Positional scoring rules

Lemma 6.10. LetF be a positional scoring rule paired with a rationalisable tie-breaking rule, and let bbe a ballot profile such thatF(b) =x. Suppose that voteri has a strategic ballotb0_i

such that F(b0_i,b(−i)) =y. Then at least one of the following must hold: (i) The position ofy in b0_i is strictly higher than inbi.

(ii) The position ofxinb0_i is strictly lower than inbi.

Proof. To makey win fromx, the score ofy should be at least as high as the score ofx. Since

F(b) =x, the score ofymust be (weakly) lower than the score ofx. To makey win, the score ofy must be increased, or the score ofxmust be decreased. For any positional scoring rule, this means that the rank of alternative y must be increased, or the rank of alternativexmust be decreased.

Theorem 6.11. Anti-plurality is susceptible to second-order risk-free manipulation.

Proof. Let F be the anti-plurality rule and let xm−1. x1. x2.· · ·. xm be the tie-breaking

rule. Now consider the preference profile pwhere pi =x1 x2 · · · xm for every i∈ N.

By applying the tie-breaking rule, we obtainF(p) =xm−1. However, this outcome is far from

ideal for every voter. In this situation, any voteri∈N has an incentive to strategically vote

bi = x1 x2 · · · xm xm−1, since F(bi,p(−i)) = x1. Obviously, if a single voter i

manipulates, this is risk-free, as no voter would have an incentive to manipulate wheni casts a strategic vote.

Theorem 6.12. Any positional scoring rule is susceptible to second-order risk-free manipulation. Proof. We prove this by induction on the number of alternatives. By Theorem 6.8, we know that this must hold for the case wherem= 3. Now assume that it holds formalternatives. LetF be a positional scoring rule form+ 1alternatives with scoring vectorscore= (score1, . . . ,scorem+1).

Take an alternativex∈X and letF−xbe the derived subrule. IfF is anti-plurality (or a scoring

rule equivalent to anti-plurality), then we can apply Theorem 6.11. So, assume that F is not (equivalent to) anti-plurality. Note that in this case, it must hold thatscore1>scorem. Hence,

F−x is a positional scoring rule. Thus, we can apply the inductive hypothesis to infer that there

exists a preference profilepovermalternatives such that there exists a voter with a second-order risk-free manipulation. Without loss of generality, let this be voter 1 with manipulative ballot

b1. LetF−x(p) =aandF−x(b1,p(−1)) =b. By Lemma 6.10, inb1the position of alternativeb

is increased, or the position of alternativeais decreased (or both).

Now consider profilep, wherexis added to the bottom of every preference order in this profile. By definition ofF−x, it holds thatF(p) =aandF(b1,p(−1)) =b. So, the manipulation of voter

1 is still successful. However, to guarantee that this manipulation is risk-free, we need an extra step. Suppose without loss of generality that voter 2 prefers c tob, whileap₁c. Then voter 2 is a potential parasitising manipulator. Since the manipulation of voter 1 of the profilepis risk-free, either some voter could already manipulate the profile pand makecwin, or voter 2 is not able to manipulate the profile (b1,p(−1))to make cwin. In the first case, the manipulation

that makescwin must also be successful under the extended profilep. Therefore, in that case voter 1’s manipulation is risk-free. In the second case, the extension of the ballot might create the opportunity for voter 2 to cast a successful strategic ballot: by putting alternativexbetween

c andb, the gap between the score ofc andb can be decreased, and the score ofc might even exceed the score of b (depending on the scoring vector). In that case, c would win and the manipulation of voter 1 would not be risk-free. To ‘secure’ voter 1’s manipulative ballot b1, she

should move alternativexright belowb. In this way, she increases the gap between the score of alternative xb and the score of any less preferred alternative, and hence she prevents any

potential parasitising manipulation. This is illustrated in Figure 6.6. We conclude that this manipulation is second-order risk-free.

p1 p2 pm b a x x x a 1 b1 p2 pm b a x x x b 1 b01 p2 pm b x a x x b

Figure 6.6: Voter 1 has risk-free manipulationb01

Suppose that the voting rule is a positional scoring rule. Let Y be set of alternatives that are at least as bad as afrom the perspective of voter i. Then if voterihas a strategic ballot in which the positions of the alternatives in Y are not changed, this ballot is a second-order risk-free manipulation.

Proposition 6.13. LetF be a positional scoring rule, bsome ballot profile withF(b) =aand

b0_i a strategic ballot for some voterisuch thatF(b0_i,b(−i)) =b. LetY ={y|ap_i y}. If for all

y∈Y,rankbi(y) =rankb0i(y), thenb

0

i is a second-order risk-free manipulation.

Proof. By Lemma 6.10, it must hold that the position of alternativeb is strictly higher inb0_i

than inb, and the score ofb0_i must strictly increase. The position of all ‘bad’ alternatives does not change. Hence, this ballot can not create an opportunity for another voter to parasitise the manipulation and we conclude that b0_i is a second-order risk-free manipulation.

It is easy to see that by Lemma 6.10, manipulation of plurality under winner-information is always second-order risk-free:

Corollary 6.14. IfF is the plurality and winner information is given, every manipulation is a second-order risk-free manipulation.