Liberal political equality does not imply proportional representation

Abstract

In their article ‘Liberal political equality implies proportional representation’, which was published in Social Choice and Welfare 33(4):617–627 in 2009, Eliora van der Hout and Anthony J. McGann claim that any seat-allocation rule that satisfies certain ‘Liberal axioms’ produces results essentially equivalent to proportional representation. We show that their claim and its proof are wanting. Firstly, the Liberal axioms are only defined for seat-allocation rules that satisfy a further axiom, which we call Independence of Vote Realization (IVR). Secondly, the proportional rule is the only anonymous seat-allocation rule that satisfies IVR. Thirdly, the claim’s proof raises the suspicion that reformulating the Liberal axioms in order to save the claim won’t work. Fourthly, we vindicate this suspicion by providing a seat-allocation rule which satisfies reformulated Liberal axioms but which fails to produce results essentially equivalent to proportional representation. Thus, the attention that their claim received in the literature on normative democratic theory notwithstanding, van der Hout and McGann have not established that liberal political equality implies proportional representation.

Appendices

Appendices

1.1 Appendix A: on positive responsiveness$_I$

Van der Hout and McGann (2009a, b) present different definitions of positive responsiveness. The notion of positive responsiveness that we discussed in the body of this paper is the one used by van der Hout and McGann (2009b) and McGann (2006). In this appendix, we will present positive responsiveness$_I$, the notion used by van der Hout and McGann (2009a) and explain why, to our minds, this notion is not normatively compelling. Here is the definition of positive responsiveness$_I$.

Definition 16

(Positive responsiveness$_I$) A seat-allocation rule f for an election ${\mathcal {E}} = (E, N, {\mathcal {A}})$ is positive responsive$_I$ iff for all profiles $\mathbf{P}$ and $\mathbf{Q}$ for which:

• (i) for all $i \in N$: if $\mathbf{P}_{ix} = 1$ then $\mathbf{Q}_{ix} = 1$, and

• (ii) for some $i \in N$: $\mathbf{Q}_{ix} = 1$ and $\mathbf{P}_{ix} = 0$,

we have: $f(\mathbf{Q})_x > f(\mathbf{P})_x$.

Now, when (i) everyone who votes for x in $\mathbf{P}$ also votes for x in $\mathbf{Q}$ whereas (ii) some vote for x in $\mathbf{Q}$ but not in $\mathbf{P}$, we say that $\mathbf{Q}$ is obtained from $\mathbf{P}$ by a change favouring x. So, when $\mathbf{Q}$ is obtained from $\mathbf{P}$ by a change favouring x, a positive responsive$_I$ f allots more seats to x in $\mathbf{Q}$ than in $\mathbf{P}$.

Although the definition of positive responsiveness$_I$ is clear enough, its conceptual underpinning is not. For, remember that in the single-vote elections under consideration, individuals can also abstain from voting. As such, when $\mathbf{Q}$ is obtained from $\mathbf{P}$ by a change favouring x, the additional support for x in $\mathbf{Q}$, relative to $\mathbf{P}$, may co-exist with stronger additional support in $\mathbf{Q}$ for other parties. Under these circumstances, it is not reasonable to require, as positive responsiveness$_I$ does, that x receives more seats in $\mathbf{Q}$ than in $\mathbf{P}$. For a concrete illustration of our qualms with the notion of positive responsiveness$_I$, consider the following two profiles for ELECT.

Profile $\mathbf{D}$. Voters 1, $2, \ldots , 8$ vote for party a, voter 10 votes for party b, all other voters abstain from voting.

Profile $\mathbf{E}$. Voters 1, $2, \ldots , 9$ vote for party a, voter $10, 11, \ldots 18$ vote for party b.

Party a receives more support in $\mathbf{E}$ than in $\mathbf{D}$. However, in $\mathbf{D}$, where the voter turnout is only $50\%$, nearly all of those who do vote, vote for a. In contrast, in $\mathbf{E}$, voter turnout is $100\%$ with only half of the voters voting for a. It seems reasonable that party a receives more seats in $\mathbf{D}$ than in $\mathbf{E}$, which conflicts with the requirements of positive responsiveness$_I$. Hence, positive responsiveness$_I$ is not normatively compelling.

Moreover, note that the proportional rule allots, to party a, 8 seats in D, and 4.5 seats in E. Hence, the profiles $\mathbf{D}$ and $\mathbf{E}$ testify that the proportional rule violates positive responsiveness$_I$. Now this is somewhat odd, as van der Hout and McGann (2009a) define the notion of positive responsiveness$_I$ with the purpose of giving an axiomatic justification of P: they seek to justify P in terms of an axiom that P does not satisfy. Although this is odd, it is not absurd. For, van der Hout and McGann only seek to define axioms whose satisfaction is sufficient for a seat-allocation rule to induce the same winning coalitions as P. To be sure, P trivially induces the same winning coalitions as P. However, this does not entail that P needs to satisfy axioms which constitute a sufficient—in contrast to a necessary—condition for inducing the same winning coalitions as P. Nevertheless, we do think that positive responsiveness$_I$ is a less compelling notion than the notion of positive responsiveness used by van der Hout and McGann (2009b) and McGann (2006), which is why we chose to work with the latter notion in the body of the paper.

1.2 Appendix B: proof of Proposition 3

For sake of completeness, we give the proof of Proposition 3. Our proof closely follows the proof of Thomson (2019: 184), but we elaborate on a couple of proof-steps for the convenience of the reader. There is one notable distinction between our proof and that of Thomson, though. In his proof, Thomson uses a theorem on Cauchy’s functional equation, which applies to functions from ${\mathbb {R}}$ to ${\mathbb {R}}$. As Proposition 3 is concerned with tallied-vote problems rather than with claims problems, we need a similar result to the one invoked by Thomson, but which pertains to functions from ${\mathbb {N}}$ to ${\mathbb {R}}$. The result that we need is the following lemma, whose proof we present for the sake of completeness.

The Cauchy lemma.

Let $V > 0$ be an integer and let $\varphi : \{0, 1, \ldots , V \} \rightarrow {\mathbb {R}}$ be a function which satisfies the following (Cauchy) equation for all $x, y \in \{0, 1, \ldots , V \}$:

$$\begin{aligned} \varphi (x) + \varphi (y) = \varphi (x + y) \end{aligned}$$

(18)

Then there exists a $\lambda \in {\mathbb {R}}$ such that:

$$\begin{aligned} \text{ For } \text{ all } x \in \{0, 1, \ldots , V \}: \varphi (x) = \lambda \cdot x \end{aligned}$$

(19)

Proof

We claim that $\lambda = \varphi (1)$ satisfies (19) and we will establish this claim by induction on $x \in \{0, 1, \ldots , V \}$.

Induction base. If $x = 0$, it follows from (18) that $2 \cdot \varphi (0) = \varphi (0)$ so that $\varphi (0) = 0$. So for $x = 0$, $\varphi (x) = \lambda \cdot x$ is satisfied for any $\lambda$ whatsoever and so in particular for $\lambda = \varphi (1)$.

Induction step. Suppose that $\lambda = \varphi (1)$ satisfies (19) for some $x \in \{0, 1, \ldots , V-1 \}$. We may then show that $\lambda = \varphi (1)$ also satisfies (19) for $x+1$, and hence establish our lemma, as follows:

$$\begin{aligned} \varphi (x + 1) = \varphi (x) + \varphi (1) = \varphi (1)\cdot x + \varphi (1) = \varphi (1)\cdot (x + 1) \end{aligned}$$

(20)

The first equality in (20) follows from (18), the second from the induction hypothesis and the last from elementary algebra. $\square$

Proposition 3 (P from no advantageous transfer) Let ${\mathcal {E}} = (E, N, {\mathcal {A}})$ be an election with $\mid {\mathcal {A}} \mid \ge 3$ and let r be a tallied seat-allocation rule for ${\mathcal {E}}$. Then r satisfies no advantageous transfer if and only if r is the proportional rule P.

Proof

It is readily verified that P satisfies no advantageous transfer. We will show that any tallied seat-allocation rule r which satisfies no advantageous transfer is the proportional rule. Let V be an integer $> 0$ and let $({\mathcal {E}}, v)$ be any tallied-vote problem such that $\sum _{j \in {\mathcal {A}}} v_j = V$. We first establish that the following four claims are true:

$$\begin{aligned} r(v)_1 + r(v)_2 = r(v_1+ v_2, 0, v_3, \ldots , v_k)_1 \end{aligned}$$

(21)

When party 1 and party 2 reallocate the votes that they receive in v, this does not affect the sum of seats received by the parties in virtue of no advantageous transfer. When 1 and 2 reallocate by giving all votes to 1 and none to 2, party 2 receives no seat in virtue of no votes no seats so that, after this reallocation, 1 get all the seats that 1 and 2 jointly received before the reallocation, which is what (21) expresses.

$$\begin{aligned} r(v)_1 = r(v_1, 0, V- v_1, 0, \ldots , 0)_1, \quad r(v)_2 = r(0, v_2, V- v_2, 0, \ldots , 0)_2 \end{aligned}$$

(22)

When all parties in ${\mathcal {A}} - \{ i \}$ reallocate the votes amongst them, the sum of seats they receive should remain the same in virtue of no advantageous transfer. In virtue of efficiency then, party i receives E minus the sum of seats allotted to the parties in ${\mathcal {A}} - \{ i \}$ before and after reallocation. This is what is expressed by (22) for $i = 1, 2$ and for the reallocation in which the parties in ${\mathcal {A}} - \{ i \}$ transfer all their votes to party 3.

$$\begin{aligned} r(v) _2 = r(0, v_2, V- v_2, 0, \ldots , 0)_2 = r(v_2, 0, V- v_2, 0, \ldots , 0)_1 \end{aligned}$$

(23)

The left-most equality of (23) follows from (22). As for the right-most equality of (23): the two vote vectors are related to one another by a reallocation of votes by party 1 and 2 so the sum of seats they receive on the basis of these vote vectors is the same in virtue of no advantageous transfer. For the vote vector displayed on the left-hand side of the equality, party 1 gets no votes so that, by no votes no seats, party 2 gets the full sum of seats. Similarly, for the vote vector displayed on the right-hand side of the equality, it is party 1 who gets the full sum. So party 1 and party 2 indeed get the same amount of seats for the vote vectors of the right-most equality of (23).

$$\begin{aligned} r(v_1 + v_2, 0, v_3, \ldots , v_k)_1 = r(v_1 + v_2, 0, V- v_1 - v_2, 0, \ldots , 0)_1 \end{aligned}$$

(24)

The truth of (24) is established similar to that of (22).

We define the function $\varphi : \{0, \ldots V \} \mapsto {\mathbb {R}}$, by letting:

$$\begin{aligned} \varphi (t):= r(t, 0, V-t, 0, \ldots , 0)_1 \end{aligned}$$

From the definition of $\varphi$, (22) and (23) we get:

$$\begin{aligned} r(v)_1 = \varphi (v_1), \qquad r(v)_2 = \varphi (v_2) \end{aligned}$$

(25)

From the definition of $\varphi$ and (24) we get:

$$\begin{aligned} r(v_1 + v_2, 0, v_3, \ldots , v_k)_1 = \varphi (v_1 + v_2) \end{aligned}$$

(26)

From (25), (26) and (21) we get:

$$\begin{aligned} \varphi (v_1) + \varphi (v_2) = \varphi (v_1 + v_2) \end{aligned}$$

(27)

Remember that V is fixed and that we are considering an arbitrary tallied-vote problem $({\mathcal {E}}, v)$ which respects the constraint that $\sum _{j \in {\mathcal {A}}} v_j = V$. So then, it follows from (27) that we have effectively established that for all $x, y \in \{0, 1, \ldots , V \}$ we have:

$$\begin{aligned} \varphi (x) + \varphi (y) = \varphi (x + y) \end{aligned}$$

(28)

It follows from (28) and the Cauchy lemma that there is a $\lambda \in {\mathbb {R}}$ such that for each $x \in \{0, \ldots , V \}$ we have:

$$\begin{aligned} \varphi (x) = \lambda \cdot x \end{aligned}$$

(29)

Following proof-steps similar used to those for establishing (25), we get:

$$\begin{aligned} \text{ For } \text{ all } \, i \in {\mathcal {A}}: r(v)_i = \varphi (v_i) \end{aligned}$$

(30)

It follows from (29) and (30) that

$$\begin{aligned} \text{ For } \text{ all } \, i \in {\mathcal {A}}: r(v)_i = \lambda \cdot v_i \end{aligned}$$

(31)

It follows from (31) and the efficiency of r that:

$$\begin{aligned} \lambda = \frac{E}{ \sum _{i \in {\mathcal {A}}} v_i } \end{aligned}$$

(32)

So that it follows from (31) and (32) that r is P. $\square$