# ECON 804 2012 Final Solutions

ECON 804
2012 Final Solutions
Please report any errors or typos that you …nd. Also let me know if there is
a simpler way to do a problem.
1a and b) By the folk theorem, the answer to both of these is yes. We know
that for all payo&curren; vectors v in the set of feasible payo&curren;s V (with dim(V ) equal
to the number of players) with vi &gt; vi , where vi is player i’s minmax value,
there exists a &lt; 1 such that for all 2 ( ; 1), there exists an SPE with payo&curren;s
v. It is easy to verify that both these vectors are feasible. Furthermore, player
1’s minmax payo&curren; is 2, while player 2’s minmax payo&curren; is 1, so in both vectors,
each player is getting more than their minmax payo&curren;.
1c) No. In the …rst period, player 2 cannot play R with probability greater
than 12 : if she did, player 1 would have a …rst-period expected payo&curren; greater
than 3 by playing D. Since player 1’s minmax value is 2, this means that in
any such SPE, player 1’s average discounted payo&curren; would be greater than 2.5.
Moreover, in the …rst period, player 2 cannot play L with probability greater
than 21 : if she did, her own …rst-period expected payo&curren; would be less than 1,
while her payo&curren; in any period cannot exceed 2. Thus her average discounted
payo&curren; would be less than 1.5. Thus, it must be that player 2 randomizes ( 12 ; 21 )
in the …rst period. But then for player 2 to achieve payo&curren; 1.5, it must be that
player 1 always plays U on the equilibrium path. This is impossible in an SPE
because it gives player 1 an average discount payo&curren; of 0.75, below her minmax
value.
2a) By symmetry, f only depends on the number of players of each type.
!
!
Let g(a; b; c)
f ( ), where
has a components equal to 1 , b components
equal to 2 , and c components equal to 3 . We must have:
- g(1; 1; 1) = x (by assumption)
- g(3; 0; 0) = x (Pareto)
- g(0; 3; 0) = y (Pareto)
- g(0; 0; 3) = z (Pareto)
- g(1; 0; 2) = g(2; 0; 1) = x (otherwise, when (a; b; c) = (1; 1; 1), type 2 would
want to deviate to avoid outcome x)
- g(0; 1; 2) = z (it cannot be x because type 2 would report type 3 to get z,
and it cannot be y because type 3 would report type 1 to get x)
- g(2; 1; 0) = x (otherwise, type 1 would report type 3 to get x)
It is easy to check that for g(0; 2; 1) and g(1; 2; 0), the only two remaining
cases, we must have one of the following:
- g(0; 2; 1) = g(1; 2; 0) = y
- g(0; 2; 1) = z and g(1; 2; 0) = x
- g(0; 2; 1) = z and g(1; 2; 0) = y.
It is easy (but cumbersome) to check that all 3 of these SCF’s are implementable in dominant strategies.
1
2b) No. There are (at least) two ways to show this.
- Because f is Pareto and all strict preferences are allowed, it must be onto.
We can then apply the Gibbart-Satterthwaite Theorem, which requires that
every player has strict preferences, universal domain within strict preferences, that there be at least 3 alternatives, and f to be onto. The theorem
says that in such cases, f is implementable in dominant strategies if and
only if it is dictatorial. But f is symmetric, so it cannot be dictatorial.
- Because every player has strict preferences, we know that any SCF
implementable in dominant strategies must be monotonic. We can
then apply the result from the …rst part of the course stating that when there
are at least 3 alternatives, any SCF with universal domain (either within
strict or all preferences), monotonicity and the Pareto property must be
dictatorial. Again, this contradicts the symmetry requirement.
3a) Bidders with the two highest bids win the object (any way of resolving
ties is …ne). Winners pay the third highest bid. Proving that bidding one’s
true value is weakly dominant follows the same reasoning as for the standard
second-price auction.
3b) No. For example, consider the auction from part a, and another auction
that is identical except that, in addition, it gives \$1 to all participants. Two
…xes are needed to make the statement correct: the utility of some type must
be …xed (usually we set the utility of the lowest type at 0), and the agents must
be risk-neutral. In that case, letting P ( ) be the probability that is among
the two highest values, we have U ( ; b) = P (b) T (b), where b is the bidder’s
reported type, and T is the transfer function. By the revelation principle, this
is maximized at = b. Taking the f.o.c. with respect to b (we can do this
since P is di&curren;erentiable, and it can be argued that T must be as well) implies
R
T 0 ( ) = P 0 ( ), so T ( ) = T (0) + 0 0 P 0 ( 0 )d 0 , which must be the same for all
auctions with the same T (0).
4a) Consider a bilateral trade setting where the buyer and the seller are
risk-neutral and have valuations that are independently drawn from a
strictly positive density over bounded supports whose interiors intersect.
Then there exists no ex post e&cent; cient and Bayesian incentive compatible mechanism that satis…es all types’participation constraints. This
theorem does not apply here because the valuation distributions do not have
strictly positive density.
4b) None. Consider the following mechanism: each agent reports their valuation. If the buyer’s is 2 and the seller’s is 1, the buyer gets the object and
pays the seller 1.5. Otherwise, the seller keeps the object, and there are no
transfers. This mechanism is clearly ex post e&cent; cient and satis…es all types’participation constraints. Furthermore, not only is it Bayesian IC, but truthtelling
is a dominant strategy.
2
5. We want to show that H v(xH ) TH
by a monotonicity constraint. We have:
H v(xH )
TH
H v(xL )
H v(xM )
TL can be replaced
TM
and
M v(xM )
TM
M v(xL )
M v(xM )
TH
H v(xM )
TL
H v(xH )
+
+
M v(xL )
TL
which is equivalent to
H v(xH )
TH
H v(xH )
TH
(
(
M )v(xM )
H
H v(xL )
H
+
M v(xL )
to the right-hand side yields:
M )(v(xM )
v(xL )) +
This implies the desired result as long as
(
or, since
H
&gt;
M
H
TL
M )(v(xM )
and v 0 (:) &gt; 0,
xM
xL
3
v(xL ))
0
H v(xL )
TL