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Arrow's Impossibility Theorem Kevin Feasel December 10, 2006 http://36chambers.wordpress.com/ arrow/

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Kevin Feasel December 10, 2006 http://36chambers.wordpress.com/arrow/. Arrow's Impossibility Theorem. The Rules. 2 individuals with 3 choices (x, y, z) --> 6 profiles for each. x > y > z, x > z > y, y > x > z, y > z > x, z > x > y, z > y > x - PowerPoint PPT Presentation

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Page 1: Arrow's Impossibility Theorem

Arrow's Impossibility Theorem

Kevin FeaselDecember 10, 2006

http://36chambers.wordpress.com/arrow/

Page 2: Arrow's Impossibility Theorem

The Rules 2 individuals with 3 choices (x, y, z) --> 6

profiles for each. x > y > z, x > z > y, y > x > z, y > z > x, z > x > y,

z > y > x 36 profiles in all for our two-person example (6 * 6)

Page 3: Arrow's Impossibility Theorem

The 36 Profiles1 2 3 4 5 6 7 8 9

3 Z 3 Y 3 Z 3 X 3 Y 3 X 3 Y 3 X 3 Y

2 Y 2 Z 2 X 2 Z 2 X 2 Y 2 Z 2 Y 2 X

1 X 1 X 1 Y 1 Y 1 Z 1 Z 1 X 1 Z 1 Z

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> > > > > > > > >

10 11 12 13 14 15 16 17 183 Y 3 Y 3 Z 3 X 3 Y 3 X 3 X 3 Y 3 Y2 X 2 X 2 Y 2 Z 2 X 2 Y 2 Y 2 X 2 Z

1 Z 1 Z 1 X 1 Y 1 Z 1 Z 1 Z 1 Z 1 X

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> > > > > > > > >

19 20 21 22 23 24 25 26 273 Y 3 Z 3 Y 3 Y 3 Z 3 Z 3 Z 3 X 3 Z

2 Z 2 X 2 Z 2 Z 2 Y 2 Y 2 Y 2 Z 2 Y

1 X 1 Y 1 X 1 X 1 X 1 X 1 X 1 Y 1 X

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> > > > > > > > >

28 29 30 31 32 33 34 35 363 X 3 Z 3 Z 3 X 3 Z 3 X 3 Z 3 X 3 X

2 Z 2 X 2 X 2 Y 2 X 2 Y 2 X 2 Z 2 Z

1 Y 1 Y 1 Y 1 Z 1 Y 1 Z 1 Y 1 Y 1 Y

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> > > > > > > > >

Page 4: Arrow's Impossibility Theorem

Explanation Of Symbols Colored Text

Blue – First choice for both

Green – Second choice for both

Red – Third choice for both

Black – Inconsistent choices (e.g., in profile 10, Z > Y for person A but Y > Z for person B)

Numbers along the left-hand side depict the #1, #2, and #3 choices for person A.

Numbers along the bottom depict the #1, #2, and #3 choices for person B.

Each profile has a number (e.g., 1 and 10 here).

How to read this: for profile 10, person A (left-hand side) likes Z > X > Y. Person B (bottom) prefers X > Y > Z.

13 Z

2 Y

1 X

1 2 3

>

103 Y2 X

1 Z

1 2 3

>

Page 5: Arrow's Impossibility Theorem

The Assumptions Completeness – All profiles must be solvable. In

this case, we have 3 choices among 2 people, so 36 total sets of preferences could arise. We must have a solution for each one of the 36.

Page 6: Arrow's Impossibility Theorem

The Assumptions Unanimity – If all individuals agree on a single

position, that position will be guaranteed. In our case, if an option is red, both individuals rank

this option as the #3 choice, so it will end up as #3 in the preference ranking setup.

Page 7: Arrow's Impossibility Theorem

The Assumptions Independence of Irrelevant Alternatives – The

relationship between X and Y should be independent of Z. In general terms, the relationship between two elements will not change with the addition of another element.

Example: in profile 25, X > Y for person A (X > Y > Z) and for person B (X > Z > Y). Adding in option Z, we assume, does not alter this relationship.

Example: in profile 26, Y > X for person A (Y > Z > X) and for person B (Z > Y > X). If we removed option Z, we would still expect Y > X for both.

253 Z

2 Y

1 X

1 2 3>

263 X2 Z

1 Y

1 2 3

>

Page 8: Arrow's Impossibility Theorem

Proving The Final Rule Non-dictatorship. We want to understand whether

we can come up with a social rule which follows the rules of completeness, unanimity, and independence of irrelevant alternatives, and which is simultaneously non-dictatorial. “Dictatorial” here means that all 36 profiles will match one person's profiles exactly. In other words, all social choices will precisely match the individual's preferences.

Page 9: Arrow's Impossibility Theorem

Solving The Problem 6 preference sets are already solved, thanks to

unanimity. In profile 1, for example, all parties agree that X > Y > Z, so X > Y > Z is the social rule. These six completed profiles have the “>” highlighted in yellow.

Page 10: Arrow's Impossibility Theorem

Solving The Problem1 2 3 4 5 6 7 8 9

3 Z 3 Y 3 Z 3 X 3 Y 3 X 3 Y 3 X 3 Y

2 Y 2 Z 2 X 2 Z 2 X 2 Y 2 Z 2 Y 2 X

1 X 1 X 1 Y 1 Y 1 Z 1 Z 1 X 1 Z 1 Z

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> X Y Z > X Z Y > Y X Z > Y Z X > Z X Y > Z Y X > > >

10 11 12 13 14 15 16 17 183 Y 3 Y 3 Z 3 X 3 Y 3 X 3 X 3 Y 3 Y2 X 2 X 2 Y 2 Z 2 X 2 Y 2 Y 2 X 2 Z

1 Z 1 Z 1 X 1 Y 1 Z 1 Z 1 Z 1 Z 1 X

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> > > > > > > > >

19 20 21 22 23 24 25 26 273 Y 3 Z 3 Y 3 Y 3 Z 3 Z 3 Z 3 X 3 Z

2 Z 2 X 2 Z 2 Z 2 Y 2 Y 2 Y 2 Z 2 Y

1 X 1 Y 1 X 1 X 1 X 1 X 1 X 1 Y 1 X

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> > > > > > > > >

28 29 30 31 32 33 34 35 363 X 3 Z 3 Z 3 X 3 Z 3 X 3 Z 3 X 3 X

2 Z 2 X 2 X 2 Y 2 X 2 Y 2 X 2 Z 2 Z

1 Y 1 Y 1 Y 1 Z 1 Y 1 Z 1 Y 1 Y 1 Y

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> > > > > > > > >

Page 11: Arrow's Impossibility Theorem

Solving The Problem In addition to this, we know that any profile with

colored text is also solved—both people agree on where to place this option. So these can all be filled in as well.

Page 12: Arrow's Impossibility Theorem

Solving The Problem1 2 3 4 5 6 7 8 9

3 Z 3 Y 3 Z 3 X 3 Y 3 X 3 Y 3 X 3 Y

2 Y 2 Z 2 X 2 Z 2 X 2 Y 2 Z 2 Y 2 X

1 X 1 X 1 Y 1 Y 1 Z 1 Z 1 X 1 Z 1 Z

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> X Y Z > X Z Y > Y X Z > Y Z X > Z X Y > Z Y X > X > X >

10 11 12 13 14 15 16 17 183 Y 3 Y 3 Z 3 X 3 Y 3 X 3 X 3 Y 3 Y2 X 2 X 2 Y 2 Z 2 X 2 Y 2 Y 2 X 2 Z

1 Z 1 Z 1 X 1 Y 1 Z 1 Z 1 Z 1 Z 1 X

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> > Z > Z > Y > Y > > > X >

19 20 21 22 23 24 25 26 273 Y 3 Z 3 Y 3 Y 3 Z 3 Z 3 Z 3 X 3 Z

2 Z 2 X 2 Z 2 Z 2 Y 2 Y 2 Y 2 Z 2 Y

1 X 1 Y 1 X 1 X 1 X 1 X 1 X 1 Y 1 X

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> > Y > Y > Z > > > X > X > Y

28 29 30 31 32 33 34 35 363 X 3 Z 3 Z 3 X 3 Z 3 X 3 Z 3 X 3 X

2 Z 2 X 2 X 2 Y 2 X 2 Y 2 X 2 Z 2 Z

1 Y 1 Y 1 Y 1 Z 1 Y 1 Z 1 Y 1 Y 1 Y

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> Z > > > Z > Z > Y > X > >

Page 13: Arrow's Impossibility Theorem

Solving The Problem There are two remaining sets of relationships, as shown

in profile 23.

a > b: in this example, Y > Z for both individuals. Y is the #2 choice for person A and Z #3, whereas Y is the #1 choice for person B and Z #2. In a case such as this, we know that Y > Z in the social preferences because both people prefer Y to Z.

a ? b: in this example, Z > X for person B, but X > Z for person A. A's preferences are X > Y > Z and B's are Y > Z > X. In this type of situation, we do not yet know if Z or X will be socially preferred because there is no agreement between individuals.

233 Z

2 Y

1 X1 2 3

>

Page 14: Arrow's Impossibility Theorem

Solving The Problem In this case, because we know that Y > Z socially,

we can put that down. But we do not know if Y > Z > X, Y > X > Z, or X > Y > Z, so we will not write these down just yet, but once things start clearing up, these rules will be used.

The key here is that Z is above and to the right of Y, which means that there is agreement that Y > Z. However, Z is above and to the left of X, meaning there is conflict. Y is also above and to the left of X, so there is another conflict, due to the fact that X > Y > Z for person A, but Y > Z > X for person B.

233 Z

2 Y

1 X1 2 3

> ? ? ?

Page 15: Arrow's Impossibility Theorem

The Single Choice If we look at profile 7, there is a single choice to be made.

Person A (left-hand column): X > Z > Y

Person B (bottom row): X > Y > Z

We know that X will be the #1 preference, but we have to decide whether Y > Z or Z > Y here. We can choose either, but let us pick that Y > Z. In other words, we support person B's preference here.

This means that the social preference will be X > Y > Z. In addition, because of the Independence of Irrelevant Alternatives axiom, any time we see a conflict between Y and Z similar to the one in profile 7, we know to choose Y > Z.

Let us fill in the chart with this new information...

73 Y

2 Z

1 X

1 2 3> X Y Z

Page 16: Arrow's Impossibility Theorem

Results1 2 3 4 5 6 7 8 9

3 Z 3 Y 3 Z 3 X 3 Y 3 X 3 Y 3 X 3 Y

2 Y 2 Z 2 X 2 Z 2 X 2 Y 2 Z 2 Y 2 X

1 X 1 X 1 Y 1 Y 1 Z 1 Z 1 X 1 Z 1 Z

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> X Y Z > X Z Y > Y X Z > Y Z X > Z X Y > Z Y X > X Y Z > Y Z X > Y Z X

10 11 12 13 14 15 16 17 183 Y 3 Y 3 Z 3 X 3 Y 3 X 3 X 3 Y 3 Y2 X 2 X 2 Y 2 Z 2 X 2 Y 2 Y 2 X 2 Z

1 Z 1 Z 1 X 1 Y 1 Z 1 Z 1 Z 1 Z 1 X

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> X Y Z > Z > Z > Y > Y > > > Y X Z >

19 20 21 22 23 24 25 26 273 Y 3 Z 3 Y 3 Y 3 Z 3 Z 3 Z 3 X 3 Z

2 Z 2 X 2 Z 2 Z 2 Y 2 Y 2 Y 2 Z 2 Y

1 X 1 Y 1 X 1 X 1 X 1 X 1 X 1 Y 1 X

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> > Y > Y > Z > > > X > X > Y

28 29 30 31 32 33 34 35 363 X 3 Z 3 Z 3 X 3 Z 3 X 3 Z 3 X 3 X

2 Z 2 X 2 X 2 Y 2 X 2 Y 2 X 2 Z 2 Z

1 Y 1 Y 1 Y 1 Z 1 Y 1 Z 1 Y 1 Y 1 Y

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> Z > > > Z > Z > X Y Z > X > >

Page 17: Arrow's Impossibility Theorem

Explanation The profiles which have been newly solved (old solutions are colored green;

new are yellow) are profiles 7, 8, 9, 10, 17, and 33. 7, 8, 17, and 33 were solved due to knowing one position already because of unanimity. 9 was solved because both individuals support Z > X, so we know that Y > Z > X there. 10 was solved because both individuals support X > Y, so we know that the result must be X > Y > Z.

In doing this, we also have a new rule: in 33, X is above and to the left of Y, and X > Y. Because of Independence of Irrelevant Alternatives, we can state that X > Y whenever X is above and to the left of Y, like in 33. This will allow us to fill in more profiles, which we will do now.

Page 18: Arrow's Impossibility Theorem

Step 2 Results1 2 3 4 5 6 7 8 9

3 Z 3 Y 3 Z 3 X 3 Y 3 X 3 Y 3 X 3 Y

2 Y 2 Z 2 X 2 Z 2 X 2 Y 2 Z 2 Y 2 X

1 X 1 X 1 Y 1 Y 1 Z 1 Z 1 X 1 Z 1 Z

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> X Y Z > X Z Y > Y X Z > Y Z X > Z X Y > Z Y X > X Y Z > Y Z X > Y Z X

10 11 12 13 14 15 16 17 183 Y 3 Y 3 Z 3 X 3 Y 3 X 3 X 3 Y 3 Y2 X 2 X 2 Y 2 Z 2 X 2 Y 2 Y 2 X 2 Z

1 Z 1 Z 1 X 1 Y 1 Z 1 Z 1 Z 1 Z 1 X

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> X Y Z > Z > Z > Y > Y > > > Y X Z >

19 20 21 22 23 24 25 26 273 Y 3 Z 3 Y 3 Y 3 Z 3 Z 3 Z 3 X 3 Z

2 Z 2 X 2 Z 2 Z 2 Y 2 Y 2 Y 2 Z 2 Y

1 X 1 Y 1 X 1 X 1 X 1 X 1 X 1 Y 1 X

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> > Y > Y > Z > > > X > X > Y

28 29 30 31 32 33 34 35 363 X 3 Z 3 Z 3 X 3 Z 3 X 3 Z 3 X 3 X

2 Z 2 X 2 X 2 Y 2 X 2 Y 2 X 2 Z 2 Z

1 Y 1 Y 1 Y 1 Z 1 Y 1 Z 1 Y 1 Y 1 Y

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> X Z Y > > X Z Y > Z X Y > X Y Z > X Y Z > Z X Y > Z X Y > X Y Z

Page 19: Arrow's Impossibility Theorem

Step 2 Results Explanation We have 7 new completed profiles: 28, 30, 31, 32, 34, 35, 36. In addition, we

have two new rules, as determined by profile 28. When X is above and to the left of Z, X > Z. Also, when Z is above and to the left of Y, Z > Y Let us see how many new solutions we can find knowing that:

Y above and to the left of Z --> Y > Z

X above and to the left of Y --> X > Y

X above and to the left of Z --> X > Z

Z above and to the left of Y --> Z > Y

Page 20: Arrow's Impossibility Theorem

Step 3 Results1 2 3 4 5 6 7 8 9

3 Z 3 Y 3 Z 3 X 3 Y 3 X 3 Y 3 X 3 Y

2 Y 2 Z 2 X 2 Z 2 X 2 Y 2 Z 2 Y 2 X

1 X 1 X 1 Y 1 Y 1 Z 1 Z 1 X 1 Z 1 Z

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> X Y Z > X Z Y > Y X Z > Y Z X > Z X Y > Z Y X > X Y Z > Y Z X > Y Z X

10 11 12 13 14 15 16 17 183 Y 3 Y 3 Z 3 X 3 Y 3 X 3 X 3 Y 3 Y2 X 2 X 2 Y 2 Z 2 X 2 Y 2 Y 2 X 2 Z

1 Z 1 Z 1 X 1 Y 1 Z 1 Z 1 Z 1 Z 1 X

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> X Y Z > Z > Z > Y X Z > X Z Y > X Z Y > Y X Z > Y X Z >

19 20 21 22 23 24 25 26 273 Y 3 Z 3 Y 3 Y 3 Z 3 Z 3 Z 3 X 3 Z

2 Z 2 X 2 Z 2 Z 2 Y 2 Y 2 Y 2 Z 2 Y

1 X 1 Y 1 X 1 X 1 X 1 X 1 X 1 Y 1 X

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> > Y > Y > Z > > Z X Y > X Z Y > Z Y X > Z Y X

28 29 30 31 32 33 34 35 363 X 3 Z 3 Z 3 X 3 Z 3 X 3 Z 3 X 3 X

2 Z 2 X 2 X 2 Y 2 X 2 Y 2 X 2 Z 2 Z

1 Y 1 Y 1 Y 1 Z 1 Y 1 Z 1 Y 1 Y 1 Y

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> X Z Y > Z Y X > X Z Y > Z X Y > X Y Z > X Y Z > Z X Y > Z X Y > X Y Z

Page 21: Arrow's Impossibility Theorem

Step 3 Results Explanation Now we have profiles 13, 14, 15, 16, 24, 25, 26,

27, and 29 filled in. And, once more, we have another rule, determined from 27: when Y is above and to the left of X, Y > X. This will lead to yet more results.

Page 22: Arrow's Impossibility Theorem

Step 4 Results1 2 3 4 5 6 7 8 9

3 Z 3 Y 3 Z 3 X 3 Y 3 X 3 Y 3 X 3 Y

2 Y 2 Z 2 X 2 Z 2 X 2 Y 2 Z 2 Y 2 X

1 X 1 X 1 Y 1 Y 1 Z 1 Z 1 X 1 Z 1 Z

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> X Y Z > X Z Y > Y X Z > Y Z X > Z X Y > Z Y X > X Y Z > Y Z X > Y Z X

10 11 12 13 14 15 16 17 183 Y 3 Y 3 Z 3 X 3 Y 3 X 3 X 3 Y 3 Y2 X 2 X 2 Y 2 Z 2 X 2 Y 2 Y 2 X 2 Z

1 Z 1 Z 1 X 1 Y 1 Z 1 Z 1 Z 1 Z 1 X

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> X Y Z > Z Y X > Y X Z > Y X Z > X Z Y > X Z Y > Y X Z > Y X Z > Y X Z

19 20 21 22 23 24 25 26 273 Y 3 Z 3 Y 3 Y 3 Z 3 Z 3 Z 3 X 3 Z

2 Z 2 X 2 Z 2 Z 2 Y 2 Y 2 Y 2 Z 2 Y

1 X 1 Y 1 X 1 X 1 X 1 X 1 X 1 Y 1 X

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> Z Y X > Y > Y > Y Z X > Y Z X > Z X Y > X Z Y > Z Y X > Z Y X

28 29 30 31 32 33 34 35 363 X 3 Z 3 Z 3 X 3 Z 3 X 3 Z 3 X 3 X

2 Z 2 X 2 X 2 Y 2 X 2 Y 2 X 2 Z 2 Z

1 Y 1 Y 1 Y 1 Z 1 Y 1 Z 1 Y 1 Y 1 Y

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> X Z Y > Z Y X > X Z Y > Z X Y > X Y Z > X Y Z > Z X Y > Z X Y > X Y Z

Page 23: Arrow's Impossibility Theorem

Step 4 Results Explanation Now profiles 11, 12, 18, 19, 22, and 23 are filled

in. And, from profile 22, we can determine that when Z is above and to the left of X, Z > X. This results in:

Page 24: Arrow's Impossibility Theorem

Step 5 Results1 2 3 4 5 6 7 8 9

3 Z 3 Y 3 Z 3 X 3 Y 3 X 3 Y 3 X 3 Y

2 Y 2 Z 2 X 2 Z 2 X 2 Y 2 Z 2 Y 2 X

1 X 1 X 1 Y 1 Y 1 Z 1 Z 1 X 1 Z 1 Z

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> X Y Z > X Z Y > Y X Z > Y Z X > Z X Y > Z Y X > X Y Z > Y Z X > Y Z X

10 11 12 13 14 15 16 17 183 Y 3 Y 3 Z 3 X 3 Y 3 X 3 X 3 Y 3 Y2 X 2 X 2 Y 2 Z 2 X 2 Y 2 Y 2 X 2 Z

1 Z 1 Z 1 X 1 Y 1 Z 1 Z 1 Z 1 Z 1 X

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> X Y Z > Z Y X > Y X Z > Y X Z > X Z Y > X Z Y > Y X Z > Y X Z > Y X Z

19 20 21 22 23 24 25 26 273 Y 3 Z 3 Y 3 Y 3 Z 3 Z 3 Z 3 X 3 Z

2 Z 2 X 2 Z 2 Z 2 Y 2 Y 2 Y 2 Z 2 Y

1 X 1 Y 1 X 1 X 1 X 1 X 1 X 1 Y 1 X

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> Z Y X > Y Z X > Z X Y > Y Z X > Y Z X > Z X Y > X Z Y > Z Y X > Z Y X

28 29 30 31 32 33 34 35 363 X 3 Z 3 Z 3 X 3 Z 3 X 3 Z 3 X 3 X

2 Z 2 X 2 X 2 Y 2 X 2 Y 2 X 2 Z 2 Z

1 Y 1 Y 1 Y 1 Z 1 Y 1 Z 1 Y 1 Y 1 Y

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

> X Z Y > Z Y X > X Z Y > Z X Y > X Y Z > X Y Z > Z X Y > Z X Y > X Y Z

Page 25: Arrow's Impossibility Theorem

Step 5 Results Explanation What happened?

In all 36 cases, person B (bottom row) matches up exactly with the social preferences. In other words, person B is our dictator.

How did this happen? We made one decision and used unanimity (in cases

where X, Y, or Z was above and to the right) and independence of irrelevant alternatives (where X, Y, or Z was above and to the left) to solve the rest.

Page 26: Arrow's Impossibility Theorem

What Does This Mean? Any complete social preference order which obeys

both unanimity and independence of irrelevant alternatives will be dictatorial (as described on slide 8).

A non-dictatorial social preference ranking would require dropping one of the three other desirable conditions. Normally, completeness and the independence of irrelevant alternatives are dropped.

Page 27: Arrow's Impossibility Theorem

Practical Results Cannot develop a rational social preference

function like what we have for individual preferences

No voting mechanism will simultaneously satisfy completeness, unanimity, independence of irrelevant alternatives, and non-dictatorship.