Voting by Committees under Constraints



∑   ∑ √pω

+ Σ ≈⅛)
p∈p⅛(p⅛)

+ Σ ≈⅛).
⅛∈P⅛(^sp,pfp)


P,≠P        B , fB t в ,

xEF р (Pi P ,P_r )

< ∑   ∑  F'W

>,'≠p 1∈Λv°<⅛')

where the equality follows from condition (??) and the inequality follows from
condition (??). Therefore,
F(P~i, Pi)PiF(P')∙, that is, F is not strategy-proof.

(b) Let {B1, ...,Bq} be a partition of K' Ç K and consider any P A".
i N, and Pi A. Since for all p = 1, ...,q the functions FBp are strategy-
proof, we have that
FBp(PBp)Rfp FBp(Pfp, P^p); that is, for all p = 1, ...q,

∑ ≈⅛) >    £   af"(j∕),

ι∈F⅞(P⅞)           p∈P⅛(Pfp ,pff)

where ufp and ufp are any pair of functions on Bp representing Pfp and
FBp, respectively. Therefore, adding up,

Q                                          Q

∑ ∑ <Tω ≥ ∑   ∑  ⅛⅛)∙

p=l   xeFbp(pbp)               p=l   уерВр(£Вр,рвР)

Hence, F{P')RiF{Pi, P-i); that is, F is strategy-proof. That {Bi,..., Bq}
is a Cylindric decomposition of Rp = K' and Rp = Bi + ... + Bq follow
g

immediately from the fact that.F (F) = Q FBp (PbF} for all P EAn.

p=l

Our strategy of proof for necessity relies heavily on invoking the Gibbard-
Satterthwaite Theorem for the case where there are more than three active
components in a section
Bp of the minimal Cylindric decomposition of the
range. This is done by proving that, then, there will be three feasible out-
comes which agents can rank as the three most-preferred, and in any relative
order (a “free triple”). But
FBp must be strategy-proof if F is (Proposition
4). If
Fbp was non-dictatorial, we could use it to construct a non-dictatorial
and strategy-proof social choice function over our free triple, which we know

23



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