1. real estate
2. buying and selling homes

Sample Problems + Solutions
1. There is one Buyer and one Seller; the Seller owns a used car. The car
is either High quality H or Low quality L. Buyer and Seller valuations
depend on quality:
VS (H) = 4 VS (L) = 1
VB (H) = 6 VB (L) = 3
It is common knowledge that 30% of Cars are High quality, 70% are
Low quality. Sellers can ask p = 5 or p = 2, conditional on their
information; Buyers can Accept or Reject, conditional on the price.
(a) Formalize this situation as a Bayesian game. What are the types
and strategies for the Buyer and Seller?
(b) Find all the Bayesian Nash equilibria and the expected payoffs
to the players, conditional on their types. Express equilibrium
strategies both as mixtures of pure behavioral strategies and as
mixed behavioral strategies.
Solution
(a) The only player who has types is the seller {L, H}
Strategies are:
Seller : {L, H} → {2, 5}
Buyer : {2, 5} → {A, R}
The Bayesian Game can be represented in the following way,
H (Pr. = 0.3)
x1
x2
x3
A,A A,R R,A
2y
-2
-2
0
4
4
0
5 (1 − y) 1
0
1
1
0
1
1
x4
R,R
0
0
0
0
L (Pr. = 0.7)
x1
x2
x3
A,A
A,R R,A
2z
1
1
0
1
1
0
5 (1 − z) 4
0
4
-2
0
-2
x4
R,R
0
0
0
0
where A, R is for example Accept a price of 2 and Reject a price
of 5. Numbers 2 and 5 at the head of the rows are the proposed
prices by the seller, who also knows the quality of the car (whether
he/she is in matrix H or L).
Let’s obtain the reaction function for each player.
Seller H
- y = 0 when some (or all) x1 , x2 and x3 are positive
- y ∈ [0, 1] only if x4 = 1
Seller L
- z = 0 only if 3x1 > x2 − 4x3
- z = 1 only if 3x1 < x2 − 4x3
- z ∈ [0, 1] only if 3x1 = x2 − 4x3
2
Buyer
This case is naturally a little more involving since there are four
possibilities. Let’s start by considering the expected payoff for
each possible strategy.
E(u)AA = 1.2y + 0.7z + 0.3(1 − y) − 0.7(1 − z)
E(u)AR = 1.2y + 0.7z
E(u)RA = 0.3(1 − y) − 0.7(1 − z)
E(u)RR = 0
(Recall that in expected terms, AA=AR+RA)
Examples of best responses are,
- x1 = 1 if either y or z are positive and 1.4z − 0.3y > 1.1
- x2 = 1 if either y or z are positive and 1.4z − 0.3y < 1.1
- x3 = 1 if at least y or z are negative, which is NOT possible
- x4 = 1 if both y or z are zero.
- A full randomization would happen if both y = z = 0 and
1.4z − 0.3y = 1.1, which clearly NOT possible.
The same type of analysis can be conducted for the cases in which
buyers randomize over 3 strategies (4 more cases) and over 2
strategies (6 more cases). It’s straightforward to show that out
if these cases, only two are possible,
1.1
+ 0.3
y
- x1 and x2 are positive when 1.2y + 0.7z > 0 and z = 1.4
1.4
- x2 and x4 are positive when y = z = 0
Checking for consistency of these reaction functions, we can find
the following Nash Equilibria.
NE in a mixture of pure behavioral strategies
i. {5; 5; RR}
3
1
3
ii. 5, 11
2
+
5,
AA
+
AR
14
14
4
4
3
NE in a mixed behavioral strategies
i. Seller
H→5
L→5
Buyer
2→R
5→R
ii. Seller
H→5
11
(2) +
L → 14
3
(5)
14
Buyer
2→A
5 → 41 (A) + 34 (R)
Payoffs conditional on type (SellerH,SellerL,Buyer)
i. (0, 0, 0)
ii. (0.25, 1, 0.55)
2. This is a variant of the used car game above (same fraction of High,
Low quality cars). All is the same except it is common knowledge that
40% of Buyers can detect when Sellers are cheating (i.e., asking 5 for
a Car of Low quality).
(a) Formalize this situation as a Bayesian game. What are the types
and strategies for the Buyer and Seller?
(b) Find all the Bayesian Nash equilibria and the expected payoffs
to the players, conditional on their types. Express equilibrium
strategies both as mixtures of pure behavioral strategies and as
mixed behavioral strategies.
(c) Why does the presence of Buyers who can detect cheating matter
for Buyers who cannot detect cheating?
4
Solution
(a) In this variant of the game, we also have two types of buyers,
the informed buyer (B,I) who knows both quality and prices and
the uninformed buyer (B,U) who only knows prices (basically our
previous characterization of the buyer).
Strategies are:
SS : {L, H} → {2, 5}
SB,I : {L2, L5, H2, H5} → {A, R}
SB,U : {2, 5} → {A, R}
(Buyers observe their own type. I just separate buyers in order to
make explicit the differences in information and to make analysis
more clear)
The informed buyer has clear best responses since information is
perfect for him. (B,I) will always accept the deal except when the
seller of a low quality car asks for a price of 5.
Hence, for the informed buyer, the only strategy profile that makes
sense is ARAA, which means accept a low quality car offered by
2, reject a low quality car offered by 5 and accept high quality
cars, no matter the price.
Therefore, we can represent the Bayesian Game in a similar way
than before but different payoffs for the seller. Expected payoffs
for the seller are: 60% of the times he/she faces an uninformed
buyer and 40% of the times he/she faces an informed buyer.
H (Pr. = 0.3)
x1
x2
x3
A,A A,R R,A
2y
-2
-2
-0.8
4
4
0
5 (1 − y) 1
0.4
1
1
0
1
5
x4
R,R
-0.8
0
0.4
0
L (Pr. = 0.7)
x1
x2
x3
A,A
A,R R,A
2z
1
1
0
1
1
0.4
0
2.4
5 (1 − z) 2.4
-2
0
-2
x4
R,R
0
0.4
0
0
Strategies in columns are only those corresponding to the
uninformed buyer (B,U)
Note that only some payoffs for the seller have changed. For example, the payoff in red have changed from 4 to 2.4. This is because,
under the strategy A,A the uninformed buyer would accept a price
5, without knowing the car is L. However, under the existence of
informed buyers, the seller knows he can get 4 only 60% of the
times, when facing an uninformed buyer, but nothing 40% of the
times, when an informed buyer rejects that offer.
Following the same solution method than before, we can see that
the NE (5, 5, ARAA, RR) no longer exists since now the low quality seller is not indifferent anymore between alternatives when the
uninformed buyer pretends to play R, R since there exists a probability to sell the car to informed buyers, hence in this case it’s
strictly preferred for the seller L to set a price of 2.
With respect to the relevant equilibrium, it changes to
3
5
7
11
5, 2 + 5, ARAA, AA + AR
14
14
12
12
expressed as a mixture of pure behavioral strategy
In mixed behavioral strategies, the equilibrium can be likewise
expressed as
Seller
H→5
11
L → 14
(2) +
3
(5)
14
6
Informed Buyer
L2 → A
L5 → R
H2 → A
H5 → A
Uninformed Buyer
2→A
7
5
(A) + 12
(R)
5 → 12
As can be seen, uninformed buyers will accept offers of 5 more
frequently than the case in which there was no informed players.
Expected payoffs in this new situation would be 0.65 for SellerH,
1 for SellerL, 0.55 for uninformed buyers and 0.85 for informed
buyers.
(b) It’s easy to see that the existence of informed buyers makes the
trade more likely and benefits sellers with high quality cars, who
gain more in expected terms because they sell more cars.
In few words, the existence of informed buyers impose some discipline to sellers of low quality cars to set prices of 2. This change
the optimal behavior of uninformed buyers but not the expected
payoffs they have.
3. The government is considering construction of a road that will benefit
three firms. Each of the firms could be a High type or a Low type. The
value of the project to a firm is 1 if the firm is of High type, 0 if the firm
is of Low type. (The probability that each firm is of High type is h,
where 0 < h < 1 and types are independent.) Firms can refuse to pay
if they report Low, but the project is completely public, so they cannot
be excluded from the benefits; formalize this as the requirement that
the government must divide the cost of the project among the firms
that report their type as High.
(a) For what values of c is there a socially efficient individually rational
incentive compatible mechanism?
(b) For each such c find such a mechanism.
7
(c) Does it matter if the government is constrained to use anonymous
mechanisms, so that all firms reporting High are charged equally?
(d) Does it matter if the government is allowed to make a profit? a
loss?
Solution There are four cases to consider: Case 1: 1 > c ≥ 0
Consider the direct mechanism in which firms report their types and
suppose we can make it socially efficient. then π(L, L, L) = 0 but
otherwise π = 1. When firm 1 is High and tells the truth its expected
utility is
h2 [1 − x1 (H, H, H)] + (h)(1 − h)[1 − x1 (H, H, L)]
+(1 − h)h[1 − x1 (H, L, H)] + (1 − h)(1 − h)[1 − x1 (H, L, L)]
When firm 1 is High and lies its expected utility is
h2 [1 − x1 (L, H, H)] + (h)(1 − h)[1 − x1 (L, H, L)]
+(1 − h)h[1 − x1 (L, L, H)] + (1 − h)(1 − h)[0 − x1 (L, L, L)]
Because Low firms can opt out, all the x1 terms in this expression are
0 so after simplifying the incentive compatibility constraint for firm 1
High becomes:
h2 [0 − x1 (H, H, H)] + (h)(1 − h)[0 − x1 (H, H, L)]
+(1 − h)h[0 − x1 (H, L, H)] + (1 − h)(1 − h)[1 − x1 (H, L, L)] ≥ 0
Rearranging gives
h2 [x1 (H, H, H)] + (h)(1 − h)[x1 (H, H, L)]
+(1 − h)h[x1 (H, L, H)] ≤ (1 − h)(1 − h)[1 − x1 (H, L, L)]
Because the other two firms can opt out, we must have x1 (H, L, L) = c
so we get
h2 [x1 (H, H, H)] + (h)(1 − h)[x1 (H, H, L)] + (1 − h)h[x1 (H, L, H)] (1)
≤ (1 − h)(1 − h)[1 − c]
Now go through the same argument for firm 2 and firm 3 to obtain
h2 [x2 (H, H, H)] + (h)(1 − h)[x2 (H, H, L)] + (1 − h)h[x2 (L, H, H)] (2)
≤ (1 − h)(1 − h)[1 − c]
8
h2 [x3 (H, H, H)] + (h)(1 − h)[x3 (H, L, H)] + (1 − h)h[x3 (L, H, H)] (3)
≤ (1 − h)(1 − h)[1 − c]
Add these three equations and keep in mind that firms that report
Low must be charged 0; we get the condition that social efficiency is
possible:
h2 c + 3h(1 − h)c ≤ 3(1 − h)2 (1 − c)
It may be helpful to think in terms of winning and losing. If firm 1 is
High and lies then:
• firm 1 wins when some other firm reports High (the road is built
and firm 1 doesn’t have to pay)
• firm 1 loses when no other firm reports High (the road is not built)
How big are the wins and losses? That depends on how much firm 1 is
being charged, which in turn depends on c. What are the expectations
of winning and losing? That depends on the probability h that each
firm is High. If c is close to 0 I don’t gain much by lying; if h is close
to 1 it is unlikely I will be caught when I lie.
Case 2: 2 ≥ c > 1 The analysis is similar. The incentive compatibility
constraints for three firms when they are High are
h2 [1 − x1 (H, H, H)] + h(1 − h)[1 − x1 (H, H, L)]
+(1 − h)h[1 − x1 (H, L, H)] ≥ h2
h2 [1 − x2 (H, H, H)] + h(1 − h)[1 − x2 (H, H, L)]
+(1 − h)h[1 − x2 (L, H, H)] ≥ h2
h2 [1 − x3 (H, H, H)] + h(1 − h)[1 − x3 (H, L, H)]
+(1 − h)h[1 − x3 (L, H, H)] ≥ h2
(4)
(5)
(6)
Add these three equations and use the fact that when exactly two
firms report High they must share the cost; this gives the condition
that social efficiency is possible:
h2 (3 − c) + 3h(1 − h)(2 − c) ≥ 3h2
9
Case 3: 3 ≥ c > 2 Here’s a direct mechanism that works: build the
road exactly when all firms report High; charge each firm c/3 if the
road is built, 0 otherwise.
Case 4: c > 3 Never build the road.
4. Find the symmetric equilibrium for the All-Pay auction when bidders
have private valuations, uniformly distributed on [0,1].
Solution
Symmetry means all bidders use the same bid function B, which we
assume to be smooth and strictly increasing.
Bidder i will get the object whenever B(si ) > B(sj ) for all j 6= i. Given
our assumptions, this means si > sj , where s are the bids.
If Bidder 1 (without loss of generality) has signal = valuation v1 and
bids B(s) then he wins when v1 > vj all j 6= 1; when he wins he gets an
item he values at v1 . Because it is an all pay auction, he always pays
B(s). Hence Bidder 1’s expected utility is:
Z
Z
Z
...
EU1 (B(s)|v1 ) =
v1 dv2 ...dvn−1 dvn − B(s)
s>vn
s>vn −1
s>v2
Simplifying.
EU1 (B(s)|v1 ) = v1 sn−1 − B(s)
The FOC is
d
EU1 (B(s)|v1 )
=0
ds
|s=v1
Differentiating and setting s = v1 gives
(n − 1)v1n−1 = B 0 (v1 )
Integrating both sides and using the assumption B(0) = 0, we get the
symmetric bidding function
n−1 n
v
n
Note that because v < 1, v n < v — in fact, if n is large then v n << v.
B(v) =
10
5. Assuming that bidders have private valuations, uniformly distributed
on [0,1], compute directly the expected revenue to the seller in the
symmetric equilibrium for:
(a) First Price Auction
(b) Second Price Auction
(c) All Pay Auction (high bidder wins, everyone pays their bid)
Does this agree with the conclusion of the Revenue Equivalence Principle? Discuss briefly.
Solution
(a) First Price Auction
The expected utility for player 1, who has a true valuation v1 and
bids s, is given by
Z
Z
Z
...
(v1 − B(s))dv2 ...dvn−1 dvn
EU1 (B(s)|v1 ) =
s>vn
s>vn −1
s>v2
EU1 (B(s)|v1 ) = sn−1 (v1 − B(s))
Now we can take FOC
(n − 1)sn−2 [v1 − B(s)] = sn−1 B 0 (s)
and evaluate it at s = v1
(n − 1)v1n−1 = v1n−1 B 0 (v1 ) + (n − 1)v1n−2 B(v1 )
since v1n−1 B 0 (v1 ) + (n − 1)v1n−2 B(v1 ) = [B(v1 )v1n−1 ]0
(n − 1)v1n−1 = [B(v1 )v1n−1 ]0
Integrating both sides and using the assumption B(0) = 0, we get
the symmetric bidding function
B(v) =
11
n−1
v
n
Now we can calculate the expected revenue to the seller. This will
be determined by the expected value of the highest bid, more preE(v M ) where v M is just the maximum expected
cisely ER = n−1
n
valuation of n independent draws from a uniform distribution defined over [0, 1].
P r(t > v M ) = P r(t > v1 , t > v2 , ..., t > vn ) = tn
The expected maximum value is
Z 1
Z 1
Z
M
M
n−1
E(v ) =
tdP r(t > v ) =
t(nt )dt =
0
0
1
ntn dt =
0
n
n+1
Hence, the expected revenue to the seller will be
ER =
n−1
n−1
E(v M ) =
n
n+1
(b) Second Price Auction
Without loss of generality, assume bidder 1 bids the highest s
being his true valuation v1 . Bidder 1 will pay the second highest
bid (B(v2 ))
Z
Z
Z
Z
(v1 − B(v2 ))dv2 ...dvn−1 dvn
...
EU1 (B(s)|v1 ) =
v2 >vn
v2 >vn −1
Z
EU1 (B(s)|v1 ) =
v2 >v3
s>v2
s
v2n−2 (v1 − B(v2 ))dv2
0
Now we can take FOC
sn−2 [v1 − B(s)] = 0
and evaluate it at s = v1
v1n−2 [v1 − B(v1 )] = 0
Hence, we get the symmetric bidding function
B(v) = v
12
The expected revenue to the seller will be determined by the expected value of the second highest bid (that we will call v M,2 ).
This is just the expected value of the highest bid conditional on
being the second highest bid, or which is the same, the maximum
expected valuation from n − 1 independent draws from a uniform
defined defined on [0, v M ].
Using the law of iterated expectations
E(v M,2 ) = E[E(v M,2 |v M )] = E[
n−1 n
n−1 M
v ]=
n
n n+1
Hence, the expected revenue to the seller will be
ER =
n−1
n+1
(c) All Pay Auction
Recall in the first exercise we found that
B(v) =
n−1 n
v
n
The expected revenue to the seller, since everybody pays, is n
times the expected bids.
Z 1
n−1 n
v ] = (n − 1)
v n dv
ER = nE[B(v)] = nE[
n
0
ER =
n−1
n+1
The Revenue Equivalence Principle holds in each case, as it should
because the assumptions are satisfied. Especially: the bidder with the
highest valuation always wins the item.
13
6. The Revenue Equivalence Principle applies to auctions with the property that the item is always assigned to the person who values it the
most. The purpose of this problem is to show that this property is not
robust to asymmetry.
Consider a first price auction with two bidders having private valuations. Assume that valuations are independently but not identically
distributed in [0,1]: Buyer i’s valuation has cdf Fi ; assume Fi is smooth
and strictly increasing and that F1 6= F2 . Let (B1 ,B2 ) be equilibrium
bid functions; assume they are also smooth and strictly increasing.
(a) Use the argument in class to solve for B1 , B2 (F2 will appear in the
formula for B1 and vice versa). Show that F1 6= F2 ⇒ B1 6= B2 .
(b) Now that you know B1 6= B2 , show that there is some pair (t1 ,
t2 ) of valuations for which the bidder with the lower value bids
more. Use this to conclude that there is positive probability that
the bidder with the lower value wins.
Solution
(a) To show that F1 6= F2 ⇒ B1 6= B2 we can show B1 = B2 ⇒
F1 = F2 Without loss of generality (and considering symmetry
arguments), the expected utility for the first bidder is given by
Z
EU1 (B1 (s), v1 ) =
(v1 − B1 (s))dF2 (v2 )
B1 (s)>B2 (v2 )
or
B2−1 (B1 (s))
Z
(v1 − B1 (s))dF2 (v2 )
EU1 (B1 (s), v1 ) =
0
which implies
EU1 (B1 (s), v1 ) = (v1 − B1 (s))F2 (B2−1 (B1 (s)))
Taking FOC
(v1 − B1 (s))f2 (B2−1 (B1 (s)))[B2−1 (B1 (s))]0 B10 (s) = F2 (B2−1 (B1 (s)))B10 (s)
14
and evaluate it at s = v1 we have
(v1 − B1 (v1 ))f2 (B2−1 (B1 (v1 )))[B2−1 (B1 (v1 ))]0 = F2 (B2−1 (B1 (v1 )))
Integrating both sides and following the lectures to integrate by
parts it can be show that
Z
−1
−1
F2 (B2 (B1 (v1 )))B1 (v1 ) = v1 F2 (B2 (B1 (v1 ))) − F2 (B2−1 (B1 (v1 )))
and dividing by F2 (B2−1 (B1 (v1 ))), we can get
Z
1
F2 (B2−1 (B1 (v1 )))
B1 (v1 ) = v1 −
−1
F2 (B2 (B1 (v1 )))
by symmetry
1
B2 (v2 ) = v2 −
−1
F1 (B1 (B2 (v2 )))
Z
F1 (B1−1 (B2 (v2 )))
Now, following our assumption that B1 (v1 ) = B2 (v2 ) and considering the case in which v1 = v2
Z
1
F2 (v1 )
B1 (v1 ) = v1 −
F2 (v1 )
and
Z
1
B2 (v2 ) = B1 (v1 ) = v1 −
F1 (v1 )
whence
1
F1 (v)
Z
0
v
1
F1 (t)dt =
F2 (v)
F1 (v1 )
F2 (t)dt
0
for all v, or equivalently
Z v
Z
F2 (v)
F1 (t)dt = F1 (v)
0
0
for all v.
15
v
Z
v
F2 (t)dt
(7)
To see that this implies F1 = F2 suppose not. Because F1 (1) =
F2 (1) = 1 it follows that
Z 1
Z 1
F2 (t)dt
F1 (t)dt =
0
0
Hence we must have F1 > F2 somewhere and F2 > F1 somewhere.
The set {x : F1 (x) > F2 (x)} is a union of disjoint intervals; let
(a, b) be one of these, so F1 (x) > F2 (x) for x ∈ (a, b) but F1 (a) =
F2 (a) and F1 (b) = F2 (b).
Differentiate equation (7) and cancel terms to get
Z x
Z x
0
0
F2 (x)
F1 (t)dt = F1 (x)
F2 (t)dt
0
0
for every x ∈ (a, b). Because F1 (a) = F2 (a) we must have
Z a
Z a
F2 (t)dt
F1 (t)dt =
0
0
so
Z
x
Z
x
F2 (t)dt
F1 (t)dt >
0
0
for every x ∈ (a, b). Hence we must have F10 (x) > F20 (x) for every
x ∈ (a, b). But F1 (a) = F2 (a) so this implies F1 (b) > F2 (b), a
contradiction. Hence we must have F1 = F2 as claimed.
Having proved B1 = B2 ⇒ F1 = F2 we’ve proved F1 6= F2 ⇒
B1 6= B2
(b) Since B1 6= B2 , it’s easy to show there is a positive probability that a bidder with a lower value wins. Assume v1 = v2 but
B1 (v1 ) > B2 (v2 ). Since we have assumed Bi are smooth and
strictly increasing there should be an > 0 such that v1 − < v2
but still B1 (v1 ) > B2 (v2 ), which means bidder 1 wins even when
having a smaller valuation for the object.
7. Begin with the Public Goods example discussed in class.
• Two firms
• Each of two possible types H, L
16
• Public project; cost = 8
• v(H) = 10, v(L) = 0
• Probability distribution µ on type profiles
µ(H, H) = µ(L, L) = µ(H, L) = µ(L, H) = .25
Assume
• Government must exactly cover cost
• No ex post subsidies/transfers between firms
(a) Show that social efficiency is not possible: there is no individually
rational incentive compatible mechanism that builds the project
exactly when the social benefit exceeds the cost.
(b) Suppose the cost were c. For what values of c would social efficiency be possible?
(c) When social efficiency is not possible, what is the maximum social
welfare?
Solution (a) By the Revelation Principle it suffices to consider direct
mechanisms. Such a mechanism is specified by
• Probabilities
π(H, H), π(H, L), π(L, H), π(L, L)
• Payments
x1 (H, H), x1 (H, L), x1 (L, H), x1 (L, L)
x2 (H, H), x2 (H, L), x2 (L, H), x2 (L, L)
Social efficiency would mean
π(H, H) = π(H, L) = π(L, H) = 1, π(L, L) = 0
The Government restrictions are
x1 (H, H) + x2 (H, H)
x1 (H, L) + x2 (H, L)
x1 (L, H) + x2 (L, H)
x1 (L, L)
x2 (L, L)
xi (· , ·)
17
=
=
=
=
=
≥
8
8
8
0
0
0
Write the incentive compatibility and individual rationality constraints
for the firms; because they are identical ex ante we ignore their names:
(IC-1H)
(.5)[10 − x1 (H, H)] + (.5)[10 − x1 (H, L)] ≥
(.5)[10 − x1 (L, H)] + (.5)[0 − x1 (L, L)]
(IR-1H)
(.5)[10 − x1 (H, H)] + (.5)[10 − x1 (H, L)] ≥ 0
(IC-1L)
(.5)[0 − x1 (L, H)] + (.5)[0 − x1 (L, L)] ≥
(.5)[0 − x1 (H, H)] + (.5)[0 − x1 (H, L)]
(IR-1L)
(.5)[0 − x1 (L, H)] + (.5)[0 − x1 (L, L)] ≥ 0
(IC-2H)
(.5)[10 − x2 (H, H)] + (.5)[10 − x2 (H, L)] ≥
(.5)[10 − x2 (L, H)] + (.5)[0 − x2 (L, L)]
(IR-2H)
(.5)[10 − x2 (H, H)] + (.5)[10 − x2 (H, L)] ≥ 0
(IC-2L)
(.5)[0 − x1 (L, H)] + (.5)[0 − x1 (L, L)] ≥
(.5)[0 − x1 (H, H)] + (.5)[0 − x1 (H, L)]
(IR-2L)
(.5)[0 − x1 (L, H)] + (.5)[0 − x1 (L, L)] ≥ 0
Analysis
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• government breaks even ⇒ x1 (L, L) + x2 (L, L) = 0
• no subsidy ⇒ x1 (L, L) = 0
• (IR-1L) ⇒ x1 (L, H) = 0
• government breaks even ⇒ x2 (L, H) = 8
• reversing roles of 1, 2 ⇒
x1 (L, L) = x2 (L, L) = x2 (H, L) = 0
x1 (H, L) = 8
• (IC-1H) ⇒
(.5)[10 − x1 (H, H)] + (.5)[10 − 8] ≥ (.5)[10]
• ⇒ x1 (H, H) ≤ 2
• (IC-2H) ⇒ x2 (H, H) ≤ 2
• ⇒ x1 (H, H) + x2 (H, H) ≤ 4
• no subsidy ⇒ x1 (H, H) + x2 (H, H) = 8
This is a Contradiction
Solution (b) Different cost?
Repeat same argument with c instead of 8
(IC-1H)
(.5)[10 − x1 (H, H)] + (.5)[10 − x1 (H, L)] ≥
(.5)[10 − x1 (L, H)] + (.5)[0 − x1 (L, L)]
(IR-1H)
(.5)[10 − x1 (H, H)] + (.5)[10 − x1 (H, L)] ≥ 0
(IC-1L)
(.5)[0 − x1 (L, H)] + (.5)[0 − x1 (L, L)] ≥
(.5)[0 − x1 (H, H)] + (.5)[0 − x1 (H, L)]
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(IR-1L)
(.5)[0 − x1 (L, H)] + (.5)[0 − x1 (L, L)] ≥ 0
(IC-2H)
(.5)[10 − x2 (H, H)] + (.5)[10 − x2 (H, L)] ≥
(.5)[10 − x2 (L, H)] + (.5)[0 − x2 (L, L)]
(IR-2H)
(.5)[10 − x2 (H, H)] + (.5)[10 − x2 (H, L)] ≥ 0
(IC-2L)
(.5)[0 − x1 (L, H)] + (.5)[0 − x1 (L, L)] ≥
(.5)[0 − x1 (H, H)] + (.5)[0 − x1 (H, L)]
(IR-2L)
(.5)[0 − x1 (L, H)] + (.5)[0 − x1 (L, L)] ≥ 0
Analysis
• government breaks even ⇒ x1 (L, L) + x2 (L, L) = 0
• no subsidy ⇒ x1 (L, L) = 0
• (IR-1L) ⇒ x1 (L, H) = 0
• government breaks even ⇒ x2 (L, H) = c
• reversing roles of 1, 2 ⇒
x1 (L, L) = x2 (L, L) = x2 (H, L) = 0
x1 (H, L) = c
• (IC-1H) ⇒
(.5)[10 − x1 (H, H)] + (.5)[10 − c] ≥ (.5)[10]
20
• ⇒ x1 (H, H) ≤ 10 − c
• (IC-2H) ⇒ x2 (H, H) ≤ 10 − c
• ⇒ x1 (H, H) + x2 (H, H) ≤ 20 − 2c
• no subsidy ⇒ x1 (H, H) + x2 (H, H) = c
• c ≤ 20/3
We are not quite done: if c ≤ 20/3 is there a mechanism?
Yes
• π(H, H) = π(H, L) = π(L, H) = 1
• π(L, L) = 0
• x1 (H, H) = x2 (H, H) = c/2
• x1 (H, L) = x2 (L, H) = c
• x1 (L, H) = x2 (L, H) = x1 (L, L) = x2 (L, L) = 0
If c > 20/3 (so efficiency is not possible) . . .
How close can we come to efficiency?
Maximize
π(H, H)(20 − c) + π(H, L)(10 − c) + π(L, H)(10 − c)
subject to the IC and IR constraints
(IC-1H)
(.5)[10π(H, H) − x1 (H, H)] + (.5)[10π(H, L) − x1 (H, L)] ≥
(.5)[10π(L, H) − x1 (L, H)] + (.5)[0 − x1 (L, L)]
(IR-1H)
(.5)[10π(H, H) − x1 (H, H)] + (.5)[10π(H, L) − x1 (H, L)] ≥ 0
(IC-1L)
(.5)[0 − x1 (L, H)] + (.5)[0 − x1 (H, L)] ≥
(.5)[0 − x1 (L, H)] + (.5)[0 − x1 (L, L)]
21
(IR-1L)
(.5)[0 − x1 (L, H)] + (.5)[0 − x1 (H, L)] ≥ 0
(IC-2H)
(.5)[10π(H, H) − x2 (H, H)] + (.5)[10π(H, L) − x2 (L, H)] ≥
(.5)[10π(L, H) − x2 (H, L)] + (.5)[0 − x2 (L, L)]
(IR-2H)
(.5)[10π(H, H) − x2 (H, H)] + (.5)[10π(H, L) − x2 (L, H)] ≥ 0
(IC-2L)
(.5)[0 − x2 (H, L)] + (.5)[0 − x2 (L, L)] ≥
(.5)[0 − x2 (H, L)] + (.5)[0 − x2 (L, L)]
(IR-2L)
(.5)[0 − x2 (H, L)] + (.5)[0 − x2 (L, L)] ≥ 0
Constraints are linear inequalities
Objective function is linear
→ this is a linear programming problem
π(H, H) appears only on left side of constraints → π(H, H) = 1
(IC-1H)
(.5)[10 − x1 (H, H)] + (.5)[10π(H, L) − x1 (H, L)] ≥
(.5)[10π(L, H) − x1 (L, H)] + (.5)[0 − x1 (L, L)]
(IC-2H)
(.5)[10 − x2 (H, H)] + (.5)[10π(L, H) − x2 (L, H)] ≥
(.5)[10π(H, L) − x2 (H, L)] + (.5)[0 − x2 (L, L)]
22
10 − x1 (H, H) + 10π(H, L) − x1 (H, L)
10 − x2 (H, H) + 10π(L, H) − x2 (L, H)
20 − c − [x1 (H, L) + x2 (L, H)]
20 − c − [π(H, L)c + π(L, H)c]
⇒ π(H, L) + π(L, H) ≤
≥
≥
≥
≥
10π(L, H)
10π(H, L)
0
0
20 − c
c
What if there is correlation?
• In this problem: correlation makes things worse
–
–
–
–
need to punish H type for lying
gain to lying: payment less
loss to lying: project not built
correlation decreases expected loss (if other is truthful)
8. Third Price Auction Solve for the symmetric equilibrium in smooth,
strictly increasing bid functions for the third price auction (highest bid
wins, pays third highest price) with I ≥ 3 bidders having private values
independently and uniformly distributed in [0, 1].
Solution Let f be the smooth strictly increasing, symmetric equilibrium strategy. Note f (0) = 0; set M = f (1). If bidder 1’s value is v1
and he bids f (w) his expected utility is
Z
Eu1 (w|v1 , f ) = (v1 − b3 )dv2 . . . dvI
where b3 is the 3rd highest bid and the integral is over configurations of
values where w wins. Because others bid their values, w wins exactly
when w > vi for i ≥ 2, and the 3rd highest bid is f (v3 ) where v3 is the
third highest value. There are I − 1 possibilities for the 2nd highest
value and I − 2 possibilities for the third highest value, so there is no
loss in assuming w > v2 > v3 > vi for all other i and multiplying by
23
(I − 1( I − 2). Hence
Z
w
Z
v2
Z
v3
Eu1 (w|v1 , f ) = (I − 1)(I − 2)
Z
Z
0
0
0
w
Z
= (I − 1)(I − 2)
v3
[v1 − f (v3 )]dv3 dv2 . . . dvI
...
0
v2
v3I−3 [v1 − f (v3 )]dv3 dv2
0
0
Combining the FOC and the equilibrium condition and differentiating
by the fundamental theorem of calculus gives
0 = (
d
|w=v1 )Eu1 (w|v1 , f )
dw
Z
v1
v3 I−3 [v1 − f (v3 )]dv3
0
Z v1
v1I−1
= (I − 1)(I − 2)[
f (v3 )dv3 ]
−
I −2
0
= (I − 1)(I − 2)
To solve this for f differentiate with respect to v1 and use the fundamental theorem of calculus again:
0=
I −1
v1 − f (v1 )
I −2
Thus
f (v1 ) =
I −1
v1
I −2
9. Sotheby’s is auctioning off Norman Lear’s vintage fly rod collection
using a sealed-bid First Price Auction. (That is, the high bidders wins
the auction and pays his/her bid.) There are two bidders; their private
values are uniformly distributed on the triangle
T = {(t1 , t2 ) : t1 + t2 ≤ 1}
(Note that values are negatively correlated.) Assume that both bidders use the same equilibrium bidding strategy B, which is continuous
weakly increasing for all t and strictly increasing for t < 1/2.
(a) Show that B(t) = B(1/2) for all t > 1/2.
(b) What is the expected value for Bidder 1 when her true type is t1
but she bids as if her type is x?
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(c) Assuming that B is smooth and strictly increasing for s < 1/2,
find B.
Suggestion: First find the conditional probability
Prob(t2 ≤ y|t1 = t∗1 )
Solution
(a) Suppose t1 > 1/2, so t2 < 1/2. Then B(t1 ) ≥ B(1/2) > B(t2 ).
Hence if Bidder 1 bids B(1/2) she will win for sure and pay B(1/2), so
it cannot be optimal to bid more and pay more. Hence B(t1 ) = B(1/2).
(b) If t∗1 < 1/2, then t2 is uniformly distributed on the interval
[0, 1 − t∗1 ]:
( 1
y if 0 ≤ y ≤ 1 − t∗1
∗
1−t1
Prob(t2 ≤ y|t1 = t∗1 ) =
1 if y ≥ 1 − t∗1
If Bidder 1 is type t1 < 1/2 and bids as if type t < 1/2 then her
expected utility is
Eu1 (t|t1 ) = [t1 − B(t)]Prob(B(t) > B(t2 ))
= [t1 − B(t)]Prob(t > t2 )
t
= [t1 − B(t)]
1 − t1
(c) To solve for B, set dEu1 (t|t1 )/dt = 0 at t = t1 .
dEu1 (t|t1 )
t
t1 − B(t)
0
= −B (t)
+
dt
1 − t1
1 − t1
Plugging in t = t1 setting = 0 and simplifying gives:
t1
t1 − B(t1 )
0
+
0 = −B (t1 )
1 − t1
1 − t1
0
= −t1 B (t) + t1 − B(t1 )
= +t1 − [t1 B(t1 )]0
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Hence
t1 B(t1 ) = t21 /2 + C
B(t1 ) = t1 /2
This answer probably looks wrong: for t < 1/2 the equilibrium bidding
strategy is the same in this environment as in the uniformly distributed
environment. The reason is that what matters is the conditional distribution of t2 conditional on a fixed type t1 , and this distribution is
uniform – just on a shorter interval. The expected utility expression
reflects that the conditional distribution depends on t1 (because it has
the length of the interval in it) but the maximum occurs at the same
place (because it only depends on the uniformity).
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