Votes and elections

Votes and elections

There may be many types of elections -- based on the types of votes used.

We will see what we can say about an electoral system without actually considering the alternative to be voted for.

First, we are given a set of all possible (affirmative) votes $V$.

Now the potential elections $E$. Where do they come from?

First, they include all single-vote elections, i.e., $V\subset E$.

An election is formed as if the voters come in and cast their votes consecutively. This is given by a sum of votes: $$u,v\in V \ \Rightarrow \ u+v\in E.$$ Furthermore, two elections are combined together in the same fashion to produce a new election. This is given by a sum of elections: $$u,v\in E \ \Rightarrow \ u+v\in E,$$

Then, the main property is that any election can be combined with any other to produce a new election. This requirement is satisfied by any binary relation.

• Unrestricted Elections -- Closedness:

$$u,v\in E \ \Longrightarrow \ u+v\in E.$$

For a given set $V$ of votes, an election is a sum (with possible repetitions) of votes: $$v=v_1 + ... + v_m,\ v_1, ..., v_m \in V.$$ Alternatively, an election is a linear combination of distinct votes with non-negative integer coefficients: $$v=p_1 v_1 + ... + p_mv_m,\ v_1, ..., v_m \in V,\ v_i\ne v_j \text{ for } i\ne j,\ p_1, ..., p_m \in {\bf Z}^+.$$

Let's examine its properties.

If three groups of voters arrive in the same order but are combined -- pairwise -- in two different ways, the result is the same election. This requirement is satisfied by any group.

• Consistency -- Associativity: For any $u,v,w\in E$, we have

$$(u+v)+w=u+(v+w).$$

Proposition. The election set $E=<V>$ is a semigroup generated by $V$.

Whether a group of voters comes to cast their votes earlier or later than another group of voters shouldn't matter. This requirement is satisfied by any Abelian group.

• Interchangeability -- Commutativity: For any $u,v\in E$, we have:

$$v+u=u+v.$$

A special, neutral vote is available, i.e., one that doesn't change the outcome of any election. Alone, it is simply a failed election. This requirement is satisfied in any monoid.

• Participation -- Zero Element: There is some element $0\in E$ such that for any $v\in E$, we have

$$v+0=v.$$

In some settings, $0$ is not allowed to be a vote. For example, an all-zero rating vote or an all-tied ranking may be or may not be allowed. We assume that an election when no voters have showed up is neutral/failed. Once again, $V$ is the set of affirmative votes: $$0\in E\setminus V.$$

These properties amount to the following.

Proposition. The election set $E=<V>$ is a commutative monoid generated by $V$.

Note that we will not be considering the $n$-vote elections apart from the $(n+1)$-vote elections. In fact, if the $(n+1)$-st vote is neutral the two are the exact same election.

Some of the elections end in failure; those are neutral elections, $$x_1+...+x_k=0.$$ Some of the elections result in votes, $$x_1+...+x_k\in V.$$ We call the former elementary elections $V$.

Example. Suppose the votes are ratings -- $1$ to $5$ stars -- of alternatives $A,B,C,...$. Then, if the first vote $v$ gives one of them $2$ points and the second $w$ gives $1$, then $(u+v)\in V$ conceivably gives $2+1=3$ points: $$v(A)=2,\ w(A)=2 \ \Rightarrow\ (v+w)(A)=2+2=4.$$ Then, $u+v\in V$. On the other hand, consider: $$v(A)=5,\ w(A)=5 \ \Rightarrow\ (v+w)(A)=5+5=10.$$ The outcome $X=u+v$ with $X(A)=5+5=10$ is not an allowed vote but it is a reasonable outcome so that $X\in E\setminus V$. Then, $u+v\not\in V$. $\square$

Example. For a ranking vote, the outcome of the election with the two identical votes: $A>B$ and $A>B$, could be chosen to be the vote $A>B$ again. On the other hand, the outcome of the election of $A>B$ combined with $A<B$ can't be a vote when ties are disallowed. Then, we have $(A>B)+(A<B)\in E\setminus V$. When ties are allowed, all outcomes are votes, $V=E$. $\square$

Example. For a selection note, suppose each vote simply picks a single winner from a list. Then, if the first vote $v$ picks $A$ and the second $w$ picks $B$, the outcome $u+v$ isn't an allowed vote but it is a necessary outcome. $\square$

Example. Another example is about voting on a budget: how to distribute a total of $ \$ 1000000$ among $100$ items, i.e., each vote is a string $(x_1,...,x_{100})$ of real numbers with the same restriction: $$x_1+...+x_{100}=1000000.$$ Then, the sum of any two such votes, $(x_1,...,x_{100})$ and $(y_1,...,y_{100})$, wouldn't be an allowed vote. $\square$

Once the outcome of each individual vote is determined, there is only one reasonable way to assign an outcome to any election.

Next, what if we have a richer algebra?

We require that any vote can be cancelled by some other vote or an election. This requirement is satisfied in any group.

• Cancellation -- Negative Element: For any $v\in V$, there is some $-v\in E$ such that

$$v+(-v)=0.$$

In some settings, it isn't possible to have a vote $-v$. For example, the negative of a rating or the “anyone but” selection may not be allowed as votes.

Proposition. The election set $E=<V>$ is an Abelian group generated by $V$ called the election group.

We may require that the outcome of a single vote coincides with the outcome of an election with this vote repeated.

• Unanimity -- Mean: For any $v\in V$,

$$v+v=v.$$

In a group, we can cancel: $$\begin{array}{ll} v+v=v & \Longrightarrow\ (v+v)+(-v)=v+(-v)\\ &\Longrightarrow\ v+(v+(-v))=v+(-v) \\ &\Longrightarrow\ 0+v=0 \\ &\Longrightarrow\ v=0. \end{array}$$ We have the theorem below.

Theorem (Triviality of Elections). The election group with Unanimity is trivial, $E=0$.

An example of Unanimity without Cancellation is produced by $V$ consisting of the three projections of ${\bf R}^3$ onto the three coordinate planes.

Rankings and selections

We consider a simple setting of ranking, i.e., ordering multiple alternatives, $$K:=\{A_1 , A_2, A_3 ,... , A_n\}.$$ The alternatives may be candidates running for elections, movie recommendations, and other votes of personal preferences.

A ranking vote $r$ is a complete ordering with ties of the alternatives. It may look like this: $$r=( A_1 < A_2 = A_3 <... < A_n),$$ or $$r= \left(A_1 < \begin{array}{cc}A_2\\A_3\end{array} <... < A_n \right).$$

In particular, for $n=3$ there are strict orderings and orderings with ties: $$R(\{1,2,3\})=\left\{ 123,132,231,321,213,312, \begin{array}{cc}1\\2\end{array}3,\begin{array}{cc}1\\3\end{array}2, \begin{array}{cc}2\\3\end{array}1,... ,1\begin{array}{cc}2\\3\end{array}, \begin{array}{cc}1\\2\\3\end{array} \right\}.$$

We denote the set of all such ranking votes by $$V=R(K):=\left\{r=\left(A_1 \begin{array}{cc}<\\=\end{array} A_2 \begin{array}{cc}<\\=\end{array} ... \begin{array}{cc}<\\=\end{array} A_n \right):\ A_k\in K,\ A_i\ne A_j \right\}.$$

We will apply the Triviality of Elections.

First, we choose the fully tied ranking is to be the neutral vote: $$0:=( A_1 = A_2 = A_3 =... = A_n).$$ Second, we choose the negative of a ranking to be the reversed ranking: $$\forall A,B\in K,\quad r:A<B \ \Longleftrightarrow\ -r:B<A;$$ or $$-\left(A_1 \begin{array}{cc}<\\=\end{array} A_2 \begin{array}{cc}<\\=\end{array} ... \begin{array}{cc}<\\=\end{array} A_n \right)\ = \ \left(A_1 \begin{array}{cc}>\\=\end{array} A_2 \begin{array}{cc}>\\=\end{array} ... , \begin{array}{cc}>\\=\end{array} A_n \right).$$

We have proven the following algebraic analog of AIT.

Theorem (Triviality of Rankings). The election group of rankings $E=<R(K)>$ is trivial.

What is the difference from the original AIT?

• In AIT, the number of voters/votes is fixed at $m$;
• AIT's pair-wise Unanimity is stronger than Unanimity we have used;
• AIT's Non-dictatorship is weaker than Interchangeability we have used;
• we replace IIA with Cancellation.

It is more typical to expect a single winner rather than a ranking in an election.

A selection vote $r$ is an ordering of a partition of the alternatives into two groups (winners and losers). It may look like this: $$s=( A_1 = A_2=...= A_k < A_{k+1} =... = A_n).$$ We can also think of a selection as a function: $$s:K\to \{0,1\}.$$

In particular, for $n=3$ there are strict orderings and orderings with ties: $$S(\{1,2,3\})=\left\{ \begin{array}{cc}1\\2\end{array}3,\begin{array}{cc}1\\3\end{array}2, \begin{array}{cc}2\\3\end{array}1,... , 1\begin{array}{cc}2\\3\end{array} , \begin{array}{cc}1\\2\\3\end{array}\right\}.$$

We denote the set of all such selection votes by $$V=S(K):=\{ \{A_1,...,A_k\}, \{A_{k+1}...,A_n \} : \ A_i\in K,\ A_i\ne A_j \}.$$

We apply the Triviality again. Selection are just special type of rankings and the meaning of the negative $-s$ of a selection $s$ is clear; it is the reversed selection: $$(-s)(A)=1-s(A).$$

We have proven the following.

Theorem (Triviality of Selections). The election group of selections $E=<S(K)>$ is trivial.

Of course, any “mixed” elections $V=<R(K)>$ with $E=S(K)$ is also trivial under these three conditions because the zero and the negatives are the same.

Disallowing ranking votes with ties won't change the result.

Disallowing selection votes with more than one winner will prevent us from having an obvious choice for a negative (there may be many) of a vote. A negative vote, that is; what about an election? For a given candidate $A\in K$, define the selection vote $s_A$ with $A$ the single winner by: $$s_A(A)=1 \text{ and } s_A(B)=0 \ \forall B\ne A.$$ Then, its negative is the election given by: $$-s_A:=\sum_{B\ne A} s_B.$$