_{x_L,x_R}(n))] x_R\). Note that, for every
\(x_L < x_R\), the cutpoint outcome
\({\hat{\theta }}_{x_L,x_R}\) is strictly increasing in
\(x_L\) and in
\(x_R\). In fact, we have
$$\begin{aligned} \frac{\partial {\hat{\theta }}_{x_L,x_R}}{\partial x_L}=\frac{F({\hat{\theta }}_{x_L,x_R})}{1+ F'({\hat{\theta }}_{x_L,x_R}) (x_R-x_L)} \quad \text { and } \quad \frac{\partial {\hat{\theta }}_{x_L,x_R}}{\partial x_R}=\frac{1-F({\hat{\theta }}_{x_L,x_R})}{1+ F'({\hat{\theta }}_{x_L,x_R}) (x_R-x_L)}. \end{aligned}$$
Both derivatives are strictly positive when
\({\hat{\theta }}_{x_L,x_R} \notin \{0,1\}\) since
F is strictly increasing. By Theorem
2, we know that for every
\({\delta }>0\) there exists a value
\(\bar{n} \in {\mathbb {R}}\) such that, for every
\(n\ge \bar{n}\),
$$\begin{aligned} \left| {\hat{\theta }}_{x_L,x_R} - X(\theta ^*_{x_L,x_R}(n)) \right| < {\delta }. \end{aligned}$$
Furthermore, let
\({\tilde{X}}(\theta ^*_{x_L,x_R}(n))\) denote the random policy outcome given the equilibrium cutpoint
\(\theta ^*_{x_L,x_R}(n)\). Given the continuity of the outcome function in the shares of votes, Lemma
5 implies that
\({\tilde{X}}(\theta ^*_{x_L,x_R}(n))\) converges in probability to its expected value. That is, for every
\(\delta >0\) and
\(\varepsilon >0\) there exists a value
\(\bar{n} \in {\mathbb {R}}\) such that, for every
\(n \ge \bar{n}\),
$$\begin{aligned} \text {Pr}\left( \left| {\tilde{X}}(\theta ^*_{x_L,x_R}(n)) - X(\theta ^*_{x_L,x_R}(n))\right| > \varepsilon \right) < \delta . \end{aligned}$$
We can now show that, for
n sufficiently large and for every choice
\(x_R\) of party
R, choosing a position
\(x_L>0\) is never a best response for party
L, as
\(x_L=0\) gives her a higher payoff. To this end, fix party
R’s choice
\(x_R\) and let
\(\bar{x}_L > 0\). Since the cutpoint outcome is strictly increasing in
\(x_L\), we have
\({\hat{\theta }}_{\bar{x}_L,x_R} > {\hat{\theta }}_{0,x_R}\). Let
\(\eta = {\hat{\theta }}_{\bar{x}_L,x_R} - {\hat{\theta }}_{0,x_R}\). We are going to split
\(\eta\) into five intervals (all equal for simplicity) and let
n increase. Two intervals will bound the distances between the two expected outcomes, given
\(\bar{x}_L\) and given 0, and the corresponding cutpoint outcomes. Two other intervals will bound the distances between the two actual outcomes and the corresponding expected outcomes. So the last interval will be the minimum distance between any two realized outcomes given the two different choices of party
L. Formally, given the results above, there exists a value
\(n_1 \in {\mathbb {R}}\) such that, for every
\(n\ge n_1\),
$$\begin{aligned} \left| {\hat{\theta }}_{\bar{x}_L,x_R} - X(\theta ^*_{\bar{x}_L,x_R}(n)) \right|< \frac{\eta }{5} \quad \text { and } \quad \left| {\hat{\theta }}_{0,x_R} - X(\theta ^*_{0,x_R}(n)) \right| < \frac{\eta }{5}. \end{aligned}$$
Moreover, for every
\(\delta >0\) there exists a value
\(n_2 \in {\mathbb {R}}\) such that, for every
\(n \ge n_2\),
$$\begin{aligned} \text {Pr}\left( \left| {\tilde{X}}(\theta ^*_{\bar{x}_L,x_R}(n)) - X(\theta ^*_{\bar{x}_L,x_R}(n)) \right|< \frac{\eta }{5} \cup \left| {\tilde{X}}(\theta ^*_{0,x_R}(n)) - X(\theta ^*_{0,x_R}(n)) \right| < \frac{\eta }{5} \right) > 1 - \delta . \end{aligned}$$
Let
\(u_L(x)\) be the payoff that party
L attains when the realized outcome is
x, and
\(U_L(x_L,x_R)\) her expected payoff of choosing
\(x_L\) given party
R’s choice
\(x_R\). Let
\(n \ge {n}_{\bar{x}_L,x_R} = \max \{n_1,n_2\}\). Note that
\(u_L \left( {\hat{\theta }}_{0,x_R} + \frac{2}{5}\eta \right) - u_L \left( {\hat{\theta }}_{\bar{x}_L,x_R} -\frac{2}{5}\eta \right) = k > 0\) is a lower bound of the gain that party
L gets choosing 0 instead of
\(\bar{x}_L\) when the outcomes realized in the two cases are no further than
\(\frac{2}{5}\eta\) from the corresponding cutpoints. Given the above discussion, this happens with probability greater than
\(1-\delta\). On the other hand,
\(u_L(1) - u_L(0) = h < 0\) is (in absolute value) an upper bound of the loss that the party can incur in every other case, which happens with probability smaller than
\(\delta\). Thus, let
\({\delta } = \frac{k}{2(k-h)}>0\). We have
$$\begin{aligned} U_L(0,x_R) - U_L(\bar{x}_L,x_R)> & {} (1-{\delta }) \left[ u_L \left( {\hat{\theta }}_{0,x_R} +\frac{2}{5}\eta \right) - u_L \left( {\hat{\theta }}_{\bar{x}_L,x_R} -\frac{2}{5}\eta \right) \right] \\&+\, {\delta } [u_L(1) - u_L(0)] = (1-{\delta })k + {\delta } h = \frac{k}{2} > 0, \end{aligned}$$
so party
L prefers to choose 0 rather than
\(\bar{x}_L\) given party
R’s choice
\(x_R\). Since this holds for every
\(x_R\) and every
\(\bar{x}_L>0\), it follows that there exists a value
\(\bar{n}_L\in {\mathbb {R}}\) such that, for
\(n\ge \bar{n}_L\), party
L has a unique best response against any choice of party
R, that is choosing
\(x_L=0\). An analogous argument proves that there exists a value
\(\bar{n}_R\in {\mathbb {R}}\) such that, for
\(n\ge \bar{n}_R\), party
R has a unique best response against any choice of party
L, that is choosing
\(x_R=1\). It follows that, for
\(n \ge \bar{n} = \max \{ \bar{n}_L, \bar{n}_R \}\), (0, 1) is the unique Nash equilibrium.
\(\square\)