Eco514|Game Theory
Lecture 8.5,More on Auctions; PS#1
Marciano Siniscalchi
October 14,1999
Introduction
These notes essentially tie up a few loose ends in Lecture 8; in particular,I exhibit examples
of ine ciencies in rst- and second-price auctions,I would also like to brie y comment on
Questions 1 and 2 in Problem Set 2.
The rst-price auction may be ine cient even with private values
Both examples I am going to show are due to Eric Maskin (to the best of my knowledge).
The rst point I wish to make is that,even in a private-values setting,asymmetries may
render a rst-price auction ine cient.
Consider a two-bidder situation in which the value of the rst bidder is uniformly dis-
tributed on [0;1],and the value of the second bidder is uniformly distributed on [0;2],We
look for a (necessarily asymmetric) equilibrium in increasing,continuous bid functions,i.e.
(using a convenient shorthand notation) maps ai,[0;i]!R+.
First,note that ai(0) = 0 in any equilibrium,For suppose w.l.o.g,that a1(0) = a > 0.
Then it must be the case that in equilibrium a wins with probability zero,so that also
a2(0) = a,But since we are assuming that bid functions are increasing and continuous,
types > 0 with < a expect to win with positive probability and pay at least a < ai( ):
contradiction.
Now,I claim that there is no such equilibrium in which a2(2) > 1,Note that the previous
argument,together with the assumptions that bid functions are increasing and continuous,
implies that positive bids win with positive probability,Hence bidder 1 will never bid above
1,her maximum value,But then,from the point of view of bidder 2,any bid a> 1 may be
improved upon by bidding a+12,Hence,bidder 2 never bids more than 1,But this means that,
for some realization of the signals close to (1,2),by continuity we will have a2(v2) < a1(v1)
even if v2 >v1; the rst-price auction assigns the object to bidder 1.
1
[Incidentally,this argument does not rule out the possibility that equilibria in non-
increasing or discontinuous bid functions might exist,Such equilibria can be ruled out
under certain conditions,see Lizzeri and Persico,Games and Economic Behavior,1999 for
details.]
The second-price auction can be ine cient with interdependent
values and more than two bidders
This example is really clever,Consider a three-bidder,interdependent-values environment,
with signals distributed according to some law on a support which includes the point (1,1,1).
The actual distribution of signals does not matter (you can rig things so you get the same
conclusion even if signals are correlated),but to keep things simple,assume that signals are
independent.
Suppose valuations are as follows:
v1(s1;s2;s3) = s1 + 12s2 + 14s3
v2(s1;s2;s3) = s2 + 14s1 + 12s3
v3(s1;s2;s3) = s3
Consider (s1;s2;s3) in a (small) neighborhood of (1;1;1),For ex-post e ciency to obtain,
the object should be assigned to either bidder 1 or 2,depending on whether s3 > 1 or s3 < 1.
That is,e cient assignment depends on s3.
But in a second-price auction,a bid depends only on a buyer’s own signal; thus,whether
bidder 1 or bidder 2 receives the object depends on the relative magnitude of s1 and s2.
Hence,for some realization of the signals in a neighborhood of (1,1,1),the outcome of the
auction will be ine cient.
[In case you were wondering,Dasgupta and Maskin o er a simple solution to this problem:
allow players to submit bids contingent on the signals of all the other players,That is,an
action for bidder 1 with signal s1 is a function f(s1)(s2;s3) to be interpreted as follows,\If
I,bidder 1,knew that my opponents’ signal were s2 and s3 respectively,then I would bid
f(s1)(s2;s3)." Every bidder is thus asked to report,not just a single bid,but a full bid
function.
The seller then computes a xpoint of these bid functions,and takes the xpoint to be the
actual vector of signals; the allocation is determined with respect to this vector,It turns out
that it is an equilibrium to report truthfully,that is,to report f(s1)(s2;s3) = v1(s1;s2;s3).]
On Problem Set 2
2
Question 1
In light of the proof of Proposition 4 in Lecture 7,choosing = A,ti(a) = ai A i should
come natural,The tricky part is how to de ne beliefs.
However,inspecting the iterative construction suggests the following,Of course,for each
player i 2 N,we can specify conditional probabilities pi( jti(a)) directly (a prior which
generates them is easy to construct).
Let K be such that AK+1 = AK,For each ai 2Ai,let ki(ai) = maxfk2f0;:::;Kg,
ai2Akig,If ki(ai) > 0,there exists ai i2 (A i) such that ai2ri( ai i) and ai i(Ak 1 i ) = 1.
Then let
pi((ai;a0 i)jti(ai;a i)) = ai i(a0 i) 8a i;a0 i2A i
Complete the description of the model by assigning pi( jti(ai;a i)) arbitrarily for ai62A1i.
The rest of the problem is now merely a matter of verifying the claims.
Question 2
This one was a bit more tricky,As we did in Lecture 5,I will de ne \belief states"
fA;0;1;2;:::;Kg with the following interpretation,in belief state A,a player is certain
that = 32,so that A is strictly dominant,In belief state 0,a player’s beliefs are consistent
with E=\the probability that = 12 is ",where 12,In belief state k > 0,a player’s
beliefs are consistent with E,plus she is certain that her opponent’s belief state is k 1.
A state !2 comprises a speci cation of the actual value of as well as of both players’
belief state,We use subscripts as follows,in state ! xy,=,Player 1’s belief state is x,
and Player 2’s belief state is y.
Finally,each player learns only her belief state,thus,t1(! xy) = f! 0x0y0,x0 = xg and
similarly for Player 2.
With this convention,de ne conditional probabilities as follows,First,for states of the
form ! Ay and ! xA,let
p1([ = 32]jt1(! Ay)) = 1 and p1([ = 32]jt1(! Ay)) = 1:
Next,for states ! 0y,let
p1([ = 12]jt1(! 0y)) = = 1 p1([ = 32]jt 1(! 0y))
and similarly for states ! x0.
Finally,assuming that probabilities for all states! xy withx;y k 1 have been assigned,
let
p1([ = 12]\[y = k 1]jt1(! ky)) = = 1 p1([ = 32]\[y = k 1]jt 1(! ky))
3
and similarly for states ! xk.This inductively completes the de nition of the posteriors,hence
of the priors of the players.
Again,you should verify (by induction) that this yields the required result.
4
Lecture 8.5,More on Auctions; PS#1
Marciano Siniscalchi
October 14,1999
Introduction
These notes essentially tie up a few loose ends in Lecture 8; in particular,I exhibit examples
of ine ciencies in rst- and second-price auctions,I would also like to brie y comment on
Questions 1 and 2 in Problem Set 2.
The rst-price auction may be ine cient even with private values
Both examples I am going to show are due to Eric Maskin (to the best of my knowledge).
The rst point I wish to make is that,even in a private-values setting,asymmetries may
render a rst-price auction ine cient.
Consider a two-bidder situation in which the value of the rst bidder is uniformly dis-
tributed on [0;1],and the value of the second bidder is uniformly distributed on [0;2],We
look for a (necessarily asymmetric) equilibrium in increasing,continuous bid functions,i.e.
(using a convenient shorthand notation) maps ai,[0;i]!R+.
First,note that ai(0) = 0 in any equilibrium,For suppose w.l.o.g,that a1(0) = a > 0.
Then it must be the case that in equilibrium a wins with probability zero,so that also
a2(0) = a,But since we are assuming that bid functions are increasing and continuous,
types > 0 with < a expect to win with positive probability and pay at least a < ai( ):
contradiction.
Now,I claim that there is no such equilibrium in which a2(2) > 1,Note that the previous
argument,together with the assumptions that bid functions are increasing and continuous,
implies that positive bids win with positive probability,Hence bidder 1 will never bid above
1,her maximum value,But then,from the point of view of bidder 2,any bid a> 1 may be
improved upon by bidding a+12,Hence,bidder 2 never bids more than 1,But this means that,
for some realization of the signals close to (1,2),by continuity we will have a2(v2) < a1(v1)
even if v2 >v1; the rst-price auction assigns the object to bidder 1.
1
[Incidentally,this argument does not rule out the possibility that equilibria in non-
increasing or discontinuous bid functions might exist,Such equilibria can be ruled out
under certain conditions,see Lizzeri and Persico,Games and Economic Behavior,1999 for
details.]
The second-price auction can be ine cient with interdependent
values and more than two bidders
This example is really clever,Consider a three-bidder,interdependent-values environment,
with signals distributed according to some law on a support which includes the point (1,1,1).
The actual distribution of signals does not matter (you can rig things so you get the same
conclusion even if signals are correlated),but to keep things simple,assume that signals are
independent.
Suppose valuations are as follows:
v1(s1;s2;s3) = s1 + 12s2 + 14s3
v2(s1;s2;s3) = s2 + 14s1 + 12s3
v3(s1;s2;s3) = s3
Consider (s1;s2;s3) in a (small) neighborhood of (1;1;1),For ex-post e ciency to obtain,
the object should be assigned to either bidder 1 or 2,depending on whether s3 > 1 or s3 < 1.
That is,e cient assignment depends on s3.
But in a second-price auction,a bid depends only on a buyer’s own signal; thus,whether
bidder 1 or bidder 2 receives the object depends on the relative magnitude of s1 and s2.
Hence,for some realization of the signals in a neighborhood of (1,1,1),the outcome of the
auction will be ine cient.
[In case you were wondering,Dasgupta and Maskin o er a simple solution to this problem:
allow players to submit bids contingent on the signals of all the other players,That is,an
action for bidder 1 with signal s1 is a function f(s1)(s2;s3) to be interpreted as follows,\If
I,bidder 1,knew that my opponents’ signal were s2 and s3 respectively,then I would bid
f(s1)(s2;s3)." Every bidder is thus asked to report,not just a single bid,but a full bid
function.
The seller then computes a xpoint of these bid functions,and takes the xpoint to be the
actual vector of signals; the allocation is determined with respect to this vector,It turns out
that it is an equilibrium to report truthfully,that is,to report f(s1)(s2;s3) = v1(s1;s2;s3).]
On Problem Set 2
2
Question 1
In light of the proof of Proposition 4 in Lecture 7,choosing = A,ti(a) = ai A i should
come natural,The tricky part is how to de ne beliefs.
However,inspecting the iterative construction suggests the following,Of course,for each
player i 2 N,we can specify conditional probabilities pi( jti(a)) directly (a prior which
generates them is easy to construct).
Let K be such that AK+1 = AK,For each ai 2Ai,let ki(ai) = maxfk2f0;:::;Kg,
ai2Akig,If ki(ai) > 0,there exists ai i2 (A i) such that ai2ri( ai i) and ai i(Ak 1 i ) = 1.
Then let
pi((ai;a0 i)jti(ai;a i)) = ai i(a0 i) 8a i;a0 i2A i
Complete the description of the model by assigning pi( jti(ai;a i)) arbitrarily for ai62A1i.
The rest of the problem is now merely a matter of verifying the claims.
Question 2
This one was a bit more tricky,As we did in Lecture 5,I will de ne \belief states"
fA;0;1;2;:::;Kg with the following interpretation,in belief state A,a player is certain
that = 32,so that A is strictly dominant,In belief state 0,a player’s beliefs are consistent
with E=\the probability that = 12 is ",where 12,In belief state k > 0,a player’s
beliefs are consistent with E,plus she is certain that her opponent’s belief state is k 1.
A state !2 comprises a speci cation of the actual value of as well as of both players’
belief state,We use subscripts as follows,in state ! xy,=,Player 1’s belief state is x,
and Player 2’s belief state is y.
Finally,each player learns only her belief state,thus,t1(! xy) = f! 0x0y0,x0 = xg and
similarly for Player 2.
With this convention,de ne conditional probabilities as follows,First,for states of the
form ! Ay and ! xA,let
p1([ = 32]jt1(! Ay)) = 1 and p1([ = 32]jt1(! Ay)) = 1:
Next,for states ! 0y,let
p1([ = 12]jt1(! 0y)) = = 1 p1([ = 32]jt 1(! 0y))
and similarly for states ! x0.
Finally,assuming that probabilities for all states! xy withx;y k 1 have been assigned,
let
p1([ = 12]\[y = k 1]jt1(! ky)) = = 1 p1([ = 32]\[y = k 1]jt 1(! ky))
3
and similarly for states ! xk.This inductively completes the de nition of the posteriors,hence
of the priors of the players.
Again,you should verify (by induction) that this yields the required result.
4
