Representative democracy and the implementation of majority-preferred alternatives

Abstract

In this paper, we contrast direct and representative democracy. In a direct democracy, individuals have the opportunity to vote over the alternatives in every choice problem the population faces. In a representative democracy, the population commits to a candidate ex ante who will then make choices on its behalf. While direct democracy is normatively appealing, representative democracy is the far more common institution because of its practical advantages. The key question, then, is whether representative democracy succeeds in implementing the choices that the group would make under direct democracy. We find that, in general, it does not. We model a population as a distribution of voters with strict preferences over a finite set of alternatives and a candidate as a strict ordering of those alternatives that serves as a binding, contingent plan of action. We focus on the case where the direct democracy choices of the population are consistent with a strict ordering of the alternatives. We show that even in this case, where the normative recommendation of direct democracy is clearest, representative democracy may not elect the candidate with this ordering.

Appendix

1.1 Proof for n = 4

In this section, we prove Proposition 3 stated in Sect. 3.2. For $n=4$, $UC(T^{\lambda })$ satisfies order consistency.

Proof

Assume majority preferences are consistent with e. We show that there is no ordering that can cover e. The key step is to recognize that we can apply Proposition 1 to rule out any ordering fewer than five transpositions from e: $eT^{\lambda }\pi$ for any $\pi$ within two transpositions, so they cannot cover e, and for those three or four transpositions away, even if they beat e, they will be defeated by at least one ordering one or two transpositions from e (which e beats). So, the only orderings that could potentially cover e are five or six transpositions away from e: $\{a_{4}a_{3}a_{1}a_{2},a_{4}a_{2}a_{3}a_{1},a_{3}a_{4} a_{2}a_{1},a_{4}a_{3}a_{2}a_{1}\}$. We rule these out one at a time:

• We cannot have $a_{4}a_{3}a_{1}a_{2}$ covers e, since $eT^{\lambda } a_{2}a_{3}a_{1}a_{4}T^{\lambda }a_{4}a_{3}a_{1}a_{2}$ for any population with majority preferences consistent with e. Proposition 1 proves $eT^{\lambda }a_{2}a_{3}a_{1}a_{4}$. We cannot have $a_{4}a_{3}a_{1}a_{2}T^{\lambda }a_{2}a_{3}a_{1}a_{4}$ since all of the orderings closer or equidistant to $a_{4}a_{3}a_{1}a_{2}$ than $a_{2} a_{3}a_{1}a_{4}$ have $a_{4}$ precedes $a_{2}$. Thus, if more than half the population were closer to or equidistant to $a_{4}a_{3}a_{1}a_{2}$, we would not have $a_{2}\succ a_{4}$ in the majority preference.

• We cannot have $a_{4}a_{2}a_{3}a_{1}$ covers e, since we must have $eT^{\lambda }a_{4}a_{2}a_{3}a_{1}$. All of the orderings closer or equidistant to $a_{4}a_{3}a_{1}a_{2}$ than e have $a_{4}$ precedes $a_{1}$. Thus, if more than half the population were closer to or equidistant to $a_{4} a_{2}a_{3}a_{1}$ than to e, we would not have $a_{1}\succ a_{4}$ in the majority preference.

• We cannot have $a_{3}a_{4}a_{2}a_{1}$ covers e, since $eT^{\lambda } a_{1}a_{4}a_{2}a_{3}T^{\lambda }$ $a_{3}a_{4}a_{2}a_{1}$ for any population with majority preferences consistent with e. Proposition 1 proves $eT^{\lambda }a_{1}a_{4}a_{2}a_{3}$. We cannot have $a_{3}a_{4}a_{2}a_{1}T^{\lambda }a_{1}a_{4}a_{2}a_{3}$ since all of the orderings closer or equidistant to $a_{1}a_{4}a_{2}a_{3}$ than $a_{3} a_{4}a_{2}a_{1}$ have $a_{3}$ precedes $a_{1}$. Thus, if more than half the population were closer to or equidistant to $a_{3}a_{4}a_{2}a_{1}$, we would not have $a_{1}\succ a_{3}$ in the majority preference.

• We cannot have $a_{4}a_{3}a_{2}a_{1}$ covers e, since $eT^{\lambda } a_{1}a_{3}a_{2}a_{4}T^{\lambda }a_{4}a_{3}a_{2}a_{1}$ for any population with majority preferences consistent with e. Proposition 1 proves $eT^{\lambda }a_{1}a_{3}a_{2}a_{4}$. We cannot have $a_{4}a_{3}a_{2}a_{1}T^{\lambda }a_{1}a_{3}a_{2}a_{4}$ since all of the orderings closer or equidistant to $a_{4}a_{3}a_{2}a_{1}$ than $a_{1} a_{3}a_{2}a_{4}$ have $a_{4}$ precedes $a_{1}$. Thus, if more than half the population were closer to or equidistant to $a_{4}a_{3}a_{2}a_{1}$, we would not have $a_{1}\succ a_{4}$ in the majority preference. $\square$

1.2 Population restrictions

A natural question to ask in this context is whether we can impose restrictions on the distribution of preferences that would guarantee order consistency. This domain restriction approach has been adopted by many social choice theorists in attempts to rule out other paradoxical outcomes; perhaps most classic is the single-peakedness restriction pioneered independently by Black (1948) and Arrow (1951) and generalized by Sen and Pattanaik (1969). Their goal was to describe a class of populations for which majority rule over alternatives would not cycle. The domain restriction they proposed requires populations to be unimodal in the sense that all members of the population, for any particular triple of alternatives, must be able to agree on an alternative that is not worst. Assuming the number of voters is odd, this condition is sufficient for transitive majority rule.

Clearly, this restriction will not be enough to assure order consistency, as the class of populations we consider in our counterexample above are indeed single-peaked in terms of preferences over alternatives. However, we can use a similar idea, that of restricting the number of modes in the distribution, in order to derive a sufficient condition for order consistency in our framework. The class of populations with transitive majority rule consistent with the ordering e can be thought of as having a “peak” or cluster of weight around e. Our sufficiency condition says that as we move away from e, we must not encounter another cluster of orderings similar to one another. In order to state this condition more formally, it will be useful to introduce some new terminology. When referring to a population with transitive majority rule consistent with e, we will call any pairwise disagreement with e an error. For example, we will say that an ordering $\pi$ that is m transpositions from e contains m errors. We can state our sufficiency condition in terms of these errors.

Proposition 5

Consider the class of populations with transitive majority rule consistent with e. Then, order consistency holds if for any set of m errors, $m\geqq 5$, we have

$$\begin{aligned} {\displaystyle \sum } \lambda \left( \pi |\pi \text { contains at least }\frac{1}{2}\text { of these }m\text { errors}\right) <\frac{1}{2} \end{aligned}$$

Proof

Suppose $e\notin$

$UC(T^{\lambda })$. We will show there must exist a set of m errors, $m\geqq 5$, such that ${\sum } \lambda (\pi |\pi$ contains at least $\frac{1}{2}$ of these m errors$)>\frac{1}{2}$. Since $e\notin$

$UC(T^{\lambda })$, we know there exists $\hat{\pi }$ such that $\hat{\pi }$ covers e. Let $\hat{\pi }$ contain m errors; we know $m\geqq 5$ in order for $\hat{\pi }$ to cover e. Since $\hat{\pi }$ and e agree on all pairwise choices outside of the m errors, we know that $f(\pi ,\hat{\pi })$ and $f(\pi ,e)$ are determined only by how many of the m errors $\pi$ contains. Those $\pi$ that have less than $\frac{1}{2}$ of the m errors have $f(\pi ,e)<f(\pi ,\hat{\pi })$. So, suppose the set of orderings that had at least $\frac{1}{2}$ of these m errors in common with $\hat{\pi }$ had mass less than $\frac{1}{2}$. Then, we would have ${\sum } \lambda (\pi |f(\pi ,e)<f(\pi ,\hat{\pi })>\frac{1}{2}$. This would imply $eT^{\lambda }\hat{\pi }$, contradicting $\hat{\pi }$ covers e. $\square$

This sufficiency condition has a straightforward intuition. If we encounter a population that contains a mass of orderings that are both (a) relatively distant from e, and (b) relatively close to one another, then we may have the type of counterexample presented above. This type of cluster of similar orderings far from e may be able to agree upon a compromise candidate which covers e, but only if together they constitute a majority. The condition rules out this possibility by assuring that no such cluster of mass greater than $\frac{1}{2}\$exists.

This is a sufficient but not necessary condition for order consistency. A gray area exists between the class of populations described in our counterexample above and the class of populations described by this sufficiency condition. For some populations that fail the condition above, the distribution of mass on orderings far from e may be too dispersed to agree upon an ordering like $\hat{\pi }$ which could beat everything that e beats. One might ask whether we could improve the sufficiency condition by restricting this set of distant orderings to fall within a certain radius of one another. Below, we provide an example that illustrates why this strategy fails.

Example 1

Why Tightening the Sufficiency Condition by Restricting the Radius of the Outlier Orderings Does Not Work.

Consider the following population, a slight variant from the example presented in Sect. 3:

Using the strategy from the proof above, we can show that $\hat{\pi } =a_{2}a_{1}a_{4}a_{3}a_{6}a_{5}a_{8}a_{7}a_{10}a_{9}$ covers e, the ordering consistent with majority preferences. We need to show that (a) $\hat{\pi }T^{\lambda }e$ and (b) for all $\pi ^{\prime }\in \Pi ,$ $eT^{\lambda }\pi ^{\prime }\Rightarrow \hat{\pi }T^{\lambda }\pi ^{\prime }$. First we will show that $\hat{\pi }T^{\lambda }e$. For the five orderings in population with weight $\frac{1}{5}(\frac{1}{2}-\varepsilon ),$ we have $f(\pi ,\hat{\pi })=1$ and $f(\pi ,e)=4$. And, we know $a_{10}a_{9}a_{8}a_{7}a_{6}a_{5}a_{4}a_{3} a_{2}a_{1}$ is closer to $\hat{\pi }$ than e, since it is maximally distant from e. Thus, $\frac{1}{2}+2\varepsilon$ of the population is closer to $\hat{\pi }$ than e, so $\hat{\pi }T^{\lambda }e.$ Now we need to show there cannot exist $\pi ^{\prime }$ such that $eT^{\lambda }\pi ^{\prime }$ but $\pi ^{\prime }T^{\lambda }\hat{\pi }.$ Suppose there did exist such a $\pi ^{\prime }$. Then, $eT^{\lambda }\pi ^{\prime }$ implies that for at least one of the orderings $\pi$ in population other than e, $\ f(e,\pi )\le f(\pi ^{\prime },\pi )$. We know there cannot exist a $\pi ^{\prime }$, $\pi ^{\prime }\ne e$, such that $f(e,a_{10}a_{9}a_{8}a_{7}a_{6}a_{5}a_{4} a_{3}a_{2}a_{1})\le f(\pi ^{\prime },a_{10}a_{9}a_{8}a_{7}a_{6}a_{5}a_{4} a_{3}a_{2}a_{1})$. So, it must be that this is true for one of the remaining five orderings. Since for any of these orderings $f(e,\pi )=4$, we must have $f(\pi ^{\prime },\pi )\ge 4$ for at least one of those five orderings $\pi$. And, the fact that $\pi ^{\prime }T^{\lambda }\hat{\pi }$ implies that we have at least one of the following two cases:

• 1. For at least one of the orderings with weight $\frac{1}{5}(\frac{1}{2}-\varepsilon )$, we have $f(\pi ^{\prime },\pi )\le f(\hat{\pi },\pi )$.

• 2. For both e and $a_{10}a_{9}a_{8}a_{7}a_{6}a_{5}a_{4}a_{3}a_{2}a_{1}$, we have $f(\pi ^{\prime },\pi )<f(\hat{\pi },\pi )$.

For case 1, we know $f(\hat{\pi },\pi )=1$, so this would imply, $f(\pi ^{\prime },\pi )\le 1$ for one of the orderings with weight $\frac{1}{5}(\frac{1}{2}-\varepsilon )$. This leads to the same violation of the triangle inequality that we reached above, since for any two orderings with weight $\frac{1}{5}(\frac{1}{2}-\varepsilon )$, we have $f(\pi _{i},\pi _{j})\le 2$. For case 2, $f(\pi ^{\prime },e)<f(\hat{\pi },e)$ implies $f(\pi ^{\prime },e)<5.$ And, $f(\pi ^{\prime },a_{10}a_{9}a_{8}a_{7}a_{6}a_{5}a_{4}a_{3}a_{2}a_{1} )<f(\hat{\pi },a_{10}a_{9}a_{8}a_{7}a_{6}a_{5}a_{4}a_{3}a_{2}a_{1})$ implies $f(\pi ^{\prime },a_{10}a_{9}a_{8}a_{7}a_{6}a_{5}a_{4}a_{3}a_{2}a_{1})<40$. But, $f(\pi ^{\prime },e)<5$ and $f(\pi ^{\prime },a_{10}a_{9}a_{8}a_{7}a_{6}a_{5} a_{4}a_{3}a_{2}a_{1})<40$ cannot both hold, since the first implies $\pi ^{\prime }$ has fewer than five errors and the second implies it has more than five errors. This leads to a contradiction. Thus, order consistency fails for this population.

This example illustrates the difficulty we encounter if we attempt to tighten the sufficiency condition for ordering consistency by imposing a radius around the orderings with common errors. Taking the basic counterexample from above, where the minority orderings all lie relatively close to another, we can move some weight to $a_{10}a_{9}a_{8}a_{7}a_{6}a_{5}a_{4}a_{3}a_{2}a_{1}$ and still arrive at $\hat{\pi }$ covers e. Thus, it is not always true that we need the minority orderings to be relatively close to one another in order to have order consistency fail.