Voting power apportionments

Abstract

I propose apportioning the United States House of Representatives so as to equalize, to the extent possible, the voting power of the individual voter. Surprisingly such an apportionment falls squarely within the traditional apportionment paradigm and is, in a very precise sense, midway between the Hill method and the Webster method of apportionment.

Appendix: computing min–max apportionments

In this appendix I will present a general algorithm for finding an apportionment that minimizes the maximum difference between rank index functions. It generalizes my earlier discussion in Edelman (2006b) and applies to the min–max voting power apportionments discussed in Sect. 5 of this paper.

By a rank index I mean a real-valued function

$$\begin{aligned} r(p,a) : {\mathbb {Q}} \times {{\mathbb {N}} } \longrightarrow {\mathbb {R}}. \end{aligned}$$

that is monotone decreasing in $a$, i.e., $r(p,a-1) > r(p,a)$. Fix a population vector $\mathbf{p}$. For ease of presentation we will assume that

$$\begin{aligned} r(p_i,a) \ne r(p_j,b) \end{aligned}$$

for all $a, b \in {\mathbb {N}}$.

One solution $\mathbf{A}_\mathbf{r}(\mathbf{p}, h)$ to the apportionment problem $(\mathbf{p}, h)$ is given by

(1) $\text {for } h=0 \text { let } \mathbf{A}_\mathbf{r}(\mathbf{p},0)=\mathbf{0}$.

(2) $\text {if } \mathbf{A}_\mathbf{r}(\mathbf{p},h)=\mathbf{A}, \text {then } \mathbf{A}_\mathbf{r}(\mathbf{p}, h+1) \text {is found by giving}\,A_i+1$

$\text {seats to the state} \, i \, \hbox {such that} \, r(p_i,A_i) > r(p_k,A_k)$

$\text {for all} \, k \ne i, \hbox {and} \, A_k \,\hbox {seats to state} \, k \,\hbox {for each}\, k \ne i$.

Let $Max_r(\mathbf{p}, \mathbf{a})=\max _i r(p_i,a_i)$ where $\mathbf{a}$ is an apportionment. Note that, from the definition of ${\mathbf{A}_\mathbf{r}}$, $Max_r(\mathbf{p}, {\mathbf{A}_\mathbf{r}})$ is a decreasing function of $h$. The following lemma generalizes a well-known property of the Adams apportionment (Balinski and Young 2001, Proposition 3.10).

Lemma 7.1

The apportionment $\mathbf{A}= \mathbf{A}_\mathbf{r}(\mathbf{p}, h)$ minimizes $Max_r(\mathbf{p}, \mathbf{a})$ over all $h$-apportionments $\mathbf{a}$.

Proof

Suppose there is an $h$-apportionment $\mathbf{a}$ so that

$$\begin{aligned} r(p_i,a_i) =Max_r(\mathbf{p}, \mathbf{a}) < Max_r(\mathbf{p}, \mathbf{A})=r(p_j, A_j). \end{aligned}$$

If $a_j \le A_j$ then $r(p_j,a_j) \ge r(p_j,A_j)$ and so $Max_r(\mathbf{p}, \mathbf{a}) \ge Max_r(\mathbf{p}, \mathbf{A})$ which is a contradiction. Hence $a_j > A_j$.

Since both $\mathbf{A}$ and $\mathbf{a}$ are $h$-apportionments there must be a state $k$ so that $a_k < A_k.$ And, since $\mathbf{A}$ assigns more seats to state $k$ than $\mathbf{a}$ does, it follows that for some $h' < h$ we have that

$$\begin{aligned} r(p_k, a_k) =Max_r(\mathbf{p}, \mathbf{A}_\mathbf{r}(\mathbf{p}, h')) \end{aligned}$$

and since $Max_r$ is strictly decreasing in $h$ we have that

$$\begin{aligned}&r(p_k,a_k) \ge r(p_j, A_j) \Longrightarrow \\&Max_r(\mathbf{p}, \mathbf{a}) \ge r(p_k, a_k) \ge r(p_j, A_j) =Max_r(\mathbf{p}, \mathbf{A}), \end{aligned}$$

which contradicts the choice of $\mathbf{a}$. $\square$

We will say that an apportionment $\mathbf{a}$ is an $h'$-extension of $\mathbf{A}_\mathbf{r}(\mathbf{p}, h)=\mathbf{A}$ if $\mathbf{a}$ is an $h'$-apportionment with $h' \ge h$ and $a_i \ge A_i$ for all $1 \le i \le s$.

Lemma 7.2

If $\mathbf{a}$ is an apportionment with

$$\begin{aligned} Max_r(\mathbf{p}, \mathbf{a}) =Max_r(\mathbf{p}, \mathbf{A}), \end{aligned}$$

where $\mathbf{A}=\mathbf{A}_\mathbf{r}(\mathbf{p}, h)$ then $\mathbf{a}$ is an $h'$-extension of $\mathbf{A}$.

Proof

Let $r(p_k,A_k) = Max_r(\mathbf{p}, \mathbf{A})$. Then we know from the definition of $\mathbf{A}_\mathbf{r}(\mathbf{p}, h)$ that

$$\begin{aligned} r(p_l, A_{l}-1) > r(p_k, A_k) \ge r(p_l, a_l) \end{aligned}$$

for all $l$. Since $r$ is monotonically decreasing in $a$ it follows that $A_l \le a_l$ for all $l$ and we are done. $\square$

There is a dual version to $\mathbf{A}_r$ apportionments. For what follows we present it at a more general level than the earlier definition.

Suppose that $\mathbf{a}$ is an $h$-apportionment. For any $h' \ge h$, define the $h'$-extension of $\mathbf{a}$, $\mathbf{J}_r^{\mathbf{a}}(\mathbf{p}, h')$ by the recursion

(1) $\text { Let } \mathbf{J}_\mathbf{r}^{\mathbf{a}}(\mathbf{p},h)=\mathbf{a}$.

(2) $\text {if } \mathbf{J}_\mathbf{r}^{\mathbf{a}}(\mathbf{p},h^*)=\mathbf{J}, \text {then } \mathbf{J}_\mathbf{r}^{\mathbf{a}}(\mathbf{p}, h^*+1) \text{ is } \text{ found } \text{ by } \text{ giving }\, J_i+1$

$\text {seats to the state}\, i\, \hbox {such that}\, r(p_i,J_i+1) > r(p_k,J_k+1)$

$\text {for all}\, k \ne i,\,\hbox { and} \, J_k\, \hbox {seats to state}\, k\, \hbox {for each}\, k \ne i$.

This definition extends the definition in Edelman (2006b) and generalizes the definition of the Jefferson apportionment.

Let $Min_r(\mathbf{p}, \mathbf{a})=\min _i r(p_i,a_i)$ where $\mathbf{a}$ is an apportionment. The following lemma generalizes a well-known property of the Jefferson apportionment (Balinski and Young 2001, Proposition3.10.)

Lemma 7.3

Let $\mathbf{a}$ be an $h$-apportionment. The apportionment $\mathbf{J}_r^{\mathbf{a}}(\mathbf{p}, h')$ maximizes $Min_r(\mathbf{p}, \mathbf{a}')$ over all $h'$-extensions of $\mathbf{a}$.

Proof

Let $\mathbf{J}=\mathbf{J}_\mathbf{r}^{\mathbf{a}}(\mathbf{p}, h')$ and $\mathbf{a}'$ be some other $h'$-extension of $\mathbf{a}$ so that

$$\begin{aligned} r(p_i,J_i)=Min_r(\mathbf{p}, \mathbf{J}) < r(p_k,a'_k)=Min_r(\mathbf{p}, \mathbf{a}'). \end{aligned}$$

Then

$$\begin{aligned} r(p_i, a'_i) \ge r(p_k,a'_k) > r(p_i,J_i) \end{aligned}$$

and so $a'_i < J_i$ because $r$ is monotone decreasing in $a$. Both $\mathbf{a}'$ and $\mathbf{J}$ are $h'$-apportionments hence there is some $l$ so that $a'_l \ge J_l +1$ and so

$$\begin{aligned} r(p_l, J_l+1) \ge r(p_l, a'_l). \end{aligned}$$

From the definition of $\mathbf{J}$ we have that

$$\begin{aligned} r(p_i,J_i) \ge r(p_l,J_l+1) \end{aligned}$$

because state $i$ received it’s $J_i^{th}$ seat before state $l$ received it’s $J_l+1^{st}$ seat. Thus

$$\begin{aligned} r(p_i,J_i) \ge r(p_l,J_l+1) \ge r(p_l,a'_l) \ge r(p_k,a'_k); \end{aligned}$$

the last inequality holds because $r(p_k,a'_k)=Min_r(\mathbf{p},\mathbf{a}')$. Contradiction. $\square$

I need one last piece of notation. Suppose that $\mathbf{a}$ is an $h$-apportionment and let $k$ be a state. By $\mathbf{a}|_k$ I mean the $(h-a_k)$-apportionment for the states with $k$ deleted.

I can now develop the algorithm that produces an apportionment that will minimize the largest difference between the rank functions, i.e., we wish to find an $h$-apportionment that achieves the value

$$\begin{aligned} \min _{\mathbf{a}} \left[ Max_r(\mathbf{p}, \mathbf{a})-Min_r(\mathbf{p}, \mathbf{a})\,\right] . \end{aligned}$$

(0)

For each $h'$, $h \ge h' \ge 0$, let $\mathbf{A}_r^{h'}=\mathbf{A}_r(\mathbf{p},h')$. If $k$ is the state that realizes $Max_r(\mathbf{p}, \mathbf{A}_r^{h'})$ let $\overline{\mathbf{J}_r^{h'}}$ be the $h$-apportionment obtained by taking $\mathbf{J}_r^{\mathbf{a}|_k}(\mathbf{p}|_k,h-a_k)$ and then assigning state $k$ the number of seats it was given in $\mathbf{A}_r^{h'}$. That is, we start by apportioning $h'$ seats according to method $\mathbf{A}_r$, set aside the the state that achieves the $Max_r(\mathbf{p}, \mathbf{A}_r^{h'})$, and extend to the $h$ apportionment using the method $\mathbf{J}_r^{\mathbf{a}|_k}$ on the rest of the states. We then have the following theorem, which extends (Edelman 2006b, Theorem 4)

Theorem 7.4

An apportionment that achieves

$$\begin{aligned} \min _{\mathbf{a}} \left[ Max_r(\mathbf{p}, \mathbf{a})-Min_r(\mathbf{p}, \mathbf{a})\,\right] . \end{aligned}$$

is found among those in the set

$$\begin{aligned} \{ \overline{\mathbf{J}_r^{h'}}| h \ge h' \ge 0\}. \end{aligned}$$

Proof

Let $\mathbf{a}'$ be an $h$-apportionment that achieves the minimum of equation $(0)$ and suppose that

$$\begin{aligned} r(p_k,a'_k) = Max_r(\mathbf{p}, \mathbf{a}'). \end{aligned}$$

By Lemma 7.1 we know that the apportionment $\mathbf{A}=\mathbf{A}_r(\mathbf{p},h)$ minimizes $Max_r(\mathbf{p}, \mathbf{a})$ over all $h$ apportionments, it must be true that for some $h' \le h$ that

$$\begin{aligned} r(p_k,a'_k) = Max_r(\mathbf{p}, \mathbf{A}_r(\mathbf{p}, h')). \end{aligned}$$

It then follows from Lemma 7.2 that $\mathbf{a}'$ is an $h$-extension of $\mathbf{A}_r(\mathbf{p}, h')$. Thus $\mathbf{a}'|_k$ is an extension of $A_r(\mathbf{p}, h')|_k$ and so by Lemma 7.3

$$\begin{aligned} Min_r(\mathbf{p}, \mathbf{a}') \le Min_r(\mathbf{p}, \overline{\mathbf{J}_r^{h'}}) \end{aligned}$$

and thus

$$\begin{aligned} Max_r(\mathbf{p}, \mathbf{a}')-Min_r(\mathbf{p}, \mathbf{a}') \ge Max_r(\mathbf{p},\overline{\mathbf{J}_r^{h'}} )-Min_r(\mathbf{p}, \overline{\mathbf{J}_r^{h'}}) \end{aligned}$$

and so are done. $\square$

This algorithm is presented for a rank index function that is monotone decreasing in $a$. I made this choice so that the theorems would be clear generalizations of my earlier work in Edelman (2006b). The astute reader will note that the rank index function that we discuss in the body of this paper, $\sqrt{\frac{a_i}{p_i}}$ is in fact monotone increasing in $a$. Fortunately the same arguments go through by the appropriate exchanges of $min$ for $max$ and $\le$ for $\ge$. I will leave the details for the reader.