April 23rd , 2015
ECON 451 • Spring 2015
Econ 451 - Homework #6
Due: April 30th , 2015
1. Consider a common value auction:
• There are a total of n bidders.
• Each bidder receives a signal ti that is uniformly distributed on [0, 1].
• Bidders are bidding on an item with the value V = t1 + t2 + · · · tn
• Player i knows only their signal ti and that all other signals are uniformly distributed over [0, 1].
(a) What is the expected value of V to player 1 given t1 .
(b) What is the expected value of V to player 1 given t1 and given that player 1 wins the auction bidding as if he has
a signal r.
(c) If the auction is a first price auction, the equilibrium bidding function is B(t) = αt. What is α? (To find this
write your equation for expected payoff for player 1 if he has signal t1 but bids as if he received a signal r, and
then maximize and find the α that makes it so that the maximum is at r = t1 .)
(d) If the auction is an all-pay auction, the equilibrium bidding function is B(t) = αtn . What is α?
Answer
• The expectation when I only know my own signal is going to be,
E [V |t1 ] = t1 + E [t2 ] + E [t3 ] + · · · E [tn ] = t1 +
1
n−1
1 1
+ + · · · + = t1 +
2 2
2
2
• The expectation when I have a signal t1 but I bid as if I have a value r and I condidition on the fact that r is the
highest,
r
r (n − 1)
r r
E [V |t1 , r > tj for j 6= i] = t1 + E [t2 |t2 < r] + E [t3 |t3 < r] + · · · E [tn |tn < r] = t1 + + + · · · + = t1 +
2 2
2
2
• To find the equilibrium bidding function we first write out the expected value assuming we both play the bidding
function B, player 1 has a type t1 but bids as if he has a type r,
U1 (B, B, t1 → r) = P (r > tj for all j 6= 1) (E [V |t1 , r > tj for j 6= i] − B(r))
r (n − 1)
= rn−1 t1 +
− αr
2
Next we maximize this with respect to r, and it will be an equilibrium if we get that r = t1 , so we need to find an
alpha such that that is the case.
αn − n(n−1)
∂U1
n (n − 1) n−1
2
= (n − 1) rn−2 t1 +
r
− αnrn−1 = 0 ⇒ t1 =
r
∂r
2
n−1
This is an equilibrium as long as the coefficient on the right side is equal to 1,
αn − n(n−1)
(n + 2) (n − 1)
2
=1⇒α=
n−1
2n
• For the all pay auction, we do exactly the same thing, except now the bid is NOT multiplied by the probability,
because we always have to pay the bid. To find the equilibrium bidding function we first write out the expected
value assuming we both play the bidding function B, player 1 has a type t1 but bids as if he has a type r,
U1 (B, B, t1 → r) = P (r > tj for all j 6= 1) (E [V |t1 , r > tj for j 6= i]) − B(r)
r (n − 1)
n−1
=r
t1 +
− αrn
2
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April 23rd , 2015
ECON 451 • Spring 2015
Next we maximize this with respect to r, and it will be an equilibrium if we get that r = t1 , so we need to find an
alpha such that that is the case.
αn − n(n−1)
∂U1
n (n − 1) n−1
2
= (n − 1) rn−2 t1 +
r
− αnrn−1 = 0 ⇒ t1 =
r
∂r
2
n−1
This is an equilibrium as long as the coefficient on the right side is equal to 1,
αn − n(n−1)
(n + 2) (n − 1)
2
=1⇒α=
n−1
2n
2. Consider the game we talked about in class where the players play the battle of the sexes game, but both players may
or may not want to meet (displayed below). In this case, player 1 believes that player 2 has signal tY2 (left rectangle)
with probability α and has signal tN
2 (right rectangle) with probability 1 − α. Similarly, player 2 believes that player 1
has signal tY1 (top rectangle) with probability β, and has signal tN
1 (bottom rectangle) with probability 1 − β. For each
pair of α, β values (α ∈ [0, 1] and β ∈ [0, 1]), find all pure strategy Nash equilibria of the Bayesian game .
Y
α
C
Y
N
1−α
C
D
2 1
0 0
β
C
D
C
2 0
0 2
D
0 1
1 0
β
D
0 0
1 2
α
C
N
1−α
C
D
0 1
2 0
1−β
D
C
C
D
0 0
2 2
1 1
0 0
1−β
1 0
0 2
D
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April 23rd , 2015
ECON 451 • Spring 2015
Solution
Represent the strategies for player 1 as si tYi s1 tN
(so the action from the yes box comes first, and the action from
i
the no box comes second). First thing we need to draw out the tables like we did in class:
CC
CD
DC
DD
C
2
2α
2-2α
0
D
0
1-α
α
1
Y
Y
N
C
D
CC
1
0
CD
2β
2-2β
C
D
CC
0
2
CD
1-β
2β
α < 2 (1 − α) ⇐⇒ α <
1 − α < 2α ⇐⇒ α >
1
3
β < 2 (1 − β) ⇐⇒ β <
CC
CD
DC
DD
C
0
2-2α
2α
2
DC
1-β
2β
DC
β
2-2β
D
1
α
1-α
0
DD
0
2
DD
1
0
1 − β < 2β ⇐⇒ β >
2
3
2
3
1
3
N
From these tables, we get the best responses:
α≤
1
3
1
3
≤α≤
2
3
2
3
≤α
1
3
β≤
1
3
≤β≤
2
3
2
3
≤β
s2
BR1
BR1
BR1
s1
BR2
BR2
BR2
CC
CD
CD
CD
CC
CD
CD
CD
CD
DC
CC
CD
CD
DC
DD
CD
DC
CD
CC
DC
DC
CD
DD
DC
DD
DC
DC
DC
DD
DC
DC
DC
CC
CC
•
CD •
•
DC
DD
CD DC DD
•
• • •
•
• •
•
• •
• •
•
• •
• • •
•
•
•
Bayesian Nash Equilibria
(CC, CD)
(CD, CD)
1
3
≤α≤
2
3
β ∈ [0, 1]
2
3
≤α
(CD, DC)
α≤
(DC, CD)
α≤
(DC, DC)
2
3
(DC, DD)
2
3
1
3
1
3
β≤
β≤
2
3
≤α
α ∈ [0, 1]
≤β
1
3
1
3
1
3
≤β
≤β≤
2
3
3. Whether candidate 1 or candidate 2 is elected depends on the votes of two citizens. The economy may be in one of
two states, A or B. The citizens agree that candidate 1 is best if the state is A and candidate 2 is best if the state is
B. Each citizen’s preferences are represented by the expected value of a Bernoulli payoff function that assigns a payoff
of 1 if the best candidate for the state wins (obtains more votes than the other candidate), a payoff of 0 if the other
candidate wins, and payoff of 12 if the candidates tie. Citizen 1 is informed of the state, whereas citizen 2 believes it is
A with probability 0.9 and B with probaiblity 0.1 Each citizen may either vote for candidate 1, vote for candidate 2,
or not vote.
(a) Construct the table of payoffs for each state of the world and draw the rectangles to get a diagram that represents
this game (the diagram should look similar to the ones we looked at in class and the one in problem #2 of this
homework).
(b) Show that the game has exactly two pure strategy Nash equilibria, in one of which citizen 2 does not vote and in
the other she votes for 1.
(c) Show that an action of one of the players in the second equilibrium is weakly dominated.
(d) Why is “swing voter’s curse” an appropriate name for the determinant of citizen 2’s decision in the first equilibrium?
Page 3/4
April 23rd , 2015
ECON 451 • Spring 2015
Solution
B
Let tA
1 be player 1’s signal when the state
of the
world is A and t1 be player 1’s signal when the state of the world is
A
B
B. Denote player 1’s strategies as s1 t1 s1 t1 .
1
2
1 1 1
A
1
2
N
1
2
2
1
2
1
2
N 1 1
0 0
0 0
0 0
1
2
2
1 0 0
B
.1 2
1
2
1
2
N 0 0
1
2
1 1
1 1
0.9
0.5
0.9
N
1
1
1
2
1
12 0.95
0.55
1
2
1
2
0
0
1N
0.9
0.55
0.95
N
1
0
1
2
21 0.45
0.05
0
0.5
0.1
0.1
2N 0.45
0.1
0.5
1
2
N
1
2
N
2
22
1
2
1
1 1
.9
11
1
1
2
N
1
0
1
2
0
N1
0.9
0.05
0.45
2
1
2
1
1
N2 0.95
0.1
0.55
N
0
1
1
2
0.9
0.1
0.5
0 0
1 1
1
2
1
2
NN
From the above tables, we get the following best response function,
s2
BR1 tA
1
1
1 or N
2
2
1
1
2 or N
2
N
BR1 tB
1
B
s1 tA
1 s1 t1
BR2
11, 21, 22, N 1, N 2, N N
1
11, 12, 1N, 2N
N
• The game table is above.
• From the best responses above, there are two pure strategy Nash equilibrium: (N 2, 1) and (12, N ).
• If citizen 2 votes for 1, then citizen 2 is indifferent between 12 and N2, so citizen 1’s strategy is weakly dominated
in the Nash equilibrium (12, N ).
• In the NE (12, N ) the two citizen always elect the correct candidate. In this equilibrium, citizen 2’s decision could
be considered ”the swing voter’s curse” because citizen 2 is less informed and therefore strategically abstains rather
than voting and potentially interfering with citizen 1’s accurate information.
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