# ML Wiki

## Arrow's Impossibility Theorem

Arrow's Impossibility Theorem is a Voting Theory theorem (sometimes called Arrow's Paradox)

In contrast to May's Theorem, in this theorem we assume 3 (and more) candidates, not 2:

• $A = \{x, y, z\}$

## Voting Theory Relations

We define the following relations:

$P$ the preference relation

• $P_i$ is a individual preference of voter $i$
• $x \ P_1 \ y$ means that voter 1 prefers $x$ to $y$
• $P$ is the global aggregated preference

$I$ is the indifference relation

• $I_i$ is an individual indifference of voter $i$
• $I_i$ is symmetric: $a \ I_1 \ b \iff b \ I_1 \ a$

$S$ is the "at least as good as" relation

• $S \equiv (P \lor I)$
• $x \ S \ y \iff x \ (P \lor I) \ y \Rightarrow {\color{grey}{(1)}} \ x \ P \ y \lor {\color{grey}{(2)}} \ x \ I \ y \lor {\color{grey}{(3)}} \ y \ P \ x$

We can express $P$ and $I$ only via $S$
1. $x \ P \ y \equiv [x \ S \ y] \land [y \ \overline{S} \ x]$
2. $x \ I \ y \equiv [x \ S \ y] \land [y \ S \ x]$

## Axioms

Axiom 1
Completeness
$\forall x, y \in A:$ either $x \ S \ y$ or $y \ S \ x$
Axiom 2
Consistency (or Transitivity)
$\forall x, y, z \in A: x \ S \ y \land y \ S \ z \Rightarrow x \ S \ z$

## Lemmas

$\forall x, y, z \in A:$

• (1) $x \ S \ x$ (reflectivity)
• (2) $x \ P \ y \Rightarrow x \ S \ y$
• (3) $x \ P y \land y \ P \ z \Rightarrow x \ P \ z$ (transitivity for $P$)
• (4) $x \ I \ y, y \ I \ z \Rightarrow x \ I \ z$ (?)
• (5) either $x \ S \ y$ or $y \ P \ x$
• (6) $x \ P \ y \land y \ S \ z \Rightarrow x \ P \ z$

### Lemma 3: Transitivity for $P$

We want to show that $\forall x, y \in A: x \ P \ y \land y \ P \ z \Rightarrow x \ P \ z$

By Axiom 2 we have transitivity for $S$:

• $x \ S \ y \land y \ S \ z \Rightarrow x \ S \ z$

$x \ S \ z \equiv x \ (P \lor I) \ y \equiv \underbrace{[x \ P \ y]}_\text{(1)} \lor \underbrace{[x \ I \ z]}_\text{(2)}$

• (1) this is what we want to show
• (2) want to show that assuming it leads to contradiction and therefore the only possible outcome is (1)

Assuming (2): $x \ I \ z$

• by Symmetry $x \ I \ z \iff z \ I \ x$
• $z \ I \ x \Rightarrow [z \ S \ x] \land [x \ S \ z] \Rightarrow z \ S \ x$

$z \ S \ x \land x \ S \ y \Rightarrow z \ S \ y$ (by transitivity of $S$)

• since $S$ is less strict than $P$ we have:
$x \ P \ y \Rightarrow x \ S \ y$ and $y \ P \ z \Rightarrow y \ S \ z$
• $z \ S \ y$ contradicts with $y \ P \ z$
$z \ S \ y$ means one of the following: $z \ P \ y$ or $z \ I \ y$, and neither of them are true

Contradiction. $\square$

### Lemma 5

$\forall x, y \in A: x \ S \ y \lor y \ P \ x$

• assume the opposite: $\exists x, y \in A: \overline{ x \ S \ y \lor y \ P \ x }$
• $\overline{ x \ S \ y \lor y \ P \ x } \equiv \overline{ x \ S \ y} \land \overline{ y \ P \ x } \equiv y \ P \ x \land y \ \overline{S} \ x$
• this is clearly a contradiction: they both can never be true at the same time

Another way to show that is by completeness of $P$ and $S$

### Lemma 6

$\forall x, y \in A: x \ P \ y \land y \ S \ z \Rightarrow x \ P \ z$

• $x \ P \ y \equiv [x \ S \ y] \land [y \ \overline{S} \ x]$ just rewriting $P$ with $S$
• $[x \ S \ y] \land [y \ S \ z] \Rightarrow x \ S \ z$ by transitivity
• $x \ S \ z \iff \underbrace{ x \ P \ z}_\text{(1)} \lor \underbrace{x \ I \ z}_\text{(2)}$
• if (2) is false then (1) must be true for the Lemma to hold

Assume $x \ I \ z$

• $x \ I \ z \equiv \underbrace{x \ S \ z}_\text{true} \land z \ S \ x$
• consider the initial hypothesis $y \ S \ z$ and $z \ S \ x$:
• by transitivity $y \ S \ z \land z \ S \ x \Rightarrow y \ S \ x$
• but $y \ S \ x$ contractions with $a \ P \ y$
• therefore $x \ P \ y$ (1) must be true

$\square$

## Conditions

The social welfare function $H$ it takes two individual rankings and produces the global (social) choice:

$H(R_1, R_2) \to R$

$H$ must satisfy the following conditions:

### Condition 1: Universality

$H$ is defined for every pair $R_1$ and $R_2$

• i.e. for each pair there should exist a solution
• we want to avoid the Condorcet Problem - a cycle

This condition is also called Unrestricted Domain

### Condition 2: Monotonicity

Monotonicity is satisfied if

• if an alternative $x$ rises or does not fall in the individual ordering
• and if $x$ was preferred to another alternative $y$ before the change in the individual orderings
• then $x$ should still be preferred to $y$ in the global ordering

### Condition 3: Independence to Third Alternatives

Independence to Third Alternatives is satisfied when

• given the set of alternatives $A = \{x, y\}$
• individual preferences over $A$ are not affected by other alternatives $z$ that are not in $A$
• when these alternatives $z$ are also considered in $A' = A \cup {z}$ it should not affect the ordering between alternatives from $A$

### Condition 4: Non-Imposition

$H$ should not be imposed

$H$ is imposed if

• for some distinct $x$ and $y$, for rankings $R_1$ and $R_2$: $x \ S \ y$
• $S$ relates to the global ordering obtained from $H$

In other words,

• $x$ is always as good as $y$ or better in the aggregated ordering
• no matter what $y$ is

Example of some imposed conditions:

• men preferred to women,
• one religion is preferred to other,
• etc

### Condition 5: No Dictatorship

$H$ should not be dictatorial

$H$ is dictatorial if

• $\exists$ individual $i$ s.t. $\forall x, y \in A: x \ P_i \ y \Rightarrow x \ P \ y$
• no matter what is the individual ranking $R_i$
• such individual $i$ is called the dictator

In other words:

• for any alternatives, the choice of the dictator determines the outcome

## Theorem

Arrow's Impossibility Theorem
All five conditions cannot be satisfied simultaneously.

## Consequences

### Consequence 1: Unanimity

Unanimity is satisfied ($N = 2$) if

• $x \ P_1 \ y$ and $x \ P_2 \ y$ then $x \ P \ y$
• or $x$ is globally preferred to $y$ if both voters vote for $x$

It follows from Non-Imposition and Monotonicity

Non-Imposition

• $H$ is imposed if $\exists x, y: \forall R_1, R_2: x \ S \ y$
• thus $H$ is not imposed if $\exists R_1, R_2: x \ P \ y$

From $R_1$ and $R_2$ we can build two rankings

• $R_1 \to R'_1$ and $R_2 \to R'_2$
• we assume only that $x$ improves his positions in these rankings:
• $x$ has the same or better position in $R' = H(R'_1, R'_2)$
• by Monotonicity if he was preferred to $y$ in $R$, he's still preferred to $y$ in $R'$

?

For example:

• $R: y < x < z \ \to \ R': x < y < z$
• $x$ improved his positions

### Consequence 2: Dictatorship

We have a dictator if:

• $\exists x, y \in A: x \ P \ y \land x \ P_1 \ y \land y \ P_2 \ x \Rightarrow [ x \ P_1 \ y \to x \ P \ y ]$
• if we have two opposite points of view of voters $a_1$ ($x$ is better) and $a_2$ ($y$ is better) and the opinion of voter $a_1$ becomes the global opinion
• then no matter how $a_2$ votes, the opinion of $a_1$ determines the global preference

Let's consider two rankings:

• $R_1: x < \ ...$ one where $x$ is on the first position
• $R_2$ any ranking with $x$ at some position - $R_2: \ ... < x < \ ...$

We can build two new rankings $R'_1$ and $R'_2$ from them:

• $R'_1 \equiv R_1$: also with $x$ on the first position
• $R'_2: \ ... < x$: by taking $R_2$ and moving $x$ to the last position
• this way we constructed two rankings with completely opposite points of view
• (this corresponds to $x \ P_1 \ y \land y \ P_2 \ x$)
• $H(R'_1, R'_2) = R'$, and by our hypothesis $x \ P \ y$ in $R'$

Now let's return from $R'_2$ to $R_2$:

• $x$ improves his positions by going from $R'_2$ to $R_2$:
• by Monotonicity he should remain the winner

I.e. no matter what is the ranking $R_2$, $R_1$ always determines the outcome

• or $x \ P_1 \ y \to x \ P \ y$

$\square$

### Consequence 3: Indifference

In a strong conflict the only possible outcome is indifference:

• if $x \ P_1 \ y \land y \ P_2 \ x \Rightarrow x \ I \ y$
• otherwise we have dictatorship

Assume $x \ I \ y$ is false. It means:

• ether $x \ P \ y$
• or $y \ P \ x$

Assume $x \ P \ y$:

• $x \ P_1 \ y \land y \ P_2 \ x \Rightarrow x \ P \ y$
• it means that the voter 1 is dictator

Same for $y \ P \ x$:

• it would mean the the voter 2 is dictator

### Consequence of Consequences

Suppose we have two ratings:

• $R_1: {\color{blue}{x < y}} < z$
• $R_2: z < {\color{blue}{x < y}}$
• i.e. both candidates agree that $x \ P \ y$

But there's a strong opposition when it comes to $z$:

• (1) $x$ vs $z$
• $x \ P_1 z \land z \ P_2 \ x$
• the only possible outcome in this case is indifference (by Consequence 3)
• $x \ I \ z$
• (2) $y$ vs $z$
• $y \ P_1 z \land z \ P_2 \ y \Rightarrow y \ I \ z$

by (1) + (2):

• $y \ I \ z \equiv z \ I \ y$
• $x \ I \ z \land z \ I \ y \Rightarrow x \ I \ y$ by transitivity of $I$ (Lemma 4)
• but it contradicts with $x \ P \ y$

Therefore, these 5 conditions cannot be satisfied simultaneously.

## Solution

How to overcome this limitation?

• Relax 1-st condition: don't need to always return the entire rankings, it's enough to return just one candidate