Electoral competition under costly policy implementation

Abstract

This paper analyzes a unidimensional electoral competition model between two policy-motivated candidates, assuming that the elected candidate has to incur an idiosyncratic policy implementation cost. Think of this as the election of chairman in an academic department, a parent-teacher association, or a condominium association board, where the election winner perceives office holding as a cost, and not benefit. We prove that in such a game, a pure strategy Nash equilibrium is guaranteed to exist even when the policy implementation costs are heterogeneous, as long as they are not very large. We also provide the equilibrium characterization, comparative statics, and welfare analysis in the case of Euclidean preferences. In particular, we show that equilibrium strategies are divergent and in general not symmetric around the expected bliss point of the median voter. A higher policy implementation cost makes a candidate propose a more extreme policy, and lowers his electoral chances. Naturally, the voter’s welfare decreases when the candidates’ implementation costs increase.

Appendices

Appendix

Proof of Proposition 1

To prove the equilibrium existence, we define a restricted game. We define a positive number $d=\max \{-k_{1},k_{2}\}$ and assume that candidate 1 is allowed to announce a policy in $[\alpha _{1},-d]$ and that candidate 2 is allowed to announce a policy in $[d,\alpha _{2}]$. The value $k_{i}\$is given by $v(\left| k_{i}-\alpha _{i}\right| )-C_{i}=v(\left| -k_{i}-\alpha _{i}\right| )$. It pins down the location beyond which candidate i would unambiguously decrease his payoff when approaching further his symmetrically located competitor. Assumption 2 guarantees that, for sufficiently small policy implementation costs: (a) a unique $k_{i}$ exists for each i; (b) it is more moderate than i’s ideal policy; and (c) it monotonically converges to zero when $C_{i}$ goes to zero.

Our aim is first to show that the restricted game admits a pure strategy equilibrium and then to argue that this equilibrium is also an equilibrium of the unrestricted version of the game, i.e., when both candidates are free to propose any policy in $\mathbb {R}$. Obviously, the restricted game exists only when the implementation costs are sufficiently small. Formally, only when $d<\min \{-\alpha _{1},\alpha _{2}\}$.

• Step 1  In the restricted game, we have that

  $$\begin{aligned} \frac{\partial \Pi _{1}\left( x_{1},x_{2}\right) }{\partial x_{1}}= & {} v^{\prime }(x_{1}-\alpha _{1})F\left( \frac{x_{1}+x_{2}}{2}\right) \nonumber \\&+\frac{ 1}{2}(-C_{1}+v(x_{1}-\alpha _{1})-v(x_{2}-\alpha _{1}))f\left( \frac{ x_{1}+x_{2}}{2}\right) . \end{aligned}$$

  We observe that $\frac{\partial \Pi _{1}\left( x_{1},x_{2}\right) }{\partial x_{1}}\mid _{x_{1}=\hat{x}_{1}}=0$ if and only if

  $$\begin{aligned} \frac{v^{\prime }(x_{1}-\alpha _{1})}{\frac{1}{2}(C_{1}-v(x_{1}-\alpha _{1})+v(x_{2}-\alpha _{1}))}=\frac{f\left( \frac{x_{1}+x_{2}}{2}\right) }{ F\left( \frac{x_{1}+x_{2}}{2}\right) } \end{aligned}$$

  (A.1)

  for some $x_{1}=\hat{x}_{1}$. Derivating the left-hand side of (A.1) with respect to $x_{1}$ yields

  $$\begin{aligned}&\frac{\partial }{\partial x_{1}}\left( \dfrac{v^{\prime }(x_{1}-\alpha _{1}) }{\frac{1}{2}(C_{1}-v(x_{1}-\alpha _{1})+v(x_{2}-\alpha _{1}))}\right) \\&\quad = \frac{v^{\prime }(x_{1}-\alpha _{1})^{2}+(C_{1}-v(x_{1}-\alpha _{1})+v(x_{2}-\alpha _{1}))v^{\prime \prime }(x_{1}-\alpha _{1})}{\frac{1}{2} (C_{1}-v(x_{1}-\alpha _{1})+v(x_{2}-\alpha _{1}))^{2}}, \end{aligned}$$

  which is strictly positive due to Assumption 2. Therefore, the left-hand side of (A.1) is strictly increasing in $x_{1}$. The right-hand side of (A.1) is strictly decreasing in $x_{1}$ (due to Assumption 1). Therefore, for any fixed $x_{2}\in [d,\alpha _{2}]$, there is at most one critical value $\hat{x}_{1}\in [\alpha _{1},-d]$ such that $\Pi _{1}\left( x_{1},x_{2}\right)$ is strictly increasing (decreasing) for any $x_{1}<\hat{x}_{1}$ ($x_{1}>\hat{x}_{1}$). In other words, $\Pi _{1}\left( x_{1},x_{2}\right)$ is strictly quasi-concave in $x_{1}$ in the restricted game for any $x_{2}\in [d,\alpha _{2}]$. By analogy and given the symmetry of the candidates’ utility functions, $\Pi _{2}\left( x_{1},x_{2}\right)$ is strictly quasi-concave in $x_{2}$ in the restricted game for any $x_{1}\in [\alpha _{1},-d]$. Therefore, by Debreu (1952) it follows that this restricted game admits a Nash equilibrium in pure strategies, $(x_{1}^{*},x_{2}^{*})$. Next, we argue that when $d\rightarrow 0$, none of the equilibrium strategies of the restricted game, $x_{1}^{*}$ and $x_{2}^{*}$, converges to zero.

• Step 2  Assume that $d\rightarrow 0$ and that the candidates play a Nash equilibrium of the restricted game, $(x_{1}^{*},x_{2}^{*})$. If $(x_{1}^{*},x_{2}^{*})\rightarrow \left( 0,0\right)$, then candidate 1 can deviate to $x_{1}=\alpha _{1}$ and get a strictly larger payoff. This is so because: (i) candidate 1’s payoff that relates to the policy is larger in profile $\left( \alpha _{1},0\right)$ than in profile (0, 0),

  $$\begin{aligned} F\left( \frac{\alpha _{1}}{2}\right) v\left( 0\right) +\left[ 1-F\left( \frac{ \alpha _{1}}{2}\right) \right] v\left( -\alpha _{1}\right) >v\left( -\alpha _{1}\right) , \end{aligned}$$

  as $v\left( 0\right) >v\left( -\alpha _{1}\right)$ and $F\left( \frac{ \alpha _{1}}{2}\right) \in (0,1)$ due to $F\left( \cdot \right)$ having full support; and (ii) candidate 1’s payoff that relates to the disutility of being in office is larger in profile $\left( \alpha _{1},0\right)$ than in (0, 0),

  $$\begin{aligned} -F\left( \frac{\alpha _{1}}{2}\right) C_{1}>-\frac{1}{2}C_{1}, \end{aligned}$$

  due to $F\left( \cdot \right)$ having full support and being symmetric around zero. Thus,

  $$\begin{aligned} \Pi _{1}\left( \alpha _{1},0\right)= & {} F\left( \frac{\alpha _{1}}{2}\right) v\left( 0\right) +\left[ 1-F\left( \frac{\alpha _{1}}{2}\right) \right] v\left( -\alpha _{1}\right) -F\left( \frac{\alpha _{1}}{2}\right) C_{1}>v\left( -\alpha _{1}\right) -\frac{1}{2}C_{1}\\= & {} \Pi _{1}\left( 0,0\right) . \end{aligned}$$

  Similarly, if $(x_{1}^{*},x_{2}^{*})\rightarrow \left( 0,b\right)$ with $b>0$ as $d\rightarrow 0$ then candidate 1 can deviate to some non-degenerate (i.e., not converging to zero) $k<0$ and get a strictly larger payoff. This is so because Assumption 1 implies that, for any $\alpha _{1}<0$, $\Pi _{1}\left( x_{1},b\right)$ is strictly decreasing with $x_{1}\in [k,0]$ for some non-degenerate $k<0$. By analogy and given the symmetry of the candidates’ utility functions, if $(x_{1}^{*},x_{2}^{*})\rightarrow \left( b,0\right)$ with $b<0$ as $d\rightarrow 0$ then candidate 2 can deviate to some non-degenerate $k>0$ and get a strictly larger payoff. Therefore, there exists $\bar{d}>0$ such that whenever $d<\bar{d}$, any Nash equilibrium of the restricted game $(x_{1}^{*},x_{2}^{*})$ is such that $(x_{1}^{*},x_{2}^{*})\in [\alpha _{1},-d)\times (d,\alpha _{2}]$ (i.e., none of the equilibrium strategies of the restricted game converges to zero). Next, we argue that if $(x_{1}^{*},x_{2}^{*})$ is an equilibrium of the restricted game, then a best response of candidate 1 in the unrestricted game to candidate 2 playing $x_{2}^{*}$ cannot be strictly to the left of his ideal policy $\alpha _{1}$. Similarly, the set of best responses of candidate 2 in the unrestricted game when candidate 1 plays $x_{1}^{*}$ does not contain policies strictly to the right of $\alpha _{2}$.

• Step 3  (By contradiction) Assume that $x_{1}<\alpha _{1}$. Then

  $$\begin{aligned} \frac{\partial \Pi _{1}\left( x_{1},x_{2}^{*}\right) }{\partial x_{1}}= & {} -v^{\prime }(-x_{1}+\alpha _{1})F\left( \frac{x_{1}+x_{2}^{*}}{2} \right) \\&+\,\frac{1}{2}(-C_{1}+v(-x_{1}+\alpha _{1})-v(x_{2}^{*}-\alpha _{1}))f\left( \frac{x_{1}+x_{2}^{*}}{2}\right) . \end{aligned}$$

  We observe that $\frac{\partial \Pi _{1}\left( x_{1},x_{2}^{*}\right) }{ \partial x_{1}}\mid _{x_{1}=\hat{x}_{1}}=0$ if and only if

  $$\begin{aligned} \frac{v^{\prime }(-x_{1}+\alpha _{1})}{\frac{1}{2}(-C_{1}+v(-x_{1}+\alpha _{1})-v(x_{2}^{*}-\alpha _{1}))}=\frac{f\left( \frac{x_{1}+x_{2}^{*} }{2}\right) }{F\left( \frac{x_{1}+x_{2}^{*}}{2}\right) } \end{aligned}$$

  for some $x_{1}=\hat{x}_{1}$. The left-hand side is strictly increasing in $x_{1}$ (by a similar reasoning to that in step 1) while the right-hand side is strictly decreasing in $x_{1}$ (due to Assumption 1). Therefore, there exists at most one $\hat{x}_{1}<\alpha _{1}$ for which $\frac{\partial \Pi _{1}\left( x_{1},x_{2}\right) }{\partial x_{1}}\mid _{x_{1}=\hat{x} _{1}}=0$. Given that

  $$\begin{aligned} \lim _{x_{1}\rightarrow \alpha _{1}^{-}}\frac{\partial \Pi _{1}\left( x_{1},x_{2}^{*}\right) }{\partial x_{1}}>0, \end{aligned}$$

  $\Pi _{1}\left( x_{1},x_{2}^{*}\right)$ is strictly increasing (decreasing) for any $x_{1}>\hat{x}_{1}$ ($x_{1}<\hat{x}_{1}$), and so $\hat{ x}_{1}$ should be a local minimum, which contradicts that $\hat{x}_{1}$ is a best response to $x_{2}^{*}$. By analogy and given the symmetry of the candidates’ utility functions, one can show that there exists at most one $\hat{x}_{2}>\alpha _{2}$ for which $\frac{\partial \Pi _{2}\left( x_{1}^{*},x_{2}\right) }{\partial x_{2}}\mid _{x_{2}=\hat{x}_{2}}=0$, and it should be a local minimum, which contradicts that $\hat{x}_{2}$ is a best response to $x_{1}^{*}$. To conclude our equilibrium existence arguments, we will show that there exists $d^{\prime }$ such that, whenever $d<d^{\prime }$, a Nash equilibrium of the restricted game $(x_{1}^{*},x_{2}^{*})$ is a Nash equilibrium of the unrestricted game too. In what follows, we define $x_{i}(x_{-i},C_{i})$ as the policy that makes candidate i indifferent between winning and losing such that $v(\left| x_{i}(x_{-i},C_{i})-\alpha _{i}\right| )-C_{i}=v(\left| x_{-i}-\alpha _{i}\right| )$.

• Step 4  Assume that in the unrestricted game the candidates play a Nash equilibrium of the restricted game, $(x_{1}^{*},x_{2}^{*})$, and that $d\rightarrow 0$. Candidate 1 has no incentives to deviate: a) to any $x_{1}\in [\alpha _{1},-d]$, $x_{1}\ne x_{1}^{*}$, because $(x_{1}^{*},x_{2}^{*})$ is a Nash equilibrium of the restricted game; b) to any $x_{1}\in (-d,x_{1}(x_{2}^{*},C_{1})]$ because $\frac{\partial \Pi _{1}\left( x_{1},x_{2}^{*}\right) }{\partial x_{1}}<0$ in this set (see arguments in Step 1); c) to any $x_{1}>x_{1}(x_{2}^{*},C_{1})$ because by definition of $x_{1}(x_{2}^{*},C_{1})$, $\Pi _{1}\left( x_{1}(x_{2}^{*},C_{1}),x_{2}^{*}\right) >\Pi _{1}\left( x_{1},x_{2}^{*}\right)$ for any $x_{1}>x_{1}(x_{2}^{*},C_{1})$; and d) to any $x_{1}<\alpha _{1}$ as we established in step 3. Therefore, candidate 1 has no incentive to deviate from $x_{1}^{*}$. By analogy and given the symmetry of the candidates’ utility functions, candidate 2 has no incentive to deviate from $x_{2}^{*}$. It follows that there exists $d^{\prime }>0$ such that whenever $d<d^{\prime }$, any Nash equilibrium of the restricted game $(x_{1}^{*},x_{2}^{*})$ is an equilibrium of the unrestricted game too.$\square$

Proof of Proposition 2

To characterize an interior equilibrium $(x_{1}^{*},x_{2}^{*})$, we notice that it must be such that: a) $\frac{\partial \Pi _{1}\left( x_{1},x_{2}^{*}\right) }{\partial x_{1}}\mid _{x_{1}=x_{1}^{*}}=0$ and $\frac{\partial \Pi _{2}\left( x_{1}^{*},x_{2}\right) }{\partial x_{2}}\mid _{x_{2}=x_{2}^{*}}=0$; and b) $\alpha _{1}<x_{1}^{*}<x_{2}^{*}<\alpha _{2}$. The first-order conditions are given by

$$\begin{aligned}&\frac{\partial \Pi _{1}\left( x_{1},x_{2}^{*}\right) }{\partial x_{1}}\, \mid \,{}_{x_{1}=x_{1}^{*}}=-F\left( \frac{x_{1}^{*}+x_{2}^{*}}{2} \right) +\frac{1}{2}(x_{2}^{*}-x_{1}^{*}-C_{1})f\left( \frac{ x_{1}^{*}+x_{2}^{*}}{2}\right) =0, \\&\frac{\partial \Pi _{2}\left( x_{1}^{*},x_{2}\right) }{\partial x_{2}} \,\mid \,{}_{x_{2}=x_{2}^{*}}=1-F\left( \frac{x_{1}^{*}+x_{2}^{*}}{2 }\right) -\frac{1}{2}(x_{2}^{*}-x_{1}^{*}-C_{2})f\left( \frac{ x_{1}^{*}+x_{2}^{*}}{2}\right) =0, \end{aligned}$$

which amount to

$$\begin{aligned} x_{2}^{*}-x_{1}^{*}= & {} 2\frac{F\left( \frac{x_{1}^{*}+x_{2}^{*}}{2}\right) }{f\left( \frac{x_{1}^{*}+x_{2}^{*}}{2} \right) }+C_{1}, \\ x_{2}^{*}-x_{1}^{*}= & {} 2\frac{1-F\left( \frac{x_{1}^{*}+x_{2}^{*}}{2}\right) }{f\left( \frac{x_{1}^{*}+x_{2}^{*}}{2} \right) }+C_{2}, \end{aligned}$$

respectively. Therefore,

$$\begin{aligned} 2\frac{F\left( \frac{x_{1}^{*}+x_{2}^{*}}{2}\right) }{f\left( \frac{x_{1}^{*}+x_{2}^{*}}{2}\right) }+C_{1}=2\frac{1-F\left( \frac{x_{1}^{*}+x_{2}^{*}}{2}\right) }{f\left( \frac{x_{1}^{*}+x_{2}^{*}}{2}\right) }+C_{2}. \end{aligned}$$

(B.1)

Assumption 1

implies that the left-hand side of (B.1) is strictly increasing in $\frac{x_{1}^{*}+x_{2}^{*}}{2}$ whereas the right-hand side is strictly decreasing in $\frac{x_{1}^{*}+x_{2}^{*}}{2}$. Furthermore, as $\frac{x_{1}^{*}+x_{2}^{*}}{2}$ goes to $+\infty$, the left-hand side of (B.1) converges to $+\infty$ whereas the right-hand side converges to a positive bounded value. As $\frac{x_{1}^{*}+x_{2}^{*}}{2}$ goes to $-\infty$, the left-hand side of (B.1) converges to a positive bounded value whereas the right-hand side converges to $+\infty$. Thus, there exists a unique value of $\frac{x_{1}^{*}+x_{2}^{*}}{2}$ for which (B.1) holds. We denote this unique value by s. Substituting s in the first-order conditions yields a unique interior equilibrium (5.1 ). Condition $\alpha _{1}<x_{1}^{*}<x_{2}^{*}<\alpha _{2}$ amounts to

$$\begin{aligned} s-\frac{F\left( s\right) }{f\left( s\right) }-\frac{C_{1}}{2}>\alpha _{1} \quad \text { and }\quad s+\frac{1-F\left( s\right) }{f\left( s\right) }+\frac{C_{2} }{2}<\alpha _{2}. \ \ \end{aligned}$$

$\square$

Proof of Proposition 3

Note that (5.2) along with the symmetry of $F\left( \cdot \right)$ around zero implies that: a) if $C_{1}>C_{2}$ then $s<0$, b) if $C_{1}<C_{2}$ then $s>0$ and c) if $C_{1}=C_{2}$ then $s=0$. Moreover,

$$\begin{aligned} \frac{\partial s}{\partial C_{1}}=-\frac{\frac{\partial }{\partial C_{1}} \left( 2\frac{F\left( s\right) }{f\left( s\right) }+C_{1}-2\frac{1-F\left( s\right) }{f\left( s\right) }-C_{2}\right) }{\frac{\partial }{\partial s} \left( 2\frac{F\left( s\right) }{f\left( s\right) }+C_{1}-2\frac{1-F\left( s\right) }{f\left( s\right) }-C_{2}\right) }=-\frac{1}{2\frac{\partial }{ \partial s}\left( \frac{F\left( s\right) }{f\left( s\right) }\right) -2 \frac{\partial }{\partial s}\left( \frac{1-F\left( s\right) }{f\left( s\right) }\right) }, \end{aligned}$$

which is always negative given Assumption 1. Similarly,

$$\begin{aligned} \frac{\partial s}{\partial C_{2}}=-\frac{\frac{\partial }{\partial C_{2}} \left( 2\frac{F\left( s\right) }{f\left( s\right) }+C_{1}-2\frac{1-F\left( s\right) }{f\left( s\right) }-C_{2}\right) }{\frac{\partial }{\partial s} \left( 2\frac{F\left( s\right) }{f\left( s\right) }+C_{1}-2\frac{1-F\left( s\right) }{f\left( s\right) }-C_{2}\right) }=\frac{1}{2\frac{\partial }{ \partial s}\left( \frac{F\left( s\right) }{f\left( s\right) }\right) -2 \frac{\partial }{\partial s}\left( \frac{1-F\left( s\right) }{f\left( s\right) }\right) }>0. \end{aligned}$$

Since $f\left( \cdot \right) >0$, the equilibrium probability of candidate i being elected $p_{i}\left( x_{1}^{*},x_{2}^{*}\right) \$ decreases with his policy implementation cost $C_{i}$ and increases with his competitor’s cost $C_{-i}$.

Furthermore,

$$\begin{aligned} \frac{\partial x_{1}^{*}}{\partial C_{1}}= & {} \frac{\partial s}{\partial C_{1}}\cdot \frac{F\left( s\right) f^{\prime }\left( s\right) }{\left( f\left( s\right) \right) ^{2}}-\frac{1}{2}, \quad \frac{\partial x_{1}^{*}}{\partial C_{2}}=\frac{\partial s}{\partial C_{2}}\cdot \frac{F\left( s\right) f^{\prime }\left( s\right) }{\left( f\left( s\right) \right) ^{2}},\\ \frac{\partial x_{2}^{*}}{\partial C_{1}}= & {} -\frac{\partial s}{\partial C_{1}}\cdot \frac{\left( 1-F\left( s\right) \right) f^{\prime }\left( s\right) }{\left( f\left( s\right) \right) ^{2}}, \quad \frac{\partial x_{2}^{*}}{\partial C_{2}}=-\frac{\partial s}{\partial C_{2}}\cdot \frac{\left( 1-F\left( s\right) \right) f^{\prime }\left( s\right) }{\left( f\left( s\right) \right) ^{2}}+\frac{1}{2}. \end{aligned}$$

Given the above observations a)-c) and Assumption 1, the following holds.

• 1. If $C_{1}>C_{2}$ then $f^{\prime }\left( s\right) >0$ and therefore

  $$\begin{aligned} \frac{\partial x_{1}^{*}}{\partial C_{1}}<0, \quad \frac{\partial x_{1}^{*}}{\partial C_{2}}>0, \quad \frac{\partial x_{2}^{*}}{ \partial C_{1}}>0. \end{aligned}$$

  Moreover, for sufficiently small (and hence similar) costs $C_{1}$ and $C_{2}$, $s\rightarrow 0$ and so $\frac{\partial x_{2}^{*}}{\partial C_{2}}>0$.

• 2. If $C_{1}<C_{2}$ then $f^{\prime }\left( s\right) <0$ and therefore

  $$\begin{aligned} \frac{\partial x_{1}^{*}}{\partial C_{2}}<0, \quad \frac{\partial x_{2}^{*}}{\partial C_{1}}<0, \quad \frac{\partial x_{2}^{*}}{ \partial C_{2}}>0. \end{aligned}$$

  For sufficiently small (and hence similar) costs $C_{1}$ and $C_{2}$, $s\rightarrow 0$ and so $\frac{\partial x_{1}^{*}}{\partial C_{1}}<0$.

• 3. If $C_{1}=C_{2}\equiv C$ then

  $$\begin{aligned} \frac{\partial x_{1}^{*}}{\partial C}<0, \quad \frac{\partial x_{2}^{*}}{\partial C}>0. \end{aligned}$$

  $\square$

Proof of Proposition 4

The median voter’s expected payoff $Ev_{MV}$ is equal to

$$\begin{aligned} Ev_{MV}\equiv & {} \int _{-\infty }^{s}\left( -\left| x_{1}^{*}-\rho \right| \right) f\left( \rho \right) d\rho +\int _{s}^{+\infty }\left( -\left| x_{2}^{*}-\rho \right| \right) f\left( \rho \right) d\rho \\= & {} \int _{-\infty }^{x_{1}^{*}}\left( \rho -x_{1}^{*}\right) f\left( \rho \right) d\rho +\int _{x_{1}^{*}}^{s}\left( x_{1}^{*}-\rho \right) f\left( \rho \right) d\rho +\int _{s}^{x_{2}^{*}}\left( \rho -x_{2}^{*}\right) f\left( \rho \right) d\rho \\&+\int _{x_{2}^{*}}^{+\infty }\left( x_{2}^{*}-\rho \right) f\left( \rho \right) d\rho . \end{aligned}$$

In the case of $C_{1}=C_{2}\equiv C$, differentiating $Ev_{MV}$ with respect to C yields

$$\begin{aligned} \frac{\partial Ev_{MV}}{\partial C}=-\frac{1}{2}\left( 1-4F\left( -\frac{C }{2}-\frac{1}{2f\left( 0\right) }\right) \right) . \end{aligned}$$

Notice that it is negative if

$$\begin{aligned} C>-2F^{-1}\left( \frac{1}{4}\right) -\frac{1}{f\left( 0\right) }. \end{aligned}$$

(D.1)

We prove next that the right-hand side of (D.1) is negative. First, due to convexity of $F\left( \cdot \right)$ for negative values,

$$\begin{aligned} \frac{\frac{1}{2}-F\left( -\frac{1}{2f\left( 0\right) }\right) }{0-\left( -\frac{1}{2f\left( 0\right) }\right) }>f\left( -\frac{1}{2f\left( 0\right) }\right) , \end{aligned}$$

which amounts to

$$\begin{aligned} \frac{1}{2}-F\left( -\frac{1}{2f\left( 0\right) }\right) >\frac{f\left( - \frac{1}{2f\left( 0\right) }\right) }{2f\left( 0\right) }. \end{aligned}$$

(D.2)

Second, due to Assumption 1,

$$\begin{aligned} \frac{f\left( -\frac{1}{2f\left( 0\right) }\right) }{F\left( -\frac{1}{ 2f\left( 0\right) }\right) }\ge \frac{f\left( 0\right) }{F\left( 0\right) } \ \ \Rightarrow \ \ \frac{f\left( -\frac{1}{2f\left( 0\right) }\right) }{2f\left( 0\right) }\ge F\left( -\frac{1}{2f\left( 0\right) }\right) . \end{aligned}$$

(D.3)

Combining (D.2) and (D.3) yields

$$\begin{aligned} \frac{1}{2}-F\left( -\frac{1}{2f\left( 0\right) }\right) >F\left( -\frac{1 }{2f\left( 0\right) }\right) \ \ \Rightarrow \ \ -2F^{-1}\left( \frac{1}{4}\right) -\frac{1}{f\left( 0\right) }<0. \end{aligned}$$

Thus, the right-hand side of (D.1) is negative and so (D.1) holds for every $C\ge 0$. This implies that $\frac{\partial Ev_{MV}}{ \partial C}<0$. $\square$

Proof of Proposition 5

If $C_{1}=C_{2}=C$ then $s_{k}=0$ for every k because

$$\begin{aligned} 2\frac{F_{k}(s_{k})}{f_{k}(s_{k})}+C_{1}=2\frac{1-F_{k}(s_{k})}{ f_{k}(s_{k})}+C_{2} \end{aligned}$$

yields $F_{k}(s_{k})=\frac{1}{2}$. Hence, $\left( x_{1}^{*},x_{2}^{*}\right) \rightarrow \left( -\frac{C}{2},\frac{C}{2}\right)$.

If $C_{1}\ne C_{2}$ then assume that there exists $\varepsilon >0$ such that for every k there exists $\hat{k}>k$ with $s_{\hat{k}}\notin [-\varepsilon ,\varepsilon ]$. Then there exists a subsequence $(\widetilde{F} _{k}\left( \cdot \right) )_{k=1}^{\infty }$ such that: a) $\widetilde{f} _{k}(0)\rightarrow +\infty$ and $\widetilde{F}_{k}(\rho )\rightarrow \widetilde{f}_{k}(\rho )\rightarrow 0$ for every $\rho <0$; and b)

$$\begin{aligned} 2\frac{\widetilde{F}_{k}(s_{k})}{\widetilde{f}_{k}(s_{k})}+C_{1}=2\frac{1- \widetilde{F}_{k}(s_{k})}{\widetilde{f}_{k}(s_{k})}+C_{2}, \end{aligned}$$

which amounts to

$$\begin{aligned} 4\widetilde{F}_{k}(s_{k})+\left( C_{1}-C_{2}\right) \widetilde{f} _{k}(s_{k})=2 \end{aligned}$$

with $s_{k}\notin [-\varepsilon ,\varepsilon ]$ for every k. But if a) holds then

$$\begin{aligned} 4\widetilde{F}_{k}(s_{k})+\left( C_{1}-C_{2}\right) \widetilde{f} _{k}(s_{k})\rightarrow 0\ne 2 \end{aligned}$$

and so b) doesn’t hold. Therefore, there exists no $\varepsilon >0$ such that for every k there exists $\hat{k}>k$ with $s_{\hat{k}}\notin [-\varepsilon ,\varepsilon ]$. Hence $s_{k}\rightarrow 0$. Notice next that for $C_{1}>C_{2}$,

$$\begin{aligned} 2\frac{F_{k}(s_{k})}{f_{k}(s_{k})}+C_{1}=2\frac{1-F_{k}(s_{k})}{ f_{k}(s_{k})}+C_{2} \end{aligned}$$

yields

$$\begin{aligned} 4\frac{F_{k}(s_{k})}{f_{k}(s_{k})}+(C_{1}-C_{2})=\frac{2}{f_{k}(s_{k})}. \end{aligned}$$

This implies that $f_{k}(s_{k})\rightarrow \frac{2}{C_{1}-C_{2}}$ and hence $(x_{1}^{*},x_{2}^{*})\rightarrow \left( -\frac{C_{1}}{2},\frac{C_{1} }{2}\right)$. Similarly, when $C_{1}<C_{2}$ we have that $(x_{1}^{*},x_{2}^{*})\rightarrow \left( -\frac{C_{2}}{2},\frac{C_{2}}{2}\right)$ . $\square$