Manipulating rules under full knowledge

Using our definition of utility, we obtain the following definition for manipulability of the aggregation rule under full knowledge.

Definition 16 (Manipulability under full knowledge). A rational aggregation ruleF is manipulable

under full knowledge by agentiwith utilityUi at profileB, containing ballotBi ifUi is top-respecting toBi and there exists a ballotBi0 such thatUi(F(B i, Bi0))> Ui(F(B i, Bi)).

A rational aggregation function F is manipulable under full knowledge if there exists an agent i, a

utility functionUi and a profileBsuch that F is manipulable under full knowledge byiat B. We will call manipulability by (agents with) a closeness-respecting utility, closeness-respecting manipu- lability.

Some characterization results

Now we do have a definition for manipulability, however this does not give us an easy way to check aggregation rules for manipulability. This is because there is an infinite number of top-respecting utility

functions and we cannot check them all. Therefore we would like to find equivalent requirements for manipulability.

Theorem 12 (Characterization of closeness-respecting manipulability for resolute aggregation rules). Let F be aresoluteaggregation rule, Ba profile containing ballot Bi.

There exists an issue j_2I and a ballotB0

i 6=Bi such thatF(Bi0,B i)j={bi,j}, andF(B)j6={bi,j}if and only if there exists a utility functionUi that is closeness-respecting toBi, such thatF is manipulable under full knowledge by agenti at profileB.

Proof. LetF be an aggregation rule,B a profile containing ballotBi. Suppose there is an issuej2I and a ballotB0

i6=Bi such thatF(B0i,B i)j ={bi,j}, andF(B)j 6={bi,j}. SinceBi06=Bi, there is aj0 such thatb0

i,j0 6=bi,j0. Ifj =j0, let

wi(j00) =

(

1 ifj00=j,

0 otherwise.

The corresponding Hamming utility is

Ui(B) =

(

1 ifbj=bi,j,

0 otherwise.

Then sinceF(B0

i,B i)j ={bi,j}andF(B)j6={bi,j}, we haveUi(F(Bi0,B i)) = 1>0 =F(B). And we have Ui(Bi) = 1>0 =Ui(Bi0), sinceb0i,j0 6=bi,j0. HenceF is manipulable under full knowledge by agenti with utilityUi, which is a Hamming utility toBi and thus closeness-respecting toBi by Theorem 11. Ifj₆=j0_,let wi(j00) = 8 > < > : 2 ifj00_=j, 1 ifj00=j0, 0 otherwise.

The corresponding Hamming utility is

Ui(B) = 8 > > > < > > > :

3 ifbj=bi,j andbj0 =bi,j0

2 ifbj=bi,j^bj0 6=bi,j0,

1 ifbj6=bi,j^bj0 =bi,j0,

0 otherwise.

Then sinceF(Bi0,B i)j ={bi,j}andF(B)j 6={bi,j}, we haveUi(F(Bi0,B i)) 2>1 F(B). And we have Ui(Bi) = 3>2 Ui(Bi0), sinceb0i,j0 6=bi,j0. HenceF is manipulable under full knowledge by agenti with utilityUi, which is a Hamming utility toBi and thus closeness-respecting toBi by Theorem 11.

For the other direction, suppose there exists a utility function Ui that is closeness-respecting to

Bi, such that F is manipulable by agent i at profile B by agent i. Then there exists a ballot

B0

i such that Ui(F(B i, Bi0)) > Ui(F(B i, Bi)) and Ui(Bi0) < Ui(Bi). Hence since Ui is closeness- respecting, we must haveAgree(Bi, F(B i, Bi0))(Agree(Bi, F(B i, Bi)). Hence there is some j such thatF(B0

i,B i)j={bi,j}, and F(B)j 6={bi,j}. ©

WhenUi is not closeness-respecting, the right-to-left direction of the previous theorem does not hold. Considerm= 2, n= 3, the majority ruleMajand

U0(B) = 8 > < > : 3 ifB= (1,0) 2 ifB= (0,1) 1 otherwise.

Now ifB= ((1,0),(1,1),(0,1)),U0(Maj(B)) =U0((1,1)) = 1<2 =U0((0,1)) =U0(Maj((0,1),B 0)).

So agent 0 can manipulate the majority rule under utilityU0. However, the onlyjfor whichMaj(B)j6=

{b0,j}isj= 1, however, for no ballotB00 we getMaj(B00,B 0)1=b0,1, since there is already a majority

for issue 1 without agent 0. Hence it is not manipulable.

Theorem 13. If F is a resolute aggregation rule, then F is monotonic and independent if and

only if F is not manipulable at any profile containing Bi by an agent with a utility function that is closeness-respecting to ballot Bi.

Proof. Let F be an aggregation rule that satisfies monotonicity and independence and let agent i

have utilityUi that is closeness-respecting to some ballotBi. Now let Bbe any profile containingBi. Now consider any alternative ballotB0i and any issuej 2I such thatF(Bi0,B i)j ={bi,j}. Now if

b0

i,j=bi,j, by independenceF(Bi,B i)j =F(Bi0,B i)j={bi,j}and ifb0i,j6=bi,j, then by monotonicity,

F(Bi,B i)j={bi,j}. Hence by Theorem 12,F is not manipulable by agenti with utilityUi. Since

Ui,Bwere arbitrary,F is not manipulable at any profile by an agent with a utility function that is closeness-respecting to some ballotBi.

For the other direction, suppose firstF is not monotonic. Then there is an agenti, issue j, a profileB containingBi, and there is an alternative ballotBi0 such thatbi,j =x,b0i,j =¬x,F(Bi,B i)j={¬x} andF(B0

i,B i)j ={x}. HenceF(Bi,B i)j=6 {bi,j}andF(Bi0,B i)j={bi,j}. Thus by Theorem 12, we have that there is aUi that is closeness-respecting toBi such that agentican manipulateF atB. Now supposeF is not independent. Then by Theorem 2, it is also not weakly independent. Hence

there is an issuej, an agenti, a profileBcontaining Bi and there is an alternative ballotBi0 such that

bi,j=b0i,j andF(B)j6=F(Bi0,B i)j. Now we can assume w.l.o.g. thatF(Bi0,B i)j={bi,j}={b0i,j} (otherwise we can switchBi andBi0). Then we haveF(Bi,B i)j 6={bi,j}, hence by theorem 12, there

is aUi that is top respecting toBi such that agentican manipulate F onB.

Hence we have thatF is weakly monotonic and weakly independent if and only ifF is not manipulable by an agent with a utility function that is closeness-respecting to some ballotB. ©

This indeed coheres with a result from Dietrich and List (2007b). Note that monotonicity and independence are very strict requirements, and indeed there are few aggregation rules that satisfy both of these. Dietrich and List (2007a) show for resolute rules that if we also require anonymity and responsiveness, these four axioms are satisfied if and only if the rule is a quota rule. The majority rule satisfies both weak monotonicity and (weak) independence hence is not manipulable by agents with a closeness-respecting utility function.

Theorem 14. For any aggregation ruleF, profileB containing ballotBi, the following is true: There

exists a top-respecting (to Bi) utility Ui such thatF is manipulable under full knowledge by agenti, if and only ifF(Bi,B i)=6 {Bi}and there is an alternative ballotBi0 such thatF(Bi,B i)6=F(Bi0,B i).

Proof. LetF be an aggregation rule,Ba profile containing ballot Bi of agenti. LetUi be such that agentican manipulate F atB. Then there is an alternative ballotB0

i such thatUi(F(Bi0,B i))>

Ui(F(Bi,B i)) andUi(Bi) > Ui(Bi0). From the first fact it follows that F(Bi,B i)6= F(Bi0,B i). From the second combined with the first, sinceUi is top respecting and set-consistent, we must have

F(Bi,B i)6={Bi}.

For the other direction, suppose there is a profileBcontainingBi and there is an alternative ballotBi0 such thatF(Bi,B i)6=F(Bi0,B i)andF(Bi,B i)6=Bi. Now supposeBi06=F(B0,B i). Then the

following utility is well defined. Ui(B) = 8 > < > : 2 ifB=Bi or B=F(Bi0,B i), 1 ifB=F(Bi,B i)orB=Bi0, 0 otherwise.

Now we have thatUiis top-respecting toBi,Ui(Bi0)< Ui(Bi)andUi(F(Bi0,B i))> F(Bi,B i), hence

F is manipulable under full knowledge by agenti.

In the case thatB0

i=F(B0,B i), the following utility is well defined.

Ui(B) = 8 > > > < > > > : 3 ifB=Bi 2 ifB=B0i (which equalsF(B0,B i)), 1 ifB=F(Bi,B i) 0 otherwise.

Now we have again thatUi is top-respecting toBi,Ui(Bi0)< Ui(Bi)andUi(F(Bi0,B i))> F(Bi,B i),

henceF is manipulable under full knowledge by agenti. ©

Corollary 2. An aggregation ruleF is manipulable under full knowledge if and only if there exists

a profile B containing ballotBi, and an alternative ballotB0

i such thatF(Bi,B i)=6 F(Bi0,B i)and

F(Bi,B i)6={Bi}.

Corollary 3. If an aggregation ruleFis non-manipulable and there areBi, Bi0,Bsuch thatF(Bi,B i)6=

F(B0

i,B i), then for all B we must haveF(B,B i) ={B}.

Proof. Suppose there are Bi, B0i,B such that F(Bi,B i)6=F(Bi0,B i), and there is a B for which

F(B,B i)6=B. Then we must have that eitherF(B,B i)6=F(Bi,B i)orF(B,B i)=6 F(Bi0,B i). HenceF is manipulable under full knowledge by Theorem 14. This proves the Corollary. ©

For the next theorem, we first need a definition.

Definition 17. An aggregation ruleF is adictatorshipwhen there is ani_2N such thatF(B) =_{Bi} for all rational profilesB.

Theorem 15. Suppose |Mod(IC)_| 3. An aggregation rule F is non-manipulable if and only if it is

either constant or a dictatorship.

Proof. Obviously dictatorships and constant rules are not manipulable. For the other direction, suppose that F is not manipulable and not constant. Since F is not constant, there are profiles

(B0, . . . , Bn 1),(B00, . . . , Bn0 1)such that

F(B0, . . . , Bn 1)6=F(B00, . . . , Bn0 1).

This implies that for somei, we must have

F(B00, . . . , Bi0 1, Bi, . . . , Bn 1)=6 F(B00, . . . , Bi0, Bi+1, . . . , Bn 1).

Now letB00i=B00, . . . , Bi0 1, Bi+1, . . . , Bn 1. SinceFis not manipulable andF(Bi,B00i)6=F(B0i,B00i), by Corollary 3, this implies thatF(B,B00i) ={B} for allB2Mod(IC). Now consider some arbitrary profileB⇤

i=B0⇤, . . . , Bn⇤ 1. We define the following function

G(B, k)l= 8 > < > : B ifl=i, B⇤ l ifl6=iandlk, Bl00 ifl6=iandl > k.

We have thatG(B, k)defines a profile. We will prove that for anyk_2I,F(G(B, k)) = _{B_} for all

B₂Mod(IC), by the induction onk.

Fork= 0,G(B,0) =B,B00

i, hence we haveF(G(B,0)) ={B}.

supposeF(G(B, k)) =_{B_}for all B ₂Mod(IC). Now if k+ 1 =i, we haveG(B, k) =G(B, k+ 1),

henceF(G(B, k)) =F(G(B, k+ 1)).

Ifk+ 16=i, letA16=A22Mod(IC)be unequal to Bk00+1 (these exist since|Mod(IC)| 3). Then by

induction hypothesis, we have

F(G(A1, k)) ={A1},

F(G(A2, k)) ={A2}.

By definition ofGwe have G(A1, k)k+1=B00k+1=G(A2, k)k+1.

Now SinceF(G(A1, k)) ={A1}6={G(A1, k)k+1}and G(A1, k+ 1) =Bk⇤+1, G(A1, k) (k+1), by Corol-

lary 3 contraposed, we must have

F(G(A1, k+ 1)) =F(G(A1, k)) ={A1},

and similarly

F(G(A2, k+ 1)) =F(G(A2, k)) ={A2}.

Thus

F(G(A1, k+ 1))=6 F(G(A2, k+ 1))

Hence by Corollary 3, for anyB we must haveF(G(B, k+ 1)) =F(B, G(A1, k+ 1) i) =B.

This shows by induction thatF(G(B, k)) =B for allB₂Mod(IC)and for allk_2I. We can conclude

from this that alsoF(B,B⇤i) =F(G(B, n 1)) =B for allB2Mod(IC), and sinceB⇤ was arbitrary,

agentiis a dictator. ©

At first sight, the result of Theorem 15 looks similar to the Gibbard-Satterthwaite Theorem. However mathematically the two results are very different. Here the impossibility is more direct because of the extreme freedom in picking a utility function that works.