Finite set of alternatives A={a,b,c,…} Set I of n players
The preference is expressed by a valuation function
vi = A -> R
vi(a): the “value” in terms of some currency that player i gives to alternative a
utility: ui := vi(a) + m
Goal: Give the object to the player with maximum value.
in a way that cannot be strategically manipulated
Strategy Set: The players bid bi (probably ≠ vi)
Allocation Algorithm: Who is going to get the object?
Payment Scheme: At what price?
player i wants to maximize his utility ui=xi(vi-p)
No Payment
p = 0€
b2 > 100€
First Price Auction
p=100 €
p=70+ε €
==> b1 > 70+ε €
definition:
Allocation Algorithm: Give the object to the player with the highest bid.
Payment Scheme: The winner pays an amount equal to the second highest bid
.
Proposition (Vickrey 1961) For every v1 ,…, vn and every v’i, let ui be the utility if he bids vi and u’i his utility if he bids v’i. Then, ui ≥ u’i. (for the utilities we assume vi as the true valuation)
Goal: The guy with the highest valuation will get the object!
Telling the truth is a dominant strategy for the players.
Remark:
choose alternative + charge payment
vi: A -> R for vi belong to Vi
set of valuations for player i Vi is a truth subset of R_A (R_A 是 Vi 的真子集)
A (direct revelation) mechanism is a social choice function f:V1×⋅⋅⋅×Vn→A and a vector of payment functions p1,…,pn, where pi:V1×⋅⋅⋅×Vn→R
A mechanism (f, p1,…,pn) is incentive compatible (or strategy proof) if for every player i, every v=(v1,…,vn) and every v’i vi(f(vi,v-i))-pi(vi,v-i) ≥ vi(f(v’i,v-i))-pi(v’i,v-i)
Example1:
f(vi,v-i)=a | f(v’i,v-i)=b |
pi(vi,v-i)=3 | pi(v’i,v-i)=5 |
vi(f(vi,v-i))=6 | vi(f(vi,v-i))=7 |
f(vi,v-i)=a | f(v’i,v-i)=b |
pi(vi,v-i)=3 | pi(v’i,v-i)=5 |
vi(f(vi,v-i))=vi(a)=6 | vi(f(v’i,v-i))=vi(b)=7 |
vi(f(vi,v-i))-p(vi,v-i) =3 | vi(f(v’i,v-i))-p(v’i,v-i)=2 |
f(vi,v-i)=a | f(v’i,v-i)=b |
pi(vi,v-i)=3 | pi(v’i,v-i)=3 |
vi(f(vi,v-i))=vi(a)=6 | vi(f(v’i,v-i))=vi(b)=7 |
vi(f(vi,v-i))-p(vi,v-i)=3 | vi(f(v’i,v-i))-p(v’i,v-i)=4 |
A mechanism (f,p1,…,pn) is called VCG mechanism if
vi(f(vi,v-i))-pi(vi,v-i) ≥ 0
pi(vi,v-i) ≥ 0