social choice session 2
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Social Choice Session 2. Carmen Pasca and John Hey. Arrow’s Impossibility Theorem. - PowerPoint PPT PresentationTRANSCRIPT
Social ChoiceSession 2
Carmen Pasca and John Hey
Arrow’s Impossibility Theorem• In social choice theory, Arrow’s impossibility theorem ...
states that, when voters have three or more discrete alternatives (options), no voting system can convert the ranked preferences of individuals into a community-wide ranking while also meeting a certain set of criteria. These criteria are called unrestricted domain, non-dictatorship, Pareto efficiency, and independence of irrelevant alternatives. (Wikipedia)
• Is this obvious or not?• This lecture we give a proof, borrowed from John Bone at
the University of York. We have also included another proof borrowed from Kevin Feasel. Use whichever you prefer.
Arrow
• Arrow asks the question: “can we aggregate individual preferences into social preferences?”
• He invokes the following conditions:• Universal Domain: is applicable to any profile.• Consistency: produces a complete, transitive ordering
of available alternatives.• Pareto: if everyone prefers x to y then so should
society.• Independence of Irrelevant Alternatives: orders x and
y on the basis only of individual preferences on x and y.
The result
• Arrow shows that if one wants these conditions then the only way to get them is through Dictatorship.
• On the one hand this seems trivial (“how can you aggregate different preferences?”)...
• ... but one should look at the result as telling us that we need to drop something if we want something that in some way represent the preferences of society – whatever that means.
• We need to drop some of the assumptions... • ... or change our way of looking at the problem.
The Story and Two of the Axioms
• We consider a very simple society with two members, Jen and Ken, and three things to order: a, b and c. Think of these as ways to organise society.
• Universal Domain: is applicable to any profile.• This means that whatever the preferences are
we should be able to aggregate them.• Pareto: if everyone prefers x to y then so
should society. • This seems unquestionable.
6
a b c a c b b a c b c a c a b c b a
a b c
a c b
b a c
b c a
c a b
c b a
11 12 13 14 15 16
21 22 23 24 25 26
31 32 33 34 35 36
41 42 43 44 45 46
51 52 53 54 55 56
61 62 63 64 65 66
Universal Domain & ParetoKen
Jen aPbbPcaPc
7
a b c a c b b a c b c a c a b c b a
a b c
a c b
b a c
b c a
c a b
c b a
11 12 13 14 15 16
21 22 23 24 25 26
31 32 33 34 35 36
41 42 43 44 45 46
51 52 53 54 55 56
61 62 63 64 65 66
Universal Domain & ParetoKen
Jen aPbbPcaPc
aPb
aPc
8
a b c a c b b a c b c a c a b c b a
a b c
a c b
b a c
b c a
c a b
c b a
11 12 13 14 15 16
21 22 23 24 25 26
31 32 33 34 35 36
41 42 43 44 45 46
51 52 53 54 55 56
61 62 63 64 65 66
Universal Domain & ParetoKen
Jen aPbbPcaPc
aPbbPc
aPc aPc
9
a b c a c b b a c b c a c a b c b a
a b c
a c b
b a c
b c a
c a b
c b a
11 12 13 14 15 16
21 22 23 24 25 26
31 32 33 34 35 36
41 42 43 44 45 46
51 52 53 54 55 56
61 62 63 64 65 66
Universal Domain & Pareto
aPb
Ken
Jen aPbbPcaPc
aPbbPc
aPc aPcbPc
10
a b c a c b b a c b c a c a b c b a
a b c
a c b
b a c
b c a
c a b
c b a
11 12 13 14 15 16
21 22 23 24 25 26
31 32 33 34 35 36
41 42 43 44 45 46
51 52 53 54 55 56
61 62 63 64 65 66
Universal Domain & Pareto
aPbaPbbPcaPc
aPbcPbaPc
bPabPcaPc
bPabPccPa
aPbcPbcPa
bPacPbcPa
aPbbPc bPc
aPc aPcaPb aPb
cPbcPbaPc aPc
bPcbPcbPa bPa
aPc aPcbPabPa
bPcbPccPa cPa
cPacPacPbcPb
aPb aPb
bPa bPacPb cPb
cPa cPa
Ken
Jen
Now the other two axioms
• Consistency: produces a complete, transitive ordering of available alternatives.
• We have a complete transitive ordering for society.• Independence of Irrelevant Alternatives: orders x and
y on the basis only of individual preferences on x and y.• This is particularly important – we are going to use it
frequently.• In the slides that follow we first impose consistency
and then repeatedly we invoke the independence of irrelevant alternatives.
12
a b c a c b b a c b c a c a b c b a
Ken
a b c
a c b
b a c
b c a
c a b
c b a
Jen 11 12 13 14 15 16
21 22 23 24 25 26
31 32 33 34 35 36
41 42 43 44 45 46
51 52 53 54 55 56
61 62 63 64 65 66
aPbaPbbPcaPc
aPbcPbaPc
bPabPcaPc
bPabPccPa
aPbcPbcPa
bPacPbcPa
aPbbPc bPc
aPc aPcaPb aPb
cPbcPbaPc aPc
bPcbPcbPa bPa
aPc aPcbPabPa
bPcbPccPa cPa
cPacPacPbcPb
aPb aPb
bPa bPacPb cPb
cPa cPa
cPb cPbcPb
cPb cPbcPb
cPb cPbcPb
aPc
Consistency & IIArequires (at least) one of these
Now what?
• We can now use an implication of what we have found.
• Note in cell 36 that bPa and cPb; and hence we can deduce that cPa.
• Let us insert that in cell 36 and explore the implications.
• This allows us to use the Independence of Irrelevant Alternatives Axiom again.
14
a b c a c b b a c b c a c a b c b a
a b c
a c b
b a c
b c a
c a b
c b a
11 12 13 14 15 16
21 22 23 24 25 26
31 32 33 34 35 36
41 42 43 44 45 46
51 52 53 54 55 56
61 62 63 64 65 66
aPbaPbbPcaPc
aPbcPbaPc
bPabPcaPc
bPabPccPa
aPbcPbcPa
bPacPbcPa
aPbbPc bPc
aPc aPcaPb aPb
cPbcPbaPc aPc
bPcbPcbPa bPa
aPc aPcbPabPa
bPcbPccPa cPa
cPacPacPbcPb
aPb aPb
bPa bPacPb cPb
cPa cPa
cPb cPbcPb
cPb cPbcPb
cPb cPbcPb
cPacPacPa
cPacPacPa
cPacPacPa
Consistency & IIAKen
Jen
Now what?
• Once again we can now use an implication of what we have found.
• Note in cell 14 that bPc and cPa; and hence we can deduce that bPa.
• Let us insert that in cell 14 and explore the implications.
• This allows us to use the Independence of Irrelevant Alternatives Axiom again.
16
a b c a c b b a c b c a c a b c b a
a b c
a c b
b a c
b c a
c a b
c b a
11 12 13 14 15 16
21 22 23 24 25 26
31 32 33 34 35 36
41 42 43 44 45 46
51 52 53 54 55 56
61 62 63 64 65 66
aPbaPbbPcaPc
aPbcPbaPc
bPabPcaPc
bPabPccPa
aPbcPbcPa
bPacPbcPa
aPbbPc bPc
aPc aPcaPb aPb
cPbcPbaPc aPc
bPcbPcbPa bPa
aPc aPcbPabPa
bPcbPccPa cPa
cPacPacPbcPb
aPb aPb
bPa bPacPb cPb
cPa cPa
cPb
cPb
cPbcPb
cPbcPb
cPb cPbcPb
cPacPacPa
cPacPacPa
cPacPacPa
bPa bPa bPa
bPa bPa bPa
bPa bPa bPa
Ken
Jen
Consistency & IIA
Now what?
• Once again we can now use an implication of what we have found.
• Note in cell 23 that bPa and aPc; and hence we can deduce that bPc.
• Let us insert that in cell 23 and explore the implications.
• This allows us to use the Independence of Irrelevant Alternatives Axiom again.
18
a b c a c b b a c b c a c a b c b a
a b c
a c b
b a c
b c a
c a b
c b a
11 12 13 14 15 16
21 22 23 24 25 26
31 32 33 34 35 36
41 42 43 44 45 46
51 52 53 54 55 56
61 62 63 64 65 66
aPbaPbbPcaPc
aPbcPbaPc
bPabPcaPc
bPabPccPa
aPbcPbcPa
bPacPbcPa
aPbbPc bPc
aPc aPcaPb aPb
cPbcPbaPc aPc
bPcbPcbPa bPa
aPc aPcbPabPa
bPcbPccPa cPa
cPacPacPbcPb
aPb aPb
bPa bPacPb cPb
cPa cPa
cPb
cPb
cPbcPb
cPbcPb
cPb cPbcPb
cPacPacPa
cPacPacPa
cPacPacPa
bPa bPa bPa
bPa bPa bPa
bPa bPa bPa
bPc bPc bPc
bPc bPc bPc
bPc bPcbPc
Ken
Jen
Consistency & IIA
Now what?
• Once again can now use an implication of what we have found.
• Note in cell 51 that aPb and bPc; and hence we can deduce that aPc.
• Let us insert that in cell 51 and explore the implications.
• This allows us to use the Independence of Irrelevant Alternatives Axiom again.
20
a b c a c b b a c b c a c a b c b a
a b c
a c b
b a c
b c a
c a b
c b a
11 12 13 14 15 16
21 22 23 24 25 26
31 32 33 34 35 36
41 42 43 44 45 46
51 52 53 54 55 56
61 62 63 64 65 66
aPbaPbbPcaPc
aPbcPbaPc
bPabPcaPc
bPabPccPa
aPbcPbcPa
bPacPbcPa
aPbbPc bPc
aPc aPcaPb aPb
cPbcPbaPc aPc
bPcbPcbPa bPa
aPc aPcbPabPa
bPcbPccPa cPa
cPacPacPbcPb
aPb aPb
bPa bPacPb cPb
cPa cPa
cPb
cPb
cPbcPb
cPbcPb
cPb cPbcPb
cPacPacPa
cPacPacPa
cPacPacPa
bPa bPa bPa
bPa bPa bPa
bPa bPa bPa
bPc bPc bPc
bPc bPc bPc
bPc bPcbPc
aPc aPc aPc
aPc aPc aPc
aPc aPc aPc
Ken
Jen
Consistency & IIA
Finally
• For the final time can now use an implication of what we have found.
• Note in cell 62 that cPb and aPc; and hence we can deduce that aPb.
• Let us insert that in cell 62 and explore the implications.
• This allows us to use the Independence of Irrelevant Alternatives Axiom again.
22
a b c a c b b a c b c a c a b c b a
a b c
a c b
b a c
b c a
c a b
c b a
11 12 13 14 15 16
21 22 23 24 25 26
31 32 33 34 35 36
41 42 43 44 45 46
51 52 53 54 55 56
61 62 63 64 65 66
aPbaPbbPcaPc
aPbcPbaPc
bPabPcaPc
bPabPccPa
aPbcPbcPa
bPacPbcPa
aPbbPc bPc
aPc aPcaPb aPb
cPbcPbaPc aPc
bPcbPcbPa bPa
aPc aPcbPabPa
bPcbPccPa cPa
cPacPacPbcPb
aPb aPb
bPa bPacPb cPb
cPa cPa
cPb
cPb
cPbcPb
cPbcPb
cPb cPbcPb
cPacPacPa
cPacPacPa
cPacPacPa
bPa bPa bPa
bPa bPa bPa
bPa bPa bPa
bPc bPc bPc
bPc bPc bPc
bPc bPcbPc
aPc aPc aPc
aPc aPc aPc
aPc aPc aPc
aPbaPb aPb
aPbaPb aPb
aPbaPb aPb
The DictatorKen
Jen
Consistency & IIA
Two questions for the break
• (1) You may recall that we started this line of logic with one of two possible ways of getting consistency.
• What do you think happens if we take the second possible way?
• (2) Why did we not pick up any intransitivities on the way? That is, for example, why did we not find something like
• aPb bPc cPa• in one of the cells?• Magic?
24
a b c a c b b a c b c a c a b c b a
a b c
a c b
b a c
b c a
c a b
c b a
11 12 13 14 15 16
21 22 23 24 25 26
31 32 33 34 35 36
41 42 43 44 45 46
51 52 53 54 55 56
61 62 63 64 65 66
aPbaPbbPcaPc
aPbcPbaPc
bPabPcaPc
bPabPccPa
aPbcPbcPa
bPacPbcPa
aPbbPc bPc
aPc aPcaPb aPb
cPbcPbaPc aPc
bPcbPcbPa bPa
aPc aPcbPabPa
bPcbPccPa cPa
cPacPacPbcPb
aPb aPb
bPa bPacPb cPb
cPa cPa
aPcaPcaPc
aPcaPcaPc
aPcaPcaPccPb
Ken
Jen
Consistency & IIA
25
a b c a c b b a c b c a c a b c b a
a b c
a c b
b a c
b c a
c a b
c b a
11 12 13 14 15 16
21 22 23 24 25 26
31 32 33 34 35 36
41 42 43 44 45 46
51 52 53 54 55 56
61 62 63 64 65 66
aPbaPbbPcaPc
aPbcPbaPc
bPabPcaPc
bPabPccPa
aPbcPbcPa
bPacPbcPa
aPbbPc bPc
aPc aPcaPb aPb
cPbcPbaPc aPc
bPcbPcbPa bPa
aPc aPcbPabPa
bPcbPccPa cPa
cPacPacPbcPb
aPb aPb
bPa bPacPb cPb
cPa cPa
aPb aPb aPb
aPb aPb aPb
aPb aPb aPb
aPcaPcaPc
aPcaPcaPc
aPcaPcaPc
Ken
Jen
Consistency & IIA
26
a b c a c b b a c b c a c a b c b a
a b c
a c b
b a c
b c a
c a b
c b a
11 12 13 14 15 16
21 22 23 24 25 26
31 32 33 34 35 36
41 42 43 44 45 46
51 52 53 54 55 56
61 62 63 64 65 66
aPbaPbbPcaPc
aPbcPbaPc
bPabPcaPc
bPabPccPa
aPbcPbcPa
bPacPbcPa
aPbbPc bPc
aPc aPcaPb aPb
cPbcPbaPc aPc
bPcbPcbPa bPa
aPc aPcbPabPa
bPcbPccPa cPa
cPacPacPbcPb
aPb aPb
bPa bPacPb cPb
cPa cPa
aPb aPb aPb
aPb aPb aPb
aPb aPb aPb
cPb cPb cPb
cPb cPb cPb
cPb cPbcPb
aPcaPcaPc
aPcaPcaPc
aPcaPcaPc
Ken
Jen
Consistency & IIA
27
a b c a c b b a c b c a c a b c b a
a b c
a c b
b a c
b c a
c a b
c b a
11 12 13 14 15 16
21 22 23 24 25 26
31 32 33 34 35 36
41 42 43 44 45 46
51 52 53 54 55 56
61 62 63 64 65 66
aPbaPbbPcaPc
aPbcPbaPc
bPabPcaPc
bPabPccPa
aPbcPbcPa
bPacPbcPa
aPbbPc bPc
aPc aPcaPb aPb
cPbcPbaPc aPc
bPcbPcbPa bPa
aPc aPcbPabPa
bPcbPccPa cPa
cPacPacPbcPb
aPb aPb
bPa bPacPb cPb
cPa cPa
cPb cPb cPb
cPb cPb cPb
cPb cPbcPb
cPa cPa cPa
cPa cPa cPa
cPa cPa cPa
aPb aPb aPb
aPb aPb aPb
aPb aPb aPb
aPcaPcaPc
aPcaPcaPc
aPcaPcaPc
Ken
Jen
Consistency & IIA
28
a b c a c b b a c b c a c a b c b a
a b c
a c b
b a c
b c a
c a b
c b a
11 12 13 14 15 16
21 22 23 24 25 26
31 32 33 34 35 36
41 42 43 44 45 46
51 52 53 54 55 56
61 62 63 64 65 66
aPbaPbbPcaPc
aPbcPbaPc
bPabPcaPc
bPabPccPa
aPbcPbcPa
bPacPbcPa
aPbbPc bPc
aPc aPcaPb aPb
cPbcPbaPc aPc
bPcbPcbPa bPa
aPc aPcbPabPa
bPcbPccPa cPa
cPacPacPbcPb
aPb aPb
bPa bPacPb cPb
cPa cPa
cPb cPb cPb
cPb cPb cPb
cPb cPbcPb
cPa cPa cPa
cPa cPa cPa
cPa cPa cPa
bPabPa bPa
bPabPa bPa
bPabPa bPa
aPb aPb aPb
aPb aPb aPb
aPb aPb aPb
aPcaPcaPc
aPcaPcaPc
aPcaPcaPc
Ken
Jen
Consistency & IIA
29
a b c a c b b a c b c a c a b c b a
a b c
a c b
b a c
b c a
c a b
c b a
11 12 13 14 15 16
21 22 23 24 25 26
31 32 33 34 35 36
41 42 43 44 45 46
51 52 53 54 55 56
61 62 63 64 65 66
aPbaPbbPcaPc
aPbcPbaPc
bPabPcaPc
bPabPccPa
aPbcPbcPa
bPacPbcPa
aPbbPc bPc
aPc aPcaPb aPb
cPbcPbaPc aPc
bPcbPcbPa bPa
aPc aPcbPabPa
bPcbPccPa cPa
cPacPacPbcPb
aPb aPb
bPa bPacPb cPb
cPa cPa
cPb cPb cPb
cPb cPb cPb
cPb cPbcPb
cPa cPa cPa
cPa cPa cPa
cPa cPa cPa
bPabPa bPa
bPabPa bPa
bPabPa bPa
bPc bPcbPc
bPc bPcbPc
bPc bPcbPc
aPb aPb aPb
aPb aPb aPb
aPb aPb aPb
The Dictator
aPcaPcaPc
aPcaPcaPc
aPcaPcaPc
Ken
Jen
Consistency & IIA
30
Universal Domain
Consistency
Pareto
Independence of Irrelevant Alternatives (IIA)
and that requires only individual rankings
Arrow’s Theorem
is the Dictator Principle
The only principle that satisfies:
i.e., where the social ordering simply replicates, across all profiles, some given individual’s ranking.
31
If, for some x and y, xPy at some profile where everyone in V ranks x above y and everyone else ranks y above x ...
.. then, for any w and z, wPz at any profile where everyone in V ranks w above z, irrespective of how anyone else ranks w and z.
Given the four conditions (U, C, P, IIA)
We have already shown this in the 2-person, 3-alternative, strict ranking case.
Generalising this to n-persons, m-alternatives is not very difficult.
Arrow’s Impossibility Theorem
• We have proved it in the context of a two-person society, but as John Bone says, it is not difficult to extend it to a larger society. This is what he does in the next three slides.
• It is not necessary for you to know this in detail, just understand the principle.
• If you would like another proof of the theorem, you might like this by Kevin Feasel which I have shamelessy downloaded from the Internet. This follows after the generalisation to n people.
33
There must be some x and y, and some individual J, such that xPy at some profile where J ranks x above y and everyone else ranks y above x.
Given the four conditions (U, C, P, IIA)
Consider any set V of individuals such that aPb at some profile where everyone in V ranks a above b and everyone else ranks b above a.
(We know there must be such a V, since from Pareto one such V is the set of all individuals.)
From the previous slide, we know that therefore aPb at any profile where everyone in V ranks a above b.
Such as ..
Suppose that V contains more than one person.
34
where V1 and V2 are subsets of V (i.e. everyone in V is in exactly one of V1 and V2 , and no-one else is).
everyone in V1
everyone in V2
a b c
c a b
everyone else b c a
aPb implies that for any c either aPc or cPb.
It follows that there is a set smaller than V, and a pair {x,y}, such that xPy at some profile where everyone in that set ranks x above y and everyone else ranks y above x.
Since we can make the same argument for any V containing more than one person, it follows that the smallest such V is an individual (let’s say J).
35
If there is any profile in which Jen ranks x above y and Ken ranks y above x,
then from IIA:
and at which ,x P y
at every profile in which Jen ranks x above y and Ken ranks y above x x P y
such as at: Jen
Ken
x y z
y z x
at which, from Pareto and Consistency: x P z
So, again from IIA, also at: x P z Jen
Ken
w x z
z w x
w P zat which, from Pareto and Consistency:
Arrow's Impossibility Theorem
Kevin FeaselDecember 10, 2006
http://36chambers.wordpress.com/arrow/
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)
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
> > > > > > > > >
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
>
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.
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.
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 X
2 Z
1 Y
1 2 3
>
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.
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.
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
> > > > > > > > >
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.
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 > >
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
>
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
> ? ? ?
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
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 > >
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.
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
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
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
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.
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
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:
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
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.
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.
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.
Arrow’s Impossibility Theorem
• So what?• Is this profound or trivial?• What implications does it have for Social Choice?• Are his conditions too strong?• What happens if we change/relax them?• We shall see in the next lecture.• And what do we do next?• Give up?