Gerrymandering in a hierarchical legislature

Abstract

We build a multiple hierarchical model of a representative democracy in which citizens elect ward representatives, ward representatives elect county representatives, county representatives elect state representatives, and state representatives elect a prime minister. We use our model to show that the policy determined by the final representative can become more extreme as the number of hierarchical levels increases as a result of increased opportunities for gerrymandering. Thus, a sufficiently large number of voters provide a district maker an advantage, enabling her to implement her favorite policy.

Appendices

Appendix A: Proofs

Proof of Lemma 1

We prove this by contradiction. Let $N = a_1 \cdot a_2 \cdot a_3 \cdots a_t \cdot a_{t+1}$, where each $a_i$, $i \in \{1,2,3,\ldots ,t, t+1 \}$ is a prime factor of N. We assume that for a given N, we can make $j+1$ decision levels, noting that $K_i > K_{i+1}$ and $K_{j+1} =1$, where $j>t$. Then, the populations of each district at each level $1,2, 3, \ldots , j, j+1$ are $\frac{N}{K_1}, \frac{K_1}{K_2}, \frac{K_2}{K_3}, \ldots , \frac{K_{t-1}}{K_{t}}, \ldots , \frac{K_{j}}{K_{j+1}}$, respectively. Thus,

$$\begin{aligned} N = \frac{N}{K_1} \cdot \frac{K_1}{K_2} \cdot \frac{K_2}{K_3} \ldots \frac{K_{t-1}}{K_{t}} \ldots \frac{K_{j}}{K_{j+1}}, \end{aligned}$$

and the number of the factors of N is $j+1$. However, N cannot be the product of more than t integers larger than 1 from its prime factorization, which is a contradiction. $\square$

Proof of Corollary 2

Equation (3) determines the winning voter for both $K_1,\ldots ,$$K_{i-1},K_i,K_{i+1},\ldots ,K_t$ and $K_1,\ldots ,K_{i-1},K_i,K^{\prime }_i,$$K_{i+1},\ldots ,K_t$. As the t factors appearing in equation (1) are the same in both cases, we only need to verify that

$$\begin{aligned} \frac{1}{2}\left( \frac{K_{i}}{K_{i+1}}+1\right)>\frac{1}{2}\left( \frac{K_i}{K_i^\prime }+1\right) \frac{1}{2}\left( \frac{K^\prime _i}{K_{i+1}}+1\right)\iff & {} \frac{K_{i}}{K_{i+1}}+\frac{K_i^\prime }{K_i^\prime }>\frac{K_i}{K_i^\prime }+\frac{K^\prime _i}{K_{i+1}} \\\iff & {} \frac{K_{i}-K_i^\prime }{K_{i+1}}>\frac{K_i-K_i^\prime }{K_i^\prime }, \end{aligned}$$

which follows from $K^{\prime }_i>K_{i+1}$. $\square$

Proof of Corollary 3

Employing (5), we obtain

$$\begin{aligned} \frac{j^*_{t+1,1}}{N} \le \frac{\left( 1+\frac{3}{4}\right) ^{t+1}}{2^{t+1}}\rightarrow 0 \end{aligned}$$

as $t\rightarrow \infty$. $\square$

Appendix B: Details of the applications

We impose Assumption 1 throughout this appendix.

1.1 Moderates’ electability

The interval we have to consider is only $\{1,2, \ldots ,\frac{N+1}{2} \}$, by symmetry. We define the district number to which $j_{t+1,1}^*$ belongs at each decision level in the cracking and packing method as:

$$\begin{aligned} \displaystyle m_1 \equiv \frac{j_{t+1,1}^*}{\frac{1}{2}(\frac{N}{K_1}+1)} = \frac{1}{2} (K_t + 1) \frac{1}{2} \left( \frac{K_{t-1}}{K_{t }} + 1 \right) \frac{1}{2} \left( \frac{K_{t-2}}{K_{t-1}} + 1 \right) \ldots \frac{1}{2} \left( \frac{K_{1 }}{K_{2 }} + 1 \right) , \end{aligned}$$

$m_2 \equiv \frac{m_1}{\frac{1}{2}(\frac{K_1}{K_2}+1)}$, $m_3 \equiv \frac{m_2}{\frac{1}{2}(\frac{K_2}{K_3}+1)}$, $\ldots$, $m_t \equiv \frac{m_{t-1}}{\frac{1}{2}(\frac{K_{t-1}}{K_t}+1)}$, and $m_{t+1} \equiv \frac{m_{t}}{\frac{1}{2}(K_t+1)} = 1$, as there are minimal majority voters or representatives with ideal points left of $j_{t+1,1}^*$. In particular, there are $\frac{1}{2}(\frac{K_{i-1}}{K_i}+1)$ voters of such type for any $i \in \{ 1, 2, 3, \ldots , t, t+1 \}$, including $j_{t+1,1}^*$ at each decision level. For convenience, we also define $m_0 = j_{t+1,1}^*$. Then, in the cracking and packing method, noting that $j_{t+1,1}^*$ is included in the $m_1$-th district at the first decision level, from the definition of $m_i$, the cardinality of the maximum minority voters with a higher index is equal to $\frac{1}{2}(\frac{K_{i-1}}{K_i}-1)$, which we insert from the first district to the $K_i$-th district (Table 5).

Table 5 Each district by the cracking and packing method at the i-th decision level

In the cracking and packing method, the numbers of district medians appear at every $\frac{1}{2}(\frac{K_{i-1}}{K_i}+1)$ position because of the cracking of all voters into minimum majorities with lower values and maximum minorities with higher values, and then packing each into one district. Here, we refer to minimum majority voters with lower values in each district as “Left wing” and maximum minority voters with higher values as “Right wing.” Now, we focus on the median of each district, especially the $m_i$-th to $\frac{K_i+1}{2}$-th districts, after the following manipulations. First, in Table 5, we remove the last-position voter in the Right wing of the $K_i$-th district and slide all voters forward one position in the Right wings of all districts. Then, as voter $K_{i-1}$ also moves forward one position, we can create a vacancy at the first position in the Right wing of the first district. Second, we insert the first-position voter $A_i+1$ in the Left wing of the second district into the vacant position instead of using voter $K_{i-1}$. Third, we slide back all voters in the Left wing from the second to $K_i$-th districts by one position and insert the removed voter from the last position in the Right wing of the $K_i$-th district into the last position in the Left wing of the $K_i$ district, which is now vacant. We refer to this series of manipulations as a cycle. Repeating the cycle, the positions of the voters in the first district become those shown in Table 6.

Table 6 Positions of the voters in the first district at the i-th decision level

By virtue of the cycle, the medians in the second to $K_i$-th districts slide and are replaced by each voter with one higher value. Noting that there are $B_i=\frac{1}{2}(\frac{K_{i-1}}{K_i}-1)$ voters in the Right wing in each district and $(\frac{K_i+1}{2}-1) B_i$ voters in the Right wings of the first to $\frac{K_i+1}{2}-1$-th districts, we can also apply the cycle in the second to $\frac{K_i+1}{2}-1$-th districts repeatedly. Then, we can slide $(\frac{K_i+1}{2}-1)B_i$ voters, and voters in the first to $\frac{K_i+1}{2}-1$-th districts in the Right wing will line up in ascending order. Consequently, voters 1 to the median of all voters line up in ascending order in the first to $\frac{K_i+1}{2}$-th districts.

Table 7 Each district median of the $m_i$-th to $\frac{K_i+1}{2}$-th districts at the i-th decision level

We now check if any voter between $j_{t+1,1}^*$ and the median of all voters is electable as the final representative. Focusing on the medians of the $m_i$-th to $\frac{K_i+1}{2}$-th districts (i.e., the median district), we extract the medians of those districts. Then, we obtain Table 7. Sliding $(m_i-1)B_i$ positions, as all voters in the Right wing of the first district to the $m_i-1$-th district have already been replaced by voter $A_i+1$ to voter $A_i+(m_i-1)B_i$, no voter is replaced before voter $m_i A_i +(m_i-1)B_i$ in the $m_i$-th district. Thus, the median of the $m_i$-th district cannot slide further; therefore, voter $m_i A_i + (m_i-1) B_i$ is unchanged after sliding by $(m_i-1)B_i$ in Table 7. Similarly, the medians of the $m_i+1$-th to $\frac{K_i+1}{2}$-th districts are also unchanged after replacing the voters in the Right wing and lining up the consecutive numbers in each district. This is true each time we slide by $B_i$ positions after sliding by $(m_i-1)B_i$ positions.

Focusing on the last column in the last row, sliding by $(\frac{K_i+1}{2}-1) B_i$ positions, in Table 7, voter $\frac{K_i+1}{2}A_i + \left( \frac{K_i+1}{2}-1 \right) B_i$ is equal to $\frac{K_{i-1}+1}{2}$. When sliding by $(\frac{K_i+1}{2}-1) B_i$ positions, all representatives in the first to $\frac{K_i+1}{2}$-th districts are lined up in ascending order at the i-th decision level. In other words, we can refer to the case in the $(\frac{K_i+1}{2}-1) B_i$-th row of the table as “full sliding.” When $i=1$ (i.e., the first decision level), voter $m_i A_i$ in the first row and first column in Table 7 is equal to $j_{t+1,1}^* = m_0$, and is the most extreme voter electable as the final representative, from the definition of $m_i$. In addition, voter $\frac{K_i+1}{2} A_i + (\frac{K_i+1}{2}-1)B_i$ in the last row and last column is equal to $\frac{N+1}{2}$, and is the median of all voters.

The same numbers may appear in Table 7 as shown in Example 3 in this section. Thus, we have all the numbers between $j_{t+1,1}^*$ and $\frac{N+1}{2}$, which means that those voters are electable as representatives of the $i+1$-th decision level.

In Table 7, we refer to the medians of the $m_i$-th district as Group $m_i$, those of the $m_i+1$-th as Group $m_i+1$, and so on. Lastly, those of the $\frac{K_i+1}{2}$-th district are Group $\frac{K_i+1}{2}$. To elect a voter as a district representative to the $i+1$-th decision level, we need to choose row l including the voter, $l \in \{ 0, 1, \ldots , (\frac{K_i+1}{2}-1)B_i \}$, from Table 7. Then, if the voter is in Group $m_i$, we district all representatives using the cracking and packing method at each decision level from the $i+1$-th to t-th levels without sliding any positions. This is because the voter is already at the same position as $j_{t+1,1}^*$’s under the cracking and packing method. If the voter is in Group $m_i+1$, she is out by one position from the $j_{t+1,1}^*$’s position. If the voter is in Group $m_i+2$, she is out by two positions, if in Group $m_i+3$, she is out by three positions, and so on. Thus, we must slide the voter by the number of positions that her group is away from Group $m_i$ at the $i+1$-th decision level to elect her as a district representative to the $i+2$-th decision level.

For simplicity, we renumber representatives at the $i+2$-th level as $\{ 1,2,3,\ldots , m_i, \ldots , \frac{K_i+1}{2}, \ldots , K_i \}$ at the $i+1$-th level. When we slide by 0 positions at the $i+1$-th decision level, each representative of Group $m_i$ can become the district median of the $m_{i+1}$-th district. When we slide by one position at the $i+1$-th decision level, each representative of Group $m_i+1$ can become the district median of the $m_{i+1}$-th district, and so on. Sliding each district representative individually and noting that $m_{i+1} = \frac{m_i}{\frac{1}{2} ( \frac{K_i}{K_{i+1}} + 1)}$, we have the same table at the $i+1$-th decision level as shown in Table 7, where i is replaced by $i+1$.

Noting that each decision level has the same structure without populations, all elected representatives are renumbered from one to $K_i$, $i=\{1, 2, 3, \ldots , t \}$ at each level. Then, at each decision level above the first, if representatives elected at the decision level below (i.e., the $i-1$-th level) are between Group $m_{i-1}$ and Group $\frac{K_{i-1}+1}{2}$, we can apply the above manipulation to the i-th level. Thus, we have the following lemma.

Lemma 2

Under Assumption 1, if a renumbered representative elected at the $i-1$-th decision level is a representative between $m_{i-1}$ and $\frac{K_{i-1}+1}{2}$ at the i-th decision level, where there are $K_{i-1}$ representatives, $i \in \{1,2,3, \ldots , t \}$, then she is electable as a district representative between the $m_{i}$-th district and the $\frac{K_{i}+1}{2}$-th to $i+1$-th decision levels.

Proof of Lemma 2

Note that $m_{i-1} = m_i A_i$ from the definition of $m_i$, as at the i-th decision level, $K_{i-1}$ representatives are elected at the $i-1$-th decision level. Thus, all representatives between $m_{i-1}$ and $\frac{K_{i-1}+1}{2}$ are lined up as district medians of the $m_{i}$-th to $\frac{K_i+1}{2}$-th districts in Table 7. Therefore, at the i-th decision level, all representatives between $m_{i-1}$ and $\frac{K_{i-1}+1}{2}$ are electable as a district representative to the $i+1$-th decision level. $\square$

Applying Lemma 2 repeatedly, we have Proposition 4.

Proposition 4

Under Assumption 1, any voter between $j_{t+1,1}^*$ and the median of all voters $\frac{N+1}{2}$ is electable as the final representative.

Proof of Proposition 4

Since $K_1> K_2> K_3> \ldots> K_t > K_{t+1} = 1$ and $\frac{1}{2}(\frac{K_i}{K_{i+1}}+1) \ge 1$, we have

$$\begin{aligned} j_{t+1,1}^* = m_0> m_1> m_2> m_3> \ldots> m_t > m_{t+1} = 1, \end{aligned}$$

from the definition of $m_i$, and we have

$$\begin{aligned} \frac{N+1}{2}> \frac{K_1+1}{2}> \frac{K_2+1}{2}> \ldots> \frac{K_t+1}{2} > \frac{K_{t+1}+1}{2} = 1. \end{aligned}$$

Thus, applying the proof of Lemma 2, both $m_i$ and $\frac{K_i+1}{2}$ in Table 7 shrink to 1 as $i \rightarrow t$. Lastly, the voter we want to elect as the final representative between $j_{t+1,1}^*$ and the voter’s median $\frac{N+1}{2}$ is elected at the final decision level according to the sandwich theorem. $\square$

In representative democracies, we should probably select the district maker from among all voters randomly if we want to avoid the policy bias of a district maker with an extreme policy position. However, this may not be so effective. We obtain the next corollary from Proposition 4.

Corollary 4

Under Assumption 1, when each voter becomes the district maker with an equal probability, the voters’ positions with the highest probability are those elected by the left and right extremists’ gerrymandering districts, $j_{t+1,1}^*$ and $N-j_{t+1,1}^*+1$, respectively.

Proof of Corollary 4

As there are N voters, the probability of each voter becoming the district maker is $\frac{1}{N}$. From Proposition 4, any district maker in $\{j_{t+1,1}^*, \ldots , \frac{N+1}{2} \}$ can elect herself as the final representative or can elect anyone in $\{ \frac{N+1}{2}+1, \ldots , N-j_{t+1,1}^*+1 \}$, by symmetry. Thus, the policies of each $\{ j_{t+1,1}^*, \ldots , \frac{N+1}{2}, \ldots , N-j_{t+1,1}^*+1 \}$ are implemented with probability $\frac{1}{N}$.

On the contrary, voters in $\{1, \ldots , j_{t+1,1}^* -1\}$ and $\{N-j_{t+1,1}^* +2, \ldots , N \}$ cannot elect themselves as the final representative by gerrymandering districts. Thus, their favorite and implementable policies by gerrymandering districts are the same as those implemented by $j_{t+1,1}^*$ and $N-j_{t+1,1}^*+1$, respectively. Therefore, $j_{t+1,1}^*$ and $N-j_{t+1,1}^*+1$ are elected as the final representative with probability $\frac{j_{t+1,1}^*}{N}$. $\square$

Example 3

Let us reconsider the example of $N=27$, $t=2$, where voters set $\mathcal {N} = \{ 1, 2, 3, \ldots , 25,26,27 \}$. In this case, the median of all voters is 14 and the most extremely liberal voter who is electable as the final representative is $j_{3,1}^* = \frac{1}{2}(3+1) \cdot \frac{1}{2}(\frac{9}{3}+1) \cdot \frac{1}{2}(\frac{27}{9}+1)=8.$ Lemma 2 and Proposition 4 state that voters between 8 and 14, who belong to either the fourth or the fifth district at the first decision level, from the definition of $m_i$, are electable as the final representative in this example.

Sliding voters: We have to slide voters by zero, one, two, three, and four positions to place voters 8, 9, 10, 11, 12, 13, 14 at the fourth and fifth district median positions, as $\frac{m_1-1}{2}(\frac{N}{K_1}-1) =\frac{4-1}{2}(\frac{27}{9}-1) = 3$, following Table 6. Then, we have Table 8.

Table 8 Placing voters between 8 and 14 on the district median at the first decision level

For instance, if we want voter 12 to be elected as a district representative in the second to final decision levels, we need to slide two positions. Then, we have the representatives $\{ 2,5,8,10,12,14,16,18,20\}$. At the second decision level, we need to slide the representatives by one position.

Now, we have the representatives $\{ 5,12,16 \}$ in the final decision level. Lastly, 12 is elected as the final representative.

Next, if we want voter 14 to be elected as a district representative in the second to final decision levels, we need to slide four positions, which is full sliding at the first decision level. Then, we have the representatives $\{ 2,5,8,11,14,16,18,20, 22 \}$. At the second level, we need to slide the representatives by one position, which is also full sliding, for representative 14 to belong to the median district.

At the final level, we have $\{ 5,14,18 \}$, and voter 14 is elected as the final representative. In this case, voters slide fully at each decision level so that all voters before voter 14 are lined up consecutively in ascending order.

Lastly, if we want voter 11 to be elected as a district representative in the second to final decision levels, we need to slide voters by one, three, or four positions. If we choose to slide by one position, we have the representatives $\{ 2,5,7,9,11,13,15,17,19\}$, and if we choose three positions, we have $\{ 2,5,8,11,13,15,17,19,21\}$. Sliding by four positions is similar to three positions because 11 is in the fourth district. When sliding by one position, at the second decision level, the representatives need to slide by one position. When sliding by three positions, they need to slide by zero positions.

As 11 appears as the district median of the fourth and fifth districts when sliding by one, three, and four positions, they achieve the same result at the first decision level. Thus, we can obtain the same result at the final decision level. $\Box$

1.2 Random districting

The next proposition, which is analogous to Proposition 2 of Gilligan and Matsusaka (2006), investigates the cases of the symmetric and upward skewed distributions of voters’ ideal points.

Proposition 5

Let Assumption 1 hold. Assuming that each districting is equally probable, the expected bias of random districting

• 1. Is zero if the voters’ ideal points are symmetrically distributed around their median, and

• 2. Is biased upward if the voters’ ideal points are skewed upward.

Proof of Proposition 5

Assume that the voters’ ideal points are in ascending order (i.e., $x_1^*<x_2^*< \ldots < x_N^*$). Let $M=(N+1)/2$. Denote by $p(x_i)$ the probability that voter i becomes the top-level representative (i.e., the policymaker or legislator).

We start by proving point 1. As a result of the symmetric setting, we must have $p(x_{M-i})=p(x_{M+i})$ and $x_M-x_{M-i}=x_{M+i}-x_M$ for all $i=0,\ldots ,M-1$. Hence,

$$\begin{aligned} E\left( x^*_{LEG}\right)= & {} \sum _{i=1}^N p(x_i) x_i=\sum _{i=1}^{M-1} p(x_{M-i})x_{M-i}+p(x_M)x_M\nonumber \\&+\sum _{i=1}^{M-1} p(x_{M+i})x_{M+i}=x_M. \end{aligned}$$

(11)

To establish point 2, we only have to replace $x_M-x_{M-i}=x_{M+i}-x_M$ with $x_M-x_{M-i}\le x_{M+i}-x_M$ for all $i=0,\ldots ,M-1$, which holds, as the distribution of ideal points is skewed upward in equation (11). $\square$

1.3 Partisan bias

Let voters who prefer party 0 to party 1 be called partisan. The proportion of voters favoring party 1 is $V = \frac{1}{N} \sum _{i=1}^{N} x_i$ and the proportion favoring party 0 is $1-V$. The party affiliation of representative k at the $t+1$-th decision level is $x_{t+1,1}^k \in \{ 0,1 \}$, corresponding to the median representative in the k-th district at the t-th decision level. The proportions of seats held by party 1 and party 0 in the final legislature are $L_t = \frac{1}{K_t} \sum _{k=1}^{K_t} x_{t+1,1}^k$ and $1-L_t$, respectively. Then, the partisan bias can be defined as $\beta _t = |L_t - V|$. If there is no bias, $L_t = V$. As a representative at the $t+1$-th decision level needs at least

$$\begin{aligned} h \equiv \frac{1}{2}\left( \frac{K_{t-1}}{K_t} +1 \right) \frac{1}{2}\left( \frac{K_{t-2}}{K_{t-1}} +1 \right) \ldots \frac{1}{2}\left( \frac{K_{1}}{K_2} +1 \right) \frac{1}{2}\left( \frac{K_{0}}{K_1} +1 \right) \end{aligned}$$

supporters at the voter’s level, as shown in the proof of Proposition 1, we obtain the next proposition of the partisan bias.

Proposition 6

Let Assumption 1 hold. Let V be the proportion of voters who support party 1 and suppose there are sufficient party 1 voters to elect at least one representative of party 1 to the $t+1$-th decision level, but not to elect all representatives: $\frac{1}{2}(\frac{K_{t-1}}{K_t}+1) \frac{1}{2}(\frac{K_{t-2}}{K_{t-1}}+1) \ldots \frac{1}{2}(\frac{K_1}{K_2}+1) \frac{1}{2}(\frac{N}{K_1}+1)< V N < K_t \frac{1}{2}(\frac{K_{t-1}}{K_t}+1) \frac{1}{2}(\frac{K_{t-2}}{K_{t-1}}+1) \ldots \frac{1}{2}(\frac{K_1}{K_2}+1) \frac{1}{2}(\frac{N}{K_1}+1)$. Then,

$$\begin{aligned} \max \beta _t = \left( \frac{ \frac{1}{K_t } \frac{K_t +1}{2}}{ \frac{j_{t+1,1}^*}{N}}-1 \right) V. \end{aligned}$$

Proof of Proposition 6

As one representative at the $t+1$-th decision level needs at least h voters, when there are $\sum _{i=1}^{K_t} x_{t+1,1}^i$ representatives at the $t+1$-th decision level, the minimum number of supporters at the voter’s level is $h \sum _{i=1}^{K_t} x_{t+1,1}^i$, which is equal to or less than the actual number of voters supporting party 1: $h \sum _{i=1}^{K_t} x_{t+1,1}^i \le \sum _{i=1}^N x_i$. Dividing this by N and multiplying by $\frac{1}{2}(K_t + 1)$, we obtain the following inequality:

$$\begin{aligned}&\frac{L_t K_t}{N} \frac{1}{2}(K_t + 1 ) \frac{1}{2}\left( \frac{K_{t-1}}{K_t} +1 \right) \frac{1}{2}\left( \frac{K_{t-2}}{K_{t-1}} +1 \right) \ldots \frac{1}{2}\left( \frac{K_{1}}{K_2} +1 \right) \frac{1}{2}\left( \frac{K_{0}}{K_1} +1 \right) \\&\quad \le \frac{1}{2}(K_t + 1 ) V. \end{aligned}$$

Using equation (3) and subtracting V from both sides, we obtain

$$\begin{aligned} \beta _t = L_t-V \le \left( \frac{\frac{1}{K_t}\frac{K_t+1}{2}}{\frac{j_{t+1,1}^*}{N}} -1 \right) V. \end{aligned}$$

$\square$

The maximum partisan bias $\beta _t$ is the ratio between the relative positions of the median seat, who is the final representative at the $t+1$-th decision level, and that at the voter level. This proposition is a generalization of the partisan bias $\beta$ calculated in Proposition 4 of Gilligan and Matsusaka (2006), and corresponds to $t=1$ in our model.