The Electoral College, battleground states, and rule-utilitarian voting

Abstract

We extend the Feddersen and Sandroni (Am Econ Rev 96(4):1271–1282, 2006) voter turnout model to include partisan districts, a battleground district, and an Electoral College. We find that expected voter turnout by a single party is highest in the battleground district, followed by the party’s majority district, which in turn is followed by the party’s minority district. Total turnout is higher in the battleground district than in the partisan districts, but the gap decreases as the level of disagreement in the partisan districts increases. Lastly, turnout in the battleground district decreases as the partisan districts become more competitive.

Appendices

Appendix A: Solving for the symmetric equilibrium

1.1 Interior solution

We restrict our search to symmetric equilibria in which $\sigma _1^{1*}=\sigma _2^{2*} =\alpha$, $\sigma _1^{2*}=\sigma _2^{1*} =\beta$, and $\sigma _1^{3*}=\sigma _2^{3*}=\gamma$. The CDF for the ratio of two independent standard uniform random variables is

$$\begin{aligned} F(z) = \left\{ \begin{array}{ll} \frac{z}{2} &{} \text {if }z \le 1;\\ 1-\frac{1}{2z} &{} \text {if }z >1.\end{array} \right. \end{aligned}$$

The payoff functions that each party seeks to maximize are $R_1(\sigma ) = wp(\sigma )-\phi (\sigma )$ and $R_2(\sigma ) = w(1-p(\sigma ))-\phi (\sigma )$. We can write $p(\sigma )=p_1p_2 + p_1(1-p_2)p_3+(1-p_1)p_2p_3$, where $p_1= F\left( \frac{(1-k)\sigma _1^1}{k \sigma _2^1} \right)$, $p_2 = F\left( \frac{k\sigma _1^2}{(1-k) \sigma _2^2} \right)$, and $p_3 = F\left( \frac{\sigma _1^3}{\sigma _2^3} \right)$. With respect to the partisan district cutoff rules ($\sigma _1^1$, $\sigma _1^2$, $\sigma _2^1$ and $\sigma _2^2$), $p_3$ is constant and is $\frac{1}{2}$ in a symmetric equilibrium. This allows the simplification $p(\sigma )=\frac{1}{2}(p_1+p_2)$ when evaluating the FOCs in the partisan districts. Taking the first order conditions and evaluating at $\sigma _1^1=\sigma _2^2=\alpha$ and $\sigma _1^2=\sigma _2^2=\beta$ gives us that

$$\begin{aligned} \frac{\partial {R_1}}{\partial {\sigma _1^1}} = \frac{\partial {R_2}}{\partial {\sigma _2^2}} = \frac{w}{2}\frac{1-k}{k}\frac{1}{\beta }f\Bigg (\frac{1-k}{k}\frac{\alpha }{\beta }\Bigg ) -\frac{1}{2}(1-k)\alpha \end{aligned}$$

and

$$\begin{aligned} \frac{\partial {R_1}}{\partial {\sigma _1^2}} = \frac{\partial {R_2}}{\partial {\sigma _2^1}} = \frac{w}{2}\frac{k}{1-k}\frac{1}{\alpha }f\Bigg (\frac{k}{1-k}\frac{\beta }{\alpha }\Bigg ) -\frac{1}{2}k\beta . \end{aligned}$$

For the third condition we first simplify $R_1$. Note that the $p_1$ and $p_2$ are constant with respect to battleground district cutoffs and $p_1=1-p_2$ in a symmetric equilibrium. Then, $p(\sigma )=p_1p_2 +(2p_1^2-2p_1+1)p_3,$ which gives us the last FOC (evaluated at $\alpha$, $\beta$, and $\gamma$):

$$\begin{aligned} \frac{\partial {R_1}}{\partial {\sigma _1^3}} = \frac{\partial {R_2}}{\partial {\sigma _2^3}} = w\frac{1}{2\gamma }\Bigg (2F^2\Bigg (\frac{1-k}{k}\frac{\alpha }{\beta }\Bigg ) - 2F\Bigg (\frac{1-k}{k}\frac{\alpha }{\beta }\Bigg )+1\Bigg )-\frac{\gamma }{4}. \end{aligned}$$

For $\frac{1-k}{k}\frac{\alpha }{\beta } >1$, the first two first order conditions are:

$$\begin{aligned} \beta = \frac{(1-k)^2}{k}\frac{2}{w}\alpha ^3,\quad \alpha = \frac{w}{2}\frac{1}{1-k}\frac{1}{\beta } \end{aligned}$$

and give

$$\begin{aligned} \alpha= & {} \sigma _1^{1*}=\sqrt{w}\left( \frac{k}{4(1-k)^3}\right) ^{\frac{1}{4}} \end{aligned}$$

(9)

$$\begin{aligned} \beta= & {} \sigma _1^{2*}=\sqrt{w}\frac{1}{(4k(1-k))^{\frac{1}{4}}} \end{aligned}$$

(10)

$$\begin{aligned} \gamma= & {} \sigma _1^{3*}=\sqrt{w}\left[ \frac{k}{1-k}-2\left( \frac{k}{1-k}\right) ^{\frac{1}{2}}+2\right] ^{\frac{1}{2}} \end{aligned}$$

(11)

The above equilibrium is the unique symmetric interior solution. To see this, suppose $\frac{1-k}{k}\frac{\alpha }{\beta } \le 1$. Re-taking the FOCs gives $\frac{\alpha }{\beta }=\sqrt{\frac{k}{1-k}}$, which implies $k\ge \frac{1}{2}$, which is a contradiction.

1.2 Boundary equilibria

Since $k, w > 0$, it follows that $\alpha , \beta , \gamma > 0$. Thus, as long as the election has positive importance, no optimal $\sigma$ will ever be on the 0 boundary. Since the ordering of the cutoffs is dependent on k, we have two cases.

1.2.1 Low levels of disagreement: $k \in (0, \frac{1}{5}]$

By Proposition 1, we know that $\sigma _1^{1*}\le \sigma _1^{3*}\le \sigma _1^{2*}$. Thus, when $k<\frac{1}{5}$, the equilibrium takes on one of four cases: (1) the interior solution; (2) only $\sigma _1^{2*}$ is on the boundary; (3) only $\sigma _1^{2*}$ and $\sigma _1^{3*}$ are on the boundary; or (4) all three cutoffs are on the boundary. The previous subsection details the interior solutions.

Case 2: Taking $\sigma _1^{2*} = 1$ and retaking FOCs for $\sigma _1^{3*}$ and $\sigma _1^{1*}$ gives boundary equilibrium of

$$\begin{aligned} {\sigma ^1_1}^*&= \left( \frac{kw}{2(1-k)^2}\right) ^{\frac{1}{3}}, \quad {\sigma _1^3}^* = \left[ \left( \frac{4wk^4}{(1-k)^2} \right) ^{\frac{1}{3}}-\left( \frac{16w^2k^2}{(1-k)}\right) ^{\frac{1}{3}} +2w\right] ^{\frac{1}{2}},\nonumber \\ {\sigma _1^2}^*&= 1 . \end{aligned}$$

The interval is:

$$\begin{aligned} \left[ \left( \frac{4wk^4}{(1-k)^2} \right) ^{\frac{1}{3}}-\left( \frac{16w^2k^2}{(1-k)}\right) ^{\frac{1}{3}} +2w\right] ^{\frac{1}{2}}<1,\quad \frac{1}{w} \le \frac{1}{(4k(1-k))^{\frac{1}{2}}}. \end{aligned}$$

Case 3: Taking both $\sigma _1^{2*}$ and $\sigma _1^{3*}$ on the boundary and retaking the FOCs yields

$$\begin{aligned} {\sigma ^1_1}^* = \left( \frac{kw}{2(1-k)^2}\right) ^{\frac{1}{3}}, \quad {\sigma _1^3}^* = 1, \quad {\sigma _1^2}^* = 1. \quad \end{aligned}$$

The interval is:

$$\begin{aligned} \frac{k}{2(1-k)^2} < \frac{1}{w}, 1 \le \left[ \left( \frac{4wk^4}{(1-k)^2} \right) ^{\frac{1}{3}}-\left( \frac{16w^2k^2}{(1-k)}\right) ^{\frac{1}{3}} +2w\right] ^{\frac{1}{2}}. \end{aligned}$$

Case 4: All three cutoffs are on the boundary when $\sigma _1^{1*}$ from Case 3 reaches 1. The interval is

$$\begin{aligned} \frac{1}{w} \le \frac{k}{2(1-k)^2}. \end{aligned}$$

1.2.2 High-levels of disagreement: $k \in (\frac{1}{5}, \frac{1}{2})$

By Property 1, the cutoffs are now ordered $\sigma _1^{1*} \le \sigma _1^{2*} \le \sigma _1^{3*}$. Setting ${\sigma _1^3}^* = 1$ and retaking the FOCs for $\sigma _1^2$ and $\sigma _1^1$ yields that

$$\begin{aligned} {\sigma ^1_1}^* =\frac{w^{\frac{1}{2}}k^{\frac{1}{4}}}{(4(1-k)^3)^\frac{1}{4}}, \quad {\sigma _1^2}^* =\frac{w^{\frac{1}{2}}}{(4k(1-k))^{\frac{1}{4}}}, \quad {\sigma _1^3}^* = 1. \end{aligned}$$

The interval is:

$$\begin{aligned} \frac{1}{(4k(1-k))^{\frac{1}{2}}} < \frac{1}{w} \le \left[ \frac{k}{1-k}-2\left( \frac{k}{1-k}\right) ^{\frac{1}{2}} +2\right] . \end{aligned}$$

When both the largest and second largest cutoffs are at the boundary, we get that

$$\begin{aligned} {\sigma ^1_1}^* = \left( \frac{kw}{2(1-k)^2}\right) ^{\frac{1}{3}},\quad {\sigma _1^2}^* = 1,\quad {\sigma _1^3}^* = 1. \end{aligned}$$

The interval is:

$$\begin{aligned} \frac{k}{2(1-k)^2} < \frac{1}{w} \le \frac{1}{(4k(1-k))^{\frac{1}{2}}}. \end{aligned}$$

All three cutoffs are on the boundary when $\sigma _1^1$ from the previous case reaches 1, which occurs when

$$\begin{aligned} \frac{1}{w} \le \frac{k}{2(1-k)^2}. \end{aligned}$$

Appendix B: Proofs of properties

1.1 Proposition 1

We first show that cutoff rules are decreasing in k for the minority and battleground districts, and increasing in k for the majority district. The derivative of the interior solutions from Eqs. 9, 10, and 11 are:

$$\begin{aligned} \frac{\partial {\sigma _1^1}^*}{\partial k}= & {} \frac{\sqrt{2w}}{2}\left( \frac{ \frac{1}{4} (1-k)^{\frac{3}{4}} k^{-\frac{3}{4}} + k^\frac{1}{4} \frac{3}{4} (1-k)^ {-\frac{1}{4}} }{(1-k)^\frac{3}{2}} \right) \\ \frac{\partial {\sigma _1^2}^*}{\partial k}= & {} -\sqrt{w} \frac{ \frac{1}{4}(4-8k) }{ (4k(1-k))^\frac{5}{4}} \\ \frac{\partial {\sigma _1^3}^*}{\partial k}= & {} \frac{\sqrt{w}}{(1-k)^2\sqrt{k}} \times \frac{\sqrt{k}-\sqrt{1-k}}{2 \left( \frac{k}{1-k}-2 \left( \frac{k}{1-k} \right) ^\frac{1}{2}+2 \right) ^\frac{1}{2}}. \end{aligned}$$

The partial derivative for the majority district, $\frac{\partial {\sigma _1^1}^*}{\partial k}$, is clearly positive, implying that the cutoff rule in the majority district is increasing in k. Since $k<\frac{1}{2}$, $\frac{\partial {\sigma _1^2}^*}{\partial k}$ is negative, implying the cutoff rule in the minority district is decreasing in k. Since $\sqrt{k}-\sqrt{1-k}<0$ for all $k<\frac{1}{2}$, it follows that the cutoff rule is decreasing for the battleground district.

We next show that the cutoff rule in the majority district is always lower. First note that all of the cutoff rules converge to $\sqrt{w}$ if $k=\frac{1}{2}$. Since the cutoff rule is increasing in k for the majority district, this must be the maximum cutoff for that district. Since the cutoff rules are decreasing in the minority and battleground district, this is the minimum cutoff for those districts. Thus, the majority district always has the lowest cutoff.

Lastly, we show that for high levels of disagreement ($k>\frac{1}{5}$), the cutoff is higher in the battleground district than in the minority district, and that for low levels of disagreement ($k<\frac{1}{5}$), the cutoff rule is higher in the minority district. With much algebra, it can be shown that the majority district has a higher cutoff rule if and only if $20k^3-24k^2+9k-1$ is positive. This polynomial is negative for k in $(0,\frac{1}{5})$ and positive for k in $(\frac{1}{5},\frac{1}{2})$.

1.2 Proposition 2

The Proposition states $\frac{1}{2}\frac{1}{2}\sigma _1^{3*}> \frac{1}{2}(1-k)\sigma _1^{1*} > \frac{1}{2}k\sigma _1^{2*}.$ First we show that $\frac{1}{2}\sigma _1^{3*}>\frac{1}{2}w^{\frac{1}{2}}>(1-k)\sigma _1^{1*}$. Some algebra will show that $\frac{1}{2}\sigma _1^{3*}>\frac{1}{2}w^{\frac{1}{2}}$ if and only if $4k^2-4k+1$ is positive, which is true for all $k \in (0,\frac{1}{2})$ since the polynomial has a global minimum of 0 at $k=\frac{1}{2}$.

Next, $\frac{1}{2}w^{\frac{1}{2}}>(1-k)\sigma _1^{1*} \Longleftrightarrow \frac{1}{4} > (1-k)k$. This inequality is assured by $k \in (0,\frac{1}{2})$. Thus we have that $\frac{1}{2}\frac{1}{2}\sigma _1^{3*} > \frac{1}{2}(1-k)\sigma _1^{1*}$.

Next,

$$\begin{aligned} (1-k)\sigma _1^{1*}> & {} k\sigma _1^{2*}\\\Longleftrightarrow & {} \frac{k^{\frac{1}{4}}(1-k)}{(1-k)^{\frac{3}{4}}}> \frac{k}{((1-k)k)^{\frac{1}{4}}}. \\\Longleftrightarrow & {} k(1-2k)> 0, \end{aligned}$$

which holds since $k \in (0, \frac{1}{2})$, and gives $\frac{1}{2}(1-k)\sigma _1^{1*} > \frac{1}{2}k\sigma _1^{2*}$, which finishes the proof of the first part of the Proposition.

The proof of Proposition 1 showed that $\frac{d\sigma _1^{3*}}{dk}>0$, so it follows that a party’s expected turnout $\frac{1}{2}\frac{1}{2}\frac{d\sigma _1^{3*}}{dk}>0$ also.

Expected turnout for the majority party is

$$\begin{aligned} \frac{1}{2}(1-k)\sigma _1^{1*} = \frac{1}{2}\left( \frac{w}{2}\right) ^{\frac{1}{2}}\big ((1-k)k)^{\frac{1}{4}}. \end{aligned}$$

Its derivative is

$$\begin{aligned} \frac{1}{8}\big (\frac{w}{2}\big )^{\frac{1}{2}}\frac{1-2k}{\left( k-k^2\right) ^{\frac{3}{4}}}>0 \end{aligned}$$

since $k \in (0, \frac{1}{2})$. Thus expected turnout for a party’s majority district is increasing in k.

Expected total turnout in a party’s minority district is

$$\begin{aligned} \frac{1}{2}k\sigma _1^{2*} = \frac{1}{2}\left( \frac{w}{2}\right) ^{\frac{1}{2}}\frac{k^{\frac{3}{4}}}{(1-k)^{\frac{1}{4}}}. \end{aligned}$$

Its derivative is

$$\begin{aligned} \frac{1}{2}\left( \frac{w}{2}\right) ^{\frac{1}{2}}\frac{\frac{3}{4}\bigg (\frac{1-k)}{k}\bigg )^{\frac{1}{4}}+\frac{1}{4}\bigg (\frac{k}{1-k}\bigg )^{\frac{3}{4}}}{(1-k)^{\frac{1}{2}}} > 0, \end{aligned}$$

and thus expected turnout for the minority party is increasing in k.

1.3 Proposition 3

Expected total turnout for each party is increasing in k for the majority and minority in the partisan districts. Thus, the total expected turnout in a partisan district is also increasing in k. Both parties’ expected turnout is decreasing in k for the battleground district, so expected total turnout is also decreasing in k.

To see that the expected total turnout in the partisan districts converges to that of the battleground district, consider:

$$\begin{aligned} \lim _{k\rightarrow \frac{1}{2}} \frac{1}{2}\frac{1}{2}\sigma _1^{3*} + \frac{1}{2}\frac{1}{2}\sigma _2^{3*} = \lim _{k\rightarrow \frac{1}{2}} \frac{1}{2}\sqrt{w}\Bigg (\frac{k}{1-k} - 2\bigg (\frac{k}{1-k}\bigg )^{\frac{1}{2}}+2\Bigg )^{\frac{1}{2}} = \frac{1}{2}\sqrt{w} \end{aligned}$$

and

$$\begin{aligned} \lim _{k \rightarrow \frac{1}{2}} \frac{1}{2}(1-k)\sigma _1^{1*} + \frac{1}{2}k\sigma _2^{1*}= & {} \lim _{k\rightarrow \frac{1}{2}} \frac{1}{2}(1-k)\sqrt{w}\frac{k^{\frac{1}{4}}}{(4(1-k)^3)^{\frac{1}{4}}} \nonumber \\&+ \frac{1}{2}k\sqrt{w}\frac{1}{(4(k(1-k))^{\frac{1}{4}}} \nonumber \\= & {} \frac{1}{4}\sqrt{w}+\frac{1}{4}\sqrt{w}= \frac{1}{2}\sqrt{w}. \end{aligned}$$

1.4 Proposition 4

The Proposition states that as k approaches $\frac{1}{2}$, the probability of winning a party’s majority district decreases, the probability of the party winning its minority district increases, and the probability a party wins the battleground district stays constant at $\frac{1}{2}$.

To see this, consider the probability of party 1 winning its majority district:

$$\begin{aligned} F\Bigg (\frac{(1-k)\sigma _1^{1*}}{k\sigma _2^{1*}}\Bigg ) = 1 - \frac{k\sigma _2^{1*}}{2(1-k)\sigma _1^{1*}} = 1 - \frac{1}{2}\Bigg (\frac{k}{1-k}\Bigg )^{\frac{1}{2}}. \end{aligned}$$

Notice that this quantity approaches $\frac{1}{2}$ as $k \rightarrow \frac{1}{2}$. Taking the derivative with respect to k gives

$$\begin{aligned} \frac{\partial F\left( \frac{(1-k)\sigma _1^{1*}}{k\sigma _2^{1*}}\right) }{\partial k} = -\frac{1}{4}\frac{1}{k^{\frac{1}{2}}{(1-k)^{\frac{3}{2}}}}, \end{aligned}$$

which is negative for $k \in (1, \frac{1}{2})$. Thus the probability that a party wins their majority district is decreasing in k.

The probability of party 1 winning district 2 is

$$\begin{aligned} F\left( \frac{(1-k)\sigma _2^{2*}}{k\sigma _1^{2*}}\right) = \frac{k\sigma _1^{2*}}{2(1-k)\sigma _2^{2*}} = \frac{1}{2}\Bigg (\frac{k}{1-k}\Bigg )^{\frac{1}{2}}. \end{aligned}$$

This quantity is increasing in k by the same argument that the previous result is decreasing in k.

Finally, the probability that candidate 1 wins the battleground district is

$$\begin{aligned} F\left( \frac{\frac{1}{2}\sigma _1^{3*}}{\frac{1}{2}\sigma _2^{3*}}\right) = F(1) = \frac{1}{2}. \end{aligned}$$